Visualize two points:  $O\equiv(0\mid 0)$ and $D\equiv(d\mid 0)$.  The two are $d$ units apart. 

Visualize a movable rod whose endpoints, $C_O$ and $S_O$, are a unit apart. $C_O$ always coïncides with $O$ and $S_O$ orbits $O$.   Visualize a movable rod whose endpoints, $C_D$ and $S_D$, are a unit apart. $C_D$ always coïncides with $D$ and $S_D$ orbits $D$.    Visualize a movable rod whose endpoints, $R_O$ and $R_D$, are $d$ units apart. $R_O$ coïncides with $S_O$, and $R_D$ coïncides with $S_D$.    I believe that such an arrangement is called a linkage. For our purpose $S_O$ (and $R_O$) can be considered as situated at $(\cos\theta\mid \sin\theta)$. $\theta$ is the parameter whose value determines the locations of all moving points in this problem.    There are two such linkages.    In the first, the rod connecting $S_O$ to $S_D$ is always parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv(d+\cos\theta\mid \sin\theta)$. Not interesting, therefore ignorable.    In the second, the rod connecting $S_O$ to $S_D$ is never parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv\left({{(d^2-1)(d-\cos\theta)}\over{d^2+1-2d\cos\theta}}\mid{{(d^2-1)(-\sin\theta)}\over{d^2+1-2d\cos\theta}}\right)$.

If one considers the direction of rotation of $OS_O$ to be positive $+$, then the following is true: If $d\gt 1$, then the direction of rotation of $DS_D$ is $+$.
If $d\lt 1$, then the direction of rotation of $DS_D$ is $-$.
If $d=1$, then $S_D\equiv (0\mid 0)$ — except when $\theta\equiv 0\mod 2\pi$, when $S_D\equiv{\left({0\over 0}\mid{0\over 0}\right)}$.

QUESTION: For a given positive value of $d$, what value of $\theta$ imparts the maximum angular velocity, $+$ or $-$, to $DS_D$?

I am indebted to Dr. Jyrki Lahtonen for his helpfulness!!!!

[Editor's note: The picture I have in mind, JL.]

The animations below have the points $S_O$ and $S_D$ marked with black dots. As the distances from the origin $O$ to $S_O$, from $S_O$ to $S_D$ and from $S_D$ to $D$ are all fixed, we can imagine them being connected by rigid rods. I took the liberty of adding those rods to the pic.

In the first animation, $d={2\over 3}$.

enter image description here

And in the second, $d={4\over 3}$ (the size of the animation is scaled down by a factor of two in comparison to the one above).

enter image description here

[Editor's note: I put the motivational speech below (from the OP) back in here, because it may be needed to make the question meet the site standards, JL]

ADDENDUM! It was requested that I clarify my question, so I shall try to eliminate abstractions in favor of concrete objects … 1. Replace the Cartesian plane with a sheet of marine plywood. 2. On the plywood, draw a line segment $d$ units long. Label its endpoints $O$ and $D$. Drill a hole at each endpoint. Fit a dowel snugly into each. 3. Take three paint mixing paddles. On two of them, draw lines $r$ units long. On one of these, label the endpoints $O$ and $S_O$; on the other, label the endpoints $D$ and $S_D$. Drill holes at all four of these endpoints. On the remaining paint mixing paddle, draw a line $d$ units long. Label its endpoints $S_O$ and $S_D$. Drill a hole at each. 4. Fit the hole $O$ onto the dowel $O$. Fit the hole $D$ onto the dowel $D$. Fit dowels into $S_O$ and $S_D$, but not completely through. 5. Fit the $S_O$ hole onto the $S_O$ dowel. Fit the $S_D$ hole onto the $S_D$ dowel. 6. Saw off any excess dowel that would interfere with the movement of the assembly. (Yes, it is supposed to move.) 7. I envision these rods as being made of untreated white pine.

NOTA BENE: For simplicity's sake, I have gotten rid of the variable $r$ and replaced it with unity.

