2d game : fire at a moving target by predicting in

问题:

Okay, this all takes place in a nice and simple 2D world... :)

Suppose I have a static object A at position Apos, and a linearly moving object B at Bpos with bVelocity, and an ammo round with velocity Avelocity...

How would I find out the angle that A has to shoot, to hit B, taking into account B's linear velocity and the speed of A's ammo ?

Right now the aim's at the current position of the object, which means that by the time my projectile gets there the unit has moved on to safer positions :)

回答1:

First rotate the axes so that AB is vertical (by doing a rotation)

Now, split the velocity vector of B into the x and y components (say Bx and By). You can use this to calculate the x and y components of the vector you need to shoot at.

B --> Bx
|
|
V

By


Vy
^
|
|
A ---> Vx

You need Vx = Bx and Sqrt(Vx*Vx + Vy*Vy) = Velocity of Ammo.

This should give you the vector you need in the new system. Transform back to old system and you are done (by doing a rotation in the other direction).

回答2:

I wrote an aiming subroutine for xtank a while back. I'll try to lay out how I did it.

Disclaimer: I may have made one or more silly mistakes anywhere in here; I'm just trying to reconstruct the reasoning with my rusty math skills. However, I'll cut to the chase first, since this is a programming Q&A instead of a math class :-)

How to do it

It boils down to solving a quadratic equation of the form:

a * sqr(x) + b * x + c == 0

Note that by sqr I mean square, as opposed to square root. Use the following values:

a := sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)
b := 2 * (target.velocityX * (target.startX - cannon.X)
          + target.velocityY * (target.startY - cannon.Y))
c := sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)

Now we can look at the discriminant to determine if we have a possible solution.

disc := sqr(b) - 4 * a * c

If the discriminant is less than 0, forget about hitting your target -- your projectile can never get there in time. Otherwise, look at two candidate solutions:

t1 := (-b + sqrt(disc)) / (2 * a)
t2 := (-b - sqrt(disc)) / (2 * a)

Note that if disc == 0 then t1 and t2 are equal.

If there are no other considerations such as intervening obstacles, simply choose the smaller positive value. (Negative t values would require firing backward in time to use!)

Substitute the chosen t value back into the target's position equations to get the coordinates of the leading point you should be aiming at:

aim.X := t * target.velocityX + target.startX
aim.Y := t * target.velocityY + target.startY

Derivation

At time T, the projectile must be a (Euclidean) distance from the cannon equal to the elapsed time multiplied by the projectile speed. This gives an equation for a circle, parametric in elapsed time.

sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
  == sqr(t * projectile_speed)

Similarly, at time T, the target has moved along its vector by time multiplied by its velocity:

target.X == t * target.velocityX + target.startX
target.Y == t * target.velocityY + target.startY

The projectile can hit the target when its distance from the cannon matches the projectile's distance.

sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
  == sqr(target.X - cannon.X) + sqr(target.Y - cannon.Y)

Wonderful! Substituting the expressions for target.X and target.Y gives

sqr(projectile.X - cannon.X) + sqr(projectile.Y - cannon.Y)
  == sqr((t * target.velocityX + target.startX) - cannon.X)
   + sqr((t * target.velocityY + target.startY) - cannon.Y)

Substituting the other side of the equation gives this:

sqr(t * projectile_speed)
  == sqr((t * target.velocityX + target.startX) - cannon.X)
   + sqr((t * target.velocityY + target.startY) - cannon.Y)

... subtracting sqr(t * projectile_speed) from both sides and flipping it around:

sqr((t * target.velocityX) + (target.startX - cannon.X))
  + sqr((t * target.velocityY) + (target.startY - cannon.Y))
  - sqr(t * projectile_speed)
  == 0

... now resolve the results of squaring the subexpressions ...

sqr(target.velocityX) * sqr(t)
    + 2 * t * target.velocityX * (target.startX - cannon.X)
    + sqr(target.startX - cannon.X)
+ sqr(target.velocityY) * sqr(t)
    + 2 * t * target.velocityY * (target.startY - cannon.Y)
    + sqr(target.startY - cannon.Y)
- sqr(projectile_speed) * sqr(t)
  == 0

... and group similar terms ...

sqr(target.velocityX) * sqr(t)
    + sqr(target.velocityY) * sqr(t)
    - sqr(projectile_speed) * sqr(t)
+ 2 * t * target.velocityX * (target.startX - cannon.X)
    + 2 * t * target.velocityY * (target.startY - cannon.Y)
+ sqr(target.startX - cannon.X)
    + sqr(target.startY - cannon.Y)
  == 0

... then combine them ...

