# How do i compute this radius?

I need to find the radius of a circle.

Consider two circles with a common center. The radius of circle #1 is r-d. The radius of circle #2 is r+d. The common center is called O.

A straight arm is rotating around O. Where the arm intersects the two circles we get two points of contact close to each other, p1 (arm in contact with circle1) and p2 (arm in contact with circle 2). You might think of p1 and p2 as the coordinates of the front wheels of a car driving around in a circle.

What is known is the instantaneous velocity (ds/dt) of the two points p1 (velocity v1) and p2 (velocity v2) in their orbit around the common center O. The distance between the two points is also known, it is 2d. I need to compute r.

Since I know v1 and v2 I can compute the quota q=v1/v2. I figure that from this quota and d it should be possible to compute the radius r. A smaller q with constant d should (I think) give larger r. A constant q with smaller d should also give larger r.

If, for example, v1 is 10 m/s, v2 is 10.001 m/s and d is 1 mm, what is the radius of the circle?

I would prefer a simple equation with r on one side of the = and v1, v2 and d on the other, if possible.

This obviously beats me, so please show me how to compute r.

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One equation is

$$v_{1} = \frac{2 {\pi}(r+d)}{T}$$

And the other

$$v_{2} = \frac{2 {\pi}(r-d)}{T}$$

You know what I mean?

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But this will not help me, will it werg22? There is no T since the velocities are instantaneous. I need to compute r. v1, v2 and d are known. Maybe I just don’t se it. Please give me an equation with r on one side of = and v1, v2 and d on the other.

It really dosen't matter what is the angular velocity function. Say the function is

$$\theta (t)$$, we have in time $$a$$, a covered arc length of

$$\int_{0}^{a} \theta (t) dt$$

The covered arc at radius k is obviously

$$k \int_{0}^{a} \theta (t) dt$$

Hence the velocity at radius $$k$$ is equal to

$$k \theta (a)$$

We thus have the equations

$$v_1 = (r + d) \theta (a)$$

and

$$v_2 = (r - d) \theta (a)$$

and hence the result

$$\frac{v_2}{r-d} = \frac{v_1}{r + d}$$

which is independent of $$\theta (t)$$

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VietDao29
Homework Helper
Ok, just to make it a little bit clearer, say, at t0, the arm cuts the 2 circles at A, and B. Then at the time (t0 + h), the arm cuts the 2 circles at A', and B', like this:

Now, you should notice that: $$\bighat{AOA} ' = \bighat{BOB} ' = \theta$$.

The arc-length of AOA' (i.e, the distance from A to A' on the bigger circle, the green one) is: $$\mbox{arclength} AOA' = R \times \bighat{AOA} ' = (r + d) \theta$$, where $$\theta$$ is in radian.

The arc-length of BOB' (i.e, the distance from B to B' on the smaller circle, the red one) is: $$\mbox{arclength} BOB' = R' \times \bighat{BOB} ' = (r - d) \theta$$, where $$\theta$$ is in radian.

Now let h ~~> 0, i.e (A' ~~> A, and B' ~~> B, but the equation $$\bighat{AOA} ' = \bighat{BOB} ' = \theta$$, of course, still hold), the instantaneous velocity at A can be calculated by:
$$v_A = \lim_{h \rightarrow 0} \frac{\mbox{arclength} AOA'}{h}$$

We now will try to find the relation between vA, and vB, we have:

$$v_A = \lim_{h \rightarrow 0} \frac{\mbox{arclength} (AOA')}{h} = \lim_{h \rightarrow 0} \frac{(r + d) \theta}{h} = \lim_{h \rightarrow 0} \frac{(r + d) \theta (r - d)}{ (r - d) h}$$ (multiply both numerator, and denominator by (r - d))
$$= \frac{r + d}{r - d} \lim_{h \rightarrow 0} \frac{\theta (r - d)}{h} = \frac{r + d}{r - d} \lim_{h \rightarrow 0} \frac{\mbox{arclength} (BOB')}{h} = \frac{r + d}{r - d} v_B$$

So we have:
$$v_A = \frac{r + d}{r - d} v_B$$

Having the above equation, can you finish the problem? :)

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The concept of derivation does not need to be restated at every problem, you know...

VietDao29
Homework Helper
The concept of derivation does not need to be restated at every problem, you know...
Well, just in case the OP was not very sure about it. But, anyway, it's not too long, and messy, right?

Ok, i solved it on my own.

Lets say that k = v1 / v2.

Then the radius of the circle is r = -dk -d / (k-1)

It works perfectly.