Transcript Lighthouse in the Sky Homework Q & A

Lighthouses in the Sky
Homework
Q&A
Junior Navigation
Chapter 1
1
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 altitudeintercept method of plotting a celestial line
of position.
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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
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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
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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
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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
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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' = 51°33.2'
distance = 51 x 60 + 33.2 = 3093.2nm
Observer B
arc = 90° - 38°16.8' = 51°43.2'
distance = 51 x 60 + 43.2 = 3103.2nm
3093.2nm is less than 3103.2nm, therefore Observer A is closer
REF: ¶ 15 - 18
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6. For the following values of Ho, determine
the radius of the resulting circle of position in
units of arc and in nautical miles.
a. 46°16.2'
b. 75°56.7'
c. 15°30.9'
Arc
Distance
43° 43.8' 2623.8nm
14° 03.3'
843.3nm
74° 29.1' 4469.1nm
Solution:
a. arc = 90° -46° 16.2' = 43° 43.8'
Distance = 43 x 60 + 43.8 = 2623.8nm
b. arc = 90° -75° 56.7' = 14° 03.3'
Distance = 14 x 60 +3.3 = 843.3nm
c. arc = 90° -15° 30.9 ' = 74° 29.1'
Distance = 74 x 60 +29.1 = 4469.1nm
REF: ¶ 15 - 18
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7. For the following distances or arcs from the
sun's GP, determine the altitude (Ho) an
observer would find.
Ho
a. 2568 nm
47° 12.0'
b. 186 nm
86° 54.0'
c. 65°39.3'
24° 20.7'
d. 15°58.8'
74° 01.2'
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
REF: ¶ 15
- 18= 15°58.8´
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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
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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
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10. 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?
2820nm 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.
REF: ¶ 22 - 27
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10. 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: ¶ 22 - 27
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11. 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
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12. 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
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Q7
Lighthouses in the Sky
End Of
Home Work Q & A
Junior Navigation
Chapter 1
16