Trajectory of a Billiard Ball and Recovery of Its Initial Velocities
Yan-Bin Jia
Department of Computer Science
Iowa State University
Ames, IA 50010, USA
[email protected]
1
Matthew T. Mason and Michael A. Erdmann
School of Computer Science
Carnegie Mellon University
Pittsburgh, PA 15213, USA
{mason+,me}@cs.cmu.edu
Billiard Trajectory
What trajectory will a billiard ball follow on a pool table? Suppose it has initial velocity v 0 and
angular velocity ω 0 . Assume no bouncing of the ball. It will first slide from its initial location x0
until the contact velocity reduces to zero at some location x1 . Then it will roll without slip to a
stop at another location x2 . In this appendix, we will show that the billiard trajectory consists of
a parabolic segment from sliding and a line segment from rolling without slip. The trajectory is
illustrated in Figure 1.
Let m and R be the mass and radius of the billiard
ball. The ball has the moment of inertia 52 mR2 . Its
angular inertia Q is 52 mR2 times the identity matrix.
ω
Denote by g > 0 the magnitude of the gravitational
v
acceleration.
R
trajectory
u
Let v and ω be the velocity and angular velocity
of
the
billiard, respectively, during the motion. The
f
velocity at its contact point with the table, also referred
to as the sliding velocity, is
Figure 2: Contact velocity u of a billiard
and contact frictional force f .
u = v + Rẑ × ω,
(1)
where ẑ = (0, 0, 1) is the upward normal of the table. Due to its cross product with ẑ, the zcomponent of ω will have no effect over the ball trajectory. So it is ignored. As mentioned earlier,
we also ignore any bouncing effect of the ball. So our analysis will make the following assumption:
v and ω have no components in the z-direction.
1.1
Sliding
The sliding velocity u lies completely in the table plane. Sliding starts when
u0 = v 0 + Rẑ × ω 0
(2)
is not zero. Denote the direction of u by the unit vector û. The ball is subject to a frictional force
f = −µs mgû, where µs is the coefficient of sliding friction. The ball’s dynamics are governed by
1
x2 at t2
Rolling
ω0
Sliding
x0
x1 at t1
v0
Figure 1: Trajectory of a billiard with initial velocity v 0 and angular velocity ω 0 .
two equations:
mv̇ = −µs mgû,
Qω̇ + ω × (Qω) = Qω̇
2
=
mR2 ω̇
5
= (−Rẑ) × (−µs mgû)
= µs mgRẑ × û.
Immediately, we obtain the ball’s acceleration and angular acceleration:
v̇ = −µs gû,
5
µs gẑ × û.
ω̇ =
2R
(3)
(4)
We derive the contact acceleration by differentiating (1) and substituting (3) and (4) in:
5
u̇ = −µs gû + µs gẑ × (ẑ × û)
2
5
= −µs gû + µs g((ẑ · û)ẑ − û)
2
7
= − µs gû,
2
(5)
since ẑ · û = 0. So we infer that u̇ is opposite to the initial contact velocity u0 . Consequently, u
will never change its direction until it becomes zero. From now on, û = u0 /ku0 k represents the
constant sliding direction. Integration of (5) yields the sliding velocity:
7
u = u0 − µs gtû.
2
(6)
Since û is constant, we easily integrate (3) and (4):
v = v 0 − µs gtû,
5
ω = ω0 +
µs gtẑ × û.
2R
Equations (7) and (8) were also derived in [1, pp. 10–11].
2
(7)
(8)
The sliding trajectory is obtained from one more round of integration — of (7):
x=
!
x
y
1
= x0 + v 0 t − µs gt2 û.
2
(9)
Sliding ends when the contact velocity u reduces to zero. From (6) this takes place at time
t1 =
2 ku0 k
·
.
