# Straight Line and its construction

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 Revision as of 13:49, 3 June 2010 (edit)← Previous diff Current revision (12:21, 29 November 2011) (edit) (undo) (Reducing length of Image Name.) (46 intermediate revisions not shown.) Line 1: Line 1: - {{Image Description + {{Image Description Ready - |ImageName=How to draw a straight line without a straight edge + |ImageName=Drawing a Straight Line |Image=S35-1.jpg |Image=S35-1.jpg - |ImageIntro=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. + |ImageIntro=The image shows the first planar {{EasyBalloon|Link=linkage|Balloon=It is defined as a series of rigid links connected with joints to form a closed chain, or a series of closed chains. Each link has two or more joints, and the joints have various degrees of freedom to allow motion between the links.Wikipedia (Linkage (mechanical))}} that drew a straight line without using a straight edge. Independently invented by a French army officer, Charles-Nicolas Peaucellier and a Lithuanian (who some argue was actually Russian) mathematician Lipmann Lipkin, it had important applications in engineering and mathematics.'''Bryant, & Sangwin, 2008, p. 34Kempe, 1877, p. 12Taimina''' - |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 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. + - |ImageDesc=== What is a straight line? == + - [[Image:Straightline.jpg|left|border|400px]][[Image:SmallGreatCircles 700.gif|right|border]] + - 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. + + =Introduction= + What is a straight line? How do you define straightness? The questions 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, why do we not assume the definition of 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 answer to the question "how do you construct a straight line without a straight edge?" - 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. + =What Is A Straight Line?--- A Question Rarely Asked.= + {{{!}} + {{!}}colspan="2"{{!}}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 can picture it, but we simply cannot articulate it. + In Euclid's book '''''Elements''''', he defined a straight line as "lying evenly between its extreme points" and as having "breadthless width." This definition is pretty useless. What does he mean by "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 {{EasyBalloon|Link=Cartesian Coordinates|Balloon=A Cartesian coordinate system specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances from the point to two fixed perpendicular directed lines, measured in the same unit of length.Wikipedia (Cartesian coordinate system)}}, the graph of $y=ax+b$ is a straight line as shown in '''Image 1'''. In addition, the shortest distance between two points on a flat plane is a straight line, a definition we are most familiar with. However, it is important to realize that the definitions of being "shortest" and "straight" will change when you are no longer on flat plane. For example, the shortest distance between two points on a sphere is the the {{EasyBalloon|Link="great circle"|Balloon=A section of a sphere that contains a diameter of the sphere, and great circle is straight on the spherical surface}}as shown in '''Image 2'''. + Since we are dealing with plane geometry here, we define straight line as the curve of $y=ax+b$ in Cartesian Coordinates. - == The quest to draw a straight line == + For more comprehensive discussion of being straight, you can refer to the book '''''Experiencing Geometry''''' by David W. Henderson. - ==='''Historical Development'''=== + - 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. + - [[Image:Img324.gif|center|border|600px]] + - The picture above shows a patent drawing of an early steam engine. It is of the simplest form with a boiler, a cylinder with piston, a beam and a pump at the other end. When the piston is at its lowest, steam is let into the cylinder from valve K and pushes the piston upwards. Afterward, when the piston is at its highest position, cold water is let in from valve E so that it cools the steam in the cylinder, causing the pressure in the the cylinder to drop below the atmospheric pressure which is on the other side the piston and pushes the piston downwards. This cycle is repeated by a link action that is not the case of the discussion here. This kind of steam engine is called "single-action" engine because power was only developed in the downward stroke, i.e. pumping water out of a mine for example. The upward stroke was made possible because of the balancing moment of the beam and since, the connection between the end of the piston rod and the beam is always in tension, a chain is used as the connection. + + Take a minute to ponder the question: "How do you produce a straight line?" Well light travels in straight line. Can we make light help us to produce something straight? Sure but does it always travel in straight line? 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. Well, another simpler method is just to fold a piece of paper and the crease will be a straight line. However, to achieve our ultimate goal (construct a straight line without a straight edge), we need a {{EasyBalloon|Link=linkage|Balloon=It is defined as a series of rigid links connected with joints to form a closed chain, or a series of closed chains. Each link has two or more joints, and the joints have various degrees of freedom to allow motion between the links.Wikipedia (Linkage (mechanical))}} and that is much more complicated and difficult than folding a piece of paper. The rest of the page revolves around the discussion of straight line linkage's history and its mathematical explanation. + {{!}}- + {{!}}align="center"{{!}}[[Image:Straightline.jpg|center|border|300px]] '''Image 1'''{{!}}{{!}}align="center"{{!}}[[Image:SmallGreatCircles 700.gif|center|border|400px]]'''Image 2''' Weisstein + {{!}}} - Anyway, the piston moves in a vertical movement and the piston rod takes only axial loading, i.e. loading 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. + = The Quest to Draw a Straight Line = + =='''The Practical Need'''== + {{{!