POV-Ray : Newsgroups : povray.general : I need math help Server Time
6 Aug 2024 19:31:23 EDT (-0400)
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From: Sir Charles W  Shults III
Subject: Re: I need math help
Date: 7 Mar 2002 18:57:42
Message: <3c87fe76$1@news.povray.org>
Treat your constant as an acceleration.  Now it becomes simple physics.
Your acceleration is 0.008009 and one time interval is one frame.  In
physics, we might say that is one second, just to make this example easy.
    Now, for t seconds, you are starting from 1 m/sec and acceleration
is -0.008009 meters per second squared.  That means that displacement (the
distance you cover would be 1/2 at^2 where a = -0.008009 and t is unknown.
We know that displacement will be 62 meters.  Solving for t gives:
                t = sqrt ( 2s / a )
    Knowing this, we plug in 2s (124 meters) and divide by the acceleration
(0.008009 meters) and get t^2.  The value is 15,482.58... we can ignore the
negative value, knowing that it means we are losing speed, not gaining it.
Taking the square root of the value yields 124.4 "seconds" or frames in your
case.

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip


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