7 µs g
(10)
When it happens, the ball has the following velocity from (7):
v 1 = v 0 − µs g ·
2 ku0 k
û
7 µs g
2
= v 0 − u0
7
5
2
=
v0 − Rẑ × ω 0 ,
7
7
(11)
by (2). Let ω 1 be the angular velocity at t1 . We take the cross product of ẑ with both sides of
v 1 + Rẑ × ω 1 = 0, obtaining
1
ω1 = z × v1 ,
(12)
R
Substitution of (10) into (9) indicates that the ball will end sliding at the position
x1 = x0 +
1.2
2 kv 0 + Rẑ × ω 0 k
·
(6v 0 − Rẑ × ω 0 ).
49
µs g
(13)
Pure Rolling
That u = 0 at time t1 implies v 1 · ω 1 = 0 via cross product of v with (1). It is possible that t1 = 0.
At t1 , since ω 1 is orthogonal to v 1 , the frictional force opposes the direction of v1 . Hence neither
ω nor v will change its direction from t1 and on. The ball begins pure rolling without slip until it
comes to rest; and u = 0 holds as it rolls.
The rolling trajectory is a straight line segment. To derive the deceleration of the ball, we adopt
Marlow’s treatment [1, p. 12] based on the principle of energy dissipation. Because the ball rolls
on a line, we can now denote its velocity and angular velocity by their magnitudes v and ω with
no ambiguity. Since v = Rω under pure rolling, the energy of the ball is
E =
=
2
1
1
mv 2 + ·
mR2 ω 2
2
2
5
7
mv 2 .
10
Differentiate the above equation:
7
mv v̇.
(14)
5
Let µr be the coefficient of rolling friction. The energy dissipation due to rolling friction is
Ė =
Ė = −µr mgv.
3
(15)
Comparing (14) and (15), we obtain the deceleration of the rolling ball:
5
v̇ = − µr g.
7
(16)
Thus the velocity magnitude varies as the ball rolls according to
5
v = kv 1 k − µr g(t − t1 ).
7
The ball will come to stop at time
t2 = t1 +
7 kv 1 k
.
5 µr g
(17)
In the vector forms, the ball velocities during the rolling over time [t1 , t2 ] are given as
5
v = v 1 − µr g(t − t1 )v̂ 1 ,
7
1
ω =
ẑ × v,
R
where v̂ 1 is the unit vector in the direction of v 1 . Integrating the velocity equation above yields
the line trajectory of the rolling ball:
x = x1 + (t − t1 )v 1 −
5
µr g(t − t1 )2 v̂ 1 .
14
(18)
Plugging (17) into the above equation, the ball will come to rest at the location
x2 = x1 +
1.3
7 kv 1 k
·
v1 .
10 µr g
(19)
Parabolic Sliding Trajectory
Under sliding, v 0 · ω 0 6= 0; hence v 0 × u0 6= 0 by (2). We take the cross products of û with both
sides of the sliding trajectory equation (9), obtaining the time when the ball reaches the location x:
t=
∆x × û
,
v 0 × û
where ∆x = x − x0 . Here, cross products are treated as scalars along the z-axis. Substitution of t
back into (9) gives rise to a pair of equations:
∆x × û
∆x × û 1
− µs gû
x = x0 + v 0 ·
v 0 × û
2
v 0 × û
2
.
(20)
The first equation of the pair is an identity over the x-coordinate. The second equation of the pair,
y = x · (0, 1)T relates the y-coordinate to the x-coordinate of every point on the trajectory. Hence
it is the implicit equation of the sliding trajectory, which is a parabola as stated in the following
claim.
4
Theorem 1 The sliding trajectory of a billiard ball that starts at the location x0 with initial velocity
v 0 and angular velocity ω 0 , all planar and v 0 · ω 0 6= 0, is part of a parabola that results from the
parabola
µs g
x2
y=
2kv 0 × ûk2
via a rotation about the origin through φ + π/2, where û = (cos φ, sin φ)T , followed by a translation
of
v 0 · û
1
x0 +
v 0 − (v 0 · û)û .
µs g
2
Proof We first determine the point on the trajectory with the maximum curvature. This would
be the vertex if the curve turns out to be a parabola. The curvature function computed from the
parametric form (9) is
µs gv 0 × û
ẋ × ẍ
=−
.
