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## 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
Date: 2013 Feb 16, 02:11 -0500

Confirming the result another way, we have:

Let R = radius of earth
Let angle = 0.103957108 degrees

Then height above surface is = R/cos(angle) - R
= 251103108.1953 / cos(0.103957108) - 251103108.1953
= 413.3194096555 inches

Circumference of earth is 2*Pi*R
= 1577727359.99983 inches

Arc Length = (360 - 2*angle)/360 * circumference of earth
= 1576816160.14673 inches

Length of string is then = 2*sin(angle)*R/cos(angle) + arc length.
= 2*tan(angle)*R + arc length
= 1577727360.99976

Subtract circumference of earth from length of string to find delta.
Delta is ~1" to an error of less than 1/10,000".  It appears that my angle is off by a tiny bit.

On Feb 15, 2013 8:52 PM, "David Fleming" <d.l.fleming.1@gmail.com> wrote:

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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