[Mathematics for Air Crew Trainees]
[From the U.S. Government Publishing Office, www.gpo.gov]
TW1.3F: |-^o/3
WAR DEPARTMENT
TECHNICAL MANUAL
MATHEMATICS FOR AIR CREW TRAINEES
September 30, 1943
NON-CIRCULATING
NTSU LIBRARY
PTM 1-900
Document Reserve
U)l. 3<-.
TM 1-900 C 1
TECHNICAL MANUAL
MATHEMATICS FOR AIR CREW TRAINEES
Changes 1 WAR DEPARTMENT,
No- 1 J Washington 25, D. C„ 6 March 1944.
IM 1-900, 30 September 1943, is changed as follows:
23. Multiplication and division of polynomials.—a. To multiply two * * * rules of addition.
* * * * * * *
Solution (Superseded): ®3 —2®2 —®—1 ____________________®2+2 ®8 — 2a?4 — a?3 — ®2 2a?3 — 4a?2 — 2®—2
./'6 — 2®4+®8 — 5®2 — 2a? — 2
Answer. *******
b. To divide one polynomial by another. *******
(7) To check division * * * to the dividend.
*******
Solution: Rearrange in order of descending powers, before dividing:
3a?8 —5®2—13®+10 by 3®—2 *******
[A. G. 300.7 (22 Feb 44).] (C 1, 6 Mar 44.)
By order of the Secretary of War :
G. C. MARSHALL, Chief of Staff.
Official :
J. A. ULIO,
Major General, The Adjutant General.
577616°—44
U. S. GOVERNMENT PRINTING OFFICE: 1144
*TM 1-900
TECHNICAL MANUAL! WAR DEPARTMENT,
No. 1-900 J Washington 25, D. C., September 30, 1943.
MATHEMATICS FOR AIR CREW TRAINEES
Paragraphs
Section I. General_________________________________________ 1-2
II. Fundamental operations.-_______________________ 3-19
III. Algebra________________________________________20-27
IV. Scales____.___________________________________28-31
V. Graphs________________________________________ 32-36
VI. Angular measurement____________________________37-40
VII. Vectors_______________________________________ 41-49
VIII. Plane geometry________________________________ 50-60
IX. Spherical geometry_____________________________61-62
X. Trigonometry__________________________________ 63-71
XI. Logarithms____________________________________72-8.4
XII. Spherical trigonometry_________________________85-94
XIII. E-6B computer__________________________________95-97
Page
Appendix I. Miscellaneous units and conversion factors____ 163
II. Tables of logarithms and trigonometric functions. 165
Index_____________________________________________________ 169
Section I
GENERAL
Paragraph
Purpose and scope________________________________________________ 1
Materials________________________________________________________ 2
1. Purpose and scope.—a. The purpose of this manual is to provide, in convenient form, a review of some topics of mathematics and related material which the air crew trainee must understand.
b. Sections I through VII include the elementary operations of arithmetic, such as addition, subtraction, multiplication, and division; percent; ratio and proportion; algebra; angular measurements; scales; the use of graphs and formulas; and the graphical solution of the more common problems involving the triangle of velocity.
c. Sections VIII to XIII, inclusive, provide additional instruction material for the required mathematical background of air crew trainees who are to qualify as pilots, navigators, and bombardiers.
(1) This includes the fundamentals and application of plane, solid, and spherical geometry, the use of logarithms, and plane and spherical trigonometry.
*This manual supersedes TM 1-900, February 26, 1943.
549535°—43——1
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ARMY AIR FORCES
(2) The E-6B aerial dead reckoning computer material is included because the computer is a device which applies the principles of mathematics learned.
(3) Sufficient material has been included to enable the trainee to solve the simpler problems of navigation and aerial bombardment numerically by vector-diagram construction, or by slide rule and computer. The solution of the oblique spherical triangle is given for the purpose of solving the astronomical triangle which is the basis of celestial navigation.
d. In mathematics, as in learning to fly, no amount of reading can replace actual practice. For this reason, many exercises have been included in each paragraph, and at the end of each section a collection of miscellaneous exercises has bpen added based on the material in that section. It is not contemplated that every student will do all of the exercises. However, an ample number of exercises has been inserted to provide an opportunity for those trainees who may feel the need for extra practice. The answers to the even numbered exercises are given to enable each student to check his own work if he wishes. Illustrative examples are profuse and should help to clarify difficult points which may arise.
e. Undoubtedly, some of the topics will seem very simple to many of the trainees. It must be remembered, however, that the mathematical proficiency demanded of a pilot not only involves an understanding of the various operations, but also the ability to perform these operations accurately and quickly, and often under trying circumstances. Therefore the time spent in practicing such a simple operation as addition, for example, will not necessarily be so much time wasted, no matter how clearly the process is understood.
2. Materials.—In addition to pencil and paper, the student will need a ruler, a protractor, graph paper, a compass, and an E-6B computer.
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MATHEMATICS FOR AIR CREW TRAINEES
3-4
Section II
FUNDAMENTAL OPERATIONS
Paragraph
Purpose and scope------------------------------------------------------- 3
Addition______________________________________________________________ 4
Subtraction------------------------------------------------------------ 5
Multiplication_________________________________________________________ 6
Division_____________________________________________________________ 7
Conversion of decimal fractions to common fractions______________ 8
Conversion of common fractions to decimal fractions-------------- 9
Addition and subtraction of fractions_________________________________ 10
Multiplication of fractions____________________________________________ 11
Division of fractions__________________________________________________ 12
Ratio and proportion___________________________________________________ 13
Positive and negative numbers________________________________________ 14
Addition of positive and negative numbers------------------------------ 15
Subtraction of positive and negative numbers___________________________ 16
Multiplication and division of positive and negative numbers----------- 17
Square root___________________________________________________________ 18
Miscellaneous exercises________________________________________________ 19
3. Purpose and scope.—The purpose of this section is to provide a review of the four fundamental operations of arithmetic: addition, subtraction, multiplication, and division. Upon these fundamental operations all other mathematical calculations are based. One or more of them must be used in solving any problem.
4. Addition.—a. Addition is the operation of finding the sum of two or more numbers. To add several numbers, place the numbers in a vertical column so that the decimal points are all in a vertical line. (When no decimal point is indicated, it is assumed to be on the right.) Then add the figures in the right-hand column and place the sum under this column. If there is more than one figure in this sum, write down only the right-hand figure and carry the others to the next column to the left.
Example: Find the sum of 30.53, 6.475, 0.00035, and 3476.
Solution:
CARRM-OVER
3 o,
6.
,53
>¥7 5 .ooo35
3¥7 6, 35/3,
o cf535 -—
Figure 1.
b. Units.-—Almost all the numbers which arise in practical arithmetic have to do with definite quantities such as 78 Jeet, 239 miles, 25 degrees, 160 miles per hour, 210 pounds, and so on. In these ex
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4-5
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amples, the words in italics, which state what the quantities are in each case, are called the units. When adding several quantities together, it is clear that the units must all be the same. For example, “the sum of 78jeet and 160 miles per hour” is a completely meaningless statement.
(1) Units are so important and occur so often that standard abbreviations have been adopted for them. A list of the correct abbreviations and the relations which exist between some of the units are given in the appendix.
(2) Example: Find the sum of 78 feet and 3 miles.
Solution: In this case, since 1 mile is the same as 5,280 feet (see app. I), then 3 miles are the same as 15,840 feet. Therefore, 78 feet and 15,840 feet may be added together to give 15,918 feet. But the student is cautioned that unless there is a relation between the various units so that all the quantities may be expressed in terms of the same units, the addition cannot be performed.
c. Symbols.—In arithmetic and in other branches of mathematics, much space and effort are saved by using symbols. Thus, in order to write “find the sum of 70.765 and 301.4,” the plus sign (+) is used and this phrase can be written simply as “70.765+301.4 = ?.” When more than two numbers are to be added, the plus sign, is repeated, for example: 70.765+301.4 + 765.84=1,138.005.
d. Exercises.
(1) 30.53 in.+6.475 in. = ?
(2) 648.03 cm.+37.895 cm.+219.921 cm.+ .08376 cm. = 905.92976 cm. Answer.
(3) 100.001+9.098 + 5678.91 = ?
(4) 897.1+0.989 + 900.76+91901.359 = 93700.208 Answer.
(5) 9876 ft. +101.109 ft.+77.007 ft. + 92.928 ft. + 94.987 ft.+60.768 ft. = ?
(6) 19.767+43.542 + 76.305+58.143 + 13.25=211.007. Answer.
(7) 11.1111 miles + 66.667 miles+1.222 miles+125.125 miles + 375.375 miles=?
(8) 78.908 + 202.202 + 62.501+0.003594 + 75=418.614594. Answer.
(9) 7.8098 + 20.202 + 6.2501+000.3594 + 7.5=?
(10) 78.808 yd.+ 98.15 yd. + 760 yd.+88199.76 yd.=89136.718 yd.
Answer.
5. Subtraction.—a. Subtraction is the operation of finding the difference between two numbers. In order to subtract one number from another, write the smaller number below the larger so that the decimal points are in a vertical column. Beginning with the right column, subtract the figures in the smaller number from the corresponding figures in the larger number above them.
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MATHEMATICS FOR AIR CREW TRAINEES
5
Example: Subtract 765.3 from 986.7.
Solution: 986. 7
765. 3
221.4 Answer.
b. If, however, the figure in the number being subtracted is larger than the figure directly above it, it is necessary to borrow one unit from the next figure to the left.
(1) Example: Subtract 765.3 from 843.1.
Solution:
7 32
765
Figure 2.
&OHHOVJ 1
(2) It is better to learn to do the “carrying over” mentally so that the preceding solution looks like this:
843. 1
765. 3
77. 8 Answer.
c. When a column has only one figure in it, zeros must be supplied in the blank spaces.
Example: Subtract 765.328 from 843.1.
Solution:
5*3/@— EXTRA
7 65,32? ZCR0S
7 7,7 7 2
Figure 3.
d. A problem in subtraction may be checked by adding the answer to the number directly above it. The sum should always be the number in the top row.
(1) Example: Check the answer to the example in c(l) above. Solution:
ADD
8 4 3.
7 6 St ^7 7, 8 4 3.
CHECK
Figure 4.
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ARMY AIR FORCES
1,205.789 in.—980.833 in. = ?
19.52-.78=18.74
760,591 — 674,892=?
73.44 cu. in. —8.7375 cu. in. = 64.7025 cu. in.
89.73-10.0045 = ?
941.7-87.372=854.328
1,004.78 miles—1,004.164 miles=?
100,433 sq. ft.-99,857 sq. ft. = 576 sq. ft.
1,000,000.3-998,757.4=?
Answer.
Answer.
Answer.
Answer.
Answer. method of
(2) Units.—As in addition, care must be exercised to be sure that the units of the two quantities in a subtraction are the same.
e. Exercises.—In each of the following exercises subtract the smaller number from the larger. The symbol used to indicate subtraction is the minus sign Therefore, an expression such as “1,205.789 — 980.833 = ?” means “What is the remainder when 980.833 is subtracted from 1,205.789?” The number following the minus sign is always subtracted from the number preceding the sign.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10) 3,756.04-2,489.7=1,266.34
6. Multiplication.—a. Multiplication is a short adding a number to itself as many times as may be indicated. The numbers multiplied together are called the factors and the result of the multiplication, is called the product. To multiply two numbers together, or in other words, to find the product of two factors, first write the factors one below the other (see example below). It is usually easier to operate with the smaller number of figures in the bottom row. Multiply the factor in the top row by the right-hand figure of the factor in the bottom row, and write this partial product directly under the second factor. If there is more than one figure in the product the same “carrying over” procedure is followed as in addition. Then multiply the factor in the top row by the second figure from the right in the second factor, and write this second partial product so that its right-hand figure is directly under the figure that was used to find it. These partial products are then added together to yield the required product.
Example: Multiply 1,653 by 247.
Solution: / 65 3 Factors
2i
/ / S
6 6 I
33 o 6
7r-
2 -
Partial products
* o 8 2 9 t —Qauu&I ^Product
Figure 5.
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MATHEMATICS FOR AIR CREW TRAINEES
6
b. When there are decimal points, they are ignored until the product has been found. Then the decimal point is inserted in the product according to the following rule: Count off the number of figures to the right of the decimal point in each factor. Then the number of figures to the right of the decimal point in the product is equal to the sum of the number of figures after the decimal point in each factor.
Example: Multiply 16.53 by 24.7.
Solution:
I 6.5 3 2 FIGURES +
2 V.7 ' FIGURE =
1 1 5 7 z 3 Figures 66/2 /
3 3 o 6 /
¥ o 8.2 ? !/ Cbnauwi "product
Figure 6.
c. When the lower factor contains zeros, the partial products corresponding to these zeros need not all be written down. Only the right hand zero is written down. However, care must be exercised to have the right hand figures of all the partial products directly below their corresponding figures in the second factor.
Example: Multiply 16.53 by 24.07.
Solution:
d. Although there is no very simple way to check a multiplication, it is good practice to anticipate the approximate size of the product before beginning a long multiplication. This is done by “rounding off” the factors to permit easy mental multiplication. Although by no means an accurate check, this will frequently catch mistakes in addition or in the location of the decimal point which would otherwise result in nonsensical answers.
Example: What is the approximate product of 15.73 multiplied by 187.04?
Solution: Round off 15.73 to 15, and 187.04 to 200. Then the
7
/o.55
/ / 5 7P
33 o &
37 7.877/
^Product
Figure 7.
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ARMY AIR FORCES
product is roughly 15 by 200=3,000. It is clear then that the product of 15.73 and 187.04 cannot be 150.6 or 6,030.3745, for example.
e. Symbols and units.—The more common symbol for multiplication is X- However, it is quite common simply to write the numbers in parentheses next to each other: (3.04) (17.78) = 3.04X17.78 = 54.0512, for example.
(1) When the same number is to be multiplied by itself, for example 3.04X3.04, this is usually indicated by a small “2” placed above and to the right of the number: 3.04X3.04 —3.042. This is read as “3.04 squared,” and 3.042 is the “square of 3.04.” If the number is to be used as a factor 3 times, then a small “3” is used: 3.04X3.04X3.04 = 3.043. This is read as “3.04 cubed,” and 3.043 is the “cube of 3.04.”
(2) Unlike addition and subtraction, multiplication of different units can be performed. The product is then expressed in a unit which is itself the product of the units of the factors.
(a) Example: Multiply 5 lb. by 7 ft.
Solution: (5 lb.) (7 ft.) = 35 lb.-ft. = 35 ft.-lb. Answer.
(&) Example: Multiply 9 ft. by 17 ft.
Solution: (9 ft.) (17 ft.) = 153 ft.Xft. = 153 (ft.)2=153 ft.2 (ft.2 is read “square feet”) Answer,
j. In arithmetic, as in other operations, there are many tricks which often simplify the work. One such trick which is useful and easy to remember is the following: To multiply any number by 25, move over the decimal point in the given number two places to the right; then divide by 4.
Example: Multiply 16.53 by 25.
Solution: Moving the decimal point over two places, 16.53 becomes 1653. Dividing this by 4: 1653/4=413.25. Therefore 16.53X25 = 413.25 Answer.
g. Exercises.—To each of the following exercises three answers have been given. Eliminate the answers which are obviously wrong by rounding off the factors and finding the approximately correct answers mentally.
(1)
600.3X42.7 =
25,632.81
1,200.62
4,273.21
(2)
180X76 =
(3)
30,740
13,680
25,580
400.785
12.45X18.3= 60.785
227, 835
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MATHEMATICS FOR AIR CREW TRAINEES
6-7
472.6
(4) 88X3.2 = . 281.6
31.7
63,743.68
(5) 1,751.2X36.4= 6,374.368
12,743.68
3,652.925
(6) 903X8.475= 7,652.925
76.52925
h. Exercises.—Perform the indicated multiplications: .0734 in.X70.34 in. = ?
831.43X71.46 = 59,413.9878
1.0073 in.X6.4 ft. = ?
8.94X9.37 = 83.7678
8,374.5X9,378.46 = ?
10,742 lbX737.2 ft. = 7,919,002.4 ft.-lb.
23.1X847.4 = ?
9,034X10.06 = 90,882.04
8.037 ft.X24.2 lb. = ?
Answer.
Answer.
Answer.
Answer.
Answer.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10) .0074X371.5=2.7491
7. Division.—a. Division is the process of finding how many times one number is contained in another. The number to be divided is called the dividend, and the number by which it is divided is called the divisor. The result of the operation, or the answer, is called the quotient.
b. When the divisor contains but one figure, the method commonly used is known as short division. To perform short division, place the divisor (one figure) to the left of the dividend, separated by a vertical line (see example below). Then place a horizontal line over the dividend. Divide the first or the first two figures of the dividend, as is necessary, by the divisor and place the quotient over the line. If the divisor does not go an even number of times, the remainder is prefixed to the next figure in the dividend and the process is repeated.
Example: Divide 4,644 by 6.
Solution. QUOTIENT
Divtsoa 7 7 V—
I u 7 i Dividend ' Do | [ Mentallv
REMAINDER
Figure 8.
9
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ARMY AIR FORCES
c. When the divisor contains two or more figures, the method used is known as long division. This is performed as follows: Place the divisor to the left of the dividend, separated by a line, and place the quotient above the dividend, as in short division. Using the divisor, divide the first group of figures of the dividend which gives a number as large or larger than the divisor (see example below). Place the first figure of the quotient above the dividend. Then multiply this figure by the divisor, and place the product below the figures of the dividend which were used for this division. Then subtract this product from the figures directly above it. The next figure in the original dividend is brought down to form a new dividend. This is repeated until all the figures of the original dividend have been used.
Example: Divide 6,646,250 by 10,634.
Solution:
QUOTIENT
/o63 4\6 64 62 5
(X / 63 8 o 4
Divisor • ’
5] Gmowv o
^Dividend
2 6 5 8 5
2 1268
5 3 17 0
5 3! 7 o o
Figure 9.
d. It is not very common in either short or long division to have the divisor go” into the last trial dividend a whole number of times. When the last trial remainder is not zero, it must be indicated in the answer.
(1) Example: Divide 4,647 by 6.
Solution:
Divisor—.
/ 7 7vAf)l2w<,.
6|*£f7
|*Z 1
REMAINDER
I 27 /
1 7»|/
Figure 10.
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MATHEMATICS FOR AIR CREW TRAINEES' 7
(2) Example: Divide 6,646,257 by 10,634.
Solution:
DIVISOR ----------7 __ _
/,
/ o 6 3 ¥)66¥6257 /
6 3 S of n / 2sTis Remainder 2 1269 /
53177 /
5 31 7 0/ 7Z
Figure 11.
e. Symbols.—“4,647 divided by 6” may be indicated in symbols in several ways. The division sign may be used: 4,647-h6. The “stroke” is more convenient to use on the typewriter: 4,647/6.
4647
Finally, the division may be indicated as a fraction: ——• The fact that 4,647 divided by 6 is 774 with a remainder of 3 may be written 3 3 4647 3
as “4647-^6-774 + 1” or “4647/6=774+f” or “^-=774+|.” 6 b b b
f. Decimal point.—To locate the decimal point in the quotient when decimal points are present in either the divisor or the dividend, move the decimal point in the divisor to the right of the right hand figure. Then move the decimal point in the dividend to the right the same number of places that the point was moved in the divisor. When dividing, be careful to place the quotient so that each figure of the quotient is directly above the right hand figure of the group of figures which were used in the dividend. Then the decimal point in the quotient will be directly above the new position of the decimal point in the dividend. It will also be helpful to remember that the number of decimal places in the quotient is equal to the difference between the number of decimal places in the dividend and divisor.
(1 ) Example: Divide 4.644 by .06.
Solution:
7 7W Om.
,o
Figure 12.
11
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ARMY AIR FORCES
(2) Example: Divide 6.646250 by 10.634.
Solution:
______________5 Cbno. / 5 ^4x42 5 o
63 8 o¥ 265 85 2/268 53/7o 5 3 / 7 o
Figure 13.
g. If decimal points are involved in a division, as a rule the remainder is not indicated when the division does not come out even. In lieu thereof extra zeros are added to the dividend, and the division is continued until the quotient has as many figures as desired.
(1) Example: Find 46.47/0.6.
Solution:
7 7,Vj a™.
/.I* w °
'extra zero
Figure 14.
(2) Example: Find 6.646250-^10.637, Solution:
>6 2 ¥82.* fl/nk
• ^6 3 7^6^ ¥ 6[2 5 0 .0 0
> 3 g 2 2 A EXTRA
2 6 ¥ o 5 ZEROS
2 / 27 ¥
5/3/0
¥25 ¥8
87 620
8 5 o¥ 6
2 52¥o
2 / 27 ¥
376 6 REMAINDER
Figure 15.
h. Mixed numbers.—In e above it was stated that u^^=774 + f-n 6 b
When the plus (+) sign is omitted, then 774%6 is called a mixed number. A mixed number is simply the sum of a whole number and a fraction written without the plus sign. To convert a mixed number to a pure fraction, multiply the whole number by the denominator of
12
MATHEMATICS FOR AIR CREW TRAINEES 7
the fractional part and add the numerator. This is the new numerator; the denominator does not change.
Example: Change 78% to a pure fraction.
Solution:
7^j- = Cfa>.
Figure 16.
i. Checks.—Any division may be checked by multiplying the divisor by the quotient and adding the remainder. The result is always the dividend.
Example: Check the answer to example in g (2) above.
Solution: .62482X10.637 = 6.64621034
6.64621034 Aremainder=6.64621034 .00003966
6.64625000 Check.
j. Units.—As in multiplication, any two different quantities may be divided, even though the units are not the same. The quotient is expressed in a unit which is itself the quotient of the units of the dividend and the divisor.
(1) Example: Divide 175 miles by 10 hours.
o , ±. 175 miles _ miles .
Solution: 777-r---=17.5 r---- Answer.
10 hours hours
In units of this type it is customary to write the denominator in the singular and to use the stroke (/) to separate the numerator from the denominator: 17.5 miles/hour, or 17.5 miles/hr. Although “miles/ hour” really means miles divided by hours it is usual to substitute the word “per” for “divided.” Hence “miles/hour” is read as miles per hour, the standard abbreviation for which is mph.
(2) Example: Divide 500 pounds by 50 square inches.
Solution: r— = 10 pounds/square inch= 10 Ib/sq. in.
50 square inches n
Answer.
(10 pounds per square inch)
k. Exercises.—Perform the indicated divisions and express the quotient as a mixed number.
(1) 894/16
(2) 755/24
13
7
ARMY AIR FORCES
(3) 31 Solution: 24/755 72_ 35 24 11 1,025/314 3111 3124 Answer.
(4) 215/72 2— J72 Answer.
(5) 4,723/353 50,10,099 11,411
(6) 6,754,000/11,411 Answer.
(7) 9,001/30
(8) 11,415/45 253| Answer.
(9) 673/37
(io: ) 1,371/38 36^ Answer.
I. Exercises.—In the following exercises, express the quotient as a decimal. Round off the decimal part of the quotient to two places.
(1) 73.01/3.4
(2) .345/.36
.958
Solution: 36/34.500
32 4
2 10
1 80
300
288
Since only two decimal places are to be obtained, the quotient is rounded off to .96. Answer.
If the figure to be thrown away is greater than or equal to 5, increase the figure on the left by 1. If the figure is less than 5, do not change the preceding figure. Thus: .953 = .95; 1.057 = 1.06; 1.053 = 1.05, etc.
(3) 13.37/.834
(4) 14.705/8.64 1.70 Answer.
(5) (157 miles)/(17.3 hours)
(6) l,942.4/.0035 554971.43 Answer.
(7) 9.63/145.4
(8) (198 miles)/(59 minutes) 3.19 miles/min. Answer.
(9) (5,280 feet)/(60 seconds)
(10) 19.437/38.6 .50 Answer.
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MATHEMATICS FOR AIR CREW TRAINEES
8
8. Conversion of decimal fractions to common fractions.—a.
•A number which consists of a decimal point followed by a sequence of figures is called a decimal fraction. Thus, .33, .9899, .00467, and .00335 are all decimal fractions. Since 33 divided by 100 is .33, then 33 9 899 497
•33=W’ Similarly, -9899=^-^ and -00467=^— Therefore, to express any decimal fraction in fractional form, write the number without the decimal point and divided it by 1 followed by as many zeros as there are figures after the decimal point in the given number.
(1) Example: Express .023678 in fractional form. Solution:
023678
6 FIGURES
25678 n /,00 0,000
6 ZEROS
Figure 17.
(2) Example: Express 4.0785 in fractional form. Solution: First express .0785 as a fraction:
0785=-7-8^
•UZ85 10,000
A A > 785 . 785
4.0785-4+10,000-410>000
54
b. As stated in paragraph 7c, a fraction such as 54/16 or is really
Answer.
just an indicated division. Very frequently in calculations it is much easier to carry a fraction along as a fraction than it is to “divide it out.” Later on, this problem will be considered in detail. At present, however, there is one very important rule, of operation on fractions which should be mastered.
(1) This rule is that both the dividend (numerator) and divisor (denominator) of any fraction may be divided or multiplied by any number (except zero), without changing the value of the fraction. For example, if the numerator (54) and the denominator (16) of the 54 .. . 54 27
fraction jg are both divided by 2, then according to this rule ^g = -g—
(2) This rule allows zeros to be added to any given decimal number. 27
since -27=^00? for example, by the rule, both numerator and denomi-270
nator can be multiplied by 10. Then 1qqq = -^70 and because the fractional value is unchanged, .27 = .270.
15
8
ARMY AIR FORCEiS
c. A fraction is said to be in its lowest terms or simplest form if there is no number which will divide both the numerator and denominator evenly. The operation of finding the simplest form of a fraction is called reduction to lowest terms or simplification.
Solution: Divide numerator and denominator by 4:
632 158
32 8
This can be simplified still more by dividing by 2:
158 97
8 =~r Answer. 4
d. Exercises:—In the following exercises express the given decimal
fractions in fractional form and then simplify. When possible, express
the simplified fraction as a mixed number.
(1) 1.875=?
(2) .9375 = ? 15 . ,Answer. lb
(3) 2.109375 = ?
(4) .125=? 1 . 8 Answer.
(5) .890625 = ?
(6) 3 ;828125 = ? O53 . 3777 Answer. b4
(7) .625=?
(8) 4.375=? ,3 . 4-g- Answer.
(9) 1.6875=?
(10) 2.3125=? „ 5 A 2= Answer. 16
e. Percentage.—Percent means a number with an understood de-50
nominator of 100. For example, 50 percent (%) means or .50.
(1) To change from percent to a decimal, divide the number of percent by 100, which is equivalent to moving the decimal point two places to the left, and omit “percent.”
(a) Example: Change 42 percent to a decimal.
Solution: 42 percent=42/100 = .42 Answer.
(6) Example: Change .9 percent to a decimal.
Solution: .9 percent=.9/100=.009 Answer.
(c) Example: Change to a decimal.
Solution: .5%=.5/100=.005 Answer.
16
MATHEMATICS FOR AIR CREW TRAINEES
8
(2) To change from a decimal to a percent, multiply the decimal fraction by 100, which is equivalent to moving the decimal point two places to the right, and add “percent” (or “%”).
(a) Example: Express .45 as a percent.
Solution: .45X100=45 percent Answer.
(6) Example: Express 6.47 as a percent.
Solution: 6.47X100=647 percent Answer.
(c) Example: Express .0048 as a percent.
Solution: .0048X100=.48% Answer.
(3) It is common to specify changes in troop strength, population, enlistments, and even changes in the physical dimensions of pistons, for example, in terms of percent. The following examples illustrate the methods of solving this type of problem:
(a) Example: If the population of the United States was 120,000,000 in 1930 and increased 7 percent during that year, what was it in 1931?
Solution: 7 percent=.07
.07X120,000,000 = 8,400,000.
Therefore the population in 1931 was 120,000,000+8,400,000 = 128,-
400,000 Answer.
(6) Example: At a certain altitude and temperature, the true air speed is 14 percent greater than the calibrated air speed. If the calibrated air speed is 172 mph., what is the true air speed?
First solution: 14 percent=.14
.14X172 mph. = 24 mph.
Therefore 172 mph.+24 mph. = 196 mph. Answer.
Second solution: 114 percent= 1.14
1.14X172 mph. = 196 mph. Answer.
(c) Example: At a certain altitude and temperature, the true air speed is 14 percent greater than the calibrated air speed. If the true air speed is 210 mph., what is the calibrated air speed?
Solution: Caution must be used in this type of problem. Percent increases as usually understood mean so many percent of the smaller number. Consequently, since 14 percent=.14,
calibrated air speed+ (.14) X (calibrated air speed)=210 mph.
Therefore (1.14) (calibrated air speed) = 210 mph.
or calibrated air speed=(210 mph.)/1.14
= 184 mph. Answer.
This kind of problem can be checked very easily. Since 14 percent of calibrated air speed =.14X184 = 26 mph., 14 percent of calibrated air speed+calibrated air speed=26 mph.+ 184 mph.=210 mph.
Check.
(4) To change a common fraction to a percent, first change the fraction to a decimal and then change the decimal to a percent.
549535°—43---2
17
8
ARMY AIR FORCES
(a) Example: Express as a percent.
