# Straight Line and its construction

(Difference between revisions)
 Revision as of 11:27, 7 July 2010 (edit)← Previous diff Revision as of 12:18, 7 July 2010 (edit) (undo)Next diff → Line 6: Line 6: |ImageDescElem=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 intuitive. We come to accept that straightness is simply straightness and its definition, like that of point and line, is simply 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 at a fixed distance from the center. This page explores the properties of a straight line and hence its construction. |ImageDescElem=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 intuitive. We come to accept that straightness is simply straightness and its definition, like that of point and line, is simply 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 at a fixed distance from the center. This page explores the properties of a straight line and hence its construction. |ImageDesc=== What is a straight line? == |ImageDesc=== What is a straight line? == - {{{!}}border="1" + {{{!}} {{!}}align="center"{{!}}[[Image:Straightline.jpg|center|border|400px]]Image 1{{!}}{{!}}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. {{!}}align="center"{{!}}[[Image:Straightline.jpg|center|border|400px]]Image 1{{!}}{{!}}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. Line 27: Line 27: ==='''The Practical Need'''=== ==='''The Practical Need'''=== - {{{!}}border="1" + {{{!}} {{!}}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. {{!}}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. {{!}}- {{!}}- Line 54: Line 54: ==='''James Watt's breakthrough'''=== ==='''James Watt's breakthrough'''=== - {{{!}}border="1" + {{{!}} {{!}}colspan="2"{{!}}James Watt found 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 [http://en.wikipedia.org/wiki/Flywheel flywheel]) and vice versa. In 1784, he invented a [http://en.wikipedia.org/wiki/Linkage_(mechanical) 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$\frac{AB}{CD} = \frac{CP}{CB}$. {{!}}colspan="2"{{!}}James Watt found 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 [http://en.wikipedia.org/wiki/Flywheel flywheel]) and vice versa. In 1784, he invented a [http://en.wikipedia.org/wiki/Linkage_(mechanical) 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$\frac{AB}{CD} = \frac{CP}{CB}$. {{!}}- {{!}}- Line 66: Line 66: ===='''Algebraic Description'''==== ===='''Algebraic Description'''==== - {{{!}}border="1" + {{{!}} {{!}}We see that $P$ moves in a stretched figure 8 and will tend to think that there should be a nice close form of the relationship between the coordinates of $P$ like that of the circle. But after this section, you will see that there is a closed form, at least theoretically, but it is not "nice" at all. {{!}}We see that $P$ moves in a stretched figure 8 and will tend to think that there should be a nice close form of the relationship between the coordinates of $P$ like that of the circle. But after this section, you will see that there is a closed form, at least theoretically, but it is not "nice" at all. {{!}}- {{!}}- Line 134: Line 134: ===='''Parametric Description'''==== ===='''Parametric Description'''==== - {{{!}}border="1" + {{{!}} {{!}}Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it would not be too bad if we could describe the motion of $P$ using the angle of ration. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of $P$'s coordinates. {{!}}Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it would not be too bad if we could describe the motion of $P$ using the angle of ration. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of $P$'s coordinates. {{!}}- {{!}}- Line 201: Line 201: ==='''The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage'''=== ==='''The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage'''=== - [[Image:Peaucellier linkage animation.gif|right|border]]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 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. + {{{!}} + {{!}}align="center"{{!}}[[Image:Peaucellier linkage animation.gif|center|border]]Image 13{{!}}{{!}}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 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. Line 208: Line 209: 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, a proof of how this linkage draws a straight line is provided. 