Questions / Comments |

Questions / CommentsI was using radius = velocity / ( 2 * sin ( turn rate / 2 ) ) after painstakingly doing the trig. But using the small angle approximation (as you have) cuts out a sine calculation. The inaccuracy is not more than 0.02 pixels by the way. *Does global delete of Math.sin in robocode project directory* --[Russ Graham]? |

/** * See <a href="http://www.robowiki.net/cgi-bin/robowiki?Trigonometry/TurnCircleRadius"> * http://www.robowiki.net/cgi-bin/robowiki?Trigonometry/TurnCircleRadius * </a> for a description of this method. * * @author David Alves * @param velocity - velocity that the robot is traveling at * @return the radius of the turning circle if the robot is traveling in a circle, turning at max velocity */ public double turnCircleRadius(double velocity){ return (Math.abs(v) * 360.0 / (10.0 - .75 * Math.abs(v))) / (2.0 * Math.PI); }

Here's the math behind this shortcut:

A robot at velocity v has a max turn rate of (10 - .75 * Math.abs(v)) degrees / tick. Therefore it takes 360 / (10 - .75 * Math.abs(v)) ticks to drive in a complete circle. After this time it will have traveled a distance of Math.abs(v) * 360 / (10 - .75 * Math.abs(v)). This is the circumference of the circle the bot is driving in, so dividing by 2 * pi gives the radius: Math.abs(v) * 360 / (10 - .75 * Math.abs(v)) / (2 * Math.PI). Note that this is not an exact answer (bots don't drive in continuous curves), but it should be within a pixel or two.

Radius at velocity 8: 114.6

Radius at velocity 7: 84.4

Radius at velocity 6: 62.5

Radius at velocity 5: 45.8 (SpinBot)

Radius at velocity 4: 32.7

Radius at velocity 3: 22.2

Radius at velocity 2: 13.5

Radius at velocity 1: 6.2

--David Alves

I was using radius = velocity / ( 2 * sin ( turn rate / 2 ) ) after painstakingly doing the trig. But using the small angle approximation (as you have) cuts out a sine calculation. The inaccuracy is not more than 0.02 pixels by the way. *Does global delete of Math.sin in robocode project directory*

--[Russ Graham]?