HomeworkQ & AJunior NavigationChapter 7The Celestial LOP

Objectives:

Understand the altitude-intercept method of plotting and the relationships between Ho, Hc and intercept. Identify the parts of the navigational triangle. Compute altitude and azimuth of a celestial body using a scientific calculator. Convert azimuth angle (Z) to azimuth (Zn)

Practical Exercise1. 2. Follow the Student Manual for guidance.

3. When the observer is closer to the GP than is the DR: a. Ho is greater than Hc. b. Hc is greater than Ho. c. intercept is away. d. the radius of the circle of position through the DR is less than that through the observer's actual position. Ref.: 16

4. In completing a sight reduction of the Sun, your results indicate Ho = 3517.4' and Hc = 3536.2'.

a. Find the value of the intercept (a).Ans: 18.8nm b. Is the intercept (a) toward (T) or away (A)? Ans: away, since Ho < Hc c. Are you closer to or further from the GP than is the DR? Ans: further from the GP Solution: Hc = 3536.2'. Ho = 3517.4' a = 18.8' = 18.8nm A Ref.: 14 - 18

5. In completing a sight reduction of the Sun, your results indicate Ho = 4345.3' and Hc = 4338.8'. Find the value of the intercept (a).Ans: 6.5nm b. Is the intercept toward (T) or away (A)? Ans: toward, since Ho > Hc

c. Are you closer to or further from the GP than is the DR? Solution: Hc = 4338.8' Ho = 4345.3' a = 6.5' = 6.5nm TRef.: 14 - 18Ans: closer to GP

6. The elevated pole of the navigational triangle is: a. the pole having the same name as the body's declination. b. the pole having the same name as the DR latitude.

c. always the North Pole. d. always the South Pole. Ref.: 21

7. The distance from the elevated pole to the reference position or DR is: a. called co-latitude. b. called co-altitude. c. sometimes greater than 90.

d. called latitude.Ref.: 24

8. Declination is: a. one side of the navigational triangle. b. the angular distance from the observer to the GP of the body.

c. always greater than 90. d. the angular distance from the equator to the GP of the body.Ref.: 25, Fig. 7 5a

9. In the navigational triangle, the distance from the DR to the GP of the body is: a. 90 - Dec. b. measured along a parallel of latitude. c. the radius of a celestial circle of position.

d. measured along a meridian of longitude.Ref.: 26

10. Azimuth (Zn) is: a. always an internal angle of the navigational triangle. b. always measured clockwise from true north. c. measured from either pole, depending on the hemisphere in which the observer is located.

d. always less than 90.Ref.: 28

11. If the observer is in the southern hemisphere and the LHA of the sun is less than 180: a. Z is north and east.

b. Z is south and east.

c. Z is south and west.

d. Z is north and west.Ref.: Figure 7-6c, Table 7-1

12. If the observer is in the northern hemisphere and the sun is west of the observer: a. Zn = 360 - Z. b. Zn = 180 + Z.

c. Zn = 180 - Z.

d. Zn = Z.Ref.: Figure 7-6a, Table 7-2

13. The two sides and angle used to solve the navigational triangle are: a. declination, azimuth angle, and altitude. b. co-latitude, co-altitude, and LHA. c. co-declination, co-altitude, and azimuth angle.

d. co-latitude, co-declination, and LHA.Ref.: 21 29, Figure 7 -6

Summary: LHA Hc Intercept Zn

a. 31641.9' 3400.8' 3.5nm T 054

b. 5843.4' 2510.5' 10.8nm A 246

c. 3918.1' 5142.7' 5.6nm T 259

d. 32203.3' 5243.4' 14.9nm A 098Ref.: 4814. Using the values given below, obtain the intercept (a) and azimuth (Zn) by calculator solution.Click ButtonTo ViewNote: Solutions use values for LHA, Lat, Dec, and Hc rounded to 5 decimal places and entered into the calculator as such. Values of arc sin and arc cos are left in the calculator at full precision and converted directly to Hc and Z.

DR L DR Lo GHA Dec Hoa. 2319.6'S 8714.2'W 4356.1' 1317.2'N 3404.3' b. 1419.5'N 15249.8'E 26553.6' 1426.8'S 2459.7' c. 2836.4'N 7050.4'W 11008.5' 1558.5'N 5148.3' d. 956.5'S 8918.5'E 23244.8' 1240.5'S 5228.5'Solution a.Solution b.Solution c.Solution d.

