# Coordinate Geometry

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Topics

• Coordinate geometry
• Archimedes trammel
• The cycloid
• Figure of the earth

Coordinate geometry, also known as analytic geometry and Cartesian 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 and vice-versa. The methods were developed by rivals Rene Descarte working in Holland and Pierre Fermat working in France in the early 1600’s, with prior art being found Mercator’s world map (1569), and in the earlier works by Oresme (circa 1350), Menaechmus (circa 350 BC), among others. Decartes argued that the only path to the understanding of laws of nature was through their mathematical description, as with Snel’s law of refraction and Galileo’s law of ballistic motion. The introduction of algebra and coordinates to geometry made it possible to quickly describe and manipulate many more curves without having to rely on special names and mechanical constructions. Coordinate geometry had a subtle but even more profound effect on our fundamental perception of curves. It broke the symmetry of Euclidean geometry where there was no concept of absolute position and only relative distances mattered. Coordinate geometry lead naturally to the asymmetric relationship of inputs and outputs so that axes could be different units like time, distance, or speed. It set the table for the development of differential calculus within a generation, and remains foundational in both applied mathematics and the philosophy of science even today.

Besides being an essential tool in our further studies, modelling with coordinate geometry provides a familiar and compelling example of how models help organize and utilize information, and how parametric models differ from non-parametric models like triangulation. We will consider three classic applications to linkages.

An important application of coordinate geometry is the formulation of equations that describe the configuration (or mobility) of mechanical systems. Engineers made huge advances in the design of machines during the industrial revolution (1770-1830), but these advances could only go so far without mathematics. The creation of machines with many interconnected moving parts like the steam engine and the Jacquard loom required a degree of precision in design and construction that had never been needed previously. Coordinate geometry, together with trigonometry, provided a language for this precision engineering.

### Solution concepts

Before we jump in, it will be helpful to consider what it means to “solve” a coordinate geometry modelling problem. Broadly, we will be building these models to predict which configurations a given mechanical system can have. This idea of prediction can be broken down into two more specific and useful questions. (1) Can the system be put into a desired configuration? (2) Can we list the possible configurations of the system? Ideally, our solution should be able to answer both of these questions, but if we can atleast find an expression that answers one of these questions, we’ll have a good starting point for answering the other.

In the problems that follow, the configurations we wish to study will correspond to curves lying in the plane. In these cases, the best form for the solution would be an explicit function of the form $$y=f(x)$$ (or $$x=f(y)$$) because these functions can efficiently answer both questions. We could test if a proposed configuration represented by the point $$(u,v)$$ lies on the curve by substituting and calculating to see if $$v = f(u)$$. On the other hand, if we want to list all possible configurations, we can vary $$x$$ over the range of interest and generate the points $$(x,f(x))$$.

Unfortunately, not all plane curves are functions, and even when they are, we may have difficulty finding an efficient formula for calculating that function. In these cases, we resort to other representations as solutions. One common representation is an implicit representation. An implicit representation of a solution curve is an equation of the form $$f(x,y)=0$$ which is for exactly those points on the solution curve we are studying, but which has not been solve to express one of the variables in terms of the other. An implicit solution can be used directly to check if a given point lies on the curve. However, an implicit solution can not directly generate other points. For example, $$x^2 + y^2 = 1$$ is an implicit representation of the unit circle. We can use it to confirm that $$(x,y) = (3/5,4/5)$$ is on the unit circle, but it does not provide a simple way to generate other solutions.

Another useful representation of a solution curve is a parametric representation, where both coordinates are treated as dependent functions of a third independent variable, as $$(x,y) = (f(\theta), g(\theta))$$. Parametric solutions can efficiently generate points on the curve by sampling values of $$\theta$$ and calculating, but they do not allow us to directly test if an arbitrary given point is on the curve because we don’t know which value of the independent parameter to which that point would correspond. Thus, the parametric representation of the unit circle $$(x,y) = (\cos(\theta),\sin(\theta))$$ is very useful for drawing the curve, but does not give us a direct way of checking if $$(x,y) = (119/169, 120/169)$$ is on the circle. And since parameterizations are not unique, the use of parametric representations complicates the comparison of different curves.

The concerns we’ve summaried here regarding the representation of solutions extends to many other kinds of problems. In may be tempting to dismiss such considerations, but keep your eyes open and you’ll see there are many instances where important questions that could be posed in terms of seemingly difficult equations were ultimately answered productively by redefining the concept of solution. Just two further examples were Cardano’s solution of cubics in terms of imaginary numbers and Fourier’s solution of the wave equation in terms of trigonometric series.

