# Cartesian geometry

( previous, home, next )

Topics

• Cartesian geometry
• Archimedes trammel
• The cycloid

• Idioms: mobility, mobility equation, Grashof condition,

Cartesian geometry, also known as analytic geometry, united the subjects of algebra and geometry by interpreting the variables of algebraic equations in terms of a coordinate system to create geometric objects. The methods were initially developed by rivals Rene Descarte and Pierre Fermat in the early 1600’s. The introduction of algebra and coordinates to geometry made the language of geometry much more powerful. Cartesian geometry made it possible to describe many more curves without having to rely on special names and mechanical constructions, lead to the development of differential calculus within a generation, and is an essential tool today in applied mathematics.

An important application of Cartesian geometry is the formulation of equations that describe the mobility of mechanical systems, and the manipulation of these equations to determine the nature of their solution. This field grew to importance during the industrial revolution. The creation of engines and machines with many interconnected metal moving parts required a degree of precision in design and construction that had never been needed previously. Cartesian geometry and trigonometry provided a language for this precision engineering.

There are many geometric results that can be used to derive mobility equations, but two are of primary importance. The first is the Pythagorian theorem: for any right triangle, the sum of the squares of the length of the legs is equal the square of the length of the hypotenuse. The second is the law of similar triangles: any two triangles are similar if their angles are equal. The ratios of the lengths of corresponding sides of simlar triangles are always equal to each other. When applied in the context of Cartesian geometry where lengths can appear signed, it is sometimes convenient to work with the squares of these ratios.

## The cycloid

At night in the summer, we sometimes see bicycles ride by with lights upon their wheels. The lights revolve in a circle around the wheel axles, but the path the lights trace appear as cheerful looping bounds down the road as the bicycles move on. These traces are called cycloids.

A cycloid is the curve traced by a bangle on the rim of a wheel. It may have been studied in ancient times, but this work, if it existed, is largely lost. It’s modern study began around 1500, and it became closely connected to many of the interesting mathematical arguments of the next two centuries, including the quadrature, the pendulum clock, optics, and the brachistochrone – it was nicknamed the “Helen of mathematics” in its time because of the wars it ignited.

To find an equation for the cycloid, we can make use of parametric coordinates. Let’s suppose we start with a wheel of radius $$R$$ rolling on a level and flat surface, and we are tracking a bangle a distance $$r$$ from the wheel axle. As a wheel rolls, its current state can be represented with a variable $$p$$ tracking the position of the axle of the wheel, and a variable $$\theta$$ tracking the angle of rotation of the wheel. As long as there is no slippage of the wheel, these two must be related – each time the wheels angle makes a full revolution of $$2 \pi$$, the wheel’s position must move forward by a factor of the circumference of the wheel. Since we assumed our wheel has radius $$R$$, one rotation of the wheel will move the axle forward $$2 \pi R$$. If, initially, $$p=0$$ and $$\theta=0$$, then $$p=\theta$$ as the wheel rolls, whatever the velocity or acceleration.

Now, let $$(x,y)$$ represent the position of a bangle on our wheel, a distance $$r$$ from the axle. Since the axle’s position can be expressed in terms of the angle as $$(p,1) = (\theta R, 1)$$, we can express the position of the bangle with trigonometry as $x(\theta) = R \theta - r \sin \theta,\quad y(\theta) = R - r \cos \theta,$ assuming the initial position is directly below the axle.

There are a number of ways we can represent a curve in Cartesian geometry, each of which has its own preferred uses and limitations. We might find an explicit function representation $$y = f(x)$$, an implicit representation $$f(x,y) = 0$$, a and parameteric representation $$x = f(\theta)$$, $$y = g(\theta)$$. The function representation may be the most useful because it can easily test if a given point falls on the curve, as well as be used to generate a set of points on the curve. Implicit representations of a curve can easily be used to test if a point falls on the curve, but does easily generate sets of solution points on the curve. Parametric representations can easily generate sets of solution points on the curve, but do not easily test if a given point is a solution or not.

## Archimedes Trammel

Another example of Cartesian modelling of a mechanical system is one you might find in a kid’s toy chest – Archimede’s Trammel, shown below. The handle end is free to be turned, but only to positions that the slider’s movement allows. But what curve, precisely is that?

