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7 Celestial LOP Homework Q

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Title: 7 Celestial LOP Homework Q


1
The Celestial LOP
Homework Q A
Junior Navigation Chapter 7
2
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)
3
Practical Exercise 1. 2. Follow the Student
Manual for guidance.
4
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
5
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
lt 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
6
  • 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 gt 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 T

Ans closer to GP
Ref. 14 - 18
7
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
8
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
9
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
10
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
11
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
12
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
13
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
14
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
15
14. Using the values given below, obtain the
intercept (a) and azimuth (Zn) by
calculator solution.
DR L DR Lo GHA
Dec Ho a. 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'
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 098
Note 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.
Click Button To View
Solution a.
Solution b.
Solution c.
Solution d.
Ref. 48
16
Solution Problem 14 a. LHA calculation GHA
4356.1 corr 360 GHA 40356.1 Lo(W)
8714.2 LHA 31641.9 Conversion of dm (deg,
min) to decimal degrees LHA 31641.9
316.69833 Lat 2319.6S 23.32667 Dec
1317.2N -13.28667 Sin Hc (cos 316.69833
cos 23.32667 cos 13.28667) (sin 23.32667
sin -13.28667) Hc 34.01267 Hc
3400.8 Cos Z sin 13.28667 (sin 23.32667
sin 34.01267) (cos 23.32667 cos
34.01267) Z S 126.36580 E Z S 126.4 E Zn
180 Z Zn 53.6 rounded to 054 Hc
3400.8 Ho 3404.3 a 3.5 3.5nm T
17
Solution Problem 14 b. LHA calculation GHA
26553.6 Lo(E) 15249.8 LHA 41843.4 corr
360 LHA 5843.4 Conversion of dm to decimal
degrees LHA 5843.4 58.72333 Lat 1419.5N
14.32500 Dec 1426.8S 14.44667 Sin Hc
(cos 58.72333 cos 14.32500 cos
14.44667) (sin 14.32500 sin 14.44667) Hc
25.17579 Hc 2510.5 Cos Z sin 14.44667
(sin 14.32500 sin 25.17579) (cos
14.32500 cos 25.17579) Z N 113. 86251 W Z
N 113.9 W Zn 360 Z 360 113.9 Zn
246.1 rounded to 246 Hc 2510.5 Ho
2459.7 a 10.8 10.8 nm A
18
Solution Problem 14 c. LHA calculation GHA
11008.5 Lo(W) 7050.4 W LHA
3918.1 Conversion of dm to decimal degrees LHA
3918.1 39.30167 Lat 2836.4N
28.60667 Dec 1558.5N 15.97500 Sin Hc (cos
39.30167 cos 28.60667 cos 15.97500) (sin
28.60667 sin 15.97500) Hc 51.71109 Hc
5142.7 Cos Z sin 15.97500 (sin 28.60667
sin 51.71109) (cos 28.60667 cos
51.71109) Z N 100.65589 W N 100.7 W Zn
360 Z 360 100.7 Zn 259.3 rounded to
259 Hc 5142.7 Ho 5148.3 a 5.6 5.6
nm T
19
Solution Problem 14 d. LHA calculation GHA
23244.8 Lo(E) 8918.5 E LHA
32203.3 Conversion of dm to decimal degrees LHA
32203.3 322.05500 Lat 956.5S
9.94167 Dec 1240.5S 12.67500 Sin Hc (cos
322.05500 cos 9.94167 cos 12.67500) (sin
9.94167 sin -12.67500) Sin Hc 0.79571 Hc
52.72259 Hc 52 43.4 Cos Z sin -12.67500
(sin 9.94167 sin 52.72259) (cos
9.94167 cos 52.72259) Cos Z 0.13753 Z S
82.09537E S 82.1E Zn 180 Z 180
82.1 Zn 97.9 rounded to 098 Hc 52
43.4 Ho 52 28.5 a 14.9 14.9 nm A
20
15. Refer to Chapter 6, Homework 6a 6b.
Complete the bottom portions of the USPS SR96
Form you started in 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 Ho are
provided below
DR L DR Lo
LHA Dec Ho a.
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'
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 a.
Click to View Solution 15 b.
Ref. 48
21
Solution 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
Ho 3006.8'N 8543.6'W 5112.8'
1050.8'N 3846.3'
Figure IM07 15a
22
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
Ho 4118.0'N 7306.8'W 32331.7'
700.8'S 3118.2'
a. Figure IM07 15b
23
Q7
The Celestial LOP
End Of Homework Q A
Junior Navigation Chapter 7
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