Collision Detection and Bouncing

Similar documents
Collision Detection and Bouncing

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

ENGR 1990 Engineering Mathematics The Integral of a Function as a Function

0 for t < 0 1 for t > 0

e t dt e t dt = lim e t dt T (1 e T ) = 1

Version 001 test-1 swinney (57010) 1. is constant at m/s.

PHYSICS 1210 Exam 1 University of Wyoming 14 February points

4.8 Improper Integrals

Minimum Squared Error

Minimum Squared Error

FM Applications of Integration 1.Centroid of Area

( ) ( ) ( ) ( ) ( ) ( y )

MTH 146 Class 11 Notes

Physics 2A HW #3 Solutions

EXERCISE - 01 CHECK YOUR GRASP

1.0 Electrical Systems

September 20 Homework Solutions

PARABOLA. moves such that PM. = e (constant > 0) (eccentricity) then locus of P is called a conic. or conic section.

Forms of Energy. Mass = Energy. Page 1. SPH4U: Introduction to Work. Work & Energy. Particle Physics:

MATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)

CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

f t f a f x dx By Lin McMullin f x dx= f b f a. 2

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction.

3 Motion with constant acceleration: Linear and projectile motion

Chapter Direct Method of Interpolation

Contraction Mapping Principle Approach to Differential Equations

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

0 time. 2 Which graph represents the motion of a car that is travelling along a straight road with a uniformly increasing speed?

A Kalman filtering simulation

RESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π

Properties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)

Think of the Relationship Between Time and Space Again

(b) 10 yr. (b) 13 m. 1.6 m s, m s m s (c) 13.1 s. 32. (a) 20.0 s (b) No, the minimum distance to stop = 1.00 km. 1.

5.1-The Initial-Value Problems For Ordinary Differential Equations

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

P441 Analytical Mechanics - I. Coupled Oscillators. c Alex R. Dzierba

Available Online :

Average & instantaneous velocity and acceleration Motion with constant acceleration

ECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN)

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment

1. VELOCITY AND ACCELERATION

f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2

Physics Worksheet Lesson 4: Linear Motion Section: Name:

The solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.

Physics 101 Lecture 4 Motion in 2D and 3D

Some Inequalities variations on a common theme Lecture I, UL 2007

An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.

More on Magnetically C Coupled Coils and Ideal Transformers

graph of unit step function t

Physics 180A Fall 2008 Test points. Provide the best answer to the following questions and problems. Watch your sig figs.

ECE Microwave Engineering

From Particles to Rigid Bodies

REAL ANALYSIS I HOMEWORK 3. Chapter 1

S Radio transmission and network access Exercise 1-2

Motion in a Straight Line

() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).

Solutions to Problems from Chapter 2

PHY2048 Exam 1 Formula Sheet Vectors. Motion. v ave (3 dim) ( (1 dim) dt. ( (3 dim) Equations of Motion (Constant Acceleration)

Mathematics 805 Final Examination Answers

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

Let us start with a two dimensional case. We consider a vector ( x,

3D Transformations. Computer Graphics COMP 770 (236) Spring Instructor: Brandon Lloyd 1/26/07 1

ME 141. Engineering Mechanics

Vidyalankar. 1. (a) Y = a cos dy d = a 3 cos2 ( sin ) x = a sin dx d = a 3 sin2 cos slope = dy dx. dx = y. cos. sin. 3a sin cos = cot at = 4 = 1

Introduction to LoggerPro

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

KINEMATICS IN ONE DIMENSION

A LOG IS AN EXPONENT.

INDEX. Transient analysis 1 Initial Conditions 1

Physics 201, Lecture 5

Suggested Practice Problems (set #2) for the Physics Placement Test

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

CHAPTER 2 KINEMATICS IN ONE DIMENSION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

M r. d 2. R t a M. Structural Mechanics Section. Exam CT5141 Theory of Elasticity Friday 31 October 2003, 9:00 12:00 hours. Problem 1 (3 points)

Some Basic Information about M-S-D Systems

Instructor: Barry McQuarrie Page 1 of 5

MOMENTUM CONSERVATION LAW

Honours Introductory Maths Course 2011 Integration, Differential and Difference Equations

Lecture 4 Kinetics of a particle Part 3: Impulse and Momentum

Solutions from Chapter 9.1 and 9.2

Homework Assignment 6 Solution Set

Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration

Motion on a Curve and Curvature

/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2

2001 November 15 Exam III Physics 191

Differential Equations

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...

An object moving with speed v around a point at distance r, has an angular velocity. m/s m

BME 207 Introduction to Biomechanics Spring 2018

Phys 110. Answers to even numbered problems on Midterm Map

T-Match: Matching Techniques For Driving Yagi-Uda Antennas: T-Match. 2a s. Z in. (Sections 9.5 & 9.7 of Balanis)

3.6 Derivatives as Rates of Change

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Circuits 24/08/2010. Question. Question. Practice Questions QV CV. Review Formula s RC R R R V IR ... Charging P IV I R ... E Pt.

