High-speed video of cue shots.

Very informative.
I notice that the shaft moves away from the point of contact/impact.
Is that due to the mass of the tip or the flex in the shaft?
Or both?

What is happening with the miss cue?

Thanks.:thumbup:

wow, thanks for links.

This may be a far reach, but how bout high speed of a person making these shots. Do from the top, back and both sides.

Maybe the same shots made by different people that use different stances and styles.

And this one I'd love to have done. A motion studying of different pool players. You know, one where you put a person in s suit to record how they do things.

Anyway, those were so cool, thanks again.

Disclaimer. This is NOT an attempt to knock the videos!!

From the video, it doesn't look like they are laying a very good stroke on the cueball.
Kind of looks like a poke stroke.

I might be way off, but it almost looks like they are bunting the ball with no follow through. Almost stopping after contact. With zero hang time for the tip on the cueball.

I only say this because of the speed at which the cueball leaves the tip, and the increasing distance between tip and cueball after the shot.
It seems like it's too much distance, too fast, especially for slow-mo.

Maybe that is a result of the slow-mo effect, but something just doesn't look right.

But either way, thank you for the videos.

Here is a website put up by Dr. Dave that covers your subject of interest. It might give you some insights toward your own experiments.
http://billiards.colostate.edu/

If you look at some of the other videos, the masse and jump shots - you'll see that whoever is performing the shot isn't hitting the CB very well. The masse shot, for example, the cue doesn't even follow through to the cloth. For the jump shot, the comes in near the top of the cue ball at a very shallow angle. Also, for the draw shot, the cue comes up after contact, where normally it'd follow through to the cloth and maybe even get some shaft-bend.

Still - interesting videos, would like to see more - with more left and right english on some shots.

DBK said:

Hi all.
I have long wanted to see what happens with cue when a cue shot. Videos, which is presented on the Internet does not fully reflect the process, but I have not my own opportunity to do such video:thumbup:. However, recently met a kindred spirit who could do it. That's what happened to date.
I tried to submit a video in English, do not judge strictly for errors in translation.

Here, my friend's article about a wave properties of the cue.

Direct link to the article

Here the video with English shot.

Here the video with the Follow and Draw shot.

Here a video about the transfer rotation between the balls.

A little later I will try to provide frame by frame transcript of videos to facilitate the timing analysis of processes.

I will appreciate your ideas. So I will be grateful if you enter the grammatical errors in the article and descriptions. If there are errors, please let me know by my e-mail address. Sorry for my English:blush:. Waiting for your opinion here on the forum or on my blog or by e-mail. Thank you.

Good luck,
Dmitry

Click to expand...

Great stuff. Thanks for posting all of the links.

Regards,
Dave

What's happening on this one?

http://www.youtube.com/watch?v=_IV8ZIz5auo&NR=1

I can't tell if was supposed to be a jump shot or a masse'.

But it's cool to see that it appears as if the cue tip is contacting the ball multiple times on the way down and then hits the table and the shaft lifts the cue ball.

It looks like a foul.:thumbup:

DBK said:

Hi all.
I have long wanted to see what happens with cue when a cue shot. Videos, which is presented on the Internet does not fully reflect the process, but I have not my own opportunity to do such video:thumbup:. However, recently met a kindred spirit who could do it. That's what happened to date.
I tried to submit a video in English, do not judge strictly for errors in translation.

Here, my friend's article about a wave properties of the cue.

Direct link to the article

Here the video with English shot.

Here the video with the Follow and Draw shot.

Here a video about the transfer rotation between the balls.

A little later I will try to provide frame by frame transcript of videos to facilitate the timing analysis of processes.

I will appreciate your ideas. So I will be grateful if you enter the grammatical errors in the article and descriptions. If there are errors, please let me know by my e-mail address. Sorry for my English:blush:. Waiting for your opinion here on the forum or on my blog or by e-mail. Thank you.

Good luck,
Dmitry

Click to expand...

Thank you so much for pointing out the videos and the article. Lots of outstanding material there! If at all possible, could you report the weights of the cue stick and cueball used in the videos?

If I may presume, concerning the amount of rotation that can be imparted on one ball by another, it can be as high as 5/14'ths or 36% of the impacting ball's spin. This would only occur with a square (full) hit where the relative tangential surface speed at the contact point is totally due to the spin of the impacting ball. (With a cut shot, the relative tangential surface speed is due to both the impacting ball's tangential velocity component, as well as any spin on it.) In order for it to be as high as 36%, there has to be enough friction to reduce the relative surface speed to zero during the collision (i.e., the balls end up rolling across each other, so to speak.) This happens often enough. The 5/14'ths figure comes from the moment of inertia of a uniformly solid sphere, which we're assuming.