Experimentation with the trammel reveals the handle’s motion traces an oval shape taller than it is wide. A natural guess, for those familiar with geometry, is that the motion traces an ellipse. But can we be certain of this?

To answer the question, we can use some geometric modelling and algebra. Let’s set the center of the trammel (where the slider grooves cross) to be the origin of our system, and let $$(x,y)$$ be the position of the handle. In Cartesian coordinates, an ellipse centered at the origin and with principle axes aligned with the coordinate axes has an equation of the form $a x^2 + b y^2 = c,$ where $$a$$, $$b$$, and $$c$$ are positive constants determining the ellipse’s shape. Every point $$(x,y)$$ on the ellipse will be a solution of this equation. If we can show that there are some constants $$a$$, $$b$$, and $$c$$, such that $$x$$ and $$y$$ always satisfy this equation, we will know the trammel traces an ellipse.

On close inspection, we see the trammel consists of two sliders trapped in perpendicular grooves. Each slider can move freely, but every time we move one slider, the other slider gets moved in response. The handle’s screw-attachments control this – the screws are always the same distance apart. So when one slider gets moved out, the other sider gets moved in to compensate for the distance change.

There are two parameters in the trammel construction that may change from one toy to another, but are never changed once the trammel is completed – the distance between the handle and the first screw, and the distance between the first and second screws. Let $$r$$ be the distance between the handle and the first screw and let $$s$$ be the distance between the two screws.

The things that can change in the trammel are the positions of the slider screws and the position of the handle. Let $$p$$ be the horizontal slider screw’s position and $$q$$ be the vertical slider screw’s position while $$(x,y)$$ is the handle’s position. There are 3 equations that constrain the motion of our 4 free variables. By the Pythagorean theorem, we always have $$s^2 = p^2 + q^2$$. And we have two similar triangles in action in the trammel, and since corresponding sides of similar triangles are always in the same proportions, we have $\frac{q}{y} = \frac{p}{p-x} = \frac{s}{s+r}.$ It is sometimes convenient to write this system as a set of polynomials, each equal to zero, in which case, \begin{align*} 0 &= s^2 - p^2 - q^2, \\ 0 &= q \,(p - x) - p y, \\ 0 &= q \,(s +r) - s y. \end{align*}

These equations completely specify the mobility of Archimede’s trammel. If this was a linear system with 4 variables and 3 equations, we expect the solution-set to be a line. However, it is a nonlinear system. Usually, the solution set to be one-dimensional in 4-dimensional space, but we cann’t say much in the abstract. For a generally polynomial system of 3 equations and four variables, solution may be a single curve or multiple curves, and it might be a bounded set or the solutions might go off to infinity. And the characteristics of the solution set might change as the parameters change.

In this particular problem, we are looking for a curve that relates the handle position to the trammel’s built-in parameters, without regard to the slider positions – something like a function that can be written $$f(x,y; r,s) = 0$$ without using $$p$$ and $$q$$. Well, the proportions can be solved for $$p$$ and $$q$$. $q = \frac{ys}{s+r}, \quad p = -\frac{sx}{r}.$ Substituting for $$p$$ and $$q$$ in the Pythagorian theorem, we find $s^2 = \left( \frac{sx}{r} \right)^2 + \left(\frac{sy}{s+r}\right)^2$ $r^2 = x^2 + \left(\frac{1}{\frac{s}{r}+1}\right)^2 y^2$ Thus, the path traced by the trammel’s handle will be an ellipse. We can go a step further, and observe that as the ratio $$s/r$$ gets smaller, the motion will become less elliptic and more circular.

## Watt’s curve

A classic applications of Cartesian modelling in mechanical engineering is in the characterization of the Watt’s linkage. James Watt was a Scottish engineer who became rich from his improvements to the design of steam engines. One of the primary uses of his engines was to pump water out of coal mines. Early pumps tended to fail regularly, not because of breakdowns in the steam-engine itself, but because of uneven wear on the pump piston. In early designs, piston shafts were linked to the steam engine though a large beam that would drive the piston up and down. But, since the beam-end traced part of a circular arc, the piston would also be rocked back and forth in a manner that would systematically wear out the piston seals or connections. Watt realized that this wear could be avoided and reliability increased if a linkage could be found that would only the piston rod up and down in a straight line. He was so proud of his solution, he had it included in his portrait. (There are actually two versions of Watt’s linkage. The original version is shown below, while the second more practical version can be seen here ).

