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Old 30th October 2004 | 13:31
  #1 (permalink)  
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Joined: Mar 2004
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From: Miami
Polar Grid Navigation

Hy,

does anybody know how to calculate this question?

At 60°N 030°W, the true heading is 090° and the Gyro is 000°. At 62°N 010°W, the True heading is 095° and it has taken 1.5 hours to travel the distance. What will the Gyro read assuming no Latitude Nut correction?

003°
345°
328°( this is the right answer)
334°
subsidence is offline  
Old 30th October 2004 | 16:42
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High Wing Drifter
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The question involves working out the transport and earth wander.

TW/hr = E/W GS * Tan(lat) / 60
EW/hr = 15 * Sin(lat)
E/W GS = change of long * Cos(lat) / duration.

Mean lat is 61
GS = (30-10) * 60 * Cos(61) / 1.5 = 388
TW = 388 * Tan(61) / 60 = 11.7
EW = 15 * Sin(61) = 13.1

Total Wander for 1.5hrs = 1.5(11.7+13.1) = 37.2

Travelling E in N hem so 360-37.2 = 322.8

So I don't get quite the same number. But as both involve a 3,2 and an 8 I wonder if there is a typo?
 
Old 31st October 2004 | 09:08
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Joined: Feb 2002
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From: Sunny Solihull
Grid/Gryo

Answer 328 is correct (no typo error), HWD calculations are correct, but you also need to ADD back in the 5 degree actual heading change from 090 to 095.

323 + 5 = 328.

Bit of a bar steward.
RichardH is offline  
Old 31st October 2004 | 09:10
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High Wing Drifter
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Well spotted
 
Old 1st November 2004 | 10:08
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Justin Abeaver
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I was always taught by the Late Great Sandy Thomson that the easiest way to do these is to draw it all out to scale, and then just measure off the angle with a protractor.

Can't remember the ins and outs cos was TOOOOOO long ago!!

But try it and see if it still works. It got me through!!
 
Old 1st November 2004 | 12:52
  #6 (permalink)  
High Wing Drifter
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The question wasn't actually a Polar grid problem. It was a gyro question.
 

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