}} + {{!}}Now having defined what a straight line is, we must figure out a way to construct it on a plane. However, the challenge is to do that 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 because it has important applications 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. + {{!}}- + {{!}}[[Image:Img324.gif|center|border|350px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 3'''Bryant, & Sangwin, 2008, p. 18 + {{!}}- + {{!}}'''Image 3''' shows a patent drawing of an early steam engine. It is of the simplest form with a boiler (lower left corner), a cylinder with piston (above the boiler), a beam (on top, pivoted at the middle) and a pump (lower right corner) at the other end. The pump was usually used to extract water from the mines but other devices can also be driven. + {{!}}- + {{!}}{{HideShowThis|ShowMessage=Click here to show how this engine works.|HideMessage=Click here to hide text|HiddenText=When the piston is at its lowest position, steam is let into the cylinder from valve K to push 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 causes 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 utilizes atmospheric pressure to cause the downward action of the piston. 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 (it is being stretched by forces at two ends) and that is why a chain is used as the connection.Bryant, & Sangwin, 2008, p. 18 Wikipedia (Steam Engine)}} + {{!}}- + {{!}}Ideally, 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 an arc of a circle. As a result, horizontal forces are created and subjected onto the piston rod. Consequently, the rate of attrition is very much expedited and the efficiency of the engine is greatly compromised. Durability is important in the design of any machine, but it was especially essential for the early steam engines. For these machines were meant to run 24/7 to make profits for the investors. Therefore, such defect in the engine posed a great need for improvements.Bryant, & Sangwin, 2008, p. 18-21 + {{!}}} - [[Image:Img325.gif|right|border|700px]][[Image:Img326.gif|border|left|250px]] + {{{!}} - Improvements came but in baby-steps. Firstly, "double-action" engines were made as shown in the picture on the right. It created power in both upward and downward stroke of the engine and two chains were used, 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 guides 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 to drive the gear. Theoretically, this solves the problem fundamentally. The piston rod is confined between the wheel at K and the gear, and it moves up-and-down. 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. + {{!}}colspan="2"{{!}}Improvements were soon developed to force the end of the piston rod move in a straight line, but these brought about new mechanical problems. The two pictures below show two improvements at the time. The hidden text explains how these improvements work and why they have failed to produce satisfactory results. + {{!}}- + {{!}}align="center"{{!}}[[Image:Img325.gif|center|border|500px]]'''Image 4'''Bryant, & Sangwin, 2008, p. 18-21{{!}}{{!}}align="center"{{!}}[[Image:Img326.gif|border|center|200px]]'''Image 5''' Bryant, & Sangwin, 2008, p. 18-21 + {{!}}- + {{!}}colspan="2"{{!}}{{HideShowThis|ShowMessage=Click to read more about these systems.|HideMessage=Hide|HiddenText=Firstly, "double-action" engines were built, part of which is shown in '''Image 4'''. Secondly, the beam was dispensed and replaced by a gear as shown in '''Image 5'''. However, both of these improvements were unsatisfactory and the need for a straight line linkage was still imperative. In '''Image 4''', 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 took turns being taut throughout 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 solve the straight line 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. + In '''Image 5''', after the beam was replaced by gear actions, 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 remained unsolved. 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.Bryant, & Sangwin, 2008, p. 18-21}} + {{!}}} - ==='''James Watt's breakthrough'''=== - [[Image:Watt James von Breda.jpg|left|400px|]] + =='''James Watt's breakthrough'''== - James Watt, whose greatest achievement was his improvement on the steam engine, had to find a mechanism to that converted the linear motion of pistons in the cylinder to the rocking 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 problem practically as illustrated by the animation below. The simplest form is that there are two radius arms that have the same lengths and point P is the midpoint of the link in the middle. 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 on the left. Point P therefore must 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 (that is moving in an arc of the circle) of the beam (or the circular motion of the [http://en.wikipedia.org/wiki/Flywheel flywheel]) and vice versa. In this way, energy in the vertical direction is converted to rotational energy of the flywheel from where is it converted to useful work that the engine is desired to do. 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 while the two hinges move in circular arcs. However, this linkage only produced approximate straight line (a stretched figure 8 actually) as shown in '''Image 7''', 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 '''Image 6'''. 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$\frac{AB}{CD} = \frac{CP}{CB}$.Bryant, & Sangwin, 2008, p. 24 + {{!}}- + {{!}}[[Image:Img327.gif|center|border|300px]]{{!}}{{!}}[[Image:Watts linkage.gif|center|border]] + {{!}}- + {{!}}align="center"{{!}}'''Image 6''' Bryant, & Sangwin, 2008, p. 23{{!}}{{!}}align="center"{{!}}'''Image 7''' Wikipedia (Watt's Linkage) + {{!}}} - [[Image:Watts linkage.gif|border|right]] + =='''The Motion of Point P'''== - [[Image:Img327.gif|center|border|500px]] + We intend to describe the path of $P$ so that we can show it does not move in a straight line (which is obvious in the animation). More importantly, this will allow us to pinpoint the position of $P$ using certain parameters we know, such as the angle of rotation or one coordinate of point $P$. This is awfully crucial 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. In addition, you can use the parametrization to create your own animation like that in '''Image 7'''. + ==='''Algebraic Description'''=== + {{{!