RE: [Algorithms] Nearest point on plane for 2D IK

[Steve W. <Ste...@im...>]

How about if you use the center of the sphere and the point at the beginning
of your ray to use as the normal for a plane passing through the center of
the sphere.  Calculate the intersection of the ray and that plane.
Calculate the point that is the sphere's radius from it's center along the
line from the center of the sphere to the intersection of the ray and the
plane.  That's not the tangent yet because we need to move that point in an
arc around the sphere toward the beginning of the ray by the same angle as
described by the center of the sphere to the intersection of the ray and the
plane to the beginning of the ray.

Just using my virtual protractor and compass.

Rockn-Roll

> -----Original Message-----
> From: David Kornmann [mailto:da...@ik...]
> Sent: Tuesday, August 29, 2000 2:43 PM
> To: gda...@li...
> Subject: Re: [Algorithms] Nearest point on plane for 2D IK
> 
> 
> 
> 
> 
> > Hmmm ... surely what you want is to find the point on the 
> line closest to
> > the center of the sphere (easy to do) and then if that is 
> less than R away
> > from the center you have definite intersection, and proceed 
> to compute the
> > points at which the line intersects the sphere, and otherwise, you
> > conveniently have a point which is in line with the nearest 
> tangential point
> > of the sphere, so you can easily find that tangential point.
> 
> Well, computing the point when the ray intersect with the 
> sphere is not a big
> deal.
> The problem comes when the ray does not intersect:
> 
>  I am not looking for the closest surface point on the sphere 
> then, but a point
> that
> can be described as follow:
> 
> - Take the plane defined with the ray and the center of the sphere.
> - Rotate the ray around its origin and along the plane until 
> it is tangent to
> the sphere.
> - Compute the tangetial point (easy).
> 
> That's what seems to me complicated and which needs to be 
> very accurate.
> 
> Any idea?
> 
> Thanks.
> 
> David Kornmann.
> --
> 
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