Jan 282011

I have been playing with target intersection prediction after I played a game called Beacon, where the enemies predict where you are going. After a bit of struggle, I got it to work accurately. This post at stackoverflow helped. I’m going to explain how it works more in depth.

Here we have a cannon on the ground, trying to hit a flying target. Let us assume the target is travelling with constant velocity, unaffected by gravity, and so is the projectile shot by the cannon.

The target will have a known velocity u. We will also know the speed of the projectile |v| in advance, and we want to find the vector v that will make it hit the target.

The idea is to transform the shooter and target to a new coordinate system where the line between them is one of the axes, match up the parallell speed, and then calculate the appropriate tangential speed for the projectile.

In order for the projectile to hit the target, the I composants of the velocities, vi and ui must be equal. If they are not, the projectile will be too fast or slow in that direction and cannot hit the target.

To calculate vi, first find the vector between A and B, and normalize it. Then project u onto AB using a dot product, giving uj. Subtracting uj from u, we get ui which is equal to vi.

Once vi has been determined, |vj| can be found by using Pythagoras theorem, since we know the magnitude |v|.

Then we can multiply the unit vector AB by |vj| to get vj, and lastly, add together vj and vi to get v


// Given: ux, uy, vmag (projectile speed), Ax, Ay, Bx, By

// Find the vector AB
ABx = Bx - Ax
ABy = By - Ay

// Normalize it
ABmag = sqrt(ABx * ABx + ABy * ABy)
ABx /= ABmag
ABy /= ABmag

// Project u onto AB
uDotAB = ABx * ux + ABy * uy
ujx = uDotAB * ABx
ujy = uDotAB * ABy

// Subtract uj from u to get ui
uix = ux - ujx
uiy = uy - ujy

// Set vi to ui (for clarity)
vix = uix
viy = uiy

// Calculate the magnitude of vj
viMag = sqrt(vix * vix + viy * viy)
vjMag = sqrt(vMag * vMag - viMag * viMag)

// Get vj by multiplying it's magnitude with the unit vector AB
vjx = ABx * vjMag
vjy = ABy * vjMag

// Add vj and vi to get v
vx = vjx + vix
vy = vjy + viy

It can be interesting to note that any vector v with vi=ui, and vj in the direction of the target, will make the projectile hit the target. For example the two purple lines in the diagram. But the only one with the correct magnitude is the v we have found.

This example is in 2D, but the same method can be used in 3D by considering the plane in which the target, cannon and target velocity lies.

Corner Cases
It is not always certain that the cannon can hit the target. This can be the case if the target is faster than the projectile, and moving away from it. For the hit to be possible u must be greater than vi at the time of fire. If u<vi the expression in the square root will be negative, and thus the root will be undefined, so you should have a check for that.

Flash example

Source code
Flashdevelop project


I have updated this article with some pseudo code, and to be much easier to understand.

Update 2, the same thing in 3D:

Unity3D example and source code


  • http://www.facebook.com/haxic Marcus Persson

    I tried to implement it, but I can only get it to work if the speed of the bullet and target fit… If either is either too slow or too fast there is a miss. Any suggestions?

    • http://Danikgames.com Danik

      Does the bullet lead the target or shoot straight at it? If you make the speed of the bullet much faster, like twice as fast as the target, do you get the same problem?
      It’s hard to tell what’s wrong without looking at the code. Could you paste it on pastebin or someplace?

  • Rodrigo legendre

    I don’t really understand how it would be applied to 3D. Particularly how to apply the 3rd axis to the equations. Can you provide an example for that?

    • http://Danikgames.com Danik

      The equations should be basically the same I believe. I can try to write a blogpost about it tonight.

    • http://Danikgames.com Danik

      I added a 3D example using Unity3D, hope it helps.

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  • peterpalacios79

    There is an assumption here. That the speed of the target is initially known. If it is not known you have to detect it. Also, the speed of the pursuer affects the angle of of take off.

  • peterpalacios79

    I apologize for my last message. This is programming, I didn’t notice this. I thought this was for robotics.

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  • Or Gal

    Any idea how to add drag to the equation? In case both the target and projectile have drag applied?

  • kip

    Holy peas, thanks so much. I’ve been trying to figure this out for an embarrassingly long time!