## Cycloids

Animation

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. If the light is on the rim of the wheel, the curve it traces is called a cycloid. The cycloid’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.

Suppose we would like to find equations for drawing a cycloid. To do this, 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. This form of the solution does lead to an elementary explicit solution of the form $$y(x)$$ because of the difficulties of inverting $$x(\theta)$$, and so we choose to stop our analysis here.

## Archimedes’s Trammel

An example of coordinate geometry applied to a mechanical system is one you might find in a kid’s toy chest – Archimedes’ Trammel, shown above. The handle end is free to be turned, but only to positions that the movement of the two sliders allows. 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 geometry 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 coordinate 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, and connected by the handle. 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.

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. Let $$(p,0)$$\$ be the horizontal slider screw’s coordinates and $$(0,q)$$ be the vertical slider screw’s position. So, we have fixed parameters $$r,s$$ and four variables $$p,q,x,y$$, and one degree of freedom in the motion of the handle, so we can expect to find 3 independent equations that constrain the motion of our 4 free variables.

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 \begin{align*} \frac{q}{y} = \frac{p}{p-x}, \quad \frac{q}{y} = \frac{s}{s+r}, \quad \text{and} \quad \frac{p}{p-x} = \frac{s}{s+r}. \end{align*} And applying the Pythagorean theorem to the two triangles in our diagram, we know \begin{align*} s^2 = p^2 + q^2, \\ (r+s)^2 = (p-x)^2 + y^2. \end{align*} So, we have 5 equations, instead of the 3 equations we expect. But if we look closely, we see that these equations are not all independent. The third equation we obtain from the similar triangles is redundant because it can be recreated from the first two. And the second Pythagorean theorem equation is derived from the first using the similar-triangle equations, so we need only consider the first. It is convenient to write the 3 independent equations as a system of polynomials, each equal to zero, to which, \begin{align*} 0 &= s^2 - p^2 - q^2, \\ 0 &= q \,(p - x) - p y, \\ 0 &= q \,(s +r) - s y. \end{align*} This system of polynomial equations completely specifies the configuration space of Archimedes’s trammel.

While complete, this system is not very useful to us at the moment. 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 an implicit solution that can be written $$f(x,y; r,s) = 0$$ without using $$p$$ and $$q$$. Well, the proportions can be solved for $$p$$ and $$q$$, giving $q = \frac{ys}{s+r}, \quad p = -\frac{sx}{r}.$ Substituting for $$p$$ and $$q$$ our system’s first equation, we find $s^2 = \left( \frac{sx}{r} \right)^2 + \left(\frac{sy}{s+r}\right)^2$ or $r^2 = x^2 + \left(\frac{1}{\frac{s}{r}+1}\right)^2 y^2.$ Thus, the path traced by the trammel’s handle will be an ellipse.

## Watt’s curve

A classic application of coordinate geometry in mechanical engineering is in the characterization of Watt’s linkage. James Watt (1736-1819) was a successful Scottish engineer famous for his improvements to steam engine design. 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’s path traced part of a circular arc, the piston would rock back and forth in a manner that would systematically wear out the piston seals.
Watt realized that the piston wear could be avoided and reliability increased if a linkage could be found that would move the piston rod up and down in a straight line. He was so proud of his solution, he had it included in his portrait!

Animated version

In Watt’s linkage, two opposite bars 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 piston 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.

