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Then, noting that h=PD-PZ or P2-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°– DZS).cos (90°– ZS)
sin R=sin DZS.sin ZS

R

sin DZS_sin

sin ZS From geometry, PZS=180°-DZS which is the azimuth of the star and the triangle is completely solved. If R is inside the triangle, PZS=DZS

Example: The navigator of an airplane estimates his position as latitude 35°20' North and longitude 140°00' West. The local hour angle to a star at the time of observation was 3 hours 30 minutes East and the declination of the star was 46°20' North, as found in the Air Almanac. Find the altitude and azimuth of the sun at the time of observation. (See fig. 138.)

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Solution: Since the declination is 46°20' north, the arc PS is 90°— 46°20'=43°40'. The arc PZ is 90°-lat. =90°— 35°20'=54°40'. The angle at P is known as the local hour angle which was given as 3 hours 30 minutes east, which means that the star is east of the observer. Since 15 degrees of longitude corresponds to 1 hour of time, 3 hours, 30 minutes=52°30' of arc. Draw a great circle arc SD through the star position and perpendicular to arc PZ. Then from triangle PDS

sin R=sin PS.sin P
. sin R=(.6905)(.7934)=.5478

R=33°13'
From triangle PDS

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cos PD-.7234

.8366 =.8647

PD=30°09' and

h=P2-PD=54°40'— 30°09'=24°31' In triangle ZDS cos ZS=cos R.cos h

cos ZS=(.8366)(.9098)=.7671

ZS=40°26' From triangle ZDS,

sin R=sin DZS.sin ZS

sin DZS=sin R

sin ZS sin DZS="5478

0.6486 =.8446

DZS=57°38' Then

angle PZS=DZS=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.

6. 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?

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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.—. 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 "hours" 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

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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.)

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.)

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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.)

904 o

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 0815 X31.2

16.5 Solution: The computer gives the answer as 154 (to the nearest third place) without regard to the decimal. The problem is approxi

1.1X30_ ivo mately --- 30=.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 6 (2) above.

Example: Multiply 315 by .11.
Solution: Set 10 on the minute scale opposite 315 on the mile scale.

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.1 or 30.

315 347.30

30

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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:
(a) .6 X 55
(6) 4.65X.17

.79 Answer. (c) 940 X.0158 (d) 30.5X.214

6.53 Answer. (e) .975X.32 (f) 1.035 X3.65

3.78 Answer. (g) .0427X22.5 (h) 61.4 X 40.6

2490 Answer. () .1265 X 91.32 6) 3.021 X.353

1.065 Answer. (k) 150X7 (1) 7.5X9.9 (m) 73X85.1 (n) 86 X 99

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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