CODEWITHSUNDEEP
javalibgdxbox2dgame-physicsprojectile
David Lasry
David Lasry

Reputation: 1407

Projectile velocity box2d

I want to launch projectiles from the bottom-right corner of the screen towards the left side of the screen. Now, I want the projectiles to fly with random velocities and angles according to the screen dimensions, just like that. Now, I know this is very simple but from some reason I can't manage to make this work.

Here is what I have tried so far:

My first try - Launch function

private void launchProjectile() {
        projectiles.peek().getBody().applyForce(projectiles.peek().getBody().getWorldVector(new Vector2(MathUtils.random(-20,-1*SCALAR_HEIGHT),
        MathUtils.random(2*SCALAR_HEIGHT,8*SCALAR_HEIGHT)).scl(MathUtils.random(3*SCALAR_HEIGHT,5*SCALAR_HEIGHT))), 
        projectiles.peek().getBody().getWorldCenter(), true);
        Gdx.app.log("System", String.valueOf(SCALAR_HEIGHT));
    }

Here is my second try - Launch function

private void launchProjectile() {
    float xVelocity;
    float yVelocity;
    xVelocity = (float) MathUtils.random(0,0)*SCALAR_WIDTH/2;
    yVelocity = (float) MathUtils.random(20,20)*SCALAR_HEIGHT;
    velocityProjectile.set(xVelocity,yVelocity); // Sets the velocity vector to the above values
    velocityProjectile.sub(projectiles.peek().getBody().getPosition());
    velocityProjectile.nor(); // Normalize the vector - Now it's fine and ready!
    // Sets the start velocity of the projectile Trajectory to the current velocity
    projectiles.peek().getBody().setLinearVelocity(velocityProjectile.scl(18+SCALAR_HEIGHT));
}

In both tries, the projectile flies way more than I need and it doens't take in consideration the screen size like it should.

Can you guys please tell me what is the right way to do this?

Thanks!!

Upvotes: 0

Views: 667

Answers (1)

iforce2d
iforce2d

Reputation: 8262

Start with this page: http://www.iforce2d.net/b2dtut/projected-trajectory

In the "How fast should it be launched to reach a desired height?" section, you can see how much vertical velocity will be required to make the projectile reach the top of the screen. So you would pick a random number less than that, to make sure it doesn't go off the top of the screen.

Next, in the "How high will it go?" section, you can see the formula to find out how many time steps it will take for the projectile to reach maximum height. It will then take the same amount of time to come back down to the starting height. For example, let's say it would take 60 time steps to reach maximum height. That means it would take 120 time steps to fall down again to the same height as it started. Then you can set the horizontal part of the launch velocity so that it cannot go outside the screen in 120 time steps.

Upvotes: 1

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