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Realistic Ballflight

Posted on 1998-10-06
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Last Modified: 2013-11-20
Hello All,

This is a math \ geometry problem, but don't let that put you off.  If we have any number crunchers in our midst, perhaps they'll like to play with this...

I am looking for a formula which will create a realistic ballflight using parameters of DistanceX and TakeOff Trajectory.  I am currently using "y = (X * (dist - X)) / (dist * Trajectory)" and find this creates a symmetrical
path over which I have sufficient control.  

However, this formula does not, of course, allow me to create a path in which the zenith is not absolutely central to the Xdisplacement.  Ideally, this would occur somewhere around 2/3(?) of the distance.

I have considered using 2 formulae to derive the path, with the calculation being processed by the secondary routine after the xdisplacement reaches 2/3 total distance, but can't make it work properly.

I am convinced (but in no way informed) that I need the formula for an 'egg' shape, cut through at an offset angle.

If anyone has any knowledge of parabolas, trajectories, ballistics or amateur golf I would love to hear your ideas.

I think we can treat the mass of the ball as a constant, and ignore any wind effect, but it would be nice to add a parameter which modifies take off velocity.

Thanks in advance.
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Question by:golfpro
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Accepted Solution

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jkr earned 800 total points
ID: 1323030
The trajectory of a constant mass in a gravitational field is always parabolic, the shape of the curve only depends on the inital parameters as velocity and start angle.

The formula is:

y = x * tan ( a) - g / ( 2* v0^2 * ( cos ( a))^2)

where a is the starting angle to the x-axis ( sorry, no alpha sign on my keyboard ;-), g is the earth gravity (9.81 m/s^2) and v0 the initial velocity.
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Expert Comment

by:jstolan
ID: 1323031
But if you want the real trajectory you do need to take into account the effect of drag.  This results in a decelleration of the object. The equation for this is

Acc = - K * V;

Where K is a constant and V the instantaneous velocity.  So the faster the object is going the greater the decelleration.
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LVL 86

Expert Comment

by:jkr
ID: 1323032
jstolan, you're absolutely right. The problem is that this would end up in solving a multidimensional differential equation of at least degree 2 (as the acceleration is a vector of the components g and Acc, and V  - also vectorial now - depends on the acceleration vector ... &%$§$ can't type integral signs here ;-), so it'd be better to assume a constant wind velocity....
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