Solution: = .75=75%.
(6) Example: Express as a percent.
o
Solution: ^=.333 . . .=33.33 . . .% or 33-^% Answer, o o
(5) All types of percentage problems may easily be solved by the “equation” method.
(a) Example: What is 65% of 220?
Solution: x is 65% of 220
a:=(.65)(220)
a =143 Answer.
Note.—“Is” always stands for “equals.” “Of” always stands for “times.” In the equation the percent must be used in the decimal form.
(6) Example: 250 is what percent of 900?
Solution: 250=x . 900
250
x=90Q=.28 = 28% Answer.
(c) Example: 65 is 13% of what number?
Solution: 65=.13z
a?=-y|=500 Answer.
. Io
Note.'—To solve these equations it is always necessary to divide by the multiplier of x.
j. Exercises.—(1) If 23 percent of a class of 1,900 cadets attend two engine training schools, find the number of cadets in this class who attended these schools.
(2) If 36 percent of a class of 1,800 cadets take their advanced training in Alabama, how many cadets does this represent? 648 Answer.
(3) A lieutenant calls a detail of 24 men. This represents 6 percent of the men in his squadron. How many men are there in the squadron?
(4) At 1,000 rpm., a propeller uses 80 percent of the horsepower developed. If the engine develops 1,500 horsepower at 1,000 rpm., how many horsepower are used by the propeller? 1,200 hp. Answer.
(5) The top speed of an aircraft at 8,000 feet is 314 mph. At 12,000 feet the top speed has increased 8 percent. What is the top speed of the aircraft at 12,000 feet?
(6) For a certain air density, the true altitude is 16 percent greater than the calibrated altitude. If the true altitude is 17,400 feet, find the calibrated altitude. 15,000 ft. Answer.
(7) At 15,000 feet altitude and at —10° C., the calibrated air speed
18
MATHEMATICS FOR AIR CREW TRAINEES
8-9
is 240 mph. The true air speed is 15 percent more than the calibrated air speed. Find the true air speed.
(8) A sample of nickel steel contained 25.61 percent of nickel and 0.17 percent of carbon. How much nickel and how much carbon would be found in a ton of this nickel steel? 512.2 lb nickel Answer.
3.4 lb carbon
(9) A machine shop employing 225 men is forced to employ 36 percent more men. What is the increase in the number of employees?
(10) If the percent of direct hits out of a bomb load is estimated at 35%, and if 7 direct hits are needed to destroy an objective,'how many bombs must be dropped? 20 bombs Answer.
9. Conversion of common fractions to decimal fractions.—a. To convert any fraction to a decimal form, simply add a decimal point on the right of the numerator and perform the indicated division as in paragraph 7g.
5
(1) Example: Express -x as a decimal fraction, o
Solution: Therefore ~=.625
o/O.UUUU o
19
(2) Example: Express as a decimal.
Solution: . /tMaXxxx Therefore —=4.75 4/19.000000 4
6. To express any fraction as a percent, first change the fraction to a decimal fraction as in a above, then change the decimal fraction to a percent as in paragraph 8.
7
Example: Express x- as a percent.
Solution: 12/7 OOOOOb ' ’ ' Therefore .583333 . . .
(The series of dots indicates that the decimal fraction may be continued indefinitely by adding 3’s.)
7
Therefore To==-583333 . . . = 58.3 percent
X Zj
c. Percent problems.—The following examples illustrate the converse (reverse) of those given in paragraph 8.
(1) Example: If 18 airplanes out of a squadron of 27 airplanes are available for combat, what percent of the squadron aircraft is available for combat? What percent is not available for combat?
18
Solution: xx=.6666 . . . = 67 percent available for combat
Z I
Answer.
100 percent—67 percent=33 percent not available for combat
Answer.
Answer.
Answer.
19
9-10
ARMY AIR FORCES
9
or — = .333 . . .=33 percent ' Answer.
£ I
(2) Example: At a pressure altitude of 20,000 feet with the air temperature at 10° below zero, the calibrated air speed is 200 knots, and the true air speed is 282 knots. What is the percent increase in the air speeds? Compare the true air speed with the calibrated air speed.
Solution: 282 knots—200 knots=82 knots
82 knots/200 knots=.41=41 percent increase
Answer.
282 knots/200 knots= 1.41 = 141 percent. Therefore the true air speed is 141 percent of the calibrated air speed Answer.
d. Exercises.—(1) The top speed of an aircraft at 8,000 feet is 304 mph. At 12,000 feet the top speed is 352 mph. What is the percent of increase in speed?
(2) The indicated horsepower of an engine is 1,500, while the actual effective horsepower is 1,275. What percent of the indicated horsepower is the actual horsepower? 85 percent
Express each of the following as a decimal fraction:
(3) |
W F2
Q
18
32
Express each of the following as a percent:
(7)|
(8) |
(9) I
7
(10) 16
Answer.
.4166
.09375
37.5 percent
43.75 percent
Answer.
Answer.
Answer.
Answer.
10. Addition and subtraction of fractions.—a. When several fractions which all have the same denominator are to be added, the addition is performed by simply adding the numerators.
„ . . 5 , 8 , 3 , 10_9
LxCLTfiplc• “i rj "I y I” •
20
MATHEMATICS FOR AIR CREW TRAINEES
10
Answer.
Answer.
a 7 5,8,3,10 5+8+3 + 10 26
Solution: -^ + ^+^+y=--------y----= y
b. When one fraction is to be subtracted from another, and both have the same denominator, the subtraction is performed by simply subtracting the numerators.
9 6
Example: y — y=?
o , 4. 9 6 9-6 3
Solution: y —- y = y = y
c. When fractions to be added or subtracted do not all have the same denominator, the fractions must first be reduced to the same common denominator. What the common denominator is does not matter very much. The reduction to the common denominator is performed by applying the rule given in paragraph 8b.
9 7 4
(1) Example: tt+ta+h”?
11 1U 5 .
Solution: By inspection it is found that all the fractions can be reduced to a denominator= 110. The first is multiplied by 10 (both numerator and denominator), the second by 11, and the third by 22. T, 9X10 _ 90 . 7X11 __ 77. , 4X22_ 88
men nxio no’10X11 110’ancl 5X22 110
. 9.7.4 90 , 77 , 88 90+77+88 255
therefore n + 10+5 ho + 11O+11O 110 110
255_255/5_51
bUt 110 110/5 22
(2) The easiest way to find a common denominator is to multiply all the different denominators together. This will frequently give a common denominator which is larger than necessary, but if the answer is then reduced to lowest terms this method will always work.
(3) Example:
Solution: 25X33=825
, 19 16_627 400_627-400_227
ihen 25 33 “825 825 825 “825
d. Exercises.
(1) - = ?
3 3
(2)
8 4 8
<3>
Answer.
Answer.
Answer.
21
10-11
ARMY AIR FORCES
W I+l+l+s+n+lHs Answer-
(5) 9^+8^+7 = ?
O o
1 5
Hint: Add x and x first.
• b 3
(6) 7| + 9| + lli=28|~ Answer.
Q 1
(7) 14|+302+4=?
(8) A dealer had 16 gallons of oil to sell. He sold 1% gallons to one customer, 2% gallons to another, 7% gallons to another, and the remainder to a fourth customer. How much did he sell to the fourth customer?
4% gal. Answer,
(9) A man can do a piece of work in 13% days. A boy can do the same piece of work in 19% days. How much longer does it take the boy to do the work?
(10) The distance from outside to outside between two holes in a steel plate is 6% inches. If one hole is 1% inches in diameter and the other is 2%4 inches in diameter, find the length of metal between the holes. 3x%4 in. Answer
11. Multiplication of fractions.—a. Two or more fractions are multiplied by multiplying the numerators together and multiplying the denominators together. The product is then a fraction the numerator of which is the product of the several numerators, and the denominator of which is the product of the denominators.
8' 17
Example: Multiply by
Solution:
~ 'J?5 ~
Figure 18.
b. To multiply mixed numbers, first change the mixed numbers to fractions, and then multiply as in a above.
2 2
Example: Multiply 15y by 19y
0 7.- ,,2 47 j in2 97
Solution: 15-N=-7r and 19y=-t 3 3 o o
47x/97_47X97_4559_ „ 14
3 X 5 3X5 15 15
Answer.
22
MATHEMATICS FOR AIR CREW TRAINEES
11-12
c. Exercises.
(1) |X5=?
(2) yXg = l Answer.
(3) 12Xj=?
9
(4) 63X2q = ?
v 7
2 20 63 of)
Solution: 63X2 g=63X jXy = 140 Answer.
2 2
(5) 12^X15^=? o O
(6) A tank holds 300 gallons of gas. If a pipe empties one-fourth of the gas in an hour, how many gallons will be left in the tank at the end of 2 hours?
Solution: %X300 = 75 gal./hr. In 2 hours, pipe will empty 150 gallons. Therefore 300—150 gallons will be left. 150 gal.
Answer.
(7) A tank is five-sixths full of gas. If one-eighth of this is drawn off, what part of the whole tank is drawn off? What part remains in the tank?
(8) The circumference of a circle is about 3% times the diameter. Find the circumference of a circle if the diameter is 14 feet; if 28 feet; if %2 foot. 44 ft; 88 ft; % ft. Answers.
(9) If a motor makes 2,100 rpm, how many revolutions does it make in % hour? In 3x%i days?
(10) An alloy used for bearings in machinery is 2%9 copper, %9 tin, and %9 zinc. How many pounds of each in 346 pounds of the alloy?
286.34 lb copper; 47.72 lb tin; 11.93 lb zinc. Answers.
12. Division of fractions.—a. To divide one fraction by another, first invert the divisor and then multiply as in paragraph 11.
Example: Divide % by
Solution:
^INVERT
3 /5 _ a Y Z7 _ 136 _ > /
3 ' n ~ JA/5 - !3S — >135 a/na-
Figure 19.
b. To divide mixed numbers, first change to a fractional form, then divide as in a above.
Example: Divide 7% by 8%.
o „2 3 23 . 35 23X 4 92 .
Solution: 7-^=8-j-=-^--r--j- =• '=.7^ Answer.
o 4L o te o XdO J.UO
23
12-13
ARMY AIR FORCES
c. Exercises.
(1) %H-10=?
(2) Y Answer.
(3) 27%h-9 = ?
(4) 5%-h2%=? 2 Answer.
(5) In the blueprint of a house % inch in the print represents 1 foot in the actual house. Find the dimensions of the rooms that measure as follows: 2% by 2% inches, 4% by 4% inches, 5% by 6 inches, 3%6 by 4 x%2 inches, respectively, on the blueprint.
(6) Two places, A and B, are 24 miles apart on a river that flows 3 miles an hour. A man can row 5 miles an hour in still water. He goes from A to B and back. Find the time for the journey.
Hint: Man’s speed down river is 8 mph. 15 hr. Answer.
(7) A car is going 11.25 miles per hour. How long will it take this car to go 468% miles?
(8) A layer of No. 8 wire, 0.162 inch in diameter, is wound on a pipe 24% inches long. How many turns o’f wire are wound on the pipe? 150.46 turns Answer.
(9) A mechanic can assemble x%2 of a motor in 1 day. How many motors can he assemble in 3% days?
(10) If a pilot flies 347 miles in 3 hours, 15 minutes, how far will he travel at the same rate in 7 hours, 45 minutes?
827.46 miles. Answer.
13. Ratio and proportion.—a. Consider two bombs, one weighing 300 pounds and the other 100 pounds. The first is three times as heavy as the second, or the second is one-third as heavy as the first. This may be expressed as “the ratio of the weight of the second bomb to the weight of the first bomb is In other words, a ratio is the quotient of two like quantities. In this example,
x. 100 lb 1
ratio —30Q lb 3
b. The statement that two ratios are equal is called a proportion. Thus, for example, if the explosive in the first bomb is 270 pounds, and the explosive in the second bomb is 90 pounds, then the ratios of
the explosives are also and
o
100 lb 90 lb
300 lb 270 lb
is called a proportion.
c. The utility of a proportion comes from the fact that if only one of the numbers is not known, it can easily be found. Suppose that two bombs are given, one weighing 450 pounds and the other weighing
24
MATHEMATICS FOR AIR CREW TRAINEES 13
150 pounds, and that the length of the first bomb is 36 inches. The length of bomb No. 2 is not known, but if the length of any bomb is “proportional” to its weight, then
weight of bomb No. 1 length of bomb No. 1
weight of bomb No. 2"“length of bomb No. 2
is the proportion expressing this fact. Now some of these quantities are known:
450 lb =______36 in.________
150 lb length of bomb No. 2
Therefore, if the proportion is true, then the length of bomb No. 2 must be 12 inches.
d. In mathematics, not only are symbols such as +, —, =, etc. used to simplify writing, but it is also convenient to introduce other symbols whenever they will shorten the work. Thus, to continue the preceding example, let
Wi=weight of bomb No. 1
w2=weight of bomb No. 2
Li—length of bomb No. 1
Z2=length of bomb No. 2
Then the proportion can be written even more simply as
Wj Tj\ 17- i o '
—and Z2=12 in.
w2 L2
The practice of using letters to represent quantities is characteristic of all mathematics.
e. There is a general rule on proportions which can be stated very briefly in terms of symbols. Let a, b, c, and d be any quantities whatever. Then if there is a proportion between these quantities such that
* a_ c l~d then also ad=bc
That is, if the terms in a proportion bo “cross-multiplied,” the products are equal:
CRO55-MULTIPLV
bxc a axd
Figure 20.
25
13
ARMY AIR FORCES
Example: Let a=15 inches, 6=60 inches, c=30 yards, and d=120 yards. Check the “cross-multiplication” rule for these quantities.
7 • o- 15 in. 30 vd. .
Solution: Since 55^ = j20 yl’ tllen
(15 in.) (120 yd) should equal (30 yd) (60 in.).
(15 in.) (120 yd) = 1,800 in.-yd=(30 yd) (60 in.) Check
f. The problem of converting from one type of unit to another can be regarded as a proportion problem.
Example: A speed of 12 miles per hour is equivalent to how many feet per second?
Solution: The known relation between the two types of units is that 1 mph. = 5,280/3,600 ft./sec.=22/15 ft./sec.
This gives the proportion
1 mph. _22/15 ft./sec.
12 mph. A ft./sec.
Solving this for x gives ®=17.6 ft./sec. Therefore, 12 mph. is equivalent to 17.6 ft./sec. Answer.
g. Exercises.-—It is customary to let x represent the unknown quantity. In the following exercises find the values of x which will make the proportions true.
(1) 5_x
1~3
(2) 5 x .
—=— in
6 3
Solution: Cross-multiplying, 6®=15 inches. If 6 ®’s are 15 inches, then obviously, one x must be one-sixth as much or 2.5 inches. In all of these problems, if the factor multiplying x is not 1, divide by the factor on both sides of the equality sign. Thus
^—l^in. is the same as 6®=15 inches. But 6x/6=aj. 6 6
Therefore ®=(15 in.)/6=2.5 in. Answer.
(3) 5 = x
3 5
(4) (|=— x=ll% Answer.
7 x
(5) .25 = 3
.75 x
x=10 Answer.
(.5) x .
26
MATHEMATICS FOR AIR CREW TRAINEES
13
(7) x_a y~b
(8) What is the value of A in the following proportions:
(a) JL7=14
45 A
(6) 3 10 *
^~25 H=7.5 Answer.
3
(c)—--
( } 4~A
(9) Two pulleys are connected by a belt. The smaller one runs at a speed of 750 rpm. and the larger at 200 rpm. What is the ratio of their speeds?
(10) An airplane travels 400 miles in 2 hours. Set up a proportion and determine how far the airplane will travel in 14 hours.
2,800 miles. Answer.
(11) If %o inch on a map represents 49 miles, how many miles are represented by 3 inches on the map?
(12) If a boat drifts down stream 40 miles in 12 hours, how far will it drift in 15 hours? 50 miles. Answer.
(13) On June 12, 1939, a pilot flew a glider plane across Lake Michigan a total distance of 92 miles in 52 minutes. He cut loose from the tow plane at 13,000 feet and descended only 5,000 feet in crossing. At the same rate of descent, how much farther could he have glided? How many more minutes would he have been in the air?
(14) A roadbed rises 3% feet in a horizontal distance of 300 feet. How many feet will the roadbed rise in 720 feet? 8 ft. Answer.
(15) If 16 gallons of gas will drive a car 288 miles, at the same rate of using gas how many gallons will it take to drive the same car from Chicago to Memphis, a distance of 564 miles?
h. Conversion exercises.—Obtain conversion factors required in the following examples from the appendix.
(1) Change 210 miles per hour to knots.
(2) How many feet per second are 32 miles per hour?
46.9 ft./sec. Answer.
(3) Express 58 centimeters in inches.
(4) Convert the following to nautical miles:
(a) 230 statute miles. 199.7 nautical miles. Answer.
(b) 34.5 statute miles. 29.9 nautical miles. Answer.
(c) 4,025 statute miles. 3,495.3 nautical miles. Answer.
27
13-14
ARMY AIR FORCES
(5) A tank containing 125 U. S. gallons of gas would contain how many British gallons?
(6) How many U. S. gallons are there in 78.5 British gallons?
94.2 U. S. gal. Answer.
14. Positive and negative numbers.—There are many quantities which by their contrary or opposite nature are best described as negative quantities in contrast to positive quantities. For example, temperatures above 0° Fahrenheit are considered as positive, whereas those below 0° are considered as negative. As a consequence it becomes necessary to consider negative and positive numbers and how to deal with them.
a. A negative number is indicated by prefixing a minus.sign ( —) in front of the number. Thus —5, —7.04, —90.003 are all negative numb ers.
Figure 21.
b. A positive number is indicated by prefixing it with a plus sign (+), if necessary. When there is no possible ambiguity the plus sign is usually omitted. Thus +5, 7.04, +63.0, 98.4 are all positive numbers.
c. It is convenient to imagine the numbers as representing distances along a straight line as follows:
________POSITIVE_______r
T ' _ T 0 H +2 +3 +4 +S
-4 -3 -2 -/ ' ' . ' '
“** NEGATIVE
Figure 22.
28
IOOC
POSITIVE^ READINGS
O*C
NEGATIVE
READINGS^
MATHEMATICS FOR AIR CREW TRAINEES
14-16
Negative distances are measured to the left and positive distances are measured to the right.
d. The signs + and — now have additional meanings. They not only indicate addition and subtraction, but positive and negative numbers as well. To distinguish the sign of operation from the sign of quality (positive or negative), the quality sign is inclosed in parentheses: 25+(+5), 25-(+5), 25+(-5), or 25-(-5). For the sake of brevity, the first and second are generally written simply as 25+5, and 25 — 5.
15. Addition of positive and negative numbers.—To add two numbers which have the same signs, add the numbers and prefix the common (or same) sign. If the numbers to be added have unlike signs, find the difference and use the sign of the larger number.
a. Example: (+6) + (—3) = ?
Solution: Since the signs are different, subtract the numbers to obtain a remainder of 3. Since the sign of the larger number is positive the sign of the remainder is also positive: 6 +(—3)=3 Answer.
b. Example: (—3) + (+2) = ?
Solution: Referring to figure 22, begin at —3 and count 2 units 1n a positive direction. The result is 1 space to the left of zero. Therefore (—3)+ (+2) = — 1.
This problem may also be done by using the rule stated in the preceding paragraph. Since the signs are unlike, subtract 2 from 3 and prefix the remainder by a minus sign since the larger number is negative. (—3) + (+2) = —1 Answer.
16. Subtraction of positive and negative numbers.—a. To subtract two numbers (positive or negative), change the sign of the number being subtracted and add the numbers as in addition (par. 15).
Example: (—3) —(—4) = ?
Solution: Changing the sign of the number being subtracted, the problem then becomes—
( —3)+( + 4) = +l Answer.
h. Exercises. (1) (-6) + (+6) = ? (2) (+2)+(+12) = + 14 Answer.
(3) (4) (—62)+(—18) = ? (+17)-(+15)=+2 Answer.
(5) ( + 32) —( — 64) = ? ■
(6) (-18)-(-64)=+46 Answer.
(7) (8) (—17)—(—15) = ? (—17.3)+(35.4)=+18.1 Answer.
(9) (10) (-17.36)-(35.4) = ? ( — 201.03) —(—10.4) =— 190.63 Answer.
29
16-17
ARMY AIR FORCES
(11) —18 + .4 = ?
(12) 20 —17.4+9=+11.6 Answer.
(13) -37.3 + 19.4 + 17.8 = ?
(14) (—175.03) + 19= —156.03 Answer.
17. Multiplication and division of positive and negative numbers.—a. If the two numbers to be multiplied have the same signs, then the product is positive. If the two numbers to be multiplied have opposite signs, then the product is negative.
(1) Example: Multiply (+3.04) by (17.8).
Solution: Since the signs are the same, the product is
+ 54.112 Answer.
(2) Example: ( + .00395)X (-345.9) = ?
Solution: Since the signs are unlike, the product is negative, or
—1.366305 Answer.
b. Exercises.—Find the product in each of the following exercises.
(1) (-1.6) (.0)
(2) (—14.4) (-12) 172.8 Answer.
(3) (12.5) (1.25)
(4) (-9) (-8) (-6) -432 Answer.
(5) (-.17) (6) (-5)
(6) (2) (3.14) (9) 56.52 Answer.
(7) (-.4) (-.4) (-.4)
(8) (~® (» «) Answer.
(9) (’%) (1.4) (1.4)
(10) (-1® (’/(») (-&) % Answer.
c. In division the quotient is positive if the divisor and dividend
have the same sign; if the divisor and dividend have opposite signs,
the quotient is negative.
Example: Divide (—15.625) by (12.5).
Solution: Since the dividend and the divisor have opposite signs, the quotient is negative.
-15.625 , o_ A
—z-x-u—= — 1.25 Answer.
1 ~ . D
d. Exercises.—Find the quotient in each of the following exercises:
(1) (-14.4)=(0.9)
(2) (-i-4.32)-i-(—4.8) 0.9 Answer.
(3) (39,483.)h-(—12.3)
(4) (1,440.)-f(0.32) 4,500 Answer.
(5) (1.679)-h(23)
(6) (—23.04)-4-(4.8) —4.8 Answer.
(7) 1,728 -144
30
MATHEMATICS FOR AIR CREW TRAINEES
17-18
(8) 390.59 —13.9 Answer.
-28.1
(9) — 72.9
-0.81
(10) 0.6118 0.007 Answer.
87.4
18. Square root.—The process of extracting square root finds many uses in solving problems. Following are the rules:
a. Begin at the decimal point and point off as many periods of two digits each as possible. Point off to the left if the number is an integer, to the right if it is a decimal. Point off to both the left and the right if there are digits both left and right of the decimal point. If the last period of the decimal has but one digit, add a cipher to complete the period.
b. Find the largest integer the square of which is equal to or less than the left hand group, and write this integer for the first digit of the root and directly over the first group of digits.
c. Square the first digit of the root; subtract its square from the first group and bring down the second group.
cl. Obtain a trial divisor by doubling the partial root already found, divide it into the remainder (omitting from the latter the right hand digit), and write the integral part of the quotient as the next digit of the root and directly over the group of digits used in determining it.
e. Annex the root digit just found to the trial divisor to make the complete divisor; multiply the complete divisor by this root digit, subtract the result from the dividend; bring down the next group for a new dividend.
j. Obtain a new trial divisor by doubling the part of the root already found, and proceed as before until the desired number of digits of the root have been found.
g. After extracting the square root of a number involving decimals, point off one decimal place in the root for every decimal group in the number.
h. If the root is exact, square it. The result should be the original number. If the root is inexact, square it and add to this result the
(1) Point off in groups of two in both directions from the decimal point.
(2) 2 is the largest integer the square of which is equal to or less than 6, so place the 2 over the 6.
remainder.
Example: Find 7681. 9213
Solution: 26 . 1 1
7681.92 13
4
46 2 81
6
276
31
18
ARMY AIR FORCES
2 76
521 5 92
1
521 5 21
5221 71 13
1
5221 52 21
1892
26. 11
26. 11
26 11
261 1
15666
- 5222
681. 7321
1892
681. 9213
(3) Square 2 and place it under the 6. Subtract.
(4) Bring down 81, the next two digits.
(5) Bring down the 2 above the 6 and double it for the trial divisor. Divide the 4 into 28, the remainder less the last digit to the right; 7 is obtained.
(6) Annex the 7 to the trial divisor and multiply by 7; 329 is obtained. This is too large. Next annex 6 and try it. This is satisfactory. Subtract the 276 from the 281. Place the 6 above the 81 group. Bring down the next group, 92.
(7) Obtain a new trial divisor by doubling the 26. Divide the 52 into the 59; 1 is obtained. Annex the 1 and multiply. Place the 1 above the 92. Subtract and bring down the next group, 13.
(8) Obtain a new trial divisor by doubling 261. Divide the 522 in the 711; 1 is obtained. Annex the 1 and multiply. Place the 1 above 13. Subtract. 1892 is obtained as the remainder.
(9) Multiply 26.11 . 26.11 and add 1892; 681.9213 is obtained.
i. Exercises. Find the square root of the following:
(1) 7625
(2) -y/289 17 Answer.
(3) 712514
(4) 7167281 . 409 Answer.
(5) 7^93
(6) 7.4387 .66234 Answer.
(7) 7983.431
(8) 710.6934 3.27008 Answer.
32
MATHEMATICS FOR AIR CREW TRAINEES 19
19. Miscellaneous exercises.—The following exercises are based on the topics in this section:
(1) If 1 cubic foot of water weighs 62.5 pounds, what is the weight of 4.18 cubic feet of water?
(2) How many cubic feet are there in 180 pounds of water? (See a above.) 2.88 cu. ft. Answer.
(3) Change the following common fractions to decimal fractions:
3. 7_. 51. 17
8’ 16’ 75’ 32’
(4) Change the following decimal fractions to mixed numbers:
1.25; 3.875; 14.375. . 1%;3%; 14% Answer-
(5) Bolts % inch in diameter and 6 inches long weigh 117 pounds per hundred bolts. What is the weight of 1,200 bolts?
(6) A certain bomber can carry a bomb load of 4,500 pounds. How many 250-pound bombs can be carried? 18 Answer.
(7) Which is larger 1%5° or %°?
(8) Divide 6% by 2%. 2% Answer.
(9) At an altitude of 5,000 feet and at 10° C., the calibrated air speed is 190 mph. The true air speed is 206 mph. What is the percent of increase in the two readings?
(10) At an altitude of 11,000 feet and at 20° C., the calibrated air speed is 210 mph. The true air speed is 242 mph. What is the percent of increase in the two readings? 15.2 percent. Answer.
(11) The top air speed of an aircraft at 10,000 feet is 325 mph. At 15,000 feet it is 335 mph. What is the percent of increase in the air speed?
(12) If 469 cadets are sent to primary schools in Georgia and this group represents 14 percent of the class, find the number of cadets in the class. ' 3,350 cadets. Answer.
(13) If 28 cadets out of a squadron of 196 are on guard duty, what percent of the squadron is on guard duty?
(14) On a certain flight a bomber used 40.5 gallons of gasoline per hour. The time of the flight was 3 hours 48 minutes. Find the amount of gasoline used. 153.9 gal. Answer.
(15) An aircraft flies a distance of 160 nautical miles. Find the distance in statute miles.
(16) The temperature reading on a centigrade thermometer was 3° C. The reading increased 2° the first hour and decreased 7° the second hour. What was the final temperature reading?
— 2° C Answer.
(17) On a certain day, 10 temperature readings were taken on a centigrade thermometer. They were 6°, —3°, —7°, —15°, — 4°, 0°, 2°, 3°, 5°, 3°. Find the average temperature reading.
549535°—43---3
33
19
ARMY AIR FORCES
Hint: Find the sum and divide by the number of readings.
(18) Find the product in each of the following:
(—6) • (—1%) • (-1%) 12 Answer.
(—2)2 • (8%) • (— %) —25 Answer.
(19) The following numbers represent the diameters of the bores on different guns: 37 mm, 3 inches, 1 inch, 155 mm, 6 inches, 75 mm. Arrange them according to size beginning with the largest one.
(20) Express a speed of 118 kilometers per hour in terms of miles per hour. 73% mph. Answer.
(21) Calculate the number of square centimeters in 1 square foot.
(22) A photographic film is designed for a picture 6 by 9 centimeters; express this in inches to the nearest quarter inch.
2% by 3% in. Answer.
(23) If 570 cadets are sent to primary schools in Florida, and this group represents 30 percent of the class, find the number of cadets in the class.
(24) If 24 cadets out of a squadron of 180 passed the high altitude test, what percent of the squadron passed the test?
13.3 percent. Answer.
(25) At a certain airdrome there are 88 aircraft, consisting of bombers and interceptor aircraft. The ratio of bombers to interceptors is 3 to 8. Find the number of each kind of aircraft.
(26) What is the diameter in inches of the bore of a 75-mm. gun? (This means the bore is 75 mm. in diameter.) 2.95 in. Answer.
(27) The following numbers represent the ranges of different aircraft: 250 nautical miles; 262 statute miles; 480 kilometers; 298 statute miles; 275 nautical miles. Arrange these distances in order of magnitude starting with the largest one.
