Geometry

The mathematics of GPS

Posted on by John

The basic idea of GPS is that if you know the distance to several satellites, you can figure out your position. But you don’t actually know, or need to know, the distance to the satellites: you know the time (according to each satellite’s clock) when the signals were sent, and you know the time (according to your clock) when the signals arrived.

The atomic clocks on satellites are synchronized with each other to within a nanosecond, but they’re not synchronized with your clock. There is some offset t between your clock and the satellites’ clocks. Presumably t is small, but it matters.

If you observe m satellites, you have a system of m equations in 4 unknowns:

|| ai − x || = ti − t

where ai is the known position of the ith satellite in 3 dimensions, x is the observer’s position in three dimensions, and ti is the difference between the time when the signal left the ith satellite (according to its clock) and the time when the signal arrived (according to the observer’s clock). This assumes we choose units so that the speed of light is c = 1.

So we have a system of m equations in 4 unknowns. It’s plausible there could be a unique solution provided m = 4. However, this is not guaranteed.

Here’s an example to suggest why there may not be a unique solution. Suppose t is known to be 0. Then observing 3 satellites will give us 3 equations in 3 unknowns. Each ti determines a sphere of radius ti. Suppose two spheres intersect in a circle, and the third sphere intersects this circle in two points. This means we have two solutions to our system of equations.

In [1] the authors thoroughly study the solution to the GPS system of equations. They allow the satellites and the observer to be anywhere in space and look for conditions under which the system has a unique solution. In practice, GPS satellites are approximately confined to a sphere (more on that here) and the observer is also approximately confined to a sphere, namely the earth’s surface, but the authors do not take advantage of these assumptions.

The authors also assume the problem is formulated in n dimensional space, where n does not necessarily equal 3. It’s complicated to state when the system of equations has a unique solution, but allowing n to vary does not add to the complexity.

I’m curious whether there are practical uses for the GPS problem when n > 3. There are numerous practical problems involving the intersections of spheres in higher dimensions, where the dimensions are not Euclidean spacial dimensions but rather abstract degrees of freedom. But off hand I cannot think of a problem that would involve the time offset that GPS location finding has.

[1] Mireille Boutin, Gregor Kemperc. Global positioning: The uniqueness question and a new solution method. Advances in Applied Mathematics 160 (2024)

Triangle circle maximization problem

Posted on by John

Let a, b, and c be the sides of a triangle. Let r be the radius of an inscribed circle and R the radius of a circumscribed circle. Finally, let p be the perimeter. Then the previous post said that

2prR = abc.

We could rewrite this as

2rR = abc / (a + b + c)

The right hand side is maximized when a = b = c. To prove this, maximize abc subject to the constraint a + b + c = p using Lagrange multipliers. This says

[bc, ac, ab] = λ[1, 1, 1]

and so ab = bc = ac, and from there we conclude a = b = c. This means among triangles with any given perimeter, the product of the inner and outer radii is maximized for an equilateral triangle.

The inner radius for an equilateral triangle is (√3 / 6)a and the outer radius is a/√3, so the maximum product is a²/6.

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Triangles to Triangles

Posted on by John

The set of functions of the form

f(z) = (az + b)/(cz + d)

with ad ≠ bc are called bilinear transformations or Möbius transformations. These functions have three degrees of freedom—there are four parameters, but multiplying all parameters by a constant defines the same function—and so you can uniquely determine such a function by picking three points and specifying where they go.

Here’s an explicit formula for the Möbius transformation that takes z1, z2, and z3 to w1, w2, and w3.

\begin{vmatrix} 1 & z & w & zw \\ 1 & z_1 & w_1 & z_1w_1 \\ 1 & z_2 & w_2 & z_2w_2 \\ 1 & z_3 & w_3 & z_3w_3 \\ \end{vmatrix} = 0

To see that this is correct, or at least possible, note that if you set z = zi and w = wi for some i then two rows of the matrix are equal and so the determinant is zero.

Triangles, lines, and circles

You can pick three points in one complex plane, the z-plane, and three points in another complex plane, the w-plane, and find a Möbius transformation w = f(z) taking the z-plane to the w-plane sending the specified z‘s to the specified w‘s.

If you view the three points as vertices of a triangle, you’re specifying that one triangle gets mapped to another triangle. However, the sides of your triangle may or may not be straight lines.

Möbius transformations map circles and lines to circles and lines, but a circle might become a line or vice versa. So the straight lines of our original triangle may map to straight lines or they may become circular arcs. How can you tell whether the image of a side of a triangle will be straight or curved?

When does a line map to a line?

It’ll be easier if we add a point ∞ to the complex plane and think of lines as infinitely big circles, circles that pass through ∞.

The Möbius transformation (az + b)/(cz + d) takes ∞ to a/c and it takes −d/c to ∞.

The sides of a triangle are line segments. If we look at the entire line, not just the segment, then this line is mapped to a circle. If this line contains the point that gets mapped to ∞ then the image of the line is an infinite circle (i.e. a line). Otherwise the image of the line is a finite circle.

