UNIT III. CS1117 VIRTUAL REALITY 7 th SEM CSE

Transcription

1 UNIT III CS1117 VIRTUAL REALITY 7 th SEM CSE

2 VIRTUAL ENVIRONMENT The Dynamics of Numbers Numerical Interpolation Linear Interpolation Non-Linear Interpolation The Animation of Objects Linear Translation Non-Linear Translation Shapes and Objects in between

3 The Dynamics of Numbers Virtual Environment A Complex Numerical DB It changes one number to another Important: Numerical Envelope of the change Example: In animating the bouncing of a ball, it s centroid is used to guide its motion Realistic motions are simulated by numerical interpol ation Numerical Interpolation Linear and Non-Linear

4 Numerical Interpolation Interpolation produces a function that matches the given data exactly. The function then can be utilized to approximate the data values at intermediate points It is also used to produce a function for which values are known only at discrete points, either from measu rements or calculations.

5 Numerical Interpolation Given data points Obtain a function, P(x) P(x) goes through the data points Use P(x) To estimate values at intermediate points For example, given data points: At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8 Find the following: At x = 4, y =?

6 Numerical Interpolation y P(x) 8? x

7 Linear Interpolation Linear Interpolation - Simplest interpolation method Apply LI(a sequence of points) A polygonal line where each straight line segment connects two consecutive points of the sequence Therefore, for every segment (P,Q) P(x) = (1 - x)p + xq where x ϵ [0,1] By varying x from 0 to 1, we get all the intermediate points between P and Q P(x) = P for x = 0 and P(x) = Q for x = 1. We get points on the line defined by P, Q, when 0 x 1

8 Non-Linear Interpolation A linear interpolation ensures that equal steps in the parameter x give rise to equal steps in the interpolated values It is often required that equal steps in x give rise to unequal steps in the interpolated values We can achieve this using a variety of mathematical techniques For example, we could use trigonometric functions or polynomials to achieve this

9 Trigonometric interpolation sin 2 (x) + cos 2 (x) = 1 If x varies between 0 and π/2 cos 2 (x) varies between 1 and 0, sin 2 (x) varies between 0 and 1 which can be used to modify the two interpolated values n1 and n2 as follows n=n1cos 2 (t)+n2sin 2 (t) for 0 t π/2

10 Trigonometric interpolation

11 Trigonometric interpolation Let n1 = 1 and n2 = 3

12 Cubic interpolation

13 Cubic interpolation

14 Cubic interpolation

15 The Animation of Objects Newton s Laws of Motion provide a useful framework to predict an object s behavior under dynamic conditions This scenario can be simulated within a Virtual Environment It can be achieved through a simple linear translation of objects

16 Modeling transformations build complex models by positioning simple components transform from object coordinates to world coordinates Viewing transformations placing the virtual camera in the world i.e. specifying transformation from world coordinates to camera coordinates Animation Uses of Translation vary transformations over time to create motion

17 Linear Translation Consider an object located at VE s origin and assigned a speed S 0 across the XY plane To simulate its sliding movement, the x & z coordinates of the object must be modified (t now t prev )V 0 The object s new velocity can be computed after it is bouncing off the boundary.

18 Linear Translation

19 Non-Linear Translation Consider an object moving along the x-axis in 1s, pause momentarily and then returns to its original position in 2s The non-linear movement of the object can be simulated by computing the x-translation as a function of time At time t1, the translation begins. At time t2, i.e.,(t1+1) it pauses momentarily and at time t3, i.e., (t1+3), it comes to rest t = T t1 while t1 T t2 T Current time t = (T t1 1)/2 while t2 T t3 t- Control parameter

20 Orientation of an object in computer graphics is often described using Euler Angles Orientation of an object in computer graphics is often described using Euler Angles Z R x R y R z Any axis order will work and could be used. Yaw, Pitch & Roll Euler Angles Y X

