Linear Targeting

As a is unknown you don't know the postion of the bullet at time t but you know it lies on a circle as it has travelled a distance v₁t from it's start. The equation of this is

(x - x₁)² + (y - y₁)² = (v₁t)²


(x₂ + v₂t cos b - x₁)² + (y₂ + v₂t sin b - y₁)² = (v₁t)²

This is a quadratic in t, so you should be able to solve it using the quadatic equation: it gives you in general two solutions. One of which may be negative so in the past, or they may be complex in which case there's no possible solution.

Once you have t you can substitute it into your original equations to get the position and so the angle.

Quote:

Through my own work I came up with this: t = ((y2 - y1) - (x2 - x1)) / (v2 cos b - v2 sin b) I would then take this t and use it to solve for my angle. However, it can't possibly be right. Two things jump out at me: if the target is not moving I will get a divide by zero, and if the target is at 45 or 225 degrees I will get a divide by zero.

You've eliminated v1 in order to get the expression for t. This means the v1 could have any value. Now, either you know your v1 and just need to find the angle, or you don't know your v1, and it is a free variable.

If the former, you've done something like this (where x and y are unkowns and a is known)

1) x = y * a
2) x = y + a

1) a = x/y
2) a = x - y

3) x/y = x - y
3) x = xy - y^2
3) x(1-y) = - y^2
3) x = y^2/(y-1)

In other words, you treated a constant as a varaible, isolated it, and subbed it back in. This doesn't actually help solve the problem as you haven't used the equations you have to isolate something that you can actually change. It's equivalent to trying to solve the problem using just one of given equations. In some cases, it does add in artificial singularities, like y = 1 in the above example, or the 45 degrees problem is your case (I think).


If, however, v1 really is an unknown, then you have 3 unknowns (v1, t and a), but only two equations, meaning something's going to be unspecified. In this case, you can vary v1 and a, so you can arbitrarily pick any t at which to satisfy your two given equations.


There is also the case where if v1 = v2, a = b, x1 = x2 and y1 = y2, then t is undefined, as no matter what t is used, the rays are coincident.


If v1 is known, what I'd suggest doing is isolate t in both equations you have, then equating them. Doing this gives me

(x2 - x1)/(v1*cos(a) - v2*cos(b)) = (y2 - y1)/(v1*sin(a) - v2*sin(b))

However actually solving this for a is rather difficult...

A better method might be to do what the first response post suggested, and find the time or point when the ray coming form (x2, y2) in direction (v1*sin(b), v1*cos(b)) intersects the circle of radius vt*t centred at (x1, y1). Knowing that you can get the necessary angle. If v1 is fixed though, there might not be such an intersection, as previous poster mentioned.

This is a solution I posted a while back:

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