Re: angle n speed of golf ball

ImageAnalyst <imageanalyst@xxxxxxxxxxxxxx> wrote in
message <3386e260-a30f-41e3-8d07-
f79e79de178b@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
On Mar 19, 8:16=A0am, "deepa bala"
<deepa_...@xxxxxxxxxxx> wrote:
i have three images of a golf ball which has been
segmented.
the 1st picture shows the ball when it is on the
ground, the
2nd picture shows the ball when the ball is on the air
and
the 3rd picture shows when the ball is in the air too
but
further then the golf ball in the 2nd picture. i have
to
calculate the golf ball speed and angle. how do i
calculate
them. please give me some idea.

deepa bala:
I assume you're looking sideways, not like what you'd
see if you were
the golfer. So you have an x,y coordinate at time point
1, time point
2, and time point 3. To simplify things, let's assume
the trajectory
is a parabola. Then you just feed those 3 coordinates
into polyfit()
to get the equations for x as a function of t (time) and
y as a
function of t. Knowing these, you can calculate the
speed and angle
at any point of the trajectory. To give you any more
hints than this
would be just doing your homework for you.
Regards,
ImageAnalyst

You do need all three picts. As ImageAnalyst points out,
you are likely expected to fit the equations of motion to
the apparent motion of the ball wrt time. So you need to
determine from the photos how far the ball displaced in x
and y for the time delta between the photos. You will be
expressing those displacements in the equations for motion
and solving for the intial speed and trajectory angle at
Time-0.

The equations for motion in x are:

Xnot + Vnot*T + (1/2)*a*T^2 which is just the intial x
position plus the x diplacement due to velocity plus the x
displacement due to acceleration. If you think about this
problem you might see that you can drop the acceleration
term. Same technique for y. Good luck.

If you show some serious work on this problem I'm guessing
you will get more meaningful help. Posting a few lines of
code (even psuedo-code) goes very far around here as far
as getting help.

Scott
.

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