Re: ATM Formulas for Apparent Angular Size

["Bell, Andrew" <AMB2@pge.com>]

Joe -- here are a couple [more] rules of thumb that might be useful for you.
I see that a number of others have already chimed in as well:

	-- one arc-second in the sky is roughly the same as one inch at a
distance of three miles [a little more precisely, 3.25 miles] [the metric
equivalent of this is one centimeter at a distance of two kilometers]

This is based on a scale factor of roughly 200,000 to 1.  There are
approximately 57.3 degrees to the unit radian (360 degrees divided by two
Pi), and so pretty close to 206,000 arc-seconds (multiply by 3600
arc-seconds per degree); that's how to work back to this scale factor to
use.  Somebody else suggested one arc-minute as one inch at 100 yards;
that's very nearly the same factor, but on an arc-minute basis.  

To use this with your example, if Jupiter's true apparent angular size is 30
arc-seconds, you could bring that one-inch disk in by a factor of 30, so
thirty arc-seconds is roughly the same thing as one-inch at a distance of
about 500 feet (i.e., one-tenth of a mile), or your two-inch image at a
distance of 1000 feet (more precisely, using 3.25 miles and then 5280 feet
to the mile, this is very nearly the same 1150 feet that Anthony Stillman
described in another message).

Then you can adjust the 1000 foot figure as appropriate for the
magnification you are using.  For example, at 200x magnification, the image
of Jupiter in your eyepiece would be about the same apparent size as that
two-inch image if viewed from a distance of five feet (more precisely:
1150/200x = 5.75 feet).

I like using this rule of thumb for objects outside the solar system, too.
For example, splitting the components of Epsilon Lyra (two arc-seconds) is
like looking at two points of light separated by two inches at three miles;
even better, it's like splitting the headlights on a truck (about ten feet)
at a distance of nearly 200 miles!

-- Andrew Bell

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	Date: Mon, 22 Oct 2001 19:37:33 -0500
	From: Joe Mayenschein <mayen1@mwt.net>
	Subject: ATM Formula

	Hi All!
	This is a little off topic but I have used up all the rest of my
	contacts to no avail, Can you guys help me out here?

	I'm terrible at math, and haven't found a java page with the formula
	as of yet. But I'm looking for a formula to convert size to
distance.
	Here's what I mean,

	say you have an image of say Jupiter, and it's 2 inches in diameter.

	Now you look up information and see tonight that Jupiter is say 30
	seconds of arc true in the sky.

	How far away would you need to place this image of jupiter so it
equals
	30 seconds of arc in size?  Then it would appear in size exactly the
same
	size in the telescope, see what I mean?

	Joe