Navigation Questions. Can you do them? - SailNet Community
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post #1 of 122 Old 02-26-2010 Thread Starter
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Navigation Questions. Can you do them?

1. As we ask these questions, lets see if we can stay away from all charts but the universal chart.
2. These are for your edification, so give a try and you may suceed.

A. Convert the following true bearings to relative bearings of a ship who is steering 302°; 009°, 158°, 328°.










Hint: R - S = T

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post #2 of 122 Old 02-26-2010
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Quote:
Originally Posted by Boasun View Post
1. As we ask these questions, lets see if we can stay away from all charts but the universal chart.
2. These are for your edification, so give a try and you may suceed.

A. Convert the following true bearings to relative bearings of a ship who is steering 302°; 009°, 158°, 328°.

Hint: R - S = T
Your formula works out to:

R = T + S

However, the army.mil site says the equation is:

RB + SH = TB or

RB = TB - SH

RB = Relative Bearing
SH = Ship's Heading
TB = True Bearing

Why the discrepancy? Am I missing something?

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post #3 of 122 Old 02-26-2010
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I don't get the question in the OP. I just head for the buoys, what could go wrong?
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post #4 of 122 Old 02-26-2010
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Ships Steering 302 True
True Bearing 009 - Relative Bearing 067
True Bearing 158 - Relative Bearing 216
True Bearing 328 - Relative Bearing 026
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post #5 of 122 Old 02-26-2010 Thread Starter
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Quote:
Originally Posted by jaschrumpf View Post
Your formula works out to:

R = T + S

However, the army.mil site says the equation is:

RB + SH = TB or

RB = TB - SH

RB = Relative Bearing
SH = Ship's Heading
TB = True Bearing

Why the discrepancy? Am I missing something?
The discrepancy is from not checking my books first before I wrote the question.
R + S = T or as you said: RB + SH = TB

Must have been a mental lapse from deminta or something...
But I did get people to start thinking...
And when I am teaching I do put errors on the board to see if my students are awake and have studied. But it wasn't suppose to happen this time.

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post #6 of 122 Old 02-26-2010
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009°T is 67°R (forward the starboard beam)
158°T is 216°R (port quarter)
328°T is 26°R (off the starboard bow)

The army.mil site is correct. What you want to do with the given information is: TB - SH = RB. If the result is negative, you can either consider it an angle towards port (clockwise) or you can add 360° to normalize it to a standard 0°-360° angle.

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post #7 of 122 Old 02-27-2010 Thread Starter
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The Difference in latitude, between two points at 39°W Longitude, is 3°13'; What is the distance in Nautical Miles?

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post #8 of 122 Old 02-27-2010
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I'm going to say 3.21NM.
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post #9 of 122 Old 02-27-2010
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193 nm
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post #10 of 122 Old 02-27-2010
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I agree with jja above. 193NM

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