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    Re: now off topic of navigation
    From: Frank Reed
    Date: 2013 Feb 8, 12:10 -0800

    Brad,

    I got 180 feet for the height and 14.3 nautical miles for the distance to the spot where the string leaves the Earth's surface, but I haven't checked my numbers.

    Although it appears as if this question has drifted far from navigation, the problem is actually very close to a traditional navigation calculation, namely, what is the distance to the geometric horizon for a given height of eye? This imaginary "string" around the Earth, if it has been pulled away at just one point (more of a rubber band now) is equivalent to the sight line from an observer at some altitude h looking out to the horizon (where the "sting" re-contacts the surface of the Earth). We do not normally have a need for the direct distance from the observer's eye to the horizon since it differs from the distance measured along the Earth's surface by only a few inches. But in this case, that's exactly what the puzzle requires. That small difference should be half an inch for the puzzle as stated. If the distance measured out to the geometric horizon from the observer's eye exceeds the distance measured along the Earth's surface from the point directly below by half an inch, then how high is the observer (how high has the string/rubber band been pulled)? And how far away is the geometric horizon (the point of contact of the string) in that case? You need the angle, a, as measured at the Earth's center that satisfies:
    tan(a)-a=q/R
    where q is half an inch and R is the radius of the Earth. If you're working on ye olde "back of the envelope", it helps to use the series expansion for tan(a) which is tan(a)=a+a^3/3+... For the angle, a, I get 14.2 minutes of arc. Then the usual formula for distance to the geometric horizon, d=1.05*sqrt(h in feet)) will give you that height of 180 feet, or you can work it out from plane trig and the series expansion for cos(a) which is cos(a)=1-a^2/2+...

    -FER


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