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, a proof of how this linkage draws a straight line is provided. - + {{!}}- - + {{!}}colspan="2"{{!}}[[Image:PL cell.png|center|border|550px]] - [[Image:PL cell.png|center|border|550px]] + {{!}}- - + {{!}}align="center" colspan="2"{{!}}Image 14 - 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 $M$. + {{!}}- + {{!}}colspan="2"{{!}}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 $M$. Since shape $APBC$ is a rhombus Since shape $APBC$ is a rhombus Line 232: Line 234: 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. 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. - + {{!}}- - + {{!}}colspan="2"{{!}}[[Image:PLcellproof2.png|border|center|550px]] - [[Image:PLcellproof2.png|border|center|550px]] + {{!}}- - + {{!}}align="center" colspan="2"{{!}}Image 15 - 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 $PN \perp OR$. + {{!}}- + {{!}}colspan="2"{{!}}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 $PN \perp OR$. Since, $\angle OCR = 90^\circ$ Since, $\angle OCR = 90^\circ$ Line 246: Line 249: 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. ∎ 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'''=== ==='''Inversive Geometry in Peaucellier-Lipkin Linkage'''=== Line 253: Line 256: ==='''Peaucellier-Lipkin Linkage in action'''=== ==='''Peaucellier-Lipkin Linkage in action'''=== - [[Image:Adapted.jpg|border|600px|center|Mr.Prim's adaptation]] + {{{!}} - 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. + {{!}}[[Image:Adapted.jpg|border|600px|center|Mr.Prim's adaptation]] - [[Image:Blowing engine.jpg|center|thumb|600px]] + {{!}}- + {{!}}align="center"{{!}}Image 16 + {{!}}- + {{!}}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. + {{!}}- + {{!}}[[Image:Blowing engine.jpg|center|border|600px]] + {{!}}- + {{!}}align="center"{{!}}Image 17 + {{!}}} ==='''Hart's Linkage'''=== ==='''Hart's Linkage'''=== - 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. + {{{!}} - [[Image:Hartlinkage3.png|border|center|600px]] + {{!}}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. - + {{!}}- - We know that $AB = CD, BC = AD$ + {{!}}[[Image:Hartlinkage3.png|border|center|600px]] + {{!}}- + {{!}}align="center"{{!}}Image 18 + {{!}}- + {{!}}We know that $AB = CD, BC = AD$ As a result, $BD \parallel AC$ As a result, $BD \parallel AC$ Line 294: Line 309: & = m(1-m)((AD)^2 - (AB)^2) & = m(1-m)((AD)^2 - (AB)^2) \end{align}[/itex] \end{align}[/itex] - + {{!}}} ==='''Other straight line mechanism'''=== ==='''Other straight line mechanism'''=== - + {{{!}} - [[Image:Circle in circle 1.png|border|center|325px]] + {{!}}[[Image:Circle in circle 1.png|border|center|325px]] - There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams above. 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 below. + {{!}}- - + {{!}}align="center"{{!}}Image 19 - [[Image:Circle in circle 2.png|border|center|300px]] + {{!}}- - 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. + {{!}}There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams above. 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 below. - + {{!}}- - [[Image:Img335.gif|border|center|300px]] + {{!}}[[Image:Circle in circle 2.png|border|center|300px]] - 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 $T$ 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. + {{!}}- - + {{!}}align="center"{{!}}Image 20 - + {{!}}- - [[Image:Ellipsograph2.png|border|center|500px]] + {{!}}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. + {{!}}- + {{!}}align="center"{{!}}Image 21 + {{!}}- + {{!}}[[Image:Img335.gif|border|center|300px]] + {{!}}- + {{!}}align="center"{{!}}Image 22 + {{!}}- + {{!}}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 $T$ 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. + {{!}}- + {{!}}[[Image:Ellipsograph2.png|border|center|500px]] + {{!}}- + {{!}}align="center"{{!}}Image 23 + {{!}}} |other=A little Geometry |other=A little Geometry |AuthorName=Cornell University Libraries and the Cornell College of Engineering |AuthorName=Cornell University Libraries and the Cornell College of Engineering

