Resolving golf ball flight forces into components to find acceleration

[Paul7]

From what I know I don't think the magnus force does affect the angular velocity, maybe the drag has some effect? or maybe there's some other forces affecting it?

 

[jamesrc]

The drag will slow the angular speed of the ball down over the course of its flight. I've read that a reasonable estimate of the resistive torque would be:
k*F_M*r, where r is the radius of the ball, F_M is the Magnus force magnitude, and k is a constant for which you can use 1/10. (So, if you were so inclined, you could say
$$\alpha = \frac{-kF_Mr}{I}$$
where I is the inertia of the golf ball.)

Note that the direction of the Magnus force is given by the cross product of the direction of the spin vector and the direction of the velocity vector. So the way you have your equations, you are assuming that ithe golf ball has "backspin" only, i.e. the direction of the spin is suck that the direction of the Magnus force is in the direction you have shown in your first post. That's fine, as long as you're aware of the assumption -- you're model does not take into account the effects of any potential "sidespin."

All of that said, I'm not sure what is going wrong with your simulation.
Your numerical integration technique is pretty crude, but as long as your time step is small enough, you should still get a reasonable simulation. I've run code based on yours in MATLAB, assuming that your library of vector functions (like Vector3D_crossproduct and Vector3D_length and Vector3D_scale) works properly (I don't know if you wrote those yourself or if they were from some predefined library). The only thing left is to make sure your initial conditions are reasonable.

P.S. are you still getting that sine wave effect? I didn't see any evidence of that. Again, it could be your sampling time.

Sorry if that's not much help.

 

[ray b]

side spin gives the hook or slice from a strait path
angle of the club head is = to the angle of flight ,higher is shorter, lower longer distance traveled

 

[Loren Booda]

How about a golf ball with extruded pips as opposed to customary inverted pips? Or a golf ball with multiple, adjacent and shallow sinusoidal channels (extending the analogy to a hard ball)?

 

[Paul7]

Thanks for that jamesrc that's just what I needed, only thing is am not too up on all this rotation stuff so am not sure what the inertia of the golf ball is. Have had a look around and it says the moment of inertia of a ball is 2/5 m r^2. Is this the same thing? Tried using this and get a value of alpha of 700 and something so that doesn't seem right.

Dont think is a problem with my vector library as have been using it for years and works fine.

Am aware that am only dealing with backspin. Will try to extend it later once get the basic flight working correctly.

My initial conditions are. Ball radius = 0.022, ball mass = 0.045, initial rpm = 3000, angle hit at = 8 degrees, initial velocity 70ms.
Think these are right, seem to be roughly same as many sites I`ve looked at.

Dont get the sine wave effect if reduce the angular velocity by a constant value. Ball just doesn't fly far enough and goes to high.

Thanks a lot for the help.

 

[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!