# Straight Line and its construction

(Difference between revisions)
 Revision as of 10:28, 22 June 2010 (edit)← Previous diff Revision as of 11:32, 24 June 2010 (edit) (undo)Next diff → Line 73: Line 73: $\begin{cases} [itex]\begin{cases} - x^2+y^2-2ax-2by+a^2+b^2=m^2 \\ + x^2+y^2-2ax-2by+a^2+b^2=m^2 \cdots \cdots Eqt 1\\ - x^2+y^2-2cx-2dy+c^2+d^2=r^2 \\ + x^2+y^2-2cx-2dy+c^2+d^2=r^2 \cdots \cdots Eqt 2\\ \end{cases}$ \end{cases}[/itex] Subtract the second equation from the first we have, Subtract the second equation from the first we have, - $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2$ + $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2\cdots \cdots Eqt 3$ Substituting $a^2+b^2=r^2$ and rearranging we have, Substituting $a^2+b^2=r^2$ and rearranging we have, Line 85: Line 85: $(-2a+2c)x-(2b-2d)y=m^2-2r^2+c^2+d^2$ $(-2a+2c)x-(2b-2d)y=m^2-2r^2+c^2+d^2$ - Since the right side of the equation is just some constant, we replace them with $k$. Hence we have, + Hence $y=\frac {-2a+2c}{2b-2d}x-\frac {m^2-2r^2+c^2+d^2}{2b-2d} \cdots \cdots Eqt 4$ - $y= \frac {-2a+2c}{2b-2d}x - \frac {k}{2b-2d}$ + Now, we could manipulate Eqt 3 to get an expression for $b$, i.e. $b=f(a,c,d,m,r,x,y)$. Next, we substitute $b=f(a,c,d,m,r,x,y)$ back into Eqt 1 and will be able to obtain an expression for $a$, i.e. $a=g(x,y,d,c,m,r)$. Since $b=\pm \sqrt {r^2-a^2}$, we have expressions of $a$ and $b$ in terms of $x,y,d,c,m$ and $r$. + + Say point $P$ has coordinates $(x',y')$, then $x'=\frac {a+x}{2}$ and $y'=\frac {b+y}{2}$ which will yield + + $\begin{cases} + x=2x'-a \cdots \cdots Eqt 5\\ + y=2y'-b \cdots \cdots Eqt 6 + \end{cases}$ + + In the last step we substitute $a=g(x,y,d,c,m,r)$,$b=\pm \sqrt {r^2-a^2}$, Eqt 5 and Eqt 6 back into Eqt 4 and we will finally have a relationship between $x'$ and $y'$. Of course, it will be a messy one but we could definitely use Mathematica to do the maths. - [[User:Xingda|Xingda]] 6/21 Now what? How is this related to P? ===='''Parametric Description'''==== ===='''Parametric Description'''==== - [[Image:JW point P 2.png|center|800px]] + [[Image:JWpara.png|center|800px]] We intend to parametrize the $P$ with the angle $\theta$. We intend to parametrize the $P$ with the angle $\theta$. Line 99: Line 107: $\begin{cases} [itex]\begin{cases} \overrightarrow {AB} = (r \sin \theta, r \cos \theta) \\ \overrightarrow {AB} = (r \sin \theta, r \cos \theta) \\ - \overrightarrow {BC} = (m \sin (\theta + ? + \beta + \alpha), m \cos (\theta + ? + \beta + \alpha)) \\ + \overrightarrow {BC} = (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha)) \\ \end{cases}$ \end{cases}[/itex] Now let $BD=l$. Then using cosine formula, we have Now let $BD=l$. Then using cosine formula, we have - $\begin{cases} + [itex]m^2+l^2-2ml\cos \alpha = r^2$ - m^2+l^2-2ml\cos \alpha = r^2 \\ + - r^2+l^2-2rl\cos \beta = m^2 \\ + - \end{cases}[/itex] + As a result, we can express \alpha and \beta as As a result, we can express \alpha and \beta as - $\alpha = \cos^{-1} \frac {m^2+l^2-r^2}{2ml}$ and $\beta = \cos^{-1} \frac {r^2+l^2-m^2}{2rl}$ + $\alpha = \cos^{-1} \frac {m^2+l^2-r^2}{2ml}$ - Since $l = \sqrt{(c-r \sin \theta)^2+(d-r \cos \theta)^2}$, $c$ and $d$ being the coordinates of point $D$, we can find $\alpha$ and $\beta$ in terms of $\theta$. + Since $l = \sqrt{(c-r \sin \theta)^2+(d-r \cos \theta)^2}$, $c$ and $d$ being the coordinates of point $D$, we can find $\alpha$ in terms of $\theta$. + + Furthermore, \begin{align} + \overrightarrow {BD} & = \overrightarrow {AD}-\overrightarrow {AB} \\ + & = (c,d) - (r\sin \theta, r \cos \theta) \\ + & = (c - r\sin \theta, d - r \cos \theta) + \end{align} + + $\therefore \beta = \tan^{-1}\frac {d-r \cos \theta}{c - r \sin \theta}$ Hence, Hence, Line 118: Line 131: \begin{align} [itex]\begin{align} \overrightarrow {AP} & = \overrightarrow {AB} + \frac {1}{2} \overrightarrow {BC} \\ \overrightarrow {AP} & = \overrightarrow {AB} + \frac {1}{2} \overrightarrow {BC} \\ - & = (r \sin \theta, r \cos \theta) + \frac {1}{2}(m \sin (\theta + ? + \alpha + \beta), m \cos (\theta + ? + \alpha + \beta)) \\ + & = (r \sin \theta, r \cos \theta) + \frac {m}{2}(\sin (\frac {\pi}{2} + \alpha + \beta), \cos (\frac {\pi}{2} + \alpha + \beta)) \\ \end{align} \end{align}[/itex] - + Now, \overrightarrow {AP} is parametrized in term of $\theta, c, d, r and [itex]m$. - [[User:Xingda|Xingda]] 6/21 Prof. Maurer, there is problem. Last time we talked, we kinda of missed out angle ?. Now I cannot find angle ? + Line 131: Line 143: + How would we find the parametric equation for point $F$ then? Well, it is easy enough. + + [[Image:JWpointF.png|center|800px]] + + $\overrightarrow {AB} = (r \sin \theta, r \cos \theta) + + \therefore \overrightarrow {AE} = \frac {e+f}{r}(r \sin \theta, r \cos \theta)$ + Furthermore $\overrightarrow {AF} = \overrightarrow {AE} + \overrightarrow {BC}$ + $\therefore \overrightarrow {AF} = \frac {e+f}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$ Take another look at Watt's modified linkage below on the left. More than twenty years later when the patent terms had expired, Phineas Crowther used Watt's simple linkage in his engine in 1800. The position of the Watt's linkage was very prominent in the diagram below. It helps to guide the piston rod that drives the gear. The Crowther engine was used to haul goals or wagons up inclines in the old fashioned mines in northeast England. For more detail in steam engine, see [http://en.wikipedia.org/wiki/Steam_engine Steam Engine][[Image:Img331.gif|border|left|680px]][[Image:Img328.gif|border|center|500px]] Take another look at Watt's modified linkage below on the left. More than twenty years later when the patent terms had expired, Phineas Crowther used Watt's simple linkage in his engine in 1800. The position of the Watt's linkage was very prominent in the diagram below. It helps to guide the piston rod that drives the gear. The Crowther engine was used to haul goals or wagons up inclines in the old fashioned mines in northeast England. For more detail in steam engine, see [http://en.wikipedia.org/wiki/Steam_engine Steam Engine][[Image:Img331.gif|border|left|680px]][[Image:Img328.gif|border|center|500px]]

