There are two cases for "shooting" at something:
- Non-Rotating Shooter
- Rotating Shooter
This article only covers the case of the non-rotating shooter. The second case is a bit more complicated. I'll post a separate article for that.
Your goal is to predict the future position of the target so that you can launch something at it and have them collide. It is given that you know the following:
- The position of the shooter when the projectile will be launched: \(\vec{P_s}\)
- The position of the target when the shooter will launch the projectile (i.e. now): \(\vec{P_T^0}\)
- The speed at which your projectiles travel: \(S_b\)
- The velocity of the target, \(\vec{v_T}\)
What you would like to find is \(\vec{P_T^1}\), the final position where the projectile will intercept the target.
We are going to have to assume that the target will travel at a constant velocity while the projectile is moving. Without this, the projectile will have to adapt after it has been launched (because it cannot predict the future position of the target if it is changing velocity in an unpredictable way). It should be pointed out that this is not limiting and in fact, still works out pretty well as long as the "edges" are handled well (target right in front of you or headed straight at you).
The Problem Space
Consider the image below:
This image shows the situation for the shooter. The target is moving with some constant velocity and the intercept position for the projectile, when launched, is unknown until the calculation is done. When the projectile is launched, it will intercept the target after \(t_B\) seconds. The projectile is launched with a constant speed. We don't know its direction yet, but we know the magnitude of its velocity, its speed will be \(S_b\).
If the target is at position \(\vec{P_T^0}\) now, it will be at position \(\vec{P_T^1}\), given by:
(1) \(\vec{P_T^1} = \vec{P_T^0} + \vec{v_T} * t_B\)
Since the projectile will have traveled for the same amount of time, it will have moved from \(\vec{P_s}\) to \(\vec{P_T^1}\) as well. In that time, it will have moved a distance of \(S_B x t_B\). Since we are talking about vector quantities here, we can write this as:
\(\mid\vec{P_T^1}-\vec{P_s}\mid = S_b * t_B\)
If we square both sides and break it into components to get rid of the absolute value:
(2) \((P_{Tx}^1 - P_{Sx})^2 +(P_{Ty}^1 - P_{Sy})^2 = S_b^2 * t_B^2\)
Breaking (1) into components as well and substituting back into (2) for the value of \(P_{Tx}^1\) and \(P_{Ty}^1\), we get the following:
\((P_{T0x} - P_{Sx} + v_{Tx}t_B)^2 + (P_{T0y} - P_{Sy} + v_{Ty}t_B)^2 = S_b^2 * t_B^2\)
For the sake of simplicity, we going to redefine:
\(P_T^0 - P_s = R \)(this is a constant)
After some algebra, this gives us the final equation:
\(t_B^2(v_{Tx}^2 + v_{Ty}^2-S_B^2) + t_B(2*R_x*v_{Tx} + 2*R_y*v_{Ty}) + (R_x^2 + R_y^2) = 0\)
This is a quadratic in \(t_B\):
\(t_b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where:
\( a =v_{Tx}^2 + v_y^2-S_B^2\)
\( b =2(R_x*v_{Tx} + R_y*v_{Ty})\)
\( c = R_x^2 + R_y^2\)
You can test the discriminant, \(b^2-4ac\):
< 0 \(\Rightarrow\) No Solution.
= 0 \(\Rightarrow\) One solution.
> 0 \(\Rightarrow\) Two solutions, pick the lowest positive value of \(t_B\).
Once you have solved the quadratic for \(t_B\), you can then substitute it back into (1) and calculate the intercept position, \(\vec{P_T^1}\).
The Code
Putting this together and covering some edge cases:
/* Calculate the future position of a moving target so that
* a projectile launched immediately can intercept (collide)
* with it.
*
* Some situations where this might be useful for an AI to
* make this calculation.
*
* 1. Shooting a projectile at a moving target.
* 2. Launching a football or soccer ball for to a player.
* 3. Figuring out the best position to jump towards in
* a platform game.
*
*
* The output value, solution, is the position that the
* intercept will occur at and the location that the
* projectile should be launched towards.
*
* The function will return false if a solution cannot
* be found. Consider the case of a target moving away
* from the shooter faster than the speed of the
* projectile and you will see at least one case where
* this calculation may fail.
*/
bool CalculateInterceptShotPosition(const Vec2& pShooter,
const Vec2& pTarget0,
const Vec2& vTarget,
float64 sProjectile,
Vec2& solution
)
{
// This formulation uses the quadratic equation to solve
// the intercept position.
Vec2 R = pTarget0 - pShooter;
float64 a = vTarget.x*vTarget.x + vTarget.y*vTarget.y - sProjectile*sProjectile;
float64 b = 2*(R.x*vTarget.x + R.y*vTarget.y);
float64 c = R.x*R.x + R.y*R.y;
float64 tBullet = 0;
// If the target and the shooter have already collided, don't bother.
if(R.LengthSquared() < 2*DBL_MIN)
{
return false;
}
// If the squared velocity of the target and the bullet are the same, the equation
// collapses to tBullet*b = -c. If they are REALLY close to each other (float tol),
// you could get some weirdness here. Do some "is it close" checking?
if(fabs(a) < 2*DBL_MIN)
{
// If the b value is 0, we can't get a solution.
if(fabs(b) < 2*DBL_MIN)
{
return false;
}
tBullet = -c/b;
}
else
{
// Calculate the discriminant to figure out how many solutions there are.
float64 discriminant = b*b - 4 * a * c;
if(discriminant < 0)
{ // All solutions are complex.
return false;
}
if (discriminant > 0)
{ // Two solutions. Pick the smaller one.
// Calculate the quadratic.
float64 quad = sqrt(discriminant);
float64 tBullet1 = (-b + quad)/(2*a);
float64 tBullet2 = (-b - quad)/(2*a);
if(tBullet1 < tBullet2 && tBullet1 >= 0)
{
tBullet = tBullet1;
}
else
{
tBullet = tBullet2;
}
}
else
{
tBullet = -b / (2*a);
}
}
// If the time is negative, we can't get there from here.
if(tBullet < 0)
{
return false;
}
// Calculate the intercept position.
solution = pTarget0 + tBullet*vTarget;
return true;
}
I have posted a working solution, with a simulation of using the above function and which you can tinker with, on github.
Article Update Log
28 Oct 2014: Initial Release
1 Nov 2014: Update discriminant description.
1 Nov 2104: Added code example.