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Hewitt,

I'm afraid that the Martelli formula as it pertains to the individual tables I through V is not obvious :(

I tried several combinations of logs but could not generate the numbers that appear in table I.

Greg Rudzinski

[NavList] Re: Martelli's Time-Sight Tables
From: Hewitt Schlereth
Date: 23 Dec 2012 20:15
Greg -

Hey, great going. Never in my life did I think of trying a moon time sight. Exotic, man.

This afternoon I looked in Vol. i of my green Bowditch and it gives Martelli' formula as:

havt = cos(L~d) – cosz / 2cosL cosd

Since havt = (1– cost ) and z = (90* – h) I rewrote the formula to:

cost = 1 – (cos(L~d) – sinh / 2cosL cosd)

With the formula now in sines and cosines, I plugged Martelli's example into a
basic scientific calculator, but got an answer that was way off.

So, I went through and wrote down the results from each operation. At which
point I noticed the final number was half the cosine of the angle I was
looking for. So, I took the 2 out of the divisor and did it again. The result
agreed with Martelli to 0.1'.

So, I think the formula is cost = 1 – (cos(L~d) – sinh / cosL cosd)

Greg, when you can spare the time, could you check me out on this? It's been a
long time since I messed around with equations.

Thanks,

Hewitt

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