## Revision as of 11:32, 24 June 2010

How to draw a straight line without a straight edge
Independently invented by a French army officer, Charles-Nicolas Peaucellier and a Lithuanian mathematician Lippman Lipkin, this is the device that draws a straight line without using a straight edge. It was one of the first such devices that draws straight line without a reference guideways and it has important applications in engineering and mathematics.

# Basic Description

What is a straight line? How do you define straightness? How do you construct something straight without assuming you have a straight edge? These are questions that seem silly to ask because they are so natural to human concept. We come to accept that straightness is simply straightness and it is assumed. However, compare this to the way we draw a circle. When using a compass to draw a circle, we are not starting with a figure that we accept as circular; instead, we are using a fundamental property of circles that the points on a circle are a fixed distance from the center. Therefore, this page explores the properties of a straight line and hence its construction.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *A little Geometry

## What is a straight line?

[[Image:SmallGreatCircles [...]

## What is a straight line?

Today, we simply define a line as a one-dimensional object that extents to infinity in both directions and it is straight, i.e. no wiggles along its length. But what is straightness? It is a hard question because we have the picture in our head and the answer right there under our breath but we simply cannot articulate it.

In Euclid's book 'Elements', he defined a straight line as "lying evenly between its extreme points" and it has "breadthless length." The definition is pretty useless. What does he mean if he says "lying evenly"? It tells us nothing about how to describe or construct a straight line. So what is a straightness anyway? There are a few good answers. For instance, in a Cartesian Coordinates, the graph of $y=ax+b$ is a straight line. In addition, we are most familiar with another definition is the shortest distance between two points is a straight line. However, it is important to realize that the definitions of being "shortest" and "straight" are different from that on flat plane. For example, the shortest distance between two points on a sphere is the the "great circle", a section of a sphere that contains a diameter of the sphere, and great circle is straight on the spherical surface.

## The quest to draw a straight line

### The Practical Need

Now having defined what a straight line is, we have to figure out a way to construct it on a plane without using anything that we assume to be straight such as a straight edge or ruler just like how we construct a circle using a compass. Historically, it has been of great interest to mathematicians and engineers not only because it is an interesting question to ponder about but also it has important application in engineering. Since the invention of various steam engines and machines that are powered by them, engineers have been trying to perfect the mechanical linkage to convert all kinds of motions(especially circular motion) to linear motions.

The picture above shows a patent drawing of an early steam engine. It is of the simplest form with a boiler(on the left), a cylinder with piston, a beam(on top) and a pump(on the right side) at the other end. When the piston is at its lowest position, steam is let into the cylinder from valve K and it pushes the piston upwards. Afterward, when the piston is at its highest position, cold water is let in from valve E, cooling the steam in the cylinder and causing the pressure in the the cylinder to drop below the atmospheric pressure. The difference in pressure caused the piston to move downwards. After the piston returns to the lowest position, the whole process is repeated. This kind of steam engine is called "atmospheric" because it utilized atmospheric pressure to cause the downward action of the piston (steam only balances out the atmospheric pressure and allow the piston to return to the highest point). Since in the downward motion, the piston pulls on the beam and in the upward motion, the beam pulls on the piston, the connection between the end of the piston rod and the beam is always in tension(under stretching) and that is why a chain is used as the connection.

Anyway, the piston moves in the vertical direction and the piston rod takes only axial loading, i.e. forces applied in the direction along the rod. However, from the above picture, it is clear that the end of the piston does not move in a straight line due to the fact that the end of the beam describes a segment of a circle. As a result, horizontal forces are created and subjected onto the piston rod. Consequently, the process of wear and tear is very much quickened and the efficiency of the engine greatly compromised. Now considering that the up-and-down cycle repeats itself hundreds of times every minute and the engine is expected to run 24/7 to make profits for the investors, such defect in the engine must not be tolerated and poses a great need for improvements.

