Video Game Math:

1
Video Game Math

• Circle-on-Circle collision detection

2
Sample Game 2D Air Hockey
3
Like a movie, the action of a video game happens
in frames.
4
Like a movie, the action of a video game happens
in frames.
5
Like a movie, the action of a video game happens
in frames.
6
Like a movie, the action of a video game happens
in frames.
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The math problem

1. In a given frame, how do we tell when the puck
   hits the paddle?

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What are we given?

• The start and end points of the pucks centerP0
  and P1
• The start and end points of the paddle (stick)s
  center S0 and S1
• The radius of the stick and puck R and r

S0
S1
P1
P0
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Points are Coordinate Pairs

• Each of the four points we are given is expressed
  as an (x,y) pair.
• In other words
• P0 (P0x, P0y)

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Circle Intersection is Easy!

• Circles overlap if
• D lt R r
• We can find D
• D2 (Px - Sx)2 (Py - Sy)2

r
P
D
S
R
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Circle Intersection isnt Enough!

• We need to look at the whole path of both objects!

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A Simpler Problem

• What if the puck is very small, and the stick
  never moves?

P1
P0
S
R
In other words, what if r 0 and S0 S1 ?
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Definitions Constructions

• Let L length of P0P1
• Lx P1x - P0x
• Ly P1y - P0y
• L2 Lx2 Ly2

P1
L
P0
S
R
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Definitions Constructions

• Let C length of P0S
• Cx Sx - P0x
• Cy Sy - P0y
• C2 Cx2 Cy2

P1
L
P0
S
C
R
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Definitions Constructions

• Let T angle P2P1S

P1
L
T
P0
S
C
R
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Definitions Constructions

• Let T angle P2P1S
• ?
• We werent given T!
• Well solve for it later.

P1
L
T
P0
S
C
R
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Equation for Line P0P1

• Introduce parameter t
• Px(t) P0x t(P1x - P0x)
• Py(t) P0y t(P1y - P0y)
• P(t) is the point ( Px(t), Px(t) )

P1
P(t)
L
T
P0
S
C
R
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Equation for Line P0P1

• Introduce parameter t
• Px(t) P0x tLx
• Py(t) P0y tLy
• P(t) is the point ( Px(t), Px(t) )

P1
P(t)
L
T
P0
S
C
R
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Parameter t is Time

• Px(t) P0x t(P1x - P0x)
• Py(t) P0y t(P1y - P0y)
• Beginning of frame
• P(0) P0
• End of frame
• P(1) P1

P1
P(t)
L
T
P0
S
C
R
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Distance Moved at Time t

• D(t) Lt
• (Lots of ways to derive this)

P1
P(t)
L
D(t)
T
P0
S
C
R
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Were Finally Ready to Solve!

• Suppose P(t) is the point of impact.
• Solve for t, the time of impact.

P1
P(t)
L
Lt
R
T
P0
S
C
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Using the Law of Cosines

• R2 (Lt)2 C2 - 2CLt cos T
• 0 L2t2 - 2tCL cos T C2 - R2

P1
P(t)
L
Lt
R
T
P0
S
C
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One more definition

• So far we have
• L2t2 - 2tCL cos T C2 - R2 0
• Let a CL cos T
• now
• L2t2 - 2at C2 - R2 0

P1
P(t)
L
Lt
R
T
P0
S
C
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Applying the Quadratic Formula

• So far we have
• L2t2 - 2at C2 - R2 0
• So

P1
P(t)
L
Lt
R
T
P0
S
C
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Examining the Discriminant

• Discriminant a2 - L2(C2 - R2)

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Examining the Discriminant

• Discriminant a2 - L2(C2 - R2)
• If Discriminant lt 0, no solutions

P1
P0
S
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Examining the Discriminant

• Discriminant a2 - L2(C2 - R2)
• If Discriminant lt 0, no solutions
• If Discriminant gt 0, two solutions
• We want the earlier solution.

P1
P0
S
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Examining the Discriminant

• Discriminant a2 - L2(C2 - R2)
• If Discriminant lt 0, no solutions
• If Discriminant gt 0, two solutions
• If Discriminant 0, one solution
• We want the earlier solution.

