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Parabolic path

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You need to look into projectile motion. Google that. You can''t have a parabolic motion with constant velocity, sorry doesn''t work that way. With no respect to air resistence, the x component has constant velocity, but the y component has an acceleration close to earth of about 9.8 m/s^2. The motion will depend on the initial angle and velocity of the object with respect to the ground.

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A simple way, use a basic parabolic y=x^2 function. Let x represent the distance it travels. Now, let's say you want to let the thing rise and fall to ground level over 8 units of distance. A simple equation to use would be: y=-x^2+16. Now, we will let distance go from -4 to 4. Not coincidentally at -4 and 4 y = 0. So to determine the height of your path at a given point you would simply plug it into your formula. So at 3 distance you would plug in the value -1 (-4+3=-1) and you would find your height is 15. At '1' distance your height would be: (-4+1=-3) = 7.

If you want the parabola to cover a unit of "z" distance then you would simply use the equation y = -x^2 + (z^2)/4. You now have a parabolic path you can easily follow at a constant velocity, and you can obviously warp the path easily by multiplying by a constant or whatnot.

Hope this helps.

EDIT: And needless to say, your object's initial position would be at -(z/2) and its final position would be at z/2. Also do you understand how to apply this to 3 dimensions? I enjoy this type of work alot and would be happy to explain.

[edited by - haro on March 24, 2004 3:51:09 AM]

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