(sqr(target.velocityX) + sqr(target.velocityY) - sqr(projectile_speed)) * sqr(t)
  + 2 * (target.velocityX * (target.startX - cannon.X)
       + target.velocityY * (target.startY - cannon.Y)) * t
  + sqr(target.startX - cannon.X) + sqr(target.startY - cannon.Y)
  == 0

... giving a standard quadratic equation in t. Finding the positive real zeros of this equation gives the (zero, one, or two) possible hit locations, which can be done with the quadratic formula:

a * sqr(x) + b * x + c == 0
x == (-b ± sqrt(sqr(b) - 4 * a * c)) / (2 * a)

回答3:

+1 on Jeffrey Hantin's excellent answer here. I googled around and found solutions that were either too complex or not specifically about the case I was interested in (simple constant velocity projectile in 2D space.) His was exactly what I needed to produce the self-contained JavaScript solution below.

The one point I would add is that there are a couple special cases you have to watch for in addition to the discriminant being negative:

• "a == 0": occurs if target and projectile are traveling the same speed. (solution is linear, not quadratic)
• "a == 0 and b == 0": if both target and projectile are stationary. (no solution unless c == 0, i.e. src & dst are same point.)

Code:

/**
 * Return the firing solution for a projectile starting at 'src' with
 * velocity 'v', to hit a target, 'dst'.
 *
 * @param Object src position of shooter
 * @param Object dst position & velocity of target
 * @param Number v   speed of projectile
 * @return Object Coordinate at which to fire (and where intercept occurs)
 *
 * E.g.
 * >>> intercept({x:2, y:4}, {x:5, y:7, vx: 2, vy:1}, 5)
 * = {x: 8, y: 8.5}
 */
function intercept(src, dst, v) {
  var tx = dst.x - src.x,
      ty = dst.y - src.y,
      tvx = dst.vx,
      tvy = dst.vy;

  // Get quadratic equation components
  var a = tvx*tvx + tvy*tvy - v*v;
  var b = 2 * (tvx * tx + tvy * ty);
  var c = tx*tx + ty*ty;    

  // Solve quadratic
  var ts = quad(a, b, c); // See quad(), below

  // Find smallest positive solution
  var sol = null;
  if (ts) {
    var t0 = ts[0], t1 = ts[1];
    var t = Math.min(t0, t1);
    if (t < 0) t = Math.max(t0, t1);    
    if (t > 0) {
      sol = {
        x: dst.x + dst.vx*t,
        y: dst.y + dst.vy*t
      };
    }
  }

  return sol;
}


/**
 * Return solutions for quadratic
 */
function quad(a,b,c) {
  var sol = null;
  if (Math.abs(a) < 1e-6) {
    if (Math.abs(b) < 1e-6) {
      sol = Math.abs(c) < 1e-6 ? [0,0] : null;
    } else {
      sol = [-c/b, -c/b];
    }
  } else {
    var disc = b*b - 4*a*c;
    if (disc >= 0) {
      disc = Math.sqrt(disc);
      a = 2*a;
      sol = [(-b-disc)/a, (-b+disc)/a];
    }
  }
  return sol;
}

回答4:

Jeffrey Hantin has a nice solution for this problem, though his derivation is overly complicated. Here's a cleaner way of deriving it with some of the resultant code at the bottom.

I'll be using x.y to represent vector dot product, and if a vector quantity is squared, it means I am dotting it with itself.

origpos = initial position of shooter
origvel = initial velocity of shooter

targpos = initial position of target
targvel = initial velocity of target

projvel = velocity of the projectile relative to the origin (cause ur shooting from there)
speed   = the magnitude of projvel
t       = time

We know that the position of the projectile and target with respect to t time can be described with some equations.

curprojpos(t) = origpos + t*origvel + t*projvel
curtargpos(t) = targpos + t*targvel

We want these to be equal to each other at some point (the point of intersection), so let's set them equal to each other and solve for the free variable, projvel.

origpos + t*origvel + t*projvel = targpos + t*targvel
    turns into ->
projvel = (targpos - origpos)/t + targvel - origvel

Let's forget about the notion of origin and target position/velocity. Instead, let's work in relative terms since motion of one thing is relative to another. In this case, what we now have is relpos = targetpos - originpos and relvel = targetvel - originvel

projvel = relpos/t + relvel

We don't know what projvel is, but we do know that we want projvel.projvel to be equal to speed^2, so we'll square both sides and we get

projvel^2 = (relpos/t + relvel)^2
    expands into ->
speed^2 = relvel.relvel + 2*relpos.relvel/t + relpos.relpos/t^2

We can now see that the only free variable is time, t, and then we'll use t to solve for projvel. We'll solve for t with the quadratic formula. First separate it out into a, b and c, then solve for the roots.