P1
P0
S
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Is t in range?

• We only collide if our time of entry is in the
  range 0,1.
• If t gt 1, impact comes too late.

S
P1
P0
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Is t in range?

• We only collide if our time of entry is in the
  range 0,1.
• If t gt 1, impact comes too late.
• If t lt 0, impact is in the past.

P1
P0
S
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Is t in range?

• We only collide if our time of entry is in the
  range 0,1.
• If t gt 1, impact comes too late.
• If t lt 0, impact is in the past.
• ... Or maybe we started intersecting.

P1
P0
S
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We still need to solve for T!

• Remember, a CL cos T
• We construct K
• By the law of cosines,
• K2 L2 C2 - 2CL cos T
• That is K2 L2 C2 - 2a

P1
L
K
T
P0
S
C
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We only need to solve for a.

• We have K2 L2 C2 - 2a
• We also know
• K2 (P2x - Sx)2 (P2y - Sy)2

P1
L
K
T
P0
S
C
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We only need to solve for a.

• We have K2 L2 C2 - 2a
• We also know
• K2 (P2x - Sx)2 (P2y - Sy)2
• We can also show
• P2x - Sx Lx - Cx
• P2y - Sy Ly - Cy

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• K2 (Lx - Cx)2 (Ly - Cy)2

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• K2 (Lx - Cx)2 (Ly - Cy)2
• Lx2 Cx2 Ly 2 - Cy2
• - 2LxCx - 2 LyCy

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• K2 (Lx - Cx)2 (Ly - Cy)2
• Lx2 Cx2 Ly 2 - Cy2
• - 2LxCx - 2 LyCy
• Lx2 Ly 2 Cx2 Cy2
• - 2LxCx - 2LyCy

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• K2 (Lx - Cx)2 (Ly - Cy)2
• Lx2 Cx2 Ly 2 - Cy2
• - 2LxCx - 2 LyCy
• Lx2 Ly 2 Cx2 Cy2
• - 2LxCx - 2LyCy
• K2 L2 C2 - 2LxCx - 2LyCy

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• We have
• K2 L2 C2 - 2LxCx - 2LyCy
• K2 L2 C2 - 2a
• 2a 2LxCx 2LyCy
• a LxCx LyCy

P1
L
K
T
P0
S
C
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Then, a bunch of algebra happens...

• We have
• K2 L2 C2 - 2LxCx - 2LyCy
• K2 L2 C2 - 2a
• 2a 2LxCx 2LyCy
• a LxCx LyCy
• AxBx AyBy AB cos T

P1
L
K
T
P0
S
C
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A Slightly Harder Problem

• The puck is a circle instead of a point.

r
P1
S
r
P0
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The Point of Impact

• The centers are exactly R r apart.

r
P1
r
R
S
r
P0
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The Point of Impact

• The pucks center lies on a circle
• radius R r

r
P1
r
R
S
r
P0
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Reduce the Problem

• Use our solution to the simpler case.

P1
R r
S
P0
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The Original Problem

• Two Moving Circles

S1
R
r
r
P1
P0
S0
R
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Change our Frame of Reference

• Just Imagine that the stick is stationary.

r
r
P1
P0
S0
R
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Change our Frame of Reference

• Consider the relative motion of the puck.

r
P0
S0
R
r
P1
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Once Again, Reduce the Problem

• Use our earlier solution.

r
P0
S0
R
r
P1
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Lets See That Again...

• How exactly do we find the relative motion?

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Lets See That Again...

• We need to subtract the motion of S from the
  motion of P.

S1
P2x P1x - (S1x - S0x) P2y P1y - (S1y - S0y)
r
P1
P0
S0
R
P2
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Now you can make an Air Hockey Game!

• (well, not quite)

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Other Problems to Solve

• Collision detection for puck vs walls.
• What happens as a result of the collisions.
• All that pesky programming stuff.

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The 3D Problem

• The math is the same, except
• Points are (x, y, z) triples.
• The objects are spheres, not circles.
• L2 Lx2 Ly2 Lz2
• a LxCx LyCy LzCz

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Questions?