Next message: Frank Reed: "US Coast Guard Academy celestial classes"
Bill wrote:
"As to distance by height, I have done many trials with Chicago buildings,
and the Michigan City (Indiana) light house, cooling tower and smokestack
with an Astra IIIB. I was amazed at my very poor results, using Bowditch
tables or trig. I won't discount my technique, but results were often 30%
off our DR/GPS position. "
I have a hunch I know what the problem is here. The bases of those buildings
are usually well below the horizon unless you're within just a few miles of
shore. I did some beach sights last week as follows (corrected for IC):
Sears Tower Altitude: 30.8'
Hancock Tower Altitude: 22.1'
Hancock-Sears Separation: 3d 7' (antennas aligned)
Michigan City Cooling Tower: 10.7'
The approximate heights of these in feet are: Sears 1450, Hancock 1127,
Cooling Tower 361. Since my horizon was 4 or 5 miles away, the lower parts of all
of these buildings were below the horizon, something like 8 minutes of arc
for the two Chicago towers were hidden below the horizon. So instead of using
the altitudes directly, use their difference: 8.7 minutes of arc.
Do we need trigonometry or tables now? No. Just memorize one number: 3438.
An angle of 8.7 minutes is a ratio of 8.7/3438 or just about 1/395. That means
that my distance from the two towers (assumed to be the same distance away,
which was roughly true) is 395 times larger than the difference in their
heights in feet. The difference in height is 323 feet so I must be about 127,000
feet or about 24 statute miles away. Notice that if I had done the
calculation for the Sears Tower without realizing that a big piece of it is hidden
beyond the curve of the Earth, I would have calculated the distance at around 31
statute miles, so we're dealing with a substantial difference here. Also note
that there are ways of making this calculation more accurate but they're
probably not worth the trouble.
-FER
42.0N 87.7W, or 41.4N 72.1W.
www.HistoricalAtlas.com/lunars
