Calculating the Acceleration of a Golf Ball Flight

[jamesrc]

You've got the right inertia formula (there's no reason to include the dimples in the golf ball in the inertia calculation since our model makes other assumptions that make that kind of thing negligible).

When I use your initial conditions with my simulation, I get a hang time of ~9.1 s, a peak of ~40m and a range of ~230 m.

That's using a constant angular acceleration. When I put in the decaying spin rate, that knocks off about 80 m of the range and halves the hang time.

I also forgot to mention before that I think you have the Magnus force twice as high as it should be (at least according to where I've seen the Magnus Force in a form like yours). So when I include that and use the decaying spin rate based on our crude model of that, I get a ~9m peak, a ~141.5 m range, and ~3.4s of hang time.

How do those results compare with yours?

One more note: you probably already realize this, but as long as you're not considering lateral forces on the ball, this is just a 2d problem. If you're going to include 3D effects later, you're approaching it the right way.

 

[Paul7]

I always thought the magnus force was too big but there doesn't seem to be anything wrong with the calculation. Have just tried dividing the force obtained by 5 and running the simulation and it works almost as required it travels about 180m and a height of 17m, this is with no spin reduction.

Can you see anything wrong with the way I am calculating the magnus force?

Fm = (2 * PI^2 * pair * v * r^4 * w) / (2 * r)

pair = 1.29 kg/m^3
initially v = 70.9 m/s
r = 0.022 m
w = 3000 rpm = 314 rad/s

therefore

Fm = (2 * PI^2 * 1.29 * 70.9 * 0.022^4 * 314) / (2 * 0.022)

= 3.01N

the weight of the ball is only 9.81 * 0.045 = 0.44N therefore this does seem an awful lot. Throughout the whole flight of the ball the lowest the magnus force gets is 1.7N so this can't be right.

Cant see anything wrong in wat I've done though?

 

[jamesrc]

Check http://carini.physics.indiana.edu/E105/spinning-balls.html for a discussion on the Magnus force; that suggests that you have an extra factor 2 in your expression for the magnus force. I don't play golf, but I thought the numbers I was getting were reasonable. I mean, 140m is a heck of a drive, isn't it?

 

[Chen]

That page uses the same formula as you do, except that it uses the rotation frequency and not the angular velocity. So the formula should be:

$$F_M = \frac{\pi pr^3v\omega}{2}$$

That would make your Magnus force about 6.3 times weaker.

 

[jamesrc]

Originally posted by Chen
That page uses the same formula as you do, except that it uses the rotation frequency and not the angular velocity. So the formula should be:

$$F_M = \frac{\pi pr^3v\omega}{2}$$

That would make your Magnus force about 6.3 times weaker.

Exactly what I was trying to say, but I looked at Paul7's posted formula too fast and did not compare them correctly. This is the formula I used in my simulation, where I got the results I posted before.

 

[Paul7]

Thanks a lot for all your help its much appreciated. Works really well now.

Tryin to do the response of the ball from the collision with the ground. Any one have any ideas on this. I need it to take into account the spin of the ball and the friction between it and the ground.

Have got the folowing formula for a collision between two bodies:

J = -vr(e+1)/[1/m1 + 1/m2 + n.(r1 X n)/ I1 + n.(r2 X n)/ I2]

where
J is the impulse in the normal direction
vr is the relative velocity along the line of action
I 1 and 2 are either impulses or more probably the inertias of the bodies, book doesn't eplain very well

v1(after) = v1(before) + [J n + (mu J) t] / m1
v2(after) = v2(before) + [-J n + (mu J) t] / m2
w1(after) = w1(before) + {r1 X [J n + (mu J) t]}/Icg
w2(after) = w2(before) + {r2 X [J n + (mu J) t]}/Icg

and here Icg is the inertia around the center of gravity i think
and mu is the coefficient of friction

As the ground has an infinite mass and no real radius am presuming that I can just assign a value of 0 to these and then get an equation of:

J = -vr(e+1)/[1/m1 + n.(r1 X n)/ I1]

v1(after) = v1(before) + [J n + (mu J) t] / m1
w1(after) = w1(before) + {r1 X [J n + (mu J) t]}/Icg


Am not entirely sure how to calculate the relative velocity in the book I have it states that for two bodies:

vr = (v1-v2) X normal

do you think that this is correct?
so for my case I just use vr = v1 X normal?
also will the inertias I1 and Icg just be 2/5 m r ^2?

 

[Chen]

What is v1, v2, w1 and w2?

As the ground has an infinite mass and no real radius am presuming that I can just assign a value of 0 to these

Actually both the radius and mass of the ground are infinite since they are a lot bigger than the radius and mass of the ball.

 

[Paul7]

v1 and v2 are the velocities of object 1 and 2 respectively and w 1 and 2 are their angular velocities

 

[Paul7]

Just realized that r1 and r2 are the vectors from the center of mass of the objects to the collision point and are not the radius's as I first thought, if that's any help!