Basic Physics
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Basic Physics
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Basic Physics - Transcript
CSCE 590E Spring 2007
Basic Physics
By Jijun Tang
Announcements
Final game demo will be held at 2:00pm,
Tuesday, May 8th
Place: Amoco hall, open to the college
Pizza and drink will be served
Second presentation will be held on April 16th
and 18th
Give progress report
Show partial demos
Discuss problems
Modify designs
The Dot Product
The dot product is a product between
two vectors that produces a scalar
The dot product between two
n-dimensional vectors V and W is
given by
In three dimensions,
The Dot Product
The dot product can be used to project
one vector onto another
α
V
W
The Cross Product
The cross product between V and W is
A helpful tool for remembering this
formula is the pseudodeterminant
The Cross Product
The cross product satisfies the
trigonometric relationship
This is the area of
the parallelogram
formed by
V and W
α
V
W
||V|| sin α
The Cross Product
The area A of a triangle with vertices
P1, P2, and P3 is thus given by
Transformations
Suppose that the coordinate axes in
one coordinate system correspond to
the directions R, S, and T in another
Then we transform a vector V to the
RST system as follows
Transformations
Homogeneous coordinates
Four-dimensional space
Combines 3 × 3 matrix and translation
into one 4 × 4 matrix
Geometry
A line in 3D space is represented by
S is a point on the line, and V is the
direction along which the line runs
Any point P on the line corresponds to a
value of the parameter t
Two lines are parallel if their direction
vectors are parallel
( )t t= +P S V
Geometry
A plane in 3D space can be defined by
a normal direction N and a point P
Other points in the plane satisfy
P
Q
N
Geometry
Distance d from a point P to a line
S + t V
P
VS
d
Geometry
Use Pythagorean theorem:
Taking square root,
If V is unit length, then V 2 = 1
Geometry
Intersection of a line and a plane
Let P(t) = S + t V be the line
Let L = (N, D) be the plane
We want to find t such that L ⋅ P(t) = 0
Careful, S has w-coordinate of 1, and V
has w-coordinate of 0
x x y y z z w
x x y y z z
L S L S L S Lt L V L V L V
+ + +
= − = −
+ +�
L S
L V
Geometry
If L ⋅ V = 0, the line is parallel to the
plane and no intersection occurs
Otherwise, the point of intersection is
( )t = −
L SP S VL V
Real-time Game Physics
Introduction
Why Physics?
The Human Experience
Real-world motions are physically-based
Physics can make simulated game worlds
appear more natural
Makes sense to strive for physically-realistic
motion for some types of games
Emergent Behavior
Physics simulation can enable a richer gaming
experience
Or it can kill a game
Approaches
Classic approaches to creating realistic
motion:
Artist-created keyframe animations
Motion capture
Both are labor intensive and expensive
Physics simulation:
Motion generated by algorithm
Theoretically requires only minimal artist input
Potential to substantially reduce content
development cost
Physics in Digital Content
Creation Software
Many DCC modeling tools provide physics
Benefits:
Export physics-engine-generated animation as keyframe
data
Enables incorporation of physics into game engines that
do not support real-time physics
Straightforward update of existing asset creation pipelines
Problems:
Does not provide player with the same emergent-
behavior-rich game experience
Does not provide full cost savings to developer/publisher
Real-time Physics in Game at
Runtime:
Enables the emergent behavior that provides player
a richer game experience
Potential to provide full cost savings to
developer/publisher
Difficult
May require significant upgrade of game engine
May require significant update of asset creation pipelines
May require special training for modelers, animators, and
level designers
Licensing an existing engine may significantly
increase third party middleware costs
License vs. Build Physics
Engine
License middleware physics engine
Complete solution from day 1
Proven, robust code base (in theory)
Most offer some integration with DCC tools
Features are always a tradeoff
Build physics engine in-house
Choose only the features you need
Opportunity for more game-specific optimizations
Greater opportunity to innovate
Cost can be easily be much greater
No asset pipeline at start of development
Engines
Commercial
Game Dynamics SDK (Havok.com)
Renderware Physics (renderware.com)
NovodeX SDK (novodex.com)
Free
Open Dynamic Engine (ODE) (ode.org)
Tokamak Game Physics SDK
(tokamakphysics.com)
Newton Game Dynamics SDK
(newtondynamics.com)
Real-time Game Physics
The Beginning: Particle
Physics
Particle Physics
What is a Particle?