(28) A detail of 33 cadets represents 15 percent of the squadron. How many cadets are there in the squadron? 220 Answer.
(29) Find the difference in temperature readings of +47° C. and -5° C.
(30) On a certain day the lowest temperature reading was —14° F. and the highest temperature reading was +19° F. Find the increase in readings. 33° Answer.
(31) Find the values of the following:
(~3)3
(-2)2 (-1)3
3(—2)3 (-1)2
2(%)2 (4)3
8(-l)2 (%)3
(32) A panel is made up of 5 plies which are % inch, % inch, % inch, % inch, and %6 inch thick, respectively. How thick is the panel?
1%4 in. Answer.
34
MATHEMATICS FOR AIR CREW TRAINEES 19-21
(33) Divide 1.5625 by 0.125.
(34) Multiply 2% %6 2% 1%. 1% Answer.
(35) Find the sum of l%+2% —%+%8.
(36) How many strips each %2 inch thick are in a laminated piece 1% inches thick? 20 Answer
(37) In a squadron of 200 cadets there are 14 cadets sick. What percent of the squadron is sick?
(38) The chord of an airplane wing is 72 inches. If the center of pressure is at a point 28 percent of the distance along the chord from the leading edge, how many inches is it from the leading edge?
20.16 inches. Answer.
L L L
(39) The p ratio for an airfoil section is jy • Find the p ratio when Zc=0.0018 and Dc=0.00008.
(40) Find the ratio of the areas of two circles having radii of 3 inches and 4 inches. (The areas are to each other as the squares of their radii.) %6 Answer.
Section III
ALGEBRA
Paragraph
Purpose and scope________________________________________________________ 20
Algebraic symbols________________________________________________________ 21
Addition and subtraction of polynomials___1______________________________ 22
Multiplication and division of polynomials_________________________•___ 23
Evaluation of algebraic expressions______________________________________ 24
Equations______________'----------------------------------------------- 25
Word problems____________________________________________________________ 26
Miscellaneous exercises__________________________________________________ 27
20. Purpose and scope.—Algebra is the basis for all work in formulas and trigonometry. In the solution of many problems of these types the first thing that is done is to write facts in the form of an equation. To be able to handle such work, equations must be understood and the necessary background of algebraic manipulation must be thoroughly mastered. The following paragraphs contain exercises with accompanying explanations designed to give the student a working knowledge of necessary fundamentals of algebra.
21. Algebraic symbols.—a. In working with general formulas, it is convenient to let numbers be represented by letters. In the actual evaluation of the formulas, the specific values are substituted for the letters.
Example: d=~ gt2 if <7=32, t=5
Solution: then c/=-gX32X25= 16X25=400 Answer.
35
21
ARMY AIR FORCES
The above formula is solved for d. If it became necessary to solve for some other letter in the formula, other operations would be required. To perform such operations, it is essential to learn some of the fundamental characteristics of algebra.
b. In algebra, because of the use of x as a letter, new signs are adopted to indicate multiplication.
(1) 4x means 4 times x. When no sign appears between letters and numbers, multiplication is indicated.
(2) 4 • x means 4 times x. When some sign of multiplication is necessary, a dot is used as indicated.
(3) In any product such as the foregoing, the coefficient of the x term is defined to be the rest of that term.
(4) When there is a product of the type x-x-X'X, it is written x\ The 4 is called the exponent of x, and x4 is called a power of x. This is simply a shorthand method of writing a product. The x is termed the base.
Example: What are the coefficients and exponents of— 4x2? x2? 4x1 xl
Solution:
4 is the coefficient and 2 is the exponent.
1 is the understood coefficient and 2 is the exponent.
4 is the coefficient and 1 the understood exponent.
1 is understood as the coefficient and exponent.
c. Rules for use of exponents.
(1) When two powers with the same base are multiplied together, the exponents are added, as shown in the following illustrations:
32-33=35=243
(9-27 = 243)
x2-x4=x6
(2) When two powers with the same base are divided, the exponents are subtracted, as the following examples show:
^=2«=8
X7 4 ____/y»4 a;3
(3) When two products are multiplied together, the coefficients are multiplied and each power is multiplied separately, as in the following example:
(a) 2x-3x2—6x3.
36
MATHEMATICS FOR AIR CREW TRAINEES 21-22
(6) 2a5z+a2z=10a36z2.
d. Terms are products or quotients separated from each other by + and — signs. For example, in the expression 6x2y—2axJr4ab, the terms are 6x2y, — 2ax, and 4ab.
e. A polynomial is a sum of two or more terms which do not involve division. To illustrate:
4x3+x2— 1 is a polynomial
a-\-b-\-c— 1 is a polynomial
Qu ' QC I 'JC,
—i— is not a polynomial
z4—2z3+-|— 7 is not a polynomial
22. Addition and subtraction of polynomials.—a. To add or subtract polynomials, the terms should be arranged so that like terms fall in columns. Then these terms should be added or subtracted according to the rules of signed (positive and negative) numbers. Only those terms that are exactly alike in all respects except for the numerical coefficient, may be added or subtracted. For example, x and x2 cannot be combined into a single term.
(1) Example: Add x3— 3a;2-}-1, x3f-x— 3, and x2+x+l.
Solution: x3— 3x2 +1
x3 +#—3.
_______X2+z+l
2z3—2x2-\-2x—1 Answer.
Note.—It is advisable, if possible, to arrange polynomials in order of decreasing exponents of the unknown.
(2) Example: Subtract x3+3x2+a:—1 from a;4+a;3—a;+2
Solution: x^-fx3 —x+2
x3-\-3x2-\~x— 1
x* —3x2—2#+3 Answer.
b. Exercises.—Add the following polynomials:
(1) 5z+5?/, 6z—8y+52, 5x— 3y~3s.
(2) x2—x— 2, 2x — 7a?2+4, 6x2+x—7. 2z—5 Answer.
(3) 2z+3, 4a;—7, x-2.
(4) 2x+2y+^z, 2^x—3~y+2z, x—y—z.
5^x—2-L/+1^z Answer.
37
22-23 ARMY AIR FORCES
(5) 126c+ac, 9ab—bc, 14bc—3ab.
(6) 2a+2c+#—5, 2a+3a;+8—3c, a—9—3c—x, 16 — 13a?—14a—c.
—9a—5c— lOz+lO Answer.
c. Exercises.—Subtract the following polynomials:
(1) 2a—9 from 7a—11.
(2) x2—3z— 5 from 2x2— 7z+9. x2—4z+14 Answer.
(3) 3a;3—5 from — 7a:3+4a;2.
(4) z2+2z—3 from 5x2—4x— 7. 4x2—6x—4 Answer.
(5) 4a2x—3a2y -\-xy from 0.
(6) 2x5—4+—5 from x4-j-2x2—3x3—xB.
— 3xB-hx4-j-x3~i-2x2-l-5 Answer.
23. Multiplication and division of polynomials.—a. To multiply two polynomials, multiply one of them by each term of the other separately, then combine the results, following the rules of addition.
Example: x3—2x2—x— 1 by z24~2.
Solution: +2x2—x — 1
x2+2
xB-\-2x4—x3— x2
2a;3+4a"2—2z—2
xB-\-2x4-\-x3+3x2—2x—2 Answer.
If three or more polynomials are to be multiplied together, multiply two of them and then multiply this product by the third, etc.
b. To divide one polynomial by another.—(1) Arrange the dividend and the divisor according to descending powers of one variable, starting with the highest powers at the left.
(2) The result of dividing the first term of the dividend by the first term of the divisor is the first term of the quotient. Ignore the rest of the terms for the time being.
(3) Multiply the entire divisor by this first term of the quotient and subtract the result from the dividend.
(4) The second term of the quotient is obtained by dividing the first term of the remainder by the first term of the divisor.
(5) Using this.second term, multiply the divisor and subtract from the remainder as before.
(6) Continue this process until the first term of the divisor cannot go into the first term of the last remainder. This remainder is written after the quotient as the numerator of a fraction, of which the divisor is the denominator.
(7) To check division, multiply the divisor and quotient. This should be equal to the dividend.
Example: Divide — 5x2— 13z+3z3+10 by 3x—2.
38
MATHEMATICS FOR AIR CREW TRAINEES 23-24
Solution: Rearrange in order of descending powers, before dividing: 3®3—5®2—13®+®10 by 3®—2
3®—2 | 3®3—5®2—13®+10 [ ®2—x—5 Answer.
3®3—2x2
—3®2—13®
—3®2+2®
—15®+10
—15® + 10
c. Exercises.—Multiply the following polynomials:
(1) (a+2) (a+5).
(2) (®+?/) (x—y). x2—y2 Answer.
(3) (3®2—12®+18) (®—4).
(4) (®2—4®—12) (®2+4®+2). ®4—26®2—56®—24 Answer.
(5) (®?—7®—12) (6®—26).
(6j (®4 —9?/2) (®3—3y\ ®7—3®4?/—9®3?/2+27?/3 Answer.
d. Exercises.—Divide the following polynomials:
(1) (a2+5«+6)=-(a+2).
(2) (®2—10®—39)-h(®+3). x—13 Answer.
(3) (2®2—13®+20)-f(2®—5).
(4) (®4—®3—4®2—®+l)-r-(®2+2®+l). ®2—3®+l Answer.
(5) (®6—t/6)h-(®2—y2Y
(6) (®7+1)-f(®+ 1). ®6—®5+®4—®3+®?—®+l Answer.
24. Evaluation of algebraic expressions.—a. To evaluate algebraic expressions, substitute the numerical values for the literal values and perform all indicated operations. In any series of operations, multiply and divide before adding and subtracting. Clear parentheses by doing work inside the parentheses first.
(1) Example: Evaluate a®2+26®—46c when a=l, 6 = 3, ®=—2, c=4.
Solution: [1 • (—2)2] + [2 • 3 • (-2)]-(4 -3-4) =
[1 • 4] + [6 • (-2)]-[12 • 4] =
4 —12—48 =—56 Answer.
(2) Example: Evaluate a[(a—6)+c(a2—c) —d(a—c2)] when a=2, 6=3, c=4, d=6.
Solution: 2[(2-3)+4(22-4)-6(2-42)] = 2[-l+4(0)-6(-14)] =
2[—l + 0+84]=2[83j = 166 Answer.
6. Exercises.—Evaluate the following if ®=3, y—2, z—5.
(1) ®(?/+2)(3y+®)
(2) 4(®2—t/)+?/2(2—?/) 40 Answer.
39
24-25
ARMY AIR FORCES
(3) (y—3x)-\-z(y—x)
(4) x^x+yj+z^+xty— 3) 37 Answer.
(5) 0[x+t/(2-7)+4]-7
2— 7*
(6) % Answer.
* y
(7)
V
25. Equations.—a. An equation is a statement of equality between any two quantities. Thus, in reality, all formulas are equations. In most equations that are used, there is an unknown quantity for which a value is sought. To find this, certain rules must be followed:
(1) Equal quantities may be added to both sides of the equation.
Example: x—4 = 7
+4=+4
x =11
(2) Equal quantities may be subtracted from both sides of the equation.
Example: x+2 = 5
—2 = —2
x = 3
(3) Both sides of the equation may be multiplied by the same quantity.
Example: k=4 A x=4.2 x=8
(4) Both sides may be divided by the same quantity (zero excepted).
Example: 3x=12
3z=12
3 “ 3
x — 4
(5) The above solutions may readily be checked and proved correct by substituting in the original equation the value found for x.
b. It is possible to shorten these operations by shifting any term from one side of the equal sign to the other and changing its sign. This is called transposition.
c. An advanced type of equation that it is necessary to know how to solve is the fractional equation. In this type one added operation must be performed before applying the preceding rules. The first
40
MATHEMATICS FOR AIR CREW TRAINEES
25
step in such equations is to find the least common denominator and with this quantity multiply both members through term by term. Example: Solve the following equation:
i-v=2-^-2)
Solution: x5x (a;—2) /T ~ .
2 g —£ g (-Lu J-A — lo)
18 • £-18 • ££=18 • 2-18(q~2).
2 6 9
9a;= 15ar=36 —2a?+4
9z-15z+2z=36+4
—4a;=40
x= —10
Answer.
d. Exercises.
fa) 5x-3 = 3z+3
(2) 3a;+5+«+3 = 0
(3) 6a;+4=a;—16
(4) 5n—4-j-6n—40=0
(5) 5z=%
(6) |-4 = 5 o
(7) ^-=3+x
4 - 5
w 2x+2 3x+2
(9) |+|=10
, . 5a;—3 3a;—
8 5 —1
(H) _1--1-1——_JL_
1 ' x+3^a;+6 a:+4
Q2) —___I_— =
’ x-2^xf-4 x—3
x=—2
n=4
a;=27
x=l
2*= 7
Answer.
Answer.
Answer.
Answer.
Answer.
Answer.
e. Formulas and exercises.—(1) The horsepower required for flying an airplane is found by the formula:
HP
DV 375’
where HP—horsepower required
total drag of the airplane in pounds
V— velocity in miles per hour and 375 is a constant
41
25
ARMY AIR FORCES
Find the horsepower required if D=250 pounds and V=285 mph. (2) The general gas law is—
PV_PXVX
T 7\
where P=initial pressure V= initial volume T=initial absolute temperature (273° + ° C.) Pi=new pressure
Vi=new volume
and 7’i=new absolute temperature
Solve the above equation for Vx.
Solution' PV_P1V1
T ~ Tx
Multiply each side of the equation by TTX:
PVTX = PXVXT
or PXVXT=PVTX
Divide each side of the equation by PXT:
T, _PVTX .
V i — p iji Answer.
(3) In exercise (2) if P=15 pounds/square inch, T=7° C., Px = 20. pounds/square inch, Vx = 450 cubic inches, and T1—27° C., find the value of V.
(4) The formula (F—32) is used in the conversion of temperature readings from the Fahrenheit scale to the centigrade scale. If the temperature reading is 86° on the Fahrenheit scale, what would the temperature reading be on the centigrade scale?
Solution:
C=^(F-32)
(7=^ (86-32) £
<7=|(54)
C— 30° Answer.
(5) (u) Using the formula in (4) above, find the Fahrenheit reading when the centigrade reading is 20°.
(6) When will the Fahrenheit and centigrade readings be equal? (Negative values may be used.)
42
MATHEMATICS FOR AIR CREW TRAINEES
25-26
(6) The formula for ’ determining the best propeller diameter for maximum efficiency is—
D 1.03t?
where D=propeller diameter in ft.
V= velocity of airplane in ft/sec.
n—revolutions/sec.
and 1.03 is a constant
Determine the propeller diameter when U=210 mph and n=30 revolutions/second.
Solution: 210 mph= ft./sec.=308 ft./sec.
3000
308 ft./sec. £l. a
D~ 1.03 (30 revolutions/sec.)-9,97 ft‘ Answer.
(7) Using the formula in exercise (6), determine the propeller diameter when V=150 mph and n=20 revolutions/second.
(8) The horsepower necessary to propel an airplane is proportional to the cube of the velocity. If 120 horsepower is required to fly an airplane at 130 mph, how many horsepower would be required to fly it at 150 mph?
■ HP^Vf
120 _(130)3
IIP2 (150)3
(130)3 . HF2—120(150)3
„p 120(150)3
2 (130)3
HP2= 184.3 hp. Answer.
(9) On the basis of the information in exercise (8), how many times would the horsepower have to be increased to double the velocity?
(10) An airplane flying 160 mph covers a certain distance in 2 hours 30 minutes. How long would it take it to cover the same distance when flying 200 mph? 2 hr. Answer.
26. Word problems.—a. One particular type of word problem is of special importance to an aviation cadet—the type involving time, rate, and distance. <
(1) The following formula always holds true in such problems: d—rt
where d—distance, r=rate, and f=time.
(2) By applying algebraic processes, this formula may also be written in the following forms:
d r~ t
t=-r
43
26
ARMY AIR FORCES
(3) Most problems of this type may be solved by employing one of the three forms of this equation.
b. In solving word problems, be sure to be specific about what the unknowns are to represent. Try to develop the equation so that a distance, time, or rate is expressed in terms both of known quantities and unknown quantities. These may then be set equal to each other. The following rules give a procedure that can be used in solving most word problems.
Rule 1. Pick one of the unknown quantities to be determined and represent it by a letter.
Rule 2. Express the unknown quantities in terms of the chosen letter, using the relationships given in the problem.
Rule 3. Find which quantities are implied to be equal by the statement of the problem, and set them equal to obtain an equation.
Rule 4. Check the exact kind of units used in expressing all quantities and be sure only like units are equated.
(1) Example: A pursuit plane flies 90 mph faster than a bomber in still air. The bomber travels 42 miles while the pursuit plane travels 56 miles. Find the average speed of each aircraft.
Solution: The unknowns to be determined are the speeds. Picking one of these (rule 1), let—
r equal average speed of the bomber in mph. Then, in terms of r, the speed of the pursuit ship can be expressed (rule 2) as r+90 equals the average speed of the pursuit ship in mph.
From the fact that time is equal to distance divided by rate, the unknown time each ship took can be expressed thus (rule 2):
42/r=time in hours for bomber.
56/r+90=time in hours for pursuit ship.
The problem states that the two times are equal, hence
(Rule 3) 42/r=56/r + 90.
This equates hours to hours (rule 4).
Cross-multiplying the equation gives
56r=42r+3,780
or 56r—42r=3,780
hence 14r=3,780
and r=270 mph., the speed of the bomber Answers.
consequently r + 90=360 mph., the speed of the pursuit ship.
(2) Example: Two airplanes start out toward each other from two
44
MATHEMATICS FOR AIR CREW TRAINEES 26
towns 570 miles apart. They meet after 2 hours.' If the speed of one is 15 mph. more than the speed of the other, find the speed of each.
Solution: Analysis of the problem reveals that the total distance traveled by the two airplanes is 570 miles. This distance may be set up in terms of the unknown rates and known times.
Let #=mph. of slower airplane. Then a:+15=mph. of other airplane. Since d=rt, set up the distances in terms of the unknown and place them equal to the known distance:
2z+2(x+15) = 570
2x+2a:+30 = 570
4x=570—30
4z=540
x=135 mph. Answers.
xfe 15 = 150 mph.
(3) Example: An airplane flew from March Field to Moffet Field at the rate of 150 mph. It returned the folio whig day at the rate of 200 mph. and required 40 minutes less time to make the trip. How far is it from March Field to Moffet Field?
Solution: The fact used in making the equation in this case is that the distance is the same both ways. The distance will not be solved for directly.
Let »=number of hours for trip from March Field to Moffett Field. 2
Then x—^=number of hours for return trip. (Note time is changed o
to hours.)
/ 2
150z=200( x—^ \ o
i rr\ 400
150a?=200;r-x—
to hours.)
/ 2\
150z=200( x—~ I \ <5 /
1 T-r\ c\r\r\ 400 150r=200r----x—
o
— 50x—---5—
o
8
*~3 g
Distance=^-150=400 miles Answer.
o
c. One other type of word problem in algebra has considerable importance in aviation. An example will be worked to show the type:
Example: Three observation airplanes, 0-1, 0-29, and 0-5, were able to map a certain region in 3 hours. A week later, plane 0-29 mapped the region alone in 8 hours. Two weeks later plane 0-1 was
45
26
ARMY AIR FORCES
sent out and it took 10 hours to do the job. Assuming all conditions equal, how long should it take plane 0-5 to map the entire region alone?
Solution: This type of problem may be solved by finding the fractional part of the job completed by each separate airplane in a certain time and setting the sum of these fractions equal to the fraction of the total job completed in this time.
Let x=number of hours necessary for 0-5 to do the job alone. In 1 hour each airplane does the following:
1 _1_ 1
8’ 10’ x
Add these and set equal to the part of the job completed together.
8+10 + x 3
This fractional equation may be solved by using the common denominator of 120x:
15z+12x+120=40z
120=132
a 3 A
z=9:j-~ Answer.
1 o
d. Exercises.—(1) A certain bomber can fly 170 mph. loaded and 190 mph. empty. If it leaves the base at 8 PM and must be back by 5 AM, how far from the base can it go if it unloads all its bombs and comes right back empty?
(2) Two airplanes start out toward each other from two towns 680 miles apart. They meet after 2 hours. If the speed of one is 20 mph. more than the speed of the other, find the speed of each.
160 and 180 mph. Answer.
(3) Three observation airplanes, 0-1, 0-2, and 0-3, were able to map a certain region in 3 hours. Plane 0-1 can do the job alone in 9 hours, and plane 0-2 can do the job alone in 6 hours. Assuming all conditions equal, how long should it take plane 0-3 to map the region alone?
(4) An airplane flew from town A to town B at a rate of 180 mph. It returned at a rate of 210 mph. and required 20 minutes less time for the return trip. Find the distance from A to B.
420 miles Answer.
(5) An airplane leaves its base and flies due east at a rate of 175 mph. Fifteen minutes later another airplane takes off in pursuit of the first airplane at a rate of 210 mph. How long does it take for the second airplane to catch the first airplane and how far from the base does this take place?
46
MATHEMATICS FOR AIR CREW TRAINEES 26-27
(6) If planes 0-1, 0—3, and 0—5 can each map a region alone in 10, 12, 15 hours, respectively, find how long it would take them to do the job together. 4 hr. Answer.
27. Miscellaneous exercises.
(1) Solve for x:
4x—3 = 5z+9
(2) Evaluate when a=2, 6 = 3, c=5:
ab[a(a—c) — a3b2(b—c)+a26—63c]
90 Answer.
(3) Three observation airplanes, 0-4, 0-7, and 0-9. wished to map a region. Each could do the work alone in 4, 6, and 8 hours, respectively. How long will it take all three airplanes to do the job together?
(4) Solve for x:
1 . _ 8 1
-4-2=-—o r=3 Answer.
Jj Jb (J
(5) Solve for x:
______Y=o
z+l z+5
(6) One airplane can fly from San Diego to Victorville in 1% hours. Another airplane can make the trip in 2 hours. If they start from opposite ends of the course at the same time and fly toward each other, after what time will they meet? • % hr. Answer.
(7) An airplane flying a course against the wind covers 160 miles in 1 hour. At a later time the airplane flies the return trip with the wind, with a ground speed of 200 miles per hour. If the round trip took 5 hours, what was the round-trip distance?
(8) Solve for x:
3a; 4 1
, "2—3^—5) —^(5a;—12) = 0 x= 13 Answer.
(9) Evaluate when x= — 1, y=4, z=2:
xy.x2-]r4 (yz-\-x)—x3-\-yz2
(10) One observation airplane can map a certain region in 1 hour. A second airplane can map the region in 1% hours. How long will it take the two to do it together. % hr. Answer.
(11) A ferry pilot travels 800 miles by airplane and 100 miles by train in 4 hours 40 minutes. Then, at the same speeds as before, he flies 700 miles in another airplane and 150 miles by train in a total of 5 hours 20 minutes. What were the speeds of the airplanes and trains?
47
28-30
ARMY AIR FORCES
Section IV
SCALES
Paragraph
Scope----------------------------------------------------------- 28
Models__________________________________________________________ 29
Maps____________________________________________________________ 30
Miscellaneous exercises_________________________________________ 31
28. Scope.—The word “scale” is used in this section as in “scale model,” the “scale of a map,” the “scale of a drawing,” and so on. The practical use of scales in connection with maps, drawings, and silhouettes is illustrated by examples and exercises.
29. Models.—a. A true scale model of an airplane, for example, is a model which has been constructed so that the ratio of the length of any part of the model to the actual length of the same part of the airplane is the same for all parts. Thus, if the wing span on the model is 5 inches, and the wing span of the actual airplane is 55 feet, then 1 inch anywhere on the model represents 11 feet or 132 inches. Then this is the scale of the model: 1 to 132, or #32. When a scale is stated simply as 1 to 132 it means that 1 unit of any kind on the model represents 132 of the same kind of units on the airplane. In other words, the model is %32 the size of the airplane.
b. Example: The U. S. Government is encouraging youths to build scale models of various aircraft. The scale to be used is the same for all aircraft: 1 to 72. The wing span of the German Heinkel bomber (He-lllK Mklll) is 76 feet. What will be the wing span of the model?
Solution: 76 feet=76X12 inches
wing span (model) _ 1 76X12 in. ~72
mu r • . i , 76X12 76 1O2/. ,
Therefore, wing span of model=—~—-=-?-= 12% in. Answer.
I A O
c. Exercises.—The following models are all constructed to the scale of 1 to 72:
(1) The model wing span of a B-18 is 15 inches. What is the actual wing span?
(2) The over-all length of a Messerschmitt (Me-110) is 36 feet. How long will the model be? 6 in. Answer.
(3) The over-all length of a model of a B-23 is 8% inches. What ' is the over-all length of the B-23 airplane?
30. Maps.—a. A map is a scale diagram to show the disposition of geographic features on the earth such as cities, roads, rivers, etc.
48
MATHEMATICS FOR AIR CREW TRAINEES
30-31
On most maps, the scale used is conveniently stated by a diagram as in figure 23. It may also be expressed as a ratio, for example: 1 to 500,000.
MILES
Figure 23.
b. Of primary interest to airmen are aeronautical charts, or maps. The sectional charts of the United States are made to a scale of 1 to 500,000. The regional charts of the United States are made to a scale of 1 to 1,000,000.
Example: What distance (miles) does 1 inch represent on a sectional chart?
Solution: Since the scale is 1 to 500,000, then 1 inch represents 500,000 inches on the earth. 1 mile=5,280 feet=5,280X 12 inches. Therefore
500,000 inches
500,000
5,280X12
miles=7.9 miles.
1 inch represents 7.9 miles. Answer.
31. Miscellaneous exercises.
(1) On a regional chart 1 inch = how many miles?
(2) The aeronautical planning chart of the United States (3060a) is drawn to a scale of 1 to 5,000,000. On this chart, a distance of 2% inches is the same as how many miles? 177.5 miles. Answer.
(3) By direct measurement, determine the scale for the map in figure 24. From Catskill to Albany is 30 miles.
(4) How far is Schenectady from Albany? (See fig. 24.) Use the scale determined in exercise (3) above. 16 miles. Answer.
(5) By direct measurement, determine the scale used for the map in figure 25. See exercise (3) above.
(6) Is Chatham located properly in figure 25? Why? (Its location is correct in figure 24.) No. Answer.
(7) The model of the German bomber Heinkel (He-177) has a wing span of 17% inches. What must be its actual wing span?
(8) The airplane models described in paragraph 295 will be used for training gunners in range determination. How far from the model should a gunner be so that it will appear to him as though the actual airplane were 600 yards away?
549535°—43-
■4
49
31
ARMY AIR FORCES
* * I
'S. £ Mechonfcs-^$ x'
I */Zfe \
Schenectady
\ JL TROY s'
W ’
/ Ravena BmI S
I \ 1 / &C ha thorn
J /
1 JlY"" SM;/mo„/
\ x-WyCATffilH, i—»»
Figure 24.
Solution: His distance from the model should be in the same ratio to 600 yards as the model scale, or as 1 is to 72:
lnnalYg==jo> or distance=6°?r yd’=8% yd.=25 ft. Answer. 600 yd. 727 72
(9) A top view silhouette of a B-23 is to be drawn as large as possible in a space 2 feet wide. The wing span of a B-23 is 91 feet. Which of the following scales should be used: 1 to 60, 1 to 100, 1 to 20, or 1 to 50?
50
MATHEMATICS FOR AIR CREW TRAINEES
31
“ 7 i
xZ'^T & fj I
' I
\ >. Vi0< I \ \ '
) Schefie€^s^y\^\3
\ v »Tf»r ,------
/ i \ / +t
I i ALBANY ! *A4&ms
J |N ] I
K 4 /
V ’ / *//
I Chatham /
X^XATSKILL /
I \____ MASS.
I | "CONN,
KINGSTON f I
) >l
I_______ i
Figure 25.
(10) Using the information in exercise (3) above, are the scales shown in figures 24 and 25 correct?
51
32-33
ARMY AIR FORCES
Section V
GRAPHS
Paragraph
Purpose_______________________________________________________________ 32
Axes and points_______________________________________________________ 33
Reading graphs________________________________________________________ 34
Examples of graphs____________________________________________________ 35
Graphic solution of algebraic equations containing two unknowns,______ 36
32. Purpose.—a. Graphs are used to represent pictorially the relationship between two quantities, that is, how one quantity varies with another quantity. The following example shows how the pressure of a gas varies with the volume at a temperature of 0° C:
l°°|----r-------------------------------------------------
00------I-------------------------------------------------
uj 60------------\------------------------------------------
a: \
\
Figure 27.
d. It is to be noted that in going from OY to the right in the X-direction, the numbers are positive, signifying the positive direction; while to the left the numbers are negative, signifying the negative direction. On the other hand, going from OX upward in the F-direc-tion, the numbers are positive, signifying the positive direction; while downward the numbers are negative, signifying the negative direction. Hence it takes two. numbers, the coordinates, to locate a single point: one of them the F-value, denoting the vertical distance, and the other, the X-value, denoting the horizontal distance. The word ordinate is sometimes used to designate the F-value, and the
b. The horizontal line is called the X-axis. The vertical line OY is called the F-axis.
c. The position of any point is fixed by reference to these two axes. Thus, the point P (fig. 27) is located by describing it as so many units horizontally from OY (in the X-direction) and so many units vertically from OX (in the F-direction); In any graph, certain units are marked off on the X-axis, and the same or different units are marked off on the F-axis. (See figs. 28 and 29.)