Solution: Problem 14 a.LHA calculation:GHA 4356.1corr +360GHA 40356.1Lo(W) 8714.2LHA 31641.9Conversion of dm (deg, min) to decimal degreesLHA 31641.9 = 316.69833Lat 2319.6S = 23.32667Dec 1317.2N = -13.28667Sin Hc = (cos 316.69833 cos 23.32667 cos 13.28667) + (sin 23.32667 sin -13.28667)Hc = 34.01267Hc = 3400.8Cos Z = [sin 13.28667 (sin 23.32667 sin 34.01267)] (cos 23.32667 cos 34.01267)Z = S 126.36580 EZ = S 126.4 EZn = 180 ZZn = 53.6 rounded to 054Hc = 3400.8Ho = 3404.3a = 3.5 = 3.5nm T

Solution: Problem 14 b.LHA calculationGHA 26553.6Lo(E) +15249.8LHA 41843.4corr 360LHA 5843.4Conversion of dm to decimal degreesLHA 5843.4 = 58.72333Lat 1419.5N = 14.32500Dec 1426.8S = 14.44667Sin Hc = (cos 58.72333 cos 14.32500 cos 14.44667)+ (sin 14.32500 sin 14.44667)Hc = 25.17579Hc = 2510.5Cos Z = [sin 14.44667 (sin 14.32500 sin 25.17579) (cos 14.32500 cos 25.17579)Z = N 113. 86251 WZ = N 113.9 WZn = 360 Z = 360 113.9Zn = 246.1 rounded to 246Hc = 2510.5Ho = 2459.7a = 10.8 = 10.8 nm A

Solution: Problem 14 c.LHA calculation:GHA 11008.5Lo(W) 7050.4 WLHA 3918.1Conversion of dm to decimal degreesLHA 3918.1 = 39.30167Lat 2836.4N = 28.60667Dec 1558.5N = 15.97500Sin Hc = (cos 39.30167 cos 28.60667 cos 15.97500)+ (sin 28.60667 sin 15.97500)Hc = 51.71109Hc = 5142.7Cos Z = [sin 15.97500 (sin 28.60667 sin 51.71109)] (cos 28.60667 cos 51.71109)Z = N 100.65589 W = N 100.7 WZn = 360 Z = 360 100.7Zn = 259.3 rounded to 259Hc = 5142.7Ho = 5148.3a = 5.6 = 5.6 nm T

Solution: Problem 14 d.LHA calculation:GHA 23244.8Lo(E) +8918.5 ELHA 32203.3Conversion of dm to decimal degreesLHA 32203.3 = 322.05500Lat 956.5S = 9.94167Dec 1240.5S = 12.67500Sin Hc = (cos 322.05500 cos 9.94167 cos 12.67500)+ (sin 9.94167 sin -12.67500)Sin Hc = 0.79571Hc = 52.72259Hc = 52 43.4Cos Z = [sin -12.67500 (sin 9.94167 sin 52.72259)] (cos 9.94167 cos 52.72259)Cos Z = 0.13753Z = S 82.09537E = S 82.1EZn = 180 Z = 180 82.1Zn = 97.9 rounded to 098Hc = 52 43.4Ho = 52 28.5a = 14.9 = 14.9 nm A

15. Refer to Chapter 6, Homework #6a & #6b. Complete the bottom portions of the USPS SR96 Form you startedin those exercises to find the intercept (a) and azimuth (Zn) for those sights by the Law of Cosines method.For reference, the DR position given in those exercises and the answers you calculated for LHA, Dec, and Hoare provided below:

DR L DR Lo LHA Dec Hoa. 3006.8'N 8543.6'W 5112.8' 1050.8'N 3846.3'

b. 4118.0'N 7306.8'W 32331.7' 700.8'S 3118.2'Click to View Solution 15 a.Summary:

a. Hc = 3848.0' a = 1.7nm A Zn = 259

b. Hc = 3116.0' a = 2.2nm T Zn = 136

Click to View Solution 15 b.Ref.: 48

Figure IM07 15aSolution for Problem 15 a.Given the data below find the intercept (a) and the azimuth (Zn) using the Law of Cosines Method.

DR L DR Lo LHA Dec Ho3006.8'N 8543.6'W 5112.8' 1050.8'N 3846.3'

Solution for Problem 15 b.Given the data below find the intercept (a) and the azimuth (Zn) using the Law of Cosines Method.

DR L DR Lo LHA Dec Ho4118.0'N 7306.8'W 32331.7' 700.8'S 3118.2'a. Figure IM07 15b

Q7End Of Homework Q & AJunior NavigationChapter 7The Celestial LOP