A sphere of finite radius with a perfectly smooth,
frictionless surface
Experiences no rotational motion (or assume the
sphere has no size)
Particle Kinematics
Defines the basic properties of particle motion
Position, Velocity, Acceleration
Location of Particle in World Space
SI Units: meters (m)
Changes over time when object moves
p ( t ) p ( t +
t )
Particle Position
zyx ppp ,,=p
Particle Velocity and
Acceleration
Velocity (SI units: m/s)
First time derivative of position:
Acceleration (SI units: m/s2)
First time derivative of velocity
Second time derivative of position
)()()(lim)(
0
tdt
d
t
tttt
t
pppV =
∆
−∆+
=
→∆
)()()( 2
2
tdt
dtdt
dt pVa ==
Newton’s 2nd Law of Motion
Paraphrased – “An object’s change in
velocity is proportional to an applied force”
The Classic Equation:
m = mass (SI units: kilograms, kg)
F(t) = force (SI units: Newtons)
( ) ( )tmt aF =
F=ma
What is Physics Simulation?
The Cycle of Motion:
Force, F(t), causes acceleration
Acceleration, a(t), causes a change in velocity
Velocity, V(t) causes a change in position
Physics Simulation:
Solving variations of the above equations over
time to emulate the cycle of motion
Example: 3D Projectile Motion
Constant Force
Weight of the projectile, W = mg
g: constant acceleration due to gravity (9.81m/s2)
Closed-form Projectile Equations of Motion:
These closed-form equations are valid, and
exact*, for any time, t, in seconds, greater than
or equal to tinit
( )initinit ttt −+= gVV )(
( ) ( ) 22
1)( initinitinitinit ttttt −+−+= gVpp
Example: 3D Projectile Motion
Initial Value Problem
Simulation begins at time tinit
The initial velocity, Vinit and position, pinit, at time tinit, are known
Solve for later values at any future time, t, based
on these initial values
On Earth:
If we choose positive Z to be straight up (away
from center of Earth), gEarth = 9.81 m/s2:
2m/s 81.9,0.0,0.0ˆ −=−= kgEarthEarthg
Concrete Example: Target
Practice
V i n i
t
F = w e i g h t = m gTarget
Projectile Launch
Position, pinit
Choose Vinit to Hit a Stationary Target
ptarget is the stationary target location
We would like to choose the initial velocity, Vinit, required to hit the target at some future time, thit.
Here is our equation of motion at time thit:
Solution in general is a bit tedious to derive…
Infinite number of solutions!
Hint: Specify the magnitude of Vinit, solve for its direction
Target Practice
( ) ( ) 22
1
inithitinithitinitinittarget tttt −+−+= gVpp
Choose Scalar launch speed, Vinit, and Let:
Where:
Solution
φφθφθ sin,cossin,coscos initinitinitinit VVV=V
( ) ( ) ( ) ( )
( ) ( )
( )θθ
φ
θθ
sincos
2
12
tan
sin ; cos
,,,,
2,,
22
2
2
,,
2
,,
,,
2
,,
2
,,
,,
+
+−+
=
−+
−±
=
−−−
−
=
−−−
−
=
xinityinitxtargetytarget
init
zinitztarget
initinit
yinitytargetxinitxtarget
yinitytarget
yinitytargetxinitxtarget
xinitxtarget
ppppA
A
V
g
ppV
AgV
AgAA
pppp
pp
pppp
pp
If Radicand in tanφ Equation is Negative:
No solution. Vinit is too small to hit the target
Otherwise:
One solution if radicand == 0
If radicand > 0, TWO possible launch angles, φ
Smallest φ yields earlier time of arrival, thit
Largest φ yields later time of arrival, thit
Rules
solution! no then ,02
12 if ,,
22
2 <
−+
− zinitztarget
initinit
ppV
AgV
AgA
969.31
Example 1
Vinit = 25 m/sValue of Radicand of tanφ equation:
Launch angle φ: 19.4 deg or 70.6 deg
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0.00 20.00 40.00 60.00
Horizontal Position (m)
Ve
rtic
al
Po
sit
ion
(m
) Projectile Launch
Position
Target Position
Trajectory 1 - High
Angle, Slow Arrival
Trajectory 2 - Low
Angle, Fast Arrival
Example 2
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0.00 20.00 40.00 60.00
Horizontal Position (m)
Ve
rtic
al
Po
sit
ion
(m
) Projectile Launch
Position
Target Position
Trajectory 1 - High
Angle, Slow Arrival
Trajectory 2 - Low
Angle, Fast Arrival
60.2Vinit = 20 m/sValue of Radicand of tanφ equation:
Launch angle φ: 39.4 deg or 50.6 deg
Example 3
290.4Vinit = 19 m/sValue of Radicand of tanφ equation:
Launch angle φ: No solution! Vinit too small to reach target!
