Straight Line and its construction

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How to draw a straight line without a straight edge
Independently invented by a French army officer, Charles-Nicolas Peaucellier and a Lithuanian (disputable) mathematician Lipmann Lipkin, this is the device that draws a straight line without using a straight edge. It was the first planar linkage that drew straight line without a reference guideway and it had important applications in engineering and mathematics.


Basic Description

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

A More Mathematical Explanation

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

What is a straight line?

[[Image:SmallGreatCircles [...]

What is a straight line?


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

For more properties on staight line, you refer to the book Experience Geometry by zzz.

The Quest to Draw a Straight Line

The Practical Need






James Watt's breakthrough


The Motion of Point P


Algebraic Description

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







Now, expanding the first two equations we have,

    x^2+y^2-2ax-2by+a^2+b^2=m^2 \cdots \cdots \\{{EquationRef2|Eq. 1}}
    x^2+y^2-2cx-2dy+c^2+d^2=r^2 \cdots \cdots \\{{EquationRef2|Eq. 2}}

Subtract the second equation from the first we have,

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

Eq. 3        



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

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

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

    x=2x'-a \cdots \cdots Eqt 5\\
    y=2y'-b \cdots \cdots Eqt 6

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

Parametric Description

Alright, since the algebraic equations are not agreeable at all, we have to resort to the parametric description. To think about, it would not be too bad if we could describe the motion of P using the angle of ration. As a matter of fact, it is easier to obtain the angle of rotation than knowing one of P's coordinates.

We will parametrize the P with the angle \theta in conformation of most parametrizations of point.

\overrightarrow {BC} = (m \sin (\frac {\pi}{2} + \beta + \alpha), m \cos (\frac {\pi}{2} + \beta + \alpha)) \\

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




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)

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


                      & = (r \sin \theta, r \cos \theta) + \frac {m}{2}(\sin (\frac {\pi}{2} + \alpha + \beta), \cos (\frac {\pi}{2} + \alpha + \beta)) \\

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


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


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

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

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

The First Planar Straight Line Linkage - Peaucellier-Lipkin Linkage

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


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


Since shape APBC is a rhombus

 AB \perp CP and CM = MP




Therefore, \begin{align}
& = OC \cdot OP\\



Since,  \angle OCR = 90^\circ


and  OC \cdot OP = ON \cdot OR


Inversive Geometry in Peaucellier-Lipkin Linkage


Peaucellier-Lipkin Linkage in action

Mr.Prim's adaptation


Hart's Linkage


We know that AB = CD, BC = AD

As a result, BD \parallel AC


Consequently, points O,P,Q are collinear

Construct rectangle EFCA

AC \cdot BD & = EF \cdot BD \\ 

For \begin{array}{lcl}

We then have  AC \cdot BD = (ED)^2 - (EB)^2 = (AD)^2 - (AB)^2.


We have \begin{align}

Other straight line mechanism

There are many other mechanisms that create straight line. I will only introduce one of them here. Refer to the diagrams above. Consider two circles C_1 and C_2 with radius having the relation 2r_2=r_1. We roll C_2 inside C_1 without slipping as show in the diagram below.



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