compass deviation by sun

[duckmanton]

Hi members, could someone please explain the formula for compass deviation via the sun.
There was an example in PBO391, page 76. Sun bearing 265 compass, Lat 39, dec 10. Variation 6W
sin ampllitude=sin 10 divide by cos 39
amplitude=W13degS
bearing=270 - 13deg=257 T
Variation=6deg W
Bearing=263 deg mag
bearing=265 comp
deviation=2 deg W.

My prodlem is that Sin 10 deg = (0.1736) divided by cos 39 (0.777) =0.2234
how do you get W13deg S

And if someone can tell me how to do degrees and divide with normal symbols would save my two typing fingers.
Thanks for any help in advance

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[Evadne]

If sin(amplitude)=0.2234, I make amplitude = arcsin(0.2234)=12.9 degrees. (Some calculators label arcsin as sin to the power of -1).


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[duckmanton]

thanks Dave, l was going to ask regards to W13deg S, but now rereading the example it states E or W according to whether Sun rising or setting,N or S according to the declination.
Thanks for imparting your knowleadge. On Sunday l am taking off from Esperance in Western Australia bound for Adelaide. Now l can run a check on the compass.
Good weather too, just had 65 knt thru there and more to come.
I think l will be able to do a post on Para anchors and Sea brake drouges, hopefully not on a Sea Saver raft that l have had to buy today as they have just condemed my 13 year old RFD

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[trev]

You need a set of Nautical Tables (Nories,Burtons etc) and look up the LOG sine and cosine of the angles, add them together (ie multiply) then antilog (sine) to get your amplitude which will need to be applied in the correct quadrant to get the true bearing.
I gather you realise that the 'amplitude' method is only good for a rising or setting body.

<hr width=100% size=1>Trev

 

[Evadne]

Have a good trip, I'll look out for your reports.
Dave

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[Juggler7823]

I think your answer is in radians. 1 radians is approximately 57 degrees. So 57*.223=12.73 degrees

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