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0.00 20.00 40.00 60.00
Horizontal Position (m)
Ve
rtic
al
Po
sit
ion
(m
) Projectile Launch
Position
Target Position
Trajectory with farthest
reach barely
undershoots target
Example 4
2063Vinit = 18 m/sValue of Radicand of tanφ equation:
Launch angle φ: 6.38 deg or 60.4 deg
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
40.00
45.00
0.00 20.00 40.00 60.00
Horizontal Position (m)
Ve
rtic
al
Po
sit
ion
(m
) Projectile Launch
Position
Target Position
Trajectory 1 - High
Angle, Slow Arrival
Trajectory 2 - Low
Angle, Fast Arrival
Real-time Game Physics
Practical Implementation:
Numerical Simulation
Beauty of Close form
Equations Presented Above
They are “closed-form”
Valid and exact for constant applied force
Do not require time-stepping
Just determine current game time, t, using system
timer
e.g., t = QueryPerformanceCounter /
QueryPerformanceFrequency or equivalent on Microsoft®
Windows® platforms
Plug t and tinit into the equations
Equations produce identical, repeatable, stable results,
for any time, t, regardless of CPU speed and frame
rate
Why Numerical Simulation?
The above sounds perfect
Why not use those equations always?
Constant forces aren’t very interesting
Simple projectiles only
Closed-form solutions rarely exist for interesting (non-
constant) forces
How should we handle wind?
We need a way to deal when there is no closed-
form solution…
Numerical Simulation
Techniques for incrementally solving the
equations of motion
when forces applied to an object are not
constant, or
when otherwise there is no closed-form
solution
Finite Difference Methods
What are They?
The most common family of numerical
techniques for rigid-body dynamics simulation
Incremental “solution” to equations of motion
Derived using truncated Taylor Series
expansions
See text for a more detailed introduction
“Numerical Integrator”
This is what we generically call a finite difference
equation that generates a “solution” over time
Finite Difference Methods
The Explicit Euler Integrator:
Properties of object are stored in a state vector, S
Use the above integrator equation to incrementally update S
over time as game progresses
Must keep track of prior value of S in order to compute the
new
For Explicit Euler, one choice of state and state derivative for
particle:
( ) ( ) ( ) )( 2
derivative state
stateprior statenew
tOtdt
dtttt ∆+∆+=∆+
SSS
pVS ,m= VFS ,=dtd
Explicit Euler Integration
F=Weight = mg Vinitpinit
Vinit = 30 m/sLaunch angle, φ: 75.2 deg (slow arrival)
Launch angle, θ: 0 deg (motion in world xz plane)
Mass of projectile, m: 2.5 kg
Target at <50, 0, 20> meters
tinit
Time p x p y p z mV x mV y mV z F x F y F z V x V y V z
5.00 10.00 0.00 2.00 19.20 0.00 72.50 0.00 0.00 -24.53 7.68 0.00 29.00
Velocity (m/s)Position (m) Linear Momentum (kg-m/s) Force (N)
mVinit
S =
∆ t = .01 s∆ t = .1 s
Explicit Euler Integration
=
=
=
−
∆+
=∆+=∆+
2900.2
0.0
0768.10
2549.72
0.0
2025.19
9000.4
0.0
7681.10
0476.72
0.0
2025.19
8000.7
0.0
5362.11
5951.67
0.0
2025.19
0.29
0.0
68.7
53.24
0.0
0.0
0.2
0.0
0.10
5.72
0.0
2.19
)()()( ttdt