}} + {{!}}We see that $P$ moves in a stretched figure 8 and will tend to think that there should be a nice {{EasyBalloon|Link=closed form|Balloon=In mathematics, an expression is said to be a closed-form expression if, and only if, it can be expressed analytically in terms of a bounded number of certain "well-known" functions. Typically, these well-known functions are defined to be elementary functions – constants, one variable x, elementary operations of arithmetic (+ – × ÷), nth roots, exponent and logarithm (which thus also include trigonometric functions and inverse trigonometric functions).Wikipedia (Closed-form expression)}} of the relationship of the $x$ and $y$ coordinates of $P$ like that of the circle. After this section, you will see that there is a closed form, at least theoretically, but it is not "nice" at all. + {{!}}- + {{!}}[[Image:JW point P.png|center|500px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 8''' + {{!}}- + {{!}}{{SwitchPreview|ShowMessage=Show derivation of the relationship of the $x$ and $y$ coordinates of $P$.|HideMessage=Hide|PreviewText=We know coordinates ${\color{Gray}A}$ and ${\color{Gray}D}$ because they are fixed.|FullText=We know coordinates $A$ and $D$ because they are fixed. Hence suppose the coordinates of $A$ are $(0,0)$ and coordinates of $B$ are $(c,d)$. We also know the length of the bar. Let $AB=CD=r, BC=m$. - Imitations were a big problems beck in the days of James Watt and inventors, when filing for a patent, 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 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 $BC$. 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 the drive the valve gears that control the opening and closing of the valves. What is the motion of point $F$? When $AB=BE$, point $F$ moves in the magnified path of point $P$. What happens when $AB \ne BE$ as illustrated? Does $F$ moves in a better quasi straight line than $P$? What is the parametric equations for the stretched 8 figure? + Suppose that at one instance we know the coordinates of $B$ as $(a,b)$, then $C$ will be on the circle centered at $B$ with a radius of $m$. Since $C$ 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 $a^2+b^2=r^2$. - [[Image:Watt1.png|left|border|800px]][[Image:Watt2.gif|center|border|600px]] + 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, + {{EquationRef2|Eq. 1}}$x^2+y^2-2ax-2by+a^2+b^2=m^2$ - 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 whose end drove the gear. The Crowther engine was used to haul goals or wagons up inclines in the old fashioned mines in northeast England.[[Image:Img331.gif|border|left|680px]][[Image:Img328.gif|border|center|500px]] + {{EquationRef2|Eq. 2}}$x^2+y^2-2cx-2dy+c^2+d^2=r^2$ - ==='''The first planar straight line linkage - Peaucellier-Lipkin Linkage'''=== - 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. + Subtract {{EquationNote|Eq. 2}} from {{EquationNote|Eq. 1}} we have, - 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.[[Image:Peaucellier linkage animation.gif|right|border]] + {{EquationRef2|Eq. 3}} $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2$ + Substituting $a^2+b^2=r^2$ and rearranging we have, - [[Image:PL cell.png|center|border|800px]] + $(-2a+2c)x-(2b-2d)y=m^2-2r^2+c^2+d^2$ - 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 $M$. + Hence {{EquationRef2|Eq. 4}} $y=\frac {-2a+2c}{2b-2d}x-\frac {m^2-2r^2+c^2+d^2}{2b-2d}$ - $\because$ shape $APBC$ is a rhombus + Now, we can manipulate {{EquationNote|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 {{EquationNote|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 $r$. - $\therefore AB \perp CP$ and $CM = MP$ + Say point $P$ has coordinates $(x',y')$, then $x'=\frac {a+x}{2}$ and $y'=\frac {b+y}{2}$ which will yield - Now, + {{EquationRef2|Eq. 5}} $x=2x'-a$ - $OA^2 = OM^2 + AM^2$ + {{EquationRef2|Eq. 6}} $y=2y'-b$ - $AP^2 = PM^2 + AM^2$ + In the last step we substitute $a=g(x,y,d,c,m,r)$,$b=\pm \sqrt {r^2-a^2}$, {{EquationNote|Eq. 5}} and {{EquationNote|Eq. 6}} back into {{EquationNote|Eq. 4}} and we will finally have a relationship between $x'$ and $y'$. Of course, it will be a messy closed form but we could definitely use Mathematica to do the maths. The point is, there is no nice algebraic form for that figure 8, though it has closed form and that is why we have to find something else.}} + {{!}}} - \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. + ==='''Parametric Description'''=== + {{{!}} + {{!}}Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it may be more manageable to describe the motion of $P$ using the angle of ratation. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of $P$'s coordinates. + {{!}}- + {{!}}[[Image:ParaP.png|center|500px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 9''' + {{!}}- + {{!}}{{SwitchPreview|ShowMessage=Show parameterization of $P$.|HideMessage=Hide|PreviewText=We will parametrize the ${\color{Gray}P}$ with the angle ${\color{Gray}\theta}$ in conformation of most parametrizations of point.|FullText=We will parametrize the $P$ with the angle $\theta$ in conformation of most parametrizations of point. - '''I should include an applet.''' + \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} - [[Image:PLcellproof2.png|border|center|750px]] + Now let $BD=l$. Then using cosine formula, we have + $m^2+l^2-2ml\cos \alpha = r^2$ - 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 $PN \perp OR$. + As a result, we can express $\alpha$ and $\beta$ as - $\because \angle OCR = 90^\circ$ + $\alpha = \cos^{-1} \frac {m^2+l^2-r^2}{2ml}$ + 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 $\theta$. - $\therefore \vartriangle OCR \sim \vartriangle ONP, \frac{OC}{OR} = \frac{ON}{OP}$ + 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, $\beta = \tan^{-1}\frac {d-r \cos \theta}{c - r \sin \theta}$ - $\therefore OC \cdot OP = ON \cdot OR$ + 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} - $\therefore ON = \frac {OC \cdot OP}{OR} =$constant, i.e. the length of $ON$(or the x-coordinate of $P$ w.r.t O[/itex]) does not change as points $C$ and $P$ move. Hence, point $P moves in a straight line. QED + Now, [itex]\overrightarrow {AP}$ is parametrized in term of $\theta, c, d, r$ and $m$.