By directly translating this description of Watt’s linkage into algebra, we end up with 5 equations involving 2 constant parameters ($$r$$ and $$\ell$$) and 6 free variables ($$x, y, u_1, u_2, v_1$$, and $$v_2$$). \begin{align*} \text{Bar 1 length:} && r^2 &= (u_1+1)^2 + v_1^2, \\ \text{Bar 2 length:} && r^2 &= (u_2-1)^2 + v_2^2, \\ \text{Bar 3 length:} && \ell^2 &= (u_2-u_1)^2 + (v_2-v_1)^2, \\ \text{Horizontal midpoint:} && 2 x &= (u_1 + u_2), \\ \text{Vertical midpoint:} && 2 y &= (v_1 + v_2). \end{align*} Since there are only 5 equations for 6 variables, we expect the system should have a single degree of freedom. However, in this form, the system is not very useful. There is no easy way to either iterate through points on the curve or to test if a given point falls on the curve. 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 similar to what we did for Archimedes’s trammel, one can show that the set of solutions can be represented implicitly with a single sixth-degree polynomial. The outline is that we will try to eliminate each of the nuisance parameters $$v_2$$, $$u_2$$, $$v_1$$, and $$u_1$$ from the system, in order so that we are left with just one equation in terms of $$x$$ and $$y$$. First, we solve the last two linear equations for $$u_2$$ and $$v_2$$ respectively, and substitute to get \begin{align} r^{2} &= v_{1}^{2} + \left(u_{1} - 1\right)^{2}, \\ r^{2} &= \left(2 y - v_{1}\right)^{2} + \left(1 - u_{1} + 2 x\right)^{2}, \\ \frac{\ell^{2}}{4} &= \left(x - u_{1}\right)^{2} + \left(y - v_{1}\right)^{2}. \end{align} Now we come to a tricky spot. Each of these is a quadratic equation in both $$u_1$$ and $$v_1$$. We could try the quadratic formula, but that could be messy. If we look close, we see that in each, the quadratic terms in both equations will be of the form $$u_1^2 + v_1^2$$. If we subtract the first from the second, we can show $$0 = x^2 + y^2 - u_1 x - v_1 y + x$$. Being linear in $$u_1$$ and $$v_1$$, we can solve for $$v_1$$ and substitute and be left with two quadratic equations in $$u_1$$. \begin{align} 0 &= x^{2} \left(1 - u_{1} + x\right)^{2} + y^{2} \left(\left(u_{1} - x\right)^{2} - \frac{\ell^{2}}{4} \right) \\ 0 &= y^{2} \left(\left(1 - u_{1} + 2 x\right)^{2} - r^{2}\right) + \left( (u_{1} - 1 - x) x + y^{2}\right)^{2} \end{align} If we use the same trick from above to eliminate the quadratic term for $$u _ 1$$ from these two equations, solve, substitute, and factor, we find $0 = 4 y^{2} \left(x^{2} + y^{2} - r^{2}\right) + \left(x^{2} + y^{2}\right) \left(\frac{\ell^{2}}{4} - r^{2} - 1 + x^{2} + y^{2} \right)^{2}.$ This can be computed more efficiently and directly using Groebner basis methods, as shown by the following script.

[Show code]
from sympy import *
u1,u2,v1,v2,a,r,L = symbols('u1,u2,v1,v2,a,r,L', positive=True)
myvars = 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).subs(a,1)
hand = 4*y**2*(x**2+y**2-r**2) + (x*x + y*y)*(L*L/4 - r*r-1+x*x+y*y)**2
shouldbeconstant = (ans/hand).factor()
assert {0} == { shouldbeconstant.diff(i) for i in myvars }

pprint( hand )


This form of the equation for Watt’s curve reveals an important pattern – the independent variable $$x$$ always appears with $$y$$ in the quadratic form $$x^2 + y^2$$. This can be exploited to make a parametric representation of the solution (see Exercises).

## Discussion

Our models of the cycloid, trammel, and Watt’s linkage are not perfect. In any real version, there are factors for which our Cartesian models have not accounted. For our cycloid model, the wheel is no perfectly circular, and the road would not be perfectly level. For our trammel model, the sliders and joints all are a little loose, and there wiggles introduce uncertainty into the parameter values we must pick.

Whether or not these sources of uncertainty are important or not depends on our intended application. If we need to build a display box for a trammel, then we can account for the uncertainty in range of motion by building a little extra space into the margins of the box around the trammel. On the other hand, if we need to predict the position of a spot of reflecting paint on the wheel of a bicycle after it has rolled a certain distance, the impact of uncertainty in the wheel circumferance strongly depends on how far down the road we need to make are prediction. This is because small errors in our estimates of the circumference accumulate as the wheel rolls. We should always keep in mind the scope of intended application of our models.

The analytic formula’s we derive are actually all parametric solutions because the geometry is characterized by a set of parameters which then appear explicitly in our solution equations. These analytic solutions capture parameter-dependencies and allow for exploratory sensitivity analysis. For example, we can use the use our cycloid solution to see how the path of the wheel light will change as we move it’s position closer to the wheel hub. This ability to do sensitivity analysis is valuable in design problems, but impossible using data-centric approaches like Snel’s triangulations, where the equations are formulated directly in terms of the data. The postulates of triangulation more of type rather than parameter – assume space is flat, or the earth is a sphere – it is not easy to relax this postulate to see the sensitivity of it’s calculations to uncertainty in the postulate.

