NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Feb 7, 18:06 -0500
Hi Norm
Circumference of a circle = Pi * D, where D is the diameter
Circumference of the Earth = 24901 miles
Inches per mile = 63360
D = 24901 * 63360 / Pi
= 502,206,216.390563"
Add an inch to the circumference
D = (24901 * 63360 + 1) / Pi
= 502,206,216.708962"
Delta D = .318309"
In order to get symmetric standoff, we split this by two
Symmetric Standoff = Delta D/2
= 0.1591545"
Are you sure I haven't made a typographical error? You had better go over this carefully before you submit this to your teacher as your own work ;-)
+++++++++
A word about your cylindrical slide rule. The moment you actually attempt repair is the moment you destroy any remaining cash value a non-functional slide rule holds.
If you are determined to truly 'fix' this, you would have to completely remove the existing scale without destroying it. Next, clean the tube upon which the scale is mounted, such that the scale mounts on the tube without adhesive AND the ends of the scale meet. The using rubber cement, apply a very thin coat to both surfaces and wait for the adhesive to dry. Apply the scale to the tube without helix errors.
Certainly, do NOT stretch the scale. The ends may then meet, yet the logarithmic scale will then be distorted and will be incapable of mathematically correct results.
A real solution would be to sell your Otis King rule and buy one that is functional from the start. Far easier and certainly less error content than a scale that has helical errors in application.
At a 66 inch scale length, you will achieve about one more decimal digit in the result than a 6" slide rule. I leave it to the cognoscenti to tell you if this is enough for celestial navigation, as opposed to a nominal 10" slide rule
Best
Brad
A belated 'Marvelous!, I just picked up an Otis-King cylindrical slide rule. It's very elegant, as yours is, One problem. The scale has detached from the cylinder and when pressed back on, does not meet itself, by about .06 in. Of course, this renders it useless as a calculator. I haven't read the entire thread of your post, but could you tell me what kind of glue you used and any suggestion for stretching the scale (humidity has been suggested. I haven't tried it yet)
My physics teacher gave me this problem: Wrap a string around the Earth, and pull it snug. Now, add an inch to the length of the string. How far off the surface will it stand?(assume it's symmetric)
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