}} + {{!}}- + {{!}}[[Image:Watt2.gif|center|border|450px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 10''' Lienhard, 1999, February 18 + {{!}}} + ==Watt's Secret== + {{{!}} + {{!}}colspan="2"{{!}}Another reason we parameterized $P$ is that Watt did not simply used that three bar linkage shown in '''Image 6''' and '''Image 7'''. Instead he used something different. To understand that, our knowledge of the parameterizaion of $P$ is crucial. Imitations were a big problems back in those days. When filing for a patent, James Watt and other inventors had to explain how their devices worked without revealing the critical secrets so that others could easily copy them. As shown in '''Image 10''', the original patent illustration, Watt illustrated his simple linkage on a separate diagram on the upper left hand corner but try looking for it on the engine illustration itself. Can you find it at all? That is Watt's secret. This is the equivalent of telling you by using the principle of 1+1 makes 2 you could get 34 x 45; the crucial step in understanding (and to make the engine work smoothly in Watt's case) is avoided. What he had actually used on his engine was the modified version of the basic linkage as show in '''Image 11'''. - ==='''Inversive Geometry in Peaucellier-Lipkin Linkage'''=== - As a matter of fact, the proof is complete once we have shown that points $O$,$C$ and $P$ are collinear and that $OC \cdot OP$ is of constant value because of inversive geometry. The above two conditions are sufficient to show that 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. QED + 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. It is important for two points to move in straight lines now is because one has to be connected to the piston rod that drives the beam, another has to 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. This is the truly famous James Watt's "parallel motion" linkage. + {{!}}- + {{!}}[[Image:Watt1.png|center|border|400px]]{{!}}{{!}}[[Image:PointF.png|center|350px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 11'''{{!}}{{!}}align="center"{{!}}'''Image 12''' + {{!}}- + {{!}}{{SwitchPreview|ShowMessage=How would we find the parametric equation for point $F$ then?|HideMessage=Hide|PreviewText=Well, it is easy enough. Refer to '''Image 12'''.|FullText=Well, it is easy enough. Refer to '''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, $\overrightarrow {AF} = \frac {e+r}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$. We now have the parameterization of point $F$ and $P$ and Watt's secret is eventually cracked.}} + {{!}}} - '''I should include an applet.''' - ==='''Peaucellier-Lipkin Linkage in action'''=== + =='''The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage'''== - [[Image:Adapted.jpg|border|1000px|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. + {{!}}align="center"{{!}}[[Image:Peaucellier linkage animation.gif|center|border]]'''Image 13''' Wikipedia (Peaucellier–Lipkin linkage){{!}}{{!}}Anyway, mathematicians and engineers had being searching for almost a century to find the solution to a straight line linkage but all had failed until 1864 when 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. '''Taimina''' - [[Image:Blowing engine.jpg|center|border|]] + - '''I should include an applet, or animation''' - ==='''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. + {{!}}colspan="2"{{!}}[[Image:PL cell.png|center|border|500px]] - [[Image:HART.png|border|center|800px]] + {{!}}- + {{!}}align="center" colspan="2"{{!}}'''Image 14''' + {{!}}- + {{!}}colspan="2"{{!}}Let's turn to a skeleton drawing of the Peaucellier-Lipkin linkage in '''Image 14'''. 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 $M$. - $\because AB = CD, BC = AD + Since shape [itex]APBC$ is a rhombus - \therefore BD \parallel AC[/itex] + $AB \perp CP$ and $CM = MP$ - Draw line $OQ \parallel AC$, intersecting $AD$ at point $P$. + Now, - $\therefore$ points $O,P,Q$ are collinear + $(OA)^2 = (OM)^2 + (AM)^2$ - Construct rectangle $EFCA$ + $(AP)^2 = (PM)^2 + (AM)^2$ + Therefore, \begin{align} [itex]\begin{align} - AC \cdot BD = EF \cdot BD\\ + (OA)^2 - (AP)^2 & = (OM)^2 - (PM)^2\\ - & = (ED + EB) \cdot (ED - EB)\\ + & = (OM-PM)\cdot(OM + PM)\\ - & = ED^2 - EB^2\\ + & = OC \cdot OP\\ \end{align} \end{align}[/itex] - $\because \begin{array}{lcl} + Let's take a moment to look at the relation [itex](OA)^2 - (AP)^2 = OC \cdot OP$. 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. - ED^2 + AE^2 & = & AD^2 \\ + {{!}}- - EB^2 + AE^2 & = & AB^2 + {{!}}colspan="2"{{!}}[[Image:PLcellproof2.png|border|center|500px]] - \end{array} + {{!}}- - [/itex] + {{!}}align="center" colspan="2"{{!}}'''Image 15''' + {{!}}- + {{!}}colspan="2"{{!}}Refer to '''Image 15'''. Let's fix the path of point $C$ such that it traces out a circle that has point $O$ on it. $QC$ is 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 $PN \perp OR$. - $\therefore AC \cdot BD = ED^2 - EB^2 = AD^2 - AB^2$ + Since, $\angle OCR = 90^\circ$ - Further - $\because \frac{OP}{BD} = m, \frac{OQ}{AC} = 1-m$ + We have $\vartriangle OCR \sim \vartriangle ONP and \frac{OC}{OR} = \frac{ON}{OP}$. - where $0Bryant, & Sangwin, 2008, p. 33-36 + {{!}}} + + =='''Inversive Geometry in Peaucellier-Lipkin Linkage'''== + As a matter of fact, the first part of the proof given above is already sufficient. Due to inversive geometry, once we have shown that points [itex]O$,$C$ 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:Adapted.jpg|border|550px|center|Mr.Prim's adaptation]] + {{!}}- + {{!}}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 '''Image 16'''. F is the inversive center and points $D$,$F$ and $B$ are collinear and $DF \cdot DB$ is of constant value. + {{!