# Find pivot point of line from known arc lengths traced by the endpoints

I have a line segment, $$AB$$, which is rotated in space around some pivot point $$P$$ on the line segment to form a new line segment $$CD$$. I believe a circle can always be fitted to coincide with points $$A$$, $$B$$, $$C$$ and $$D$$.

The position of the pivot point $$P$$ is unknown, but the arc lengths traced around the constructed circle from $$A$$ to $$C$$ and from $$B$$ to $$D$$ are known. The length of the line segment $$|AB| = |CD|$$ is also known.

I would ultimately like to determine the transformation (translation and rotation) between the midpoint of $$AB$$ (denoted $$M_{AB}$$) and the midpoint of $$CD$$ (denoted $$M_{CD}$$).

It seems that, to do this, it will probably be necessary to calculate the position of $$P$$ (e.g. by distance $$|PM_{AB}|$$) and also the angle $$\angle APC = \angle BPD$$. But I am not sure how to do this.

I have tried using various intersecting chord theorems (e.g. The measure of an angle formed by two chords that intersect inside a circle is equal to the arithmetic mean of the measures of the two intercepted arcs) but, without knowing the radius of the circle, none of them seem to be much use.

• The exact location of the pivoting point $P$ cannot be determined. Would you like to settle for its approximate position? Furthermore, could you tell us the size of $\angle CPA$, whether it is large or very small?
– YNK
Commented Apr 13, 2023 at 8:42
• It depends how approximate, but I suspect that it may not be useful for the real-world application of this problem (modelling skid steering). $\angle CPA$ should always be quite small. Commented Apr 20, 2023 at 12:10
• While waiting for you to respond to my comment, I have tested the calculation procedure I have devised using MS-Excel. I have achieved an accuracy up to 13 decimal places for a turning angle of $5^o$ with only 2 iterations. I will post my answer as soon as possible for you to go through and decide whether it is useful for the mentioned real-world application. I will also make available the Excel sheet for downloading.
– YNK
Commented Apr 20, 2023 at 16:22

## 2 Answers

As OP has correctly guessed, the four points $$A$$, $$B$$, $$C$$, and $$D$$ lie on a circle, which we will call the constructed circle hereinafter. Let its radius be $$r$$. Please note that, to keep our answer short, we will not be providing proofs of this observation and several others to come.

In $$\mathrm{Fig.\space 1}$$ we reproduce OP’s sketch after adding to it the two arcs of the constructed circle. For brevity, we denote lengths of the shorter and the longer arc as $$a_{AC}$$ and $$a_{BD}$$ respectively. We also let $$AB=CD=d$$, $$AP=CP=b$$, and $$\measuredangle CPA=\phi$$. Our aim is to derive an equation expressing $$r$$ in terms of the three knowns lengths $$a_{AC}$$, $$a_{BD}$$, and $$d$$.

Just for a moment, assume that we know the position of $$A$$, $$B$$, $$C$$, $$D$$, and $$P$$ in 2D-space. Then, the simple geometrical construction shown in $$\mathrm{Fig.\space 2}$$, can be used to locate the center of the constructed circle. In other words, when perpendicular bisectors of segments $$CB$$ and $$BD$$ are constructed, they intersect at the sought center, which we denote by $$O$$.

Furthermore, following statements can be shown to be true. $$\quad\enspace\space \text{i.\enspace The segment BD is parallel to the segment AC.}$$ $$\quad\enspace \text{ii.\enspace Line NT, the perpendicular bisector of BD, passes through P.}$$ $$\quad\space \text{iii.\enspace Line NT is also the perpendicular bisector of AC.}$$ $$\quad\space \text{iv.\enspace Line NT, is the angle bisector of \angle CPA, \angle COA, and \angle DOB.}$$

Let $$\measuredangle COA = \omega$$ and $$\measuredangle DOB = \theta$$. Now, we need to state or derive several equations making use of geometrical properties of the configuration shown in $$\mathrm{Fig.\space 2}$$. Let us start by doing some angle chasings. According to statement (iv), we have, $$\measuredangle CPN=\dfrac{\phi}{2}\qquad\measuredangle CON=\dfrac{\omega}{2}\qquad\measuredangle DOB=\dfrac{\theta}{2}.$$

If we extend $$CO$$ to cut the arc $$BD$$ at $$E$$, we obtain, $$\measuredangle DOE=\dfrac{\theta-\omega}{2}.$$

Since $$\angle DOE$$ and $$\angle DCE$$ are angles subtended by the same chord $$DE$$ at the center and the circumference of the constructed circle respectively, we can write, $$\measuredangle DCE=\dfrac{\theta-\omega}{4}.$$

Since $$\angle CPN$$ is an exterior angle of the $$\triangle COP$$, we shall write, $$\measuredangle CPN=\dfrac{\phi}{2}=\measuredangle COP+\measuredangle PCO=\dfrac{\theta+\omega}{4} \qquad\rightarrow\qquad \phi=\dfrac{\theta+\omega}{2}. \tag{1}$$.

