Any of the physicists out there know how much force is lost from the break if the cue ball strikes with a 9/10 hit versus a full ball? What about 2/3, etc? Basically I'm wondering how much difference a perfectly full hit makes versus a faster contact...
Question about break force
- Thread starter Crispy Fish
- Start date
the force on the ball u hit will be exactly the same, only the force distribution will be different. if u hit left of topball, the rightballs will have more force, and the left balls will have less.
first:
F= m.v² (mass of the cue u play with, and speed of the stroke will make the force that u will give to the cb)
then again:
F= m.v² (so mass of cb , and speed (dependent of the resisitivity of the cloth (this one dependent on material specifications, humidity, windspeed and temperature) ) of the cb will make the force given from the cb to the topball)
then this force will be distributed to the other balls that are on contact with it. if u hit it perfectly straight, and cb center alined, the balls on the left and the right will get same amount of force. (the lower the line of the ball, the lower his force will be = (from more balls on same row, some force will be devided more, kinetic and static forces from contact between the balls upwards if they arent perfectly against eachother)
if u hit it of center: the left balls wont get the same amount of force as the right balls..
not sure if i am correct, but i probably am.... so u need to be more specific on your question.. wich ball are u refering too? where does the cb start?
first:
F= m.v² (mass of the cue u play with, and speed of the stroke will make the force that u will give to the cb)
then again:
F= m.v² (so mass of cb , and speed (dependent of the resisitivity of the cloth (this one dependent on material specifications, humidity, windspeed and temperature) ) of the cb will make the force given from the cb to the topball)
then this force will be distributed to the other balls that are on contact with it. if u hit it perfectly straight, and cb center alined, the balls on the left and the right will get same amount of force. (the lower the line of the ball, the lower his force will be = (from more balls on same row, some force will be devided more, kinetic and static forces from contact between the balls upwards if they arent perfectly against eachother)
if u hit it of center: the left balls wont get the same amount of force as the right balls..
not sure if i am correct, but i probably am.... so u need to be more specific on your question.. wich ball are u refering too? where does the cb start?
If 9/10 means 1/10 of 1 and 1/4" inches off center, I estimate there is about a 2-3% loss in terms of the force being applied to the front ball.Crispy Fish said:
I expect it is a function of the curvature of the ball and so the sine function. As the hit becomes narrower, the loss of force for each 1/10th becomes greater.
How this angle change effect the other collisions and how they spread the pack is a bit of a wild guess.
But I think 1/10 off center won't make a big difference except that it may cause a scratch more often.
If you really want to know, you might try to get ahold of Mr. Bob Jewett and the team at www.sfbilliards.com. I hear those guys have done some really cool, interesting experiments regarding billiard physics over the years. Hope this sends you in the right direction.
Hi Crispy Fish.
Here is a nice compact formula that I derived....
y = cos^2 [sin-1 (1-x)]**
where...
y = percentage of CB's original momentum transferred to rack (in the direction of original CB trajectory)
x = percentage of head ball hit (0.50 being half ball hit, 0.25 quarter ball hit, etc.)
cos = cosine function
sin-1 = inverse sine function
So for example, if you hit the head ball with a half ball hit, then you'd deliver approximately 75% of the CB's momentum to the rack [(cos 30deg)*(cos 30deg) = 0.75, where sin-1 (1-0.5) = 30 degrees]. If you hit the OB with a 90% hit, then you'd deliver 99% of the CB's momentum.
So it seems from the above equation, if you hit most of the head ball (at least half ball hit), then you'd transfer more than 3/4 of the CB's original momentum to the rack. So if you hit the rack dead on at 15mph, it would be equivalent to hitting the head ball with half ball hit at 20mph. (If you can guarantee hitting the head ball dead on at 15mph, but you are totally erratic at 20mph, then just break at 15mph. However, if you can guarantee at least a half ball hit at 20mph, then swing away!...if you don't mind the CB flying around the table.)
**This equation is an approximation. It assumes totally elastic collisions and conservation of momentum. Let me know if anyone wants my derivation. I did this derivation pretty hastily, so i don't know if it's completely true.
Here is a nice compact formula that I derived....
y = cos^2 [sin-1 (1-x)]**
where...
y = percentage of CB's original momentum transferred to rack (in the direction of original CB trajectory)
x = percentage of head ball hit (0.50 being half ball hit, 0.25 quarter ball hit, etc.)
cos = cosine function
sin-1 = inverse sine function
So for example, if you hit the head ball with a half ball hit, then you'd deliver approximately 75% of the CB's momentum to the rack [(cos 30deg)*(cos 30deg) = 0.75, where sin-1 (1-0.5) = 30 degrees]. If you hit the OB with a 90% hit, then you'd deliver 99% of the CB's momentum.
