|
|
|
|
|||||
|
||||||
From: Bill (no email)
Date: Tue Jun 06 2006 - 20:56:40 EDT
> 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
|