Improvements came but in baby-steps. Firstly, "double-action" engines were made, part of which is shown in the picture on the right. Atmospheric pressure acts in both upward and downward strokes of the engine and two chains were used (one connected to the top of the arched end of the beam and one to the bottom), both of which will take turns to be in tension in one cycle. One might ask why chain was used all the time. The answer was simple: to fit the curved end of the beam. However, this does not fundamentally solved the problem and unfortunately created more. The additional chain increased the height of the engine and made the manufacturing very difficult (it was hard to make straight steel bars and rods back then) and costly. Secondly, beam was dispensed and replaced by a gear as shown on the left. Consequently, the piston rod was fitted with teeth(labeled k) to drive the gear. Theoretically, this solves the problem fundamentally. The piston rod is confined between the guiding wheel at K and the gear, and it moves only in the up-and-down motion. However, the practical problem was still there. The friction and the noise between all the guideways and the wheels could not be ignored, not to mention the increased possibility of failure and cost of maintenance due to additional parts. Therefore, both of these methods were not satisfactory and the need for a linkage that produces straight line action was imperative.

### James Watt's breakthrough

James Watt, whose greatest achievement was his improvement on the steam engine, had to find a mechanism that converted the linear motion of pistons in the cylinder to the semi circular motion of the beam(or the circular motion of the flywheel) and vice versa. In 1784, he invented a three member linkage that solved the linear motion to circular problem practically as illustrated by the animation below. In its simplest form, there are two radius arms that have the same lengths and a connecting arm with midpoint P. Point P moves in a straight line. However, this linkage only produced approximate straight line (a stretched figure 8 actually) as shown on the right, much to the chagrin of the mathematicians who were after absolute straight lines. There is a more general form of the Watt's linkage that the two radius arms having different lengths like shown in the figure in the middle. To make sure that Point P still move in the stretched figure 8, it has to be positioned such that it adheres to the ratio

$\frac{AB}{CD} = \frac{CP}{CB}$

### The Motion of Point P

We intend to described the path of P using parametric equations.

#### Algebraic Description

We know coordinates $A$ and $D$. Hence let the coordinates of $A$ be $(0,0)$, coordinates of $B$ be $(c,d)$. We also know the length of the bar. Let $AB=CD=r, BC=m$.

Suppose that at one instance we know the coordinates of $B$ as $(a,b)$, then $P$ will be on the circle centered at $B$ with a radius of $m$. Since $P$ is on the circle centered at $D$ with radius $r$. Then the coordinates of C have to satisfy the two equations below.

$\begin{cases} (x-a)^2+(y-b)^2=m^2 \\ (x-c)^2+(y-d)^2=r^2 \end{cases}$

Now, since we know that $B$ is on the circle centered at $A$ with radius $r$, the coordinates of $B$ have to satisfy the equation $a^2+b^2=r^2$.

Therefore, the coordinates of $C$ have to satisfy the three equations below.

$\begin{cases} (x-a)^2+(y-b)^2=m^2 \\ (x-c)^2+(y-d)^2=r^2 \\ a^2+b^2=r^2 \end{cases}$

Now, expanding the first two equations we have,

$\begin{cases} x^2+y^2-2ax-2by+a^2+b^2=m^2 \cdots \cdots Eqt 1\\ x^2+y^2-2cx-2dy+c^2+d^2=r^2 \cdots \cdots Eqt 2\\ \end{cases}$

Subtract the second equation from the first we have,

$(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2\cdots \cdots Eqt 3$

Substituting $a^2+b^2=r^2$ and rearranging we have,

$(-2a+2c)x-(2b-2d)y=m^2-2r^2+c^2+d^2$

Hence $y=\frac {-2a+2c}{2b-2d}x-\frac {m^2-2r^2+c^2+d^2}{2b-2d} \cdots \cdots Eqt 4$

Now, we could manipulate Eqt 3 to get an expression for $b$, i.e. $b=f(a,c,d,m,r,x,y)$. Next, we substitute $b=f(a,c,d,m,r,x,y)$ back into Eqt 1 and will be able to obtain an expression for $a$, i.e. $a=g(x,y,d,c,m,r)$. Since $b=\pm \sqrt {r^2-a^2}$, we have expressions of $a$ and $b$ in terms of $x,y,d,c,m$ and $r$.

