pISSN 0374-4914 eISSN 2289-0041

## Research Paper

New Phys.: Sae Mulli 2022; 72: 208-223

Published online March 31, 2022 https://doi.org/10.3938/NPSM.72.208

## Motions of a Billiard ball After a cue Stroke

Hyeong-Chan Kim*

School of Liberal Arts and Sciences, Korea National University of Transportation, Chungju 27469, Korea

Correspondence to:*E-mail: hckim@ut.ac.kr

Received: November 19, 2021; Revised: January 19, 2022; Accepted: January 21, 2022

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License(http://creativecommons.org/licenses/by-nc/3.0) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

In this study, I investigate the collision between a cue and a ball in billiard games. First, I study the collision process in detail. Then, I find the velocity and the angular velocity of the ball after the collision. A head-on collision maximizes the efficiency of kinetic energy transfer from the cue to the ball. Furthermore, I determine the squirt angle, the deviation of the ball from the intended direction. The main factor determining the squirt angle is the end-mass, the mass of the end part of the cue, that the cue responds instantly to the collision. Because this end-mass is much lighter than the ball’s, the squirt angle is tiny. The angle slowly decreases as the cue-ball mass ratio increases.

Keywords: Impulse with friction, Billiards, Coefficient of restitution, Squirt

In 1835, Coriolis [1] studied the mathematical analysis of billiards for the first time. To my knowledge, it takes more than one century for another work to appear for billiard games based on physics [2]. In Refs. [3,4], the authors considered the effect of friction on collisions of billiard balls in 2-dimensions. In Ref. [5], Han studied various aspects of billiard physics. Mathavan used a fast camera to analyze the ball motions in the billiard games [6] and studied the motion of the ball under cushion impact [7]. The billiard models were also used to develop a robotics system [8].

A cue stroke on a billiard ball makes the ball run along the line of cue incidence. This ordinary point of view in billiards games is correct for a head-on collision only between the ball and the cue. Coriolis [1] calculated the outcome of the collision based on a coefficient of restitution (COR) when the cue strikes the ball's center with zero impact parameter. Here, the COR is defined based on the head-on collision. The impact parameter b represents the distance from the ball's center to the incident line of the cue. For collisions with a non-vanishing impact parameter, he had tried the following ansatz: “the fraction of kinetic energy loss is the same as that without the impact parameter.” This ansatz must be refined by using the impact dynamics nowadays.

When the impact parameter does not vanish, b ≠ 0, the ball can be given a topspin/backspin or a side-spin by striking it above/below or left/right side of its center. Moore [2] presented part of the analysis on the stroke with impact parameter. To avoid miscue, the phenomenon that the cue slides on the ball so that the ball deviates highly from the intended direction, the impact parameter should satisfy [10]1

μstatictanϕ=b/R1(b/R)2,

where R and µstatic denote the radius of the ball and the coefficient of static friction between the cue-tip and the ball, respectively. When a cue strikes above/below from its center, the ball may run along the intended path with topspin/backspin.

However, when a cue strikes off the center of the ball, it heads off along a path parallel to the sum of the normal and the friction forces. In this oblique collision, a situation commonly encountered in billiards, the ball does not follow the line of incidence but deflects from the direction by a few degrees. Billiard players commonly call this deflection “squirt” [11,12] because the ball squirts away from its intended path. Anecdotal evidence indicates that one can decrease the squirt angle by using a light cue-tip. In Ref. [13], Cross presented experimental data for the squirt angle. He also presented part of the theoretical resolution based on the bouncing ball model [14].

However, I fail to find literature which deals with a complete theoretical analysis on the squirt based on the impact dynamics [15] which considers the effect of friction during the collision. Naturally, there is no mathematical description of the resulting motions of the ball after the impact based on the fundamental principle of physics, which is the important part of the billiard play. In this work, I analyze the collision between cue and ball based on impact dynamics. I find the state of the ball just after the collision: the squirt, the central velocities, and the angular velocities. Here, I assume that the kinetic friction coefficient and the (energetic) coefficient of restitution (ECOR) between the ball and the cue are independent of their horizontal/vertical relative speed, respectively. The two assumptions are well known to hold when their relative speed is not very high. For the most part in this work, I also assume that the player holds the cue very lightly so that the hand's force does not affect the ball during the impact and does the stroke horizontally.

In Sec. II, I consider the collision between a billiard ball and a rigid cue stick and show that the assumption “rigidity” fails to explain the smallness of the squirt angle. By taking into account the flexibility of the cue, I recalculate the motions of the billiard ball in Sec. III. I summarize the results in Sec. IV. Most of the mathematical calculations are given in 4 appendices.

In this section, I study the squirt angle of a billiard ball after a stroke of a rigid cue-stick. I show that the squirt angle of the ball stroked by the rigid cue is too large to be compatible with the physical motion of a billiard ball. However, an ordinary cue-stick consists of wood, flexible to impacts orthogonal to the symmetry axis. I consider the effect of this flexibility in the continuing sections.

In figure 1, a cue stick approaches a billiard ball of radius R with impact parameter b=Rtanϕ. The radius of the cue-tip is about d~6 mm. So the contact point of the cue-tip does not generally coincide with the central axis of the cue. The normal force FN acts along a line from the contact point to the center of the ball, while the friction force f acts at right angles to FN. The resultant force F= F N+f acts at an angle θ to the normal-direction, where tanθ=FN/f defines an effective coefficient of friction between the cue-tip and the ball. If the tip slides on the ball, tanθ=μk, where μk is the coefficient of sliding friction. If the tip grips the ball, then μ<μk. If the ball initially static, it will exit from the cue along the direction parallel to F at an angle

ψ=ϕθ

from the line of incidence of the cue. The angle ψ commonly called the squirt angle describes the undesirable deflection of the ball from the intended path.

The experimental data in Ref. [13], for well-chalked tips, show that the squirt angle increases almost linearly for impact parameter b<0.5 and shows a bit of non-linearlity for larger impact parameters b=Rsinϕ~0.5R.

The sidespin obtained from the stroke increases almost linearly with the impact parameter.