It follows from triangles $$ONC$$ and $$PNC$$, which are sharing the side $$NC$$, that, $$NC=r\sin\left(\dfrac{\omega}{2}\right) = b\sin\left(\dfrac{\phi}{2}\right) \qquad\rightarrow\qquad b=r\dfrac{\sin\left(\dfrac{\omega}{2}\right)}{ \sin\left(\dfrac{\phi}{2}\right)}\tag{2}$$

Consider the isosceles triangle $$COD$$, where it is possible to deduce that, $$2r\cos\left(\dfrac{\theta-\omega}{4}\right)=d\qquad\rightarrow\qquad r=\dfrac{d}{2\cos\left(\dfrac{\theta-\omega}{4}\right)}. \tag{3}$$

Considering the two arcs $$AC$$ and $$BD$$ of the constructed circle, we obtain, $$\omega=\dfrac{ a_{AC}}{r} \qquad \text{and}\qquad \theta=\dfrac{ a_{BD}}{r}. \tag{4}$$

When we replace $$\theta$$, and $$\omega$$ of (3) using their values given in (4), we complete the derivation of the sought equation. $$r=\dfrac{d}{2\cos\left(\dfrac{a_{AC} - a_{BD}}{4r}\right)}. \tag{5}$$

As we see from (5), it is impossible to derive an equation, in which $$r$$ is its subject. Since our hopes of expressing $$r$$ explicitly in terms of $$a_{AC}$$, $$a_{BD}$$, and $$d$$ has been now dashed, we have to resort to other methods to achieve our goal. The first of the two methods we would like to suggest is the following iterative scheme based on Newton-Raphson method, which could be launched with an educated guess for $$r$$.

$$\left\{\begin{array}{l} {\large{r_0=}\max\large{\left\{\dfrac{d}{2},\space \dfrac{a_{AC} + a_{BD}}{6}\right\}}}\\ {\large{r_{n+1}=r_n d \left(\dfrac{1+\beta\tan\left(\beta\right)}{{2r_n\cos\left(\beta\right)+d\beta\tan\left(\beta\right)}}\right), \enspace}} \text{where} \large{\enspace \beta=\dfrac{a_{BD} - a_{AC}}{4r_n}} \end{array}\right. \tag{6}$$

Once you have $$r$$, you can calculate the values of $$\omega$$, $$\theta$$, $$\phi$$, and $$b$$ in that order using (4), (1), and (2) respectively. This method delivers accurate enough values of all these unknown parameters for values of the turning angle $$\phi$$ less than $$170^o$$. With these parameters, you are in a position to locate of $$C$$ and $$D$$, which pave the way to determine the transformation of the midpoint $$M_{CD}$$ of $$CD$$ using the method described in @bubba’s answer.

We have managed to test the iteration scheme (6) using a MS-Excel worksheet functions. The corresponding worksheet is available for downloading at Dropbox. OP is invited to try it out by inputting his/her values of $$d$$, $$a_{AC}$$, and $$a_{BD}$$.

Another method to determine $$r$$ can be devised by replacing the cos-term of the denominator on the right-hand side of (5) by its series expansion. After expanding and simplifying, we can omit the higher order terms of the series because their contribution is negligible. The formula given below was obtained by taking only the terms of order five and less into account.

$$6144{\large{r}^4}-3072{\large{dr}^3}-192{\large{\left(a_{\small{BD}}- a_{\small{AC}}\right)}^2r^2}+{\large{\left(a_{\small{BD}}- a_{\small{AC}}\right)^4}}=0 \tag{7}$$

What we have in (7) is a quartic equation and, therefore, we need to use an equation solver like WolframApha or a numerical method such as Newton-Raphson method to obtain the value of $$r$$.

• @rossng The Excel sheet is now available for downloading. The link is given in the answer. Try it out.
– YNK
Commented Apr 22, 2023 at 17:22
• This is a fantastically detailed answer. Thank you for the effort put into this! Commented May 1, 2023 at 14:19

I assume you know $$A, B, C, D$$. Then $$M_{AB} = \tfrac12(A+B)$$ and similarly $$M_{CD} = \tfrac12(C+D)$$. Then the translation is just $$M_{CD}- M_{AB}$$.

If $$U$$ and $$V$$ are unit vectors in the directions $$AB$$ and $$CD$$ respectively, then the rotation angle is the angle between $$U$$ and $$V$$. The simplest way to calculate this is as $$\cos^{-1}(U \cdot V)$$.

• Thanks. I know $A$ and $B$, but not $C$ or $D$. I don't think I have any way to calculate $C$ or $D$ without knowing the radius of the constructed circle and/or the location of the pivot point $P$. Commented Apr 12, 2023 at 11:48