
Re: Equation driven curves
Harold Brunt Sep 20, 2010 7:32 AM (in response to Sreenath Varma)If I understand it correctly the formula is basically a circle x^2 + y^2 = R^2 plus the z (height) component. If you solve for x and y you get x(t) = cos(t) and y(t)= sin(t) with t being the postion of the rotation. Then add the z component of z(t) = sin(t) for the basic formulas.

Re: Equation driven curves
Sreenath Varma Sep 21, 2010 12:53 AM (in response to Harold Brunt)_ _
rCosU + a
P(u) = rSinU + b U belongs to [0,2pi]
_ _This is the parametric equation of a circle with centers a,b and radius r.
If you require a helix an additional Z parameter will come into picture denoted by hU, where h is the height
of the helix._ _
rCos(2*pi*n*U) + a
P(u) = rSin(2*pi*n*U) + b U belongs to [0,1]
_ hU _We multiply U with (2*pi*n) where n is the number of revolution.
In Z direction, we multiply "U" by the height of the helix (: h in this case)
Circle is nothing, but a helix in 2D.But how do you solve for wave spring. What are the laws governing the shape.
Please bear with my matrix. No Office tools are installed in mine, so had to get around with notepad.

Re: Equation driven curves
Josh Brady Sep 21, 2010 6:19 AM (in response to Sreenath Varma)Perhaps if you will plot z as a function of t the answer may make itself quite plain to you.

Re: Equation driven curves
Sreenath Varma Sep 21, 2010 7:22 AM (in response to Josh Brady)Yup, I got it right after I posted my query.
Thanks.


