# Series P: Piecemeal

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 1347735.  Sat May 09, 2020 3:41 am In the tedium of lockdown I decided to check the maths on the 'for a piece of string around the equator increase the height above ground by 16cm and find the increase in length of the string' question. So here it is... The metre was originally defined in 1793 as one ten-millionth of the distance from the equator to the North Pole along a great circle, so the Earth's circumference is approximately 40000km; at the equator the distance is closer to 40075km. C= 40075km = 40075000m C = πD = 2πr => D = C/π π ≈ 3.14159265359 D = 40075000m/3.14159265359 = 12756268.6888m C = 2πr = πD => r = D/2 r = 12756268.6888m/2 = 6378134.3444m r + 0.16m = r* 6378134.3444m + 0.16m = 6378134.5044m = r* C* = 2πr* C* = 2 x 3.14159265359 x 6378134.5044m = 40075001.0053m C*-C = 40075001.0053m - 40075000m = 1.0053m Rather more interesting is what happens when you compare that result to the result for a sphere of circumference 1m... C= 1m C = πD = 2πr => D = C/π π ≈ 3.14159265359 D = 1m/3.14159265359 = 0.31830988618m C = 2πr = πD => r = D/2 r = 0.31830988618m/2 = 0.15915494309m r + 0.16m = r* 0.15915494309m + 0.16m = 0.31915494309m = r* C* = 2πr* C* = 2 x 3.14159265359 x 0.31915494309m = 2.0053m C*-C = 2.0053m - 1m = 1.0053m I know this, and that it holds for all the circumferences I've tested to date, but I still don't understand it really except to say For Increase in radius x and Increase in circumference = i C* = 2π(C/2π + x) => C*-C = [2π(C/2π + x)]-C => i = 2πC/2π + 2πx –C => i = C + 2πx –C => i = 2πx but it still seems intuitively wrong.

1347807.  Sat May 09, 2020 5:12 pm

 PDR wrote: Yes, they made us do that at school and it still feels wrong!

It's just simple algebra for sure but the last time I posted anything like this I got some feedback to the effect that some found it unintelligible. With that in mind I posted the 'big print' version in the hope that some readers might be able to follow the path through it.

Call it home schooling if you like!

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