## Revision as of 12:18, 7 July 2010

How to draw a straight line without a straight edge
Independently invented by a French army officer, Charles-Nicolas Peaucellier and a Lithuanian (disputable) mathematician Lipmann Lipkin, this is the device that draws a straight line without using a straight edge. It was the first planar linkage that drew straight line without a reference guideway and it had 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 intuitive. We come to accept that straightness is simply straightness and its definition, like that of point and line, is simply 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 at a fixed distance from the center. 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 [...]

## What is a straight line?

 Image 1 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 width." 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 the 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 a 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. For more properties on staight line, you refer to the book Experience Geometry by zzz. Image 2

## The Quest to Draw a Straight Line

### James Watt's breakthrough

 James Watt found 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$\frac{AB}{CD} = \frac{CP}{CB}$. Image 6 Image 7

### The Motion of Point P

We intend to described the path of $P$ so that we could show it does not move in a straight line (which is obvious) and more importantly to pinpoint the position of $P$ using certain parameter we know such as the angle of rotation or one coordinate of point $P$. This is awfully important in engineering as engineers would like to know that there are no two parts of the machine will collide with each other throughout the motion.

#### Algebraic Description

We see that $P$ moves in a stretched figure 8 and will tend to think that there should be a nice close form of the relationship between the coordinates of $P$ like that of the circle. But after this section, you will see that there is a closed form, at least theoretically, but it is not "nice" at all.
Image 8
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,

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

Subtract Eq. 2 from Eq. 1 we have,

 $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2\cdots \cdots$ Eq. 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 $y=\frac {-2a+2c}{2b-2d}x-\frac {m^2-2r^2+c^2+d^2}{2b-2d} \cdots \cdots$ Eq. 4

Now, we could manipulate Eq. 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 Eq. 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

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

In the last step we substitute $a=g(x,y,d,c,m,r)$,$b=\pm \sqrt {r^2-a^2}$, Eq. 5 and Eq. 6 back into Eq. 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

 Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it would not be too bad if we could describe the motion of $P$ using the angle of ration. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of $P$'s coordinates. Image 9 We will parametrize the $P$ with the angle $\theta$ in conformation of most parametrizations of point. $\begin{cases} \overrightarrow {AB} = (r \sin \theta, r \cos \theta) \\ \overrightarrow {BC} = (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha)) \\ \end{cases}$ 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 $m$. Image 10 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. Image 11 How would we find the parametric equation for point $F$ then? Well, it is easy enough. Image 12 $\overrightarrow {AB} = (r \sin \theta, r \cos \theta) \therefore \overrightarrow {AE} = \frac {e+f}{r}(r \sin \theta, r \cos \theta)$ Furthermore $\overrightarrow {AF} = \overrightarrow {AE} + \overrightarrow {BC}$ Therefore, $\overrightarrow {AF} = \frac {e+f}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$

### The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage

 Image 13 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 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, a proof of how this linkage draws a straight line is provided. Image 14 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 $M$. Since shape $APBC$ is a rhombus $AB \perp CP$ and $CM = MP$ Now, $(OA)^2 = (OM)^2 + (AM)^2$ $(AP)^2 = (PM)^2 + (AM)^2$ Therefore, \begin{align} (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. Image 15 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 $PN \perp OR$. Since, $\angle OCR = 90^\circ$ We have $\vartriangle OCR \sim \vartriangle ONP, \frac{OC}{OR} = \frac{ON}{OP}$, and $OC \cdot OP = ON \cdot OR$ 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.

### Peaucellier-Lipkin Linkage in action

 Image 16 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. Image 17

 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. Image 18 We know that $AB = CD, BC = AD$ As a result, $BD \parallel AC$ Draw line $OQ \parallel AC$, intersecting $AD$ at point $P$. Consequently, 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} For $\begin{array}{lcl} (ED)^2 + (AE)^2 & = & (AD)^2 \\ (EB)^2 + (AE)^2 & = & (AB)^2 \end{array}$, We then have $AC \cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2$. Further, due to $\frac{OP}{BD} = m, \frac{OQ}{AC} = 1-m$ where $0 We have \begin{align} OP \cdot OQ & = m(1-m)BD \cdot AC\\ & = m(1-m)((AD)^2 - (AB)^2) \end{align}

### Other straight line mechanism

 Image 19 There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams above. 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 below. Image 20 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. Image 21 Image 22 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 $T$ 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. Image 23

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# 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.

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