The study of linkages also provides a counter-point to the standard theory of linear algebra. One of the things we would discover on a deeper exploration of applied algebraic geometry is that even when we have done all of our modelling correctly, knowning that a specific configuration is a solution of our algebraic equations does not ensure that it is physically relevant. Examples are found in the 4-bar linkage, which, for certain given lengths, can be assembled in 2 different ways. Both assemblies satisfy the same set of configuration equations, but once they are assembled, the mobility is further constrained and there is no way to move from one to the other!

## The figure of the earth

The general approach to geometric modelling we’ve illustrated above for mechanical linkages can be extended to areas where calculus is needed. An example from the 1600’s is the derivation of an equation that can be used to estimate the shape of the earth. This derivation is a less intuitive than those above, but will be useful in the coming lectures.

The hypothesis that earth is a ball with no edges traces back at least as far as the pre-Socratic Greek philosophers(c. 500 BC) but was first confirmed by the Magellan-Elcano circumnavigation completed in 1522. But with the increased scientific sophistication of the age of enlightenment, it became a matter of great debate if the earth’s figure was actually spherical, and if not, was it more like a lemon sinched around the equator or an onion squached at the poles? How would you go about determining the figure of the earth? If the earth’s cross-section is not a circle, how can we tell? It was easy to answer this question for the planet Jupiter – astronomers can directly observe the figure of Jupiter through a telescope. Giovanni Cassini did this and determined that Jupiter was wider than tall. But we can not do the same for the planet we live on. This puzzle forces us to take a closer examination of what we mean by “figure” and “shape”.

Our descriptive language of the differences between the shapes of a lemon and onion suggests a way forward. An important local characteristic of shape is curvature – how unlike a straight line a curve is. A circle has the same curvature all the way around as you go, but other shapes like a lemon become more or less curved and for some shapes even reverse the direction of curvature. If we can measure curvature at different locations, then we might be able to stitch togethor those local curvatures determine the overall shape.

Curvature can be thought of as the rate of direction change per unit of distance travelled. Suppose we start at some point $$p$$ on a planar curve. Draw a tangent vector to the curve at the point we are starting. Now, we walk along the curve, keeping track of the distance $$s$$ we have travelled. As long as the curve we are walking along is smooth enough that we can measure changes in the tangent vector to the curve, and we have a universal reference frame like the night sky, we can also recording the angle $$\lambda(s)$$ between the tangent vector to the curve at our current location and the tangent vector at our start location. The local curvature, then, can be thought of as $$d\lambda/ds$$. On a circle, this implies that the curvature is constant everywhere and proportional the reciprocal of the radius. As we might expect, the bigger the circles get, the smaller their curvatures. On a surface, we can get the same results by tracking changes in the direction of an oriented unit normal vector.

Now, let’s get down to specifics. In order to apply these curvature ideas, particularly back in the 1600’s when this was first done, we have to have a relatively simple parameterizable model that can accomodate both lemon and onion shapes. The only model widely known to mathematicians of the time was that of ellipsoids of revolution – bodies defined by rotating an ellipse around one of it’s axes. All of spheres, lemon, and onion shapes could be constructed by rotating various ellipses. Of course, the earth is not a perfect ellipsoid of revolution – it has mountains and plateaus and valleys and icecaps – but that’s a natural start.

We can represent an ellipsoid of revolution in terms of an ellipse $x^2 + (1+\epsilon)y^2 = r^2$ where $$r$$ is the radius at the equator and $$\epsilon$$ represents the distortion of the ellipse from a perfect circle. When the ellipse is rotated around the vertical axis, it creates a lemon (prolate ellipsoid) when $$\epsilon < 0$$ and an onion (oblate ellipsoid) when $$\epsilon > 0$$.