Say point $P$ has coordinates $(x',y')$, then $x'=\frac {a+x}{2}$ and $y'=\frac {b+y}{2}$ which will yield

$\begin{cases} x=2x'-a \cdots \cdots Eqt 5\\ y=2y'-b \cdots \cdots Eqt 6 \end{cases}$

In the last step we substitute $a=g(x,y,d,c,m,r)$,$b=\pm \sqrt {r^2-a^2}$, Eqt 5 and Eqt 6 back into Eqt 4 and we will finally have a relationship between $x'$ and $y'$. Of course, it will be a messy one but we could definitely use Mathematica to do the maths.

#### Parametric Description

We intend to parametrize the $P$ with the angle $\theta$.

$\begin{cases} \overrightarrow {AB} = (r \sin \theta, r \cos \theta) \\ \overrightarrow {BC} = (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha)) \\ \end{cases}$

Now let $BD=l$. Then using cosine formula, we have $m^2+l^2-2ml\cos \alpha = r^2$

As a result, we can express \alpha and \beta as

$\alpha = \cos^{-1} \frac {m^2+l^2-r^2}{2ml}$

Since $l = \sqrt{(c-r \sin \theta)^2+(d-r \cos \theta)^2}$, $c$ and $d$ being the coordinates of point $D$, we can find $\alpha$ in terms of $\theta$.

Furthermore, \begin{align} \overrightarrow {BD} & = \overrightarrow {AD}-\overrightarrow {AB} \\ & = (c,d) - (r\sin \theta, r \cos \theta) \\ & = (c - r\sin \theta, d - r \cos \theta) \end{align}

$\therefore \beta = \tan^{-1}\frac {d-r \cos \theta}{c - r \sin \theta}$

Hence,

\begin{align} \overrightarrow {AP} & = \overrightarrow {AB} + \frac {1}{2} \overrightarrow {BC} \\ & = (r \sin \theta, r \cos \theta) + \frac {m}{2}(\sin (\frac {\pi}{2} + \alpha + \beta), \cos (\frac {\pi}{2} + \alpha + \beta)) \\ \end{align}

Now, $\overrightarrow {AP}$ is parametrized in term of $\theta, c, d, r$ and $m$.

Imitations were a big problems beck in those days. When filing for a patent, James Watt and other inventors, had to explain how their devices work without revealing the critical secrets so that others could easily copy them. As seen in the original patent illustration on the bottom right, Watt illustrated his simple linkage on a separate diagram but we couldn't find it in anywhere in the illustration. That is Watt's secret. What he had actually used on his engine was the modified version of the basic linkage as show on the left. The link $ABCD$ is the original three member linkage with $AB=CD$ and point $P$ being the midpoint of $BC$. A is the pivot of the beam fixed on the engine frame while D is also fixed. However, Watt modified it by adding a parallelogram $BCFE$ to it and connecting point $F$ to the piston rod. We now know that point $P$ moves in quasi straight line as shown previously. The importance for two points move in a straight line is that one has to be connected to the piston rod that drives the beam, another will convert the circular motion to linear motion so as to drive the valve gears that control the opening and closing of the valves. It turns out that point F moves in a similar quasi straight line as point P.

How would we find the parametric equation for point $F$ then? Well, it is easy enough.

$\overrightarrow {AB} = (r \sin \theta, r \cos \theta) \therefore \overrightarrow {AE} = \frac {e+f}{r}(r \sin \theta, r \cos \theta)$

Furthermore $\overrightarrow {AF} = \overrightarrow {AE} + \overrightarrow {BC}$

$\therefore \overrightarrow {AF} = \frac {e+f}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$

Take another look at Watt's modified linkage below on the left. More than twenty years later when the patent terms had expired, Phineas Crowther used Watt's simple linkage in his engine in 1800. The position of the Watt's linkage was very prominent in the diagram below. It helps to guide the piston rod that drives the gear. The Crowther engine was used to haul goals or wagons up inclines in the old fashioned mines in northeast England. For more detail in steam engine, see Steam Engine

