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Launch Angle Calculations Giving Wrong Result

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#1 nerdboy64   Members   -  Reputation: 515

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Posted 27 July 2014 - 03:44 PM

Recently I've been working on a top-down tank game in which you aim your shots with the mouse. Until now, the cannon would point directly at the mouse vector and not account for the force of gravity on the shots. I just changed it to use a formula that should give the required launch angle, but it isn't producing the correct result. I have tried two formulas: this one, from Wikipedia's page on projectile trajectories, and the one from this thread. Each produces different results, but both fail to hit the target vector. No actual projectiles exist yet, so instead I am drawing the expected trajectory using y0 + v0 * sin(a) * t + g * t^2 for the Y and x0 + v0 * cos(a) * t for the X. The trajectory aims in the right horizontal direction, but always either over- or undershoots the target.

 

One of three things could be happening here:

  • Both formulas are wrong and will never give the right angle.
  • One or both of the formulas is correct, but I've gotten units and/or degrees and radians mixed up.
  • The angle formulas are correct but the method I'm using to draw the trajectory is wrong.

 

I'm working on getting some screenshots to more accurately describe the problem.
 
My code:
 
Both angle calculation methods and trajectory prediction method:

//This is the formula from Wikipedia
	public static double calculateLaunchAngle(Vector3f position, Vector3f target, float velocity){
		double g = -9.81F;
		
		double range = Math.sqrt(Math.pow(position.x - target.x, 2) + Math.pow(position.z - target.z, 2));
		
		double height = target.y - position.y;
		double tangent = Math.pow(velocity, 2) + Math.sqrt(Math.pow(velocity, 4) - (g * ((g * range * range) + (2 * height * velocity * velocity))));
		tangent /= (g * range);
		double a = Math.atan(tangent);
		
		return a;
	};
	
	//This is the forume from the gavedev.net forums
	public static double calculateLaunchAngle2(float range, float height, float velocity){
		double g = -9.81F;
		
		double tangent = (velocity - Math.sqrt(Math.pow(velocity, 2) - (2 * g * (height + ((g * range * range) / (2 * velocity * velocity)))))) / (g / velocity);
		double a = Math.atan(tangent);
		
		return a;
	};
	
	//This method calculates the expected position of the shot at the given time in seconds
	public static Vector2f calculateShotTrajectory(double y0, double angle, double velocity, double time){
		double g = -9.81F;
		double height = y0 + (velocity * Math.sin(angle) * time) + (g * time * time);
		double distance = velocity * Math.cos(angle) * time;
		
		return new Vector2f((float) distance, (float) height);
	}

Implementation (gunX, gunY, and gunZ are the location of the breech of the gun relative to the world. I didn't use the muzzle because that would have required a lot more trig.):

gunAngle = -(float) Math.toDegrees(calculateLaunchAngle(new Vector3f(gunX, gunY, gunZ), target, velocity));
		//Elevation and depression limits disabled to more accurately view the trajectory
/*if(gunAngle < maxDepression) gunAngle = maxDepression;
		if(gunAngle > maxElevation) gunAngle = maxElevation;*/
		
		trajectoryPoints.clear();
		float t = 0;

//getting the trajectory in 2D and rotating it with some trig
		Vector2f v2f = new Vector2f(0, 10);
		for(t = 0; v2f.x < hDist || v2f.y > 0; t += 0.1F){
			v2f = calculateShotTrajectory(gunY, Math.toRadians(gunAngle), velocity, t);
			Vector3f v3f = new Vector3f(posX + (v2f.x * (float)Math.sin(Math.toRadians(targetAngle))), v2f.y, posZ + (v2f.x * (float)Math.cos(Math.toRadians(targetAngle))));
			trajectoryPoints.add(v3f);
		}
		trajectoryPoints.add(new Vector3f(posX + (hDist * (float)Math.sin(Math.toRadians(targetAngle))), target.y, posZ + (hDist * (float)Math.cos(Math.toRadians(targetAngle)))));

Edited by nerdboy64, 27 July 2014 - 05:20 PM.

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#2 Buckeye   Crossbones+   -  Reputation: 4430

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Posted 27 July 2014 - 07:00 PM

With regard to the wikipedia formula: are you checking both roots? It appears you're only calculating plus sqrt and not minus sqrt. Also, you need to check the value of the term v4 - g(...). Taking the sqrt of a negative number isn't ideal.


Please don't PM me with questions. Post them in the forums for everyone's benefit, and I can embarrass myself publicly.


#3 nerdboy64   Members   -  Reputation: 515

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Posted 27 July 2014 - 09:14 PM

With regard to the wikipedia formula: are you checking both roots? It appears you're only calculating plus sqrt and not minus sqrt. Also, you need to check the value of the term v4 - g(...). Taking the sqrt of a negative number isn't ideal.

I had been switching between the two to see if it made a difference and, while it changed the result, it was still wrong.

 

I actually solved this myself a little while ago, and when I figured out the problem I felt like a complete idiot. After rewriting the angle calculation to clean it up a bit, I went back over the method to predict the trajectory. Turns out I was doing y0 + (v0 * sin(angle) * t) + (g * t^2), when the last term should have been ... + 0.5(g * t^2). That's first month of high school physics stuff, and I'm rather disappointed that it took me that long to catch it.


My website: Brass Watch Games

Come check out Shipyard, my first real game project (Very WIP): Game Website Development Journal

Shipyard is a 2D turn-based strategy with a sci-fi theme, in which you build ships from individual parts rather than being given a selection of predefined models.






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