The normal vector perpendicular to a point on this ellipse can be found by treating this ellipse as a single contour line of the function $$F(x,y) = x^2 + (1+\epsilon) y^2$$. The gradient can be determined by differentiation, $$\nabla F = [ 2 x, 2 (1+ \epsilon ) y ]$$, and at each point, the gradient vector will be perpendicular to the ellipse curve forming a contour of constant value. Let $$\lambda$$ be the angle between the normal vector and the equatorial plane, e.g. $$\lambda$$ is the celestial latitude of a position $$(x,y)$$. Then from elementary trigonometry, $\tan \lambda = \frac{(1 + \epsilon)y}{x}.$ We can use this equation to represent both $$x$$ and $$y$$ parametrically as functions of $$\lambda$$. Solving for $$y$$ and substituting into our equation for the ellipse, $\begin{gather*} x^2 + \frac{1}{1 + \epsilon} x^2 \tan^2 \lambda = r^2, \\ x(\lambda) = r \cos(\lambda) \sqrt{\frac{1+\epsilon}{1+\epsilon \cos^2 \lambda}}. \end{gather*}$ Similarly, solving our tangent equation for $$x$$ and substituting leads to $y(\lambda) = r \sin(\lambda) \sqrt{\frac{1 }{(1+\epsilon)(1 + \epsilon \cos^{2}(\lambda ))}}.$

Now, having position and latitude, we would like to relate latitude as a function of arclength to get curvature. However, we can already see it is going to be very hard to solve for latitude in terms of arclength. Instead, let’s flip things around – reciprocal curvature describes the change in arc length as a function of change in latitude. If we parameterize the arclength $$s$$ as a function of the latitude $$\lambda$$, standard calculus references tell us $\frac{ds}{d\lambda} = \sqrt{ \left( \frac{dx}{d\lambda} \right)^2 + \left( \frac{dy}{d\lambda} \right)^2 }.$ If we calculate the derivatives, substitute, and simplify, we find the reciprocal curvature $\frac{ds}{d\lambda} = \sqrt{\frac{(1 + \epsilon) r^2}{\left(1 + \epsilon \cos^2(\lambda)\right)^3}}.$

This formula is closely related to one obtained by Maupertius in 1736 for variation in the earth’s radius as a function of the latitude and elliptic eccentricity.

## Historical background regarding the earth’s figure

The story of the debate over the figure of the earth begins with one of the most practical applied-mathematics problems of the day – ship navigation, and the longitude problem in-particular. As nations grew and trade expanded during the Renaissance, shipwrecks became an important source of financial and personal risk. The risks became larger after 1492, as explorers discovered new trade routes consisting of long, expensive, and uncertain ocean crossings. Accurate navigational charts were being drawn with the help of trigonometry, but could not help ships unable to determine their own positions in open ocean. Latitude could be approximated based on observations of the stars, but longitude could not because the earth’s daily rotation blurred the most accessible celestial references.

In 1656, Christiaan Huygens, a remarkable Dutch natural philosopher whom we will encounter again later, patented the first pendulum clock. One of the uses Huygens proposed for his clock was the determination of longitude. If you know what time it was in New York when it was noon in Paris, you would know the time-difference between the two, and by multiplication, the longitude of New York relative to Paris. Huygen’s own clocks were not precise enough to practically solve the longitude problem – John Harrison’s chronometer designs would eventually do the job a century later. But the idea already intrigued many, and in 1670, when the recently created Académie des Sciences in Paris decided to send it’s second scientific expedition to Cayenne in French Guyana, clock-testing was part of the plan. Two years later, Jean Richer reported back that the length of a seconds-pendulum was not the same in Cayenne and Paris! In Cayenne, the seconds-pendulum was slightly shorter than its 24.84 centimeters in Paris. Parisian clocks would lose about 2 minutes and 28 seconds each day when used in Cayenne.

This one piece of modest data was the pebble that started an avalanche of debate. Giovanni Cassini, the new influential Director of the Paris observatory, interpreted Richer’s observation in the context of Huygen’s pendulum theory (see later). There, the slower period implied that gravity was weaker in Cayenne, probably because earth was smaller around the equator than it was around the poles. Alternatively, in 1683, Robert Hooke had hypothesized that the earth was shaped like an onion rather than a sphere. Isaac Newton ran with this idea in the first edition of Principia (1687). He argued that the slowing of the pendulum was due to the spin of the earth counteracting gravitational forces, and that the slowing would coincide with a bulging of the earth around the equator. Cassini’s and Newton’s contradictory positions, proposed by influential scientists on opposite sides of the English channel, made the figure of the earth a hot topic.

