Re: Instantaneous radius of a rotation

Atreides wrote:
I guess you want the "extrinsic curvature" of a space curve. That
should point you in the right direction.

Thanks for the tip. Actually I was aware of this formula but not in the
concept of this study.

Well, I wasn't aware of this formula, but I knew the name of the
concept. That's putting our heads together. ;-)

The formula doesn't exactly leap off the page in intuitive appeal, does
it?

(http://en.wikipedia.org/wiki/Curvature#Curvature_of_space_curve)

This helps me find the radius. According to my tests, this formula is
producing correct results.

Next question I need to answer is the vector pointing to the center of
instantaneous rotation to find the direction of the centrifugal force.

Hmm... If you are simply trying to find the vector acceleration acting
on an observer transiting a given trajectory, you are going about this
is a roundabout way.

The idea that the trajectory has an instantaneous radius of curvature
and that the acceleration can be calculated from the instantaneous
radius and velocity vectors, is probably correct (you are effectively
fitting the trajectory to the best approximation in two time
derivatives). But it is unnecessary. If the trajectory is u(t), the
acceleration is u"(t), and the required force is mu"(t) . The reaction
(centrifugal) force is just opposite this.

Is that what you were after?

.

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