Even though the experiment was performed with an unusual cue that produces large squirt angles, the qualitative characteristics will be the same as those of the ordinary cue.

Let me analyze the physical situation in detail. I choose the coordinate system so that I can deal the collision between the ball and the cue-stick easily. If I stroke the cue-stick horizontally, I can ignore the friction between the ball and the table. Then, the only important thing is the relative position between the cue-stick and the ball. In this case, without loss of generality, I locate the coordinatés origin at the impact point between the ball and the cue. Besides, the ball's surface at the contact point is parallel to the (x,z) plane as in Fig. 1. The center of the ball is now located on the y-axis initially and the cue-stick moves on the (x,y)-plane with the angle ϕ with respect to the y-coordinate in clock-wise direction.

Figure 1. (Color online) The collision between the cue and the ball.

Let me call the cue-part of the impact point Ccue and the ball-part Cball. Then, the displacement vector from the ball's center to the impact point is

RI(XI,YI,0)=(0,R,0)=Ry^.

The displacement vector from the center of mass of the cue to the impact point is, along the incidence line of the cue,

rc=(xc,yc,0)=rc(sinϕ,cosϕ,0), 0ϕ<45o.

In an ordinary billiard games with the kinetic friction coefficient μ~0.7, to avoid cue-miss, the impact point should satisfy ϕ<35o. In this work, I choose the upper bound of ϕ to be ϕ=45o which is very close to the Coriolis' limit of the impact parameter bmax=0.7R<Rsin45o0.707R. If the cue is a rigid body, rc corresponds to the distance from the the cue-tip to the center of mass of the cue, which is 0.7 times the length of the cue roughly.

Now, let me write the equation of motions satisfied by the cue and the ball. Let the masses of the cue and the ball be Mc, M, their velocities be U, V, and their angular velocities be ω, Ω. The initial values for the cue and the ball becomes Ui=Ui(sinϕ,cosϕ,0), Vi=0, and ωi=0=Ωi. Then, the equation of motions for of the center of masses for the cue and the ball are

McdUdt=F, MdVdt=F,

where M and Mc are the masses of the ball and the cue, respectively. Here F=(F,F,0) represents the force of the cue acting on the ball. Because the cue moves on the (x,y)-plane, there is no force along the z-direction. Here, the subscript and represent the parallel x-direction and orthogonal y-direction to the impact plane, respectively.

The force exerted by the cue on the ball has two origins: One is the normal force F to the ball's surface related to the cue-tip elasticity and the other is the friction force F parallel to the surface. So the force F is generally not parallel to the incidence line of the cue. Adding and integrating the two equations above, I get the following law of conservation of momentum:

McU+MV=Mc U i,

where the right-hand side denotes the initial momentum of the cue. Let the rotational inertia of the cue and the ball be Ic=νcMcrc2 and I=νMR2, respectively. For the case of a ball of uniform density ν=2/5 and for an object with a skewed mass such as a cue, it is roughly νc0.136 [13], respectively. Then, the angular equation of motion of the cue is

Icdωdt=rc×(F)=rc(cosϕFsinϕF)z^.

Here, rc is the displacement (4) from the center of mass of the cue to the impact point Ccue.

The rotational equation of motion of the ball is

IdΩdt=RI×F=RF z^.

As one can see from this equation, both the cue and the ball rotate around their own z-axis after the collision.

Let me describe the collision between the cue and the ball in detail. When a player strikes a ball off the center to give the ball a spin, first of all, the cue-tip starts to slip on the ball's surface. The relative sliding velocity Uisinϕ fails to disappear instantly unless the friction force diverges. The normal force acting on the cue-tip makes elastic energy accumulate as the tip contracts. The friction force between the tip and the ball reduces the relative sliding velocity. The slip disappears at a moment while the cue-tip is contracting due to the frictional force. This slip period is responsible for the squirt phenomena of the ball. Later, the tip grips the ball and moves together. As I know, the maximum static frictional force always equals to or is greater than the kinetic frictional force. Therefore, once the slip stops, an additional one will not happen. At some point, the cue-tip contracts to its maximum, and the elastic energy accumulated until then will restore the shape of the tip and push the ball out. By the end of the contact period, the friction force is no longer enough to hold the ball firmly, so the ball bounces off the compressed tip. In Appendices A and B, I solve the equation of motions (5)-(8) for the processes to get the resulting movement of the ball just after the stroke. The ball's velocity just after the collision is given in Eq. (B7):

V0=Q01+Q0Uicosϕ(1γ)β1+e*G+μctanϕ,V0=Q01+Q0Uicosϕ(1γ)βμc1+e*G+μctanϕγ,

where e* is the ECOR, Q0=Mc/M is a mass ratio of the cue to the ball, and G, μc, and γ are the notation introduced by Stronge [15] constructed from the initial configuration of the ball and the cue:

μcβ×β, γβ×2ββ1,G=(1+μctanϕ)2γ11μ/μc1(μctanϕ)2,

where

β×=sinϕcosϕνc(Q0+1),β=1+Q0ν(Q0+1)+cos2ϕνc(Q0+1),β=1+sin2ϕνc(Q0+1).

Here, νc0.136. In the resulting form, the kinetic friction coefficient between the ball and the cue appears only in G.

As seen in the figure 1, the squirt angle of the ball, by using tanθ=V0/V0, is given by

ψ=ϕθ=ϕarctanμc(1+(γ11)μctanϕ1+e*G+μctanϕ).

In figure 2, I depict the squirt angle for e*=3/4 and μ=0.6.

In calculating the squirt angle, I assumed that the cue is a rigid body. When the cue-ball mass ratio is about Mc/M~3 and the cue strikes half the radius (ψ=30o0.523 rad.), the squirt angle is about 16o. This value is very different from the actual squirt angle that appears in billiard games. However, I can find qualitative tendencies for the squirt angle. The squirt angle vanishes when Q00 and ϕ=0 and increases with ϕ and Q0. In addition, the squirt angle in this figure is quite close to the experimental data for rigid iron cue given in Ref. [13].

