Straight Line and its construction

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How to draw a straight line without a straight edge

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]

How to draw a straight line without a straight edge

Field: Geometry
Created By: Cornell University Libraries and the Cornell College of Engineering

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 properties of straight lines and more importantly and interestingly, the answer to the question "how do you construct something straight 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, a section of a sphere that contains a diameter of the sphere, and great circle is straight on the spherical surface.

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

For more properties on straight line, you can refer to the book Experiencing Geometry by David W. Henderson.

Image 1

Image 2 ^[6]

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^[7]

Image 3 shows a patent drawing of an early steam engine. It is of the simplest form with a boiler (on the left), a cylinder with piston (above the boiler), a beam (on top) and a pump (on the right side) at the other end. The pump was usually used to extract water from the mines.

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.^[8] ^[9]

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.^[10]

Improvements were made. 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.
Image 4^[11]	Image 5 ^[12]
[Hide] 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.^[13]

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 $\frac{AB}{CD} = \frac{CP}{CB}$ .^[14]

Image 6 ^[15]	Image 7 ^[16]

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 $A$ 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,

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

$\begin{cases} x=2x'-a \cdots \cdots \\ y=2y'-b \cdots \cdots \end{cases}$	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 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 $P$ 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 ^[18]

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 totally 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 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 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
[Hide] 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, $\overrightarrow {AF} = \frac {e+r}{r}(r \sin \theta, r \cos \theta) + (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha))$

The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage

Image 13 ^[19]	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. 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 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 in Image 13. 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 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. ∎^[20]

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 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 Image 17. 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.^[21]

Hart's Linkage

After the Peaucellier-Lipkin Linkage was introduced to England in 1874, Mr. Hart of Woolwich Academy ^[22] 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 $0<m<1$

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 ^[23]
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 ^[24].

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. ^[25]

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.

Teaching Materials

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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.

Notes

↑ Wikipedia (Linkage (mechanical))
↑ Bryant, & Sangwin, 2008, p. 34
↑ Kempe, 1877, p. 12
↑ Taimina
↑ Wikipedia (Cartesian coordinate system)
↑ Weisstein
↑ Bryant, & Sangwin, 2008, p. 18
↑ Bryant, & Sangwin, 2008, p. 18
↑ Wikipedia (Steam Engine)
↑ Bryant, & Sangwin, 2008, p. 18-21
↑ Bryant, & Sangwin, 2008, p. 18-21
↑ Bryant, & Sangwin, 2008, p. 18-21
↑ Bryant, & Sangwin, 2008, p. 18-21
↑ Bryant, & Sangwin, 2008, p. 24
↑ Bryant, & Sangwin, 2008, p. 23
↑ Wikipedia (Watt's Linkage)
↑ Wikipedia (Closed-form expression)
↑ Lienhard, 1999, February 18
↑ Wikipedia (Peaucellier–Lipkin linkage)
↑ Bryant, & Sangwin, 2008, p. 33-36
↑ Ferguson, 1962, p. 205
↑ Kempe, 1877, p. 18
↑ Bryant, & Sangwin, 2008, p.44
↑ Bryant, & Sangwin, 2008, p.42-44
↑ Cundy, & Rollett, 1961, p. 240

References

Bryant, John, & Sangwin, Christopher. (2008). How Round is your circle?. Princeton & Oxford: Princeton Univ Pr.
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

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

For more properties on straight line, you can refer to the book Experiencing Geometry by David W. Henderson.

