The Circumference of the Earth

I often find it amazing that people long ago, made many discoveries that we would find difficult without today’s equipment and computing power. If asked “how long ago was the size of Earth accurately determined?”, what would you guess? Somewhere between 1500 CE and today? What about between 500 CE and today? It turns out that calculation was made almost 2300 years ago, around 245 BCE.

Eratosthenes

Eratosthenes was a Greek mathematician, astronomer, philosopher, poet, and music theorist.Β He was born in 276 BCE and died in 194 BCE. He noticed that in Syene (now Aswan Egypt), that a well with vertical sides did not cast any shadow midday at the summer solstice (when the sun is at its highest point in the sky as the earth revolves about the sun). However, in Alexandria, a vertical pole stuck into the ground, cast a shadow at the same time of the day and year. This happens because the earth is spherical and the direction of the sun at different latitudes is different.

So Eratosthenes measured the length of the pole above the ground in Alexandria and the length of the shadow. How can we use these measurements?

The Calculations

First let’s measure the angle the shadow in Alexandria makes with the stick. To get the same answer as Eratosthenes, I will assume the height of the stick above the ground in Alexandria was 1 meter and the length of the shadow was 12.63 centimeters:

From trigonometry, the tangent of the angle πœƒ, is the opposite side divided by the adjacent side. So if we use the inverse tangent on 0.1263m/1m (the angle whose tangent is 0.1263), we get πœƒ = 7.2Β°. What does this angle have anything to do with the earth? Quite a lot actually.

Now let’s draw the earth underneath the stick:

As the sun is very far away, we can approximate the rays from the sun as being parallel at any spot on earth, so the line between the center of the earth and the well in Syrene is parallel to the edge of the stick’s shadow in Alexandria. This means that the angle between the stick and the well at the earth’s center is the same, 7.2Β°.

Now there are 360Β° to span the entire circumference of the earth, so the curved distance between Alexandria and Syrene divided by the earth’s circumference will be the same as 7.2Β°/360Β° = 0.02. But what is that curved distance.

Evidently, there were people called “bematists” who were trained to measure large distances by counting their steps. Well Eratosthenes employed them to measure the distance between Alexandria and Syrene. Looking at the map, the Nile River gets in the way of this at several places, but I’ll assume they found a way to handle this.

Converting their units to meters, they measured the distance to be 800 kilometers. So 800/(earth’s circumference) = 0.02, which means that earth’s circumference = 40,000 km.

This is a very accurate estimate of the earth’s circumference. As the earth is not a perfect sphere, the circumference measured around the poles is 40,008 km using satellite data and all sorts of equipment and calculation power not available to Eratosthenes. The circumference around the equator is 40,075 km.

Using the formula for the circumference of a circle, we can also get an estimate of the earth’s diameter: C = πœ‹d ⟹ d = 40000/πœ‹ = 12,732 km compared to the technology derived value of 12,714 km.

I am impressed!

An Equation Transformation Example

One of the many skills maths students develop in high school is the ability to change the position of a graph. I sometimes need to remind my students that this is not just “busy” work. It is a skill used frequently in technical fields such as science and engineering. What follows is an example from celestial mechanics.

The Two-Body Problem

Calculating orbits and their characteristics is usually a computer-intensive exercise. For example, how far from the earth is an orbiting satellite at a particular time. However, a good first approximation of this is to pretend we are in the unrealistic universe where only two point masses exist. In this universe, the shape of orbits can be perfectly modelled with equations called conic sections. Why they are called conic sections is a bit beyond the scope of this post, but please free free to look that up.

The orbital shape I want to describe here is the ellipse. The equation of an ellipse which is centered at the origin is:\[\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1\] where 2a is the length of the ellipse in the x direction and 2b is the length of the ellipse in the y direction. If a > b, the line along the a intercepts is called the major axis of the ellipse and the line along the b intercepts is called the minor axis:

\[\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1\]

This is a possible shape of an orbit about one of the masses in our perfect two-mass universe. In reality, orbits of say satellites around the earth, approximate this shape but other factors like other objects in the solar system and the earth not being a point mass with an uneven distribution of its mass, make the orbital shape slightly different than an ellipse. Not only that, in our perfect universe, the orbit path is exactly contained in a plane and can be completely drawn on flat paper. In reality, orbits also go below and above the average plane of the orbit.

Now one of the masses, say mβ‚‚, is on the ellipse. The other mass, m₁, is not at the center of the ellipse, the origin in the above figure. So where is it?

An ellipse has two special points associated with it call focal points. These points have the property that any point on the ellipse has a total distance between it and the focal points is a constant. This constant is 2a if the longest dimension of the ellipse is along the x-axis. Otherwise, this constant length is 2b.

Total length of the red line = 2a regardless of where (x,y) is.