Also, since the amount of squirt for a genuine cue is much smaller than this angle, I notice that the main assumption I took in this section,“the cue behaves as if it is a rigid body” must be incorrect. Because this result does not represent the physical billiard balls, I stop here. In the next section, I break out the assumption and see how I can build a more physically acceptable cue model.

I show that a rigid cue model fails to explain the squirt in Sec. II. Now, I models the flexibility of the cue by dividing the cue into two parts, the body and the so-called cue-end, which I explain below. Then, I show that the model successfully explains the squirt of the billiard ball after the collision between the cue and the ball. I also display how the ball moves/rotates after the collision.

When a cue strikes a billiard ball, the instant that the cue is in contact with the ball is about 0.002 seconds. The vibrating frequency of the end part of the cue is about f100 Hz.2 Thus, the period for the vibration is about 0.01 seconds. Therefore, the contact instance for striking the billiard ball is much less than the vibration period and there is not enough time for the impact acted on the cue-tip to spread over the entire cue. The actual functioning of the billiard cue, then, can be modeled by an end-part of the cue including the cue-tip (the so-called cue-end) as shown in Fig. 3. When one makes a shaft from any material, one needs to pay attention to its vibration frequency to get a better performance.

Let me calculate the mass mc and the center of mass location of the cue-end. Consider a typical shaft of length l~70,cm and mass Mshaft=120,g. Both ends of the shaft have circular forms of radii d~6 mm and a~10.5 mm at its cue-tip and the opposite side, respectively. Regarding the shaft as a part of a cone, the volume of the shaft becomes Vshaft=πl3(a2+ad+d2)~153.cm3. Therefore, the mass density of the shaft becomes ρshaft=Mshaft/Vshaft0.782g/cm3. If I look at the super-slow-motion video [16], the end part of the cue vibrates as soon as it hits a billiard ball. The swinging length is about ten times the diameter of the cue-tip3. Since the cue-tip radius is about d=6 mm, I estimate the cue-end length to be x~12 cm. Then, the volume of the cue-end, by using the above volume formula with the replacement lx and ad+(ad)x/l, becomes V=πx[d2+(ad)dx/l+(ad)2x2/(3l2)]~15.4,cm3. Multiplying the shaft density to the cue-end volume, the mass of the cue-end (usually called end-mass) is approximated to be 1/10 of the mass of the shaft:

mc12 g.

Note that this end-mass is crucially dependent on the cue-end length x. If the cue-end is 25% longer than the previous value so that x=15 cm, I get 65% higher end-mass mc19.8 g, which will lead to much different squirt angle as I will show later in this section. Of course, this value will also depend on the type and the elasticity of cue.

The center of mass of the cue-end is located at

rc=y¯=V ydVV0.52x

from the end of the tip roughly, where y runs from zero to x. Therefore, since the position of the center of mass is not significantly different from the center position of the cue-end (0.5 l), I can regard the center of mass is located at the center without a big error. Therefore, the rotational inertia of the cue-end, Ic, is

Ic=νcmcrc2; νc13.

Here, νc is obtained from the rotational inertia of a uniform rod. When I calculate numerical values for the cue-end, I will use the values mentioned here.

The body (the other part of the cue than the cue-end) plays the role of acting a force (indicated as F in the figure) on the cue-end along the line of the cue. If necessary, one can also consider the effect of vibration of the cue-end. Of course, because the collision ends before the cue-end performs one period of oscillation, one can approximate the vibration as a frictional force [13].

I choose the coordinate system to be the same as that in the previous section. Let the speed of the body be Vcue. If the central velocity of the cue-end is U, I can express the velocity of the cue Vcue to be the line-of-cue part of the cue-end velocity:

VcueVcueU^i= U Ui Ui2Ui=(Ucosϕ+Usinϕ)U^i.

Here Vcue(0)=Ui is the velocity of the cue before the collision. The mass of the body is Mcmc. Let F denote the force of this body acting on the cue-end. The equation of motion for the body is

(Mcmc)dVcuedt=F.

The equation of motion of the cue-end, because it receives F from the body and gives F to the ball, is

mcdUdt=FF.

The ball of mass M gains force F from the cue-end. Therefore, the equation of motion of the ball is

MdVdt=F.

Let ω denote the angular velocities of the cue-end rather than the cue itself. Then, the rotational equation of motion for the cue-end and the ball are

Icdωdt=rc×(F), IdΩdt=RI×F.

The force F does not contribute to the rotational motion because this force always points toward the center of the cue-end. The rotational inertia of the cue-end is Ic=νcmcrc2.

In Appendix (C), I solve the equation of motions (15)-(18) to get the motion of the ball just after the collision. The resulting velocity of the ball is given in Eq. (C8):

V0=p0M(1+e*G+μctanϕ),V0=μcp0M1+e*G+μctanϕγ,

where

p0Mϵ(1γ)(1+ϵ)Uicosϕβ, ϵmcM.

Here, I introduce a small mass-ratio ϵ between the end-mass and the ball.

Here, μc, γ, and G can be obtained from the previous formula (10) with the replacement βkβk with k=,,×. Note that this result is the same as the velocity in the previous section with the replacements Q0mc/M, βkβk, where the constant βk are

β1+mνM+mcos2 ϕmc1νc(1mcMc)tan2 ϕ,β1mcos2 ϕmc1mcMc tan2 ϕνc,β×mmc1νc+1mcMccosϕsinϕ.

where mMmc/(M+mc). The constant βk can be obtained from βk defined in the previous section by adding a term considering the cue-structure. The rotational angular velocity of the ball is also given in the appendix:

RΩ0=μcp0νM1+e*G+μctanϕγz^.

The squirt angle ψ=ϕθ takes formally the same form as Eq. (12) except that the new values for γ, μc, and G should be used.

Rather than writing down the formula explicitly, I concentrate on how the flexibility of the cue modifies the motions of the ball and the squirt angle. To consider the most dramatic situation first, I consider the limit ϵ0, i.e., the mass of the cue-end goes to zero. In the limit, using m/mc1ϵ, and mc/Mc=ϵ/Q0, I get

ββ01+νcνccos2ϕ,ββ01+νcνcsin2ϕ,β×β×01+νcνccosϕsinϕ.