53
Y Y
5 "ABSCISSA-(4,5)
4 IM 200
3 150
2 Io 100
51-3-1) . jo
F-------1- I 50
1 • - t n ft I g y y
-4 -3 -2 -1 1 2 3 4 A -20-15-10-5 5"'io"l5 20'* X
-50
- 2 -100
- 3 -150
- 4 -200
♦-Y l-Y
Figure 28. Figure 29.
33-34
ARMY AIR FORCES
word abscissa is sometimes used to designate the X-value. In figure 28, the point R is located as being 4 units horizontally from OY (4 units in the ^-direction or X=4) and 5 units vertically from OX (5 units in the F-direction or Y= 5). Symbolically, the point is described by its coordinates (4, 5) meaning X=4, and F=5. The coordinates of the point S' are (—3, 1) meaning X= — 3, and Y=l: that is, 3 units to the left of OY in the W-direction and 1 unit above OX in the F-direction.
e. This practice of plotting the position of a point by coordinate axes is used in locating the latitude and longitude of a point in a Mercator Chart.
34. Reading graphs.—To illustrate the procedure, suppose it is desired to know the calibrated air speed corresponding to an indicated air speed of 150 mph on the meter the calibration curve of which is shown in figure 30. From “150” on the horizontal axis, move up
^rnTTrni 1111111 II 111, r, ,t,, ......
£ :::::::::::zz::z::zEEEEEEEEEEEE±||EEE
2 2oo
ft. ::±zEEE---:-::::::::Er??F-:;H-$r?:S
• 200* 3OCf
TEMPERATURE (®C)
Figure 31.—Pressure and temperature of a gas at constant volume.
1200, hence 1:30 PM becomes 1330 hour and 5:55 PM becomes 1755 hour. The use of this system decreases the chances for making errors and for this reason it has been adopted for use in the U. S. Army Air Forces.
d. Sunset graphs.—These graphs enable the pilot to determine the time of sunset for any position on the earth. The 24-hour system of keeping time is used in sunset graphs.
(1) Instructions for use.— ((a) Enter the top or bottom scale with proper date.
(5) Move vertically down or up to the curve for observer’s latitude (observer’s position).
(c) Move horizontally to the right or left and read local civil time of sunset on vertical scales at the side.
(2) Find the sunset time for November 1 at latitude 30° N. (Follow instructions for use.) 1721 hr. Answer.
55
34
ARMY AIR FORCES
(3) Find the sunset time for May 15 at latitude 50° N.
(4) Find the sunset time for May 20 at latitude 30° N.
1848 hr. Answer.
(5) Find the sunset time for June 10 at latitude 10° N.
(6) Find the sunset time for February 10 at latitude 40° N.
1736 hr. Answer.
(7) Find the sunset time for October 20 at latitude 30° N.
TiOO trarrm H+HI HHW+H4-l4mi4W4 H-I414H-H ITTITffl
th
iFrELtM Ek^ F ' ! ; ,.-±:
i i+v-Agw+idhF/i+E+pif
*" j£E|EEEEEEE±5FF^fOFF|EEEtE;ffi+EEE
2 'ooo Fit
'«* k| ||i||teWE|| illlfc
SUNSmpIOhp?
Is»a IfWffimtffimWiSWwIWHllWfemR
JANI MARI MAY I JUL 1 SEP 1 NOVI JANI
DAY
Figure 32. —Sunset graph.
e. Fuel consumption graph.—This graph (fig. 33) shows the relationship between the air speed and the fuel consumption.
(1) At an air speed of 180 mph., the fuel consumption is_______
gallons/hour.
(2) At an air speed of 168 mph., the fuel consumption is_______
gallons/hour.
(3)____If the fuel consumption is 53 gallons/hour, the air speed is mph.
56
MATHEMATICS FOR AIR CREW TRAINEES
34-35
6°llllin II III IRH ril 11HI H1TII Fl II1111111 rTTTTTTTI
g so FUEL CONSUMPTION ^;+p-E
Ho z|:zzz:z::zz:|:z:±::zzz::z:::::z:::::iz:z::z:z:
3 z±zzz::z:zzzz|:z-|zz:zz-::::::zzzz::::::::z:z:
'® ++++nj+|++++H=+++n+
<*jlll III II HII III II II II IIII Mt
0 HO 80 120 160 200
AIR SPEED (mph)
Figure 33. —Typical fuel consumption curve.
(4 ) If the fuel consumption is 39 gallons/hour, the air speed is mph.
35. Examples of graphs.—a. The graphs that will be used and constructed will consist of a succession of points plotted with reference to coordinate axes and connected by a smooth line forming a straight line or a curve. The coordinates are obtained either from a formula or from empirical data (that obtained by observation).
b. Graphing from an equation.—(1) In algebraic work, various expressions suitable for graphing are employed, especially the equations. Consider the equation y—2x=5, or y=2xf-5, and draw its graph. Since it is of the type ax+by=c (a, b, c some known constants), the graph of which is always a straight line, the graph of ?/=2fc+5 should be a straight line. If a number, for example 3, is substituted for x in the quantity 2x+5, then the quantity 2r+5 (or y) takes on the value 11. In a similar manner, many more pairs of numbers can be obtained which satisfy the above equation. For example:
when x= 3, y= 11,2(3) +5 = 11
x= 2, ?/= 9, 2(2) +5=9
x= l,y= 7,2(1) +5=7
x= 0, t/= 5,2(0) +5=5
x=-l, y= 3, 2(-l)+5=3 x=-2, y= 1, 2(-2)+5 = l z=-3, y=-l, 2(—3)+5 = —1, etc.
57
35
ARMY AIR FORCES
Construct a table of these values, set up a pair of coordinate axes with suitable scales, and plot the points. Always remember that the two numbers describing the location of a point will satisfy the equation being graphed. After the points are plotted, notice that
♦Y
io /
8 /1
« /
/
y/ 2 ;
_V <--1 * .| A1- |-X=2 j--4--1-S—» > Y
-5 -4 -1/-2 -1 1 2 5 4 5
/ -2
Figure 34.
they all lie in a straight line. Draw a straight line through the points. Every point on this line has a F-value and an X-value satisfying the equation.
(2) A freely falling object will fall a distance d feet in t seconds as given by: 1
Use 32 for g, and the formula then becomes: d=16f2. This formula is of a different type from those discussed above. After calculating a few values which satisfy this equation and drawing a graph representing the formula, figure 35 is obtained.
d
300 /
/ t_____d_
250 . I 0 0
/ 1 16
200 /d-16t2 2 64
/ 3 144
150 j 4 256
100 • / ' ‘
50 /
0 1.....—।----1-- —।-----> f
1 2 3 4 5
Figure 35.
58
(2x+ 5)
* >
i •
• i
* »
-3 -1
-2 1
-1 3
O 5
1 7
2 9
5 11
i i
i <
• •
MATHEMATICS FOR AIR CREW TRAINEES 35
Exercise.—From the graph—
How far will an object fall in 2% seconds?
Check by calculating this distance from the formula.
c. Exercises.—Plot the following points on a coordinate system of graph paper:
(1) (2, 3); (5, 7); (8,0); (0,5).
(2) (—2, 5); (-7, 8); (-5, 0); (-2, 3).
(3) (—2, -4); (0, -5); (-4, -7); (-2, -3).
(4) (1, -5); (2, -4); (5, -8); (2, -7).
(5) Three corners of a rectangle are at (1, 4), (4, 8), and (9, —2). Find the coordinates of the other corner.
(6) Compute a table of values and draw a graph of the following:
(a) y=2x; y—3x=5.
(6) y=6;y—x=0.
(c) y—7 — — 2z; 2y—3z=9.
(d) x— 3; 2 g—1-
1
(e) y=^x2-, y2=4x.
(/) d=5—f2.
(g) 2m=16+8/~t2.
(7) From the graph in figure 35, read off—
(a) The distance that an object will fall in 1% seconds and in 3% seconds.
(5) How long it will take an object to fall 50 feet? 100 feet? 225 feet?
(8) The velocity of sound in air depends on the temperature of the air. By use of the following data, draw a graph showing how the velocity varies with the temperature.
Velocity (ft. per sec.) _ 1, 030
Temperature (F.)_____ —30°
1,040 1,060 1,080 1,110
-20° 0° 20° 50°
1, 140 80°
1, 170 110°
From the graph, find the velocity if the temperature is 35°; 10.5°; —25°; 120°.
1095; 1070.5; 1035; 1180 Answers.
(9) The effective disk area of a propeller depends on the diameter of the propeller. By use of the table below, draw a graph showing how the effective area A in square feet varies with the diameter in feet.
59
35-36
ARMY AIR FORCES
A (sq. ft.) _ _ 4. 2 6. 5 9. 4 12. 7 16. 6 21
D (ft.) — 8 10 12 14 16 18
From the graph, find the area if the diameter is 9 feet; 12.5 feet; 14% feet.
36. Graphic solution of algebraic equations containing two unknowns.—a. It is sometimes necessary to find a pair of numbers that will satisfy two equations at the same time. One method of doing this is to graph each equation on one set of coordinate axes and find the intersection of the curves.
Example: Find the values of x and y which will satisfy the following two equations simultaneously:
2x—y= 3 •
3a?+2y=8
C\ 'Y /B
\5 ■ O /
\ " /
* \ '7
3 ■ \ /
3 \ V
2 \ /
1...VP (2,1)
-X ■ .. - 1-1 । ■ । --i / i \i-1-1——► Y
-4 -3 -2 -1 1/ 2 \3 4
-1 / \
/ \ °*
*2 / \ *
■y
/-5 \D
a/ _y
Figure 36.
Solution: By graphing each equation separately on the same pair of coordinate axes, the line AB is obtained, every point of which has coordinates satisfying the equation 2x—y=3; and the line CD is obtained, every point of which has coordinates satisfying the equation 3z+2?/=8.
The intersection of CD and AB is the point P the coordinates of which (2, 1) are the only ones satisfying both of the given equations.
b. This graphical method usually gives approximate results only,
60
MATHEMATICS FOR AIR CREW TRAINEES 36
because of errors in measuring line segments when determining certain coordinates.
c.~ If the lines are parallel, it is obvious that no coordinate values will satisfy both of the given equations.
d. Exercises.—Solve graphically—
(1) x+?/=3
x—2y=Q
(2) 2y—3x=0
4?/+3a;=-18 (-2, -3) Answer.
(3) se-H2?/=4
3x—y=6
(4) 5y—3=0
10?/+3a;=4 (—%) Answer.
e. Under certain circumstances, curves other than straight lines are plotted in pairs and their intersection found.
qV'4,8) e L
v\ 7 A
\ X" /
\ / \ 3' >
»Z \
-
>z Zv
Scole • 1cm • 10 mph
Figure 55.
MATHEMATICS FOR AIR CREW TRAINEES
45-46
north, making it 10 cm. long (1 cm. = 10 mph.; hence 10 cm. = 100 mph.).
(c) Draw the wind vector away from 110°, making it 2% cm. long (2% cm.=25 mph.) and from the head of the air speed heading vector.
(d) Air speed heading vector+wind vector=ground speed track vector (Va+Vw=Vff).
(e) The angle between the ground speed track vector and the north line determines the course or track. The length of the ground speed track vector, according to the scale used, determines the ground speed.
(/) The drift angle is the angle between the air speed heading vector and the ground speed track vector. Since this is a case of left drift, the drift angle of 14° is subtracted from the heading of 56° to obtain the course 42°.
b. Exercises.—(1) On a secret maneuver, the pilot has been ordered to fly on a heading of 90° until further orders. Air speed of the airplane is 180 mph. After 1 hour of flight, motor trouble develops and a landing must be made. He is told over his radio that wind has been 40 mph. from 135°. What has been his track and ground speed?
(2) The pilot observes an enemy scouting party. To avoid being seen, he changes his heading (as point “A”) and flies on a heading of 315° with air speed 150 mph. If wind is 30 mph. from 45°, what will be his new course and ground speed? C = 304°, GS = 153 Answer.
(3) Because of an error in plotting his course, a pilot finds himself over unfamiliar territory. His air speed has been 160 mph., his heading has been 180°, and he verifies that wind throughout his flight has been 20 mph. from 315°. What has been his track and ground speed?
(4) Heading has been 30° and air speed of plane is 180 mph. If wind is 60 mph. from 240°, what has been the course (or track) and ground speed? C=37°, GS=234 Answer.
(5) Figure 54 was carefully drawn, but was not properly labeled. Give two different and equally reasonable sets of facts which it might be intended to represent.
46. Type II.—The type of problem next to be considered is one that must be worked out before the flight is started. As a flight is being planned, a certain course is desired. A line showing the desired course is laid off on the map and its azimuth (course angle) is measured. Wind direction and speed are accurately observed and reported at frequent intervals. The air speed of the airplane to be used is known. The heading which the airplane must have in flight in order to make good the desired course, may be determined. Heading and ground speed over the actual course may be read from a vector diagram.
73
46
ARMY AIR FORCES
a. Example.—Course to be flown is 308°. Air speed of the airplane is 110 mph. and wind velocity is 15 mph. from 230°. Find the heading and ground speed.
Note.—See back of manual for figure 57.
b. Solution: First, the desired course is indicated by a line of indefinite length drawn in the direction 308°. Since the parallelogram method is to be used to find the unknown parts of the vector diagram, the wind velocity vector originates at the same point as the course line. Wind velocity, 15 mph., is represented by a line 1.5 cm. long from the direction 230°. Since air speed, 110 mph., is known but heading is not, the compass is set with a radius of 11 cm. (11 cm. = 110 mph.), and, using the head of the wind velocity vector as center, strike an arc which will intersect the course line. The line drawn from this point to the head of the wind vector has the length and direction
Scole: 1 Inch = 40 mph
Figure 56.
of the air speed heading vector. The segment of the course line cut off by the intersecting arc is the length representing the ground speed. The parallelogram is completed in such a way that all three vectors originate from the origin point with the wind velocity vector and the air speed heading vector as sides and the ground speed track vector as the diagonal.
c. The solution of the same problem by the triangle method is given next.
(1) Draw the north line.
(2) Draw the wind velocity vector from 230°, making it 1.5 cm. long.
(3) Draw a line of indefinite length in the direction 308°, indicating the desired course.
74
MATHEMATICS FOR AIR CREW TRAINEES
46
(4) Draw the air speed heading vector by placing one end of a ruler on the head of the wind vector and mark where a vector 11 cm. long will touch the course line.
(5) The segment of the course line cut off by the head of the air speed heading vector represents the ground speed.
(6) The angle between the air speed heading vector and the north line measured clockwise determines the heading.
-
Scale* 1cm = 10mph
3 ° 8 . XL 2 3 0 0
/(A---'
Figure 58.
d. Exercises.—(1) Course to be flown is 270°. Air speed of the airplane is 120 mph., and wind is 40 mph. from 45°. ' Find the required heading and the ground speed.
(2) Wind is 30 mph. from 180°. Desired course is 45°, and air
75
46-48
ARMY AIR FORCES
speed of the airplane is 140 mph. What should be the heading, and what will be the ground speed? H=54°, GS=160 Answer.
(3) Air speed of the airplane is 125 mph. What will be the required heading to fly a course 135° if wind is blowing 25 mph. from 225°? What will be the ground speed?
(4) Course to be flown is 225°, and wind is 30 mph. from 90°. If air speed of the airplane is 160 mph., find the required heading and the ground speed. H = 217°, GS=180 Answer.
(5) Wind is 45 mph. from 10°, and air speed of the airplane is 165 mph. If the course to be flown is 150°, what will be the required heading and what will be the ground speed?
47. Type III .—If the air speed, heading, ground speed, and track (angle of actual course flown) are known, the wind velocity can be found. These cases arise when the first two factors are obtained from instrument reading and the last two are computed by timing the flight between two landmarks on a map.
a. Plot both the air speed heading vector and the ground speed track vector from the given data.
b. Draw a vector with tail at the arrow end of the air vector and arrow at the arrow end of the ground vector. This is the wind vector.
c. By drawing a north line through the tail of the wind vector, one can measure the azimuth of the wind vector, that is, get the wind direction. The length represents the wind speed.
d. Note that the sum of the wind vector and the air vector is the ground vector; that is, Vw-}-VA= VG.
e. Example: An airplane’s air speed is 155 mph. and its heading is 240°. By computation from a chart, it is found that the ground speed is 170 mph. and the track is 251°. What is the wind velocity?
From the vector diagram the wind velocity is 33 mph. from 129°.
Answer.
Note.—See back of manual for figure 59.
48. Summary.—In any vector triangle, there are six quantities involved: the length and the direction of each of the three vectors.
a. It can be seen from the preceding examples that when any four of the six quantities are given, the other two can always be found. However, the triangle cannot be solved if less than four of the quantities are given.
b. The two unknown quantities can always be found by using Vwf~VA=VG, or briefly WAG.
c. Exercises.— (1) If an airplane with an air speed of 180 mph. flies with a heading of 160° in a 20-mph. wind blowing from 215°, what are its track and ground speed?
76
MATHEMATICS FOR AIR CREW TRAINEES 48-49
(2) If an airplane must fly a 600-mile course, which is 350° in 4 hours, while a 40-mph. wind is blowing from 25°, what must its air speed be? AS=184, H=357° Answer.
(3) If an airplane with an air speed of 310 mph. must fly a 200° course in a 35-mph. wind blowing from 180°, what must its heading be, and what will its ground speed be?
(4) An airplane’s air speed is 160 mph., its heading 10°. The wind velocity is 28 mph. from. 116°. What are the track and ground speed?
GS=170, T=l° Answer.
(5) An airplane with an air speed of 130 mph. must fly a true course of 127° in a 25-mph. wind from 221°. What must be the heading, and what will the ground speed be?
(6) An airplane with an air speed of 150 mph. and a heading of 65° flies a course of 80° with a ground speed of 140 mph. Find the wind velocity. WS=40, WD = 357° Answer.
(7) An airplane’s air speed indicator reads 205 mph. and its compass 67°. The navigator notes that the airplane passes over two landmarks which are given on his map. The distance between them is 310 miles and the direction of the line between them is 76°. If the airplane took 1 hour, 40 minutes to fly from one to another, what is the wind velocity?
(8) An airplane with an air speed of 245 mph. must fly a course of 310° and return to the same base along the same line. A 280° wind is blowing 28 mph. How long should the pilot fly out along the line if he has 3 hours’ fuel? What heading must he take in and out?
T=1 hr. 39 min.; H in=133°; H out=307° Answer.
(9) A pilot flying an airplane which has an air speed of 200 mph. takes off from field A to fly to field B which lies 800 miles and 315° from field A. A wind of 30 mph. is blowing from O°. The pilot miscalculates his course and lands at field C to get his bearing. C is 565 miles from both A and B, and is due north of A, and due east from B. How much time did he lose by not flying a direct line?
49. Miscellaneous exercises.—(1) If an airplane with an air speed of 160 mph. and a heading of 312° flies in a 35-mph. wind from 20°, what are the track and ground speed?
(2) An airplane with an air speed of 135 mph must fly a course of 93° in a 30-mph. wind from 225°. What must be the heading, and what will the ground speed be? H=102°; GS=153 Answer.
(3) An airplane’s heading is 35°. A 25-mph. wind is blowing 95°. If the airplane’s air speed is 120 mph., what are its ground speed and track?
(4) An airplane’s air speed is 210 mph. A 45-mph. wind is blowing 195°. If the airplane must fly a course of 248°, what must the heading
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be and what will the ground speed be? H=238°; GS= 180 Answer.
(5) An airplane must fly a 75° course and maintain a ground speed of 165 mph. A 40-mph. wind is blowing from 160°. What must the air speed and heading be?
(6) In a 30-mph. wind from 350°, an airplane flies with a heading of 235° and an air speed of 170 mph. What are the track and ground speed? T=226°; GS=185 Answer.
(7) An airplane must fly a 23° course in a 25-mph. wind from 272°. If the airplane’s air speed is 140 mph., what must the heading be? What will the ground speed be?
(8) An airplane must fly from A to B in 4 hours 30 minutes. These fields are 630 miles apart and the direction of the line between them is 210°. If the wind is 32 mph. from 105°, what must the heading and air speed be? H = 196°; AS=136 Answer.
(9) An airplane with an air speed of 135 mph. and a heading of 355° passes over a landmark, and 1 hour 20 minutes later passes over another which is 200 miles and 10° from the other. What is the wind velocity?
(10) An airplane with an air speed of 170 mph. and a heading of 177° flies directly over a road the direction of which is 165°. The pilot sees smoke blowing from a chimney from 275°. What are the ground speed and wind speed? GS=178;WS=37 Answer.
(11) An airplane with an air speed of 190 mph. must fly along a course of 310° and return along the same line. Total flying time is to be 2 hours 30 minutes. If the wind is 40 mph. from 190°, how long should the pilot fly out along the line before turning back?
(12) A pilot flying an airplane with an air speed of 175 mph. must land at either field A or field B as soon as possible. Field A is 117 miles and 30° from the airplane, while field B is 141 miles and 195° from the airplane. If the wind velocity is 40 mph. from 335°, to which field should the pilot fly? Arrives at B 6 minutes sooner. Answer.
(13) An airplane with an air speed of 150 mph. takes off from a field and flies with a heading of 305°. A 35-mph. wind is blowing from 200°. Forty minutes later an airplane with an air speed of 190 mph. takes off from the same field to overtake, the first airplane. What heading must the pilot fly, and how long will it take him to overtake the other airplane?
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Section VIII
PLANE GEOMETRY
Paragraph
Purpose and scope_________________________________________________________ 50
Basic definitions____________।____________________________________________ 51
Elements of plane geometry------------------------------._________________ 52
Constructions with ruler and compass_____*________________________________ 53
Special properties of triangles___________________________________________ 54
Relationships between- two triangles_______________________________________ 55
Special properties of circles_____________________________________________ 56
Special properties of quadrilaterals______________________________________ 57
Special properties of miscellaneous figures_________________________________ 58
Solid geometry—definitions and properties of some geometric solids________ 59
Intersection of a sphere and a plane-------------------------------------- 60
50. Purpose and scope.—The purpose of this section and the one following is to provide a working knowledge of the properties of certain geometric figures. Certain useful ruler-compass constructions are also to be included.
51. Basic definitions.—a. Plane geometry.—That branch of mathematics which deals with figures on flat or plane surfaces.
b. Axiom.-—A. mathematical statement whose truth is accepted without proof.
Examples: (1) If equals are added to equals, the sums are equal.
(2) If equals are multiplied by equals, the products are equal. These axioms are applicable to many branches of mathematics.
c. Postulate.—A geometric statement accepted to be true without proof.
Examples: (1) Two straight lines cannot intersect at more than one point.
(2) Two lines in the same plane must either be parallel or they must intersect.
Postulates apply specifically to geometry.
d. Proposition.—A statement of a geometric truth.
e. Theorem.—A proposition to be proved.
Example: The sum of the angles of a triangle is equal to 180°. A proof is the logical argument which shows a proposition is true.
/. Corollary.—A theorem that is similar in content to a previously proved theorem. A corollary often follows as an added conclusion to a proved theorem with very little added proof.
Example:
Theorem: If a triangle is isosceles, the angles opposite the equal sides are equal.
Corollary: An equilateral triangle is equiangular.
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52. Elements of plane geometry.—a. Point and line.—The two undefined elements of plane geometry. A line has no definite length. A segment is a portion of a line with a definite length.
b. Straight line.—A line in a certain direction.
c. Broken line.—A line made up of straight line segments, not all of which are necessarily in the samg direction.
d. Curved line.—A line no portion of which is a straight line segment.
e. Parallel lines (//).—Are lines that lie in the same plane and do not meet however far extended.
/. Angle.—The opening formed by two straight lines drawn from the same point, called the vertex.
Figure 60.—Straight line.
Figure 61—Broken line.
Figure 62.—Curved line.
i
(1) The angle is described as angle BAC ( A BAC), Z CAB, or Z 1. The vertex of the angle is A. AC and BA are the sides. The size of an angle does not depend on the length of the sides, but on how much side AB must revolve about A to take the position of side AC.
(2) In labeling an angle the letter at the vertex must be in the middle. Thus, Z BAC is correct and Z ABC is incorrect.
g. Angles, classification of.—Angles may be classified as follows: Acute angle.—An angle more than 0° and less than 90°.
(1)
(2)
(3)
(4)
(5)
Right angle.—An angle of 90°.
Obtuse angle.—An angle more than 90° and less than 180°.
Reflex angle.—An angle more than 180° and less than 360°.
Straight angle.—An angle of 180°.
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MATHEMATICS FOR AIR CREW TRAINEES
52
h. Perpendicular.— If two lines meet at right angles, they are said to be perpendicular.
/ 90° 9O°\
90° I
Figure 64.—Perpendicular lines.
i. Polygon.—Any closed figure bounded by line segments.*
Figure 66.—Quadrilateral—A polygon of four sides.
j. Circle.—A closed curve, all points of which are equally distant from a point within, called the center.
I D/om eTer j j
\ J>/ / I
\?Z /fr
~ ——c
Figure 71.—Bisecting an angle.
(1) With A as center and any convenient radius, strike an arc intersecting the sides of the angle at M and N.
(2) With M and N as centers and any radius greater than half MN, describe arcs intersecting at P.
(3) Draw line AP, which bisects the angle BAC.
e. Parallel to a line through a point.— To construct a parallel to a line L though a point P, figure 72:
P/ L
____________L
X7m
Figure 72. —Constructing a parallel to a line through a point.
(1) Through P, draw any line cutting L at A.
(2) With A as a center and any radius, strike an arc cutting AP at N and L at M.
(3) With P as center and the same radius, strike an arc cutting AP at R.
(4) Take as radius the distance from N to M and with R as center, strike an arc cutting the arc previously drawn ((3) above) at /S'.
(5) Draw line PS.
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/. Perpendicular at a point.—To construct a perpendicular at a given point P in a line L, figure 73:
-----------Le—/!_(_
Figure 73. —Constructing perpendicular at a point.
(1) With P as center and any radius, strike off arcs on each side of P intersecting L at M and N.
(2) With a larger radius and M and N as centers, strike off arcs intersecting at Q.
(3) Draw line PQ.
g. Perpendicular from a point.—To construct a perpendicular to a line from a point P outside the line at L, figure 74:
P
Figure 74. —Constructing perpendicular from a point.
(1) With P as center and radius large enough to intersect L, strike off an arc intersecting L at points M and N.
(2) With radius greater than one-half MN and M as center, strike off an arc on opposite side of L from P. Using same radius and N as center, strike off arc intersecting last arc at Q.
(3) Draw line PQ.
h. Equilateral triangle.—To construct an equilateral triangle, figure 75:
Figure 75. —Constructing equilateral triangle.
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MATHEMATICS FOR AIR CREW TRAINEES 53
(1) Draw a line L of indefinite length.
(2) With any point P on L as center and any radius, strike off an arc intersecting L at M.
(3) With M as center and same radius, strike off an arc intersecting first arc at Q.
(4) Draw lines PQ and QM.
In an equilateral triangle, each angle is 60°.
i. Sum of two angles.—To find the sum of two angles 1 and 2, figure 76:
* —
(a) (h)
o X’ L
/ (C.)
Figure 76. —Finding sum of two angles.
(1) With A and B as centers and any radius, describe arcs cutting the sides of the angles at M, N, P, and Q, figures 76 (a) and (b).
(2) With same radius and 0—one end point of line L, figure 76 (c), as center—strike an arc intersecting L at R.
(3) With R as center and MN as radius describe an arc intersecting the first at T.
(4) With T as center and the distance PQ as radius, strike off an arc intersecting the first arc at B.
(5) Draw line OS.
7^ I / I
1____1 \ 1----
(A) ’/ / \ (b)
p3O° I
(<=)
Figure 77.—Constructing 75° angle.
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j. Adding angles.—Adding angles consists of copying the angles so that they have a common vertex and a common side between them. This construction, coupled with other of the basic constructions, makes it possible to construct many specific angles.
Example: Construct a 75° angle:
(1) Construct a perpendicular (90°), figure 77 (a), to a line and bisect one of the angles. This gives a 45° angle.
(2) Construct an equilateral triangle, figure 77 (b), and bisect one of the angles. This gives a 30° angle.
(3) Add the 45° and 30° angle, figure 77 (c).
k. Constructing triangle, two sides and included angle given.—To construct a triangle when two sides b and c, figure 77 (a) and the included angle 1, figure 77 (b) are given.
= /
Z 1 1
A I 5 ‘B
' r
g .." < V*
/ h \
----------------b--------\ B
Figure 99.—Trapezoid.
pendicular distance between the parallel sides. The area of a trapezoid is equal to half the product of the sum of the two parallel sides b and b' by the altitude.
A=l/2h(b+b')
Example: Find the area of a trapezoid with altitude of 15 feet and bases of 25 and 31 feet.
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57-58
A=X 15(254-31)
. _15-5J5
~ 2
840 . ,
= -=- = 420 sq. ft.
b. Exercises.—(1) A baseball diamond is a square 90 feet on a side. Find the area.
