# Reflection angle of golf ball

I'd like to begin by ignoring friction and center of gravity at the moment if that is possible...and throw out any other variables until I get the larger picture down and then I'd like to ask for a discussion including those items.....

The picture shows a block moving towards a ball...similar to a putter getting ready to strike a ball.

The putter has and angle to it and when it strikes the ball I'd like to be able to calculate the reflection angle of the ball. Did I do this correct?

Thanks.

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mathman
Based on the picture, the block hits the ball at 5 deg. from horizontal, so that's the direction the ball will go.

What do mean by reflection angle?

The block is moving horizontal towards the ball. The block is angled back 5 degrees so the ball should 'bounce' off the face at some angle. I believe this angle to be double the angle of the face.

Danger
Gold Member
Where's the water? Every time that I hit a golf ball, it lands in water. If I hit one in my bedroom, it will find its way to the toilet...

Sorry... I'll go home now...

A.T.
Did I do this correct?
Nope. Mathman is correct that it will be 5deg in the
initial rest frame of the ball (ground frame). Your 10 deg is the angle in the rest frame of the box, assuming perfectly elastic collision and negligible acceleration of the box during the collision.

So if an angled block titled back 5 degrees hits a ball moving horizontal then the ball will shoot up in a vector that is 5 degrees?

For some reason I feel like this is similar to a laser hitting a 45 degree mirror and the result is a reflection of double that or 90 degrees. That is why I thought the vector of the ball bouncing off the block would be double the angle of the block.

A.T.
For some reason I feel like this is similar to a laser hitting a 45 degree mirror and the result is a reflection of double that or 90 degrees.
That is only true in the rest frame of the mirror. Here, you have the same thing in the rest frame of the box, assuming perfectly elastic collision and negligible acceleration of the box during the collision..

I'm confused by your terminology as I am a layman when it come to physics......

question:
are you saying that when a horizontal aimed laser hits a mirror angled at 45 deg that the light would bounce off at 90 degrees but if a 45 degree angled block traveling horizontal smashed into a ball that the ball would bounce off at 45 degrees? I am not arguing here I am trying to understand physics so I thank you for your explanation.

A.T.
Which terms are the problem?
are you saying that when a horizontal aimed laser hits a mirror angled at 45 deg that the light would bounce off at 90 degrees
In the reference frame where the mirror is at rest.
but if a 45 degree angled block traveling horizontal smashed into a ball that the ball would bounce off at 45 degrees?
In the reference frame where the ball was initially at rest, assuming no friction.

I was confused with 'reference frame' but I think I have it now.

Let's forget about reference frames for a second and we only measure angles from the horizontal vector. The angled block smashes into the stationary ball with the block having an angle of 45 deg. The ball should then bounce off at 45 deg from the horizontal assuming such things as friction are eliminated for now?

A.T.
Let's forget about reference frames for a second and we only measure angles from the horizontal vector.
The angle of the velocity vector from the horizontal depends on the reference frame, so you cannot "forget about reference frames", if you ask about that angle. You have to specify in which reference frame you want to measure that angle.

If a 45 degree block is moving towards a ball perfectly horizontal that ball is going to bounce off at an angle.

My question is: what is the angle that the ball bounces off at in relationship to that perfectly horizontal motion of the block?

A.T.
in relationship to that perfectly horizontal motion of the block
You mean relative to the block? Relative to the block it's 90deg.

But your sketch suggested that you are interested in the frame of the ground, where the ball is initially at rest, and the block moves. Relative to the ground it's 45deg.

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What is the bounce angle of the ball in the attached picture. I am looking for the angle of the yellow angle. Thanks.

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A.T.
What is the bounce angle of the ball in the attached picture. I am looking for the angle of the yellow angle. Thanks.
See post #13.

your post 13 doesn't make sense to someone ignorant to physics language. My latest attachment should be simple for anyone without your knowledge in this field to understand as well as I.

If you don't mind posting the angle I was looking for in the previous diagram I would appreciate it. The angle I am looking for is the ANGLE in yellow.

Thanks

A.T.
If you don't mind posting the angle I was looking for in the previous diagram I would appreciate it.
See post #2.