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Categories: Math Images In Progress | Geometry | Cornell University Libraries and the Cornell College of Engineering

@@ Line 23: / Line 23: @@
 For more properties on straight line, you can refer to the book '''''Experiencing Geometry''''' by David W. Henderson.
 {{!}}-
-{{!}}align="center"{{!}}[[Image:Straightline.jpg|center|border|400px]] '''Image 1'''{{!}}{{!}}align="center"{{!}}[[Image:SmallGreatCircles 700.gif|center|border]]'''Image 2''' <ref>Weisstein</ref>
+{{!}}align="center"{{!}}[[Image:Straightline.jpg|center|border|300px]] '''Image 1'''{{!}}{{!}}align="center"{{!}}[[Image:SmallGreatCircles 700.gif|center|border|400px]]'''Image 2''' <ref>Weisstein</ref>
 {{!}}}
@@ Line 60: / Line 60: @@
 {{!}}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<math>\frac{AB}{CD} = \frac{CP}{CB}</math>.<ref>Bryant, & Sangwin, 2008, p. 24</ref>
 {{!}}-
-{{!}}[[Image:Img327.gif|center|border|400px]]{{!}}{{!}}[[Image:Watts linkage.gif|center|border]]
+{{!}}[[Image:Img327.gif|center|border|300px]]{{!}}{{!}}[[Image:Watts linkage.gif|center|border]]
 {{!}}-
 {{!}}align="center"{{!}}'''Image 6''' <ref>Bryant, & Sangwin, 2008, p. 23</ref>{{!}}{{!}}align="center"{{!}}'''Image 7''' <ref>Wikipedia (Watt's Linkage)</ref>
@@ Line 72: / Line 72: @@
 {{!}}We see that <math>P</math> 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).<ref>Wikipedia (Closed-form expression)</ref>}}  of the relationship of the <math>x</math> and <math>y</math> coordinates of <math>P</math> 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|600px]]
+{{!}}[[Image:JW point P.png|center|500px]]
 {{!}}-
 {{!}}align="center"{{!}}'''Image 8'''
@@ Line 140: / Line 140: @@
 {{!}}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 <math>P</math> using the angle of ratation. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of <math>P</math>'s coordinates.
 {{!}}-
-{{!}}[[Image:ParaP.png|center|600px]]
+{{!}}[[Image:ParaP.png|center|500px]]
 {{!}}-
 {{!}}align="center"{{!}}'''Image 9'''
@@ Line 183: / Line 183: @@
 {{{!}}
-{{!}}colspan="2"{{!}}Another reason we parameterized <math>P</math> is that Watt did not simply used that three bar linkage shown in '''Image 6''' and '''Image 7'''. Instead he used something totally different. To understand that, our knowledge of the parameterizaion of <math>P</math> 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 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 in '''Image 11'''. The link <math>ABCD</math> is the original three member linkage with <math>AB=CD</math> and point <math>P</math> being the midpoint of <math>BC</math>. 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 <math>BCFE</math> to it and connecting point <math>F</math> to the piston rod. We now know that point <math>P</math> 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.
+{{!}}colspan="2"{{!}}Another reason we parameterized <math>P</math> is that Watt did not simply used that three bar linkage shown in '''Image 6''' and '''Image 7'''. Instead he used something totally different. To understand that, our knowledge of the parameterizaion of <math>P</math> 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 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 in '''Image 11'''.
+The link <math>ABCD</math> is the original three member linkage with <math>AB=CD</math> and point <math>P</math> being the midpoint of <math>BC</math>. 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 <math>BCFE</math> to it and connecting point <math>F</math> to the piston rod. We now know that point <math>P</math> 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|600px]]{{!}}{{!}}[[Image:PointF.png|center|550px]]
+{{!}}[[Image:Watt1.png|center|border|400px]]{{!}}{{!}}[[Image:PointF.png|center|350px]]
 {{!}}-
 {{!}}align="center"{{!}}'''Image 11'''{{!}}{{!}}align="center"{{!}}'''Image 12'''