The other mass, m₁, is at one of these focal points. The coordinates of these points can be found using the Pythagorus theorem. Looking at the blue triangle above, you can see that \[c=\sqrt{a^{2} -b^{2}}\]

An Example

Suppose we want to analyze the relationship between two masses that are in orbit. In orbital mechanics, the shape of an orbit is given by its eccentricity, e. An elliptical orbit has an eccentricity between 0 and 1. By the way, a perfectly circular orbit has an eccentricity of 0. So let’s say we have an orbit with the major axis length 2a = 10 and minor axis length 2b = 6. This is an orbit with an eccentricity of 0.8. If we want to know the position of mβ‚‚ with respect to m₁, we need a coordinate system. A convenient one would be a Cartesian system aligning the major axis of the orbit along the x-axis and centering the ellipse at the origin. This way we can immediately write down the equation of the orbit:\[\frac{x^{2}}{25} +\frac{y^{2}}{9} =1\]

as a = 5 and b = 3. This means that the focal points are \[c=\sqrt{25 -9} =4\]from the center:

An example orbit

Two of the more basic pieces of information we would like to know (especially if this is a satellite orbiting the earth) is what is the distance r between m₁ and mβ‚‚ and what is the direction of mβ‚‚ with respect to m₁. We can define the direction as the angle πœƒ of the line connecting the two masses with the positive x -axis. This is actually the reference used in much of celestial mechanics. This angle is called the true anomaly.

Well this is nice, but since we want to find relationships of mβ‚‚ with respect to m₁, it would be even more convenient to put the origin of our coordinate system at m₁. How do we do that?

Transformations to the Rescue

Somewhere in your year 10 or 11 maths (grades 10 or 11 in the States), you took the equation of a standard parabola, y = xΒ², and replaced the x with xh to get y = (xh)Β². This moved the parabola h units to the left or right depending on the sign of h.

Effect of replacing x with (xh)Β²

It turns out that in any relationship between x and y, replacing the x with xh has the exact same effect on its graph. So in our example, if we move the ellipse 4 units to the left, m₁ would be on the origin of our coordinate system.

\[\frac{(x+4)^{2}}{25} +\frac{y^{2}}{9} =1\]

We can now calculate r and πœƒ a bit more easily than before the transformation. Let’s look at where mβ‚‚ is in the above figure. Here, mβ‚‚ is at (-4,3). The following process would work with any point, but at this point, the numbers are “nicer”.

The figure below shows the answers. The distance r can be found using the Pythagorus theorem but once you see that the two sides are 3 and 4, then this is the standard 3-4-5 right triangle:

The solution

In this perfect universe, other orbital shapes are possible: circles, parabolas, and a hyperbolic. These have their own standard equations but they all can be transformed to move them to any place you wish on the coordinate system to make your calculations easier.

Modelling and Graphs

This post presents a simplified engineering modelling and design example. I offer this as a motivation to students struggling with graphing topics like linear, quadratic, cubic, etc equations. This is not just busy work. It is the beginning of developing skills you will need to work in technical professions like engineering.

Why Modelling?

Engineers use mathematical models for many reasons. Some of these reasons are:

  1. Easier and cheaper to analyse and design than the physical thing being modelled
  2. Cheaper to tweak to do “what if” analysis
  3. Necessary when the physical object needs to be encoded, eg a control system for an aircraft – the control system needs to know how the aircraft will react to its controls.

The Example

Let’s use a mathematical model in a familiar scenario.

The Spring

Early automobiles used a suspension that just consisted of leaf springs to support the entire carriage. Later, coil springs for each wheel were used. With just springs, what is the reaction of the carriage (the part that people sit in) to a bump in the road? I think you can imagine that it would look like this:

This is a plot of a cosine function which you may have seen already. This is a model of what would happen to a car with just springs. But it’s not a good model since it assumes that the oscillations would go on forever. Of course, in real life, the springs would lose energy and there would be wind resistance as well. But engineers typically start out with a simple model and add complexity as they continue analysing it.

This particular function I just plotted is x = 5cos(3t), where x is the position of the carriage and t is time. The “5” corresponds to the strength of the bump in the road, and the “3” relates to how weak or stiff the spring is.

The letter k is usually used to represent the spring stiffness. The higher k is, the stiffer the spring and the faster the oscillations experienced by the people in the car. Changing the “5” just changes the extreme values of the curve, but changing the “3” has a more interesting effect. Below are three graphs showing the effect of changing k:

Red: k = 1
Blue: k = 3
Green:k = 5

The Damper

As I mentioned before, just using a cosine function to model the spring suspension is not good because there is natural damping that occurs due to energy loss in the spring (heat) and resistance due to the air around the car. The real reaction of the carriage would look something like this:

Spring with damper

But even this would not be comfortable as the car would be frequently bouncing as it goes over bump after bump (but certainly more comfortable than no suspension at all!). So more damping would be desireable.

