Difference between revisions of "Maximum Escape Angle"

From Robowiki
Jump to navigation Jump to search
(Rewriting first part for clarity)
Line 31: Line 31:
 
</pre>
 
</pre>
 
We don't know the value of <code>C</code>, but we can take the worst scenario where
 
We don't know the value of <code>C</code>, but we can take the worst scenario where
<code>C = PI/2</code (<code>sin(C) = 1</code>) to get a Maximum Escape Angle of  
+
<code>C = PI/2</code> (<code>sin(C) = 1</code>) to get a Maximum Escape Angle of  
 
<code>A = asin(Vr/Vb * 1) = asin (Vr/Vb)</code>.  
 
<code>A = asin(Vr/Vb * 1) = asin (Vr/Vb)</code>.  
  

Revision as of 09:47, 13 November 2007

When firing, the Maximum Escape Angle (MEA) is the largest angle offset from zero (i.e., Head-On Targeting) that could possibly hit an enemy bot, given the Game Physics of Robocode.

Calculation

Let's assume a triangle with sides a, b and c and angles (vertices) A, B, and C. A is the angle opposite to a, B is opposite to b, and C is opposite to c. The Law of sines says that:

                  A
                  /\
                 /  \
             b  /    \  c
               /      \
              /________\ 
            C     a      B

     a/sin(A) = b/sin(B) = c/sin(C)

Now let's say that your bot is in the vertex A and the enemy bot is in the vertex C. We will fire a bullet with angle A to hit the bot in vertex B. We know the value of b (it is the distance D from your bot to the enemy). We don't know c, but we know that it will be the distance traveled by the bullet. Also, we know that a will be the distance traveled by the enemy bot. If we put a, b, and c as a function of time, we have:

b = D
c = Vb * t (Vb is the bullet speed)
a = Vr * t (Vr is the enemy bot velocity)

Now, using the Law of sines:

   a/sin(A) = c/sin(C) 
-> Vr*t / sin(A) = Vb*t / sin(C) 
-> sin(A) = Vr/Vb * sin(C) 
-> A = asin(Vr/Vb * sin(C))

We don't know the value of C, but we can take the worst scenario where C = PI/2 (sin(C) = 1) to get a Maximum Escape Angle of A = asin(Vr/Vb * 1) = asin (Vr/Vb).

With a maximum Robot velocity of 8.0, a theoretical Maximum Escape Angle would be asin(8.0/Vb). Note that the actual maximum depends on the enemy's current heading, speed, and Wall Distance.

See Also