# Exercises

1. Construct an explicit solution of the cycloid model of the form $$x(y)$$.
2. Over what domain and range are the explicit solution defined?
3. Write a python function to check if a given point $$(u,v)$$ falls on a cycloid curve with known parameter values $$R$$ and $$r$$, with an absolute error of no more than $$10^{-2}$$. Use your function to test the following 3 points when $$R = 2$$ and $$r=3$$: $$(-25.288,0.073)$$, $$(0,1.770)$$, $$(27.5,4)$$.
2. 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.
3. A circle of radius $$a$$ is placed inside a larger circle of radius $$b$$. Initially, the small circle is at rest at the bottom of the large circle. Let’s set to origin of a Cartesian coordinate system to be the point where these two circles touch, with the centers of both circles on the y-axis. Now, we roll the little circle inside the big circle, and trace the path of the point $$(x,y)$$ on the little circle that was initially touching the big circle. If $$\theta$$ is the angle between the center of the small circle and it’s initial location, find a parametric equation for $$x$$ and $$y$$ in terms of $$\theta$$. What relationship must be satisfied between $$a$$ and $$b$$ for the point to end up where it started after exactly one roll around the big circle?

4. In Archimedes’s trammel, it is physically reasonable to $$r=0$$, but our implicit solution equation does not appear to make sense. Can you fix our model so it can handle this special case correctly?

5. Archimedes’s trammel is sometimes also called an “ellipsograph” because it can be used to draw ellipses. Suppose we want to make a trammel that can be used to draw an ellipse that is 5 units tall and 2 units wide. What values of $$r$$ and $$s$$ would you use?

6. 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 loop, hang it over the thumb tacks, and draw the curve of the farthest the loop can reach. Show using coordinate geometry that this curve is also an ellipse, as long as as $$L > 2 d.$$ (This construction, first found in the work of Hagia Sophia architect Anthemius of Tralles in 6th century Byzantium, was independently rediscovered by Kepler and contributed to his theory of the planets.)

7. The implicit 6th degree equation obtained above for Watt’s linkage is not very easy to plot. This can be partially solved by choosing a different parameterization and exploiting the linkage-solution’s 4-fold symmetry.
1. Take $$\alpha = x^2 + y^2$$ and find a parametric solution of the form $$(\pm x(\alpha), \pm y(\alpha,x))$$.
2. When you write a program to plot Watt’s linkage using this parameterization, how well does it work?
8. There is actually a second of Watt’s linkage. The the second more practical version can be seen here. Draw and label a diagram of this second version of Watt’s linkage and construct a system of equations to determine the mobility of the the top of the piston rod.

9. A simple gasoline engine design consists of a piston, a crank shaft, and a connecting rod. The piston’s center is directly over the crank shaft’s center of rotation, and straight up and down. The connecting rod of length $$r$$ between the piston and the crank shaft is connected to the crank shaft a distance $$a$$ from the center of rotation. Find the function $$h(\theta)$$ for the height of the piston as a function of the angle of rotation of the crank shaft. Assume $$\theta = 0$$ is when the crank shaft is at its lowest point, and that $$h(0) = 0$$.

10. 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 rope 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 rope 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.
11. 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 $$(p(x,a,b), q(x,a,b))$$.
12. Gray’s oograph consists of a handle turning in a circle around a pivot point, and a slider connected to a point on the handle by a connecting rod of fixed length. A pen is passed through a point on the connecting rod between the slider and the handle. Find an equation for the curve drawn by the pen.

13. Stanley’s oograph is similar to Gray’s, but rather than being connected to a slider, the connecting rod slides through a pivot point a fixed distance from the handle’s pivot point. Find an equation for the curve drawn by the pen.

14. A hexagonal trammel is similar to Archimedes’s trammel, except that it consists of 3 sliders moving along the 3 axes of a regular hexagon. Model the motion of a point on the handle of this trammel.

15. 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?
16. The Peaucellier-Lipkin consists of 4 bars of length $$b$$ forming a rhombus, and an isoceles triangle formed by two longer bars of length $$m$$ joined to opposite corners of the linkage. If the peak of the triangle is fixed to the origin, and the inner corner of the rhombus is fixed to move on a circle of radius $$r$$ centered at the midpoint between the origin and the inner corner of the rhombus, find a set of equations for the positions of the three other corners.

17. 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.
18. Small changes of the lengths of components in a mechanical linkage can sometimes lead to big changes in the range of motion of the linkage. 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 $$a =-1$$.
19. (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.

20. (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.
21. Show that the intersection between a cylinder and a plane is an ellipse.

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

23. Show that the intersection between a cone and a plane is an ellipse.

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

25. Design your own linkage of rods and pivots that turns circular motion into parabolic motion.

26. Find the shape of a curve over which a square could roll without slipping and keeping it’s center level as it goes.

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