21 Problem with Euler Angles Rotations not uniquely defined Ex: (z, x, y) [roll, yaw, pitch] = (90, 45, 45) = (45, 0, -45) takes positive x-axis to (1, 1, 1) Cartesian coordinates are independent of one another, but Euler angles are not Remember, the axes stay in the same place during rotations Gimbal Lock Term derived from mechanical problem that arises in gimbal mechanism that supports a compass or a gyro Second and third rotations have effect of transforming earlier rotations, we lose a degree of freedom ex: Rot x, Rot y, Rot z If Rot y = 90 degrees, Rot z is equivalent to -Rot x With Euler Angles, knowing the transformations for the entire rotation does not help us with the intermediate rotations. Each is a unique set of three rotations about the x,y and z axes

22 Quaternions Invented by Sir William Hamilton (1843) Do not suffer from Gimbal Lock Provide a natural way to interpolate intermediate steps in a rotation about an arbitrary axis Are used in many position tracking systems and VR software support systems

23 Quaternions You can think of quaternions as an extension of complex numbers where there are three different square roots of -1. q = w + i x + j y + k z where i = (-1), j = (-1), k = (-1) i*j = k, j*k = i, k*i = j j*i = -k, k*j = -i, i*k = -j You could also think of q as a value in four-dimensional space, q = [w, x, y, z] Sometimes written as q = [w, v] where w is a scalar and v is a vector in 3-space

24 Shape and Object Inbetweening Shape Inbetweening It is a technique in the cartoon industry to speed up the process of creating art works The inbetween objects are derived from two key images drawn by a skilled animator The images are called as Key Frames The key frames shows an object in two different positions The number and position of inbetweenig images determines the dynamics of final animation

25 Shape Inbetweening

26 Object Inbetweening By inbetweening the z-coordinate, the above technique can be applied to 3D contours We cannot expect to interpolate between two different complex objects and geometrically consistent Given suitable geometric definitions, it is possible to transform one object into another and create subtle animation

27 Simple Pendulums, Springs & Flight Dynamics of an Aircraft

28 3. 7 Simple Pendulums Simple pendulums exhibit approximate simple harmonic motion Consists of heavy concentrated mass suspended by extensible cord Restricted to swing of up to ±14 o Given for Simulation M Mass of the pendulum L Cord Length Analyzing the forces Mass displaced by angle of θ radians Restoring force m sin θ attempts to reestablish equilibrium

29 3. 7 Simple Pendulums Contd When a Pendulum is displaced by an angle θ from the vertical position, a restoring force msinθ attempts to reestablish equilibrium. The future position of the pendulum is determined by computing its velocity and acceleration from this force

30 3. 7 Simple Pendulums Contd Acceleration a produced as a result of restoring force is given by a msin m / g a g sin.. ( 3.72) If angle of swing does not exceed ±14 o Sin θ is approximately equal to θ radians Therefore,.. ( 3.71) a g.. ( 3.73)

31 3. 7 Simple Pendulums Contd For smaller angles, Therefore, a g At any point in time t, Pendulum has displacement d d L d L Pendulum has Velocity v Pendulum has Acceleration a.. ( 3.74).. ( 3.75)

32 3. 7 Simple Pendulums Contd Therefore at time t+ Δt, The new displacement d is given by, d ' d vt.. ( 3.76) Acceleration is given by, a g d L.. ( 3.77) New Velocity v is given by, v ' v at.. ( 3.78)

33 3. 7 Simple Pendulums Contd If equations for displacement, velocity and acceleration are iterated The pendulums motion is simulated The Pendulum is animated by rotating the object description through angle relative to the upright position where θ = d / L

34 3. 8 Springs Object of mass m attached to the free end of the spring o Spring bounces up and down with Simple Harmonic motion Law s that describe active forces o Newton s Second Law of motion o Hooke s Law According to Hooke s law the tension T in a spring is given by e T.. ( 3.81 ) l Where λ is the modulus of the spring and e is the extension

35 3. 8 Springs Contd Simulation of Motion of an object attached to a Spring / Elastic We can analyze the dynamic forces acting upon the object If a mass m is attached to a light spring it will oscillate with simple harmonic motion

36 3. 8 Springs Contd During up and down motion, motion is controlled by ma mg T Where, m is the Object s Mass a is the acceleration of the Object g is the acceleration due to Gravity At any time t, Mass has position y Velocity v & Acceleration a.. ( 3.82 )