Then, the βk dependent parameters become

μctanϕ, γ1, G1cos2ϕ

In the introduction, I have presented an upper bound on ϕ to avoid cue-miss based on the static friction coefficient. The bound for the angle ϕ is refined further to Eq. (B4), which simply gives ϕ<π4=45o at the present case. Putting these results to Eq. (12) and using βk, I obtain

ψ=ϕθ=0.

In other words, in the limit of zero end-mass, the squirt disappears.

In the present formulation, the end-mass mc represents the property of the cue. The product of the density of the material and the volume of the cue-end gives the end-mass. Noting the definition of the cue-end, the stiffness of the end part of the shaft determines the cue-end length. The stiffer the shaft, the longer the cue-end, the heavier the end-mass. When the shaft is too stiff, I may set the end-mass to be the same as that of the cue, mc=Mc.4 On the other hand when the shaft is highly flexible, I have mcMc. To find the velocity of the ball after the collision, let me examine Eq. (20). I find that the absolute size of p0 is ill-defined because both the numerator and the denominator go to zero in the limit ϵ0 because γ1. Therefore, to find the zeroth order velocity of the ball, I need to calculate the results up to the first order in ϵ. Using the end-mass (13) and the mass of the ball M210 g, I get

ϵmcM12210=2351.

This justifies the perturbative calculations around ϵ=0. In Appendix D, I show the calculations in detail. The resulting velocity and the angular velocity of the ball, to the 0th order in ϵ, become

V0=1+e*1+Q01+ν1sin2ϕUi,RΩ0= Uisinϕν1+e*1+Q01+ν1sin2ϕz^.

Therefore, most importantly, the velocity of the ball is proportional to the initial velocity of the cue Ui. The velocity is maximized when the cue hit the center of the ball with vanishing impact parameter, ϕ=0. The two formula for the ball's velocity and the ball's angular velocity are quite simple but I cannot find them in any previous literature.

The resulting squirt angle ψ=ϕθ to the first order in ϵ, by using Eq. (D1), becomes

ψϵtanϕ1+νc11+1Q0+1ν1+Q01+ν1sin2ϕ1+e*.

As you can see from this equation, the squirt angle depends on various factors. For example, νc is a value related to the structure of the cue-end, ν is dependent on the mass distribution of the ball, and e* is the ECOR for the frontal collision between the cue and the ball. Also, Q0 corresponds to the mass ratio between the cue and the ball. The squirt angle decreases as this mass ratio increases. That is, the squirt angle decreases as the mass of the cue increases with given end-mass as shown in the central panel of Fig. 4. However, as can be seen by comparing with the number (1+ν)/ν~7/2, if the mass ratio is higher than a certain value (approximately Q03), the effect of this mass ratio is not significant. On the other hand, if the mass ratio decreases, this squirt angle can be noticeably increased, especially if the cue mass is less than that of the ball. Finally, the squirt angle simply proportional to ϵ=mc/M, the mass ratio between the cue-tip and the ball.

In the left panel of Fig. 4, I plot the squirt angle for several values of ϵ for Q0=18/7 and e*=3/4. It is obvious that the squirt angle is proportional to ϵ, the mass ratio between the ball and the cue-end. In addition, one can notice that the squirt angle increases almost linearly for small impact parameter b. For larger impact parameter b~.5, non-linear behaviors just appears. In the right panel of Fig. 4, I show the squirt angle for various e*. It is obvious that the squirt angle varies faster for a larger ECOR.

For a well-made cue, (mc/M2/35), if the impact parameter is less than 0.55R (~ the maximum value for an ordinary billiard play), the squirt angle is less than 2 degree. For a squirt angle of 1 degree, in the case of the carom billiards table, an error of approximately 49.6 mm occurs when I strike the ball from one end to the other end in the long axis direction (2844 mm). Since the radius of the billiard ball is about 30.75 mm, this makes an error of about 0.81 thickness (roughly 0.1 thickness per point), which is close to our ordinary experience in billiard game. In playing billiards, personal cues are especially meaningful because the player notices this squirt angle of his own cue.

In 2001, Ron Shepard [12] calculated the squirt angle by using the law of momentum conservation after assuming that the terminal velocity of the cue-end is the same as the terminal velocity of the impact point5. If this assumption is correct, I can write the squirt angle in an easier form,

tanψbνR1(b/R)21+1/ϵ+(b/R)2/ν.

This formula presents a result qualitatively similar to that in Fig. 4 regarding the mass-ratio between the cue-end and the ball. However, it gives a larger numerical squirt angle. For example, if ϵ=0.1, this formula gives the squirt angle ψ=5.32o for b=0.5R, but the formula in Eq. (27) gives ψ=2o even though it has additional dependences on e* and Q0. The present result allows additionally to take into account the effect of the cue-ball mass ratio, the effect of the change of the elasticity, and the effect of the structural change of the cue-tip on the squirt angle.

In this work, I have studied the impact of a cue on a billiard ball. I have assumed that the friction coefficient and the ECOR are independent of the relative speed between the ball and the cue. Respecting the flexibility of a cue, I model it as a sum of two parts: the body and the cue-end. The cue-end is the short `end-part of the cué which responses to the impact from the ball instantly. The cue-end is much lighter than the ball. This fact suppress the squirt angle to O(ϵ), where ϵ(1) denotes the mass ratio between the cue-end and the ball. The squirt angle increases with the impact parameter b=Rsinϕ, where R denotes the radius of the ball. For small bR, it increases linearly. For b~R/2, non-linear behaviors start to appear, which is consistent with the experimental data in [13].

Before closing this work, I summarize the ball motion struck by the cue with the impact parameter b=Rsinϕ. A player strikes the cue horizontally on the height of the center of the ball. Because the squirt angle is small I may use the approximation ϵ1 and write the (angular) velocities only to the first non-vanishing order:

V0=1+e*1+Q01+ν1sin2ϕUi,RΩ0= Uisinϕν1+e*1+Q01+ν1sin2ϕz^.