(2) One parallel side of a trapezoid is 15 feet, and its altitude is 6 feet. If the area is 75 square feet, find the other parallel side.
10 ft. Answer.
(3) The diagonal of a rectangle is 37 units. Find the area if one side is 12 units.
(4) Find the side of a square whose area is equal to the sum of the areas of three squares whose sides are 8, 9, and 12 units.
17 units. Answer.
(5) Find the area of a trapezoid with altitude 6 and bases 7 and 13.
(6) Find the altitude of a parallelogram with a 12-foot base, if the area is 120 square feet. 10 ft. Answer.
58. Special properties of miscellaneous figures.—a. An ellipse is the path of a point whose distance from two fixed points is a constant sum. The orbits of the earth and other planets are ellipses. Ellipses are also used in making machine gears.
Figure 100.—Ellipse. *■
b. A parabola is the path of a point which is equidistant from a line and a fixed point not on the line. The parabola has the property that if a light is placed at its fixed point, all of the light is reflected in parallal rays. Therefore parabola reflectors are used for searchlights. The reverse process is also true and therefore reflecting telescopes use
549535°—43---7
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parabolic mirrors. The paths of projectiles in a vacuum are parabolas. Also the graphs of many scientific formulas are parabolas.
Figure 101.—Parabola.
c. A hyperbola is the path of a point whose distances from two fixed points have a constant difference.
Figure 102—Hyperbola.
Hyperbolas were used during World War I in sound ranging to locate distant invisible enemy guns.
d. The ellipse, parabola, and hyperbola each have an equation and the curves may be obtained by graphing the equation. This was done in the section on graphing for the parabola.
59. Solid geometry-—Definitions and properties of some geometric solids.—a. Polyhedron.-—A geometrical solid formed by portions of planes called faces, whose lines of intersection are called edges and whose points of intersection are called vertices.
(1) Prism.—A polyhedron generated by a plane polygon moving
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MATHEMATICS FOR AIR CREW TRAINEES 59
through space always parallel to a fixed plane and in such a way that the vertices of the polygon move along straight lines. The bases are
• Figure 103.—Triangular (a) and pentagonal (6) prisms.
congruent polygons, and the faces are parallelograms. The volume of a prism is the product of the area of the base times the altitude (the perpendicular distance between the polygon bases). V=Bh where B represents the area of the base.
Example: Find the volume of a square based pyramid with altitude 6 inches and side of base 4 inches.
B=4-4 —16 sq. in.
V—Bh= 16X6 = 96 cu. in.
(2) Pyramid.—A polyhedron which has a polygon for a base and the other faces are triangles all of which meet at a common vertex.
Figure 104,—Pyramid.
The volume of a pyramid is equal to one .third the product of the area of the base by the altitude (the perpendicular distance from the common vertex to the base).
V=±Bh o
Example: Find the volume of a square-based pyramid whose altitude is 6 feet and whose base edge is 4 feet.
V=\Bh B=4 • 4=16 o
1 96
16.6=^ = 32 cu. ft.
3 o
99
Cb)
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b. Cylinder.—A surface generated by a closed plane curve moving along a straight line parallel to a given plane. The altitude of a
cylinder is the perpendicular distance between the parallel bases. The volume of a cylinder is equal to the product of the area of the base by the altitude.
V=Bh
The most commonly used cylinder is the circular cylinder.
Example: Find the volume of a circular cylinder whose altitude is 5 inches and the radius of the base is 7 inches.
B=7rr2=3.14X7X7
B= 153.86 sq. in.
V=Bh= 153.86X5 = 769.3 cu. in.
c. Cone.—A surface generated by all projection lines from a fixed point to a plane closed curve. The fixed point should not be in the same plane as the curve. A right circular cone is made by revolving
Figure 106.—Right circular cone.
a right triangle about one of its sides as an axis. It is the most commonly used cone. The volume of a cone is one-third the product of the base area by the altitude.
100
Figure 105.—Cylinder.
MATHEMATICS FOR AIR CREW TRAINEES
59
V=Uh o
Example: Find the. volume of a cone with 4 feet altitude and circular base with radius 8 feet.
B=7rr2=3.14X8X8
= 3.14X64 = 200.96 sq. ft.
V=|X2OO.96X4 O , *
= 267.95 cu. ft.
Note the similarity in the formulas for the volumes of prisms and cylinders, also pyramids and cones.
d. Sphere.—A closed surface whose points are all equally distant from a fixed point inside the surface called the center.
(1) Aline segment (R, fig. 107) which joins any point on the surface to the center is a radius of the sphere.
(2) A line segment (D, fig. 107) which passes through the center and has its end points on the surface is a diameter of the sphere.
(3) The volume of a sphere is given by the formula:
4 S=|7rr3 o
(4) The surface area of a sphere is given by the formula:
S=47rr2
Example: Find the surface area and volume of a sphere of 4-foot radius,
101
Figure 107.—Sphere.
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S=47rr2=4X3.14X4X4
=200.96 sq. ft.
4 4
V=7rr3=-X3.14X4X4X4 o o
=267.95 cu. ft.
e. Exercises:—(1) Find the volume of a—
(a) Sphere whose radius is 3 inches.
(6) Hemisphere whose radius is 5 inches.
261.67 cubic inches. Answer.
(c) Cylinder whose altitude is 10 inches and circular base with' radius 3% inches.
(d) Prism with base area of 25 square inches and altitude of 2 feet.
50 cubic inches. Answer.
(e) Cone whose altitude is 12 inches and whose base is a circle of 9-inch diameter.
(f) Prism with rectangular base (sides 10 and 7 inches) and altitude
11 inches. 770 cubic inches. Answer.
(g) Pyramid with triangular base (5=6 inches, h=4% inches) and altitude of 11 inches.
(2) Find the surface area of a—
(a) Sphere of 5 units radius.
(6) Sphere of 12 units diameter. 452.16 square units. Answer.
(c) Sphere with a great circle area of 252 square inches.
(3) Find the radius of a— / 22\
(a) Sphere whose surface area is 2,404 square inches (use 7r=-y-k
(6) Sphere whose volume is 113% cubic inches (use tt=^-)•
3 ins. Answer.
60. Intersection of a sphere and a plane.—a. Circular intersections:—If a plane intersects a sphere, the intersection is a circle.
102
Figure 108.
MATHEMATICS FOR AIR CREW TRAINEES
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(1) When a plane which intersects a sphere passes through the center of the sphere, the circle cut out is called a great circle of the sphere. The center of the great circle is the center of the sphere.
(2) When the intersecting plane does not pass through the center of the sphere, the circle cut out is a small circle of the sphere.
b. Arcs on sphere.-—Through any two given points on a sphere, an arc of a great circle may be drawn.
(1) When the two points are not extremities of a diameter, one, and only one, great circle can be drawn.
(2) When the two points are the extremities of a diameter, every circle through them is a great circle.
c. Tangent planes and lines.—A plane which intersects a sphere at just one point is the tangent to the sphere at that point.
(1) A tangent plane is perpendiular to the radius of the sphere drawn to the point of tangency.
(2) Every line in the tangent plane and through the point of tangency is a tangent line to the sphere.
Section IX
SPHERICAL GEOMETRY
Paragraph
Basic definitions_______________________________________________________________ 61
Physical applications of spherical geometry_____________________________________ 62
61. Basic definitions.—a. Quadrant.—One-fourth of a great circle, for example, arc AB, figure 109, is a quadrant.
b. Distance between two points on a sphere is measured along the great circle which connects them, for example arc BD connecting points B and D, figure 109.
47+
\ dV 1 /
\ \ / /
\ \ i /
c
Figure 109—Sphere.
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c. Dihedral angle.—The angle formed by two intersecting planes. A dihedral angle is measured numerically by the plane angle formed by one perpendicular in each plane drawn to a common point on the edge of the dihedral angle. Z.BOD measures the dihedral angle between the great circle A through D, and the great circle A through B.
d. Axis of a circle.—The diameter of a sphere, perpendicular to a circle of that sphere, is called the axis of that circle. The axis of the earth is generally taken as the diameter perpendicular to the equator, that is, as the axis of the equator.
e. Pole of a circle.—The poles of a circle are the points at which the circle’s axis meets the sphere. Points A and C are poles of circle BDEF.
Note.—To draw the great circle which has a given point as pole, measure off a quadrant’s distance on the surface of the sphere from the given point.
Example: The distance f rom either the north pole or the south pole ' to the equator is a quadrant.
f. Spherical angles.—A spherical angle is formed by two minor arcs of great circles having a common end point. Arcs AB and AD, figure 109, form a spherical angle. It is measured as follows:
(1) It equals numerically the angle formed by the tangents to the arcs at the vertex of the angle.
(2) It equals numerically the dihedral angle formed by the planes of its sides.
Note.—The sides of a spherical angle are arcs of great circles and each of these great circles lies in a plane. These planes determine the dihedral angle.
(3) It equals numerically the arc intercepted on the great circle of which its vertex is a pole.
(4) It is apparent that any angle on the earth’s surface is in reality a spherical angle. For small distances these angles are sometimes considered as plane angles, since the great circles determining the sides are almost straight lines if the distance is small enough.
g. Spherical polygon.—A closed figure formed by three or more minor arcs of great circles on a sphere is called a spherical polygon.
h. Spherical triangles.—A spherical polygon of three arcs is a spherical triangle. Each of the arcs is called a side. The figure determined by arcs DE, DC, and CE, figure 109, is a spherical triangle.
(1) Properties of spherical triangles:
(a) Any side of a spherical triangle is less than the sum of the other two sides. (The sides are arcs of great circles and are measured in degrees. The number of degrees is the same as the number of degrees of the central angle subtended at the center of the sphere in the plane of the circle.)
(6) The sum of the angles of a spherical triangle is greater than
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61-62
180° and less than 540°. Any angle of a spherical triangle must be less than 180°.
(c) The sum of the sides of a spherical triangle is less than 360°. A very important application of the spherical triangle is" met in celestial navigation.
(2) All triangles whose sides are distances between points on the earth’s surface are spherical triangles. Again, for very short distances the triangle formed may be considered a plane figure.
(3) Examples: (a) If two angles of a spherical triangle are 65° and 105° what can be said about the third side?
Solution: 65° plus 105° equals 170° and there must be more than 180°. Therefore the third angle must be more than 10°. Also, since every angle of a spherical triangle is less than 180°, the third angle must be less than 180°.
(5) If two sides of a spherical triangle are 35° and 76°, what can be said about the third side?
Solution: The 35° side plus the third side must be greater than the 76° side, since the sum of two sides is greater than the third side. Therefore the third side must be greater than 41°. Also the third side must be less than the sum of 76° and 35° or less than 111°.
(4) Exercises.— (a) If two sides of a spherical triangle are 70° and 90°, what can be said about the third side?
(5) If two angles of a spherical triangle are 100° and 70°, what can be said about the size of the third angle?
Greater than 10° but less than 180°. Answer.
(c) Is it possible to construct a spherical • triangle with angles 75°, 65°, 35°? Why?
(d) May a side or an angle of a spherical triangle equal 180°?
(e) If two sides of a spherical triangle are 125° and 160°, what can be said about the third side?
Greater than 35° and less than 75°. Answer.
62. Physical applications of spherical geometry.—a. Arbitrary landmarks.—For practical purposes, the earth may be considered to have a spherical surface. Upon this surface, for reasons of determining position, the following real and imaginary landmarks are set up:
(1) Equator.—A great circle on the earth’s surface perpendicular to the usual “earth’s axis.” This is of course an imaginary landmark.
(2) Latitude.—The distance in degrees of arc on a great circle north or south of the equator. These degrees may be converted.to nautical miles by remembering that 1 minute of arc (of a great circle) is equal to one nautical mile.
(3) Meridian.—Any great circle passing through the North and South Poles.
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(4) Prime Meridian.—The meridian passing through Greenwich, England. It is chosen as the reference or zero point from which all other meridians are measured.
(5) Longitude.—The angular distance from any meridian to the Prime Meridian. The angles run both east and west of the Prime Meridian and never exceed 180° in either direction. It should be noted that the longitudinal distance between two meridians in nautical miles at the equator is not the same as at some point above or below the equator. One minute of arc of longitude varies from 1 nautical mile at the equator to 0 nautical miles at the poles.
Note.—The number of degrees of arc between two meridians is equal to the number of degrees of the dihedral angle formed by the two meridians, and is also equal to the number of degrees in the spherical angle formed at the pole by the meridians.
b. Determination of distances.—(1) Example: Find the distance in nautical miles between the following places, which have the same longitude:
(a) 12°15' N and 30°ll' N.
Solution: The difference in latitude between the two places is obtained by subtraction.
30°ll' N
12°15' N
17 °56' = difference in latitude
To change degrees latitude to minutes latitude, multiply by 60.
17X60=1,020
____________56
1,076' = difference
in minutes. Using the fact that 1 minute of latitude equals 1 nautical mile, the distance is 1,076 nautical miles.
Note.—In the foregoing subtraction it was necessary to borrow a degree in order to subtract the 15'.
(6) 15°16' N and 27°23' S.
Solution: Since these are on opposite sides of the equator, it is necessary to add the latitudes.
15°16' N
27°23' S
42°39' = difference in latitude
(42X60)-f-39'=2559 minutes of latitude=2,559 nautical miles.
(2) Exercises.—Find the distance in nautical miles between each of the following places that have the same longitude, if the latitudes are:
(a) 14°14' N and 76°30' N.
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MATHEMATICS FOR AIR CREW TRAINEES 62
(5) 43°12' N and 25°28' N 1,064 nautical miles. Answer.
(c) 14°27' N and 16°14' S.
(d) 25°11' N and 76° 59' S. 6,130 nautical miles. Answer.
c. Time differences.—(1) Since the earth turns on its axis approximately once each day, it is apparent that 360° of turning must correspond to 24 hours of time. For convenience, the earth’s surface is divided into 24 time belts of roughly 15° longitude each: that is, 15° of longitude corresponds to 1 hour of time. It is also evident that 1° longitude corresponds to 4 minutes of time. The reference time belt again starts at Greenwich, England. Sometimes for further convenience, the time belts do not directly follow the meridians but take geographical and population factors into consideration. In the United States we have four time belts, making a difference in time between the East and West Coasts of 3 hours.
(2) True time is not determined by belts, but must be determined for each separate location from the exact longitudinal position.
(a) Example 1: Find the exact time difference between 15°13' E and 74°43' E.
Solution: The difference in longitude is obtained by subtraction.
74°43' E
15°13' E
59°30' or 59.5° difference in longitude
Since each degree corresponds to 4 minutes time, multiplying 59.5 by 4 gives 238 minutes time difference. Changing to hours and minutes, the time difference is 3 hours 58 minutes.
(6) Example 2: Find the exact time difference between 17°10' W and 21°35' E.
Solution: Since the longitudes are on opposite sides of the Prime Meridian, they must be added.
17°10' W
21°35' E
38°45' or 38%° = difference in longitude
Each degree corresponds to 4 minutes difference in time. Multiplying 38%° by 4 gives 155 minutes as the time difference. Changing to hours and minutes the time difference is 2 hours 35 minutes.
(c) Exercises.—Find the exact difference in time between the following:
1. 15°13' E and 120°13' E.
2. 18°15' E and 91°30' E. 4 hours 52 minutes. Answer.
3. 12°30' E and 47°30' W.
4- 58°10' E and 17°35' W. 5 hours 3 minutes. Answer.
107
63-64
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Section X
TRIGONOMETRY
Paragraph
Purpose and scope________________________________________________________ 63
Development and basic formulas------------------------------------------- 64
Functions of 30°, 45°, and 60° angles------------------------------------ 65
Complementary angles----------------------------------------------------- 66
Use of tables____________________________________________________________ 67
Methods of solving right triangles--------------------------------------- 68
Interpolation____________________________________________________________ 69
Use of right-triangle methods in solving obtuse triangles---------------- 70
Miscellaneous exercises__________________________________________________ 71
63. Purpose and scope.—Trigonometry is based on the properties of similar triangles. It is applied whenever angles enter into the solution of the problem and is important in navigation and bombing. This section is limited to a discussion of the properties of right triangles.
64. Development and basic formulas.—a. Take any acute angle DAE, figure 10, and from one side drop any number of perpendiculars to the other side; a series of similar triangles are formed. A A is the same angle and the series of angles at C, C', C", etc. are all 90°, hence all angles of the triangles are equal.
a XfA------------J-----------J----------- E
M c cz czz
Figure 110.—Similar triangles formed by dropping perpendiculars.
b. From geometry it is known that the corresponding sides are in BC B'C' B"C"
proportion; that is AJE=1AB'= AB" ’ an(^ S0 ^or ot'^ier sides-
AB AC AB
Similarly each of the ratios and has a series of other
ratios equal to it. These ratios of the sides remain unchanged as long as the angle remains unchanged, but they change as the angle
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, ™ , .. AB AB BC BC AC , AC
changes, lhe six possible ratios and are
therefore the functions of angle A, because each is completely determined by angle A.
c. Names of functions.—(1) By definition, the following ratios, figure 111, in a right triangle have been given names.
8
'Ji
a. §
A t_________________ c
Zfc/fclC
Figure 111.—Standard lettering used for identification of angle functions.
. side opposite a sin A = =----—=~
hypotenuse c
. side adjacent b cos A= <--------- =-
hypotenuse c
A side opposite a tan zx • i i • i
side adjacent o . . side adjacent b cot A = =J-------7—=-
side opposite a
A _ hypotenuse __ c SeC side adjacent b csc hypotenuse _c side opposite a
(2) The foregoing relations must be learned by the student. It should also be noted here that the lettering in figure 52 is standard and variations thereof are not used.
65. Functions of 30°, 45°, and 60° angles.—By using geometry, it is rather easy to find given functions for 30°, 45°, and 60° angles.
a. The functions of the 45° angle may be found as follows: Example: In the isosceles right triangle ABC, figure 112,
A=45°, B=45°, a=b
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A 8 y?5 c/ z a
/45*
A Z------------5------------’c
Figure 112—Isosceles right triangle.
Solution: Since a2+62=c2, it follows
2a2=c2 or c=af2
This gives the sides of the triangle in terms of one unknown and therefore the following is true:
^rir$=^=nh=0-7p71
b. In a like manner, the other five functions may be found. The functions of the 30° and 60° angle may be found as follows:
Example: In the equilateral triangle ABD, figure 113, BC is the perpendicular bisector of the base AD and also the bisector of A ABD.
8
/o \
r \
c/ a \
AZ---------------L-------------XD
Figure 113.—Equilateral triangle.
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Solution: From a theorem in plane geometry
6=xc
/I \2 1 9
and a2=c2—b2—c2 —( -oc ) =c2 — xc2=-.c2
\2 / 4 4
or
A
From the foregoing values of a, b, and c the functions of a 60° angle may be found as follows:
sin 60°=-=^2~=^1=^1=.8660
c ---- 2c 2
c
cos 60° = ~=i = .5000
c 2
(1) cos 45°
(2) tan 45° 1.000 Answer.
(3) cot 45°
(4) sec 45° 1.141 Answer.
(5) esc 45°
(6) tan 60° 1.732 Answer.
(7) cot 60°
(8) sec 60° 2.000 Answer.
(9) Find all six functions of 30°.
66. Complementary angles.—a. Two angles whose sum is 90° are said to be complementary and either is the complement of the other. By inspection of the previously defined functions, the following relationships are seen to exist.
sin A = a —=cos c B=cos (90° -A)
cos A— b -=sm c B=sin (90° -A)
tan A— a -r=cot b B=cot (90° -A)
cot A= b . -=tan a 7?=tan (90° -A)
sec A= c T=csc b 5=esc (90° -A)
esc A= c -=sec B=sec (90° -A)
Ill
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(1) The functions of one angle are equal to the conamed functions of the complementary angle, hence the term cofunction.
Example:
sin 65° = cos (90—65) = cos 25°
cot 20° = tan (90-20) = tan 70°
(2) Therefore, any function of an angle between 45° and 90° may be found by taking the conamed function of the complementary angle which is between 0° and 45°. Thus we need never have a direct table of functions beyond 45°.
b. Exercises.—Express the following functions as functions of angles less than 45°.
cot 63° cos 72° sin 18° Answer.
sin 45° cos 45° Answer. tan 46°
tan 86° esc 78° sec 12° Answer.
cos 51° sin 39° Answer. sin 51°
sec 72° cot 89° tan 1° Answer.
67. Use of tables.—a. The functions of angles, other than those discussed previously, are not so easily computed. More advanced methods are employed in securing these functions for other angles. Table II of appendix II has been computed by these methods, which will be used. The table gives the values of the functions to four decimal places for every degree from 0° to 90°. All such values are only approximate as the fourth decimal place has been rounded off to the nearest integer. As previously explained, cos 45°=sin 45°, cos 46°= sin 44°, etc. Hence the column of sines from 0° to 45° is the same as the column of cosines from 45° to 90°. Thus, in finding the functions of angles from 0° to 45°, read from the top down; in finding the functions of angles from 45° to 90°, read from the bottom up.
b. Exerciser— (1) From table II (app. II) find the values of the
following: sin 5° cot 5° =11.4301 cot 82°= cot 45° =
2419 cot 82°= . 1405
sin 35° — _ tan 75°= 3.7321 tan 17°=.
. 1219 tan 19° = sin 90°= 1. 0000
cos 72°—_ sin 0° = . 0000 cos 90°=.
cos 7° — . 9925 cos 15° = sin 3° — . 0523
sin 75°= . tan 42°= . 9004 tan 46°=.
(2) Given the following values from the table find the corresponding angles:
sin A=.2588 ZA=_______ sin A= .6428 ZA=40°
sin B=.9205 ZB=67° tan B= .4245 ZB=------
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MATHEMATICS FOR AIR CREW TRAINEES
67-68
sin D=.6947 cos B=.7771 cos 4.=.5736 ZD= cot 0=2.4751 ZO=22° ZB= cos B= .7431 ZB= Z4.=39° sin A= .0175 AA=1° tan A =1.4826 ZB=
68. Methods of solving right triangles.—a. Given any two sides or one side and one angle, other than the 90° angle, all other parts of a right triangle, including the area, can be found. In working with problems where one angle and a side are given there is a choice of two functions to use. In one of these, multiplication is all that is necessary; but in the other, division by a 4- or 5-place decimal becomes necessary.
(1) Example:
Given: Figure 114, where ZH=20° and c=50 units. Find side a.
B a -----------------b-----------------------'c
Figure 114.—Reference triangle for example (1).
Solution: By definition: Method 1
A a sm A=-c c sin A=a 50 sin 20°=a (50) (.3420) =a a=V7 A
Method 2
A C esc A= -a
a esc A=c
a esc 20°=50
50
2.9238“ a
a=17.1
From this comparison it is evident that the proper choice of functions will save a great deal of unnecessary work. (The easier method is to start with the unknown in the numerator.) When any problem of this type is encountered there is always one function that can be chosen to assure the multiplication rather than the division operation.
(2) Example:
Given: -4=40°, 6=20 units. Find side a.
549535°—43----8
113
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B
C S
A Z---------------------le
Figure 115.—Reference triangle for example (2).
Solution: In this case side b is known and side a is to be determined, so the fraction should be r-b
By definition: tan A=y or b tan A=a
Therefore u=20 tan 40°= (20) (.8391) = 16.782
b. Another type problem is one where two or three sides are known and the angle is unknown. In this kind a division is necessary.
Example:
Given: a=5.5 and c=ll. Find angle A.
B
s' a
A ----------------------b--------------c
Figure 116.—Reference triangle for example in paragraph 6.
Solution: By definition: sin A=—
Therefore sin 21=-%=.50000
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MATHEMATICS FOR AIR CREW TRAINEES 68-69
Looking this up in the table it is found that sin 30°=.5000, therefore ZA=30°.
c. Exercises.
Z+=18°, c=18. Find a
ZA=47°, 6=29. Find c 42.57 Answer.
Z+=32°, a=42. Find 6
ZA=75°, 6=37. Find a 138.09 Answer.
ZA=29°, c=72. Find 6
Z+=55°,a=49. Find c 59.82 Answer.
a=18, 6=45. Find ZA to nearest degree
a=12, c=27. Find ZA to nearest degree 26° Answer.
An airplane at an elevation of 17,000 feet sights an airfield. Measurements to the nearest boundary show the angle of depression to be 7°. How far in a straight line is the airplane from the field and what is the horizontal distance from a point directly above the boundary?
Seen from a point on the ground, the angle of elevation of an airplane is 64°. If the airplane is 10,000 feet above the ground, how far is it in a straight line from the observer?
11,300 feet. Answer.
69. Interpolation.—a. Interpolation is a method of estimating the value of functions of angles which are not given in the tables or estimating the angle, given the function which is not listed in the tables.’ Briefly, it is a process that assumes a straight line difference between two values such that sin of 27.5° has a value halfway between 27° and 28°, and may be found by adding one-half their difference to the function of 27°.
(1) Example: Find the sin of 47.5°
Solution: sin 48°=. 7431
sin 47°=. 7314
Differences 0117
The desired function is .5 of the way from .7314 to .7431. Therefore
.5X.0117=.0059 and
sin 47.5°=.7314+.0059 = .7373 Answer.
(2) Example: Find the sin of 18°20'.
Solution: sin 19°=. 3256
sin 18°=. 3090
Differences 0166
20 1
The angle 18°20' is s or r of the way between the two angles.
t) U o
115
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Therefore:
|X-0166 = .0055 D
sin 18°20' = .3090+ .0055 = .3145 Answer.
(3) Example: Find cot of 34.2°.
Solution: cot 34°= 1.4826
cot 35°= 1.4281
Difference= .0545 .2X.0545= .0109
Therefore: cot 34.2°=1.4826 —.0109 = 1.4717 Answer.
(4) It will be noted that in the third example the process varies slightly. It is necessary to subtract the difference from the value of the smaller angle. This is true in the case of all cofunctions because their values decrease as the angle increases. The process varies slightly when an angle is desired from a given function.
(5) Example: Find the angle whose sin is .4295
Solution: sin 26°=.4384
sin X = .4295 sin 25° =.4226
Difference between sin 25° and sin 26°=.0158 -
Difference between sin 25° and sin X =.0069
From this the desired angle is of the way from 25° to 26°.
Changing this to a decimal, it becomes .44 or rounding off to tenths 0.4.
Therefore the desired angle is 25.4°.
b. Exercises.—(1) Find the required functions of the following angles:
(a) sin 18.4° . (d) cot 53°20' .7446 Answer.
(6) cos 23.7° .9156 Answer. (e) sin 27°49'
(c) tan 72.9° (/) tan 44° 50' .9943 Answer.
(2) Find the angles whose functions are: (a) sin A =.7590 (d) cot I? = 1.0990 (6) cos 7? =.8028 ZR=42.3° AB=36.6° Answer. (e) tan A=l.0490 (c) tan D=.O7OO (/) sin 5=.381O ZB=22.4 Answer. Answer.
(3) Find the unknown sides and angles in the following right triangles by trigonometry:
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MATHEMATICS FOR AIR CREW TRAINEES 69-70
(a) 6=15, sin B= .5195
(6) (c) a=7, tan B=.3799 c=12, cos A=.6211 6 = 2.66, c=7.49 Answer.
(<*) (e) a=10, Zri=42.7° 6=9, ZA=29.1° 6=10.84, c= 14.75 Answer.
a) (ff) a=7, 6=12 6=90, c=lll a=3.25°, c=13.89 Answer.
(A) a=17, c=94 a=10.4°, 6=.9246 Answer.
(4) In securing a telephone pole, the guy wire should be fastened
to the pole 12 feet from the ground, and secured to the ground 8 feet from the pole. Find the angle the guy wire makes with the pole and with the ground. How much wire is needed if 2 feet is allowed on each end?
(5) How far from the center of a circle of 12-inch radius will a tangent meet a diameter (extended) with which it makes an angle of, 10.7°? 18.4 ft. Answer.
56.31°?
33.69°?
70. Use of right triangle methods in solving obtuse triangles.—a. When two sides and the included angle or the angles and a side of an obtuse triangle are known the rest of the triangle including the area may be found.
(1) Example: Find sides BD and AD of the obtuse triangle shown in figure 117, angle DAB being 30°, angle ABD 100°, angle BDA 50°, and side AB 9 inches.
B
Xi\ ^X 1 \
9.X 1 \
X^^o |
A --------------7------* D-
Figure 117.—Reference triangle for solving obtuse triangle problems.
Solution: In this case, drop a perpendicular from the obtuse angle to the base forming two right triangles, ABC and BDC. From triangle ABC, obtain:
BC
sin 30°= X and BC= (.5)(9)=4.5 in.
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cos 30°= y-’ and AC=(.866O) (9) = 7.7940 in.
In a like manner:
BD,
CSC 50°= yr’ and BD= (1.3054) (4.5) = 5.8743 in. Answer. CD
cot 50°= and <7Z>=(.8391) (4.5) = 3.7760 in.
AC+CD=AD, and AD=7.7940+3.7760= 11.5700 in. Answer.
(2) Example: An airplane is flying a heading of 120° with an air speed of 240 mph. The wind velocity is 40 mph. from 90°. What are the ground speed, track, and angle of drift?