37 At time t+δt, 3. 8 Springs Contd New position y is given by y ' y vt.. ( 3.83 ) Extension e is given by, e y l.. ( 3.84 ) Given, e ma mg l Acceleration a is equal to a g e lm.. ( 3.85 ).. ( 3.86 )

38 3. 8 Springs Contd New velocity v of the mass is equal to: v ' v at.. ( 3.87 ) If the value of v is substituted in Eqn (3.83), the trace of y is simple harmonic Value of y can be substituted to subject of an object to this motion Useful feature Model can be dynamically modified while running Any parameters can be interactively interchanged Result : Instantaneous change in motion

39 Springs - Restitution No loss in the model Simple Harmonic motion continues indefinitely If the motion is required to decay at zero, Restitution term k d can be introduced in 3.87 v ' kdv at.. ( 3.88 ) When, k d = 1, the motion continues unchecked k d < 1, the motion decays away

40 Springs Moving Springs Spring of length l connected to a wheel which can travel over a surface Mass m placed over the spring top When spring s restitution is less than unity (k< 1) and mass placed on the spring Spring oscillates and settle down to state of Equilibrium Mass will oscillate only when subjected by external force Introducing a force Roll the wheel over an undulating terrain As wheel rises and falls Spring is compressed and stretched modulating the spring tension transmitted to the mass

41 Springs Moving Springs Contd Consider the image given below: If a mass m is attached to a light spring it will move with Simple Harmonic motion and settle down to a state of Equilibrium As the Wheel rises and falls, the mass attached to the free end of the spring will subject the spring to extra forces caused by accelerating the mass

42 Springs Moving Springs Contd h Extra string compression Initially set to 0 (Zero) Difference between terrain height sampled at t + Δt and t Introduced into the spring extension equation as follows: e y l h.. ( 3.89 ) Where, y is the height of the spring before adding mass The new equation then becomes y ' y vt e y l h

43 Springs Moving Springs Contd a g e lm v ' kdv at.. ( 3.90 ) h h new h Where, h new is the next value of h sampled at t + Δt This algorithm produces very realistic results

44 Springs Elastic Structures Use of FDD s by animators to form flexible objects If elasticity is introduced Control points of FDD s must be moved with suitable dynamics Fabrics modeled as a matrix of mass points connected together in a mesh Each mesh connected to a neighboring mass using springs Force acting on the springs related to the masses they act upon (Use Newton s Second law) Analysis Consider the dynamic simulation of an elastic thread fixed at both ends

45 Springs Elastic Structures Contd..

46 Springs Elastic Structures Contd..

47 Springs Elastic Structures Contd.. Analysis Contd Represented approximately by a sequence of mass points m Connected to their neighbors with springs of length l Imagine the middle of the thread raised and allowed to fall under gravity According to Hooke s law The spring s tension is proportional to the applied force Position s of the three mass points are specified by (x i-1, y i-1 ),, (x i, y i ) and (x i+1, y i+1 )

48 Springs Elastic Structures Contd.. Analysis Contd The left handed and right handed l L, l R will be: ) ( ) ( i i i i L y y x x l ) ( ) ( i i i i R y y x x l The left handed and right handed spring extensions e L, e R will be: l l e L L l l e R R Identify the horizontal and vertical components of the forces. The spring s angles are given by :

49 Springs Elastic Structures Contd.. Analysis Contd cos ( i xi 1 x ) / l L sin ( i yi 1 y ) / l L cos ( x x ) / l i1 i R sin ( y y ) / l i1 i R As force exerted by spring is proportional to its extension and stiffness S, The horizontal forces acting on the i th mass are (Use Newton s second law of motion) ma h S e R cos e cos L

50 Springs Elastic Structures Contd.. Analysis Contd Therefore, L i i L R i i R h l x x l l l x x l l m S a 1 1 g l y y l l l y y l l m S a L i i L R i i R v 1 1 Horizontal Acceleration Vertical Acceleration Determine the x and y coordinates that satisfies horizontal and vertical acceleration Use Euler s method for numerical integration Assume that the mass points move small distances in time

51 Springs Elastic Structures Contd.. Analysis Contd Acceleration is a measure of rate of change of velocity Change of velocity is a measure of rate of change of displacement with time Horizontal acceleration can be defined in terms of horizontal velocity as follows: Where, a h v t is any point in time h t t v t t Δt is small increment in t h

