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Re: Aldebaran occultation
From: Brad Morris
Date: 2017 Mar 7, 13:07 -0500

Frank

I went to Table XXVI Logarithms of Numbers (Bowditch 1849) and using 1740 as the independent variable, got

24055 as the dependent variable

6.464 + .24055 =

6.70455 (!!)

Arcsin (10^(6.70455-10))= 1.7411

For an error term less than 1" (!!)

Neat trick!

That answers part two of my question to Lars. I am much more interested in part one, which is the general form of his occultation reduction.

Brad

On Mar 7, 2017 12:40 PM, "Frank Reed" <NoReply_FrankReed@navlist.net> wrote:

"You mentioned taking the sine of very small numbers which the logarithmic tables didn't cover. I note the log (sin (1'.74)) = 6.704. I worked that in reverse asin (10^(6.704-10)) by calculator to verify. Did you obtain this intermediate result by calculator or by a method available 100 years ago?"
This is actually easy, and the "method" has been known for centuries. The sine or tangent of a small angle is equal to the angle. Of course that angle has to be converted from minutes of arc to a pure number (also known as "radians") by dividing it by 3438 (or 60·180/pi for full accuracy). This small angle equivalence is accurate to about 1 part in 10,000 for angles up to 5°, and it's accurate to better than one part in a million for angles smaller than 0.5°. If you want to jump straight to the logarithm (+10) of the function, it's easy to show that you can calculate it from
logsin(A) = log(A) + 6.464
when A is the angle in minutes of arc.
Just a reminder: you can calculate the angular size of any object in minutes of arc from
angle(m.o.a.) = 3438 · (size across the line of sight) / (distance out, along the line of sight).
Frank Reed

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