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> I'll forward your "problem" to where I work, where I can look at the chart and
> see where they came up with their answer.  Sorry I can't help sooner.  I'll
> try to have an answer to you around 00h00 (UTC) 8 Jun.

Thanks Pete

I initially worked it without a chart using rectangular to polar conversion:

dLat 4.9
dlon 8.3
Mean Lat 41d 13' 27"
Conversion factor, lon to nm
= mean lat cosine = .752137015
.752137015 * 8.3' lon = 6.242737222 nm

After R to P conversion:
Distance = 7.936105344 nm
True = 051d 52' 16.3"

C (psc)  056
D        +04 E
M        060
V        -15W
T        045

One angle of the oblique triangle
= 051d 52' 16.3" - 045d = 006d 52' 16.3"

One adjacent leg = 7.936105344 nm
The other = time * speed = 52 min * 9.2 = 7.9733333 nm

Using the law of cosines the drift leg = .954121933 nm
Using the law of sines to derive the other angles and doing a bit of
geometry, I come up with set of 136d 00.4'

Plotting it graphically on the chart, on a plotting sheet, and in a computer
drawing program, my results agree within +/- .05 nm and +/- 1d, so I am at a
loss.

Bill

   

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