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The following Figure shows the function from configuration space (Torus) to operational space (Annulus).
Fig1

There is a naturally defined continuous function from configuration space $(\theta_A, \theta_B)$ to operational space $f(\theta_A, \theta_B)$. If any of $\theta_A$ and $\theta_B$ varies, chalk at the end of rod B draws an 'arc', Fig0.

Supposing we have an arc in operational space and we want to find its inverse arc in configuration space. Here are two simple examples: Fig2 Fig3

The problem is how can we find the arc of the configuration space (i.e. $\theta_A=f(\theta_B)$ for the following two operational space?: Fig4

All I have approached is a little knowledge:

$1-$ the coordinate $(x,y)$ of the end point can be determined through:

$x={l_A}{\cos\theta_A}+{l_B}{\cos(\theta_A+\theta_B)}$ and $y={l_A}{\sin\theta_A}+{l_B}{\sin(\theta_A+\theta_B)}$.

$2-$ The polar equation of the left picture is $r^2−2{l_B}r\cos(\theta)+{l_B}^2={l_A}^2$. But $\theta$ is in operational space, nothing to say about $(\theta_A,\theta_B)$ in configuration space (?)

How it is possible to find the relation between $\theta_A$ and $\theta_B$ such that results in the mentioned arcs?

I don't have any clue how to solve it. I would highly appreciate any helps. Thank you.

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    $\begingroup$ I think there should be some software for this kind of problems, but I need to solve it by Mathematics and LEARN. Thank you in advance for any help. $\endgroup$
    – user200918
    Commented Jan 23, 2015 at 13:49
  • $\begingroup$ Using the transformation $(\theta_A, \theta_B) \mapsto (x, y)$ given, the origin of the $(x, y)$ coordinates in operational space is common center of the circles bounding the gray annulus. In that case, the circular curve $p_1$ given cannot have that polar equation since $r = 2a\cos \theta$ is equivalent to $(x - a)^2 + y^2 = a^2$, a circle through the origin. $\endgroup$ Commented Jan 24, 2015 at 22:51
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    $\begingroup$ Source of the problem: Exercise 4.38., Introduction to Topology by C. Adams. $\endgroup$
    – user200918
    Commented Jan 26, 2015 at 1:47
  • $\begingroup$ The first arc (the circle) has a pretty solution: $\theta_B = - \theta_A$, and $\theta_A$ goes from $0$ to $2\pi$. $\endgroup$
    – D. Thomine
    Commented Jan 26, 2015 at 9:51
  • $\begingroup$ D. Thomine - I think in above-half-circle of $p_1$, $\theta_A$ goes from $0$ to $\pi$ while $\theta_B$ goes from $0$ to $2\pi$. Then again the same in the below-half-circle. Why it has to be $\theta_B = - \theta_A$ ? $\endgroup$
    – user200918
    Commented Jan 26, 2015 at 10:13

2 Answers 2

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I'll write $(x, y)$ or $(r, \theta)$ for the coordinates of the point in the operational space, respectively in cartesian or polar coordinates.

First important remark : the map $(\theta_A, \theta_B) \mapsto (x,y)$ shears and then flatten the torus onto the annulus. The shearing part is bijective, and the "flattening" is two-to-one, except on the boundary of the annulus where it is one-to-one. Hence, there will be in general two possible choices of $(\theta_A, \theta_B)$ for any given position of the chalk. If $(\theta_A, \theta_B)$ works, then by symmetry, so does $(2\theta-\theta_A, -\theta_B)$.

To make things a bit clearer, I'll write $a$, $b$ instead of $l_A$, $l_B$.

I - Finding $\theta_B$

By Al-Kashi's theorem (the law of cosine),

$$r^2 = a^2+b^2+2ab\cos(\theta_B).$$

Among the two possible set of parameters for a given position, there is always exactly one with $\theta_B \in [0, \pi]$. I choose to compute this position. Then:

$$\theta_B = \arccos \left( \frac{r^2-a^2-b^2}{2ab} \right),$$

and:

$$\sin(\theta_B) = \frac{\sqrt{(r-a+b)(r+a-b)(a+b-r)(a+b+r)}}{2ab}.$$

I think this formula can also be deduced from the law of sine and Heron's formula for the area. If you choose the other value of $\theta_B$, which lies in $[-\pi, 0]$, then $\sin(\theta_B)$ becomes the opposite of the value above.

II - Finding $\theta_A$

Now, let's develop the equations giving $x$ and $y$:

$$\begin{cases} x & = & a \cos (\theta_A) + b \cos (\theta_A+\theta_B) \\ y & = & a \sin (\theta_A) + b \sin (\theta_A+\theta_B) \end{cases}$$

becomes:

$$\begin{cases} x & = & [a + b \cos (\theta_B)] \cos (\theta_A) - b \sin(\theta_B) \sin(\theta_A) \\ y & = & b \sin (\theta_B) \cos (\theta_A) + [a + b \cos (\theta_B)] \sin (\theta_A) \end{cases}$$

