 # NavList:

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

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Re: HR-1 working re-creation and new questions
From: David Fleming
Date: 2013 Feb 15, 17:50 -0800

Let me begin by saying I get numbers that agree with your result.

Let me define:

s = half arc length of earth where string is above ground
a= half string lenth not touching earth
h= heigth of string at max above earth

hopefully you can draw appropriate picture for these variables.

1) delta = 2(a-s) the amount string exceeds sphere circumference is delta

2) theta = s/R 1/2 angle at center of sphere

3) tan(theta) = a/R

4) a^2 + R^2 = (R+h)^2 Pythagorean Theorem

Given above:

expand tan(theta)= theta + (1/3)*theta^3 +... = a/r

ignore other than first two terms on LHS and transpose first term to RHS and substitue 2) for theta on RHS

(1/3)*theta^3 = (1/R)*(a-s) = delta/2R

thus given delta and R we compute theta.

then a = R tan(theta) we compute a

from 4) we have h = sqroot(a^2 + R^2) - R

or h = (1/2)*a^2/R ( factor R from sqroot and expand sq root in power series)

using your value for radius of earth you'll get h = 4.13.36 inches for delta=1".

It is interesting that in first problem which shoud read for string having circular symmetry height does not depend on size of the ball. A string one inch longer than a cue ball or one inch larger than the earth is 1/2Pi above the surface. While for second case which has mirror symmetry about the point you pull on the string the answer depends on the size of the sphere. And the answer is surprisingly large.

Dave F
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