# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Distance off with Chicago buildings**

**From:**Bill B

**Date:**2005 Nov 14, 15:48 -0500

Frank Been playing with your observations over the past month and keep scratching my head. Using the Bowditch formula: d = square root ((tanA/.0002419)^2 + (H-h/.7349)) - (tanA/.0002419) where d is distance in nautical miles, A is sextant angle after IC, H is height of object in feet, and h is height of eye in feet. (If I recall you have some concerns about the refraction correction values uses in the formulas.) I used height of eye as 11.167 feet, and building heights +30 feet to account for the elevation of their bases above water level. Converting to statute miles: Sears: 24.3 sm John Hancock: 24.2 sm Using the separation angle of 3d 07' and a distance between Sear and Hancock of 1.5 sm, I did a sanity check with the Law of Sines, and came up short. Can you help me locate the spot you used on the beach? Was it between Indiana Harbor and the Gary works (chart distance of 20.7 sm to Sears and 21.4 sm to Hancock); or east of Gary, near the old Bethlehem/Burns Harbor Plant (approx. 24 sm)? Thanks Bill > 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.