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

x(m) t(sec ) Homework #2. Ph 231 Introductory Physics, Sp-03 Page 1 of 4

Transcription:

Collision Deecion nd Bouncing Collisions re Hndled in Two Prs. Deecing he collision Mike Biley mj@cs.oregonse.edu. Hndling he physics of he collision collision-ouncing.ppx If You re Lucky, You Cn Deec he Collision Direcly (i.e., s closed-form soluion If You re Lucky, You Cn Compue he Collision Direcly XLEFT (x,y (vx,vy XRIGHT We need o nswer he quesion: Wh will cuse he ll o sop flying freely hrough he ir? The nswer is i will hi wll or he floor. From our rules of physics, he nex ounce will hppen one of he soluions o hese equions: XRIGHT RDIUS x v x XLEFT RDIUS x v x XRIGHT RDIUS x v x XLEFT RDIUS x v x YBOTTOM RDIUS y vy g The nex ounce will hppen one of he soluions o hese equions: XRIGHT RDIUS x v XLEFT RDIUS x v x x YBOTTOM RDIUS y vy g ( 0 g vy y YBOTTOM RDIUS vy vy g( y YBOTTOM RDIUS 3, 4 g YBOTTOM Noe: g < 0. How do you know which of he 4 vlues is he one h you should use? In he Generl Cse, here re Two Types of Collision Deecion Deecing Collisions Beween Two Ojecs. Discree B. Do s mny fs rejecions s you cn. Coninuous. Do hierrchicl fs rejecions 3. Discree: compre ll edges of Ojec gins ll fces of Ojec B B 3. Coninuous: cree pseudo-edges y connecing respecive poins in Ojec cross he ime sep, hen compre ll hese pseudo-edges of Ojec gins ll fces of Ojec B

Try o Simplify he Inersecion Tes Brek he Scene ino Grid Try o Simplify he Inersecion Tes-- Bounding Sphere Discree: You only hve o do inersecion ess gins ojecs h live in he sme grid squre. Try o Simplify he Inersecion Tes- Bounding Spheres Try o Simplify he Inersecion Tes-- Bounding Box R Ymx R C C These spheres overlp if: Disnce(C,C < R +R Xmin Xmx To void he squre roo: Disnce (C,C < (R +R Ymin Try o Simplify he Inersecion Tes- Bounding Boxes Quickly compre wo ojecs y fiing ech wih ounding ox nd hen compring he wo ounding oxes. Try o Simplify he Inersecion Tes- -- Two Types of Bounding Boxes xis-ligned Bounding Box (BB Check for overlp y looking for overlp in jus X, hen jus Y, hen jus Z These oxes do no overlp if: Xmx < Xmin Ymx < Ymin Xmx < Xmin Ymx < Ymin These oxes do overlp if: rirry-oriened Bounding Box (OBB This is igher fi round he ojec, u he overlp comprison is more involved Xmx Oregon > Xmin Se Universiy && Ymx > Ymin && Xmx > Xmin && Ymx > Ymin

Hierrchy of Bounding Boxes Discree -- You Cn Esily Tell if Poin is Inside Convex Polyhedron ssume h ll surfce normls poin ouwrds (usul convenion. Use he Disnce-from--Poin-o--Plne formul for ech fce of he polyhedron. If ll disnces re negive, he poin is inside he convex polyhedron Wh if he polyhedron is no convex? Pu Convex Hull round he polyhedron nd es gins h. If he poin is no inside he convex hull, hen i is no inside he polyhedron eiher. If i is inside he convex hull, hen more deiled nlysis is needed. P Disnce from Poin o Plne nˆ Discree nd Coninuous Compring n Edge on Ojec gins Fce on Ojec B P nˆ d Q Q P The equion of he line segmen is: P( P P 0 The equion of he plne is: x,y,z Q x,q y,q (n x,n y,n z 0 z which expnds ou o ecome he more fmilir x + By + Cz + D = 0 The disnce from he poin P o he plne is sed on he plne equion: dpq nˆ The do produc is nswering he quesion How much of (P-Q is in he norml direcion?. Noe h his gives signed disnce. If d > 0., hen P is on he sme side of he plne s he norml. P 0 If poin P wns o e poin in he plne, hen: P,P,P x y z Q x,q y,q (n,n,n x y z 0 z If we susiue he prmeric expression for P ino he plne equion, hen he only hing we don know in h equion is. Knowing * will le us compue he (x,y,z of he cul inersecion using he line equion. If * hs zero in he denominor, hen h ells us h *=, nd he line mus e prllel o he plne. This gives us he poin of inersecion wih he infinie plne. We would now use he mehod covered few slides go o see if P lies inside he ringle in quesion. n Is Poin inside Tringle? S Le: n( RQ ( SQ If Find Discree Inerference Do inry serch cross he ime sep unil you find he ime of collision P Q R If ( nn,( nn, nd( nn q r s re ll posiive, hen P is inside he ringle QRS n ( RQ ( PQ q n ( SR ( PR r n ( QS ( PS s 3