The line between z1 and  z2 can be parameterized by

z1 + t(z2 − z1)

where t is real. So the image of this line will be a line if and only if

z1 + t(z2 − z1) = −d/c

for some real t. So solve for t and see whether you get a real number.

Note that if the point that is mapped to ∞ lies inside the line segment, not just on the line, then the image of that side of the triangle is infinitely long.

Examples

To keep things as simple as possible without being trivial, we’ll use the Möbius transformation f(z) = 1/z. Clearly the origin is the point that is mapped to ∞. The side of a triangle is mapped to a straight line if and only if the side is part of a line through the origin.

First let’s look at the triangle with vertices at (1, 1), (1, 4), and (5, 1). None of the sides is on a line that extends to the origin, so all sides map to circular arcs.

Next let’s move the second point from (1, 4) to (4, 4). The line running between (1, 1) and (4, 4) goes through the origin, and so the segment along that line maps to a straight line.

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Golden ellipse

Posted on by John

A golden ellipse is an ellipse whose axes are in golden proportion. That is, the ratio of the major axis length to the minor axis length is the golden ratio φ = (1 + √5)/2.

Draw a golden ellipse and its inscribed and circumscribed circles. In other words draw the largest circle that can fit inside and the smallest circle outside that contains the ellipse.

Then the area of the ellipse equals the area of the annulus bounded by the two circles. That is, the area of the green region

equals the area of the orange region.

The proof is straight forward. Let a be the semimajor axis and b the semiminor axis, with a = φb.

Then the area of the annulus is

π(a² − b²) = πb²(φ² − 1).

The area of the ellipse is

πab = πφb².

The result follows because the golden ratio satisfies

φ² − 1 = φ.

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Areal coordinates and ellipse area

Posted on by John

Barycentric coordinates are sometimes called area coordinates or areal coordinates in the context of triangle geometry. This is because the barycentric coordinates of a point P inside a triangle ABC correspond to areas of the three triangles PBC, PCA and PAB.

(This assumes ABC has unit area. Otherwise divide the area of each of the three triangles by the area of ABC. We will assume for the rest of this post that the triangle ABC has unit area.)

Areal coordinates take three numbers two describe a point in two dimensional space. Why would you do that? It’s often useful to use an overdetermined coordinate system. The benefit of adding one more coordinate is that you get a coordinate system matched to the geometry of the triangle. For example, the vertices of the triangle have coordinates (1, 0, 0), (0, 1, 0), and (0, 0, 1), regardless of the shape of the triangle.

Here is an example of a theorem [1] that is convenient to state in terms of areal coordinates but that would be more complicated in Cartesian coordinates.

First we need to define the midpoint triangle, also called the medial triangle. This is the triangle whose vertices are the midpoints of each side of ABC. In terms of areal coordinates, the vertices of this triangle are (0, ½, ½), (½, 0, ½), and (½, ½, 0).

Now let P be any point inside the midpoint triangle of ABC. Then there is a unique ellipse E inscribed in ABC and centered at P.

Let (α, β, γ) be the areal coordinates of P. Then the area of E is

\pi \sqrt{(1 - 2\alpha)(1 - 2\beta)(1 - 2\gamma)}

Because P is inside the medial triangle, each of the areal coordinates are less than ½ and so the quantity under the square root is positive.

Finding the equation of the inscribed ellipse is a bit complicated, but that’s not necessary in order to find its area.

Related posts

[1] Ross Honsberger. Mathematical Plumbs. 1979

Ceva, cevians, and Routh’s theorem

Posted on by John

I keep running into Edward John Routh (1831–1907). He is best known for the Routh-Hurwitz stability criterion but he pops up occasionally elsewhere. The previous post discussed Routh’s mnemonic for moments of inertia and his “stretch” theorem. This post will discuss his triangle theorem.

Before stating Routh’s theorem, we need to say what a cevian is. Giovanni Ceva (1647–1734) was an Italian geometer, best known for Ceva’s theorem, and for a construction in that theorem now known as a cevian.

A cevian is a line from the vertex of a triangle to the opposite side. Draw three cevians by connecting each vertex of a triangle to a point on its opposite side. If the cevians intersect at a point, Ceva’s theorem says something about how the lines divide the sides. If the cevians form a triangle, Routh’s theorem find the area of that triangle.

Routh’s theorem is a generalization of Ceva’s theorem because if the cevians intersect at a common point, the area of the triangle formed is zero, and then Routh’s area equation implies Ceva’s theorem.

Let A, B, and C be the vertices of a triangle and let D, E, and F be the points where their cevians intersect the opposite sides.

Let x, y, and z be the ratios into which each side is divided by the cevians. Specifically let x = FB/AF, y = DC/BD, and z = EA/CE.

Then Routh’s theorem says the relative area of the green triangle formed by the cevians is

\frac{(xyz - 1)^2}{(xy + y + 1)(yz + z + 1)(zx + x + 1)}

If the cevians intersect at a point, the area of the triangle is 0, which implies xyz = 1, which is Ceva’s theorem.