Here, Ui denotes the speed of the cue before the impact.

The simple formula was absent in previous literature. The magnitude of this angular velocity satisfies R|Ω|=V/ν. Note that the size of this angular velocity takes its maximum

RΩ0max=Ui2ν1+e*1+Q01

at sinϕM=ν(1+Q01). As the mass ratio Q0 grows, the angle ϕM decreases. For an ordinary cue-ball mass ratio such as Q0=18/6 or Q0=3, the angle becomes sinϕM=0.745 or sinϕM=0.730, respectively. Both are located outside of the bound of the impact parameter bM(=RsinϕM)>bmax(0.7R). Therefore, the angular speed increases with ϕ. If the cue is massive so that Q011, I have sinϕM=2/5~0.632. In this case, the impact parameter bM<bmax. There appears a region of ϕM<ϕ<π/4 where the angular speed achieved by the stroke decreases with ϕ. An interesting observation is that the angular speed never depends on the magnitude of the friction coefficient between the cue and the ball. In other words, the friction coefficient determines only the limiting value that prevents the cue-miss.

Let me find the angular velocity of the ball when a horizontally struck cue hits the point RI=(h,y,lR) on the ball with the impact parameter b to have a vertical rotation. The simplest way to do this job is i) to rewrite the above rotational angular velocity by using the impact parameter, the vector from the center of the ball toward the impact point, and the vector representing the direction of the cue stroke, and then ii) generalize the equation. This process is possible because the instance the cue strikes the ball is so short that neither gravity nor the friction between the ball and the table plays a role. First, the impact parameter satisfies

b=h2+(lR)2, y=R2b2, b=Rsinϕ.

When the cue hits the point of the same height as the center of the ball, one gets l=R and h=b. Let the unit vector along the impact direction of the cue be V^cue. Then, the direction of the ball rotational becomes

z^= R I× V ^ cueb.

Putting this to the above formula for the angular velocity, I get the general form for the angular velocity.

The total kinetic energy of the ball, summing the kinetic translational energy and the rotational energy, is

K=MUi221+e*1+Q01+ν1sin2ϕ21+sin2ϕν.

The kinetic energy of the ball monotonically decreases with ϕ. Therefore, the efficiency of kinetic energy transferal from the cue to the ball happens for the head-on collision.

The squirt angle decreases as the mass ratio Q0=Mc/M increases. The minimum value at Q0 becomes

ψψtanϕ,ϵνc1+νc1+1ν1+ν1sin2ϕ1+e*.

Comparing this to the value with Q0=18/7, I get

ψ18/7ψ2518+1ν25/18+ν1sin2ϕ1+e*1+1ν1+ν1sin2ϕ1+e*.

Putting numerical data, ν=2/5, e*=3/4 and comparing the maximal squirt angle at ϕ=π/4, this ratio becomes 1.075. In other words, the squirt angle for an ordinary cue is 7.5% higher than that for a very massive cue. Even though a massive cue presents better performance for a given cue speed, players should choose appropriate mass for their arm because a heavier cue makes them have difficulty in performing cue stroke properly. The squirt angle is small. However, when striking a ball at a distant target ball, even a small squirt angle of 1° creates a nontrivial error on the thickness of more than half radius. For the details of the squirt angle, please refer to the formula (27) or Fig. 4.

Consider the case that a player holds his cue tightly when he strikes a billiard ball. He does not apply extra force at the instance of the collision. Mass of the arm moving with the cue added to the cue mass effectively. Therefore, the parameter Q0=Mc/M increases. This increase makes the maximum angular velocity of the ball also increase even for a smaller impact parameter. Then, from the above velocity formula and Eq. (27), as discussed in the previous paragraph, I find that the ball's (angular) velocity increases. But the squirt angle decreases 7.5% maximally. In this sense, higher Q0 improves the properties of the ball motion. However, a natural demerit happens: the player cannot use the snaps of the fingers and the wrist. In general, the squirt angle may also depend on the acceleration of the cue, which dependence requires additional research.

To deal the collision between the cue and the ball, I need to consider the relative motion of the two impact points. The velocity of the impact point of the cue, Ccue, is U+ω× r c and the velocity of the impact point of the ball, Cball, is V+Ω× R I. The relative velocity v between the impact point of the ball and that of the cue-tip can be obtained by subtracting the former from the latter:

v(v,v,0)=VU R I×Ω+ r c×ω.

Conversely, the relative velocity of the cue-tip (Ccue) with respect to the impact point of the ball is v and the cue-tip will be contracted by the reaction force F of the ball to the cue-tip. Because the cue-tip is more resilient than the ball, the elastic phenomena happening during the collision are mainly due to the cue-tip.

The change of the relative velocity, applying the equation of motions (5), (7), and (8), is described by

dvdt=dUdt+dVdt+rc×dωdtRI×dΩdt=1mFr c×(r c×F)IcR I×(R I×F)I.

Here, the mass parameter m is

mMMcM+Mc.

Using the vector relation rc×(rc×F)=rc2F+(rcF)rc, I get

dvdt=1m1+mrc2Ic+mRI2IF(r cF)r cIc(R IF)R II.

Let me write the amount of impulse that the cue acts on the ball for a short period dt along the direction parallel to and perpendicular to the plane of impact to be

dpFdt=MdV, dpFdt=MdV,

respectively. Then, I write the equation of motion (A3) for the relative velocity by means of the impulses:

dv=βdpβ×dpm, dv=β×dp+βdpm.

In describing the collision process, I follow the construction of Stronge [15] and βk are given in Eq. (11). These constants satisfy

β,β>0, 0γ1; γβ×2ββ.

Because the friction force does not change its direction on the (x,z)-plane and the constant βk does not change, one can sum Eq. (A5) over the changes of the impacts dp and dp.

The impact pp orthogonal to the impact plane increases monotonically during the collision where F0. Therefore, I describe the collision process with this parameter rather than time. I omit the subscript because p will be used widely here.

