Straight Line and its construction
From Math Images
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==='''The Motion of Point P'''=== | ==='''The Motion of Point P'''=== | ||
We intend to described the path of P using parametric equations. | We intend to described the path of P using parametric equations. | ||
+ | |||
+ | ===='''Algebraic Description'''==== | ||
[[Image:JW point P.png|center|900px]] | [[Image:JW point P.png|center|900px]] | ||
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<math>y= \frac {-2a+2c}{2b-2d}x - \frac {k}{2b-2d}</math> | <math>y= \frac {-2a+2c}{2b-2d}x - \frac {k}{2b-2d}</math> | ||
- | <font color=blue> Now what? </font> | + | <font color=blue> [[User:Xingda|Xingda]] 6/21 Now what? How is this related to P?</font> |
- | + | ===='''Parametric Description'''==== | |
+ | [[Image:JW point P 2.png|center|800px]] | ||
+ | |||
+ | We intend to parametrize the <math>P</math> with the angle <math>\theta</math>. | ||
+ | |||
+ | <math>\begin{cases} | ||
+ | \overrightarrow {AB} = (r \sin \theta, r \cos \theta) \\ | ||
+ | \overrightarrow {BC} = (m \sin (\theta + ? + \beta + \alpha), m \cos (\theta + ? + \beta + \alpha)) \\ | ||
+ | \end{cases}</math> | ||
+ | |||
+ | Now let <math>BD=l</math>. Then using cosine formula, we have | ||
+ | <math>\begin{cases} | ||
+ | m^2+l^2-2ml\cos \alpha = r^2 \\ | ||
+ | r^2+l^2-2rl\cos \beta = m^2 \\ | ||
+ | \end{cases}</math> | ||
+ | |||
+ | As a result, we can express \alpha and \beta as | ||
+ | |||
+ | <math>\alpha = \cos^{-1} \frac {m^2+l^2-r^2}{2ml}</math> and <math>\beta = \cos^{-1} \frac {r^2+l^2-m^2}{2rl}</math> | ||
+ | |||
+ | Since <math>l = \sqrt{(c-r \sin \theta)^2+(d-r \cos \theta)^2}</math>, <math>c</math> and <math>d</math> being the coordinates of point <math>D</math>, we can find <math>\alpha</math> and <math>\beta</math> in terms of <math>\theta</math>. | ||
+ | |||
+ | Hence, | ||
+ | |||
+ | <math>\begin{align} | ||
+ | \overrightarrow {AP} & = \overrightarrow {AB} + \frac {1}{2} \overrightarrow {BC} \\ | ||
+ | & = (r \sin \theta, r \cos \theta) + \frac {1}{2}(m \sin (\theta + ? + \alpha + \beta), m \cos (\theta + ? + \alpha + \beta)) \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | <font color=blue> [[User:Xingda|Xingda]] 6/21 Prof. Maurer, there is problem. Last time we talked, we kinda of missed out angle ?. Now I cannot find angle ? </font> | ||
+ | |||
+ | |||
+ | Imitations were a big problems beck in those days. When filing for a patent, James Watt and other inventors, had to explain how their devices work without revealing the critical secrets so that others could easily copy them. As seen in the original patent illustration on the bottom right, Watt illustrated his simple linkage on a separate diagram but we couldn't find it in anywhere in the illustration. That is Watt's secret. What he had actually used on his engine was the modified version of the basic linkage as show on the left. The link <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. The importance for two points move in a straight line is that one has to be connected to the piston rod that drives the beam, another will convert the circular motion to linear motion so as to drive the valve gears that control the opening and closing of the valves. It turns out that point F moves in a similar quasi straight line as point P. | ||
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- | 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 | + | Take another look at Watt's modified linkage below on the left. More than twenty years later when the patent terms had expired, Phineas Crowther used Watt's simple linkage in his engine in 1800. The position of the Watt's linkage was very prominent in the diagram below. It helps to guide the piston rod that drives the gear. The Crowther engine was used to haul goals or wagons up inclines in the old fashioned mines in northeast England. For more detail in steam engine, see [http://en.wikipedia.org/wiki/Steam_engine Steam Engine][[Image:Img331.gif|border|left|680px]][[Image:Img328.gif|border|center|500px]] |
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Now, | Now, | ||
- | <math>OA^2 = OM^2 + AM^2</math> | + | <math>(OA)^2 = (OM)^2 + (AM)^2</math> |
- | <math>AP^2 = PM^2 + AM^2</math> | + | <math>(AP)^2 = (PM)^2 + (AM)^2</math> |
<math>\begin{align} | <math>\begin{align} | ||
- | \therefore OA^2 - AP^2 & = OM^2 - PM^2\\ | + | \therefore (OA)^2 - (AP)^2 & = (OM)^2 - (PM)^2\\ |
