Solving spherical triangles

Posted on by John

This post is a side quest in the series on navigating by the stars. It expands on a footnote in the previous post.

There are six pieces of information associated with a spherical triangle: three sides and three angles. I said in the previous post that given three out of these six quantities you could solve for the other three. Then I dropped a footnote saying sometimes the missing quantities are uniquely determined but sometimes there are two solutions and you need more data to uniquely determine a solution.

Todhunter’s textbook on spherical trig gives a thorough account of how to solve spherical triangles under all possible cases. The first edition of the book came out in 1859. A group of volunteers typeset the book in TeX. Project Gutenberg hosts the PDF version of the book and the TeX source.

I don’t want to duplicate Todhunter’s work here. Instead, I want to summarize when solutions are or are not unique, and make comparisons with plane triangles along the way.

SSS and AAA

The easiest cases to describe are all sides or all angles. Given three sides of a spherical triangle (SSS), you can solve for the angles, as with a plane triangle. Also, given three angles (AAA) you can solve for the remaining sides of a spherical triangle, unlike a plane triangle.

SAS and SSA

When you’re given two sides and an angle, there is a unique solution if the angle is between the two sides (SAS), but there may be two solutions if the angle is opposite one of the sides (SSA). This is the same for spherical and plane triangles.

There could be even more than two solutions in the spherical case. Consider a triangle with one vertex at the North Pole and two vertices on the equator. Two sides are specified, running from the pole to the equator, and the angles at the equator are specified—both are right angles—but the side of the triangle on the equator could be any length.

ASA and AAS

When you’re given two angles and a side, there is a unique solution if the side is common to the two angles (ASA).

If the side is opposite one of the angles (AAS), there may be two solutions to a spherical triangle, but only one solution to a plane triangle. This is because two angles uniquely determine the third angle in a plane triangle, but not in a spherical triangle.

The example above of a triangle with one vertex at the pole and two on the equator also shows that an AAS problem could have a continuum of solutions.

Summary

\begin{tabular}{|l|c|c|} \hline \textbf{Case} & \textbf{Plane} & \textbf{Spherical} \\ \hline SSS & 1 & 1 \\ SAS & 1 & 1 \\ SSA & 1 or 2 & 1 or 2 \\ AAS & 1 & 1 or 2 \\ ASA & 1 & 1 \\ AAA & $\infty$ & 1 \\ \hline \end{tabular}

Note that spherical triangles have a symmetry that plane triangles don’t: the spherical column above remains unchanged if you swap S’s and A’s. This is an example of duality in spherical geometry.

The Navigational Triangle

Posted on by John

The previous post introduced the idea of finding your location by sighting a star. There is some point on Earth that is directly underneath the star at any point in time, and that location is called the star’s GP (geographic position). That is one vertex of the navigational triangle. The other two vertices are your position and the North Pole.

Unless you’re at Santa’s workshop and observing a star nearly directly overhead, the navigational triangle is a big triangle, so big that you need to use spherical geometry rather than plane geometry. We will assume the Earth is a sphere [1].

Let a be the side running from your position to the GP. In the terminology of the previous post a is the radius of the line of position (LOP).

Let b be the side running from the GP to the North Pole. This is the GP’s lo-latitude, the complement of latitude.

Let c be the side running from your location to the North Pole. This is your co-latitude.

Let A, B, and C be the angles opposite a, b, and c respectively. The angle A is known as the local hour angle (LHA) because it is proportional to the time difference between noon at your location and noon at the GP.

Given three items from the set {a, b, c, A, B, C} you can solve for the other three [2]. Note that one possibility is knowing the three angles. This is where spherical geometry differs from plane geometry: you can’t have spherical triangles that are similar but not congruent because the triangle excess determines the area.

If you know the current time, you can look up the GP coordinates in a table. The complement of the GP’s latitude is the side b.

Also from the current time you can determine your longitude, and from that you can find the LHA (angle A).

As described in the previous post, the altitude of the star, along with its GP, determines the LOP. From the LOP you can determine the arc between you and the GP, i.e. side a. We haven’t said how you could determine a, only that you could.

