[Gllug] Public IPs - When are they appropriate

[David Damerell]

On Thursday, 15 Nov 2001, Stephen Harker wrote:
>I'm afraid the Earth is 12756 km in diameter, not 10000 km, and it is
>not a sphere, but an equatorially oblate spheroid (surface topography 
>notwithstanding).

's why I said something like "assume it's a sphere 10000km in
diameter, which it's not". We don't really care about a factor of two
or so. However, if people get pedantic, we should probably point out
their silly errors;

>The surface area of spheroids is
>4 * pi * cuberoot(r1 ^ 2) * cuberoot(r2 ^ 2) * cuberoot(r3 ^ 2)
>where r1, r2 and r3 are the perpendicular axial radii (x, y, z axes).

This looks plausible enough; in particular, if the object is a sphere
(and so r1 = r2 = r3), it's 4 * pi * r^2, which is right.

>In an equatorially oblate spheroid, r1 = r2, so you can simplify to
>4 * pi * cuberoot(r1 ^ 4) * cuberoot(r2 ^ 2)
>where r1 is the equatorial diameter and r2 is the polar diameter.

Unfortunately, the diameter is not the radius, giving us an answer
four times as large as it needs to be!

>Given a maximum polar oblateness of +/- 35 km, we can say the SA is
>roughly

... such a tiny difference as to suggest it's insignificant.

>4 * pi * cuberoot(12756000 ^ 4) * cuberoot(12721000 ^ 2) square metres
>= 2.04 * 10^15 square metres

... which we now know is 4 times too large; the right answer is 5.1 x
10^14. Curiously, the surface area of a sphere of that diameter is
also 5.1 x 10^14. My 'rough and ready' figure was 3.14 x 10^14, which
is out by less than two.

So; as anyone could have told you, the whole 'oblate spheroid' thing
was a pointless exercise in pedantry which made no difference
whatsoever compared to assuming the Earth is a sphere; and the answer
thus obtained was actually out by a larger error than mine based on a
blatant guess, owing to an elementary mistake.

-- 
David Damerell <damerell at chiark.greenend.org.uk> flcl?

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