Go to Fun_Math Content Table Three Famous Problems Greek Circle Squarer Later Squarer

Kochansky published this result in 1685.

OA = OB = r, and AD = 3r

BC is paralllel to AD, and angle BOC = 30 degrees.

Then CD = 3.141533..

***************** Kochansky.dwg** ***************

**To create this drawing : **
** Load pi_construction.lsp (load "pi_construction")**

Then from command line, type ** Kochansky **

Jacob de Gelder (1765- 1848) published this result in 1849.

This is based on the approximation 355/113 = 3 + 4^{2}/(7^{2} + 8^{2}).

This decimal portion is to be constructed.

CD = 1, CE = 7/8, AF = 1/2

FG is parallel to CD, and HF is parallel to GE.

Then AH = 4^{2}/(7^{2} + 8^{2}).

****************** Gelder.dwg** *****************

**To create this drawing : **
** Load pi_construction.lsp (load "pi_construction")**

Then from command line, type ** Gelder **

Hobson published this result in 1913.
He constructed the approximate value of square root of pi,which is the true "circle squaring".
p^{1/2} = 1.77245...
His constructed length is 1.77246...,which is very,very close!!

OA = 1.0, OD = 3/5, OF = 3/2, and OE = 1/2

Draw the semicircles DGE , AHF with DE and AF as diameters.

The perpendicular to AB through O intersects These semicircles at G & H.

Then GH = 1.77246...

****************** Hobson.dwg** *****************

**To create this drawing : **
** Load pi_construction.lsp (load "pi_construction")**

Then from command line, type ** Hobson **

More topics to be added.

Go to Fun_Math Content Table Three Famous Problems Greek Circle Squarer Later Squarer

All questions/suggestions should be sent to Takaya Iwamoto

Last Updated Nov 22, 2006

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