Inverse trajectory determination-most ancient of computer probs.

#1 Not sure about your equations there......I don't have game gems 3, but a quick derivation gives me slightly different results....

The vertical location at time, t is given by the standard equation:

y = ut + 1/2*a*t^2

If the projectile is launched at an angle theta, the initial vertical velocity, u is equal to v*sin(theta). Where v is the initial speed. So:

y = v*sin(theta)*t + 1/2*g*t^2

The horizontal distance travelled, d, is (ignoring drag!) given by:

d = v*cos(theta)*t
=> t = d / (v*cos(theta))

substituting for t gives:

y = v*sin(theta)*d/(v*cos(theta)) + 1/2*g*d^2/((v*cos(theta))^2)
  = d*tan(theta) + 1/2*g*d^2/((v^2)*cos(theta)^2)

if we multiply the last term by (sin(theta)^2 + cos(theta)^2 == 1), we get

y = d*tan(theta) + g*d^2/(2*v^2) + g*d^2/(2*v^2)*tan(theta)^2

this is just a quadratic equation of the form:

ax^2 + bx + c

where:

x = tan(theta)
y = vertical distance between source and target
d = horizontal distance between source and target
a = g*d^2/(2*v^2)
b = d      // this is different to what you have
c = a - y  // this is different to what you have

As you are already doing, use the quadratic formula to get the roots, these will give you two possible firing angles, just choose one.

The major problem I can see with your code is that when you compute the distance using VectorDist, you should ignore the Z component, as you want the XY plane distance only.

Once you have your firing angle, you need to convert it to some initial velocities:

// theta = firing angle as determined above
// phi = angle to target on horizontal plane
phi = atan(target.x - src.x, target.y - src.y);
x = v*cos(theta)*sin(phi);
y = v*cos(theta)*cos(phi);
z = v*sin(theta);

#2 Wow, this is great. Thanks for posting this explanation. Great refresher since I havent messed wtih projectile physics in the last 3 years =-D

EDIT: just noticed - his quadratic formula seems odd - shouldnt it be (( -%b + % val)/(2*%a));

#3 @ Chris - You should update your profile to "Programmer - Advanced"

#4 Hrm. Ok I'm having trouble implementing this function. I get the theroy behind it, but the number just dont seem... right. heres my code block:

//////////////////////////////////////////////////////////
	//constants:
	%g = -9.81;
	%v = 20;
	%attackerPos = VectorAdd(%attacker.getPosition(), "1 1 1");
	%targetPos = %target.getPosition();
	//////////////////////////////////////////////////////////
	//nice to know stuff:
	%y = getWord( %targetPos,2) - getWord( %attackerPos,2);
	%d = VectorDist( VectorDot(%attackerPos, "1 1 0"), VectorDot(%targetPos, "1 1 0")) ;
	
	//ax^2 + bx + c
	%a = %g*%d*%d/(2*%v*%v);
	%b = %d;     
	%c = %a - %y;
	
	//////////////////////////////////////////////////////////
	//use quadratic formula:
	%discrim =  msqrt( (%b*%b) - (4 * %a * %c) );
	%x1 = (( -%b + %discrim)/(2*%a));
	%x2 = (( -%b - %discrim)/(2*%a));
	%theta1 =  mAsin(%x1/    ( msqrt( 1 + %x1*%x1 )) );
	%theta2 = mAsin(%x2/( msqrt( 1 + %x2*%x2 ))  );//mATan(1, %x2);

	//grab the largest angle, so we can see it obviously during testing
	if( (%theta2 > %theta1) )
	{
		%theta = %theta2;
	}
	else
	{
		%theta = %theta1;
	}
	//////////////////////////////////////////////////////////
	// theta = firing angle on z
	// phi = angle to target on horizontal plane - this seems a'ok
	%phi = mAtan(	getWord(%targetPos, 0) - getWord(%attackerPos, 0), getWord(%targetPos, 1) - getWord(%attackerPos, 1));
	%x = %v*mcos(%theta)*msin(%phi);
	%y = %v*mcos(%theta)*mcos(%phi);
	%z = %v*msin(%theta);
	
	echo( "Theta:" SPC %theta SPC "1:" SPC %theta1 SPC "2:" SPC %theta2 );
	echo( "X:" SPC %x SPC "Y:" SPC %y SPC "Z:" SPC %z );
	echo( "A:" SPC %a SPC "B:" SPC %b SPC "C:" SPC %c );

         %initVel  = %x SPC %y SPC %z;

The prblem is it seems to aim fine when they are on similar elevations - but once they cange elevations, I get crazy z values (like 29) which sends it flying over their head!

#5 Look at chris second equation. it assumes a height of zero at time=0. add another constant for initial height and allow the result propagate through the rest of your equations.

now someone answer my datablock questions!

