Comparing vector elements by percentage

Angle=[5 10 20 60 180 190 195 300 310], these angles determine the possible directions of robots. If there are 2 robots starting with the same angle or plus minus 5%(5 degrees) they will make collision. I'd like to have the result in on vector such that:[2 2 1 1 1 2 2 1 1]. Any help?

5 Comments

yousef - why will two robots collide if they have similar directions? Won't their starting position and speeds also impact whether they collide or not?
All robots start at the same time with the same speed.Thanks
Is there any mistake here?
Angle=[5 10 20 60 180 190 200 300 310];
collision=[];
for i=1:9
targetValue = Angle(i);
tolerance = 0.05 * targetValue;
differences = abs(Angle-targetValue) ;
inToleranceIndexes = differences < tolerance;
withinTolerance = Angle(inToleranceIndexes)
collision(i)= Angle(inToleranceIndexes)
end
end
Hello,
the codeline
collision(i)= Angle(inToleranceIndexes)
won't work, if inToleranceIndexes has more than one value==1. Then Angle(inToleranceIndexes) i.e. is [300 310] and you can't assign a vector to a single element as you do with collision(i).
But a question in general: You say tolerance is 0.05*targetValue, that means your tolerance increases with increasing start-angle. Is this really how it is supposed to be?
Best regards, Stefan
Hi,the tolerance is plus minus 5 degrees.The tolerance is fixed and can not be changed.I have another try,please check it.
  • Angle=[5 10 20 60 180 190 200 300 310];
  • collision=[];
  • for i=1:9
  • c=[];
  • targetValue = Angle(i);
  • tolerance = 0.5 * targetValue;
  • differences = abs(Angle-targetValue) ;
  • inToleranceIndexes = differences < tolerance;
  • withinTolerance = Angle(inToleranceIndexes);
  • c=[c;inToleranceIndexes];
  • end
  • collision=[collision;c];

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Answers (1)

As mentioned in your duplicate question there will be no collision unless you have different starting points. And all your code does is tell you that each angle interferes/collides with no other angle except itself, which is pretty useless.

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