So it seems from the above equation, if you hit most of the head ball (at least half ball hit), then you'd transfer more than 3/4 of the CB's original momentum to the rack. So if you hit the rack dead on at 15mph, it would be equivalent to hitting the head ball with half ball hit at 20mph. (If you can guarantee hitting the head ball dead on at 15mph, but you are totally erratic at 20mph, then just break at 15mph. However, if you can guarantee at least a half ball hit at 20mph, then swing away!...if you don't mind the CB flying around the table.)
**This equation is an approximation. It assumes totally elastic collisions and conservation of momentum. Let me know if anyone wants my derivation. I did this derivation pretty hastily, so i don't know if it's completely true.
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If I shoot a half ball shot on any ball, it seems like both the ob and the cb will travel about the same distance after contact. If that is correct, it would seem that half the energy from the linear movement is transfered to the ob and half remains with the cue ball. Following that line of reasoning, it would seem that a half ball hit on the break shot would only send half of the energy into the rack and let the cue ball retain the other half.
I didn't major in physics, so I don't know any formulas to back up this idea. Just something based on observation and practical application. If my thinking is wrong, please correct me.
Steve
I didn't major in physics, so I don't know any formulas to back up this idea. Just something based on observation and practical application. If my thinking is wrong, please correct me.
Steve
Secaucus Fats
Banned
pooltchr said:
That's what I have observed as well, and like you I do not have a background in physics.
Here is a link to an article from Jim Meador. (Jim was a regular contributor to RSB and ran his own billiards website "Biliards World". He died way too soon, and will always be missed by those who got to know him thru his posts and articles.)
http://www.billiardworld.com/cntrlspd.html
I think Solartje said it all but I'm still trying to comprehend the mysteries he's laid bare for us.Crispy Fish said:
A simpler way of figuring it is to take the offset (1/10 for a 9/10 hit), multiply it by itself (square it), subtract the result from one, and then take the square root of this. This happens to be the cosine of the "cut angle".
So for a 3/4 ball hit, you'll get Sqrt[1 - (1/4)(1/4)] = Sqrt[1 - 1/16] = 0.97 of the force of a full ball hit. This isn't much of a loss because the cosine of small angles changes slowly with changes in the angle. Here are some other values:
.9 ball hit = .99 of full head on force
.7______ = .95
.5______ = .87
.3______ = .71
.1______ = .44
This is a bit of a simplification as it ignores friction forces (which are relatively small) and differences in the amount of energy transfered based on how much mass the cueball "sees" when it collides with the pack. The greater the difference between the cueball's mass and how much it sees, the less the amount of energy that is transfered. And I would guess that how much mass it sees is a function of approach angle (ie, shooting from the head spot vs from the side rail). I have no idea which approach angle yields the most or least mass.
Jim
I agree that the CB and OB does travel approximately the same distance after contact when hit with a half ball, but that assumes the CB is rolling during contact. My calculations assume the CB is NOT rolling (90 degree rule). A rolling CB has angular momentum, which combined with its linear momentum causes it to travel further after contact than if there was no roll. (Remember, a half ball hit doesn't mean the CB and OB depart at a 45 degree angle. A half ball hit corresponds to a 30 degree trajectory, meaning more velocity would be transferred to the OB than the CB. You can't argue with physics and geometry.pooltchr said:
)If you want to transfer equal velocity to the OB and CB at impact, then the contact angle should be 45 degrees. This would correspond to a [1 - sin(45deg)], or a 29.3% hit between CB and OB. Again, this assumes the CB is NOT rolling.
I didn't major in physics either, so my derivations might have all kinds of errors. But it's always nice to double check theoretical computations with practical observation and experience.pooltchr said:
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Crispy Fish said:
Imho...
I'm of the Pi rule on breaks. Pi is 7/22. When you break... (changed my mind, I'm not going to explain this)
Just work on your stroke for your breaks. If need be, "hire" somebody to rack for you, so that you can "work" on it.
Part of life is learning how to learn... You've started the first part of your quest.
Thanks for your question
it's an excellent start.
Hi Jim,Jal said:
I think you may have got the square inversed accidentally.
I think the property he really needs to know here is kinetic energy such that ke = m x v x v/2 . So on a half ball hit, the balls both end up with 0.7071 of th original CB velocity (assuming non rotating) and when w square 0.7071 we get 0.5. So each ball has half the original kinetic energy.
So if, according to the cosine of 1/10th we get 0.99, then we should square this which equals less. eg. Approx 0.98 And so for a 3/4 ball, I don't have a calculator here but would be about .95 x 0.95 which is about 90% of K.E. transfered to the OB.
Or 2% loss in Kinetic energy, but just 1% loss in momentum.
Anyway, that's what seems to make sense to me.
Force cannot be calculated as we don't know the accelleration rate i the collision, but it would be a pretty spiffy number. Stick your finger in front of the the break ball to get some idea
This point has been discussed extensively in RSB and was part of a very recent column in Billiards Digest. You may want to look that column up.pooltchr said:
If the cue ball is rolling smoothly on the cloth and strikes an object ball half full, the two balls will travel about the same distance after impact.
If the cue ball is sliding at impact (stun shot) and the cut angle is 45 degrees (slightly fuller than 1/4 ball) the two balls will travel about the same distance after impact.