Before solving, though, remember that we want the best solution where t is smallest, but we need to make sure that t is not negative (you can't hit something in the past)

a  = relvel.relvel - speed^2
b  = 2*relpos.relvel
c  = relpos.relpos

h  = -b/(2*a)
k2  = h*h - c/a

if k2 < 0, then there are no roots and there is no solution
if k2 = 0, then there is one root at h
    if 0 < h then t = h
    else, no solution
if k2 > 0, then there are two roots at h - k and h + k, we also know r0 is less than r1.
    k  = sqrt(k2)
    r0 = h - k
    r1 = h + k
    we have the roots, we must now solve for the smallest positive one
    if 0<r0 then t = r0
    elseif 0<r1 then t = r1
    else, no solution

Now, if we have a t value, we can plug t back into the original equation and solve for the projvel

 projvel = relpos/t + relvel

Now, to the shoot the projectile, the resultant global position and velocity for the projectile is

globalpos = origpos
globalvel = origvel + projvel

And you're done!

My implementation of my solution in Lua, where vec*vec represents vector dot product:

local function lineartrajectory(origpos,origvel,speed,targpos,targvel)
    local relpos=targpos-origpos
    local relvel=targvel-origvel
    local a=relvel*relvel-speed*speed
    local b=2*relpos*relvel
    local c=relpos*relpos
    if a*a<1e-32 then--code translation for a==0
        if b*b<1e-32 then
            return false,"no solution"
        else
            local h=-c/b
            if 0<h then
                return origpos,relpos/h+targvel,h
            else
                return false,"no solution"
            end
        end
    else
        local h=-b/(2*a)
        local k2=h*h-c/a
        if k2<-1e-16 then
            return false,"no solution"
        elseif k2<1e-16 then--code translation for k2==0
            if 0<h then
                return origpos,relpos/h+targvel,h
            else
                return false,"no solution"
            end
        else
            local k=k2^0.5
            if k<h then
                return origpos,relpos/(h-k)+targvel,h-k
            elseif -k<h then
                return origpos,relpos/(h+k)+targvel,h+k
            else
                return false,"no solution"
            end
        end
    end
end

回答5:

Following is polar coordinate based aiming code in C++.

To use with rectangular coordinates you would need to first convert the targets relative coordinate to angle/distance, and the targets x/y velocity to angle/speed.

The "speed" input is the speed of the projectile. The units of the speed and targetSpeed are irrelevent, as only the ratio of the speeds are used in the calculation. The output is the angle the projectile should be fired at and the distance to the collision point.

The algorithm is from source code available at http://www.turtlewar.org/ .

// C++
static const double pi = 3.14159265358979323846;
inline double Sin(double a) { return sin(a*(pi/180)); }
inline double Asin(double y) { return asin(y)*(180/pi); }

bool/*ok*/ Rendezvous(double speed,double targetAngle,double targetRange,
   double targetDirection,double targetSpeed,double* courseAngle,
   double* courseRange)
{
   // Use trig to calculate coordinate of future collision with target.
   //             c
   //
   //       B        A
   //
   // a        C        b
   //
   // Known:
   //    C = distance to target
   //    b = direction of target travel, relative to it's coordinate
   //    A/B = ratio of speed and target speed
   //
   // Use rule of sines to find unknowns.
   //  sin(a)/A = sin(b)/B = sin(c)/C
   //
   //  a = asin((A/B)*sin(b))
   //  c = 180-a-b
   //  B = C*(sin(b)/sin(c))

   bool ok = 0;
   double b = 180-(targetDirection-targetAngle);
   double A_div_B = targetSpeed/speed;
   double C = targetRange;
   double sin_b = Sin(b);
   double sin_a = A_div_B*sin_b;
   // If sin of a is greater than one it means a triangle cannot be
   // constructed with the given angles that have sides with the given
   // ratio.
   if(fabs(sin_a) <= 1)
   {
      double a = Asin(sin_a);
      double c = 180-a-b;
      double sin_c = Sin(c);
      double B;
      if(fabs(sin_c) > .0001)
      {
         B = C*(sin_b/sin_c);
      }
      else
      {
         // Sin of small angles approach zero causing overflow in
         // calculation. For nearly flat triangles just treat as
         // flat.
         B = C/(A_div_B+1);
      }
      // double A = C*(sin_a/sin_c);
      ok = 1;
      *courseAngle = targetAngle+a;
      *courseRange = B;
   }
   return ok;
}

回答6:

Here's an example where I devised and implemented a solution to the problem of predictive targeting using a recursive algorithm: http://www.newarteest.com/flash/targeting.html

I'll have to try out some of the other solutions presented because it seems more efficient to calculate it in one step, but the solution I came up with was to estimate the target position and feed that result back into the algorithm to make a new more accurate estimate, repeating several times.

