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I'm designing a game where objects have to move along a series of waypoints. The object has a speed and a maximum turn rate. When moving between points p1 and p2 it will move in a circular curve like so:

enter image description here

Angle a is the angle the object must rotate through. Therefore we can calculate the time it will take to rotate through this angle as t = a / turn rate.

However, if the object's current speed means it will cover distance d in quicker than this time, it must slow down on this corner or it will miss p2. I need to calculate the maximum speed it can take this corner at, which means calculating the distance of the curve d (so I can calculate the maximum corner speed as d / t).

I think I've figured out b = 2a, but to determine d I need to know the radius of the circle. How can I determine it from this information?

AshleysBrain
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  • Do we know the distance between $p_1$ and $p_2$? – awllower Jan 24 '13 at 14:28
  • Yes, they're both known. – AshleysBrain Jan 24 '13 at 14:28
  • Since we know $b$, we could find the radius by noting that $p_1 p_2 b$ forms an equilateral triangle. Thus, denoting the distance between $p_1$ and $p_2$ by $\Pi$, we find the radius to be $\rho = \Pi *arcsin(a)/2$. – awllower Jan 24 '13 at 14:31
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    Note that looking at your diagram carefully, (and the arrows showing the direction of movement at the start and end of the arc) the angle you have marked as $a$ is only one half of the total angle the object has to turn through. – Shard Jan 24 '13 at 20:52
  • @awllower: I think you mean $1/\sin a$ where you say $\arcsin a$. It doesn't make sense to take the arcsine of an angle. –  Jan 25 '13 at 08:16
  • @ℝⁿ. Indeed, thanks for the indication. – awllower Jan 30 '13 at 03:47

2 Answers2

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[Sorry, figure-fu misplaced]

The angle between the line $P_1 P_2$ and the radius is $\frac{\pi}{2} - \alpha$, where $\alpha$ is the angle you seem to call $a$. By symmetry, the angle at $P_2$ is the same, so the angle $\beta$ at the center is just $2 \alpha$. By the law of cosines, if $d$ is the linear distance between $P_1$ and $P_2$: $$ d^2 = 2 r^2 (1 - \cos (2 \alpha)) $$ I'm sure this can be simplified further...

vonbrand
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    I think it would be polite to use the same variable names that the asker used. $d$ is already the length of the arc, so let's call the straight-line length of the chord $c$. Then we have $c^2=2r^2(1-\cos2a)=4r^2\sin^2a$, so $r=c/(2\sin a)$. –  Jan 25 '13 at 08:14
  • Sorry about that. – vonbrand Jan 25 '13 at 08:15
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Is "Then we have $c^2=2r^2\left(1−\cos^2a\right)=4r^2\sin^2a$, so $r=\frac c{2\sin a}$" correct? I mean, what about exponents, are they correct? Maths looks bad!

Would not it be: $ c ^ 2 = 2 R ^ 2 \left( 1 - \cos ^ 2 a \right) = 4 R^2 \sin ^ 2 a $?

So then, would not it be: $ R = \frac c { \sin ^ 2 a } $?

Mohsen Shahriari
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Laura
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