& = (OM-PM)\cdot(PM + PM)\\ | & = (OM-PM)\cdot(PM + PM)\\ | ||
& = OC \cdot OP\\ | & = OC \cdot OP\\ | ||
\end{align}</math> | \end{align}</math> | ||
- | Let's take a moment to look at the relation <math>OA^2 - AP^2 = OC \cdot OP</math>. Since the length <math>OA</math> and <math>AP</math> are of constant length, then the product <math>OC \cdot OP</math> is of constant value however you change the shape of this construction. | + | Let's take a moment to look at the relation <math>(OA)^2 - (AP)^2 = OC \cdot OP</math>. Since the length <math>OA</math> and <math>AP</math> are of constant length, then the product <math>OC \cdot OP</math> is of constant value however you change the shape of this construction. |
- | |||
[[Image:PLcellproof2.png|border|center|750px]] | [[Image:PLcellproof2.png|border|center|750px]] | ||
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==='''Inversive Geometry in Peaucellier-Lipkin Linkage'''=== | ==='''Inversive Geometry in Peaucellier-Lipkin Linkage'''=== | ||
- | As a matter of fact, the first part of the proof given above is already sufficient. Due to | + | 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 <math>O</math>,<math>C</math> and <math>P</math> are collinear and that <math>OC \cdot OP</math> is of constant value. Points <math>C</math> and <math>P</math> are inversive pairs with <math>O</math> as inversive center. Therefore, once <math>C</math> moves in a circle that contains <math>O</math>, then <math>P</math> will move in a straight line and vice versa. ∎ See [[Inversion]] for more detail. |
- | |||
- | |||
==='''Peaucellier-Lipkin Linkage in action'''=== | ==='''Peaucellier-Lipkin Linkage in action'''=== | ||
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[[Image:Blowing engine.jpg|center|border|]] | [[Image:Blowing engine.jpg|center|border|]] | ||
- | |||
==='''Hart's Linkage'''=== | ==='''Hart's Linkage'''=== | ||
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Construct rectangle <math>EFCA</math> | Construct rectangle <math>EFCA</math> | ||
+ | |||
<math>\begin{align} | <math>\begin{align} | ||
- | AC \cdot BD = EF \cdot BD\\ | + | AC \cdot BD & = EF \cdot BD \\ |
- | & = (ED + EB) \cdot (ED - EB)\\ | + | & = (ED + EB) \cdot (ED - EB) \\ |
- | & = ED^2 - EB^2\\ | + | & = (ED)^2 - (EB)^2 \\ |
\end{align}</math> | \end{align}</math> | ||
<math>\because \begin{array}{lcl} | <math>\because \begin{array}{lcl} | ||
- | ED^2 + AE^2 & = & AD^2 \\ | + | (ED)^2 + (AE)^2 & = & (AD)^2 \\ |
- | EB^2 + AE^2 & = & AB^2 | + | (EB)^2 + (AE)^2 & = & (AB)^2 |
\end{array} | \end{array} | ||
</math> | </math> | ||
- | <math>\therefore AC \cdot BD = ED^2 - EB^2 = AD^2 - AB^2</math> | + | <math>\therefore AC \cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2</math> |
Further | Further | ||
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<math>\begin{align} | <math>\begin{align} | ||
- | \therefore OP \cdot OQ = m(1-m)BD \cdot AC\\ | + | \therefore OP \cdot OQ & = m(1-m)BD \cdot AC\\ |
- | & = m(1-m)(AD^2 - AB^2) | + | & = m(1-m)((AD)^2 - (AB)^2) |
\end{align}</math> | \end{align}</math> | ||
- | |||
- | |||
Revision as of 14:25, 21 June 2010
- 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.
How to draw a straight line without a straight edge |
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Basic Description
What is a straight line? How do you define straightness? How do you construct something straight without assuming you have a straight edge? These are questions that seem silly to ask because they are so natural to human concept. We come to accept that straightness is simply straightness and it is assumed. However, compare this to the way we draw a circle. When using a compass to draw a circle, we are not starting with a figure that we accept as circular; instead, we are using a fundamental property of circles that the points on a circle are a fixed distance from the center. Therefore, this page explores the properties of a straight line and hence its construction.A More Mathematical Explanation
- Note: understanding of this explanation requires: *A little Geometry
What is a straight line?
Today, we simply define a line as a one-dimensional object that extents to infinity in both directions and it is straight, i.e. no wiggles along its length. But what is straightness? It is a hard question because we have the picture in our head and the answer right there under our breath but we simply cannot articulate it.