For the first estimate I "fire" at the target's current position and then use trigonometry to determine where the target will be when the shot reaches the position fired at. Then in the next iteration I "fire" at that new position and determine where the target will be this time. After about 4 repeats I get within a pixel of accuracy.

回答7:

I just hacked this version for aiming in 2d space, I didn't test it very thoroughly yet but it seems to work. The idea behind it is this:

Create a vector perpendicular to the vector pointing from the muzzle to the target. For a collision to occur, the velocities of the target and the projectile along this vector (axis) should be the same! Using fairly simple cosine stuff I arrived at this code:

private Vector3 CalculateProjectileDirection(Vector3 a_MuzzlePosition, float a_ProjectileSpeed, Vector3 a_TargetPosition, Vector3 a_TargetVelocity)
{
    // make sure it's all in the horizontal plane:
    a_TargetPosition.y = 0.0f;
    a_MuzzlePosition.y = 0.0f;
    a_TargetVelocity.y = 0.0f;

    // create a normalized vector that is perpendicular to the vector pointing from the muzzle to the target's current position (a localized x-axis):
    Vector3 perpendicularVector = Vector3.Cross(a_TargetPosition - a_MuzzlePosition, -Vector3.up).normalized;

    // project the target's velocity vector onto that localized x-axis:
    Vector3 projectedTargetVelocity = Vector3.Project(a_TargetVelocity, perpendicularVector);

    // calculate the angle that the projectile velocity should make with the localized x-axis using the consine:
    float angle = Mathf.Acos(projectedTargetVelocity.magnitude / a_ProjectileSpeed) / Mathf.PI * 180;

    if (Vector3.Angle(perpendicularVector, a_TargetVelocity) > 90.0f)
    {
        angle = 180.0f - angle;
    }

    // rotate the x-axis so that is points in the desired velocity direction of the projectile:
    Vector3 returnValue = Quaternion.AngleAxis(angle, -Vector3.up) * perpendicularVector;

    // give the projectile the correct speed:
    returnValue *= a_ProjectileSpeed;

    return returnValue;
}

回答8:

I've seen many ways to solve this problem mathematically, but this was a component relevant to a project my class was required to do in high school, and not everyone in this programming class had a background with calculus, or even vectors for that matter, so I created a way to solve this problem with more of a programming approach. The point of intersection will be accurate, although it may hit 1 frame later than in the mathematical computations.

Consider:

S = shooterPos, E = enemyPos, T = targetPos, Sr = shooter range, D = enemyDir
V = distance from E to T, P = projectile speed, Es = enemy speed

In the standard implementation of this problem [S,E,P,Es,D] are all givens and you are solving either to find T or the angle at which to shoot so that you hit T at the proper timing.

The main aspect of this method of solving the problem is to consider the range of the shooter as a circle encompassing all possible points that can be shot at any given time. The radius of this circle is equal to:

Sr = P*time

Where time is calculated as an iteration of a loop.

Thus to find the distance an enemy travels given the time iteration we create the vector:

V = D*Es*time

Now, to actually solve the problem we want to find a point at which the distance from the target (T) to our shooter (S) is less than the range of our shooter (Sr). Here is somewhat of a pseudocode implementation of this equation.

iteration = 0;
while(TargetPoint.hasNotPassedShooter)
{
    TargetPoint = EnemyPos + (EnemyMovementVector)
    if(distanceFrom(TargetPoint,ShooterPos) < (ShooterRange))
        return TargetPoint;
    iteration++
}

回答9:

I made a public domain Unity C# function here:
http://ringofblades.com/Blades/Code/PredictiveAim.cs

It is for 3D, but you can easily modify this for 2D by replacing the Vector3s with Vector2s and using your down axis of choice for gravity if there is gravity.

In case the theory interests you, I walk through the derivation of the math here:
http://www.gamasutra.com/blogs/KainShin/20090515/83954/Predictive_Aim_Mathematics_for_AI_Targeting.php

回答10:

Basically , intersection concept is not really needed here, As far as you are using projectile motion, you just need to hit at a particular angle and instantiate at the time of shooting so that you get the exact distance of your target from the Source and then once you have the distance, you can calculate the appropriate velocity with which it should shot in order to hit the Target.

The following link makes teh concept clear and is considered helpful, might help: Projectile motion to always hit a moving target

回答11:

I grabbed one of the solutions from here, but none of them take into account movement of the shooter. If your shooter is moving, you might want to take that into account (as the shooter's velocity should be added to your bullet's velocity when you fire). Really all you need to do is subtract your shooter's velocity from the target's velocity. So if you're using broofa's code above (which I would recommend), change the lines

  tvx = dst.vx;
  tvy = dst.vy;

to

  tvx = dst.vx - shooter.vx;
  tvy = dst.vy - shooter.vy;

and you should be all set.