(21)
κ=
3
kẋk
kv 0 − µs gtûk3
We need only find the minimum of the denominator, or equivalently,
the minimum speed of the billiard since v = v 0 −µs gtû according to (7). û
v∗
Minimization of
v · v = v 0 · v 0 − 2µs g(v 0 · û)t + µ2s g2 t2
x∗
yields the parameter value
t∗ =
v 0 · û
.
µs g
(22)
x0
Substitution of t∗ into (9) then yields
Figure 3: Parabolic sliding
v 0 · û
1
x = x0 +
v 0 − (v 0 · û)û .
trajectory with vertex x∗ .
µs g
2
Here, x∗ is the translation we need.
The orientation of the curve is indicated by the velocity vector (7) at t∗ determined as v ∗ =
v 0 −(v 0 · û)û, which is tangent to the curve. See Figure 3. In other words, the tangent is orthogonal
to û = (cos φ, sin φ). Furthermore, since
1
(v 0 · û)2 > 0,
(x∗ − x0 ) · û =
2µs g
∗
the curve bends away from u. This implies that φ + π/2 is the desired rotation. Applying the
rotation to the implicit equation (i.e., second in (20)) yields the ordinate as a quadratic function of
the abscissa. Thus the curve is a parabola with vertex at x∗ and axis of symmetry in the direction
−û.
Finally, we decide the canonical form y = ax2 of the parabola. Its curvature 2a at the vertex
must be equal to the curvature value (21) at t∗ . So we obtain
a =
=
=
µs gkv 0 × ûk
2kv 0 − (v 0 · û)ûk3
µs g kv 0 × ûk
·
2 kv 0 × ûk3
µs g
.
2kv 0 × ûk2
5
1.4
Straight Sliding Trajectory
When v0 ⊥ ω 0 , the ball slides along a straight trajectory. It is more convenient to treat the case
in terms of two orthogonal directions î and ĵ such that î × ĵ = ẑ. The direction î is of v 0 if v 0 6= 0
and otherwise it is of ω 0 × ẑ. For convenience, we write ω 0 = ω0 ĵ, v k = vk î, for k = 0, 1, and
xl = xl î for l = 0, 1, 2. Note that v0 ≥ 0. Without loss of generality, we let î point to the right.
Then ĵ points inward to make ẑ upward. Hence ω < 0 for a counterclockwise rotation; and ω > 0
for a clockwise rotation. The contact velocity and the path of the ball are also collinear with i.
The initial sliding velocity by (2) becomes
u0 = u0 î = (v0 − Rω0 )î.
(23)
If v0 < Rω0 , û0 opposes the initial velocity which consequently increases under sliding friction.
Recall that t1 and t2 are the times at which sliding and rolling ends, respectively. Equations (13),
(11), (10), (19), and (17) now respectively become:
x1 − x0 =
v1 =
t1 =
x2 − x1 =
t2 =
2
|v0 − Rω0 |(6v0 + Rω0 ),
49µs g
1
(5v0 + 2Rω0 ),
7
2 |v0 − Rω0 |
·
,
7
µs g
1
|5v0 + 2Rω0 |(5v0 + 2Rω0 ),
70µr g
1
t1 +
|5v0 + 2Rω0 |.
5µr g
(24)
(25)
(26)
(27)
(28)
Suppose the ball’s velocity v becomes zero at time t3 and location x3 = x3 î. If this happens during
sliding, then we infer from (7) that v 0 and û are in the same direction, namely, v0 ≥ Rω0 . In this
case, from (7) and (9) we have
t3 =
x3 − x0 =
v0
,
µs g
1 v02
.
2 µs g
(29)
(30)
The rest follows from a case-based analysis that measures Rω0 in terms of v0 . The five cases
are illustrated in Figure 4.
(1) Rω0 < − 52 v0 . Hence Rω0 > v0 . This implies 0 < t3 < t1 by (26) and (29), and therefore
x1 < x3 . From (23) v 0 and u0 are in the same direction. The ball will reverse its motion at
x3 = v02 /2µs g before sliding ends. We also have x2 < x1 by (27).
(2) Rω0 = − 52 v0 . The ball has zero velocity and thus zero angular velocity when sliding ends.
