Rope puzzle

so what if it was round the circumferance of an elephants belly would it still need to be 6' longer to hover 1'?
Roughly, assuming that the elephant's belly was pretty much circular.
 
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So we have the approximate equation
1/3 x^3 + O(x^5) = 1/3 w^3
=> x^3 ( 1 + O(x^2) ) = w^3
=> x ( 1 + O(x^2) ) = w
=> x = w ( 1 + O(x^2) ) = w ( 1 + O(w^2) ) = w + O(w^3).
Using more terms of the series of tan we get the relation

2 2 29 4 14 6 8
x = w ( 1 - -- x - ---- x - ---- x + O(x ) )
15 1575 3375
____________
3 / x^3 / 3
= w \ / ------------
\/ tan(x) - x

x0 = w + O(w^3)
x1 = w - 2/15 w^3 + O(w^5)
x2 = w - 2/15 w^3 + 3/175 w^5 + O(w^7)
. . .
2 3 3 5 2 7 16 9 11
x = w - -- w + --- w - ---- w - ------ w + O(w )
15 175 1575 202125

Therefore the equation 1 - sin(x)/x = d has the solution

_____ 3 321 2 3197 3 445617 4 1766784699 5 6
x = \/ 6 d ( 1 + -- d + ---- d + ------ d + -------- d + ------------ d + O(d ) )
20 5600 112000 27596800 179379200000



Piece of **** if you ask me..... :rolleyes:
 
Be honest Zampa, thats a made up equation isnt it and bears no relation whatsoever to the question :D
 
f*ck ya zampa, i fell for it. i can count money thats as far as my maths skills go :eek:
 
Be honest Zampa, thats a made up equation isnt it and bears no relation whatsoever to the question :D

Only those in the know on here will know whether its made up or not...



:evil:
 
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