Difference between revisions of "Maximum Escape Angle"

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When firing, the largest angle offset from zero (i.e., [[Head-On Targeting]]) that could possibly hit an enemy bot, given the [[Game Physics]] of [[Robocode]].
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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 ==
 
== Calculation ==
  
Let's assume a triangle with sides <code>a</code>, <code>b</code> and <code>c</code>, and angles (vertices) <code>A</code>, <code>B</code>, and <code>C</code>, where <code>A</code> is the angle opposed to <code>a</code>, <code>B</code> to b, and <code>C</code> to <code>c</code>. The [http://en.wikipedia.org/wiki/Law_of_sines Law of sines] says that:
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Let's assume a triangle with sides <code>a</code>, <code>b</code> and <code>c</code> and angles (vertices) <code>A</code>, <code>B</code>, and <code>C</code>. <code>A</code> is the angle opposite to <code>a</code>, <code>B</code> is opposite to b, and <code>C</code> is opposite to <code>c</code>. The [http://en.wikipedia.org/wiki/Law_of_sines Law of sines] says that:
  
 
<pre>
 
<pre>
    a/sin(A) = b/sin(B) = c/sin(C)
 
 
 
                   A
 
                   A
 
                   /\
 
                   /\
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               /________\  
 
               /________\  
 
             C    a      B
 
             C    a      B
 +
 +
    a/sin(A) = b/sin(B) = c/sin(C)
 
</pre>
 
</pre>
  

Revision as of 02:29, 12 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</code (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