@@ Line 263: / Line 266: @@
 {{!}}[[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 '''Image 17'''. 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.<ref>Ferguson, 1962, p. 205</ref>
+{{!}}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 '''Image 17'''. 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.<ref>Ferguson, 1962, p. 205</ref>
 {{!}}}
@@ Line 311: / Line 316: @@
 =='''Other Straight Line Mechanism'''==
 {{{!}}
-{{!}}[[Image:Circle in circle 1.png|border|center|325px]]{{!}}{{!}}[[Image:Circle in circle 2.png|border|center|300px]]{{!}}{{!}}[[Image:Img335.gif|border|center|300px]]
+{{!}}[[Image:Circle in circle 1.png|border|center|200px]]{{!}}{{!}}[[Image:Circle in circle 2.png|border|center|200px]]{{!}}{{!}}[[Image:Img335.gif|border|center|200px]]
 {{!}}-
 {{!}}align="center"{{!}}'''Image 19'''{{!}}{{!}}align="center"{{!}}'''Image 20'''{{!}}{{!}}align="center"{{!}}'''Image 21''' <ref>Bryant, & Sangwin, 2008, p.44</ref>
@@ Line 321: / Line 326: @@
 {{!}}colspan="3" align="center"{{!}}'''Image 22'''
 {{!}}-
-{{!}}colspan="3"{{!}}Interestingly, if you attach a rod of fixed length to point <math>C</math> and <math>P</math> and the end of the rod <math>T</math> will trace out an ellipse as seen in '''Image 22'''. Why? Consider the coordinates of <math>P</math> in terms of <math>\theta</math>, <math>PT</math> and <math>CT</math>. Point <math>T</math> will have the coordinates <math>(CT \cos \theta, PT \sin \theta)</math>. Now, whenever we see <math>\cos \theta</math> and <math>\sin \theta</math> together, we want to square them. Hence, <math>x^2=CT^2 \cos^2 \theta</math> and <math>y^2=PT^2 \sin^2 \theta</math>. Well, they are not so pretty yet. So we make them pretty by dividing <math>x^2</math> by <math>CT^2</math> and <math>y^2</math> by <math>PT^2</math>, obtaining <math>\frac {x^2}{CT^2} = \cos^2 \theta</math> and <math>\frac {y^2}{PT^2} = \sin^2 \theta</math>. Voila again! <math>\frac {x^2}{CT^2} + \frac {y^2}{PT^2}=1</math> and this is exactly the algebraic formula for an ellipse. <ref>Cundy, & Rollett, 1961, p. 240</ref>
+{{!}}colspan="3"{{!}}Interestingly, if you attach a rod of fixed length to point <math>C</math> and <math>P</math> and the end of the rod <math>T</math> will trace out an ellipse as seen in '''Image 22'''. Why? Consider the coordinates of <math>P</math> in terms of <math>\theta</math>, <math>PT</math> and <math>CT</math>. Point <math>T</math> will have the coordinates <math>(CT \cos \theta, PT \sin \theta)</math>.
+Now, whenever we see <math>\cos \theta</math> and <math>\sin \theta</math> together, we want to square them. Hence, <math>x^2=CT^2 \cos^2 \theta</math> and <math>y^2=PT^2 \sin^2 \theta</math>.
+Well, they are not so pretty yet. So we make them pretty by dividing <math>x^2</math> by <math>CT^2</math> and <math>y^2</math> by <math>PT^2</math>, obtaining <math>\frac {x^2}{CT^2} = \cos^2 \theta</math> and <math>\frac {y^2}{PT^2} = \sin^2 \theta</math>. Voila again! <math>\frac {x^2}{CT^2} + \frac {y^2}{PT^2}=1</math> and this is exactly the algebraic formula for an ellipse. <ref>Cundy, & Rollett, 1961, p. 240</ref>
 {{!}}}

Straight Line and its construction

From Math Images

Revision as of 12:31, 15 July 2010

Contents

Introduction

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

The Quest to Draw a Straight Line

The Practical Need

James Watt's breakthrough

The Motion of Point P

Algebraic Description

Parametric Description

The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage

Inversive Geometry in Peaucellier-Lipkin Linkage

Peaucellier-Lipkin Linkage in Action

Hart's Linkage

Other Straight Line Mechanism

Conclusion---The Take Home Message

Teaching Materials

About the Creator of this Image

Related Links

Additional Resources

Notes

References

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