A
i
D l.----------------------------1------------------
c A
Figure 118.-—Heading-wind-traek triangle.
Solution: Problems of this nature are of frequent occurrence. Two sides of a triangle and the included angle BAG, figure 118, are known, since the direction of both the heading BA and the wind AC are given. To solve this type of triangle, extend the wind vector AC and drop a perpendicular to it from B. Note that the perpendicular is opposite the known angle. In the triangle ABD,
BD=AB sin 30°=240 (,5000) = 120.0
AD=AB cos 30°=240 (,8660)=207.8
Then,
CD=DA—CA=207.8—40= 167.8
BD 120 „ '
tan EBCD——■=—=.7151
CD 167.8
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MATHEMATICS FOR AIR CREW TRAINEES
70
ZBC'D=35°34'
BC=BD ■ cosec ABCD
BC=120 (1.7192)=206.3 mph. Ground speed.
ZABD=90o-30°=60°
Z(7BD=90o-35°34' = 54°26'
AABC= AABD- ZCRD==60o00'-54o26,==5°34'
Angle of drift.
The track is ~ ACBD=\W~M°2V =A2N34' Track.
(3) Example: (a) In the heading-wind-track problem, figure 118, the known angle in the triangle, is frequently obtuse. In this case
J?
/ B 'A \
/ 125°^^ X
K— .. -..... -...< .TyNk
A (a) c
7y«
' T
Figure 119.—Heading-wind-track triangle with obtuse angle.
some difficulty may be experienced in deciding how to draw the perpendicular necessary to form the right triangle required for the solution of the problem. Figure 119 shows the procedure to use when the angle is obtuse. The air speed is 200 mph. and the heading 75°. The wind is 50 mph. from 310°. The angle between the air speed vector and wind vector is 125°. If the air speed vector is extended
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70-71 ARMY AIR FORCES
and a perpendicular dropped to it from C, a right triangle ADC is formed which can be easily solved for ground speed and track.
DC= wind X sin 55°
wind X cos 55°
AD=AB+BD
CD
tan ECAD=-------=angle of drift
AD
AC=AD sec ECAD, the ground speed Track is simply heading (75°)+ ECAD.
(6) If the wind vector had been extended and a perpendicular dropped from A to E, a different right triangle would have been formed which could be solved in the same manner.
b. Exercises.—(1) An airplane at an altitude of 3,800 feet is sighted from two stations simultaneously. The angles of elevation are 58° and 47° respectively. Find the distance, between the stations if the airplane is between the stations.
(2) By observing landmarks, a pilot calculates his ground speed as 188.5 mph. and his track as 221.5°. The wind is known to be 35.6 mph. from 211.5°. What should his air speed indicator read?
223.6 mph. Answer.
71. Miscellaneous exercises.—(1) A target was sighted at an angle of depression of 18°. The bomber was flying at an altitude of 18,000 feet. How far will the plane have to travel to be directly over the target?
(2) Find the angles of a right triangle whose legs are 7 and 9 units long, respectively? 52.1° and 37.9°. Answer.
(3) An observer sighted a squadron of airplanes at an angle of elevation of 17.7°. They were flying at the time over a landmark that was known to be 5 miles away from the observer. What was the altitude of the airplanes in feet?
(4) In the right triangle shown in figure 120, find the other sides if—
\c a K
-----------------\A
c b
Figure 120.—Reference triangle for exercise (4).
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MATHEMATICS FOR AIR CREW TRAINEES
71
(a) c=26, A=37.7°.
(6) 6=42.4, 21=32.3°.
(c) a=3.12, B=5.4°.
(5) In the right triangle shown in figure 121, find the missing sides and angles if—
B
a
C b A
Figure 121.—Reference triangle for exercise (5).
(<0 »=!<>,‘=23 B = 21.3«J Answ^
(6) a=95, 6=37, c= 103.16
(6) An airplane is directly over the middle of a lake and the angle subtended by the lake (angular width) is 13.4°. If the airplane is 2 miles high, find the width of the lake in feet. 2,481 ft. Answer,
(7) Find AD in figure 122.
B
* X
$ N. ^x
90° X.
I-------------------------X D
C A
Figure 122.—Triangles to be solved in exercise (7).
(8) Find AC in figure 123. 1,682 ft. Answer.
(9) An airplane flying a level course passes directly over a landmark. One minute later the landmark is observed again and has an angle of depression of 30°, the same size as the angle of elevation. If the plane is flying at the rate of 240 mph., find its altitude.
121
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(10) A man standing 500 feet from a building observes a flag pole on the top. The angle of elevation of the bottom of the pole is 27°
A
D 5“oo. B C
Figure 123.—Triangle to be solved in exercise (8).
and that of the top is 29.5°. Find the height of the pole and of the building. 28.2 feet and 254.75 feet. Answers.
1 °
(11) Find AA, correct to nearest^ and find AC in figure 124.
B
/ o
/ 0
a aoo d c
Figure 124.—Triangle to be solved in exercise (11).
(12) The estimated trajectory of a bomb dropped from 20,000 feet will intersect the target 2 miles away. What will be the angle of
Base
Figure 125.—Plan view of search problem.
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MATHEMATICS FOR AIR CREW TRAINEES
71-73
depression from the plane to the target at the time the bomb is dropped? 62.2° Answer.
(13) On a certain observation maneuver a squadron of 18 airplanes, figure 125, was ordered to search an area of ocean starting from a base on the coast and fanning out radially from the base. Visibility for this maneuver was estimated to be 20 miles. A 5-mile overlap was necessary for optimum efficiency. Assuming a straight coast line, at what distance from their base would the operation begin to be ineffective?
Section XI
LOGARITHMS
Paragraph
Purpose___________________________________________________________________ 72
Introduction to logarithms________________________________________'_____c_ 73
Logarithms_______________________________„________________________________ 74
Mantissa__________________________________________________________________ 75
Use of tables_____________________________________________________________ 76
Interpolation_____________________________________________________________ 77
Negative characteristics__________________________________________________ 78
Multiplication and division_______________________________________________ 79
Powers of numbers_________________________________________________________ 80
Extraction of roots_______________________________________________________ 81
Application to right triangles____________________________________________ 82
Evaluating formulas with logarithms--------------------------------------- 83
Oblique triangles_________________________________________________________ 84
72. Purpose.—The use of logarithms presents a convenient method for performing arithmetical calculations with ease and rapidity. It will be seen that logarithms are exponents, or powers, of ten, and hence that the rules for the combination of exponents hold for them. Therefore calculations involving the operations of multiplication, division, raising to powers, and extraction of roots can be performed with a sufficient degree-of accuracy and with little effort.
73. Introduction to logarithms.—a. Find the values corresponding to the positive integral powers of 10:
10°= 1
lO^lO
102=100
103= 1,000
104= 10,000 etc.
b. This tabulation suggests the question: For any number greater than one, say 50, is there a number which may be used as an exponent so that 10 to that power will equal 50? From the table, since 50 is between 10 and 100, it seems reasonable to suppose that this exponent is between 1 and 2 (or 1 plus a decimal if it exists). At a later stage
123
73-75,
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in mathematics, it is proved that such a number does exist and that its value is between 1 and 2. Again, since 576.3 is between 100 and 1,000, note that the exponent of 10 should be between 2 and 3; that is, 102+=576.3. These exponents of 10 are called logarithms.
74. Logarithms.—The logarithm of any number to the base 10 is that exponent which causes 10 to that power to be equal to the given number. Thus, since 103= 1,00 0, the definition states that log 1,000 = 3. These two equations are equivalent; they are merely two different ways of stating the same fact (the relation between 3, 10, and 1,000). It happens, however, that the logarithmic statement is more convenient for computational purposes. Again, since 10b+=50, it can be said that log 50=1.+ (some decimal fraction).
a. Logarithms (and all numbers for that matter) may be considered to be composed of an integral (or whole numbered) part and a decimal fraction part. The integral part of the logarithm is called the characteristic and the decimal part, the mantissa. Consider now the numbers 600, 105.3, 732, 986.2; referring to the tabulation in paragraph 72, each of these four will be 2 plus a decimal part, that is, their logs all have 2 as their characteristic, as also is the case for all numbers between 100 and 1,000.
b. This indicates the following rule which MUST BE MEMORIZED: Any number greater than one has a logarithm whose characteristic is one less than the number of digits to the left of the decimal point.
c. The consideration of logarithms of numbers less than one will be deferred until later.
d. Exercise.—(1) What are the characteristics of the logs of the following numbers: 62; 385; 78.34; 67.823; 4; 5.78; 10,000; 7,683; 17?
(2) State between which powers of 10 the numbers whose logs have the given characteristics lie: 0; 1; 2; 3; 4; 5; 6; 7; 8.
e. When a particular number is given, say 27.38, there are given two things; a certain set or sequence of digits, and the location of the decimal point. It has been seen how intimately the characteristic of a log of a number is associated with the position of the decimal point in the number. One might therefore expect (and it is true) that the mantissa or decimal part of a logarithm is dependent upon, or determined by, the sequence of digits in the number. In other-words, two numbers such as 27.38 and 273.8 will have logarithms whose mantissas are the same; their logs will differ only in the characteristics.
75. Mantissa.—Mantissas in general are found to be unending decimals similar to the decimal equivalents of -y/2 and tt. The accuracy desired for a computation will determine the number of places to be used in the mantissa table; that is, if results to only four significant
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figures are desired, mantissas to four decimals are used, etc. The tables of mantissas have been computed by processes developed in more advanced mathematics; and, because of the mantissa and characteristic properties already mentioned, the table needs to be computed for only a certain range of numbers. It is in this fact that the number 10 finds its justification and practical value as the base for a system of logarithms.
76. Use of tables.—A four-place logarithms table is given in appendix II. Make it a definite policy, when finding the logarithm of a number, to write the characteristic first. Only after this is done, begin looking for the mantissa in the table.
(1) Example: Find the logarithm of 30. Write log 30=1.— as determined by the characteristic rule. The sequence of digits is 3,0,0,0. The first two digits 3,0, are found in the column headed at the top by “N.” Follow down this column to the number by 30. It is known that the third digit, the one after the decimal point, is 0. Looking in the vertical column headed “O”, the number 47712 is found opposite 30. This number is the mantissa of the logarithm of 30, that is, log 30 = 1.4771. This means according to the discussion that, to four significant places, 1014771 = 30.00.
(2) Example: Find log 7,640. First, log 7640=3.—. Then find 76 in the column headed “N.” In the row corresponding to 76, go over to the column headed 4; the number appearing there is 8831, the desired mantissa. Hence log 7,640=3.8831.
77. Interpolation.—a. If the fifth digit of a number is not zero, interpolation must be used to find the logarithm’s mantissa.
Example: Find the logarithm of 76,430. Since 76,430 is between 76,400 and 76,500, the mantissa of its logarithm will be between the mantissas of the logs of 76,400 and 76,500: in fact, one could estimate 3
it to be Jq of the difference between the mantissas given in the table beyond the smaller one appearing there. Previously it was found that log 76,400=4.8831; the difference in the table or (tabular differ-3
ence) to the next mantissa is 6 in the last place. Then -^qX6=1.8: and rounding off to 2, one finds the mantissa of log 76,430 to be .8833. Then log 76,430=3.8833. This relation may be expressed as a formula, thus:
. .. . . Last significant digit X tabular difference
Added value=------------------------------------------
b. If given a logarithm and the number which has it as its logarithm are to be found, apply the previous process in reverse order.
(1) Example: log N—1.9206, find N.
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77-78 ARMY AIR FORCES
Since the characteristic determines the decimal point, leave it to the last. First find the sequence of digits corresponding to a mantissa of 9206. Hunting in the interior of the tables, one finds this to be 833. Then the rule for characteristics states that there must be two digits to the left of the decimal point. Hence 2V=83.3.
(2) Example: Log TF=2.5748, find W. One finds that the mantissa .5748 is between .5740 and .5752. Corresponding to the smaller of these, there is the sequence of digits 375. In this case the- adding factor and difference are known; they are 8 and 12. Then using the equation in a above, one gets,
8X10
Last digit= =6%=7, when rounded off. Hence JF=375.7,
lA
the decimal point being determined by the rule.
c. Exercises.—Find the following logarithms:
(1) log 3,730
(2) log 46.8 (3) log 68.5 1.6702 Answer.
(4) log 273.5 (5) log 9 2.4370 Answer.
(6) log 81.32 (7) log 7.008 1.9102 Answer.
(8) log 760.1 (9) log 5010 2.8809 Answer.
Find the number which has the given log. (1) log B=1.7810 (2) log +=2.9576 (3) log £=1.8790 +=907.0 Answer.
(4) log (7=4.3521 (5) log £=3.3333 (7=22.50 Answer.
(6) log D=2.6969 (7) log £=.5724 £=497.6 Answer.
(8) log £=.6008 £=3.988 Answer.
(9) log £=7.1660
78. Negative characteristics.—a. Take for granted the definition (which can be justified) that a negative exponent means the reciprocal of the number affected by the same numerical but positive exponent. By this definition:
8—~8’ and 10’_2="102=100=:0-01,
b. Now construct a table of negative powers of 10:
10°= 1
10-! = .1
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MATHEMATICS FOR AIR CREW TRAINEES 78
10—2=.O1
10—3=.001
10—4=.0001 etc.
Note that the numbers on the right are positive, decreasing toward zero as the exponents decrease. This indicates, first, that no real power of ten is a negative number, which means further that negative numbers have no real logarithms. The table indicates, secondly, what might be expected for the logarithms of numbers between zero and one.
c. For example, the number .06 is between .1 and .01, and one expects the power of 10 giving .06 to be between (—1) and (—2). This logarithm could be given either as — 1 minus a decimal fraction or as —2 plus a decimal fraction. The latter method is preferred and must always be used because the mantissa, as determined by the sequence of digits, in this case is the one which appears in the tables.
d. The numbers .3, .45, and .1893 are all between .1 and 1; and thus by the preceding table of negative powers of 10, the characteristic of their logarithms is (—1). From the information gained from the tabulation in paragraph 786, it is possible to state the following rule which must be memorized: The characteristics of the logarithms of numbers between zero and one is negative, and numerically one greater than the number of zeros between the decimal point and the first nonzero digit.
Example: The characteristic of log .0058 is (—3), that of log .68 is (—1), and that of log .00003792 is (—5). Find log .0068. Its characteristic is (—3), the sequence of digits is 6, 8, 0, and therefore the mantissa is .8325. Hence
log .0068 = 3.8325
with the minus sign above the characteristic, which emphasizes that only the characteristic is negative, the mantissa always being positive. Another convenient way of writing negative characteristics is by adding and subtracting 10 to the characteristic found by the rule. Example: log .0068 = 7.8325—10. The mantissa is found, by interpolation if necessary, exactly as before.
e. If given a logarithm with a negative characteristic and asked to find the number which has this for a logarithm, the decimal point is located by applying the foregoing rule.
Examples: (1) Log A?=8.6475—10, find N. The given mantissa is .6474 for which the sequence of digits of N is 4441. The given characteristic is (—2); then, by the rule, there is one zero between the decimal point and the first nonzero digit. Hence
N=.0441
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78-79 ARMY AIR FORCES
(2) If log $=9.6075 — 10, proceding in the same manner, one finds $=.405.
/. Exercises.— (1) Give the characteristics of the logarithms of the following numbers: .69, .0038, .6007, .00005, 65.3, 3.12, .00312.
(2) Find the logs of the above numbers.
79. Multiplication and division.—a. General.—It was found that logarithms are exponents, and hence the laws of operation upon logarithms are the same as those governing the combination of exponents. If a, m, and n are any three numbers whatsoever, two of these rules are:
(1) an.am=am+m
(2) an=an~m
am
Thus, briefly stated, multiplication is accomplished by adding exponents, division by subtracting the exponent of the denominator from that of the numerator.
b. Multiplication.
34.35=34+^39
102-103=102+3=105
68=68-5=63
65
103=103-4=10-1=0.1
104
Examples: (1) Multiply 60 by 3 using logarithms. It is found that log 60=1.7782, that is, 60= 101-7782; and log 3 = .4771 or 3 = 104771. Hence, 60X3 = 10b 7782X 10-4771 = 102-2553.
(2) Find 47X51X.08. Let 47X51X.08=2V
log 47= 1.6721
log 51= 1.7076 log .08= 8.9031—10
Add
log N= 12.2828—10
A7=191.8
c. Division.
Examples: (1) To perform divisions, subtract the log of the denominator from the log of the numerator: Find N in N— "£££•
log 200=2.3010+ ,
log 40=1.602 ljSubtiact
log Ny= .6989
A7=5.00
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MATHEMATICS FOR AIR CREW TRAINEES 79-80
If the log of the denominator is larger than that of the numerator, the subtraction will yield a negative number. Incorporate the negative part entirely into the characteristic, thus keeping the mantissa positive, by increasing and decreasing the log of the numerator by ten (or any appropriate number).
20
(2) Divide 20 by 30; write yj=2V, then
log 20=1,3010=11.3010-101. bf
log 30=1.4771= 1.4771
log N= 9.8239-10
N= .6667
d. Exercises.—Perform the indicated operations:
(1) (68) (74)
(2) (34) (.08) (65) 176.8 Answer.
(3) (.038) (65) (17.35)
(4) 726=132 5.5 Answer.
(5) 230=25
(6) (.00062) (300.5) .1863 Answer.
(7) .037=.027
(68)(35.2) (98) . (&) (24) (3 5) 2792 Answer.
(9) (27)(24)
(10)(100)
(.028) (1.032) ‘°751 Answer.
80. Powers of numbers.—a. Rule.—Powers of numbers are
evaluated by applying to logarithms another law of exponents. This
rule is that, for any numbers, a, m, and n, for example
= = amn
or, briefly stated, powers are obtained by multiplying the exponents. For example:
(72)3 = 76
(215)3 = 2116 (103)16=104-6
b. Example: (1) Find 63. Say 63=tV, then since log 6 —.7782, the problem becomes (10-7782)3 = 2V. Applying the aforegoing law for exponents, one gets 1O2-3346=A7, and vV=216.
549535°— 43---9
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This process is written more briefly in the following form: log 6= .77821A, u. , log 63=3 /Multiply
2.3346
63=216.0
(2) Find (.3X29)°=N
log. 3=9.4771-101 ,.,
log 29=1.4624 f
10.9395-10=.9395
4^39S}Multiply .
56370
37580
log 2V= 4.32170
N= 20,980
81. Extraction of roots.—a. First, express an indicated root of a number as an exponent with the use of fractional exponents. Hence a« = A/a; the indicated root appears as the denominator of the fractional exponent.
Example:
V7=71/2
V^29=(.29)1/s
^162=(162)1Z7==162/7
b. To find a root of a number, write the number with appropriate fractional exponent, and then proceed as in the preceding section.
Example: (1) Find the square root of five.
Write V5=51/2
then log 5 = .699O
and 51/2= (IQ.6990)1/2
Thus log log 5 = .3495.
Therefore 51/2=2.236
(2) Find the cube root of 64. Write ^64 then log 64=1.8062
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MATHEMATICS FOR AIR CREW TRAINEES
81
131
multiply by 1/3 log 641/3=-^ log 64 = .6O21 o
641/3=4.000.
/2 . . /2\1/2
Find a/3^5 ^'ls’s ( 3 ) '
log 2=10.3010-10+ K+ ,
log 3= .4771 ! Subtract
log |=9.8239-10
o
log (0 = % log 1=4.9119—5^multiplying by 0
/2\1/2
=.8165
c. If it is necessary to multiply a logarithm with a negative characteristic by a fractional exponent, it is desirable to so change the given characteristic that its negative part is exactly divisible by the denominator of the fraction.
Example: Find 7-3018.
Find log .3018=1.4797.
Write this negative characteristic of (—1) as (29—30), thus
log .3018 = 29.4797-30
log (.3018)1/3=9.8266 —10
(.3018)1/3=.6708
d. Exercises.—Solve by logarithms:
(1) 32.758X8.3759
(2) (1.74)17 12260 Answer.
(3) (4.4377)3X.9746
(4) 717 ' 1.604 Answer.
(5) 777 + 715
(6) 7-0000067 0.2661 Answer.
(7) (1.42)12
zoa 3/".002396 ..A
(8) a/ 0-noA-- 0.06449 Answer.
183X.031/2
( } 34X75
81-82
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(10)
173—142
(U)
7.01 173+142
47150 Answer.
82. Application to right triangles.—a.
Logarithm table.—(1) Table IV of appendix II gives common logarithms of circular or trigometric functions and will be used to solve trigonometric problems by logarithms. It is to be understood that a (—10) must be supplied whenever a characteristic of 8 or 9 is given in a log column. Interpolation is applied as in common logarithms.
(2) Corresponding to 18°, one finds log sin 18°=9.4980—10. For 62°, it is found that log tan 62°=.2743. Again, by interpolation loo-cos 7.7°=9.9961 —10.
b. Solving right triangles.—The following process gives the unknown parts of a right triangle: By use of the definition of the sine, cosine, tangent, and cotangent functions, write an equation which involves just one missing part; then solve this equation for the missing part, and perform the indicated arithmetic by logarithms.
Example 1: Given A=62°10', a=78; find B, b, and c.
0
z £
t/Z
Z ii
/gz°/o'_________.
A b C
Figure 126.—Reference triangle for example 1.
(1) B is found by subtraction: B=90-62°10'=27°50'
(2) cot A=->
a
b=a cot A.=78 cot 62°10' log 78= 1.8921 1 . , ,
log cot 62°10'= 9.7226 — 10J Ac c
log 6 = 11.6147—10 6=41.19
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82-83
(3) sin
a _ 78
C sin A Sin 62°10'
log 78 = 11.8921-101 S b log sin 62°10'= 9.9466 —10/&Dtact
log c= 1.9455
c=88.2
Example 2: Given a=40, c=59.3; find A, B, and b. (Since the Pythagorean relation1 * is not well adapted to logarithms, find the angles first, and then find b by a trigonometric relation.)
A a 40
(1) smd-c-593
log 40=11.6021-101G k+ .
log 59.3= 1.7731 /Subtract
log sin A= 9.8290—10
A =42.4°
B=90°—42.4°=47.6°
(2) cos
b=c cos A = 59.3 cos 42.4°
(. Log 59.3= 1.7731 1 Ad ,
log cos 42.4°= 9.8683-10/Aaa
log 6=11.6414 — 10
6=43.79
83. Evaluating formulas with logarithms.—a. There has been given a number of formulas for calculating areas and volumes of various objects. By solving the equation of the formula for a missing part, logarithms can often be used to compute the missing part when the others are given.
Example 1: Find the volume of a sphere whose radius is 6 feet.
Use tt=3.1416, or log tt=.4971
4
4 V=X(6)3 o
1 The relation of the hypotenuse to the legs of a right triangle, namely that the square of the hypotenuse
is equal to the sum of the squares of the legs.
133
83
ARMY AIR FORCES
log 6 - . 7782],Multiply
log 63 = 2. 3346]
log 4 = . 6021 Add
log tt = . 4971J
3.4338% , . .
log 3 = .477!/Subtract
log V=2. 9567
U= 905.2 cubic feet
Example 2: A right circular cone has an altitude of 18 inches and a volume of 364 cubic inches. What is the radius of its base?
V= }iirr2h and solving for r.
2=3U, irk
r= /fy3/3u\^ y irk \ 7Th/
r /3X364\^
\ 18tt )
log 3— .4771] * jj
log 364=2. 56111Aaa
3.0382% ,. ,
log denom. = l. 7524%ubtract
1. 2858]
1 ^Multiply
2 J
log r= . 64290 r=4.394 in. log 7r= . 4971 log 18=1. 2553
log denom. = l. 7524
b. Exercises.—Using the conventional triangle, solve for the missing parts of the following right triangles, finding sides to four significant digits and angles to tenths of degrees or minutes.
(1) A=23.5°, c=627
(2) B=76°15', c=93.4
(3) A=60°, 6=4
(4) B=68°, a=73
(5) B=3.5°, 6=2
A—13°45', a=22.20, 6=90.7 Answer.
-4=22°, 6 = 180.7, c= 194.9 Answer.
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MATHEMATICS FOR AIR GREW TRAINEES
83-84
(6) a=21.9, c=91.9
A=13°47'l" B=76°13' 6=89.25 Answer.
(7) 5=18.3,c=30.75
(8) A tin can has base diameter 4.5 inches and height 5 inches. What is its cubic capacity? 79.52 cubic in. Answer.
(9) What is the radius of a sphere whose volume is 700 cubic feet?
(10) What is the volume of a sphere whose radius is twice that of the sphere in the preceding problem? 5,600 Answer.
(11) Find the angle at the vertex of the cone discussed in example 2, a above.
84. Oblique triangles.—a. It is often possible to find the missing parts of oblique triangles by constructing auxiliary lines, generally altitudes, which will allow the use of the theory developed for right triangles.
Example 1: Find side b of the triangle shown in figure 127.
8 i
/20. n ।
*--------Z------D------- C
Figure 127.—Reference triangle for example 1.
Construct the altitude, BD, to the side AC, and two right triangles are formed: ABD and BDC.
AD
(1) In triangle ABD, cos A =-, and AD=c cos A=20.95 cos
c
20.5°.
log 20.95= 1.3212 1
log cos 20,5°= 9.9716—10jdd
log AD =11.2928-10
AD =19.62
DC
(2) In triangle BDC, cos C=—^-i and DC=a cos C—15.62 cos 27.5°.
log 15.62= 1.1937 1
log cos 27.5°= 9.9479- 10JAdd
log DC =11.1416-10
DC =13.85
(3) 5=AD+DC=19.62 + 13.85, 6=33.47
Example 2: Find sides a and b of the triangle shown in figure 128.
135
84
ARMY AIR FORCES
Drop a perpendicular from B to the line AC, extended, meeting it at D.
AD
(1) In the right triangle ABD. cos A=-^~, and AD=c cos A= 73.34 cos 32.4°.
log 73.34= 1.8653 1
log cos 32.4°= 9.9265—10[dd
log AD =11.7918 — 10
AD =61.91
8
% 3^-^*******^-*^^^^
A b C ” ~ o
Figure 128.—Reference triangle for example 2.
(2) In the right triangle ADB, ■ a BD sin A=-----------------------, and
c
BD=C sva +=73.34 sin 32.4° log 73.34= 1.8653 1
log sin 32.4° 9.7290—10 fAdd
log BD =11.5943—10
BD =39.29
(3) In right triangle ADB, ZZZB+=90°-32.4°=57.6°
In right triangle BCD
ZPBC=57.6°—28.2°=29.4° ZBG£=90°-24.4°=60.6° , »z/n CD cot BCD=jfij
CD=BD cot ABCD G£=39.29 cot 60.6° log 39.29= 1.5943 1
log cot 60.6°= 9.7509—10jAdd log CD= 11.3452-10
C7?=22.14
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MATHEMATICS FOR AIR CREW TRAINEES
84
, .. . n Hl' ,
(4) sin 60.6=BC ’
39.29
jDC sin 60.6°
log 39.29 = 11.5943-101 I
1 • aa n nnim in>Subtl'aCt
log sm 60.6 = 9.9401 —10j
log BC= 1.6542
BC=45.1
(5) tt=BC=45.10, 6=AP-OP=61.91-22.14=39.77.
b. Exercises.—(1) Two angles and the included side of a triangle are
35°, 25°, and 18° inches. What are the missing parts?
(2) Find the area of a triangle whose sides are 25, 29, and 31.
339.8 Answer.
(3) Find the missing parts of the triangle shown in figure 129.
Figure 129.—Triangle for exercise (3).
/\ C = 30-3 0
(4) Find the length of the altitude to the 29 side in exercise (2).
23.43 Answer.
(5) Find the angles of the triangles in exercises (4) and (5).
(6) An airplane has an air speed of 217.6 mph. and a heading of 110°. A 47.5 mph. wind is blowing from 20°. Find the ground speed and track. 122.3°, 222.7 mph. Answer.
(7) An airplane flies over a straight road. Two observers on the road 15 miles apart, find its angles of elevation at the same time and record them as 14° and 10.3°. How high is the airplane?
Section XII
137
SPHERICAL TRIGONOMETRY
Paragraph
Purpose and scope____________________________________________________ 85
Definitions and geometric properties of a sphere--------------'------ 86
Special rules for solving spherical triangles------------------------ 87
Napier’s rules______________________________________________--------- 88
Formula exercises-----.---------------------------------------------- 89
General laws_________________________________________________________ 90
Solution of right spherical triangles------------------------------- 91
Exercises____________________________________________________________ 92
The astronomical triangle-------------------------------------------- 93
Exercises---------------------------------------------------------- 94
85-86
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85. Purpose and scope.—a. Spherical trigonometry is presented for the purpose of providing the necessary mathematical background in the study of celestial navigation.
b. The scope includes the basic elements of the right and oblique spherical triangles together with the application of spherical triangles to the solution of the astronomical triangle as used in celestial navigation. The paragraphs below contain explanations with illustrations designed to give the student a working knowledge of necessary fundamentals of spherical trigonometry.
86. Definitions and geometric properties of a sphere.—a. In order to solve right spherical triangles it is necessary that new terms be introduced.
b. The definitions and properties which follow should be learned thoroughly as they are essential to an understanding of spherical trigonometry.