There is no pure rolling phase with x1 = x2 = x3 and t1 = t2 = t3 = v0 /µs g.
6
ω0
(2)
(1)
(4)
(3)
− 25 v0
0
ω1
ω3
v0
(5)
u0
Rω0
v0
x0
t3
0
x3
v=0
x1
u=0
x2
v=ω=0
t1
t2
(1)
ω0
ω1 > ω0
ω0
v0
v0
u0
u0
x0
0
0
t1 = t2 = t3
x2 = x3
x1
x0
x1 = x2 = x3
t1
t2 = t3
(3)
(2)
ω0
ω0
0 < ω1 < ω0
v0
v0
u0 = 0
x0 = x1
x2 = x3
t1 = 0
u0
x0
t2 = t3
0
(4)
x2 = x3
x1
t1
t2 = t3
(5)
Figure 4: Five scenarios (1)–(5) of straight billiard motion based on Rω0 relative to v0 shown in the upper
left corner.
7
(3) − 52 v0 < Rω0 < v0 . So the initial sliding velocity u0 > 0. In this case, t3 > t1 ; that is, the
ball’s velocity reduces to zero after sliding ends. This implies that t2 = t3 . The frictional
force increases ω during sliding.
(4) Rω0 = v0 . In this case, t1 = 0. The ball starts pure rolling right away.
(5) Rω0 > v0 . Here v 0 and u0 are in opposite directions. The ball’s velocity increases while
angular velocity decreases until v = Rω when pure rolling starts.
2
Recovering Initial Velocities from a Billiard Trajectory
From the video of a billiard motion we can reconstruct its trajectory, and estimate the locations
x0 and x1 at which the ball begins the motion and ends sliding. Based on the information, we
would like to solve for the ball’s initial velocity v 0 and angular velocity ω 0 assumed to have only xand y-components.1 Both velocities are in the xy-plane. The implicit equation for the trajectory,
whether a parabolic segment or a line segment, is known, from fitting, for example. When the
trajectory is a line segment, the final resting location x2 of the ball is also needed to determine the
initial velocities.
2.1
Parabolic Segment
We infer that v 0 · ω 0 6= 0. Applying standard coordinate transformation to its equation, we can
determine that the parabola is generated from a rotation of y = ax2 , a > 0, through θ followed
by a translation of x∗ = (x∗ , y ∗ ). Here, a, θ, and x∗ are known. The sliding trajectory has the
parametric form (9) dependent on v 0 , ω 0 and x0 . Take the cross product of ẑ with the initial
contact velocity u0 = v 0 + Rẑ × ω 0 :
ẑ × u0 = ẑ × v 0 + R((ẑ · ω 0 )ẑ − ω 0 )
= ẑ × v 0 − Rω 0 .
This leads to
1
ẑ × (v 0 − u0 ).
(31)
R
We will first determine the direction û of u0 , and then v 0 and u0 . Afterward, ω 0 will be known
from (31). Let û = (cos φ, sin φ). Under Theorem 1, φ + π/2 = θ. We immediately have
ω0 =
û = cos θ −
π
π
, sin θ −
2
2
= (sin θ, − cos θ).
(32)
Next, we recover the initial velocity v0 . Denote p = x∗ − x0 . Theorem 1 states that
v 0 · û
1
p=
v0 − (v 0 · û)û .
µs g
2
Take the dot products of û with both sides of the above equation, and rearrange the terms:
(v 0 · û)2 = 2µs gp · û.
1
(33)
We cannot recover the z-components of v 0 and ω 0 since they do not affect the trajectory in the horizontal plane.
8
Meanwhile, Theorem 1 also states that
µs g
.
2kv 0 × uk2
a=
Hence
kv 0 × uk2 =
µs g
.
2a
(34)
Add up equations (33) and (34):
kv 0 k2 = (v 0 · û)2 + kv 0 × ûk2
1
= µs g 2p · û +
2a
This gives us the magnitude of the initial ball velocity:
s
kv 0 k =
ξ
x
v0
β
x1
(35)
(1, 2aξ)
(cos β, sin β) = p
.
1 + 4a2 ξ 2
Figure 5: Determining the orientation of initial ball velocity.
