Difference between revisions of "Maximum Escape Angle"

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(revised text from old RandomTargeting page into its own article)
 
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* [[Precise Maximum Escape Angle]] - Some bots use a more sophisticated calculation for [[Maximum Escape Angle]], using [[Precise Prediction]].
 
* [[Precise Maximum Escape Angle]] - Some bots use a more sophisticated calculation for [[Maximum Escape Angle]], using [[Precise Prediction]].
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[[Category:Robocode Theory]]

Revision as of 01:44, 12 November 2007

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.

Calculation

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

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

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 ofC, 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