Re: [ATM] Conic height

["Richard F.L.R. Snashall" <rflrs@rcn.com>]

Vladimir Galogaza wrote:

 > In this case general (ASCII) expression for the conic is
 >
 > z(x)=x^2/{R*[1+sqrt(1-(K+1)*(x^2/R^2)]}            (1)
 >
 > R stands for paraxial radius of curvature
 > K is conic constant ( -1 for parabola, 0 for circle)
 >
 > For given R and K and zone radius xo= yn
 > corresponding conic ordinate z(xo)=yc is
 > obtained from the conic expression (equation) (1)
 >
 > tangent to the conic z(x) in the point (xo,yo) is:
 >
 > t(x,xo,K) = d1z(xo)*(x-xo) + z(xo,K)
 >
 > normal to this tangent in the point (xo,yo) is:
 >
 > n(x,xo,K) = (-1/d1z(xo))*(x-xo) + z(xo)   ( this is the line you are 
looking
 > for)

  From this and the similar triangles, yn/yc = ( S + Q ) / Q, where Q is
the (-1/d1z(x0) * (0 - xo) term above and S is the z(xo) term above.
But that means:

   yn/yc = 1 / ( (K+1)/(R*sqrt(1-(K+1)*(yc^2/R^2))) ) + K*R/(K+1)

or, squaring (and introducing all manner of additional roots):

   ( yn/yc - K*R/(K+1) )^2 = ((K+1)/R)^2 / (1 - (K+1)*(yc^2/R^2))

and (multiplying through by yc^2 * (1 - (K+1)*(yc^2/R^2)):

   (1 - (K+1)*(yc^2/R^2)) * (yn - (K*R/(K+1))*yc )^2 = ((K+1)/R)^2 * yc^2

which is quartic in yc (assuming I haven't made a mistake, of course).

-- 

Rick S.

http://users.rcn.com/rflrs


_______________________________________________
ATM mailing list http://www.atmlist.net/