The initial ball velocity is in the direction
v̂ 0 =
In the local frame, the tangent vector at a point
(ξ, aξ 2 ) on the parametrized parabola has the direction (1, 2aξ). Let β ∈ (− π2 , π2 ) be the tangential angle
at x0 with respect to the ξ-axis such that
x0
(
1
.
2a
The direction v̂0 of v 0 must be tangent to the curve
at x0 , as shown in Figure 5. We project the vector
−p = x0 −x∗ onto the direction ξ̂ = (cos θ, sin θ) of the
“x-axis” of the local frame of the parabola, obtaining
the abscissa
ξ = −p · ξ̂.
θ
x∗
µs g 2p · û +
(cos(θ + β), sin(θ + β)),
if (x1 − x0 ) · ξ̂ > 0,
−(cos(θ + β), sin(θ + β)), otherwise.
Combining this with the magnitude (35), we have determined the initial ball velocity:
v0 =
s
1
v̂ 0 .
µs g 2p · û +
2a
(36)
Now we use û and v 0 to derive the magnitude u0 of the initial contact velocity u0 . This will
give us u0 = u0 û, and subsequently ω 0 from (31). Since u0 = v0 + Rẑ × ω 0 , equation (13) can be
rewritten as
2 u0
x1 − x0 =
(7v 0 − u0 ).
(37)
49 µs g
9
Take the cross products of û with both sides of the above equation:
(x1 − x0 ) · û =
2 u0
(7(v 0 · û) − u0 ) .
49 µs g
Rewrite the above as a quadratic equation in u0 :
u20 − 7(v 0 · û)u0 +
49
µs g(x1 − x0 ) · û = 0.
2
The solution to the equation is
q
7
u0 =
v 0 · û ± (v 0 · û)2 − 2µs g(x1 − x0 ) · û .
2
(38)
In (38), the sign ‘+’ is chosen if v 0 · û ≤ 0. If v 0 · û > 0, both signs ‘+’ and ‘−’ are possible.
In this case, we need to verify the solution using (37). It can be shown by contradiction that a
unique solution exists as long as v 0 and ω 0 are not orthogonal (which is true given the parabolic
trajectory).
2.2
Line Segment
When the sliding trajectory is a line, we infer that v 0 and ω 0 must be orthogonal to each other.
Let the unit vector î be in the direction of x2 − x0 , and ĵ such that î × ĵ = ẑ. Write v 0 = v0 î,
ω 0 = ω0 ĵ, and xk = xk î for k = 0, 1, 2. Equations (24) and (27) hold for x1 − x0 and x2 − x1 ,
respectively.
To solve for v0 and ω0 , first from (27) we obtain
5v0 + 2Rω0 = C,
where
 p
 70µr g(x2 − x1 ),
C=
 p
(39)
if x2 − x1 ≥ 0;
(40)
− 70µr g(x1 − x2 ), if x2 − x1 < 0.
From (24) a constant D ≡
49
2 µs g(x1
− x0 ) is rewritten as follows:
D = |v0 − Rω0 |(6v0 + Rω0 )
5 5
C
C
= v0 − + v0 6v0 + − v0
2
2
2
2
7
C 7
C
= v0 − v0 +
2
2 2
2

1
49
2
2

4 v0 − 4 C , if Rω0 ≤ v0 ;
=
 1 C 2 − 49 v 2 , if Rω > v .
0
0
4
4 0
by (39)
(41)
(42)
Hence we solve the above for v0 :
√
C 2 + 4D, if Rω0 ≤ v0 ;
v0 =
 1 √C 2 − 4D, if Rω > v .
0
0
7


1
7
10
(43)
The initial angular velocity follows from (39):
ω0 =
1
(C − 5v0 ).
2R
(44)
We need to hypothesize Rω0 ≤ v0 and Rω0 > v0 respectively. Then check if the values of v0 and
ω0 according to (43) and (44) satisfy the original hypothesis. Multiple solutions may exist.
References
[1] Wayland C. Marlow. The Physics of Pocket Billiards. Marlow Advanced Systems Technologies,
1994.
11