Danger
Gold Member
I'm just going to interject something here that won't help at all, but might put things into perspective. The first statement in the original post rules out reality. There is no way to get even an approximate solution without including all physical factors.
I was somewhat serious in mentioning my lack of success with golf balls, but I am a pool player of some accomplishment (World Championship player 5 years in a row, not that I ever got beyond the 5th round.) The first thing that I ever tell a new student is that the geometry looks good on paper, but it instantly goes out the window when physics cuts in. If you gently hit the cueball with 1/2 tip-width of top and moderate follow-through from in front of a side pocket toward the centre diamond of the next rail up on the opposite side, it will rebound to the corner pocket. If you belt it in the exact same direction, it will come up about 6" short. If you put a shitload of opposite spin on it, it will come right back to you (given a good enough rag material). When you hit an object ball with the cueball, it goes in the same direction as the cueball before veering off in the "reflection" angle. The harder you hit it, the more you undercut it. That effect is multiplied for combination shots. An exercise that I do to practise my signature move that has gotten me out of more than one pickle: I place the cueball on the headspot and shoot it straight down through the footspot to the centre diamond of the far end with some significant sidespin. The ball goes up the centre of the table in a straight line, then comes back and drops into the side pocket. It works the other way around too; shoot in at an angle and have it come out straight. (I love to break a snooker rack that way; it just totally screws the opponent. )

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256bits
Gold Member
your post 13 doesn't make sense to someone ignorant to physics language. My latest attachment should be simple for anyone without your knowledge in this field to understand as well as I.

If you don't mind posting the angle I was looking for in the previous diagram I would appreciate it. The angle I am looking for is the ANGLE in yellow.

Thanks
This is similar to a passanger walking on a train corridor and you are asked to say what his speed is. To an observer stationary on the ground platform, with the train moving relativtive to him at 100mph, and the passanger 's walking speed relative to the train at 5mph, the platform observer can say the passenger is moving at 95mph or 105mph depending on which way he is walking. Of course, to an observer on the train, the passenger is moving at 5mph.

It's the same thing for the angle that the motion of the ball makes.
Certainly if you throw a ball against a surface, neglecting all other factors, it acts just like a mirror with angle of reflection = angle of incidence, and only if you as the observer are stationary wrt the surface ( ie same velocity). In this case the angles are measured wrt the surface, and you being stationary wrt the surface find the same angles.

If you the observer have a velocity wrt the surface, than the angle of rebound measured by you comes out different, as a direct result of your velocity.

Danger
A.T.
This is similar to a passanger walking on a train corridor and you are asked to say what his speed is. To an observer stationary on the ground platform, with the train moving relativtive to him at 100mph, and the passanger 's walking speed relative to the train at 5mph, the platform observer can say the passenger is moving at 95mph or 105mph depending on which way he is walking. Of course, to an observer on the train, the passenger is moving at 5mph.
To bring the frame dependence of velocity orientations into this, one might consider the passenger walking up stairs on the train, that are inclined by 45°. Relative to the train he moves at 45° from the horizontal. Relative to the ground it's a different angle from the horizontal.

Danger
Danger
Gold Member
To bring the frame dependence of velocity orientations into this, one might consider the passenger walking up stairs on the train, that are inclined by 45°. Relative to the train he moves at 45° from the horizontal. Relative to the ground it's a different angle from the horizontal.
Believe it or not, I've never even thought of that occurrence. (I don't do this for a living.) Thanks for bringing it up.

A.T.
Believe it or not, I've never even thought of that occurrence. (I don't do this for a living.) Thanks for bringing it up.
Here is a nice clip. At 0:20 you see the ball flying a rather flat parabola in the reference frame of the car. At 0:30 you see it moving vertically in the reference frame of the ground.

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Danger
Gold Member
Here is a nice clip. At 0:20 you see the ball flying a rather flat parable in the reference frame of the car. At 0:30 you see it moving vertically in the reference frame of the ground.
Oh, yeah... I saw that episode. (I try to see all of them, and I've even kept a couple. Can't get enough of that cement truck blowing up or Kari in swimwear. ) Back to your train thing, though... I just extrapolated from that to something that I''d love to present to a math geek: model the relative motion of someone walking up and down a spiral staircase on a train as observed by someone on an aeroplane flying at 86° horizontally and 3.5° vertically removed from said train. I hate math, and I don't care about the answer... I just want a video of his facial expressions as he tries to figure it out.