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    $\begingroup$ Voting to close for unclear what you're asking. I don't understand the question at all. $\endgroup$ – user122283 May 4 '14 at 15:55
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    $\begingroup$ @SanathDevalapurkar Yea, I don't know if it's "standard" notation to write it like that but it's certainly greek to me. $\endgroup$ – MCT May 4 '14 at 16:11
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    $\begingroup$ What does $O(0 \mid 0)$ mean? What about $S_O(r \cos \theta | r \sin \theta)$? I have no idea what the question is asking. $\endgroup$ – user61527 May 5 '14 at 3:53
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    $\begingroup$ I beg to disagree. Not necessarily with the decision to put on hold, but its reason. Yes, the notation is non-standard, and the question does not show effort. It is also mistagged, as circle does not really describe what's being asked. But fer crying out loud it is perfectly clear what IS being asked. Don't look at the formulas - look at the text and draw the picture in your imagination. See the machine of four rods with three joints and two hinged+fixed endpoints swirling in the plane? $\endgroup$ – Jyrki Lahtonen May 5 '14 at 5:26
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    $\begingroup$ This question lacks key information. Most importantly, what have you tried? Also, where did you encounter the problem? That information helps other people write answers that are as helpful as possible. $\endgroup$ – Carl Mummert May 5 '14 at 9:59

I'm breaking my promise not to answer this as no one else bites. This is largely just a sketch of an argument explaining why the maximum is obtained at $\theta=0$, when $S_O$ is as close to $D$ as possible, moving tangentially. A calculus solution by differentiation is possible, but I don't want to do that.

Here's a variant of the animation. This time $d\approx 1.54$. I added the linkage discarded by the OP in red, because it aids understanding what is going on. Furthermore I added a green `rubber band' connecting the points $D$ and $S_O$.

enter image description here

As expected, the `uninteresting' red dot, call it $S_D'$, revolves about the pivot $D$ with exactly the same angular rate as the point $S_O$ about the pivot $O$. This is a consequence of the fact that $DS_D'S_OO$ is always a parallelogram. But let's look at the picture more closely. We see that the black dot $S_D$ is always the mirror image of $S_D'$ with respect to the green line. The mirror symmetry is a geometrically obvious consequence of the fact that the points $S_D$ and $S_D'$ are the two points on the plane with the prescribed distances to $D$ and $S_O$.

Therefore the angular velocity of the green line (about $D$) is the average of the angular velocities of $S_D$ and $S_D'$ (the green line bisects the angle $\angle S_DDS_D'$). So $$ \omega_{S_D}=2\omega_{green}-\omega_{S_D'}. $$ As $\omega_{S_D'}$ is constant this means that $\omega_{S_D}$ is maximized simultaneously with $\omega_{green}$. But $\omega_{green}$ is the angular velocity of $S_O$ about $D$. We can apply the principle: angular velocity = tangential component of the orbital speed vector divided by the distance. As the orbital speed of $S_O$ is constant it happens that the tangential component is maximized at the point of the closest approach, so the numerator reaches its maximum and the denominator reaches its minimum simultaneously at $\theta=0$. Therefore the ratio reaches its maximum at the closest approach of $S_O$ to $D$.

  • $\begingroup$ You need to be a bit careful in interpreting the signs of the angular velocities. In the last paragraph I treated clockwise direction as positive, so $\omega_{S_D'}<0$. As we see, when $d<1$ the sign of $\omega_{green}$ varies, but is positive at the point $\theta=0$, so the conclusion holds. $\endgroup$ – Jyrki Lahtonen May 12 '14 at 9:52
  • $\begingroup$ If you are turning the crank of the paint mixer by hand (older version of the question - see the edit history), then you can probably notice that maximum torque is require at that point, when the mixer paddle at $S_D$ is moving at its greatest speed. $\endgroup$ – Jyrki Lahtonen May 12 '14 at 10:16
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    $\begingroup$ Dr. @Jyrki Lahtonen: Once again, TYVM! When I first became interested in such linkages and noticed the acceleration, $+$ or $-$, of the driven (as opposed to the driving) pivot, I set up a display (similar to the three that you contributed) and hooked up three or four in tandem. The result was spectacular: Although the display moved quite slowly — I was only bumping the value of $\theta$ a degree at a time. — the final pivot in the chain would suddenly spring to life like the business end of a rat trap. $\endgroup$ – Senex Ægypti Parvi May 12 '14 at 22:11

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