}}- + {{!}}[[Image:Blowing engine.jpg|center|border|600px]] + {{!}}- + {{!}}align="center"{{!}}'''Image 17''' + {{!}}- + {{!}}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 in the middle of '''Image 17'''. Prim's machine was driven by a steam engine.Ferguson, 1962, p. 205 + {{!}}} + + + =='''Hart's Linkage'''== + {{{!}} + {{!}}After the Peaucellier-Lipkin Linkage was introduced to England in 1874, Mr. Hart of Woolwich Academy Kempe, 1877, p. 18 devised a new linkage that contained only four links which is the blue part as shown in '''Image 18'''. The next part will prove that 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|550px]] + {{!}}- + {{!}}align="center"{{!}}'''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} [itex]\begin{align} - \therefore OP \cdot OQ = m(1-m)BD \cdot AC\\ + AC \cdot BD & = EF \cdot BD \\ - & = m(1-m)(AD^2 - AB^2) + & = (ED + EB) \cdot (ED - EB) \\ + & = (ED)^2 - (EB)^2 \\ \end{align} \end{align}[/itex] - '''what is wrong with the formatting? Also I need animation''' + For \begin{array}{lcl} + (ED)^2 + (AE)^2 & = & (AD)^2 \\ + (EB)^2 + (AE)^2 & = & (AB)^2 + \end{array} + , + we then have $AC \cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2$. - ==='''Other straight line mechanism'''=== + Further, let's define $\frac{OP}{BD} = m, hence \frac{OQ}{AC} = 1-m$ - 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. + where $0Bryant, & Sangwin, 2008, p.44 + {{!}}- + {{!}}colspan="3"{{!}}There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to '''Image 19'''. Consider two circles [itex]C_1$ and $C_2$ with radius having the relation $2r_2=r_1$. We roll $C_2$ inside $C_1$ without slipping as show in '''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. In 1801, James White patented his mechanism using this rolling motion. It is shown in '''Image 21''' Bryant, & Sangwin, 2008, p.42-44. + {{!}}- + {{!}}colspan="3" align="center"{{!}}[[Image:Ellipsograph2.png|border|center|500px]] + {{!}}- + {{!}}colspan="3" align="center"{{!}}'''Image 22''' + {{!}}- + {{!}}colspan="3"{{!}}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 as seen in '''Image 22'''. Why? Consider the coordinates of $P$ in terms of $\theta$, $PT$ and $CT$. Point $T$ will have the coordinates $(CT \cos \theta, PT \sin \theta)$. - 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. + 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. Cundy, & Rollett, 1961, p. 240 + {{!}}} - [[Image:Ellipsograph2.png|border|center|800px]] + =Conclusion---The Take Home Message= + We should not take the concept of straight line for granted and there are many interesting, and important, issues surrounding the concepts of straight line. A serious exploration of its properties and constructions will not only give you a glimpse of geometry's all encompassing reach into science, engineering and our lives, but also make you question many of the assumptions you have about geometry. Hopefully, you will start questioning the flatness of a plane, roundness of a circle and the nature of a point and allow yourself to explore the ordinary and discover the extraordinary. |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 Line 176: Line 324: |SiteURL=http://kmoddl.library.cornell.edu/model.php?m=244 |SiteURL=http://kmoddl.library.cornell.edu/model.php?m=244 |Field=Geometry |Field=Geometry - |FieldLinks=*)http://kmoddl.library.cornell.edu/model.php?m=244 + |FieldLinks=*http://kmoddl.library.cornell.edu/model.php?m=244 - *)http://dlxs2.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=Kemp009 + *http://dlxs2.library.cornell.edu/cgi/t/text/text-idx?c=math;cc=math;view=toc;subview=short;idno=Kemp009 - *)http://kmoddl.library.cornell.edu/tutorials/04/ + *http://kmoddl.library.cornell.edu/tutorials/04/ - *)http://www.howround.com/ + *http://www.howround.com/ - |References=How to draw a straight line: a lecture on linkages, Alfred Bray Kempe, Ithaca, New York: Cornell University Library + *http://en.wikipedia.org/wiki/Wikipedia:Citing_sources - + =Notes= - How round is your circle?, John Bryant and Chris Sangwin, Princeton, Princeton University Press + - |ToDo=I need to change the size of the main picture and maybe some more theoretical description what a straight line here. + |References=#Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr. - |InProgress=Yes + #Cundy, H.Martyn, & Rollett, A.P. (1961). Mathematical models. Clarendon, Oxford : Oxford University Press. + #Henderson, David. (2001). Experiencing geometry. Upper Saddle River, New Jersey: Prentice hall. + #Kempe, A. B. (1877). How to Draw a straight line; a lecture on linkage. London: Macmillan and Co.. + #Taimina, D. (n.d.). How to Draw a Straight Line. Retrieved from The Kinematic Models for Design Digital Library: http://kmoddl.library.cornell.edu/tutorials/04/ + #Ferguson, Eugene S. (1962). Kinematics of mechanisms from the time of watt. United States National Museum Bulletin, (228), 185-230. + #Weisstein, Eric W. Great Circle. Retrieved from MathWorld--A Wolfram Web Resource: http://mathworld.wolfram.com/GreatCircle.html + #Wikipedia (Steam Engine). (n.d.). Steam Engine. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Steam_engine + #Wikipedia (Watt's Linkage). (n.d.). Watt's Linkage. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Watt%27s_linkage + #Wikipedia (Cartesian coordinate system). (n.d.). Cartesian coordinate system. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Cartesian_coordinate_system + #Wikipedia (Linkage (mechanical)). (n.d.). Linkage (mechanical). Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Linkage_(mechanical) + #Wikipedia (Closed-form expression). (n.d.). Closed-form expression. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Closed-form_expression + #Lienhard, J. H. (1999, February 18). "I SELL HERE, SIR, WHAT ALL THE WORLD DESIRES TO HAVE -- POWER". Retrieved from The Engines of Our Ingenuity: http://www.uh.edu/engines/powersir.htm + #Wikipedia (Peaucellier–Lipkin linkage). (n.d.). Peaucellier–Lipkin linkage. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Peaucellier%E2%80%93Lipkin_linkage + |InProgress=No + |ImageSize=280px }} }}