Watt’s linkage is a special-case of a common mechanism used in mechanical engineering – the 4-bar linkage. A 4-bar linkage is exactly that – 4 bars pinned together such that the angles between each adjacent pair of bars can change. A 4-bar linkage often has only 3 bars, with the last two pivot points connected to some larger static structure that keeps their positions fixed.

Despite being the simplest non-trivial linkages, 4-bar linkages are surprisingly complicated objects mathematically, and worth further study in their own right, but we will avoid the digression.

In the case of Watt’s linkage, two opposite bar in the linkage are of equal length, and the piston rod is connected to the midpoint of connecting rod. Specifically, suppose two pivots are fixed at $$(-1,0)$$ and $$(1,0)$$ respectively, and that bars connected to these pivots each have length $$r$$. Let $$(u_1,v_1)$$ be the location of the free end of the first, while $$(u_2,v_2)$$ be the location of the free end of the second. These two bars are connected by a third bar of length $$\ell$$, and the pison rod is connected to the point $$(x,y)$$ that is the midpoint of this third bar. Based on the opposite convexities of the circles described by the two opposite equal bars, Watt argued that the motion of the midpoint $$(x,y)$$ would be linear. However, it was almost 100 years before this conjecture was sorted out by Samual Roberts.

The 5 equations that determine the position of the midpoint are given as \begin{align*} r^2 &= (u_1+1)^2 + v_1^2, \\ r^2 &= (u_2-1)^2 + v_2^2, \\ \ell^2 &= (u_2-u_1)^2 + (v_2-v_1)^2, \\ x &= (u_1 + u_2)/2, \\ y &= (v_1 + v_2)/2. \end{align*} A natural approach to solving this system for midpoint location $$(x,y)$$ is to parameterize $$u_1$$ and $$v_1$$ in polar coordinates, and calculate the midpoint as a function of the angle. However, this approach becomes clumsy because of the repeated need to calculate the intersections of two circles. With some clever algebra (today, we use Groebner bases, a tool you may study elsewhere), one can show that $$(x,y)$$ solves the single sixth-degree equation

$0 = 4 y^{2} \left(x^{2} + y^{2} - r^{2}\right) + \left(x^{2} + y^{2}\right) \left(\frac{\ell^{2}}{4} - r^{2} - 1 + x^{2} + y^{2} \right)^{2}.$

[Show code]
from sympy import *
u1,u2,v1,v2,a,r,L = symbols('u1,u2,v1,v2,a,r,L')
sys = [ \
(u1 + u2) - 2*x, \
(v1 + v2) - 2*y, \
(u1 - a)**2 + v1**2 - r**2, \
(u2 + a)**2 + v2**2 - r**2, \
(u2 - u1)**2 + (v2 - v1)**2 - L**2, \
]
ans = (groebner(sys, (v1,v2,u1,u2,x))[-1]*q).collect((q,a),factor).subs(q,1)
pprint(ans)

To determine the parameter conditions when Watt’s linkage mobility is approximately straight around its midpoint, we can apply a series solution method. We can check by substitution and inspection that $$(x,y) = (0,0)$$ lies on the curve. Now, using a substitution $y \approx a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5,$ and picking the $$a_i$$’s so that coefficients vanish one by one, we can show that when $$\ell = 2 \sqrt{1 - r^2}$$, then the motion of the piston link is approximately give by $y(x) = a_1 x + a_5 x^5 + O(x^6), \quad a_1 = \pm \frac{r}{\sqrt{1-r^2}}, \quad a_5 = \frac{-(1 + a_1^2)^3}{8 a_1 r^2 (1 - r^2)}.$ This shows that Watt’s linkage motion is approximately a straight-line motion to fifth-order when the linking bar length $$\ell$$ is picked right.