Let me write the initial values. The initial orthogonal impact p must be zero because the impact begins at the instance that the cue-tip touches the ball. At the moment, the velocities of the cue normal and parallel to the impact plane are Uicosϕ and Uisinϕ, respectively. Before the collision, the ball is static. Therefore, the initial value of the relative velocity v is

v(0)=v0=Uicosϕ, v(0)=v0=Uisinϕ.

The velocity and the angular velocity change according to the equations (5) and (7), respectively. The changes of the center of mass velocities of the cue and the ball satisfy

McdU=dp=MdV, McdU=dp=MdV.

The angular velocities of the cue and the ball, according to the equation, follow

Icdωz=rc(cosϕ,dpsinϕ,dp), IdΩz=Rdp.

Because the mass, the rotational inertia, the angle ϕ are independent of time, I can obtain the velocities once I know the total impact after the collision. In general, it is unclear which is first between the moment when the maximal compression happens and the moment when the slip stops. The order must be dependent on the initial values. In general collisions, one may analyze the maximal compression first case separately from the slip stop first case. In the case of the billiards, as I mention in III, the slip disappears almost instantaneously when the cue touches the billiard ball. So in this work, I consider only the former.

Let me explain the collision process in the order of time.

• If the impact parameter b=Rsinϕ0 when the cue-tip touches the ball, it enters the slip state. The cue collides with the ball with a vertical velocity of Uicosϕ. At the same time it slides on the surface of the ball with a velocity Uisinϕ. Because the friction force is finite, the slip of the cue never stops instantaneously. The existence of the slip state is the origin of the squirt when striking off the center of the ball.

If the kinetic friction coefficient is μ, the amount of friction force f on the ball is proportional to the magnitude of the normal force (=F). So the amount of impulse parallel to the impact surface by the frictional force satisfies

dp=μ dp.

Here, the kinetic friction coefficient μ is assumed to be independent of the speed at which the cue slides and the vertical pressure, which holds when the sliding is not too fast on most surfaces. If the friction coefficient is large enough and the cue-miss does not happen, the sliding state will stop soon. Then, the cue and the ball will change to the ‘stick state’. Let the amount of the vertical impact at the moment of this change be p=ps. In other words, I mark the moment when slip stops and changes to ‘stick state’ as ps.

During the slip, from the equation (A5), the horizontal and the vertical components of the relative velocity are

dvdp=μββ×m,dvdp=μβ×+βm,0p<ps.

The initial value of v is negative, because of Eq. (A7). If I want the sliding speed |v| to decrease, the kinetic friction must be large enough to satisfy dv>0, i.e.,

μμcβ×β.

In ordinary collisions, if the friction coefficient is not large enough, the slip can reverse its direction due to the frictional force. But there is no such phenomenon in the collision between the cue and a billiard ball if the friction coefficient is large enough to stop slip so that a cue-miss has not occurred. Then, the cue grabs the ball soon, i.e., v(ps)=0. Integrating Eq. (A11), the relative velocity at p=ps becomes

0=v(ps)=v0+μμc1β×psm,v(ps)=v0+1μγμcβpsm.

At p=ps, I have v=0 and dv=0. Then, I can write the vertical impulse ps in terms of the the initial values and the kinetic friction coefficient:

ps=mv0β×(μ/μc1).

The vertical part of the relative velocity at this moment is

v(ps)=v011γμ/μc μ/μc 1μc γtanϕ.

Here, its initial value is v0=Uicosϕ. The constants γ and μc are determined from Eqs. (A6) and (A12).

The amount of the horizontal impulse, p, delivered to the ball during the slip state, by using Eq. (A10), is

p(ps)=μps=μmUisinϕβ×(μ/μc1).

As expected, this amount is proportional to the cués horizontal velocity. It vanishes if the friction coefficient vanishes. In that case, the ball gets only the vertical velocity and will proceed in the y-direction, which is very different from the direction of the cue.

The vertical relative speed may increase or decrease depending on the initial condition. Since I expect the ball to separate over time, I generally expect the vertical relative velocity to increase continuously, then satisfies

dv>0,,μc>γμ.

Combining this with Eq. (A12), I obtain a constraint on the kinetic friction coefficient μ:

1μμc<γ1.

Under specific circumstances, this vertical relative speed may initially decrease. In this case, it is said that a ‘jam’ occurred. When the ‘jam’ occurs, γ>μc/μ holds. At the moment the slip stops, the ‘jam’ also ends. Under ordinary circumstances, if the sliding state ends during the compression of the cue-tip, dv=0 and v0<v(ps)<0 must be satisfied at p=ps. That is, the cue-tip is still undergoing compression at the time. Then, from Eqs. (A14) and (A15), the initial value must satisfy

v0v0=tanϕ1μcμ/μc11/γμ/μc

if I want the sliding state stops during the compression. If this equation is not satisfied, the cue does not go into the ‘stick staté during the compression process, but makes a cue-miss or becomes stuck during the restoration process. Conversely, this equation determines the lower bound for γ:

γμ/μc1μctanϕ+μμc1.

The constraint (A19) will replace the cue-miss condition (1).

Next, I find the energy used to compress the cue-tip while the cue slides on the ball. Suppose that the particle is moved by the displacement x along the direction of the force F from t=0 to t=tf. Then, the work done by the force during this time is

Wf=0xFdx=0tf Fvdt=0pf vdp,dp=Fdt.

If the normal velocity changes linearly with respect to the vertical impulse p, the work done by the force during pi<p<pj is

W=pipjvdp= vj+ vi 2(pjpi)

where vi and vj are the corresponding values of normal velocity.

From this equation, the energy absorbed by the cue-tip compression during the slip state, 0pps, is

W(ps)=(v(ps)+v0)ps2=v0ps+1γμμcβps22m.

The energy absorbed in this way changes the shape of the cue-tip near the impact point and is saved as deformation energy of the cue-tip. At the later stage of the collision, the stored energy will be released into the forms of kinetic energy of the ball, heat, and sound.

• After the slip has stopped p>ps, the stick state begins. Now, the impact point of the ball moves at the same speed as that of the cue horizontally. The cue-tip and the ball will stick together due to friction. Since the ball and cue tips are attached, there is no relative horizontal speed. Therefore, from the expression (A5), I get

dv=0  dp=μcdp, p>ps.

