Hi Dave and others: Let me see if I understand your calculations.
First, knowing that 1 arcminute is the limit of visibility, you calculated that 1 arcminute of width (of a line or point) is equal to a width of .0035 inches at one foot. or .35 inches at 10 ft. and so on. The table which you so kindly provided (the second one with all the 1/16 inch increments) gives line widths in inches for different distances. I think it would be more useful for a sundial designer to have the table in a more useable form that lists the maximum distance of visibily for different widths. (I picked 1/16 inch increments because these are the sizes of hardware like rods, cables and spheres available) Would you mind please checking my math? Line width (inches) & maximum distance of visibility(feet) 1/16" = 17.9 ft. (from 1/16/.0035) 1/8" = 35.7 ft. (from 1/8/.0035) 3/16" = 53.6 ft. etc. 1/4" = 71.1 ft. 5/16" = 89.3 ft. 3/8" = 107.1 ft. 7/16" = 125 ft. 1/2" = 142.9 ft. 9/16" = 160.7 ft. 5/8" = 178.6 ft. 11/16" = 196.4 ft. 3/4" = 214.3 ft. 13/16" = 232 ft. 7/8 " = 250 ft. 1" = 285.7 ft. (example: a line one inch thick is visible from 285 feet away) thanks Dave, John >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 > >