In Euclid's book 'Elements', he defined a straight line as "lying evenly between its extreme points" and it has "breadthless length." The definition is pretty useless. What does he mean if he says "lying evenly"? It tells us nothing about how to describe or construct a straight line. So what is a straightness anyway? There are a few good answers. For instance, in a Cartesian Coordinates, the graph of is a straight line. In addition, we are most familiar with another definition is the shortest distance between two points is a straight line. However, it is important to realize that the definitions of being "shortest" and "straight" are different from that on flat plane. For example, the shortest distance between two points on a sphere is the the "great circle", a section of a sphere that contains a diameter of the sphere, and great circle is straight on the spherical surface.
The quest to draw a straight line
The Practical Need
Now having defined what a straight line is, we have to figure out a way to construct it on a plane without using anything that we assume to be straight such as a straight edge or ruler just like how we construct a circle using a compass. Historically, it has been of great interest to mathematicians and engineers not only because it is an interesting question to ponder about but also it has important application in engineering. Since the invention of various steam engines and machines that are powered by them, engineers have been trying to perfect the mechanical linkage to convert all kinds of motions(especially circular motion) to linear motions.
The picture above shows a patent drawing of an early steam engine. It is of the simplest form with a boiler(on the left), a cylinder with piston, a beam(on top) and a pump(on the right side) at the other end. When the piston is at its lowest position, steam is let into the cylinder from valve K and it pushes the piston upwards. Afterward, when the piston is at its highest position, cold water is let in from valve E, cooling the steam in the cylinder and causing the pressure in the the cylinder to drop below the atmospheric pressure. The difference in pressure caused the piston to move downwards. After the piston returns to the lowest position, the whole process is repeated. This kind of steam engine is called "atmospheric" because it utilized atmospheric pressure to cause the downward action of the piston (steam only balances out the atmospheric pressure and allow the piston to return to the highest point). Since in the downward motion, the piston pulls on the beam and in the upward motion, the beam pulls on the piston, the connection between the end of the piston rod and the beam is always in tension(under stretching) and that is why a chain is used as the connection.
Anyway, the piston moves in the vertical direction and the piston rod takes only axial loading, i.e. forces applied in the direction along the rod. However, from the above picture, it is clear that the end of the piston does not move in a straight line due to the fact that the end of the beam describes a segment of a circle. As a result, horizontal forces are created and subjected onto the piston rod. Consequently, the process of wear and tear is very much quickened and the efficiency of the engine greatly compromised. Now considering that the up-and-down cycle repeats itself hundreds of times every minute and the engine is expected to run 24/7 to make profits for the investors, such defect in the engine must not be tolerated and poses a great need for improvements.
Improvements came but in baby-steps. Firstly, "double-action" engines were made, part of which is shown in the picture on the right. Atmospheric pressure acts in both upward and downward strokes of the engine and two chains were used (one connected to the top of the arched end of the beam and one to the bottom), both of which will take turns to be in tension in one cycle. One might ask why chain was used all the time. The answer was simple: to fit the curved end of the beam. However, this does not fundamentally solved the problem and unfortunately created more. The additional chain increased the height of the engine and made the manufacturing very difficult (it was hard to make straight steel bars and rods back then) and costly. Secondly, beam was dispensed and replaced by a gear as shown on the left. Consequently, the piston rod was fitted with teeth(labeled k) to drive the gear. Theoretically, this solves the problem fundamentally. The piston rod is confined between the guiding wheel at K and the gear, and it moves only in the up-and-down motion. However, the practical problem was still there. The friction and the noise between all the guideways and the wheels could not be ignored, not to mention the increased possibility of failure and cost of maintenance due to additional parts. Therefore, both of these methods were not satisfactory and the need for a linkage that produces straight line action was imperative.
James Watt's breakthrough
James Watt, whose greatest achievement was his improvement on the steam engine, had to find a mechanism that converted the linear motion of pistons in the cylinder to the semi circular motion of the beam(or the circular motion of the flywheel) and vice versa. In 1784, he invented a three member linkage that solved the linear motion to circular problem practically as illustrated by the animation below. In its simplest form, there are two radius arms that have the same lengths and a connecting arm with midpoint P. Point P moves in a straight line. However, this linkage only produced approximate straight line (a stretched figure 8 actually) as shown on the right, much to the chagrin of the mathematicians who were after absolute straight lines. There is a more general form of the Watt's linkage that the two radius arms having different lengths like shown in the figure in the middle. To make sure that Point P still move in the stretched figure 8, it has to be positioned such that it adheres to the ratio
The Motion of Point P
We intend to described the path of P using parametric equations.