From Eq. (A5), the change of the normal relative velocity in the absence of slip is

dv=(1γ)β,dpm, p>ps.

Also, the horizontal impulse cumulated during this stick state is, by integrating Eq. (A23),

p(p)=μps+μc(pps), p>ps.

• Because of the vertical velocity, the cue-tip continues to compress, accumulating energy. At the moment of maximal compression, the vertical compression stops v=0. Let the amount of impulse at the moment be p=pc. Then, ps<pc because vertical compression stops during the tip and the ball are stuck. Integrating the expression in Eq. (A24) from p=ps to p=pc and setting p=pc one gets

0=v(pc)=v(ps)+(1γ)β(pcps)m.

Of course, the relative speed in the horizontal direction vanishes, v(pc)=0, because they are stuck. During this additional compression period, from the above equation, an additional vertical impulse

pcps=mv(ps)(1γ)β

acts to the ball.

Therefore, the amount of impulse up to the moment of maximum compression, in terms of the initial values using the equation (A14), is

pc=m(v0)(1γ)β1+μctanϕ.

Here, tanϕ=v0/v0. For later convenience, I write v0 in terms of ps/pc,

v0=pc(1γ)βm(1+μctanϕ)=(1γ)βpcm(μctanϕ1+μctanϕ1)=(1γ)βpcm(γ(μ/μc1)1γpspc1).

By using Eq. (A21), I get the energy cumulated by the normal force during the stick state, because v(pc)=0, to be

W(pc)W(ps)=v(pc)+v(ps)2(pcps)=β(1γ)2m p c p s 2.

Summing this with the energy in Eq. (A22), the total energy absorbed during the whole compression process including the ‘slip state’ is

W(pc)=W(ps)+[W(pc)W(ps)]=v0ps+(1γμμc)β,ps22m(1γ)β(pcps)22m=(1γ)βpc22m1(μ/μc1)γ1γ ps pc 2.

Here, I use Eq. (A28) and

pspc=(1γ)μctanϕγ(μ/μc1)(1+μctanϕ).

• As soon as the vertical velocity of the cue becomes the same as that of the ball, the compression stops, and the restoration process begins. During this process, the shape of the cue-tip returns to its original form, and the elastic energy stored during the compression process is released to push the ball. Let's call the terminal value of the vertical impulse obtained in the restoration process as pf and find the restored kinetic energy.

Since vertical compression stops while the cue-tip and the ball are attached, ps<pc is satisfied. In this case, the tip and the ball remain attached until the ball is finally separated at p=pf. I integrate the equation (A24) for pcppf to get the terminal vertical relative velocity,

v(pf)=(1γ)βm(pfpc).

The terminal relative velocity parallel to the impact plane is v(pf)=0 because it is in the stick state.

The energy recovered in the restoration process from the elasticity of the cue-tip, by using Eq. (A21), is

W(pf)W(pc)=v(pf)+v(pc)2(pfpc)=(1γ)βpc22m p f p c 12.

Also, the impulse in the horizontal direction cumulated from the initial slip state to the final restoration period is calculated from Eq. (A25):

p(pf)=μcpc(μ μc 1) ps pc + pf pc .

So far, I have looked into the collision process of the cue and the ball in detail. Stronge [15] defined the (energetic) coefficient of restitution (ECOR) from the ratio of the elastic energy absorbed in the compression process to that recovered in the restoration process as

e*2=W(pf)W(pc)W(pc).

Therefore, the ECOR, by using Eq. (A30) and (A33), is

e*2=(pfpc1)21(μ/μc1)γ1γpspc2.

From this result, I can find the ECOR once I know the initial values and pf.

Conversely, given the ECOR, the ratio pf/pc can be found by arranging Eq. (A36):

pfpc=1+e*1(μ/μc1)γ1γ ps pc 2.

Here, since pf>pc, a positive radical was taken. In addition, from Eq. (A32) for a given ECOR, the ratio of the vertical velocity is

v(pf)v(0)=e*G,

where G is the modification factor of the Newtonian COR determined by the initial value

G1+μctanϕ1(μ/μc1)γ1γpspc2.

The impulse ps=0 and v0=0 if the cue strikes the center of the ball, which gives exactly the same result as the existing Newtonian COR. However, if the impact parameter does not vanish and the cue does not strike the center of the ball, the ECOR is different from the existing COR by the proportionality factor G.

In this section, I determine the vertical relative velocity after the collision using a correctly defined ECOR.

First, let's put together several equations to find the relative speed after the impact. The initial value of the relative velocity, as shown in Eq. (A7), is v(0)=Uisinϕ,, v(0)=Uicosϕ I find the ratio of impulses at each moment by using the expressions~ (A31), (A39) and (A37), to be

pspc=(1γ)μctanϕγ(μ/μc1)(1+μctanϕ), pfpc=1+e*G1+μctanϕ.

In addition, the impulse pc at the moment of maximal contraction is

pc=p0(1+μctanϕ); p0mUicosϕ(1γ)β.

Putting these values to Eq. (A39), I get

G=(1+μctanϕ)2γ11μ/μc1(μctanϕ)2.

In addition, a bound for γ is given from Eq. (A20):

μ/μc1μctanϕ+μμc1γ1.

If a ‘jam' state does not happen, the upper limit of γ is reduced to μc/μ,(<1). However, the upper limit of γ is 1 because a ‘jam' may occur in some situations.

As you can see in the expression (B3), the value of G behaves monotonically with respect to γ. Then, the range of G for the bound (B4) becomes

1sin2ϕ(ν+1)(Q0+1)1/2G1sin2ϕ(ν+1)(Q0+1)1.

Therefore, the modification factor G is always not smaller than 1. In addition, both the upper and the lower bounds increase monotonically with ϕ.

Given G, I determine the terminal relative velocity after the impact from their initial values:

v(pf)=e*Gv0=e*GUicosϕ, v(pf)=0.

