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> For what it's worth, I've got heights for the Sears Tower and the Hancock  in
> Chicago. Both buildings have a main roof level and a secondary "mechanical
> penthouse" roof level above it creating a sort of head-and-shoulders look. For
> the Sears, I find that the main roof level is 1450 feet above the level of
> Lake  Michigan. For the Hancock, the main roof level is 1122 feet above the
> lake. I  estimate observational errors of +/-2 feet. For both buildings, I
> estimate that  the higher, smaller roof is 20 feet above the main roof.

Frank

Thank you

Do you think you were observing the main roofline or penthouse? (20 ft =
approx 0.5' at  23.08 nautical miles.)

I had reworked a post sent earlier, but it was based on the assumption of
Sears at 1454 plus 30 base above water level.  At that distance, I figure 4
ft 1 inch = .1', so the results were relatively meaningless without height
above water level.  Below were the last set of figures.

> Given: Sextant angle 30.8 minutes, height of eye 15 ft, Sears 1454 feet, GPS
> distance 23.08 nautical miles.
> Assumed:  Base of Sears 30ft above water level.
>
> T15 Bowditch formula would indicate observed angle should be 0d 26.4'
>
> Using my modified constants of 0.0002001 and 0.6079 in the T15 equation, the
> observed angle should be  0d 28.1'

Frank

Blew the buffer and forgot to account for dip in the following;
>
> 30.8' is 4.4' over stock Bowditch
> 30.8' is 2.7' over Bowditch with proposed values for constants

Accounting for dip (30.8'- 3.8'= 27.0'):

27.0' is 0.6' over 26.4' stock Bowditch
27.0  is 0.9' under using 27.9' calculated with .15' lift per mile
27.0' is 1.1' under 28.1 using my proposed values (.002001 and .6079)

With new values for height, I will have to recalculate.

Bill

   

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