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    <p>Well, if I understood you correctly, you wish to make some "asteroids" for your application.</p> <p>This is not specific to Android, but you probably need to define an asteroid as an entity in your application and when you need the asteroids you just create a random number of them with random positions (you may wish to check if there is already an asteroid or other object in that position to avoid collisions).</p> <p>Other than that, you just need to give each asteroid a velocity (in a 2D plane, an X and Y velocity) and update that accordingly in a loop as your application progresses.</p> <p>This is a simple example, but here goes:</p> <pre><code>//To make things easier, let's assume you have an Entity class, from which every game object is inherited public abstract class Entity { // Fields used to know object position private float x; private float y; // Fields used to calculate object motion private float x_speed; private float y_speed; ... // You would probably have a generic method to draw every entity - details are not relevant to your question, but you should draw the object taking it's x and y coordinates into account here public void draw() { ... } // Generic function to update the object's position regarding its speed public void updatePosition() { this.x += this.x_speed; this.y += this.y_speed; } ... } //Let's say you have an Asteroid class, which represents each asteroid public class Asteroid extends Entity { // Just add a constructor to set it's initial position and speed public Asteroid(float initial_x, float initial_y, float ini_x_speed, float ini_y_speed) { this.x = initial_x; this.y = initial_y; this.x_speed = ini_x_speed; this.y_speed = ini_y_speed; } } </code></pre> <p>From here on, you would just have to create a random number of Asteroid objects, with random positions and on your application's main loop call the updatePosition and draw methods for each entity.</p> <p>EDIT: Oh, and don't forget to "clear" what you've drawn at each loop cycle, so you won't see the already drawn objects in their old positions. :)</p>
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