Because I care deeply about the sanity of the more geeky elements here on ARRSE, I thought you might like to try this. A tightly fitting belt is 'worn' around the Earth's equator. If you added an extra 1 (one) metre of length to that belt, pinched it at one point, and lifted the belt up until all the slack was gone, how high above the Earth's surface would the pinched point then be? In other words, find the value of 'h' in the diagram below: Correct answers will earn the approbriation of their peers.

Please will you get on with tidying up the house and not wasting your time and others on this forum? Mrs D

Actually I think someone is extracting the urine! Funny though. For what it's worth, is it 1 metre divided by pi? I'll probably end up looking stupid but worth a bash.

If you want real geekiness, check out this presentation: mms://wm.microsoft.com/ms/msnse/0511/25766/print_xps_2005_MBR.wmv msr

Fcuk me - I've got a stalker. I guess I really have arrived on the ARRSE scene. Right - where's Flash? He's the expert on these matters...

Hmmmm... that's actually a tricky question... I'm pretty sure WelshBloke's answer is wrong though. He is thinking of a similar belt around the earth problem which is not 'pinched at one point' but which is raised above the ground right around the earth. In this case the height of the belt would be 1 / 2.Pi which is about 16 centimeters. Not the 1 / PI which WelshBloke said. Have to have a think about this... Tricam.

Can you show us a working for that? Did you have to find a number for the radius of the earth? Tricam

Can't believe I actually tried to answer this... Anyway, bear with me... Circumference of earth = 2*pi*r metres where r is the radius at the equator But, the circumference of belt = 2*pi*r + 1 metre The tightend belt will from a 'pinch' at h metres above the earth Fig.1 The straight distance to the horizon will be the distance from the 'pinch' to the horizon = B The curved distance to the horizon is from a point directly below the pinch to the horizon following the curvature of the earth. = d So, the length of the belt = 2*pi*r +1 = distance round the earth - 2*curved distance to horizon + 2*straight distance to horizon Equation A: 2*pi*r + 1 = 2*pi*r - 2d + 2B The distance of the pinch to the horizon (straight and curved) can be calculated using the formulas stated here. Expressing d and B (as shown in Fig.1 above) in these terms d = r cos^-1(r/(r+h) B = root((h(2r+h)) Put this into equation A above 2*pi*r + 1 = 2*pi*r - 2[r cos^-1(r/(r+h)] + 2[root((h(2r+h))] or 1 = - 2[r cos^-1(r/(r+h)] + 2[root((h(2r+h))] 1/2 = - [r cos^-1(r/(r+h)] + [root((h(2r+h))] ...erm? .....erm.....? ...and now I havn't a clue how to calculate h.

hmmmmm Disagree with 121.6m If ther is 1m of excess material the band wrap the earth and if pinched at the surface there will be an excess of 1m 50 cm up and 50 cm back down. so with a triangle with approaching 0 degrees angle h=50 cm now the band is released, this will naturally cause the band to form a equilateral triangle. Due to the size of the earth we can make the assumpation that the base is effectively flat. As the 50cm for each will increase as the angle increases then h should also increase Not sure of the math but it must be between 50cm and 1m uneducated guess is 1m

Fack!! Well done G3Ops... that was seriously impressive... with reference to Darth's original post - you have my approbation... you are the king of the geeks.... Its nice to finally see a real world application for maclaurin series etc after struggling with them for years... well, as real world as putting a belt around the earth gets!!! For those trying to follow his answer the one bit I didn't get at first is where he gets... h = r(sec x - 1) To get this equation consider the right angle triangle that has 'the point of the pinch', 'where the belt touches the earth's surface' and 'center of the earth' as its three vertices... cos(angle) = adjacent / hyopteneuse for right angled triangles... cos(x) = r / (h+r) do some manipulation on this and recalling sec(x) = 1 / cos(x) gives you h = r(sec x - 1). G3Ops - I tried it a different way... cause I'm shite at maclarin etc. I had to use trial and error to solve the equation... I used Matlab to keep plugging in values for h until it solved the equation... the interesting thing is that even though you only used the 1st, 2nd terms of the expansions your answer was still within 0.1m of my 'true' answer... (in fact it was probably even more accurate than 0.1m but that's as far as I went). Thanks for wasting my evening Darth... Tricam.

Gawd you had me panicing for a while .... 3 years at uni doing maths and ........ I didn't know didly sqwat about the answer Maybe it needed a weather pattern moving in a NE direction

I have to say that this pretty much sums up my grasp of knowledge about planets n' stuff! It certainly explains why toast always lands butter-side down.