(1) Sphere.—A sphere is a closed surface in three dimensions all of whose points are equidistant from a fixed point called the center. Usually the earth is considered as a true sphere, although this is not strictly true.
(2) Circle.—A plane intersects the surface of a sphere in a circle.
c j
Figure 130.—Spherical triangles.
(3) Great and small circles.—A plane passing through the center of a sphere intersects the surface of a sphere in a great circle. Other circles on the sphere are called small circles.
(4) Hemispheres.—A plane through the center of a sphere divides the sphere into two equal parts, called hemispheres.
(5) Spherical triangle.—A spherical triangle is that portion of the surface of a sphere which is bounded by three great circular arcs.
(6) Great circle distance.—Distance on a great circle of a sphere is usually measured in arc units, for example, degrees, minutes, and seconds.
(7) Spherical triangle rules.—Referring to triangle ABC, figure 130, the following rules, which were given in section IX, apply:
(a) (a + 6+c) is less than 360°.
(6) (A+jB+G) is greater than 180° but less than 540°.
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MATHEMATICS FOR AIR CREW TRAINEES
86-87
(c) If the length of two sides of a spherical triangle are equal, the opposite angles are equal, and conversely.
87. Special rules for solving spherical triangles. — This section deals only with right spherical triangles except for a special solution of an oblique triangle by right triangle formulas. As in the case of plane right triangles, the right angle is lettered C. A right spherical triangle can be solved if any two parts other than the right angle are given, even if the two given parts are angles.
a. The nature of the angles and arcs of a spherical triangle can be made clear by a study of figure 131.
/ 12/7 Vy
/ oi/w* fiy* I ■A _
' —TV
I -7 xyZ / /
V' X/ b
____A
Figure 131.—Spherical triangle showing nature of angles and arcs.
(1) Radii drawn to A, B, and C, form three planes OBC, OBA, and OAC.
(2) The plane DEF is drawn perpendicular to the line OA and thus angles OED and OEF are right angles.
(3) The dihedral angle between planes OAB and OAC is angle DEF and is also angle A of the spherical triangle.
(4) This is true because the measure of a dihedral angle between two planes is the angle between two lines in the two planes which meet at the vertex and are perpendicular to the vertex line, as EF and DE.
(5) The distance OD in the figure was made unity or 1 for convenience.
b. In the right triangle OED
. n DE n OE r.r,
sin C=-^-=DE, and cos C=—^- = OE
Since DFE was drawn perpendicular to OA, it is perpendicular to both plane OAC and plane OAB. Plane OBC is perpendicular to plane
139
87
ARMY AIR FORCES
OAC since the spherical triangle is a right spherical triangle with the right angle at C. Therefore DF is perpendicular to both EF and OF. In triangle OBF—
DF sin a=— -=DF
OF cos a=—=OF
c. From triangle OEF—
OE=OF-cos h
and since OF=cos a
. and OE—cos c
then cos c=cos a«cos b
d. From triangle OEF—
and since and then EF=0F-sin b = OE-tiiii b OF= cos a OE= cos c EF=cos a-sin 6 = cos c-tan b
so cos-a sin 5=cos c-tan b
and cos a-sin b cos c= 7 tan b
but sin b tan b= 7 cos b
therefore cos b-cos a=cos c
e. Since angle DEF= angle A . a DF tan A ftp
But in triangle ODF
sin a= DF .EA-=DF
so tan A= sin a
cos a-sin b
but sin a — tan a
cos a
therefore tan A= tan a sin b
/. In triangle DEF— DF
sin A DE
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MATHEMATICS FOR AIR CREW TRAINEES
87-88
but in triangle AED
DE sm c=—j—=DE
and since
Then
or
ZZF=sin a
sin A
sin a sin c
sin A-sin-c=sin a
g. In triangle DEF—
. EF cos c. tan b , , ,
cos A==-r7,-.=---r------= cot c . tan b
Ue sm c
h. In a similar manner, by drawing a plane perpendicular to OB so that a triangle with angle EDF=B can be drawn, a similar development would give five more formulas, making a total of 10 formulas for the solution of the right spherical triangle. These formulas are as follows:
(1) sin a=sin c-sin A
(2) tan 6=tan c-cos A
(3) tan a=sin 6-tan A (4) cos A=sin B-cos a
(5) cos c=cos a-cos b
(6) sin 6=sin c-sin B
(7) tan a=tan c-cos B
(8) tan 6=sin a-tan B
(9) cos 71=sin A-cos b
(10) cos c=cot A-cot B
88. Napier’s rules.—a. General.—The ten formulas just given are required in the solution of right spherical triangles. They may be obtained from two simple rules which were discovered by John Napier. These rules which follow serve as an aid in remembering the formulas.
b. Rules.—Consider the right spherical triangle ABC, figure 132, lettered as usual. All parts of the triangle except <7=90° are arranged in circular order, figure 133, as they occur in the triangle, using
Figure 132.—Reference right spherical triangle. Figure 133.—Parts of triangle shown in figure 132 arranged in circular order.
90°—A, 90°—c, 90°—B instead of merely the letters A, c, B as in figure 132. With the circle labeled as in figure 132, looking at the
141
B
& / go°~ B 3
/ / \ 90 ° b /
X/ 9° A
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ARMY AIR FORCES
circle, figure 133, any part may be called the middle part; the two parts next to it, adjacent parts, and the remaining two, opposite parts. Napier’s rules state:
(1) The sine of a middle part equals the product of the tangents of the adjacent parts.
Thus:
sin 6=tan a-tan (90°—A)
but tan (90°—A)=cot A (by definition)
therefore sin 6=tan a-cot A
(2) The sine of the middle, part equals the product of the cosines of the opposite parts. Thus:
sin (90°—A)=cos (90°—B)-cos a
but cos (90°—B)=sin B (by definition)
therefore sin (90°—A)=sin 7?-cos a
but sin (90°—A)=cos A (by definition)
therefore cos A=sin B-cos a
c. An aid in memorizing.—These rules may be easily remembered if it is observed that the letter “i” is common in the words “sine” and “middle”; the letter “a” is common in the words “tangent” and “adjacent”; and the letter “o” is common to the words “cosine” and “opposite.”
89. Formula exercises.—a. Examples.—In these exercises the object is to obtain the formulas to be used in the solutions, solved for the unknown part.
Example 1: Given a and c, find b.
Solution: Of the three parts, a, b, and 90°—c, the part 90°—c is the middle part in the circular diagram while a and b are opposite parts;
Hence sin (90°—c)—cos a-cos b
or cos c=cos a-cos b
Since b is the unknown:
cos b =
cos c
cos a
Similarly,
sin A
sin a sin c
and cos B=tan a-cot c
Example 2: Given B and b, find A.
Solution:
. cos B sm A =-----r
cos b
b. Exercises.—Obtain the formulas solved for the unknown part
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MATHEMATICS FOR AIR CREW TRAINEES
89-91
necessary for the solution of the following right spherical triangles, where the parts given are indicated.
(1) a and b (6) b and B
(2) a and A (7) c and A
(3) a and B (8) c and B
(4) b and c (9) A and B
(5) b and A (10) Given a and c, find B
90. General laws.—a. Obtuse angles.—In the solution of right spherical triangles, obtuse angles will be encountered. Obtuse angles are greater than 90° and less than 180°. Since trigonometric tables give only the functions of acute angles, it is necessary to learn certain rules which follow:
b. Rules.— (1) Acute angles will be called first quadrant angles or angles in the first quadrant.
(2) Obtuse angles will be called second quadrant angles or angles in the second quadrant.
(3) If a trigonometric function of an obtuse angle is involved in the solution, use the same function of its supplementary angle. For example, instead of cos 110°, use cos (180°—110°)=cos 70°.
c. Laws of quadrants.—The following laws of quadrants must be mastered by the student:
(1) An angle and its opposite side are in the same quadrant.
(2) If two of the three sides, a, 6, c, are in the same quadrant, the third is in the first quadrant; if two are in different quadrants, the third is in the second quadrant.
91. Solution of right spherical triangles.—a. Suggestions. The following suggestions will be found useful in the solution of any right spherical triangle.
(1) Take any two given parts and solve for an unknown part by use of Napier’s rules. .
(2) Use the law of quadrants to determine in what quadrant the unknown parts lie.
Figure 134.—Reference spherical triangle for example 1.
143
B f
/ \ a ------* (dA (b)
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ARMY AIR FORCES
b. Examples: (1) Solve the right spherical triangle ABC if B=113°40' and a-23°50'.
Solution:
(a) Using Napier’s rules with the parts, a, b, 9()°—B,
sin a=tan 6-tan (90°—B) = tan 6-cot B
, sin a tan b=——J, cot B
Since B=113°40', use cot (180°-113°40')=cot (66°20')
log sin 23°50' = 19.6065-201Subtract
log cot 66°20'=9.6418-10 J
log tan 6 = 9.9047—10 b= 180— °42°40' = 137°20'
The supplementary angle to 42°40' was used since b and B are in the same quadrant.
(6) To find c, take the parts 90°—c, a, 90°—B giving sin (90°—B) = tan a-tan (90°—c)
cos B=tan a-cot c and cot c—v—— tan a
log cos 66°20' = 19.6036-201Subtract
log tan 23°50' = 9.6452—10 J
log cot c=9.9584—10 c= 180° —47°44'= 132°16'
a is in the first quadrant; b is in the second quadrant. Hence c is in the second quadrant.
(c) To find A, take the parts A, a, 90°—B, giving
sin (90°—A)=cos a-cos (90°—B) cos A.=cos a-sin B
log cos 23°50, = 9.9613-10]Add
log sin 66°20'=9.9618-10]
log cos A—19.9231—20 A=33°6'
A is in the first quadrant since a is.
(2) Solve the right spherical triangle if a=97° and 6= —102°.
Solution: Since b is in the second quadrant, B is also. Similarly, A is in the second quadrant while c is in the first quadrant. Complete the solution.
92. Exercises.
a. a=87°13' 6=95°42'
b. a=131°19' 6=102°43'
144
MATHEMATICS FOR AIR CREW TRAINEES
92-93
c. a=84°27' 5=108°30'
d. 6=112°11' c=64°46'
e. b=U°2T A=103°2'
/. A=97°35'
B=47°14'
g. C=67°44' A=ii2°39' h. a=89°14' _B=75°13' i. c=68°21'
B=122°23' j. ' /z \ s' s' X
/ \ x | rK \
I z \ \ \ Ux 1
lx \ O\|^ Y / \tx. xj
"■pT------7 y 71
I X \v\ \ \ n 17 I
i \z vk k \o/ s I
1 \\ \ y/ /
\ -------------------' /
\ ' /
\ \l /
Figure 135.—The astronomical triangle.
b. Line of position.—The PZS triangle is shown in figure 136. A sight is taken on a star and its altitude above the horizon is measured and compared to the computed altitude based on an assumed position. This provides data to plot a “line of position” {LOP} on a navigation chart. Two such lines intersecting provide a “fix” or position of the plane on the chart.
54953J °—43----10
145
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ARMY AIR FORCES
c. Determination of azimuth.—The PZS triangle is oblique but a perpendicular can be dropped from N to PZ or to PZ extended, as in figure 136. Triangle SDZ is a right triangle and so is triangle SDP and either may be solved by the 10 right triangle formulas or by
Figure 136.—Determination of line of position.
Napier s rules. If the latitude and longitude of the position are assumed (the usual procedure), then PZ is known, angle ZPS, called the hour angle of the star, is known and side PS is known. Then the altitude of the star may be computed and so may its azimuth, which is angle PZS. From the right triangle PDS:
Figure 137.—Reference spherical triangle for determination of azimuth.
Z
/ V
I )
? A / \ L
\/
Co PS CoZSD
sin P=cos (90°—PN)-cos (90°-P) (See fig. 137) sin P=sin PN-sin P
From triangle PDS,
cos PS— cos P«cos PD
pnq PS?
or cos PD=-------which gives PD
cos R b
146
D p
I I X 'Z-
\/f__+ .+++
i /p yD
(a) N/
5 Z.
MATHEMATICS FOR AIR CREW TRAINEES
93
Then, noting that h—PD—PZ or PZ—PD
sin (90° — ZS=cos R-cos DZ
cos ZS=cos R-cos h (this is 90° altitude).
From triangle DZS
sin R=cos (90°-Z*ZS)-cos (90°-ZS)
sin R=sin DZS-sin ZS
• 7) = 30°09'
and In triangle ZDS h=PZ— PD=54°40' — 30°09' = 24°31 cos ZS= cos A-cos h cos ZS= (.8366)(.9098) = .7671 ZS=40°26'
From triangle ZDS,
sin F=sin DZS-sin ZS
Then • nvc sin F sin DZS-——7777 sm ZS sin DZS=-^^.= .8446 .6486 PZS=57°38' angle PZ*S'=DZ*S'=57°38'
Now the arc ZS is 90°—altitude so that the altitude is 90°—altitude= 40°26' or altitude=90°—40°26'=49°34' and the azimuth, which is angle PZS is 57°38'.
94. Exercises.—a. At a moment when the declination of the sun was 12°00', the navigator observes the altitude of the sun to be 40°00' and its azimuth to be 131°20'. Find the latitude of the observer.
b. An observation was made on a star from latitude 32°30' S. The declination of the star was —20°20' and its hour angle was 3 hours 12 minutes east. What were the altitude and azimuth of the star?
Section XIII
Purpose________________
Numerical calculations
Vector triangles_______
E-6B COMPUTER
Paragraph
-------------------------------------- 95
---------------------------------------- 96
---------------------------------------- 97
148
MATHEMATICS FOR AIR CREW TRAINEES 95-96
95. Purpose.—The purpose of this section is to familiarize the bombardier or navigator trainee with the use of the E-6B aerial dead reckoning computer in solving numerical calculations; distance, rate, time problems; and the wind triangles involved in simpler navigation and bombardiering problems.
96. Numerical calculations.—a. Description.—The E-6B computer is composed essentially of two main parts. On one side is a transparent dial which is used in solving wind triangle problems. On the other is a fixed scale and rotating plate for performing numerical calculations. On this side, the outside (fixed) scale is labeled “miles.” The one adjacent to it on the dial is labeled “minutes”; the “hourg” scale, just inside it, is often used in conjunction with it. The miles and minutes scales are similar in arrangement.
b. Reading mile or minute scale.—These two scales constitute a “circular slide rule,” which is a mechanical device for adding or subtracting logarithms in multiplication and division problems. The numbers are arranged according to the values of their logarithms. However, since only the mantissas are added, the slide rule gives no information about the decimal point; computations are performed without any regard for decimal points and the size of the answer is
Li I
Figure 139—Location of 117 on E-6B computer.
approximated, by methods described in (2) below, to locate the decimal point. It is to be noted, also, that it is usually possible to carry computations only to three significant figures; greater accuracy must be obtained by some other means. In case a number of more than three digits is encountered in a problem, it must usually be rounded off to three digits. However, numbers may be read fairly well to four digits on certain parts of the scale.
(1) Since the decimal point is to be disregarded in setting a number, 1.25, 1250, .00125, etc., would all be set at the same point. The first two digits of the number to be set are usually given on the dial; the third must be located by means of the small lines between the numbers, some of which represent one-tenth, some two-tenths, some five-tenths, and some the entire distance between the successive third digits.
Example 1: Locate 1.17 on the mile scale.
Solution: Locate 11 on the mile scale. Each short line after it represents one-tenth of the distance between 11 and 12; 117 would be the seventh short line. (See fig. 139.)
149
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ARMY AIR FORCES
Example 2: Locate 4,680 on the mile scale.
Solution: Locate 45 on the scale. Since the next number is 50, 46 would be the first long line after 45. The short line after that would be 465, since it is half way to 47. 468 would be three-fifths of the way from 465 to 47. (See fig. 140.)
468
Figure 140.—Location of 468 on E-6B computer.
Example 3: Locate .0904 on the mile scale.
Solution: Locate 90. The next line is 91. A point four-tenths of the way between these lines is approximated to give 904 (See fig 141.)
gO/*L( I
\yALLU-Lu/
Figure 141.—Location of 904 on E-6B computer.
(2) To locate the decimal point in the answer, approximate the numbers involved with quantities which can be computed mentally. This gives an idea of the magnitude of the answer by which the decimal point can be located.
Example: Find the location of the decimal point in the answer to .0815X31.2
16.5
Solution: The computer gives the answer as 154 (to the nearest third place) without regard to the decimal. The problem is approxi-.1X30
mately —jy—= .1X2—.2. Therefore, the answer is .154.
c. Multiplication.—To multiply two numbers, set 10 on the minute scale opposite either of the numbers to be multiplied. Locate the other number on the minute scale. The product is directly above it on the mile scale.
(1) Since the mantissa of the logarithm of 10 is zero, this is merely a method of adding logarithms. The decimal point is located as described in b (2) above.
Example: Multiply 315 by .11.
Solution: Set 10 on the minute scale opposite 315 on the mile scale.
150
MATHEMATICS FOR AIR CREW TRAINEES
96
Opposite 11 on the minute scale read the product, 347. (See fig. 142.) The answer is obviously 34.7 since the product is approximately
300X.l or 30.
VHX.
Figure 142.—Multiplying 315 by 11 on E-6B computer.
(2) To multiply more than two numbers, find the product of the first two, as previously described, and multiply this by the next number. Continue this process until all of the numbers have been used.
Example: Multiply 3.1 by 1.9 by 27.
Solution: Set 10 on the minute scale opposite 31; opposite 19 on minute scale, read 589 on mile scale. Mark this point and set 10 on minute scale opposite it. Read product of 159 on mile scale opposite 27 on minute scale. Answer is 159.
(3) Exercises.—Multiply the following:
(«) (6) .6X55 4.65 X.l 7 .79 Answer.
(c) (d) 940X.0158 30.5 X.214 6.53 Answer.
G) (/) .975X.32 1.035X3.65 3.78 Answer.
0) (A) .0427X22.5 61.4X40.6 2490 Answer.
(*) 0) .1265X91.32 3.021 X.353 1.065 Answer.
(*) (0 (m) (ft) 150X7 7.5X9.9 73X85.1 86X99
d. Division.—To divide one number by a second, set the second on the minute scale opposite the first on the mile scale. Above 10 on the minute scale read the quotient on the mile scale.
(1) This is merely a method of subtracting logarithms. The decimal point is located as previously described.
Example: Divide 30.5 by .90.
Solution: Set 90 on the minute scale opposite 305 on the mile scale. Opposite 10 on the minute scale, read 339 on the mile scale. (See fig. 143.) The quotient is thus 33.9.
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ARMY AIR FORCES
, lilpi I5 6 7
9 10
Figure 143.—Dividing 305 by 90 on E-6B computer.
(2) Exercises.—Perform the following divisions: (a) 2.4=7
(6) 19.4=11 1.764 Answer.
(C) .405=1.62
( solutions of distance, t r ’
rate, time problems are based merely on multiplication or division. However, in some cases r is given in miles per hour, while t is to be found, or is given, in minutes to obtain greater accuracy. This requires the use of the factor 60 (60 minutes per hour) in many problems in addition to the values of d and r, or d and t. For this
154
MATHEMATICS FOR AIR CREW TRAINEES
96
reason, a heavy arrowhead-shaped mark is placed on the dial of the computer at 60. This is called the speed index. If t is given in minutes, 7-7 is the number of hours which must be used in the formula. 60
vt
The formula then reads d— This can then be changed to the pro-d t
portion ■j’=gQ’ and all problems where time is given in minutes can be solved by the proportion method.
Example 1: Given ground speed=180 knots, time of flight=35 minutes. Find distance traveled.
Solution: The equation is Set the speed index (60) oppo
site 180 on the mile scale. Opposite 35 on the minute scale read 105, the distance in nautical miles.
Example 2: Given ground speed = 150 knots, distance to travel=420 nautical miles, find time required to fly distance.
490 150
Solution: The equation is “y-=='gQ_‘ Set speed index (60) to 150 on mile scale. Opposite 420 on mile scale read 168 minutes. Within the minute scale, 168 minutes can be read as 2 hours 48 minutes on the hours scale.
Example 3: Given distance traveled=240 nautical miles, elapsed time — 1 hour 20 minutes or 80 minutes. Find ground speed.
Solution: The equation is
240 r
80 ~^60
Set 80 on minute scale or
1:20 on hour scale opposite 240 on mile scale, site speed index.
Exercises:
x 91
(a) Solve for x in
(6) Solve for x m-y-=294
Read 180 knots oppo-
x=70.7.
Answer.
(c) Using the formula of g above for time of turning in radius or action, find t if T=4 hours 20 minutes, $i=168, $2=142 mph.
(«’) The formula for determining ground speed by use of a B-3 drift yy
meter is GS—~rX. 177, $$=ground speed in knots, ZZ=altitude in V
feet, f=time elapsed in seconds. If Z/=3,650 feet and f=2.9 seconds, find the ground speed by the proportion method.
$$=223 knots. Answer.
(e) Change 241 miles per hour to the equivalent number of feet per second.
155
96-97
ARMY AIR FORCES
(/) Change 168 feet per second to the equivalent number of miles per hour. 246 miles per hour. Answer.
(g) An airplane with a ground speed of 216 knots flies for 2 hours 18 minutes. How far does it fly?
(A) An airplane flies 412 miles in 1 hour 48 minutes. What is its average ground speed? 229 miles per hour. Answer.
(i) An airplane with a ground speed of 127 knots must fly 568 nautical miles. How long will it take?
i. Conversion of distance and rate units (see par. 14, TM 1-205).— On the mile scale of the computer are found three indices on the upper end, one marked “NAUT.” (nautical miles), a second marked “STAT.” (statute miles), and a third marked “KM” (kilometer). If 1.0 is set opposite NAUT., it is seen that 1.15 is then opposite STAT., and 1.85 is opposite KM. These are the conversion factors from nautical miles to the other two units, or knots to statute miles per hour or kilometers per hour. When corresponding numbers are placed opposite these indices on the minute scale, the conversion factors are automatically set, and converted values can be read under the proper indices on the minute scale.
Example 1: Given a distance of 220 nautical miles, find the equivalent distance in statute miles and kilometers.
Solution’. Set 220 on dial opposite index marked “NAUT.” Opposite indices marked “STAT” and “KM” read 253 statute miles and 407 kilometers, respectively.
Example 2: Given a speed of 180 miles per hour, find equivalent speed in knots and kilometers per hour.
Solution: Set 180 opposite “STAT” index. Opposite “NAUT” and “KM” indices, read 156 knots and 289 kilometers per hour, respectively.
j. Exercises.—(a) Convert 223 miles per hour to knots and kilometers per hour. 194 knots or 358 kilometers per hour. Answer.
(6) An airplane with a ground speed of 163 knots must fly 93 statute miles. How long will it take?
(c) An airplane flies 47 statute miles in 13 minutes 30 seconds. What is its ground speed in knots? 182 knots. Answer.
97. Vector triangles.—a. Solving vector diagrams by means of the vector face of the B-6B computer is essentially the same as solving them by construction. However, the method is distinguished by the fact that few if any construction lines need be drawn, since they are already incorporated in the computer.
b. The straight lines on the computer all radiate from a common point (not on the computer). The curved lines are equally spaced concentric circles, all with centers at the point from which the straight
156
MATHEMATICS FOR AIR CREW TRAINEES
97
lines radiate, and the figures along the center line indicate the number of units distance of each circle from the center. Figures at various points of the face show the number of degrees between the center line and radiating lines.
c. The following rules are important:
(1) When the heading is given, the center line is used as the heading, and the wind vector is plotted from the center of the disk down the line.
(2) If the course is given it is used as the center line, and the wind vector is plotted from above to the center of the disk.
(3) Drift is always measured from heading to the course.
d. It should be noted that, having determined the drift angle to be right, the heading is found by subtracting the right drift angle from the course. If the drift angle is left, it is added to the course to give the heading.
Example 1:
Given: Wind force and direction 30 knots from 315°.
Heading (true) 165°.
Air speed (true) 180 knots.
Find: Track.
Ground speed.
Solution by E-6B computer: (1) When using the E-6B computer to solve problems in which heading is given, the center line of the chart is used to represent the heading line. Hence, air speeds will be measured along this center line. The track (course) line will be one of the radiating lines to the right or left of the center line, and ground speed will be indicated along this radiating line. The drift, being measured from heading to track (course), will be read from the center line to the right or left to the track (course) line. In this type of problem the wind must be plotted from the center of the plotting disk straight down the center line of the chart. The tail of the wind arrow is at the center of the disk.
(2) Turn compass rose to read 315°, the direction from which the wind is blowing, at “true index.” Plot wind arrow from center of plotting disk straight down center line of chart 30 units according to scale of center line. Set heading (165°), at true index and slide chart to read air speed (180 knots) at center mark of transparent disk. At end of wind arrow, read from radiating track (course) lines and speed circles of chart the drift (4° left) and ground speed (206 knots). Opposite 4° left on drift and variation scale read track (161° true).
(3) In solving vector triangles on the E-6B computer, it infre
157
97
ARMY AIR FORCES
quently occurs that the end of one of the vectors falls off the transparent disk, thus preventing solution unless %-hour vector lengths are used instead of hourly lengths. When this occurs, the proper procedure is to divide the lengths of all known vectors by two before setting up on the computer. All vectors resulting from the solution will be read at one-half their true values. Throughout this procedure,
VP”* N
I ^————7 180
I Y ______________L 170 7 /
\ y w I »6o /
\ praw vector / /
Figure 145.—Drawing wind vector—E-6B computer.
angles must not be reduced to one-half value. If any vector falls off the plotting disk when reduced to half-hour lengths, the problem may be worked by reducing all vectors to quarter-hour lengths.
e. Exercises.—Find ground speed and track in each case.
(1) Heading=310°, air speed=170 knots, wind speed=40 knots, wind direction = 185°.
(2) Heading=6°, air speed = 285 m. p. h., wind speed = 44 m. p. h., wind direction =127°. GS=310, T= 359° Answer.
(3) Heading=273°, air speed = 126 knots, wind speed = 33 knots, wind direction = 147°.
(4) Heading=355°, air speed = 320 m.p.h., wind speed = 51 m.p.h., wind direction=292°. 6^=300, T=4° Answer.
158
MATHEMATICS FOR AIR CREW TRAINEES
97
(5) Heading=96°, air speed = 137 knots, wind speed=28 knots, wind direction=81°.
(6) Heading=4°, air speed=185 mph., wind speed=47 mph., wind direction = 148°.
Example 2:
/+MX.X
XXX |65oa Up /X. |
o -------X® ftNJ
/ /pRlFT= 4° X^ \
/ / (found) X \
/ X—~~ ~ ~——A \
II \ \ 1
II V 11
I L-—’—~ I
I \ Ground \ 7 /
\ \ vector J \ II
\ \ (found) Air vector / / \ \ J Xq,ven) / J
\+X^ fi///
xx11—+Z
Figure 146.—Solution for course and ground speed—E-6B computer.
Given: Wind force and direction. 40 knots from 90°.
Course to be made good, 215°.
Air speed (true), 160 knots.
Find: Heading (true).
Ground speed.
Solution by E-6B computer: (1) When using the E-6B computer to solve problems in which course to be made good or track is given, the simplest procedure is to use the center line as the course (track) line. Ground speed will then be measured along the center line. The heading line will be one of the radiating lines to the right or left of the center line, and air speed will be indicated along the radiating line. The drift, being measured from heading to track, will be read from the radiating heading line to the right or left to the course (track) line which will be the center line of the chart. In the solving problems in which
159
97
ARMY AIR FORCES
the center line represents course (track), the wind must be plotted to the disk center from a point on the chart center line vertically above it. The head of the arrow is at the center of the disk.
(2) Set compass rose to direction from which wind is blowing (90° true) and plot wind to center from a point 40 units above it. Set course (215° true) opposite true index, and move chart until tail of wind arrow falls on true air speed (160 knots) as indicated on concentric circles. Read ground speed (180 knots) at center of disk. Read
I N D
y / _______2oq_______ n.
/ / ( 190 \ \
/ I "y/lND ~ 4-0 I8o " \ \
I I-------* <160 I
Figure 147.—Wind setting—E-6B computer.
drift (12° R.) at tail of wind arrow. Since the drift is 12° R., a true heading of 215° 12° or 203° must be flown to make good the given course.
g. Exercises —Find heading and ground speed in each case.
(1) Course=35°, air speed = 187 knots, wind speed=42 knots, wind directi on=73°.
(2) Course=312°, air speed=210 mph., wind speed = 31 mph., wind direction = 358°.
6tS=187, 77=318° Answer.
(3) Course=63°, air speed=173 knots, wind speed=28 knots, wind direction = 340°.
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MATHEMATICS FOR AIR CREW TRAINEES
97
(4) Course= 154°, air speed=245 mph., wind speed=53 mph., wind direction=12°.
(5) Course=237, air speed=310 mph., wind speed=47 mph., wind direction =149°.
(6) Course=3°, air speed ==135 knots, wind speed=27 knots, wind direction=42°. GS= 113, H— 10° Answer.
Example 3:
Given: Wind force and direction; 20 knots from 260°.