## Current revision

Drawing a Straight Line
Field: Geometry
Image Created By: Cornell University Libraries and the Cornell College of Engineering
Website: Model: S35 Peaucellier Straight-line Mechanism

Drawing a Straight Line

The image shows the first planar linkage that drew a straight line without using a straight edge. Independently invented by a French army officer, Charles-Nicolas Peaucellier and a Lithuanian (who some argue was actually Russian) mathematician Lipmann Lipkin, it had important applications in engineering and mathematics.[2][3][4]

# Introduction

What is a straight line? How do you define straightness? The questions 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, why do we not assume the definition of 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 answer to the question "how do you construct a straight line without a straight edge?"

# What Is A Straight Line?--- A Question Rarely Asked.

 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 can picture it, but we simply cannot articulate it. In Euclid's book Elements, he defined a straight line as "lying evenly between its extreme points" and as having "breadthless width." This definition is pretty useless. What does he mean by "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 as shown in Image 1. In addition, the shortest distance between two points on a flat plane is a straight line, a definition we are most familiar with. However, it is important to realize that the definitions of being "shortest" and "straight" will change when you are no longer on flat plane. For example, the shortest distance between two points on a sphere is the the "great circle"as shown in Image 2. Since we are dealing with plane geometry here, we define straight line as the curve of $y=ax+b$ in Cartesian Coordinates. For more comprehensive discussion of being straight, you can refer to the book Experiencing Geometry by David W. Henderson. Take a minute to ponder the question: "How do you produce a straight line?" Well light travels in straight line. Can we make light help us to produce something straight? Sure but does it always travel in straight line? 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. Well, another simpler method is just to fold a piece of paper and the crease will be a straight line. However, to achieve our ultimate goal (construct a straight line without a straight edge), we need a linkage and that is much more complicated and difficult than folding a piece of paper. The rest of the page revolves around the discussion of straight line linkage's history and its mathematical explanation. Image 1 Image 2 [7]