Mathematicians and engineers have being searching for almost a century to find the solution to the straight line linkage but all had failed until 1864, a French army officer Charles Nicolas Peaucellier came up with his "inversor linkage. Interestingly, he did not publish his findings and proof until in 1873, when Lipmann I. Lipkin, a student from University of St. Petersburg, demonstrated the same working model at the World Exhibition in Vienna. Peaucellier acknowledged Lipkin's independent findings with the publication of the details of his discovery in 1864 and the mathematical proof.

Take a minute to ponder the question: "How do you produce a straight line?" We all know, or rather assume, that light travels in straight line. But does it always do that? Einstein's theory of relativity has shown (and been verified) that light is bent by gravity and therefore, our assumption that light travels in straight lines does not hold all the time. Another simpler method is just to fold a piece of paper and the crease will be a straight line.

Now, the linkage that produces a straight line motion is much more complicated than folding a piece of paper but the Peaucellier-Lipkin Linkage is amazingly simple as shown on the left and right. In the next section, I will provide a proof of how this linkage indeed draws a straight line.

Let's turn to a skeleton drawing of the Peaucellier-Lipkin linkage. It is constructed in such a way that $OA = OB$ and $AC=CB=BP=PA$. Furthermore, all the bars are free to rotate at every joint and point $O$ is a fixed pivot. Due to the symmetrical construction of the linkage, it goes without proof that points $O$,$C$ and $P$ lie in a straight line. Construct lines $OCP$ and $AB$ and they meet at point $M$.

$\because$ shape $APBC$ is a rhombus

$\therefore AB \perp CP$ and $CM = MP$

Now,

$(OA)^2 = (OM)^2 + (AM)^2$

$(AP)^2 = (PM)^2 + (AM)^2$

\begin{align} \therefore (OA)^2 - (AP)^2 & = (OM)^2 - (PM)^2\\ & = (OM-PM)\cdot(PM + PM)\\ & = OC \cdot OP\\ \end{align}

Let's take a moment to look at the relation $(OA)^2 - (AP)^2 = OC \cdot OP$. Since the length $OA$ and $AP$ are of constant length, then the product $OC \cdot OP$ is of constant value however you change the shape of this construction.

Refer to the graph above. Let's fix the path of point $C$ such that it traces out a circle that has point $O$ on it. $QC$ is the the extra link pivoted to the fixed point $Q$ with $QC=QO$. Construct line $OQ$ that cuts the circle at point $R$. In addition, construct line $PN$ such that $PN \perp OR$.

$\because \angle OCR = 90^\circ$

$\therefore \vartriangle OCR \sim \vartriangle ONP, \frac{OC}{OR} = \frac{ON}{OP}$

$\therefore OC \cdot OP = ON \cdot OR$

$\therefore ON = \frac {OC \cdot OP}{OR} =$constant, i.e. the length of $ON$(or the x-coordinate of $P$ w.r.t $O$) does not change as points $C$ and $P$ move. Hence, point $P$ moves in a straight line. ∎

### Inversive Geometry in Peaucellier-Lipkin Linkage

As a matter of fact, the first part of the proof given above is already sufficient. Due to inversive geometry, once we have shown that points $O$,$C$ and $P$ are collinear and that $OC \cdot OP$ is of constant value. Points $C$ and $P$ are inversive pairs with $O$ as inversive center. Therefore, once $C$ moves in a circle that contains $O$, then $P$ will move in a straight line and vice versa. ∎ See Inversion for more detail.

