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Subject: Re: How to get angle and speed to reach a movable point



H.-B.,

thanks for your contributions.

Using my equation set, I found a similar equation
like yours, an analytical solution, derived by
cos^2(alpha)+sin^2(alpha) = 1 from
f1(alpha,t) = 0
f2(alpha(t) = 0 .

--By H.-B.B--
(B'^2-A'_max^2)*t^2 + 2*t*(B-A).B' + (B-A)^2 = 0

c2*t^2 + c1*t + c0 = 0

The coefficient c2 is zero for /B'/=/A'/=A'_max.
This is geometrically not a singularity, but the
quadratic equation degenerates into to a linear
equation.

It CAN be handled numerically by a more sophisticated
quadratic equation solver like this:
http://www.fho-emden.de/~hoffmann/quadequ04062002.pdf

Using the simple 'p-q-formula' , your quadratic
equation would deliver division by zero THOUGH there
is nowhere a singularity.

Why is this solution not reasonable ?
IMO, because signed variables are squared.

Best regards --Gernot Hoffmann


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How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>Re: How to get angle and speed to reach a movable point =>

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