52 Springs Elastic Structures Contd.. Analysis Contd Velocity at time t + Δt is given by: v h t t vht tah It can also be defined as: v h t t t xt t The new horizontal position for the mass point is given by: x t t xt tv (t) h

53 Springs Elastic Structures Contd.. Analysis Contd When t = 0, The horizontal velocity and acceleration are zero: v h t 0 0 t 0 0 a h When this algorithm is implemented We discover that the central mass bounces around with a realistic sense of elasticity There is no opportunity for energy loss The system refuses to decay

54 3. 9 Flight Dynamics of an Aircraft Simulating the dynamic behavior of a 100 ton aircraft Non trivial exercise Numerical models used to simluate forces are complex Illustration Develop a simple flight model used to fly an aircraft In a Virtual Environment Aerodynamics Characteristics are affected by Weight, Speed, Altitude, Centre of Gravity, Angle of Attack, Flight Surface Geometry and Ground Proximity effects Forces transmitted during take off and landing Determined by Mechanical, Hydraulic and Pneumatic Assemblies

55 3. 9 Flight Dynamics of an Aircraft Different types of aero engines possess different dynamic Characteristics: Thrust profiles, Capacities Fuel Burning Differing Temperature / Efficiency Characteristics Differing response patterns Atmospheric Model reflects variation of air density and temperature with height Wind speed affects a plane s true ground speed Wind Shear creates highly dangerous flying conditions

56 3. 9 Flight Dynamics of an Aircraft

57 3. 9 Flight Dynamics of an Aircraft Traditional methods of representing the axial frame of reference for an Aircraft Right Handed rule Z axis directed downwards First Order Model for Aircraft Approximation Assumptions Point without Mass Travelling with speed v Body fixed at z axis Velocities v x and v y in the x and y axis respectively Heading is specified by angle between z axis and WCS

58 A Simple Aircraft Model A First Order Flight model for an Aircraft assumes that the craft is a point without mass, with instantaneous velocities v, v x, v y This Plan elevation shows the definition of the heading angle θ

59 Given A Simple Aircraft Model Aircraft s position (x, y, z) Speed v Acceleration a Heading Ø New position (x, y, z ) after time Δt can be computed as follows Incremental distance Δd moved forward is given by: d vt 1 2 at 2 The corresponding increments Δd and Δz in the x and z axes are given by:

60 A Simple Aircraft Model x d sin z d cos Then, x ' x x z ' z z To compute a new altitude, air speed and heading Assume an agent (autopilot) provides new Demand altitude y a Demand speed v d Demand Heading Ø d

61 A Simple Aircraft Model Altitude New altitude y can be computed by determining the altitude error y err y err y d y New vertical velocity v y becomes, v y ' k v y err V des v y ' v cli Where, k v is some gain factor -V des is the maximum descent rate V cli is the maximum climb rate

62 A Simple Aircraft Model Altitude Contd If y err is below threshold value, altitude acquired is disabled y v y t New altitude y becomes, y ' y y

63 A Simple Aircraft Model Acceleration New acceleration a can be computed by determining the air speed error v err v err v d v New acceleration in the direction of travel then becomes, a ' k v a err A dec a ' A acc Where, k v is some gain factor -A dec is the maximum deceleration permitted A acc is the maximum acceleration rate permitted

64 A Simple Aircraft Model Acceleration Contd If v err is below threshold value, airspeed acquired is disabled. New airspeed is given by: v ' v at

65 A Simple Aircraft Model Heading We establish the Error between the current heading and demand heading err d New lateral velocity v x becomes, v ' x k err V v x V ' Where, k v is some gain factor V is the maximum lateral velocity M is the maximum turn rate

66 A Simple Aircraft Model Heading Contd The change in heading is given by, tan 1 v x v The new heading is given by, ' ' After time Δt the aircraft has, Position (x, y, z ) Speed v Acceleration a Heading Ø

67 A Simple Aircraft Model After time Δt the aircraft has, Position (x, y, z ) Speed v Acceleration a Heading Ø These become the new values of (x, y, z), v, a and Ø respectively