Difference between revisions of "Maximum Escape Angle"

From Robowiki
Jump to navigation Jump to search
(revised text from old RandomTargeting page into its own article)
 
(The work done at Maximum Escape Angle/Precise Positional needs to be seen! No pages were linking to it yet.)
 
(16 intermediate revisions by 6 users not shown)
Line 1: Line 1:
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]].
+
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 [[Robocode/Game Physics|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:
+
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>a/sin(A) = b/sin(B) = c/sin(C)</pre>
 
  
<pre>
+
[[Image:LawOfSines.png|center]]
                  A
 
                  /\
 
                /  \
 
            b  /    \  c
 
              /      \
 
              /________\
 
            C    a      B
 
</pre>
 
  
 
Now let's say that your bot is in the vertex <code>A</code> and the enemy bot is in the vertex <code>C</code>. We will fire a bullet with angle <code>A</code> to hit the bot in vertex <code>B</code>. We know the value of <code>b</code> (it is the distance <code>D</code> from your bot to the enemy).  
 
Now let's say that your bot is in the vertex <code>A</code> and the enemy bot is in the vertex <code>C</code>. We will fire a bullet with angle <code>A</code> to hit the bot in vertex <code>B</code>. We know the value of <code>b</code> (it is the distance <code>D</code> from your bot to the enemy).  
Line 29: Line 20:
 
-> A = asin(Vr/Vb * sin(C))
 
-> A = asin(Vr/Vb * sin(C))
 
</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>.  
  
With a maximum Robot velocity of 8.0, a theoretical [[Maximum Escape Angle]] would be <code>asin(8.0/Vb)</code>. Note that the actual maximum depends on the enemy's current heading, speed, and [[Wall Distance]].
+
With a maximum Robot velocity of 8.0, a theoretical Maximum Escape Angle would be <code>asin(8.0/Vb)</code>. Note that the actual maximum depends on the enemy's current heading, speed, and [[Wall Distance]].
  
 
== See Also ==
 
== See Also ==
  
* [[Precise Maximum Escape Angle]] - Some bots use a more sophisticated calculation for [[Maximum Escape Angle]], using [[Precise Prediction]].
+
* [[Maximum Escape Angle/Precise Positional]] - A more precise method of calculating Maximum Escape Angle, that doesn't require movement simulation.
 +
* [[Maximum Escape Angle/Precise]] - A sophisticated calculation for Maximum Escape Angle, using [[Precise Prediction]].
 +
 
 +
{{Targeting Navbox}}
 +
[[Category:Robocode Theory]]
 +
[[Category:Terminology]]

Latest revision as of 13:39, 5 September 2012

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:

LawOfSines.png

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