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Body roundness index

Posted on by John

Body Roundness Index (BRI) is a proposed replacement for Body Mass Index (BMI) [1]. Some studies have found that BRI is a better measure of obesity and a more effective predictor of some of the things BMI is supposed to predict [2].

BMI is based on body mass and height, and so it cannot distinguish a body builder and an obese man if both have the same height and weight. BRI looks at body shape more than body mass.

The basic idea behind Body Roundness Index is to draw an ellipse based on a person’s body and report how close that ellipse is to being a circle. The more a person looks like a circle, higher his BRI. The formula for BRI is

BRI = 364.2 − 365.5 e

where e is the eccentricity of the ellipse.

Now what is this ellipse we’ve been talking about? It’s roughly an ellipse whose major axis runs from head to toe and whose minor axis runs across the person’s width.

There are a couple simplifications here.

  1. You don’t actually measure how wide someone is. You measure the circumference of their waist and find the diameter of a circle with that circumference.
  2. You don’t actually measure how high their waist is [3]. You assume their waist is at exactly half their height.

It’s conventional to describe an ellipse in terms of its semi-major axis a and semi-minor axis b. For a circle, a = b = radius. But in general an ellipse doesn’t have a single radius and a > b. You could think of a and b as being the maximum and minimum radii.

So to fit an ellipse to our idealized model of a person, the major axis, 2a, equals the person’s height.

a = h/2

The minor axis b is the radius of a circle of circumference c where c is the circumference of the person’s waist (or hips [3]).

b = c / 2π

As explained here, eccentricity is computed from a and b by

 

e = \sqrt{1 - \frac{b^2}{a^2}}

As an example, consider a man who is 6 foot (72 inches) tall and has a 34 inch waist. Then

a = 36
b = 17/π = 5.4112
e = √(1 − b²/a²) = 0.9886
BRI = 364.2 − 365.5 e = 2.8526

Note that the man’s weight doesn’t enter the calculation. He could be a slim guy weighing 180 pounds or a beefier guy weighing 250 pounds as long as he has a 34 inch waist. In the latter case, the extra mass is upper body muscle and not around his waist.

thin ellipse graph

Related posts

[1] Diana M. Thomas et al. Relationships Between Body Roundness with Body Fat and Visceral Adipose Tissue Emerging from a New Geometrical Model. Obesity (2013) 21, 2264–2271. doi:10.1002/oby.20408.

[2] Researchers argue over which number to reduce a person to: BMI, BRI, or some other measure. They implicitly agree that a person must be reduced to a number; they just disagree on which number.

[3] Or waist. There are two versions of BRI, one based on waist circumference and one based on hip circumference.

Drawing with a compass on a globe

Posted on by John

Take a compass and draw a circle on a globe. Then take the same compass, opened to the same width, and draw a circle on a flat piece of paper. Which circle has more area?

If the circle is small compared to the radius of the globe, then the two circles will be approximately equal because a small area on a globe is approximately flat.

To get an idea what happens for larger circles, let’s a circle on the globe as large as possible, i.e. the equator. If the globe has radius r, then to draw the equator we need our compass to be opened a width of √2 r, the distance from the north pole to the equator along a straight line cutting through the globe.

The area of a hemisphere is 2πr². If we take our compass and draw a circle of radius √2 r on a flat surface we also get an area of 2πr². And by continuity we should expect that if we draw a circle that is nearly as big as the equator then the corresponding circle on a flat surface should have approximately the same area.

Interesting. This says that our compass will draw a circle with the same area whether on a globe or on a flat surface, at least approximately, if the width of the compass sufficiently small or sufficiently large. In fact, we get exactly the same area, regardless of how wide the compass is opened up. We haven’t proven this, only given a plausibility argument, but you can find a proof in [1].

Note that the width w of the compass is the radius of the circle drawn on a flat surface, but it is not the radius of the circle drawn on the globe. The width w is greater than the radius of the circle, but less than the distance along the sphere from the center of the circle. In the case of the equator, the radius of the circle is r, the width of the compass is √2 r , and the distance along the sphere from the north pole to the equator is πr/2.

Related posts

[1] Nick Lord. On an alternative formula for the area of a spherical cap. The Mathematical Gazette, Vol. 102, No. 554 (July 2018), pp. 314–316

Ptolemy’s theorem

Posted on by John

Draw a quadrilateral by pick four arbitrary points on a circle and connecting them cyclically.

inscribed quadrilateral

Now multiply the lengths of the pairs of opposite sides. In the diagram below this means multiplying the lengths of the two horizontal-ish blue sides and the two vertical-ish orange sides.

quadrilateral with opposite sides colored

Ptolemy’s theorem says that the sum of the two products described above equals the product of the diagonals.

inscribed quadrilateral with diagonals

To put it in colorful terms, the product of the blue sides plus the product of the orange sides equals the product of the green diagonals.

The converse of Ptolemy’s theorem also holds. If the relationship above holds for a quadrilateral, then the quadrilateral can be inscribed in a circle.

Note that if the quadrilateral in Ptolemy’s theorem is a rectangle, then the theorem reduces to the Pythagorean theorem.

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