On Fri, 11 Feb 2000, John Carmichael wrote:
> Since the calculation of arcseconds is a bit complicated, what would help
> would be some sort of precalculated table which would give the maximum
> readability distance for different shadow widths. For example, the table
> might say that a 1/4 inch wide shadow is easily visible from 100 ft. or an
> 1/8 inch cable is visible from 50 ft. and so on. (These numbers are not
> correct as this is just an example)
>
> I wish one of the dialist/optomologists who answered Ross's question could
> come up with such a table, with shadow widths of 1/16 in to 1 inch in 1/16th
> inch increments. Then we wouldn't have to do the calculations. This would
> be very useful to us.
Not too difficult... Here are two versions:
First, the required feature width (1 min) and character height (5 min) in
inches, for 20/20 vision at a few distances. Obviously, scale up (or down)
as needed (nearly exactly linear in this small angle region; for 100 ft,
use 10 times the width for 10 ft, etc.):
1 min 5 min
10 ft 0.035 0.175
20 ft 0.070 0.349
50 ft 0.175 0.873
Second, your suggestion of inches by 1/16", in arc-minutes:
10 ft 20 ft 50 ft
1/16 1.8 0.9 0.4
2/16 3.6 1.8 0.7
3/16 5.4 2.7 1.1
4/16 7.2 3.6 1.4
5/16 9.0 4.5 1.8
6/16 10.7 5.4 2.1
7/16 12.5 6.3 2.5
8/16 14.3 7.2 2.9
9/16 16.1 8.1 3.2
10/16 17.9 9.0 3.6
11/16 19.7 9.8 3.9
12/16 21.5 10.7 4.3
13/16 23.3 11.6 4.7
14/16 25.1 12.5 5.0
15/16 26.9 13.4 5.4
16/16 28.6 14.3 5.7
Dave
