1. Game Physics and Projectiles Soon Tee Teoh CS 134

2. Interactive Systems and Events • In a game, the player interacts with the game through input devices: mouse, keyboard, joystick, steering wheel etc. • The OS manages user input • A user input generates an Interrupt at the hardware level • Interrupt gets converted into event, and put in queue at the OS level • The application controls how to handle this event stream

3. Game Loop and Events • In a game, we need to run the game loop continuously. • At the same time, we need to receive input from the user, for example, the user has pressed a button, or moved the joystick. • How to handle this? • Several options: • Polling: Periodically query input devices to find out if event has occurred. • Non-blocking: If nothing has happened, then query function exits • Waiting: Wait until event happens • Blocking: Event query function will wait until event happens • The Callback abstraction • This uses the polling method • The program registers some functions to be called whenever an event occurs. • When event occurs, the registered callback function is called to handle the event. • Otherwise, if no event occurs, then the idle function is called to run the game loop. • Alternatively, game loop incorporates event handling while (1) { ProcessEvents(); UpdateAnimation(); Render(); } Game Loop

4. Game Loop and Physics • The Game Loop is a function that is called continuously. • For example, it can be the following code (same as previous page): • Alternatively, put game loop in idle function, which is called whenever there is nothing else going on (callbacks handle events, so game loop does not): • Thus, a game loop may look like: • How to update the position of an object? • If we know its position and velocity/acceleration etc. and we know the time elapsed since the last game loop, we can update the object’s position. while (1) GameLoop(); void idleFunc() { GameLoop(); } Update position, velocity, acceleration etc. according to physics model. Also, perform collision detection. Perform collision response if necessary. t may be different each time void GameLoop() { float t = TimeSinceLastCall(); for (i=0;i<nCharacters;i++) Characters[i].AI(); for (i=0;i<nObjects;i++) Objects[i].UpdateStatus(t); } need t as argument

5. Example: Object at Constant Velocity • Suppose in the previous game loop, an object is at position P. • Suppose that the object is traveling at constant velocity V. • Suppose that between the previous game loop and the current game loop, t amount of time has elapsed. • Then, the new position of the object is: P + Vt

6. Vector and Its Components • A vector has a magnitude and direction. • Example: Force and velocity are vectors. • Often, the vector is in 2D or 3D space, so we express the vector as (x,y) or (x,y,z). • We also express a vector pictorially as an arrow. • Vector addition: • Just add the components • Pictorially, sum the arrows • Splitting vector to components: • For convenience, we often consider a vector in terms of orthogonal components. • For example, we split vector V into Va and Vb. So, V = Va + Vb V = (2,1) V1 + V2 V2 V1 Vb V Va

7. Projectiles • At the starting point, an object is fired with a certain speed in a certain direction. • After that, there is no force acting on the object except gravity. • Given the initial position and velocity of the object, we wish to calculate its path. We also wish to calculate some other useful numbers.

8. Projectile Displacement Over Time v0 y q x Suppose that the projectile starts at (0,0). Suppose that the initial velocity of the projectile is v0. X is the horizontal direction, and Y is the vertical direction. Then, the distance in the x and y direction over time t is: x(t) = (v0 cos q) t y(t) = (v0 sin q) t – (gt2)/2 Here, g is the gravitational acceleration.

9. Projectile Velocity Over Time v0 y q x We can also calculate the velocity over time: Velocity in the x direction Vx(t) = v0 cos q Velocity in the y direction Vy(t) = v0 sin q – gt Overall velocity V(t) = sqrt(v02 – 2gtv0sin q + g2t2)

10. Projectile Maximum Height v0 y h q x We can also calculate the maximum height gained by the projectile: Maximum height h = (v02 sin2q )/ (2g)

11. Projectile Time to Impact • To calculate time to impact, we consider three cases: • Case 1: Target is of the same height as projectile launch. • Case 2: Target is higher than projectile launch. • Case 3: Target is lower than projectile launch.

12. Projectile Time to Impact Case 1: T = (2v0 sin q)/g Case 2: T = (v0 sin q)/g + sqrt(2(h-b)/g) h b Case 3: T = (v0 sin q)/g + sqrt(2(h+b)/g) h b

13. Projectile Total Horizontal Displacement Total horizontal displacement R = v0 T cos q R h b R h b R