Sun Shot Reduction

[Tom Wilson]

Try WolfPack, under computations there is a solar reduction program. C. Ghilani – good stuff, and free.

T.W.

http://www.personal.psu.edu/cdg3/free.htm

[Retired PLS]

What azimuth did you come up with?

[John Harmon]

What state and what SPC zone

[Norman_Oklahoma]

> What state and what SPC zone
Lat/Long puts it in western Kentucky.

[rochs01]

I could try it in my HP41CX. I know that’s right 😉

[charles-l-dowdell]

I couldn’t get my HP 75C program to load into the 75C computer memory, and my old 86B uses the older input data, Equ. of time, etc., so that one would have taken a lot of conversions. Anyway, I used Spade that I have on my Windows computers and came up with a line bearing of N19°42’15″W, but I’m not that familiar with that program, so I may not even be in the ballpark. Spade doesn’t give the options to enter the GHA, Decl, Semi-Diameter, so I assumed that it must all be taken care of in the program. My 75C program takes all the input data you provided, which is the reduction program I am most familiar with.

[charles-l-dowdell]

SPC zone does not even come into play for determining the azimuth or bearing of the line observed.

[Harold]

I used my trusty HP48gx wit a TDS survey gx card in it.
Results:
Ephemeris method
Shooting left trailing edge
North latitude
West longitude
uT1 truncated at two decimal places in the seconds numbers
Back sight circle 0d 00’00”

AZ = 340d 17′ 48.6″

Hope this helps and is useful. I started using GPS only last year, so the hour angle method of direction determination is still fairly fresh on my mind.

[charles-l-dowdell]

Harold, we were within 0°00’04” with our calc’s.

[Harold]

:good:

[Retired PLS]

My HP48 program has the same results as yours.
340° 17′ 48.6″

[ashton]

I used the Naval Observatory’s MICA program and came up with an azimuth to the center of the sun of 189 39 22.82. After applying the semidiameter I obtained 189 23 18.31. Applying the angle between the line and the sun yielded 340 23 8.31.

The only parameter requested by the MICA program but not stated in by the original poster was the elevation; I used 0.

[Jon Payne]

Thanks for the help!

I was hoping someone had the old HP programed for the solution.

I was off in the minutes range. I was making a bone headed mistake and using the wrong math function + or – depending on leading or trailing edge being sighted.

I am getting agreement with the several solutions you guys provided now.

[ashton]

Just so no one suspects the MICA program, when I repaired my correction for semidiameter, I got 340°17’50”, in agreement with the others.