1 Homework Q & A Junior Navigation Chapter 1 Lighthouses in the Sky

2 Objectives: ■ Define the terms altitude, circle of position, geographical position, intercept, celestial line of position. ■ Given altitude, determine the radius of a circle of position and vice versa. ■ State why accurate time is important in celestial navigation. ■ Describe, in general terms, the altitude- intercept method of plotting a celestial line of position.

3 1. In celestial navigation, altitude is the: a. angle between your line of sight to a celestial body and your line of sight to the horizon. b. distance between your position and the geographical position of a celestial body. c. latitude of the observer. d. angle between true north and the GP of a celestial body, measured from your DR position. REF: ¶ 11

4 2. If you were at latitude 40°N and sailed due west from Europe to America, as you proceed you would see Polaris: a. sink gradually lower in the sky. b. appear gradually higher in the sky. c. remain at essentially the same position in the sky. d. remain at a constant altitude of 50°. REF: ¶ 11

5 3. The center of a celestial circle of position is always at the: a. observer's position. b. North Pole. c. center of the earth. d. GP of the celestial body. REF: ¶ 12

6 4. The geographic position of a celestial body is: a. your dead-reckoning position. b. the same as an estimated position. c. a position on the surface of the earth directly below the celestial body. d. the position of a celestial body measured as an angle above the visible horizon. REF: ¶ 12

7 5. Observer A measures the altitude of the sun bearing 110° True as 38°26.8'. Observer B measures the altitude of the sun at the same instant as 38°16.8', and the sun's bearing is about the same. Which observer is closer to the GP of the sun, Observer A or B? Solution: Observer A arc = 90° - 38°26.8' = 53°33.2' distance = 53 x = nm Observer B arc = 90° - 38°16.8' = 53°43.2' distance = 53 x = nm nm is less than nm therefore Observer A is closer REF: ¶

8 6. For the following values of Ho, determine the radius of the resulting circle of position in units of arc and in nautical miles. ArcDistance a. 46°16.2' b. 75°56.7' c. 15°30.9' Solution: a. arc = 90° -46° 16.2' = 43° 43.8' Distance = 43 x = nm b. arc = 90° -75° 56.7' = 14° 03.3' Distance = 14 x = 843.3nm c. arc = 90° -15° 30.9 ' = 74° 29.1' Distance = 74 x = nm REF: ¶ ° 43.8'2623.8nm 14° 03.3' 843.3nm 74° 29.1'4469.1nm

9 7. For the following distances or arcs from the sun's GP, determine the altitude (Ho) an observer would find. Ho a nm b. 186 nm c. 65°39.3' d. 15°58.8' Solution: a. arc = 2568 nm / 60 = 42.8° = 42°48.0´ Ho = 90° – 42°48.0´ = 47°12.0´ b. arc = 186 nm / 60 = 3.1° = 3°06.0´ Ho = 90° – 3°06.0´ = 86°54.0´ c. arc = 65°39.3´ Ho = 90° – 65°39.3´ = 24°20.7´ d. arc = 15°58.8´ Ho = 90° – 15°58.8´ = 74°01.2´ REF: ¶ ° 12.0' 86° 54.0' 24° 20.7' 74° 01.2'

10 8. Two observers on opposite sides of the same circle of position make observations of the sun. They will: a. measure different altitudes of the sun that add up to 90°. b. measure the same altitude of the sun. c. observe the same true bearing of the sun. d. measure bearings of the sun that differ by 90°. REF: ¶ 13

11 9. Knowing the time of a sight to the nearest second is important because: a. the altitude of the body will change by about 15° per hour. b. the latitude of the geographical position of the body will change by about 0.25' per second. c. the longitude of the geographical position of the body will change by about 15° per hour. d. you have to take the motion of your boat into account. REF: ¶ 21

An observer is at L 36°N, Lo 158°W. The sun's GP is at L 11°S, Lo 158°W. How far is the observer from the GP of the sun? REF: ¶ nm North of Sun's GP What is the true bearing of the sun from the observer? Sun bears 180° True Solution The GP of the body and our Lo are the same, so the azimuth is either 000° or 180°. Since we are in north latitude and the GP is in south latitude, the azimuth is 180°. We are located at 36° North and the GP is 11° South. The arc between our position and the GP is 36° + 11° = 47° and the distance is 47° × 60 = 2820 nm.

An observer is at L 36°N, Lo 158°W. The sun's GP is at L 11°S, Lo 158°W. How far is the observer from the GP of the sun? Solution: ∆° = L 36°N – (L 11°S) = 47° 47° x 60nm/° = 2820nm North of Sun's GP What is the true bearing of the sun from the observer? Sun bears 180° True REF: ¶

The difference between Hc and Ho is: a. called the intercept. b. the radius of a circle of position. c. the distance between the DR position and the GP of a body. d. the same as 90° minus the altitude of a body. REF: ¶ 37

A celestial line of position is a: a. straight line from the center of a celestial body to the center of the earth. b. line from the GP of a celestial body to your position. c. straight-line approximation of a portion of a circle of position. d. line from the DR position to the GP of a celestial body. REF: ¶ 39

16Q7 End Of Home Work Q & A Junior Navigation Chapter 1 Lighthouses in the Sky