Thanks again and looking forward to any further offerings by you and your friends.

Jim

LAMas said:

It looks like a foul.:thumbup:

Click to expand...

Yes I know but I wanted to know the context.

DBK said:

Thanks Jim,

Cue weight about 710 grams (25 ounces), the weight of Russian balls about 280 grams (10 ounces), the tip diameter about 12.3 mm, the tip of the first video is poor Triangle (note it vibration after impact), the second part of the video a new Moori M .

Sorry for my English. You mean that 36% is the maximum possible part of the spin that can be imparted on one ball by another in this experiment? If so, why and where it has been proven?

And please explain what is 5/14'ths ?:blush: Maybe in PM. Thanks. Dmitry.

Click to expand...

And thank you for the feedback on on the cue stick/cue ball, Dmitry. If I get some time, I'd like to maybe take some measurements off the videos to check a few things out.

I don't know what your background is in physics, but you can find a derivation of that ~36% (5/14) figure in one of Dr. Dave's technical proofs on this page (see TP A.27 - Spin Transfer):

http://billiards.colostate.edu/technical_proofs/index.html

If I can try to summarize while leaving out most of the math, consider a friction force acting tangentially on the surface of a solid sphere of uniform density (e.g., a pool ball ). Suppose whatever is exerting that force is essentially immovable, for instance, the bed of a billiard table - but this discussion is not limited to that.

As a result, the surface of the ball, say, at the point of contact, will undergo a change of surface speed. (Of course, if it was at rest to begin with, then the change will be from zero to some non-zero speed.) The change in surface speed will be expressed in two ways: a change in velocity of the ball as a whole (translational velocity in the direction of the force), and a change in spin rate, with the change of surface speed at the point of contact, due to the change in spin rate, also moving in the direction of the force.

Given that the ball has mass m, and moment of inertia of (2/5)mR^2, it turns out that the total change in surface speed, let's call it d_Vs (delta Vs), is divided between changes of translational velocity and spin as follows: The change in velocity is (2/7)d_Vs, while the change in spin rate multiplied by the radius of the ball is (5/7)d_Vs. (The change in surface speed due to a change in spin is equal to the radius multiplied by the change in spin rate.)

Suppose the thing exerting the force is no longer immovable, but is another ball. At the point of contact, you initially have some relative surface speed, call it Vs. Since the friction force acts to reduce (or eliminate) this surface speed, and since it acts equally but in opposite directions on the two balls, the relative surface surface speed is reduced (or eliminated) by the friction twice as fast, so to speak. That is, both balls react to the friction and participate equally to shed surface speed.

So if a certain amount of relative surface speed is gotten rid of, call it d_Vs again, only half as much as in the previous case of the immovable agent will be attributable to each individual ball. In other words, instead of a ball undergoing a change of translational velocity of (2/7)d_Vs, it'll now undergo a change of (1/7)d_Vs. Likewise, the change in surface speed due to a change in spin rate will now be (5/14)d_Vs, instead of (5/7)d_Vs.

In a head on collision, all of the relative tangential surface speed (Vs) is due to whatever spin one or both balls happen to have. If this is reduced to zero during the collision, and one ball was at rest to begin with, then it follows that that ball will end up with 5/14 of Vs as spin. Since Vs was completely due to the spin on the moving ball, it'll have picked up 5/14 of that ball's spin. And 5/14 x 100 = 35.7%.

As an example, with a softly hit straight shot, it's been observed that maximum throw (maximum induced sideways velocity component) occurs when using about half of maximum sidespin (english). This corresponds to striking the cueball at 1/4 of a ball radius off-center. (Maximum throw takes place when the relative surface speed is reduced to zero just at the very end of the collision.) In this case, then, you can get about 36% x 1/2 or 18% of the maximum spin you can put on the cueball (for a softly hit cueball, that is). When hit harder, maximum throw occurs at lesser offsets.

As I indicated, I don't know your background and I may have just bored the heck out of you. But if you want more elaboration on those 2/7 and 5/7 numbers mentioned earlier, let me know, here or in a PM. They come quite directly (i.e., not much math) from the reaction of a ball to a force and the accompanying torque when the force acts off-center.

As an aside, I don't think there's any reason to apologize for your English. You write quite well.

Jim

@Jal Not gonna lie...my head just assploded a little bit.

The more you post, the better your english is
Great videos as I already said.