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Topic: Calculating Pi using polygon sides sum / radius
Replies: 40   Last Post: Aug 2, 2013 5:27 PM

 Messages: [ Previous | Next ]
 JT Posts: 1,448 Registered: 4/7/12
Re: Calculating Pi using polygon sides sum / radius
Posted: Aug 1, 2013 4:13 AM

Den torsdagen den 1:e augusti 2013 kl. 04:30:12 UTC+2 skrev William Elliot:
> > > > If you create a hexagon with the radius of one, and recursively double the
>
> > > > number of vertices of the hexagon(have same radius as side length)
>
> > > > Maybe this is a valid way to calculate Pi?
>
> > > Given a regular n sided polygon, with distance from center to the vertices
>
> > > of r, the length of one side is 2r.sin pi/n and the total length of all the
>
> > > sides is 2rn.sin pi/n =
>
> > > Now lim(n->oo) 2r.pi.(sin pi/n)/(pi/n)
>
> > > = lim(n->oo) 2r.pi.(sin pi/n)/(pi/n) = 2pi.r
>
> > > For n = 6, sin pi/6 = 1/2, so 2*6r(1/2) = 6r is a crude first approximation
>
> > > for the circumference of a circle of radius r, giving approximately 3 for pi.
>
> > Well if you iterate in a loop double up the vertices you do a lot better
>
> > then 3, smooth.
>
> You go iterate off to a computer newsgroup. This is math newsgroup.
>
> Use lim(n->oo) 2^(n+1)r.pi.(sin pi/2^n) and don't be so dumb as to
>
> limp around on a google goose. Get a real bird to fly the newsgroups.
>
>
>
> pi = lim(n->oo) n.sin pi/n = lim(n->oo) 6*2^n sin pi/6*2^n

Well i do not intend to use any trigonometric functions, and honestly i always thought the ancients, did geometry *the correct way* not losing precision. By double up vertices and using fractions.

Date Subject Author
7/30/13 JT
7/30/13 JT
7/30/13 Peter Percival
7/30/13 JT
7/30/13 Richard Tobin
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7/30/13 Peter Percival
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7/31/13 William Elliot
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7/31/13 Peter Percival
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8/1/13 Virgil
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