# Exercises

1. Show that the intersection between a cylinder and a plane is an ellipse

2. Find the equation for a torus using a 3-dimensional Cartesian coordinate system.

3. In a flat area of sand on a beach, a piling with circular cross-section of radius $$1$$ foot is driven vertically into the ground. A string has been wrapped tightly around this piling at ground level, with one end free. Grab the free end and walk around the piling to unwind the rope, while keeping the string taut.
1. Find an equation for the path the free end of the rope traces in the sand as you unwind it. (Hint: This curve is not an Archimedean spiral.)
2. Show that the curve is well-approximated by an Archimedean spiral asymptotically as the length of unwound rope gets larger using a plot.
4. The Conchoid of Nicomedes was a curve traced by a linkage mechanism where one pin is fixed to the vertical axis, while the second pin can move freely along the x axis. (see here) This conchoid was designed to help divide an angle into thirds, solving one of the classical Greek puzzles of geometry.
1. What are the two physical length parameters that are determined by the linkage construction and control the curves shape. Call these parameters $$a$$ and $$b$$.
2. Using $$x$$ to represent the horizontal position of the movable pin, and $$(p,q)$$ representing the point of the linkage, find a pair of parametric curves for the Conchoid in the forms $$(p(x,a,b), q(x,a,b))$$.
5. In class we found that Archimedes’ trammel traced out an ellipse, based on the separation between the pins and the handle. But it is not the only way to construct a perfect ellipse. Imagine two thumb tacks placed on cardboard a distance $$d$$ apart. Take a string of length $$L$$, tie it in a circle, loop it over the thumb tacks, and draw the curve of the farthest the string can reach. Show using Cartesian geometry that this curve is also an ellipse, as long as as $$L > 2 d$$

6. A curve traces the intersection point of two lines. The first line is flat, starts at height 1, and moves down with constant speed. The second line is through the origin, starts off vertical, and rotates at a constant angular speed. Both lines reach the x-axis at the same time.
1. Find an equation for the curve tracing the intersection.
2. Use L’Hopital’s rule to find the intersection point of the curve with the x-axis.
7. (Hard) A pair of sophomore students are moving a large mural painting on plaster board into a new apartment, but to get to the room, you have to turn the rectangular mural around a tight corner between two hallways. The mural is about as tall as the hallway, so tilting it only makes the fit harder.
1. What 3 variables control whether or not you can get the couch
2. Find an equation for the possible width of the first and second hallways for which you will be able to get the couch to your apartment.
8. Plot the cycloid path in the following cases.
1. Plot the path when R=1 and r = 0.6 for 4 revolutions.
2. Imagine that in our cycloid problem, we could extend the position of the bangle beyond the rim of the wheel.
Plot the path when R=1 and r = 1.5 for 3 revolutions.
9. Take a 1-foot bar 1 and connect it to the origin. Take a second bar three feet in length and connect the free end of the first bar 1 foot from one end. Fix the free end of the long bar furthest from the pivot point so that it always lies on the negative x-axis, but slides freely. Let $$(x,y)$$ be the position of the opposite free end of the long bar.
1. Find a system of equations that can be used to determine possible positions $$(x,y)$$.
2. (Hard) Reduce your system to a single polynomial equation involving only $$x$$ and $$y$$.
3. Observe that your solution curve is symmetric across the y-axis, even though the true solution does not have any negative values for $$x$$. What happened?
10. Consider the polynomial system $$0 = a - 4 x^2 + 2 v^2 + y^2$$, $$0 = x^2 - v$$ for the variables $$v,x,y$$.
1. Use substitution to find a single equation relating $$x$$ and $$y$$ that is independent of $$v$$.
2. Show that if $$a$$ is large, then there are no solutions of this system.
3. Show that for any constant value of $$a$$, the solution set is bounded.
4. Show the solution set contains two unconnected curves when $$a = 1$$.
5. Show the solution set is a simply connected curve when $$a =-1$$.
11. (Hard) Suppose you have the equations for a plane and a helix, both in 3 dimensions. How many solutions are there to this equation? Discuss their enumeration and calculation.

12. Transform the equation for Watt’s curve from Cartesian to polar coordinates. Use this and symmetry to plot the curve when $$r=5/7$$ and $$\ell = 2 \sqrt{1-r^2}$$.

13. (Hard) Explain Philon’s construction of a cube roots in modern language.

( previous, home, next )