A function that has a damping shape is the exponential function
x = adt where the “a” and the “d” are some positive constants for a particular equation. As will be seen when you study calculus, a convenient “a” to use is the irrational number e. Don’t worry if you haven’t seen e before. It is just a number approximately equal to 2.71828. I say approximately because e, like πœ‹, is irrational so it has a non-repeating decimal part. The d indicates how strong the damping effect is; the larger d is, the stronger the damping.

So as an example, the plot of x = 5e-0.5t looks like:

x = 5e-0.5t

The “5” in his example is just to provide the initial bump. So it would be nice if we could combine the action of the spring with a damping force that a shock absorber would provide. This can be done by multiplying the shock absorber model
x = 5e-0.5t and the spring model x = 5cos(3t) together:
x = 5e-0.5tcos(3t) where we just need the one “5” for the initial bump. The plot of this looks like:

x = 5e-0.5tcos(3t)

Much better but room for improvement. Depending on the type of car we are designing, we may want a stiff response where the damping is strong (like in a sports car where one wants to “feel the road”) or a softer one for a family car for example.

Below are two plots changing the damping strength and using k = 3:

Blue: d = 0.5
Red: d = 1.0

A carriage following the red curve would be a better ride. The damping can be made stronger for a sports car.

When the car is actually built and testing reveals that the car responds to bumps as was predicted by the engineer’s model, well that’s a very good feeling!

Why Study Maths?

The engineer in this discussion used several skills that you are studying now:

  1. Drawing graphs of equations
  2. How to change the shape of those graphs my changing the values in the equation
  3. Circular (trigonometric) and exponential functions
  4. Calculus
  5. And all of the above is based on algebra.

The calculus skill is not apparent in the discussion so far, but I will explain that further in the next section.

Creating a Model

We can generalise the model for the suspension of a car as x = Aedtcos(kt) where A is the size of the initial bump in the road, d is the stiffness of the shock absorbers, and k is the strength of the springs. How did I know to use the exponential and cosine functions? They are the solution to what are called differential equations which are based on calculus.

When developing a model, especially for dynamic (moving) systems, they start with basic physics. Frequently, Newton’s second law is the starting point:\[F = ma\] where F is the force applied or exerted by a mass m, accelerating (or decelerating) by a. Now, we want a model that predicts x, the position of the car carriage. Many applications of calculus involve the rate of change of something. The rate of change of position is called velocity which you are familiar with. And the rate of change of velocity is called acceleration. Newton’s second law explains why you feel your back press against the seat only when you accelerate (or against the seatbelt when you decelerate). The rate of change of position, that is velocity, is mathematically called the first derivative of position. The rate of change of velocity, that is acceleration, is the first derivative of velocity or the second derivative of position. So Newton’s second law is really an equation about the second derivative of position. Equations that have derivatives in them, are called differential equations and these are very familiar to engineers.

Notation-wise, a single dot is shown above a variable to indicate a first derivative and two dots represent a second derivative. So another way to show Newton’s second law is \[F=m\ddot{x}\] as acceleration is the second derivative of the position, x.

Now without going into the detail, the forces exerted on the carriage by the springs and the shock absorbers are kx and dαΊ‹ respectively. So if you replace the F in Newton’s second law with these forces, you get the differential equation\[kx+d\dot{x}=m\ddot{x}\]

An engineer uses calculus to solve this equation for the position x as a function of the time t, that is x(t). Given certain initial conditions like the size of the bump experienced by the car, and the values of k and d, a solution to this equation can be x = Aedtcos(kt). This is where I got the cosine and the exponential function from at the beginning.

My logo

Let me explain a bit about my logo. The foreground words are self-explanatory except for the πœ‹ symbol used in place of the “T”. πœ‹ is the Greek letter pi. Greek symbols are used extensively in maths and πœ‹ is the most common one used. You will see it used many times in subsequent posts.Β  The background equations not only have math equations, but symbols representing all sorts of math applications: astronomy, biology, chemistry, genetics, nuclear physics, electronics, … . I think this is a very appropriate logo for me. What do you think?

Why can we do maths?

Ever wonder why humans can do math? What was the evolutionary pressure that gave us the ability to do calculus? It turns out that our math abilities are a by-product of our language abilities. A good book that explains this is “The Maths Gene” by Keith Devlan. Check it out!

Introducing the DavidTheMathsTutor Blog

Hi all, Well after a bit of learning how to set up a blog site and addressing the security concerns and a lot of trial and error, I’ve finally set up the blog site. I’m sure some tweeking will occur in the early days, and I’m open to comments to improve the site. I will post my first official math related post soon. Hopefully, the facebook page will be automatically updated.