# Collision of Billiard Balls

Snap with your fingers a mare’s nose – she will wag her tail.

Kozma Prutkov

Absolutely elastic collision is the collision model where the total kinetic energy of the system is preserved. Since the deformations of the bodies are neglected, it is considered that energy is not lost on deformation, and that the interaction is immediately distributed over the entire body. The example of the collision of billiard balls is often drawn to illustrate the case of absolutely elastic collision. However, is the deformation of the balls during their collision, in fact, that negligible, and is the time of their interaction, in fact, that short?

The time of interaction of billiard balls and their deformation, in addition to direct measurement, can be determined from the formulas describing the collision of arbitrary balls of mass M, radius r, with elastic modulus E and Poisson’s ratio γ.

Thus, the time of interaction , and the maximum convergence of balls , where .

The maximum convergence is a conventional term for the difference between the sum of the radii of the balls and the minimum distance between their centers during the collision. These formulas can be found in the world famous textbook on Physics by Lev Landau and Evgeny Lifshitz (Course of Theoretical Physics, vol. 7: Theory of Elasticity, p. 50).

For calculating τ and h0 we need to know the Poisson’s ratio and the modulus of elasticity (Young’s modulus). For most plastic materials, Poisson’s ratio is between 0.3 and 0.35. In his article, “Modeling the Effects of Velocity, Spin, Frictional Coefficient, and Impact Angle on Deflection Angle in Near-Elastic Collisions of Phenolic Resin Spheres,” (http://arxiv.org/ftp/physics/papers/0402/0402008.pdf), Sam Crown gave a value of Poisson’s ratio for a billiard ball γ = 0.34 and Young’s modulus E = 5.84 GPa, without specifying, however, his method for finding out these values.

Currently, there exist two basic methods of measuring the modulus of elasticity (Young’s modulus). The first way is a direct measurement of sample deformation under compression or extension. The second way is to calculate the modulus of elasticity using the speed of sound propagation in the sample , where c is the speed of sound, and ρ is the density of the material.

For measuring the modulus of elasticity using the speed of sound, we used a billiard ball with a diameter of 60 mm, which on both sides was filed down on a lathe to the size of 50 mm. Two piezoelectric elements and contact groups for them were set on the filed planes. Actually, this simple construction is evident from the assembly drawing and a photograph.

If you lightly hit the flat end of the ball, both of the piezoelectric elements will react with characteristic signals with a small delay in time. The time interval between points A and B is 27–30 µs and corresponds to the time needed for the compression wave to propagate from one end of the ball to the other (L = 50 mm).

The speed of sound, calculated according to these values, amounts to about 1760 m / s. It is easy to calculate that with a density of this particular billiard ball of about 1768 kg/m3, the modulus of elasticity will be 5.48 GPa.

For measuring the modulus of elasticity (Young’s modulus) using the direct method (that is, the immediate strain), a sample with a diameter of 32.82 mm and a length of 50.20 mm was made from a billiard ball on a lathe.

The modulus of elasticity was determined using the INSTRON 3369 Tester, which allows for any measurements associated with compression, bending, and breaking of the sample. The maximum compression load is 5 tons. As it turned out, this load was not enough to destroy the sample.

Here is a chart obtained in the process of compressing the sample in the strain range that did not exceed 0.2 percent.

The modulus of elasticity, determined from two tests was found to be 5.30 Gpa. This is slightly less than the modulus of elasticity of 5.48 GPa, as measured by the speed of sound in the material, and correlates well with the magnitude of 5.84 GPa which Sam Crown mentions in his work.

Having determined the modulus of elasticity for a billiard ball as being E = 5.4 GPa, the Poisson’s ratio γ = 0.34, mass M = 0.282 kg, the radius r = 0.068 m, the speed of the cue ball V = 7 m / s, and using the above formulas, we obtain the interaction time of the balls τ = 320 µs and the maximum convergence of the balls h0 = 0.76 mm (which corresponds to a deformation of 0.38 mm for each of the balls).

For confirming these results, high-speed video of the process of billiard ball collision was carried out. The video was performed using a digital high-speed monochrome camera Phantom V12, at a frequency of 49,000 frames per second. The average contact time of billiard balls with a diameter of 68 mm, obtained from four takes, was about 244–268 µs with an impact velocity of about 7 m/s.

Another way of determining the length of the collision could be to measure the time of the closed state of the contact group, attached to the billiard balls.

However, the inability to control the speed of the balls before the collision may give only an approximate value of contact time, which amounted to about 280–300 µs.

Having determined the position of the balls before and after their collision on the basis of high-speed video, we can find the value of their closest convergence h0 = 0.6–0.7 mm, as well as deformation of each of the balls ∆х = 0.3–0.35 mm.

Given that the calculated time of the interaction between the balls and the magnitude of their maximum convergence agrees well with the results of high-speed video and measurements using the contact group, we can consider the values of the modulus of elasticity E = 5.3–5.5 GPa, the interaction time of the balls τ = 250–300 µs, and the degree of their deformation ∆х = 0.3–0.35 mm quite reliable at the collision velocity V = 7 m/s.

