Polar Grid Navigation
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Joined: Mar 2004
Posts: 34
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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°
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°
Guest
Posts: n/a
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?
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?
Guest
Posts: n/a
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!!
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!!





