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(2) Erercises.- Perform the following divisions:
1.764 Answer. (c) 405-:-1.62 (d) 385-+ 106.
3.63 Answer. (e) 2.69:-.73 (f) 4.55:13.55
.336 Answer. (9) 2.807-: .905 (h) 19.85:-63.6
.312 Answer. () 143.2:-40.91 (1) 14.86: 13.58
1.095 Answer. (k) 311-14 (1) 89:-3.5 (m) 376:17 (n) 22.5:-5 (0) 1.73:-0.14
e. Problems involving multiplication or division.--Solve the following problems:
(1) An airplane has 220 gallons of fuel in its tanks at a certain moment. If the rate of consumption is 72 gallons per hour, what is the remaining flight time?
(2) An airplane consumes 175 gallons of gasoline in 2 hours 04 minutes flying time. What is the rate of consumption?
84.7 gallons per hour. Answer. (3) An airplane climbs to 3,350 feet in 1 minute 40 seconds. What is its average rate of climb in feet per minute?
(4) For a certain type of antiaircraft gun, an average of 6.25 seconds elapses between successive firings. How long will it take to shoot 43 shells?
4 minutes 29 seconds. Answer.
(5) The rate of consumption of an airplane is 105 gallons of fuel per hour, and the fuel remaining is 271 gallons. What is the remaining flight time?
(6) An airplane consumes 165 gallons of fuel in an elapsed time of 1 hour 40 minutes. What is the rate of fuel consumption?
(7) The rate of consumption of fuel of an airplane was 87 gallons per hour and the time of flight was 3 hours 18 minutes. Find the fuel consumption.
f. Proportion.-Write the proportion as an equation involving two fractions. First, consider the fraction in which both parts are known. Set the denominator on the minute scale opposite the numerator on the mile scale. If the known in the other fraction is the denominator, locate this on the minute scale and read the unknown opposite it on the mile scale. If the known is the numerator, locate it on the mile scale and read the unknown opposite it on the minute scale.
(1) Each of the four quantities in the proportion (three known and one unknown) holds the same position on the computer as in the equation. Numerators (on the mile scale) are above the corresponding denominators (on the minute scale). Thus, any three quantities in a proportion are set in their same positions as in the equation, and the unknown is read in its corresponding position.
Example: (a) Solve the following for x:
Solution: Set 26 on the minute scale opposite 515 on the mile scale. Opposite 23 on the minute scale, read 455. (See fig. 144.) Thus X=4.55.
FIGURE 144.-Solving the proportion 1:2.3::51.5:26 on the E-6B computer.
16 (6) Solve for x in
Solution: Set 92 on the minute scale opposite 16 on mile scale. Opposite 71 on mile scale, read 408. Thus x=
=4.08. (2) Solution of a proportion by the method outlined constitutes simultaneous multiplication and division. Setting numerator and denominator of one fraction opposite each other divides the two. Reading the unknown opposite the third number constitutes multiplying or dividing the quotient, depending on whether the third number is in the denominator or numerator. Problems involving these simultaneous operations may thus be solved by the proportion method.
Solving this as a proportion, x=1.028
(6) Change 76 miles per hour to the equivalent number of feet per second.
Solution: To change miles per hour to feet per second, it is necessary to first multiply by 5,280 (to change miles to feet) and then
76.5280 divide by 3,600 (to change hours to seconds). Hence, x=
3,600 Write this as a proportion:
76 3,600 76 30
This type of problem, or the reverse type, may thus always be solved by setting 44 on the minute scale and 30 on the mile scale. Then miles
per hour and feet per second are read on the mile scale and minute scale, respectively, the unknown being read opposite the known.
g. Solving formulas by proportion.— The technique for solving problems in simultaneous multiplication and division may be applied to evaluating formulas. The formula is first written in proportion form, then solved by the method just described. Example: The time (t) for turning back in a radius of action problem
S.XT is given by t= where T=fuel hours, Si=rate of departure,
Si+S2 Sz=rate of closure. (See sec. XI, ch. 1, TM 1-205.) Find t if Si= 155, Sz=135, T=5 hours.
135 Solution: Write the formula as TS+S2
290° Solving by the proportion method, t=2.33 hours=2 hours 20 minutes. h. Distance, rate, time problems (see par. 170, TM 1-205).—Using
d the fundamental formula, d=rt, or r=
solutions of distance, 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
In this case, 5
> or t=
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
t minutes, is the number of hours which must be used in the formula. 60
rt The formula then reads d= This can then be changed to the pro
60 d portion
T-60 be solved by the proportion method.
Example 1: Given ground speed=180 knots, time of flight=35 minutes. Find distance traveled.
d 180 Solution: The equation is
Set the speed index (60) oppo
35 60 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.
420 150 Solution: The equation is
Set speed index (60) to 150
t 60 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 minụtes or 80 minutes. Find ground speed.
240 Solution: The equation is
Set 80 on minute scale or
80 60 1:20 on hour scale opposite 240 on mile scale. Read 180 knots opposite speed index, Exercises:
91 (a) Solve for x in
19.35 8.05 (6) Solve for x in
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, Si=168, S=142 mph. (d) The formula for determining ground speed by use of a B-3 drift
H meter is GS X.177, , H= feet, t=time elapsed in seconds. If H=3,650 feet and t=2.9 seconds, find the ground speed by the proportion method.
GS=223 knots. Answer. (e) Change 241 miles per hour to the equivalent number of feet per second.
() Change 168 feet per second to the equivalent number of miles
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?
(h) 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