Hi,

Francois LE COAT writes :
> SharkD writes :
>>> t in [0,2*Pi], y in ]-1,1[ then bell curve is
>>>
>>>     "cos^2(t)+sin^2(t)=(4+ath(-y))^2"
>>>
>>> ath() is hyperbolic arctangent means inverse of hyperbolic tangent th().
>>>
>>> Does it suits you ?
>>
>> Is this a correct transformation?
>>
>> 0 = tanh(sqrt(cos(t)^2 + sin(t)^2) - 4) + y
> 
> y = th( 4 - sqrt(x^2 + z^2) ) with x in [-8,8] and z in [-8,8]
> 
> That's correct, it looks like :-)

It corresponds to the following surface :

<http://www.ief.u-psud.fr/~lecoat/bell.nb>

that you can easily play with :

<http://www.wolfram.com/products/player/>

and looks to me as a bell curve, don't you think so ?

Best regards,

-- 

Author of Eureka 2.12 (2D Graph Describer, 3D Modeller)
http://eureka.atari.org/
http://fon.gs/eureka/

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