If Find Coninuous Inerference Connec ll poins cross ime nd look for he minimum in n inersecion wih he oundry The Physics of Collisions -- Definiions Line of Impc If he ojecs velociies re prllel o he Line of Impc, his is Direc Impc If he ojecs velociies re perpendiculr o he Line of Impc, his is Tngenil Impc cominion of he wo is clled n Olique Impc Direc Impc Tngenil Impc Olique Impc (oh direc nd ngenil The Physics of Collisions Fundmenl Quniies The Physics of Collisions Conservion of Momenum The momenum of n ojec is defined s is mss muliplied y is velociy: Momenum The energy of n ojec is defined s one hlf of is mss muliplied y is velociy squred: Energy In collision, he ol momenum fer he impc is equl o he ol momenum efore he impc. lwys. where he primes refer o velociies fer he impc This is referred o s he Conservion of Momenum Lw Momenum is lwys conserved hrough ny collision Conservion of Momenum s Explined y Newon s Crdle The Physics of Collisions Coefficien of Resiuion In collision, energy is conserved in he enire sysem, u no necessrily in he form of velociies. (I cn ecome permnen deformion, he, ligh, ec.. This loss of velociy is expressed s he Coefficien of Resiuion (COR. The COR, e, is how much less he relive velociies of he ojecs re fer impc hn hey were efore impc: hp://www.grnd-illusions.com/clog/newons_crdle.hml nd, of course, where would ny opic e wihou kiens? hp://dsc.discovery.com/videos/myhusers-newons-crne-crdle/ v v e( v v (he negive sign is here o indice he ounce hp://www.uzzfeed.com/koolnewslog/cs-for-he-love-of-physics-dgl 4

The Physics of Collisions Comining Momenum nd Resiuion Lws The Physics of Collisions wih Immovele Ojecs Sring wih hese wo equions: v v e( v v Tre he wo iniil velociies s inpus nd solve for he wo resuling velociies. This gives: em( vv v m m em( vv v m m To re he cse of mss eing n immovele ojec, such s he ground or solid wll, solve for he resuling velociies king he limi: lim em( vv lim v m m m em( v v lim m m m m m m m 0 v e( v v v ev Since mss is immovele, is velociy is zero, so h s pos-collision velociy is: m Collisions Experimenlly Deermining he Coefficien of Resiuion Velociies re hrd o mesure live, u disnces re no. So, drop he ojec from heigh h, nd mesure is ounce o heigh h : Before he ounce: fer he ounce: v gh 0 v gh 0 v gh v gh v e v Collisions Some Coefficiens of Resiuion of Blls Bounced on Concree Surfce Bll Meril CoR rnge golf ll 0.858 ennis ll 0.7 illird ll 0.804 hnd ll 0.75 wooden ll 0.603 seel ll ering 0.597 glss mrle 0.658 ll of ruer nds 0.88 hollow, hrd plsic ll 0.688 v gh e v gh h h hp://hyperexook.com/fcs/006/resiuion.shml The Physics of Collisions Tolly Plsic Collisions em( vv v m m em( vv v m m If e=0, hen he wo ojecs sick ogeher nd end up wih he sme resuling velociy: v v m m Very dmged B 5 mph One of my Jury Duies: Two vehicles collide. One is very dmged, he oher hrdly ll. Wh hppened? Who s righ? Hrdly dmged How much velociy chnge did ech cr undergo? (Δv is n pproxime mesure of dmge. v = 0. mph v = 5 mph =. m/sec m =.0 m =.0 e =.30 0 mph v ev.(.3 v 4.9 m/ sec m m 3 e.(.3* v.5 m/ sec m m 3 Δv = 4.9 0. = 4.9 m/sec =.0 mph Δv =.5. = 9.7 m/sec =.7 mph 5

The Physics of Collisions Tolly Elsic Collisions Wh hppens when e=? The wo fundmenl equions re Rerrnging: v v e( v v ( v v m v v m v v v v v v The Physics of Collisions Elsic Collisions Then, muliplying he wo ogeher gives: Or: m v v m v v v v v v This shows h energy is conserved when he Coefficien of Resiuion is.0 The Physics of Collisions Olique Impcs Voxelizion noher wy o do Collision Deecion? Olique Impcs re hen hndled y using vecor mh o deermine he direc nd ngenil velociy componens wih respec o he Line of Impc. The direc componens re chnged using he equions we jus derived. The ngenil componens re lef unchnged. (This ssumes no fricion. The new componens re hen comined o produce he resuling velociy vecors. Rndy Ruwendl 6