The new linkage caused considerable excitement in London. Mr. Prim, "engineer to the House", utilized the new compact form invented by H.Hart to fit his new blowing engine which proved to be "exceptionally quiet in their operation." In this compact form, $DA=DC$, $AF=CF$ and $AB = BC$. Point $E$ and $F$ are fixed pivots. In the diagram above. F is the inversive center and points $D$,$F$ and $B$ are collinear and $DF \cdot DB$ is of constant value. I left it to you to prove the rest. Mr. Prim's blowing engine used for ventilating the House of Commons, 1877. The crosshead of the reciprocating air pump is guided by a Peaucellier linkage shown at the center. The slate-lined air cylinders had rubber-flap inlet and exhaust valves and a piston whose periphery was formed by two rows of brush bristles. Prim's machine was driven by a steam engine.

After Peaucellier-Lipkin Linkage was introduced to England in 1874, Mr. Hart of Woolwich devised a new linkage that contained only four links which is the blue part as shown in the picture below. Point $O$ is the inversion center with $OP$ and $OQ$ collinear and $OP \cdot OQ =$ constant. When point $P$ is constrained to move in a circle that passes through point $O$, then point $Q$ will trace out a straight line. See below for proof.

$\because AB = CD, BC = AD \therefore BD \parallel AC$

Draw line $OQ \parallel AC$, intersecting $AD$ at point $P$.

$\therefore$ points $O,P,Q$ are collinear

Construct rectangle $EFCA$

\begin{align} AC \cdot BD & = EF \cdot BD \\ & = (ED + EB) \cdot (ED - EB) \\ & = (ED)^2 - (EB)^2 \\ \end{align}

$\because \begin{array}{lcl} (ED)^2 + (AE)^2 & = & (AD)^2 \\ (EB)^2 + (AE)^2 & = & (AB)^2 \end{array}$

$\therefore AC \cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2$

Further

$\because \frac{OP}{BD} = m, \frac{OQ}{AC} = 1-m$ where $0

\begin{align} \therefore OP \cdot OQ & = m(1-m)BD \cdot AC\\ & = m(1-m)((AD)^2 - (AB)^2) \end{align}

### Other straight line mechanism

There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams below. Consider two circles $C_1$ and $C_2$ with radius having the relation $2r_2=r_1$. We roll $C_2$ inside $C_1$ without slipping as show in the diagram on the right. Then the arch lengths $r_1\beta = r_2\alpha$. Voila! $\alpha = 2\beta$ and point $C$ has to be on the line joining the original points $P$ and $Q$! The same argument goes for point $P$. As a result, point $C$ moves in the horizontal line and point $P$ moves in the vertical line.

In 1801, James White patented his mechanism using this rolling motion. Its picture is shown on the right. Interestingly, if you attach a rod of fixed length to point $C$ and $P$ and the end of the rod $T$ will trace out an ellipse. Why? Consider the coordinates of $P$ in terms of $\theta$, $PT$ and $CT$. Point $P$ will have the coordinates $(CT \cos \theta, PT \sin \theta)$. Now, whenever we see $\cos \theta$ and $\sin \theta$ together, we want to square them. Hence, $x^2=CT^2 \cos^2 \theta$ and $y^2=PT^2 \sin^2 \theta$. Well, they are not so pretty yet. So we make them pretty by dividing $x^2$ by $CT^2$ and $y^2$ by $PT^2$, obtaining $\frac {x^2}{CT^2} = \cos^2 \theta$ and $\frac {y^2}{PT^2} = \sin^2 \theta$. Voila again! $\frac {x^2}{CT^2} + \frac {y^2}{PT^2}=1$ and this is exactly the algebraic formula for an ellipse.

# About the Creator of this Image

KMODDL is a collection of mechanical models and related resources for teaching the principles of kinematics--the geometry of pure motion. The core of KMODDL is the Reuleaux Collection of Mechanisms and Machines, an important collection of 19th-century machine elements held by Cornell's Sibley School of Mechanical and Aerospace Engineering.

# References

How to draw a straight line: a lecture on linkages, Alfred Bray Kempe, Ithaca, New York: Cornell University Library

How round is your circle?, John Bryant and Chris Sangwin, Princeton, Princeton University Press

I need to change the size of the main picture and maybe some more theoretical description what a straight line here.