If we replace $\cos(\theta_B)$ by its expression we found at the beginning, we see that each column has norm $r$. So, by dividing by $r$, we get:

$$\begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix} = \begin{pmatrix} \frac{a+b\cos(\theta_B)}{r} & -\frac{b \sin(\theta_B)}{r} \\ \frac{b \sin(\theta_B)}{r} & \frac{a+b\cos(\theta_B)}{r} \end{pmatrix} \begin{pmatrix} \cos(\theta_A) \\ \sin(\theta_A) \end{pmatrix}$$

Since the 2x2 matrix is a rotation matrix, it is enough to transpose it to get its inverse, whence:

$$\begin{pmatrix} \cos(\theta_A) \\ \sin(\theta_A) \end{pmatrix} = \begin{pmatrix} \frac{a+b\cos(\theta_B)}{r} & \frac{b \sin(\theta_B)}{r} \\ -\frac{b \sin(\theta_B)}{r} & \frac{a+b\cos(\theta_B)}{r} \end{pmatrix} \begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix}.$$

From here, finding $\theta_A$ is just a matter of computing an angle from its sine and cosine.

III - Parametrization of the circle

If you have an arc in operational space and you want to compute a time-dependent parametrization $(\theta_A, \theta_B) (t)$ which follows the arc you want, all you have to do is to compute pointwise $\theta_B$ and $\theta_A$ with the recipe above. However, it can be quite tedious, especially for the special case where everything can be made explicit.

I won't deal with your arc $p_2$, as I don't have an equation for this curve. That said, let's have a look at $p_1$. It is a circle of radius $a$ and center $(b, 0)$. Since it makes a full turn around the origin, the parameter $\theta_A$ must also make a turn. I'll assume that one can do so monotonically, i.e. one can take $\theta_A (t)= t$. Let $f(t)$ be the value of $\theta_B$ at time $t$. Then:

$$\begin{cases} x (t) & = & a \cos (t) + b \cos (t+f(t)) \\ y (t) & = & a \sin (t) + b \sin (t+f(t)) \end{cases}$$

But since $(x,y)$ belongs to the circle,

$$(x-b)^2+y^2=a^2$$

Replace $x$ and $y$ by the functions of $t$, develop, simplify a lot, and get:

$$b(1-\cos(f(t)+t)) + a(\cos(f(t))-\cos(t)) = 0,$$

which, with the help of some trigonometric formula, yields:

$$\sin \left(\frac{f(t)+t}{2} \right) \left[ b \sin \left(\frac{f(t)+t}{2} \right) - a \sin \left(\frac{f(t)-t}{2} \right) \right] = 0.$$

The equation above has one obvious solution: $f(t) = -t$. Hence, if you input $(\theta_A, \theta_B) (t) = (t, -t)$, you will draw the curve $p_1$. The second branch is more delicate to compute. By developping the sine, we get:

$$(a+b) \sin \left(\frac{t}{2} \right) \cos \left(\frac{f(t)}{2} \right) = (a-b) \cos \left(\frac{t}{2} \right) \sin \left(\frac{f(t)}{2} \right),$$

whence:

$$\tan \left(\frac{f(t)}{2} \right) = \frac{a+b}{a-b} \tan \left(\frac{t}{2} \right).$$

Finally, we get:

$$f(t) = 2 \arctan \left( \frac{a+b}{a-b} \tan \left(\frac{t}{2} \right) \right),$$

which is well-defined for $t \neq \pi$ (but that is not a problem, as $f(\pi) = \pi$ necessarily).

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  • $\begingroup$ If $\theta_B \in [0, \pi]$ then must be $s(r) = +1$ for $r^2 < a^2+b^2$, as well (?) $\endgroup$
    – user200918
    Commented Jan 28, 2015 at 6:14
  • $\begingroup$ Thank you D. Thomine. It's really beautiful! Thank you. $\endgroup$
    – user200918
    Commented Jan 28, 2015 at 7:03
  • $\begingroup$ @ Ali.E .: Good catch. The sine is nonpositive for the other choice of $\theta_B$, but always nonnegative for the choice I made. $\endgroup$
    – D. Thomine
    Commented Jan 28, 2015 at 9:29
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In complex numbers, your planimeter measures $f(\theta_A, \theta_B) = A e^{i \theta_A} + B e^{i(\theta_A + \theta_B)} $. There doesn't seem to be any restriction on the angle so $0 \leq \theta_A, \theta_B < 2\pi$. Maybe it will be easier to use $\theta_{B'} = \theta_A + \theta_B$ and the configuration space will still be the same torus "twisted".

Your image will be the Minkowski sum of the two circles of radius $A$ and $B$ and the result is that $f$ is a 2-1 map from the torus to an annulus. except at the edges where it's 1-1. It's like loooking at a donut with radii $A,B$ from the top.

enter image description here

How to compute $f^{-1}$ ?

The $f(\theta_A + \delta , \theta_{B'} + \delta) = e^{i\delta}f(\theta_A , \theta_{B'} ) $. This means rotating the annulus is the same as moving diagonally in the donut in the NE diagonal direction $(1,1)$.

The radius can be computed from the Law of cosines: $|f(\theta_A, \theta_B)| = A^2 + B^2 + 2AB \cos \theta_B $


A reasonable thing to do would be draw the torus in 3D and plug into the equation of your circle of interest.

$(\theta,\phi) \mapsto ( (A + B \sin \theta) \cos \phi, (A + B \sin \theta ) \sin \phi, B \cos \theta) $

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