To obtain the velocities of the cue and the ball, I calculate the terminal impulse parallel to the impact plane. First, I put Eqs. (A14), (A26), (A31), (A37), (A39) to Eq. (A34). Then, I get

pf=μcp01+e*G+ μc tanϕγ.

I get the velocity V0 of the ball after the impact by dividing the terminal impulse the ball by the mass of the ball, since the ball was static before the impact. In other words, V0=pf/M and V0=p(pf)/M. Using Eqs. (B1), (B2), and (B6), I get

V0=mMUicosϕ(1γ)β1+e*G+μctanϕ,V0=mMUicosϕ(1γ)βμc1+e*G+μctanϕγ.

In order to describe the change of the cue-end velocity Ui according to the amount of impulse received by the ball, I apply Eq. (14) to Eqs. (15) and (16) and get

[mc+(Mcmc)cos2ϕ]dU+(Mcmc)cosϕsinϕ,dU=dp,(Mcmc)cosϕsinϕ,dU+[mc+(Mcmc)sin2ϕ],dU=dp.

Here dp and dp are defined in Eq. (A4).

Rewriting the equation with respect to dU and dU, I get

dU=[mc+(Mcmc)sin2ϕ]dp+(Mcmc)cosϕsinϕ,dpMcmc,dU=(Mcmc)cosϕsinϕ,dp[mc+(Mcmc)cos2ϕ]dpMcmc.

Using this result and Eq. (14), I get

dVcue=cosϕ,dp+sinϕ,dpMc.

Writing the velocity change of the cue-end as a component,

dVcue=cos2ϕ,dp+cosϕsinϕ,dpMc,dVcue=sinϕcosϕ,dp+sin2ϕ,dpMc.

The relative velocity of the impact points between the cue tip and the ball is also given by the result obtained in the previous section (A1), and the change of this relative velocity, by using Eqs. (15), (16), and (17), is

dvreldt=dUdt+dVdt+rc×dωdtRI×dΩdt=(Mcmc)mcd Vcuedt+1mFrc×(rc×F)IcRI×(RI×F)I.

Here, the mass m represent not the mass (A2) in the previous section but is determined by the cue-end and the ball by

mMmcM+mc.

Using the vector relation rc×(rc×F)=rc2F+(rcF)rc and Eq. (C1), I remove F in the equation. Then, I write the change of the relative velocity dv and dv by using the impulse (A4) acting on the ball during a short period of time dt:

dv=βdpβ×dpm, dv=β×dp+βdpm.

The constant βk in Eq. (21) can be obtained formally from the previous βk by adding a term considering the cue-structure and still satisfies

β,β0, γ=(β× )2ββ1.

Here νc1/3. In addition, because the mass of the cue-end is much smaller than that of the ball, I have m/mc=(1+mc/M)1~1mc/M.

Since the friction force does not change direction in the (x,z) plane and the values of βk hardly change during the collision, I can find the velocity v by integrating with respect to p and p. The initial value of the relative velocity of the impact point Cball of the ball to that of the cue is given by Eq. (A7) as before. Now all the calculations done in the previous sections will hold with the same form by replacing them with βkβk. Therefore, after putting βkβk in the result of the previous section, I get formally the same form for G, μc, and γ in Eqs. (B3), (A12), and (B4).

Next, I write pf and pp(pf) using the initial values. From Eq. (B1), pf becomes

pf=p0(1+e*G+μctanϕ), p0mUicosϕ(1γ)β.

In addition, from Eq. (B6), I have

pf=μcp0(1+e*G)+ μc tanϕγ.

The velocity of the ball after the collision is, from V0=pf/M,,V0=p(pf)/M,

V0=p0M(1+e*G+μctanϕ),V0=μ0p0M1+e*G+μctanϕγ.

Integrating Eq. (A9) I get the rotational angular velocity of the ball,

RΩ0=μcp0νM1+e*G+μctanϕγz^.

After the impact, let θ denote the angle between the velocity of the ball and the y-axis. Then, from tanθ=V0/V0,

tanθ=V0V0=μc(1+(γ11)μctanϕ1+e*G+μctanϕ).

Let me calculate the quantities to their first order to get the velocity. First, I write βk to the first order of ϵ:

ββ0(1ϵA);A1νc(1+ν)ν(1+νc)cos2ϕνctan2ϕQ0(1+νc),ββ0(1ϵA);A1νc(1+νc)sin2ϕνccot2ϕQ0(1+νc),β×β×0(1ϵA×);A×1+νc(1+νc)Q0.

Putting these into Eqs. (A12) and (C5) I get

μctanϕ,[1ϵ(A×A)]=tanϕ1ϵνc(1+νc)cos2ϕ1Q0 +1+νν,

and

γ1ϵνc1+νc1sin2ϕcos2ϕ1+1Q0+sin2ϕν.

Because the squirt angle is very small I may use the approximation ϵ1 and write the velocity only to the first non-vanishing order, from Eq. (C6),

p0MUicos3ϕ1+Q01+ν1sin2ϕ.

The velocity and the angular velocity of the ball in Eq. (26) can be obtained by putting these results to Eqs. (C8) and (C9).

Putting these results to Eq. (C10) to obtain the angle between the velocity of the ball and the y-axis, I find

tanθtanϕ1ϵνc(1+νc)(1+e*)1ν+1ν+1+1Q0e*cos2ϕ.

From this, one can get the squirt angle ψ.

This work was supported by the National Research Foundation of Korea grants funded by the Korea government NRF-2020R1A2C1009313. The author thanks to Dr. Youngone Lee for helpful discussions.

1 I refine this formula later in this work.

2 I have obtained this value by measuring the frequency of sound from the cue when it hits a billiard ball by using a smartphone application ‘spectrum’. This value will be dependent on cue.

3 This is a rough estimation. The length of the cue-end depends on the type of wood, the method construting the shaft, etc.

4 Even if mc = Mc, minor differences exist from the rigid cue because the shaft does not compose the whole cue. The structural difference is reflected by the numerical value of νc.

5 If one can ignore the rotational motion of the cue tip, the translational velocity of the cue-end and the final speed of the impact point will be the same.

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