dtttt SSS
∆ t = .2 s
=
=
=
2895.2
0.0
0768.10
2549.72
0.0
2.19
8510.4
0.0
1536.10
0476.72
0.0
2.19
6038.7
0.0
5362.11
5951.67
0.0
2.19
Solution form-Closed Exact,
Truncation Error
The previous slide highlights values in the numerical solution
that are different from the exact, closed-form solution
This difference between the exact solution and the numerical
solution is primarily truncation error
Truncation error is equal and opposite to the value of terms
that were removed from the Taylor Series expansion to
produce the finite difference equation
Truncation error, left unchecked, can accumulate to cause
simulation to become unstable
This ultimately produces floating point overflow
Unstable simulations behave unpredictably
Controlling Truncation Error
Under certain circumstances, truncation error
can become zero, e.g., the finite difference
equation produces the exact, correct result
For example, when zero force is applied
More often in practice, truncation error is
nonzero
Approaches to control truncation error:
Reduce time step, ∆ t
Select a different numerical integrator
Explicit Euler Integration –
Truncation Error
=
==∆
=
==∆
=
==∆
0005.0
0.0
0.0
2895.2
0.0
0768.10
-
2900.2
0.0
0768.10
0.01s)t(Error Truncation
049.0
0.0
0.0
8510.4
0.0
1536.10
-
9000.4
0.0
1536.10
0.1s)t(Error Truncation
1962.0
0.0
0.0
6038.7
0.0
5362.11
-
800.7
0.0
5362.11
0.2s)t(Error Truncation
exactnumerical
exactnumerical
exactnumerical
Truncation ErrorLets Look at Truncation Error (position only)
Explicit Euler Integration –
Truncation Error
(1/∆t) * Truncation Error is a linear (first
order) function of ∆ t: explicit Euler
Integration is FirstOrderAccurate in time
This accuracy is denoted by “O(∆ t)”
0.00
0.20
0.40
0.60
0.80
1.00
0.00 0.05 0.10 0.15 0.20 0.25
∆t(secs)
(1/
t )
* T
run
ca
tio
n E
rro
r
Explicit Euler Integration -
Computing Solution Over Time
0.00
10.00
20.00
30.00
40.00
50.00
0.00 20.00 40.00 60.00
Horizontal Position (m)
Ve
rtic
al
Po
sit
ion
(m
) Projectile Launch
Position
Target Position
Closed-Form
Explicit Euler
The solution proceeds step-by-step, each time
integrating from the prior state
Time p x p y p z mV x mV y mV z F x F y F z V x V y V z
5.00 10.00 0.00 2.00 19.20 0.00 72.50 0.00 0.00 -24.53 7.68 0.00 29.00
5.20 11.54 0.00 7.80 19.20 0.00 67.60 0.00 0.00 -24.53 7.68 0.00 27.04
5.40 13.07 0.00 13.21 19.20 0.00 62.69 0.00 0.00 -24.53 7.68 0.00 25.08
5.60 14.61 0.00 18.22 19.20 0.00 57.79 0.00 0.00 -24.53 7.68 0.00 23.11
M M M M M
10.40 51.48 0.00 20.87 19.20 0.00 -59.93 0.00 0.00 -24.53 7.68 0.00 -23.97
Velocity (m/s)Position (m) Linear Momentum (kg-m/s) Force (N)
Finite Difference Methods
The Verlet Integrator:
Must store state at two prior time steps, S(t) and S(t-∆ t)
Uses second derivative of state instead of the first
Valid for constant time step only (as shown above)
For Verlet, choice of state and state derivative for a particle:
pS = aFS == mdtd /22
( ) ( ) ( ) ( )
derivative state
2
22
2 stateprior 1 stateprior state new
)(2
∆+∆−−=∆+ tdt
dtttttt SSSS
Verlet Integration
a=<0,0,-g>p
S =
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