The problem is not how much energy you are getting into the rack, it's how much energy and motion you keep off the cue ball. You want to keep the cue ball from scratching. Scratching is much worse than not making a ball. Players with bigger arms than brains often forget this. If the cue ball hits a cushion after the break, you did something wrong (assuming the cue ball didn't stop dead and then get kissed to the cushion).Crispy Fish said:
Bob is 100% correct here. The main reason to hit full on the head ball, with a centerball hit, is for a complete transition of energy from the CB, into the rack, and total control of the CB. The best break happens when the energy is completely released from the CB, and the CB squats in the middle of the table. As an example, it is already difficult enough to strike the CB exactly dead center, and difficult enough to strike the head ball square...at almost any speed. When you add variables like extreme power, body movement, sidespin, topspin, or draw, on the CB, it becomes even more difficult to efficiently release the energy in the CB (to spread the rack out as best as possible), and still control the CB (because anything other than a dead ball hit will still leave energy on the CB). Like Bob stated...the best break happens when the CB does not hit a side rail (assuming no collision by an OB).
Scott Lee
www.poolknowledge.com
Scott Lee
www.poolknowledge.com
I'm not sure what you mean Colin. My formula is nearly the same as jsp's, but in a different form. We both find the cosine of the cut angle since this is the speed of the cueball along the line between the centers of the two balls at impact. Using his x (fraction of the object ball hit), 1-x is the sine of the cut angle. He then uses the inverse sine function to get the cut angle, and takes the cosine of this. My formula computes the cosine using cosine=Sqrt[1-(1-x)(1-x)]. These are equivalent. But he then squares the cosine for some reason.Colin Colenso said:
But why is the cosine relevant here?
We know that at the moment just before impact the cueball's velocity can be analyzed (arbitrarily but conveniently) into two components: one in the direction of the line between their centers, which is vcos(phi), and one in the direction tangential to their surfaces at the contact point, which is vsin(phi), where phi is the cut angle. If we ignore the friction effects, we can completely ignore the tangential component vsin(phi), as far as the interaction with the object ball is concerned. The only thing the object ball "sees" is the cueball coming at it with a speed of vcos(phi) and from the direction of the line between their centers. This is true whether the cueball is colliding with a single ball or the head ball of the rack.
I think you mean a 45 degree hit, or slightly thicker than a quarter ball hit?Colin Colenso said:
I agree that kinetic energy might be a more appropriate thing to consider but the original poster's question was about force. The impulse acting between the cueball and the head ball of the pack acts along their line of centers (ignoring friction). It's the average force multiplied by the contact time and is equal to the change in the momentum of the cueball along the line of centers. (It's also equal to the sum of the momentum picked up by the balls in the rack.) Since the contact time is nearly the same at all cut angles, the momentum of the cueball in this direction, mvcos(phi), is the appropriate indicator of the force that will develop. Just how much force is generated depends on the effective mass of the pack, which in turn depends on the approach angle, I think. If you hit the head ball at the same contact point so that the approach angle into the rack is the same, but vary the starting location of the cueball so that you vary the fullness of the hit, the mass of the pack that the cueball sees will be the same in each case. So at least here we can say, I think, that the force that develops will vary exactly, or very closely, with the cosine of the cut angle on the head ball.
Colin Colenso said:
I agree, but as you indicated, this would be true for a collision with a single object ball. The collision with a pack is different because the mass of the pack is not equal to the mass of the cueball.
If you still think I'm wrong, please fire away. I could be having a bad trig day.
Jim
I know what you mean (get rid of as much energy as possible) but it's impossible to completely transfer all of it. The mass of the pack is greater than the cueball which causes it to inevitably rebound. (Haven't collected my quota of nits today.)Scott Lee said:
Jim
Good question. I'm glad someone noticed.Jal said:
At first, my equation was just a single cosine, without squaring it. But that corresponds to a momentum vector that is in the direction of the cut angle. If the head ball was hit at half ball, then this vector would be oriented at 30 degrees from verticle (verticle being the original CB trajectory oriented straight through the rack). Because of this cut angle, not all of the momentum will be transferred to the rack the same way a verticle momentum vector would with the same magnitude. This is where I make an approximation. I just take the vertical component of that vector, which is another cosine of the cut angle. And hence, you get cosine squared.I don't like using kinetic energy regarding ball to ball collisions, since kinetic energy is just a scalar quantity and not a vector. I'd rather use momentum, which is a vector containing both magnitude and direction. You can simplify things by assuming ball to ball collisions are perfectly elastic such that conservation of momentum holds. And since the masses of the cue balls are identical, you can factor out the mass terms and all you have are equations using velocity vectors.Jal said:
Hmmm... perhaps I used the wrong terminology. I didn't mean to confuse anybody.Jal said:
Very interesting information in here. Basically it sounds like the rack should still spread quite well even if you don't get a full contact, but at the loss of cue ball control. Good to know, and thanks for all the detailed responses!