#6 @John:

I'm not sure about this line:

%d = VectorDist( VectorDot(%attackerPos, "1 1 0"), VectorDot(%targetPos, "1 1 0"));

I assume you are trying to calculate the horizontal distance between %attackerPos and %targetPos. VectorDot will return a scaler, VectorDist expects two vectors. Also, the "1 1 0" vector you are using with VectorDot does not have length=1 so you are not getting the correct distance anyway.

Try using:

%x = getWord( %targetPos,0) - getWord( %attackerPos,0);
%y = getWord( %targetPos,1) - getWord( %attackerPos,1);
%d = msqrt(%x*%x + %y*%y);

You will need to change the name of your '%y' variable to eg. '%z'.

Why are you using:

%theta1 =  mAsin(%x1/    ( msqrt( 1 + %x1*%x1 )) );

instead of:

%theta1 =  mAtan(%x1, 1);

You will also want to check (before doing the square root) that your discriminant is not negative. If it is, there is no angle that would cause the projectile to hit the target. Likely cause is insufficient initial velocity.

#7 @Chris

Aha, excellant. Thats exactly what I needed. I had forgotten what dot product is actually. (blushes). Haven't had or used linear algebra in quite a while. =-)

I was using the sin ident instead of the atan x , 1 becaues i didnt realize that property of it.

However, now I know.

Thanks tons for all your work on this.

#8 Oh wow, i didnt realize this thread was alive...

Did y'all get this to work, as mine works fine, at least for what I need. Glad to see others using it. I should post my new code. The major difference from what I can remember is that the dist does need to be computed on the XY plane.

#9 Yeah this above code works after the changes Chris suggested. Furthermore iI added motion tracking by just adding the targets velocity to the initialVelocity of the projectile. I can post this in full if someone wants it, though its about 3 lines changed =-)

#10 Hi!

I was looking for something like this, but I had a different point of view. I thought I would have a given theta, say 45

#11 If anyone get's something like this working, you should post it as a resource.

#12 Hi Luis,

It is certainly a better idea to choose a theta angle, then solve for initial velocity. You can make a pretty good guess at the appropriate firing angle based on the angle between source and destination positions. To account for gravity, you'd generally make the firing angle a bit larger than the angle between the two objects.

eg. If the angle between the two is 65 degrees, you might choose to fire the projectile at 75 degrees. eg.

theta = mAtan(target.z - src.z, d);
// add a little to firing angle to account for gravity
theta += 0.1745;

It's then fairly straightforward to derive the equation to compute initial velocity:

//////////////////////////////////////////////////////////
// calculate difference vectors
%dx = getWord(%targetPos,0) - getWord(%attackerPos,0);
%dy = getWord(%targetPos,1) - getWord(%attackerPos,1);
%dz = getWord(%targetPos,2) - getWord(%attackerPos,2);

// d = distance to target on horizontal (xy) plane
%d = msqrt(%dx*%dx + %dy*%dy);

// phi = angle to target on horizontal (xy) plane
%phi = mAtan(%dx, %dy);

//////////////////////////////////////////////////////////
// determine firing angle
%theta = mAtan(%dz, %d);
%theta += 0.1745;  // add 10 degrees to account for gravity

//////////////////////////////////////////////////////////
// determine initial speed
%tanTheta = mTan(%theta);
%v = msqrt(%g*%d*%d * (1 + %tanTheta*%tanTheta) / (2*(%dz - %d*%tanTheta*%tanTheta)));

%vx = %v*mCos(%theta)*mSin(%phi);
%vy = %v*mCos(%theta)*mCos(%phi);
%vz = %v*mSin(%theta);

[edited for small error in formula - thanks, Luis!]

#13 Yes, ok, that was the idea :)

However, you need a minimum shooting angle. If you don't use it, and your target is a few good angles below you (say it's you, 'o' , then a plain, then a slope and the enemy, 'x'):

O
--------------------------------
----------------------------------
------------------------------------    X
--------------------------------------------------

and you use just Atan + 0.17 for the shooting angle, you will most probably shoot the floor before the slope. Anyway, that kind of targets are difficult to hit. You need a wide shooting angle.

And just a little error, isnt' it...

%v = msqrt(%g*%d*%d * (1 + %tanTheta*%tanTheta) / (2 * (%dz - %d*%tanTheta*%tanTheta)));

without that '2' the initial velocity is too high.

#14 I know there are other derivations of this general formula in game gems 3 and I'm sure elsewhere. I'll take a look this week and at least post how to solve for the initial velocity.

#15 Ok, I know its been a while, sorry, but here's the Vi solution:

initial velocity = ( x/ cos (theta) ) * squareroot ( grav/ ( 2 * ( x * tan(theta) - y) ) )

havent tested it, but this is what game gems claims!


-s