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HomeworkQ & A

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Chapter 7Chapter 7

The Celestial LOPThe Celestial LOP

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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)

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

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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

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4. In completing a sight reduction of the Sun, your results indicate Ho = 35°17.4' and Hc = 35°36.2'.

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 = 35°36.2'.Ho = 35°17.4'a = 18.8' = 18.8nm A

Ref.: ¶ 14 - 18

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5. In completing a sight reduction of the Sun, your results indicate Ho = 43°45.3' and Hc = 43°38.8'.

a.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 = 43°38.8'Ho = 43°45.3'a = 6.5' = 6.5nm T

Ref.: ¶ 14 - 18

Ans: closer to GP

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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

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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

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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

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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

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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

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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

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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

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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

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Summary:

LHA Hc Intercept Zn

a. 316°41.9' 34°00.8' 3.5nm T 054°

b. 58°43.4' 25°10.5' 10.8nm A 246°

c. 39°18.1' 51°42.7' 5.6nm T 259°

d. 322°03.3' 52°43.4' 14.9nm A 098°

Ref.: ¶ 48

14. Using the values given below, obtain the intercept (a) and azimuth (Zn) by calculator solution.

Click ButtonTo View

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.

DR L DR Lo GHA Dec Hoa. 23°19.6'S 87°14.2'W 43°56.1' 13°17.2'N 34°04.3'b. 14°19.5'N 152°49.8'E 265°53.6' 14°26.8'S 24°59.7'c. 28°36.4'N 70°50.4'W 110°08.5' 15°58.5'N 51°48.3'd. 9°56.5'S 89°18.5'E 232°44.8' 12°40.5'S 52°28.5'

Solution a.

Solution b.

Solution c.

Solution d.

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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 Hoa. 30°06.8'N 85°43.6'W 51°12.8' 10°50.8'N 38°46.3'

b. 41°18.0'N 73°06.8'W 323°31.7' 7°00.8'S 31°18.2'

Click to View Solution 15 a.

Summary:

a. Hc = 38°48.0' a = 1.7nm A Zn = 259°

b. Hc = 31°16.0' a = 2.2nm T Zn = 136°

Click to View Solution 15 b.

Ref.: ¶ 48

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Chapter 7Chapter 7

The Celestial LOPThe Celestial LOP