Algebraic Description
We know coordinates and . Hence let the coordinates of be , coordinates of be . We also know the length of the bar. Let .
Suppose that at one instance we know the coordinates of as , then will be on the circle centered at with a radius of . Since is on the circle centered at with radius . Then the coordinates of C have to satisfy the two equations below.
Now, since we know that is on the circle centered at with radius , the coordinates of have to satisfy the equation .
Therefore, the coordinates of have to satisfy the three equations below.
Now, expanding the first two equations we have,
Subtract the second equation from the first we have,
Substituting and rearranging we have,
Since the right side of the equation is just some constant, we replace them with . Hence we have,
Xingda 6/21 Now what? How is this related to P?
Parametric Description
We intend to parametrize the with the angle .
Now let . Then using cosine formula, we have
As a result, we can express \alpha and \beta as
and
Since , and being the coordinates of point , we can find and in terms of .
Hence,
Xingda 6/21 Prof. Maurer, there is problem. Last time we talked, we kinda of missed out angle ?. Now I cannot find angle ?
Imitations were a big problems beck in those days. When filing for a patent, James Watt and other inventors, had to explain how their devices work without revealing the critical secrets so that others could easily copy them. As seen in the original patent illustration on the bottom right, Watt illustrated his simple linkage on a separate diagram but we couldn't find it in anywhere in the illustration. That is Watt's secret. What he had actually used on his engine was the modified version of the basic linkage as show on the left. The link is the original three member linkage with and point being the midpoint of . 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 to it and connecting point to the piston rod. We now know that point moves in quasi straight line as shown previously. The importance for two points move in a straight line is that one has to be connected to the piston rod that drives the beam, another will convert the circular motion to linear motion so as to drive the valve gears that control the opening and closing of the valves. It turns out that point F moves in a similar quasi straight line as point P.
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.
Let's turn to a skeleton drawing of the Peaucellier-Lipkin linkage. It is constructed in such a way that and . Furthermore, all the bars are free to rotate at every joint and point is a fixed pivot. Due to the symmetrical construction of the linkage, it goes without proof that points , and lie in a straight line. Construct lines and and they meet at point .
shape is a rhombus
and
Now,
Let's take a moment to look at the relation . Since the length and are of constant length, then the product is of constant value however you change the shape of this construction.
Refer to the graph above. Let's fix the path of point such that it traces out a circle that has point on it. is the the extra link pivoted to the fixed point with . Construct line that cuts the circle at point . In addition, construct line such that .
constant, i.e. the length of (or the x-coordinate of w.r.t ) does not change as points and move. Hence, point moves in a straight line. ∎
Inversive Geometry in Peaucellier-Lipkin Linkage
As a matter of fact, the first part of the proof given above is already sufficient. Due to inversive geometry, once we have shown that points , and are collinear and that is of constant value. Points and are inversive pairs with as inversive center. Therefore, once moves in a circle that contains , then will move in a straight line and vice versa. ∎ See Inversion for more detail.
Peaucellier-Lipkin Linkage in action
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, , and . Point and are fixed pivots. In the diagram above. F is the inversive center and points , and are collinear and 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.
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 is the inversion center with and collinear and constant. When point is constrained to move in a circle that passes through point , then point will trace out a straight line. See below for proof.
Draw line , intersecting at point .
points are collinear
Construct rectangle
Further
where
Other straight line mechanism
There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams below. Consider two circles and with radius having the relation . We roll inside without slipping as show in the diagram on the right. Then the arch lengths . Voila! and point has to be on the line joining the original points and ! The same argument goes for point . As a result, point moves in the horizontal line and point moves in the vertical line.
In 1801, James White patented his mechanism using this rolling motion. Its picture is shown on the right. Interestingly, if you attach a rod of fixed length to point and and the end of the rod will trace out an ellipse. Why? Consider the coordinates of in terms of , and . Point will have the coordinates . Now, whenever we see and together, we want to square them. Hence, and . Well, they are not so pretty yet. So we make them pretty by dividing by and by , obtaining and . Voila again! and this is exactly the algebraic formula for an ellipse.
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.
Related Links
Additional Resources
- )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://kmoddl.library.cornell.edu/tutorials/04/
- )http://www.howround.com/
References
How to draw a straight line: a lecture on linkages, Alfred Bray Kempe, Ithaca, New York: Cornell University Library
How round is your circle?, John Bryant and Chris Sangwin, Princeton, Princeton University Press
Future Directions for this Page
I need to change the size of the main picture and maybe some more theoretical description what a straight line here.
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