# The Quest to Draw a Straight Line

## The Practical Need

 Now having defined what a straight line is, we must figure out a way to construct it on a plane. However, the challenge is to do that 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 because it has important applications 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. Image 3[8] Image 3 shows a patent drawing of an early steam engine. It is of the simplest form with a boiler (lower left corner), a cylinder with piston (above the boiler), a beam (on top, pivoted at the middle) and a pump (lower right corner) at the other end. The pump was usually used to extract water from the mines but other devices can also be driven. When the piston is at its lowest position, steam is let into the cylinder from valve K to push 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 causes 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 utilizes atmospheric pressure to cause the downward action of the piston. 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 (it is being stretched by forces at two ends) and that is why a chain is used as the connection.[9] [10] Ideally, 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 an arc of a circle. As a result, horizontal forces are created and subjected onto the piston rod. Consequently, the rate of attrition is very much expedited and the efficiency of the engine is greatly compromised. Durability is important in the design of any machine, but it was especially essential for the early steam engines. For these machines were meant to run 24/7 to make profits for the investors. Therefore, such defect in the engine posed a great need for improvements.[11]
 Improvements were soon developed to force the end of the piston rod move in a straight line, but these brought about new mechanical problems. The two pictures below show two improvements at the time. The hidden text explains how these improvements work and why they have failed to produce satisfactory results. Image 4[12] Image 5 [13] Firstly, "double-action" engines were built, part of which is shown in Image 4. Secondly, the beam was dispensed and replaced by a gear as shown in Image 5. However, both of these improvements were unsatisfactory and the need for a straight line linkage was still imperative. In Image 4, 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 took turns being taut throughout 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 solve the straight line 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. In Image 5, after the beam was replaced by gear actions, 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 remained unsolved. 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.[14]

## James Watt's breakthrough

 James Watt found a mechanism that converted the linear motion of pistons in the cylinder to the semi circular motion (that is moving in an arc of the circle) of the beam (or the circular motion of the flywheel) and vice versa. In this way, energy in the vertical direction is converted to rotational energy of the flywheel from where is it converted to useful work that the engine is desired to do. 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 while the two hinges move in circular arcs. However, this linkage only produced approximate straight line (a stretched figure 8 actually) as shown in Image 7, 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 Image 6. 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$\frac{AB}{CD} = \frac{CP}{CB}$.[15] Image 6 [16] Image 7 [17]

## The Motion of Point P

We intend to describe the path of $P$ so that we can show it does not move in a straight line (which is obvious in the animation). More importantly, this will allow us to pinpoint the position of $P$ using certain parameters we know, such as the angle of rotation or one coordinate of point $P$. This is awfully crucial 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. In addition, you can use the parametrization to create your own animation like that in Image 7.

### Algebraic Description

 We see that $P$ moves in a stretched figure 8 and will tend to think that there should be a nice closed form of the relationship of the $x$ and $y$ coordinates of $P$ like that of the circle. 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 ${\color{Gray}A}$ and ${\color{Gray}D}$ because they are fixed. We know coordinates $A$ and $D$ because they are fixed. Hence suppose the coordinates of $A$ are $(0,0)$ and coordinates of $B$ are $(c,d)$. 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 $C$ will be on the circle centered at $B$ with a radius of $m$. Since $C$ 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, Eq. 1        $x^2+y^2-2ax-2by+a^2+b^2=m^2$ Eq. 2        $x^2+y^2-2cx-2dy+c^2+d^2=r^2$ Subtract Eq. 2 from Eq. 1 we have, Eq. 3         $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2$ 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 Eq. 4         $y=\frac {-2a+2c}{2b-2d}x-\frac {m^2-2r^2+c^2+d^2}{2b-2d}$ Now, we can 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 Eq. 5         $x=2x'-a$ Eq. 6         $y=2y'-b$ 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 closed form but we could definitely use Mathematica to do the maths. The point is, there is no nice algebraic form for that figure 8, though it has closed form and that is why we have to find something else.

### Parametric Description

 Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it may be more manageable to describe the motion of $P$ using the angle of ratation. 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 ${\color{Gray}P}$ with the angle ${\color{Gray}\theta}$ in conformation of most para [...] 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 [19]

## Watt's Secret

 Another reason we parameterized $P$ is that Watt did not simply used that three bar linkage shown in Image 6 and Image 7. Instead he used something different. To understand that, our knowledge of the parameterizaion of $P$ is crucial. Imitations were a big problems back in those days. When filing for a patent, James Watt and other inventors had to explain how their devices worked without revealing the critical secrets so that others could easily copy them. As shown in Image 10, the original patent illustration, Watt illustrated his simple linkage on a separate diagram on the upper left hand corner but try looking for it on the engine illustration itself. Can you find it at all? That is Watt's secret. This is the equivalent of telling you by using the principle of 1+1 makes 2 you could get 34 x 45; the crucial step in understanding (and to make the engine work smoothly in Watt's case) is avoided. What he had actually used on his engine was the modified version of the basic linkage as show in Image 11. 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. It is important for two points to move in straight lines now is because one has to be connected to the piston rod that drives the beam, another has to 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. This is the truly famous James Watt's "parallel motion" linkage. Image 11 Image 12 Well, it is easy enough. Refer to Image 12. Well, it is easy enough. Refer to 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, $\overrightarrow {AF} = \frac {e+r}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$. We now have the parameterization of point $F$ and $P$ and Watt's secret is eventually cracked.