TRUE
z 7 N
/ / I 1
I Fl 180 I
U\ \ / /
\ <\ -4rV^o / /
\^\/ZT GROUNP y /
\*OA \ 1CrT0R VECTOR / /
\*X Wi \ -*Xaw*MS0) / /
\ \\ r*^(FOMw°) 7 7
X. ^*****^ 7
Figure 148.—Solution for heading and ground speed—E-6B computer
Course to be made good, 340° true. Scheduled ground speed, 150 knots.
Find: Heading.
Solution by E-6B computer. Set compass rose (fig. 149) to direction from which wind is blowing (260° true) and plot wind to disk center from point 20 units above it, marking head of wind arrow at center of disk. Set course to be made good (340°) opposite true index. Move chart until ground speed (150 knots) is under center of disk. At tail of wind arrow read air speed that must be maintained (154 knots) and drift (7° R.). The drift correction to be applied to course
549535°— 43--11
161
97
ARMY AIR FORCES
will be —7°; therefore, the true heading will be 340°—7° = 333° true.
h. Exercises. Find heading and air speed, given scheduled ground speed, course, wind speed, and wind direction:
(1) £*8=185, <7=110°, TF*8=35, +£=172°
(2) £*8=232, <7=351°, IF*8=47, W=33°
+*8=269, £=358° Answer.
(3) £*8=129, <7=186°, IF*8=38, W=243°
I K iTl' IN O Fx
r
/\ \ / 3 / p *1 \ \
/ O / | \ \
\ \
I » I -&2HH O) 1 I
I - I
P I /
\ \ I 45«ouno / /
\ \ A'a | VECTOR / /
Figure 149—Solution for heading and air speed—E-6B computer.
(4) £*8=189, <7= 27°, JU*8=42, W=165°
+£=160, £=37°
(5) £*8=165, <7=223°, 1F*8=21, JU£=267°
(6) £*8=215, <7=271°, JF*8=48, W=316°
+*8=252, £=279°
Answer.
Answer.
162
Appendix I
MISCELLANEOUS UNITS AND CONVERSION FACTORS
1. Formulas.
a. Area of circle, radius=r, is 7rr2(7r=3.1416).
b. Area of rectangle, width=w, and length=Z, is wl.
c. Area of triangle, base=6, and altitude=A, is bh/2.
d. Volume of sphere, radius=r, is % 7rr3(7r=3.1416).
e. Volume of prism, area of base=A, height=A, is Ah.
2. Conversion factors. 66 nautical miles = 76 statute miles=122 kilometers (approximate)
1 centimeter (cm.) 1 foot (ft.) = 0.393700 inch (in.) = 12 inches (in.) =30.4801 centimeters (cm.)
1 inch (in.) 1 kilometer (km.) =2.54001 centimeters (cm.) = 1,000 meters (m.) = 0.62137 statute mile (mile, stat.) = 0.53959 nautical mile (mile, naut.) = 3280.83 feet (ft.)
1 meter (m.) = 100 centimeters (cm.) = 1,000 millimeters (mm.) = 3.28083 feet (ft.) =39.3700 inches (in.)
1 millimeter (mm.) 1 nautical mile (naut.) = 0.039370 inch (in.) = 6,080.20 feet (ft.) = 1.151553 statute miles (miles, stat.)
1 statute mile = 1.853249 kilometers (km.) = 5,280 feet (ft.) = 1.60935 kilometers (km.) = .868393 nautical mile
1 U. S. gallon (gal.) =231 cubic inches (cu. in.) = .13368 cubic foot (cu. ft.) = .83310 British gallon
1 cubic foot (cu. ft.) = 1,728 cubic inches (cu. in.) = 7.4805 United States gallons (gal.)
1 kilometer per hour=0.62137 statute mile per hour
(km./hr.) 1 knot = 0.53959 knot = 1 nautical mile per hour
163
APPENDIX
= 1.853249 km./hr.
= 1.151553 mph. or miles/hr.
1 statute mile per hour= 1.4666 feet per second (mph. or mile/hr.) = 1.60935 km./hr.
= 0.868393 knot
3. Temperature scales.
Freezing: 0° C. = 32° F. =273° K. (Absolute) Boiling: 100° C.=212° F.=373° K. (Absolute) C°=%(F°—32).
F°=%C° + 32.
K°=C°+273.
4. Time equivalents.
GMT . .
—___—- London
/< MIDNIGHT /\--------C«iro
New York(EVT) A 2/00 I 0300\
Chicago (CWT) A. I A
Denver (MWT) -1800 • 0600-] Next day
Son Francisco (PWT) r" -J
\ tsoo ooooy
X, MIDDAY FX—Tokyo
Figure 150.
Note.—Time given for Tokyo is standard time. Difference in standard time between New York and Tokyo is 14 hours.
5. Densities.
Gasoline (aviation) weighs 45 Ib./cu. ft. or 6 lb./gal.
Oil (aviation) weighs 56 Ib./cu. ft. or 7.5 lb./gal.
Water weighs 62.4 Ib./cu. ft. or 8.34 Ib./gal.
Air (dry) weighs .0765 Ib./cu. ft. at 15° C. (59° F.) and standard atmospheric pressure.
164
APPENDIX
Appendix II
TABLES OF LOGARITHMS AND TRIGONOMETRIC FUNCTIONS
Table I.—Common logarithms
N 0 1 2 3 4
10 0000 0043 0086 0128 0170
11 0414 0453 0492 0531 0569
12 0792 0828 0864 0899 0934
13 1139 1173 1206 1239 1271
14 1461 1492 1523 1553 1584
15 1761 1790 1818 1847 1875
16 2041 2068 2095 2122 2148
17 2304 2330 2355 2380 2405
18 2553 2577 2601 2625 2648
19 2788 2810 2833 2856 2878
20 3010 3032 3054 3075 3096
21 3222 3243 3263 3284 3304
22 3424 3444 3464 3483 3502
23 3617 3636 3655 3674 3692
24 3802 3820 3838 3856 3874
25 3979 3997 4014 4031 4048
26 4150 4166 4183 4200 4216
27 4314 4330 4346 4362 4378
28 4472 4487 4502 4518 4533
29 4624 4639 4654 4669 4683
30 4771 4786 4800 4814 4829
31 4914 4928 4942 4955 4969
32 5051 5065 5079 5092 5105
33 5185 5198 5211 5224 5237
34 5315 5328 5340 5353 5366
35 5441 5453 5465 5478 5490
36 5563 5575 5587 5599 5611
37 5682 5694 5705 5717 5729
38 5798 5809 5821 5832 5843
39 5911 5922 5933 5944 5955
40 6021 6031 6042 6053 6064
41 6128 6138 6149 6160 6170
42 6232 6243 6253 6263 6274
43 6335 6345 6355 6365 6375
44 6435 6444 6454 6464 6474
45 6532 6542 6551 6561 6571
46 6628 6637 6646 6656 6665
47 6721 6730 6739 6749 6758
48 6812 6821 6830 6839 6848
49 6902 6911 6920 6928 6937
50 6990 6998 7007 7016 7024
51 7076 7084 7093 7101 7110
52 7160 7168 7177 7185 7193
53 7243 7251 7259 7267 7275
54 7324 7332 7340 7348 7356
5 6 •7 8 9 u. d.
0212 0253 0294 0334 0374 4.2
0607 0645 0682 0719 0755 3.8
0969 1004 1038 1072 1106 3.5
1303 1335 1367 1399 1430 3.2
1614 1644 1673 1703 1732 3.0-
1903 1931 1959 1987 2014 2.8
2175 2201 2227 2253 2279 2.6
2430 2455 2480 2504 2529 2.5
2672 2695 2718 2742 2765 2.4
2900 2923 2945 2967 2989 2.2
3118 3139 3160 3181 3201 2.1
3324 3345 3365 3385 3404 2.0
3522 3541 3560 3579 3598 1.9
3711 3729 3747 3766 3784 1.8
3892 3909 3927 3945 3962 1.8
4065 4082 4099 4116 4133 1.7
4232 4249 4265 4281 4298 1.6
4393 4409 4425 4440 4456 1.6
4548 4564 4579 4594 4609 1.5
4698 4713 4728 4742 4757 1.5
4843 4857 4871 4886 4900 1.4
4983 4997 5011 5024 5038 1.4
5119 5132 5145 5159 5172 1.3
5250 5263 5276 5289 5302 1.3
5378 5391 5403 5416 5428 1.3
5502 5514 5527 5539 5551 1.2
5623 5635 5647 5658 5670 1.2
5740 5752 5763 5775 5786 1.2
5855 5866 5877 5888 5899 1.1
5966 5977 5988 5999 6010 1.1
6075 6085 6096 6107 6117 1.1
6180 6191 6201 6212 6222 1.0
6284 6294 6304 6314 6325 1.0
6385 6395 6405 6415 6425 1.0
6484 6494 6503 6513 6522 1.0
6580 6590 6599 6609 6618 1.0
6675 6684 6693 6702 6712 .9
6767 6776 6785 6794 6803 .9
6857 6866 6875 6884 . 6893 .9
6946 6955 6964 6972 6981 .9
7033 7042 7050 7059 7067 .9
7118 7126 7135 7143 7152 .8
7202 7210 7218 7226 7235 .8
7284 7292 7300 7308 7316 .8
7364 7372 7380 7388 7396 .8
165
APPENDIX
Table I.—Common logarithms—Continued
166
N 0 1 2 3 4 5 6 7 8 9 u. d.
55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 .8
56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 .8
57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 .8
58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 .7
59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 .7
60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 .7
61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 .7
62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 .7
63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 .7
64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 .7
65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 .7
66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 .7
67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 .6
68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 .6
69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 .6
70 8451 8457 8563 8470 8476 8482 8488 8494 8500 8506 .6
71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 .6 .
72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 .6
73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 .6
74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 .6
75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 . 6
76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 .6
77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 .6
78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 . 6
79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 .5
80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 .5
81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 .5
82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 .5
83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 .5
84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 .5
85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 .5
86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 .5
87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 .5
88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9480 .5
89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 .5
90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 .5
91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 .5
92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 .5
93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 .5
94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 .5
95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 .5
96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 .5
97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 .4
98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 .4
99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 .4
APPENDIX
Table II.—Trigonometric functions for every degree from 0° to 90°
Angle Sin Cos Tan Cot Sec Csc
0° 0. 0000 1. 0000 0. 0000 1. 0000 90°
1° . 0175 . 9998 . 0175 57. 2900 1. 0002 57. 2987 89°
2° . 0349 . 9994 . 0394 28. 6363 1. 0006 28. 6537 88°
3° . 0523 s 9986 . 0524 19. 0811 1. 0014 19. 1073 87°
4° . 0698 . 9976 . 0699 14. 3007 1. 0024 14. 3356 86°
5° . 0872 . 9962 . 0875 11. 4301 1. 0038 11. 4737 85°
6° . 1045 . 9945 . 0151 9. 5144 1. 0055 9. 5668 84°
7° . 1219 . 9925 . 1228 8. 1443 1. 0075 8. 2055 83°
8° . 1392 . 9903 . 1405 7. 1154 1. 0098 7. 1853 82°
9° . 1564 . 9877 . 1584 6. 3138 1. 0125 6. 3925 81°
10° . 1736 . 9848 . 1763 5. 6713 1. 0154 5. 7588 80°
11° . 1908 . 9816 . 1944 5. 1446 1. 0187 5. 2408 79°
12° . 2079 . 9781 . 2126 4. 7046 1. 0223 4. 8097 78°
13° . 2250 . 9744 . 2309 4. 3315 1. 0263 4. 4454 77°
14° . 2419 . 9703 . 2493 4. 0108 1. 0306 4. 1336 76°
15° . 2588 . 9659 . 2679 3. 7321 1. 0353 3. 8637 75°
16° . 2756 . 9613 . 2867 3. 4874 1. 0403 3. 6280 74°
17° . 2924 . 9563 . 3057 3. 2709 1. 0457 3. 4203 73°
18° . 3090 . 9511 . 3249 3. 0777 1. 0515 3. 2361 72°
19° . 3256 . 9455 . 3443 2. 9042 1. 0576 3. 0716 71°
20° . 3420 . 9397 . 3640 2. 7475 1. 0642 2. 9238 70°
21° . 3584 . 9336 . 3839 2. 6051 1. 0711 2. 7904 69°
22° . 3746 . 9272 . 4040 2. 4751 1. 0785 2. 6695 68°
23° . 3907 . 9205 . 4245 2. 3559 1. 0864 2. 5593 67°
24° . 4067 . 9135 . 4452 2. 2460 1. 0946 2. 4586 66°
25° . 4226 . 9063 . 4663 2. 1445 1. 1034 2. 3662 65°
26° . 4384 . 8988 •. 4877 2. 0503 1. 1126 2. 2812 64°
27° . 4540 . 8910 . 5095 1. 9626 1. 1223 2. 2027 63°
28° . 4695 . 8829 . 5317 1/8807 1. 1326 2. 1301 62°
29° . 4848 . 8746 . 5543 1. 8040 1. 1434 2. 0627 61°
30° . 5000 . 8660 . 5774 1. 7321 1. 1547 2. 0000 60°
31° . 5150 . 8572 . 6009 1. 6643 1. 1666 1. 9416 59°
32° . 5299 . 8480 . 6249 1. 6003 1. 1792 1. 8871 58°
33° . 5446 . 8387 . 6494 1. 5399 1. 1924 1. 8361 57°
34° . 5592 . 8290 . 6745 1. 4826 1. 2062 1. 7883 56°
35° . 5736 . 8192 . 7002 1. 4281 1. 2208 1. 7434 55°
36° . 5878 . 8090 . 7265 1. 3764 1. 2361 1. 7013 54°
37° . 6018 . 7986 . 7536 1. 3270 1. 2521 1. 6616 53°
38° . 6157 . 7880 . 7813 1. 2799 1. 2690 1. 6243 52°
39° . 6293 . 7771 . 8098 1. 2349 1. 2868 1. 5890 51°
40° . 6428 . 7660 . 8391 1. 1918 1. 3054 1. 5557 50°
41° . 6561 . 7547 . 8693 1. 1504 1. 3250 1. 5243 49°
42° . 6691 . 7431 . 9004 1. 1106 1. 3456 1. 4945 48°
43° . 6820 . 7314 . 9325 1. 0724 1. 3673 1. 4663 47°
44° . 6947 . 7193 . 9657 1. 0355 1. 3902 1. 4396 46°
45° . 7071 . 7071 1. 0000 1. 0000 1. 4142 1. 4142 45°
Cos Sin Cot Tan Csc Sec Angle
167
APPENDIX
Table III.—Common logarithms of trigonometric functions for every degree from 0° to 90°
Angle Sine log Tangent log Cotangent log Cosine log
0° 0. 0000 90°
1° 8. 2419 8. 2419 1. 7581 9. 9999 89°
2° 8. 5428 8. 5431 1. 4569 9. 9997 88°
3° 8. 7188 8. 7194 1. 2806 9. 9994 87°
4° 8. 8436 8. 8446 1. 1554 9. 9989 86°
5° 8. 9403 8. 9420 1. 0580 9. 9983 85°
6° 9. 0192 9. 0216 . 9784 9. 9976 84°
7° 9. 0859 9. 0891 . 9109 9. 9968 83°
8° 9. 1436 9. 1478 . 8522 9. 9958 82°
9° 9. 1943 9. 1997 . 8003 9. 9946 81°
10° 9. 2397 9. 2463 . 7537 9. 9934 80°
11° 9. 2806 9. 2887 . 7113 9. 9919 79°
12° 9. 3179 9. 3275 . 6725 9. 9904 78°
13° 9. 3521 9. 3634 . 6366 9. 887 77°
14° 9. 3837 9. 3968 . 6032 9. 9869 76°
15° 9. 4130 9. 4281 . 5719 9. 9849 75°
16° 9. 4403 9. 4575 . 5425 9. 9828 74°
17° 9. 4659 9. 4853 . 5147 9. 9806 73°
18° 9. 4900 9. 5118 . 4882 9. 9782 72°
19° 9. 5126 9. 5370 . 4630 9. 9757 71°
• 20° 9. 5341 9. 5611 . 4389 9. 9/30 70°
21° 9. 5543 9. 5842 . 4158 9. 9702 69°
22° 9. 5736 9. 6064 . 3936 9. 9672 68°
23° 9. 5919 9. 6279 . 3721 9. 9640 67°
24° 9. 6093 9. 6486 . 3514 9. 9607 66°
25° 9. 6259 9. 6687 . 3313 9. 9573 65°
26° 9. 6418 9. 6882, . 3118 9. 9537 64°
27° 9. 6570 9. 7072 . 2928 9. 9499 63°
28° 9. 6716 9. 7257 . 2743 9. 9459 62°
29° 9. 6856 9. 7438 . 2562 9. 9418 61°
30c 9. 6990 9. 7614 . 2386 9. 9375 60°
31° 9. 7118 9. 7788 . 2212 9. 9331 59°
32° 9. 7242 9. 7958 . 2042 9. 9284 58°
33° 9. 7361 9. 8125 . 1875 9. 9236 57°
34° 9. 7476 9. 8290 . 1710 9. 9186 56°
35° 9. 7586 9. 8452 . 1548 9. 9134 55°
36° 9. 7692 9. 8613 . 1387 9. 9080 54°
37° 9. 7795 9. 8771 . 1229 9. 9023 53°
38° 9. 7893 9. 8928 . 1072 9. 8965 52°
39° 9. 7989 9. 9084 . 0916 9. 8905 51°
40° 9. 8081 9. 9238 . 0762 9. 8843 50°
41° 9. 8169 9. 9392 . 0608 9. 8778 49°
42° 9. 8255 9. 9544 . 0456 9. 8711 48°
43° 9. 8338 9. 9697 . 0303 9. 8641 47°
44° 9. 8418 9. 9848 . 0152 9. 8569 46°
45° 9. 8495 . 0000 . 0000 9. 8495 45°
log Cosine log Cotangent log Tangent log Sine Angle
Note: Log sec 2=log cos x, log esc x= —log sin x.
168
INDEX
Paragraph Page
Acute triangle______________________________________________ 54a (5) 88
Addition:
Of angles, graphic__________________________________________ 59 98
Of fractions_______________________________________________ 10 20
Of numbers___________________________________________________ 4 3
Of polynomials______________________________________________ 22 37
Of positive and negative numbers____________________________ 15 29
Air navigation system of measuring directions_______________ 385 62
Air speed:
Calibrated______________________________________________ 34 54
Definition______________________________________________ 446 (1) 70
Meter, calibration graph________________________________ 34a 54
Vector_______________________.>_________________________ 44a 70
Algebra:____________________________________________________ 20-27 35
Addition of polynomials________________________________ 22a 37
Division of polynomials_________________________________ 236 38
Evaluation of algebraic expressions_____________________ 24a 39
Multiplication of polynomials_______________________.___ 23a 38
Subtraction of polynomials-_______________________________ 22a 37
Symbols___________________________________________________ 21a 35
Word problems______________________________________________ 26a 43
Angle, angles: «
Addition of, graphic_______________________________________ 53z 85
Bisecting of_______________________________________________ 53d 83
Central_________________________________________________ 56a (6) 92
Classification of__________________________________________ 52^ 80
Complementary_______________________________________________ 66 111
Construction of____________________________x____________ 38e 64
Copying of_________________________________________________ 536 82
Corresponding_______________________________________________ 55a 90
Definition of___________________________________________ 38, 52/ 62, 80
Laying off with protractor_________________________________ 38c 64
Obtuse spherical____________________________________________ 90 143
Spherical________________7------------------------------ 61/ 104
Units of measurement________________________________________ 38 62
Arc:
Circle determined by_________________________________________ 56 92
Major. _____________________________________________________ 56 92
Minor....__________________________________________________ 56 92
On sphere___________________________________________________ 60 102
Area:
Of circle_________________________________________________ 56c 94
Of triangle------------------------------------------------- 54c 88
Astronomical triangles, solution of_____________________________ 93 145
Axes and points________________________________________________ 33a 52
Axiom_______________________________________________________ 21, 51 35, 79
Bisecting an angle--------------------------------------------- 53c 83
169
INDEX
Paragraph
Calibrated air speed________,_______________________________ 34
Centigrade scale____________________________________________ 25e (4)
Central angle---------------------------------_------------- 56o (g)
Circle:
Area---------------------------------------------------- 5g
Circumference_________________________________________ 5g
Constructions___________________________________________ 5g
Definition_______________________________________________ 52J
Definition of functions of__________________________
Great..---------------_•------------------_■________60a, 86a
Properties of, special______________________________ 5g
bmall----------------------------------------------- 56a, gga
Circular functions______________________________________ -ga
Circumference________________________________________ __ 5 g-
Common denominator____________________________ _________ jqc
Complementary angles_____________________________________ gg
Computer E-6B____________________________________________ 95-97
Concentric circles___________________________________ _ 5ga (g)
Cone--------------------------------------------------------59c
Congruent triangles________________________________________ 55^
Constructing triangle when:
Three sides are known_____________________
53 in
Two angles and included side are known__________________ .53/
Two sides and included angle are known_______________ 53^
Constructions with rule and compass___j.__________________ ,53
Conversion factors____________________________________ _ j
Corollary_________________________________________________ -jy
Cosecant__________'___________________________________ 63-71
Cosine------------------------------------------------’ ’ 63-71
Cotangent_____________________________________________ 63-71
Course________________________________________________ _ gg
Cylinder-------------------------------------------------- 5g&
Decimal fractions_________________________________________
Decimal point_____________________________________________ 4a g^ yy
Definitions of:
Circular functions__________________________
Earth’s landmarks_______________________
Logarithmic terms___________________________
Plane geometric functions__________________________
Quadrilateral___________________________
Solid geometric functions__________________________
Spherical geometric functions
Spherical triangle______________________
Trigonometric functions____________________________
Dividend______________________________
Division______________________________
Fractions______-________________________
Long division___________________________
Polynomials_____________________________
Short division______________________________
Divisor_______________________________
Page 54 42 92
92 92
92 81 92
102, 138 92
92, 138 92
94
21 111 149
93 100
90
87 86
86 82
163 79
108 108
108
65
100 15
3, 7, 11
92 105 124
79 90 98
103 138
109
9, 11 9
23
10
38 9
9
56a
62a
74
51
55a
59
61
86
64c
7a, f
7a
12a
7c
236
7b
7a
170
INDEX
Paragraph
Page
Drift________________________________________________________ 39c
Drift angle______________________________________39c, e; 45a (2) (/)
Ellipse_____-_________________________________________________ 58a
Equal parts, dividing line into_______________________________ 53m
Equations, algebraic___________________________________________ 25
Axioms in solving---------------------------------------- 25a
Formation of______________________________________________ 8e
Graphic solutions of equations containing two unknowns.. 36
Graphing from equations__________________________________ 355
Equator_____________________________________________________ 62a (1)
Equilateral triangle________________________________________ 54a (3)
Construction of________________________________________ 535
Exercises:
Addition----------------------------------------------- 44
Air navigation_________________________________________ 40a, b
Algebra______________________225, c; 23c, d; 245, 254, e; 264; 27
Astronomical triangle---------------------------------- 94
Division_______________________________________________ 174
Equations, numerical___________________________________ 22
E-6B computer
96c (3), d (2), h; 97 f, h
Fractions_____________________________8d, f; 9d; lOd; 11c; 12c
Fundamental numerical operations_____________________ 19
Graphs________________________________________ 35c, 364, 36/
Logarithms____________________________________ 76, 77, 81, 83
Longitude____________________________________________ 625 (2)
Multiplication___________________________________6g, h; 175
Plane geometry______________________
53o, 544, 55d, 56d, 575
Plane trigonometry______________________________ 355, 68c, 71
Positive and negative numbers______________________ 165; 175, d
Proportion_____________;_______________________________ 13a
Scales. _■_____________________________________________29c, 31
Solid geometry_________________________________________ 59e
Spherical geometry_____________________________________89, 92
Spherical trigonometry_________________________________ 615 (4)
Subtraction. _________________________________________ 5c
Time differences_______________________________________ 52c
Triangle of velocity____________________________44a, 45, 46, 47
Vectors_________________________________________________ 434
Extraction of roots by logarithms-----------------------------81
Factors________________________________________________________ 6a
Rounding off______________________________________________ 64
Fahrenheit scale____________________________________________ 25e (4)
Figures, miscellaneous, properties of-------------------------- 58
65
65, 66, 73
97
87 40
40
16 60
57 105
88 84
4
66 137,38,39, 39,39,41, 1.41, 46, 47
148
30 37 1 151,152, | 159, 162 116,18,20, [21, 23, 24
33
59, 61, 62 f 125,130, [ 133
106
8, 9, 30 [88,90,92, [ 94, 97
1 57,115, I 120
29, 30, 30
24
48, 49 102 142, 144
105
6 80
f 70,71, t 73, 76
69
130
6
7
42
97
171
INDEX
Formulas:
Evaluating with logarithms
Trigonometric______________
45-degree angle functions_____
Fractions:
Addition of_________________________________
Conversion of_________________________________
Division of_________________________________
Multiplication of___________________________
Reduction to lowest terms________________________
Subtraction of______________________________
Fuel consumption graph___________________________
Functions:
Paragraph
83
64
65
Page 133 108 109
10a 20
8a, 9a 15, 19
12a 23
Ila 22
8c 16
105 21
34e 56
Of circles-------------------------------------------- 56a 92
Of 30, 45, and 60-degree angles_______________________ 65 jqq
Trigonometric-----2___________________________________ 64c 109
Geometry—See Plane Geometry, Solid
Geometry, and Spherical Geometry
Graphs---------------------------------------------------- 32_36 52
Air-speed meter calibration___________________________ 34a
Axes and points__________________________________________ 33a -2
Construction______ _ ____
Examples-------------- . ' “ A ' ’ 71 . ’' 35 37
From equations_________________r______________________ 355 57
Pressure-temperature_____________________________________ 345 5-
Reading-------------------------------------------- _ 34
Sunset------------------------------------------------- 34d 55
Used to solve equations containing two unknowns_______ 36 60
Great circle--------------------------------------------- 60a (1) i03
Distance---------------------------------------------- 865 (6) 138
Greenwich meridian____________________________________________ 62c 107
Ground speed:
Definition____________________________________________ 44
Vector-------------------------------------- 45a, b; 46; 47. 48
Hemispheres---------------------------------------------- 866 (4)
Horsepower_________________________________________ 2>-g
Hyperbola____________________________________________ __ _ -6
Hypotenuse, square of, right triangles____________________ 54
Indicated airspeed____________________________________ _ 34
Interpolation:
Logarithmic tables _______________________________________ 77
Trigonometric tables______________________________________ 69
Isosceles triangles______________________________ _______53a (2)
Landmarks, arbitrary, earth 6?a
Latitude------------------------------------------------- 62a
Line, Lines___________________________________________________ 52
Logarithms:
70
71,73,73,
76, 76
138
41
92
88
54
125
115
82
105
105
80
Application to:
Division______________________________ _
Extraction of roots_________________
Evaluation of formulas_______________________
Multiplication_______________
79 128
81 130
83 133
79 128
172
INDEX
Logarithms—Continued.
Applications to—Continued. Paragraph Page
Oblique-triangle problems----------------------------- 84 135
Powers of numbers_____________________________________ 80 129
Right-triangle problems______________________________ 82 132
Characteristic:__________________________________________ 74a 124
Negative______________________________________________ 78 126
Interpolation____________________________________________ 77 125
Introduction to___________________________________________ 73 123
Mantissa__________________________________________________ 75 124
Powers of numbers_________________________________________ 80 129
Purpose___________________________________________________ 72 123
Tables of_____________________________________________ App. II 165
Longitude__________________________________________________ 62a (5) 106
Mantissa----------„---------------------------------------- 75 124
Maps_______________________________________________________ 30a 48
Materials needed___________________________________________ 2 2
Meridian___________________________________________________ 62a (3) 105
Mixed numbers___________________________________________________ 7h 12
Models________________________________________________________ 29a 48
Multiplication:_________________________________________________ 6a 6
Of fractions_____________________________________________ 11a 22
Of polynomials_________________________________________ 23a 38
Of positive and negative numbers_________________________ 17a 30
With E-6B computer________________________________________ 96 149
With logarithms___________________________________________ 79 128
Napier’s rule__________________________________________________ 88 141
Oblique triangles solved:
By right-triangle methods--------------------------------- 70 117
With logarithms----------------------------------------- 84 135
Obtuse spherical triangle, solution of__________________________ 90 143
Parabola_____________________________________________________ 585 97
Parallelogram______________________________________________ 57a (2) 95
Percent________________________________________________________ 8e 16
Perpendicular:---------------------------------------------- 52h 81
At a point, construction of_____________'_____________ 53/ 84
From a point, construction of--------------------------- 53. S\ \ \ ^X. ^X
Z> \ \ ' ' / ' 7C>, Sb/ S?\^' X. \ \ \ X
yX\ i'o XX XX X\ ■ X \
Z\ X £ St 1 X\ N\ X\ X "X \
X s / Z/X^X x \ X \
#x fx/ V i \ \ k js
A" /x // /°°\ XkXK x s \ oAT J9 P
x, S X/ Fp X>Sx x \ /.v. ^iXxxX y /
A / 2|56« s
V" X""F°°
\^ X/ \ o. /y /
z / \V / 4X xF
X/ u/ X/' ' / X\>X
XX ,hX / *^