After a simple algebraic transformation, we can write the equation for the time of interaction between the balls and the size of their closest convergence in the following form:

and respectively, where c is the speed of sound in the material.

Analyzing the equations written in this form, we can see that the duration of the collision is mostly determined by the size of the colliding balls and the speed of sound in the material, rather than by the speed of the balls themselves. Thus, a decrease in the speed of convergence of balls by two and a half times will increase the duration of the collision of balls by only 17 percent. At the same time, the degree of deformation of the balls already significantly depends not only on their size and the speed of sound, but also on their speed of collision. It is also easy to see that a transition from balls with a diameter of 68 mm to balls with a diameter of 60 mm will reduce both the time of collision, and the degree of deformation by 12 percent. The proof of this is a high-speed video of the collision between balls with a diameter of 60 mm at a speed of 2.8 m/s.

One of the possible extensions of this study could be the comparison of the characteristics of modern billiard balls from various manufacturers with the properties of ivory balls.

A very interesting article. I too hav spent much time looking into impakt times. I dont know what to think about the importance of the speed of sound. I suspekt that static type strain tests might not be a reliable guide to dynamic impakts. My own calculations were based on a mixture of my own static tests and dynamic tests. The impakt time can be over 1 millisecond at very slow speed. Ivory ball shood be interesting. I hav allways wondered how it iz that an ivory ball impakt sounds shorter and harder than any plastic ball, yet ivory iz softer, and ivory impakt time iz much longer. I dont know if speed of sound correlates with % loss of sound — ie with heat conductivity. But i dont trust formulas, certainly i dont like hertzian stuff. I will hav to study this topic. And i certainly would be very interested in tests on ivory.

mac.

The preliminary comments on perfict elasticity need some comment. There iz no such thing az PE. In fakt i think that experimentation would confirm that az materials-objekts-collizions get more elastik the curve of rezults would not head towards 1.00 or 100.00% etc — i think that the curve would head towards say 0.90. Koz at the end of impakt the objekt(s) must be ringing like a bell — and i think that a good investigator would soon establish that the ringing kannot be less than a certain % of the initial energy — praps say 10%. Skoolkids would be better served by being introduced to reality than by impossible silly mathLand definitions. What to call this reality, this Theorem — why not….

THE BELL BARRIER or,

THE BELL CONSTANT.

mac.

Thank you for your comments. I have given your questions and doubts to the author of the article .

By the way, I am ready to publish your own calculations, which you mentioned in the first comment, or link to your calculations on the network.

Good luck,

Dmitry

When good quality (Aramith Tournament Professional) snooker balls (52.5 mm diameter, 142 gm) collide head on at a reasonably fast speed (15 Km/hr) how much heat is generated at the point of impact?

jack — I think zero heat at kontakt. There would be lots of heat at kontakt if there woz slippage.

But i suppoze that there iz more than zero heat at konakt, koz of lots of local nonelastik deformation at the kontakt (or just skindeep).

Momentum iz konserved, but energy aint. If the coefficient of impakt iz 0.94 then praps the total energy loss iz ???% And ?% of the energy loss iz lost in the skin in the kontakt area. I guess that it iz impossible to kalkulate, what with most of the energy loss going into sound (in the air) and ringing (of the balls). Measuring the temp of the kontakt wouldnt help much, koz u dont know how deep that temp iz.

mac.

Sorry for the delay with the comment. I was looking for frames with thermal imager, which shooted by A.Sorokin. Here we see how the cloth is heated under the ball, as well shaft is heated by friction on the hand, but in the contact point of the tip and the ball is no heating fixed. I think that at the point of contact balls also do not have much heat.

Amazing. I do not understand how u kan hav a very hot hotspot on the BED (lots of heat and/or temp) yet very little heat or temp on the BALL ie the area of ball initially touching the bed. The friktion force and slippage on BALL and BED must be the same, hencely the heat must be the same, and the initial temp must be similar — but they aint. Amazing.

Allso az u say there iz allmost zero heat and temp on the ball for the area of kontakt of QTIP. Kleerly this must be koz there iz very little slippage of qtip on ball (ie zero slippage = allmost zero heat).

And az u say there will be very little heat and temp in a ball’to’ball kontakt area koz there will never be much slippage.

Thanx for the footage.

mac.

The heat and temp of the BED will of course here be due mainly to hysteresis in the bedkloth. Only a small part of the energy of the cue iz transferring to the ball — so the rest iz going into noise and vibration of the table and krushing of the bedkloth.

There iz no qball’to’bed slippage, u kan see the qball rolling off with allmost zero backspin. Hencely the heat and temp on the ball iz due to touching the hot area of bedkloth, not due to friktion.

The hotspots further along (the qball’s track) on the bed are due to hysteresis in the bedkloth due to ball bounce, and due to energy loss in the bedkloth due to rolling rezistance in the bedkloth. Az there iz zero slippage then there iz zero heat on the ball other than heat due to touching (ie conduction) the hot areas of bedkloth.

mac.