 Image 13 [20] Anyway, mathematicians and engineers had being searching for almost a century to find the solution to a straight line linkage but all had failed until 1864 when 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. Taimina Image 14 Let's turn to a skeleton drawing of the Peaucellier-Lipkin linkage in Image 14. 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(OM + 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 Image 15. Let's fix the path of point $C$ such that it traces out a circle that has point $O$ on it. $QC$ is 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 $\vartriangle OCR \sim \vartriangle ONP and \frac{OC}{OR} = \frac{ON}{OP}$. Moreover $OC \cdot OP = ON \cdot OR$. 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. ∎[21]

## 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.

 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 Image 16. F is the inversive center and points $D$,$F$ and $B$ are collinear and $DF \cdot DB$ is of constant value. Image 17 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 in the middle of Image 17. Prim's machine was driven by a steam engine.[22]

 After the Peaucellier-Lipkin Linkage was introduced to England in 1874, Mr. Hart of Woolwich Academy [23] devised a new linkage that contained only four links which is the blue part as shown in Image 18. The next part will prove that 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, let's define $\frac{OP}{BD} = m, hence \frac{OQ}{AC} = 1-m$ where $0 We finally have \begin{align} OP \cdot OQ & = m(1-m)BD \cdot AC\\ & = m(1-m)((AD)^2 - (AB)^2) \end{align}which is what we wanted to prove.

## Other Straight Line Mechanism

 Image 19 Image 20 Image 21 [24] There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to Image 19. 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 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. In 1801, James White patented his mechanism using this rolling motion. It is shown in Image 21 [25]. Image 22 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 as seen in Image 22. 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. [26]

# Conclusion---The Take Home Message

We should not take the concept of straight line for granted and there are many interesting, and important, issues surrounding the concepts of straight line. A serious exploration of its properties and constructions will not only give you a glimpse of geometry's all encompassing reach into science, engineering and our lives, but also make you question many of the assumptions you have about geometry. Hopefully, you will start questioning the flatness of a plane, roundness of a circle and the nature of a point and allow yourself to explore the ordinary and discover the extraordinary.

# 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.

# Notes

2. Bryant, & Sangwin, 2008, p. 34
3. Kempe, 1877, p. 12
4. Taimina
5. Wikipedia (Cartesian coordinate system)
7. Weisstein
8. Bryant, & Sangwin, 2008, p. 18
9. Bryant, & Sangwin, 2008, p. 18
10. Wikipedia (Steam Engine)
11. Bryant, & Sangwin, 2008, p. 18-21
12. Bryant, & Sangwin, 2008, p. 18-21
13. Bryant, & Sangwin, 2008, p. 18-21
14. Bryant, & Sangwin, 2008, p. 18-21
15. Bryant, & Sangwin, 2008, p. 24
16. Bryant, & Sangwin, 2008, p. 23
18. Wikipedia (Closed-form expression)
19. Lienhard, 1999, February 18
21. Bryant, & Sangwin, 2008, p. 33-36
22. Ferguson, 1962, p. 205
23. Kempe, 1877, p. 18
24. Bryant, & Sangwin, 2008, p.44
25. Bryant, & Sangwin, 2008, p.42-44
26. Cundy, & Rollett, 1961, p. 240

# References

1. Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr.
2. Cundy, H.Martyn, & Rollett, A.P. (1961). Mathematical models. Clarendon, Oxford : Oxford University Press.
3. Henderson, David. (2001). Experiencing geometry. Upper Saddle River, New Jersey: Prentice hall.
4. Kempe, A. B. (1877). How to Draw a straight line; a lecture on linkage. London: Macmillan and Co..
5. Taimina, D. (n.d.). How to Draw a Straight Line. Retrieved from The Kinematic Models for Design Digital Library: http://kmoddl.library.cornell.edu/tutorials/04/
6. Ferguson, Eugene S. (1962). Kinematics of mechanisms from the time of watt. United States National Museum Bulletin, (228), 185-230.
7. Weisstein, Eric W. Great Circle. Retrieved from MathWorld--A Wolfram Web Resource: http://mathworld.wolfram.com/GreatCircle.html
10. Wikipedia (Cartesian coordinate system). (n.d.). Cartesian coordinate system. Retrieved from Wikipedia: http://en.wikipedia.org/wiki/Cartesian_coordinate_system
13. Lienhard, J. H. (1999, February 18). "I SELL HERE, SIR, WHAT ALL THE WORLD DESIRES TO HAVE -- POWER". Retrieved from The Engines of Our Ingenuity: http://www.uh.edu/engines/powersir.htm

   |colspan="2"|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 can picture it, but we simply cannot articulate it.


In Euclid's book Elements, he defined a straight line as "lying evenly between its extreme points" and as having "breadthless width." This definition is pretty useless. What does he mean by "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 as shown in Image 1. In addition, the shortest distance between two points on a flat plane is a straight line, a definition we are most familiar with. However, it is important to realize that the definitions of being "shortest" and "straight" will change when you are no longer on flat plane. For example, the shortest distance between two points on a sphere is the the "great circle"as shown in Image 2.

Since we are dealing with plane geometry here, we define straight line as the curve of $y=ax+b$ in Cartesian Coordinates.

For more comprehensive discussion of being straight, you can refer to the book Experiencing Geometry by David W. Henderson.

Take a minute to ponder the question: "How do you produce a straight line?" Well light travels in straight line. Can we make light help us to produce something straight? Sure but does it always travel in straight line? 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. Well, another simpler method is just to fold a piece of paper and the crease will be a straight line. However, to achieve our ultimate goal (construct a straight line without a straight edge), we need a linkage and that is much more complicated and difficult than folding a piece of paper. The rest of the page revolves around the discussion of straight line linkage's history and its mathematical explanation. |-

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