If you know two sides (in our case a and b) and the angle opposite one of the sides (in our case A) you can solve for the rest.

Adding detail

This post is more detailed than the previous, but still talks about what can be calculated but now how. We’re adding detail as the series progresses.

To motivate future posts, note that just because something can in theory be computed from an equation, that doesn’t mean it’s best to use that equation. Maybe the equation is sensitive to measurement error, or is numerically unstable, or is hard to calculate by hand.

Since we’re talking about navigating by the stars rather than GPS, we’re implicitly assuming that you’re using pencil and paper because for some reason you can’t use GPS.

Related posts

[1] To first approximation, the Earth is a sphere. To second approximation, it’s an oblate spheroid. If you want to get into even more detail, it’s not exactly an oblate spheroid. How much difference does all this make? See this post.

[2] In some cases there are two solutions for one of the missing elements and you’ll need to use additional information, such as your approximate location, to rule out one of the possibilities. More on when solutions are unique here.

Line of position (LOP)

Posted on by John

The previous post touched on how Lewis and Clark recorded celestial observations so that the data could be turned into coordinates after they returned from their expedition. I intend to write a series of posts about celestial navigation, and this post will discuss one fundamental topic: line of position (LOP).

Pick a star that you can observe [1]. At any particular time, there is exactly one point on the Earth’s surface directly under the star, the point where a line between the center of the Earth and the star crosses the Earth’s surface. This point is called the geographical position (GP) of the star.

This GP can be predicted and tabulated. If you happen to be standing at the GP, and know what time it is, these tables will tell your position. Most likely you’re not going to be standing directly under the star, and so it will appear to you as having some deviation from vertical. The star would appear at the same angle from vertical for ring of observers. This ring is called the line of position (LOP).

An LOP of radius 1200 miles centered at a GP at Honolulu

The LOP is a “small circle” in a technical sense. A great circle is the intersection of the Earth’s surface with a plane through the Earth’s center, like a line of longitude. A small circle is the intersection of the surface with a plane that does not pass through the center, like a line of latitude.

The LOP is a small circle only in contrast to a great circle. In fact, it’s typically quite large, so large that it matters that it’s not in the plane of the GP. You have to think of it as a slice through a globe, not a circle on a flat map, and therein lies some mathematical complication, a topic for future poss. The center of the LOP is the GP, and the radius of the LOP is an arc. This radius is measured along the Earth’s surface, not as the length of a tunnel.

One observation of a star reduces your set of possible locations to a circle. If you can observe two stars, or the same star at two different times, you know that you’re at the intersection of the two circles. These two circles will intersect in two points, but if you know roughly where you are, you can rule out one of these points and know you’re at the other one.

 

[1] At the time of the Lewis and Clark expedition, these were the stars of interest for navigation in the northern hemisphere: Antares, Altair, Regulus, Spica, Pollux, Aldeberan, Formalhaut, Alphe, Arieties, and Alpo Pegas. Source: Undaunted Courage, Chapter 9.

Lewis & Clark geolocation

Posted on by John

I read Undaunted Courage, Stephen Ambrose’s account of the Lewis and Clark expedition,  several years ago [1], and now I’m listening to it as an audio book. The first time I read the book I glossed over the accounts of the expedition’s celestial observations. Now I’m more curious about the details.

The most common way to determine one’s location from sextant measurements is Hilare’s method [2], developed in 1875. But the Lewis and Clark expedition took place between 1804 and 1806. So how did the expedition calculate geolocation from their astronomical measurements? In short, they didn’t. They collected data for others to turn into coordinates later. Ambrose explains

With the sextant, every few minutes he would measure the angular distance between the moon and the target star. The figures obtained could be compared with tables show how those distances appeared at the same clock time in Greenwich. Those tables were too heavy to carry on the expedition, and the work was too time-consuming. Since Lewis’s job was to make the observations and bring them home, he did not try to do the calculations; he and Clark just gathered the figures.

The question remains how someone back in civilization would have calculated coordinates from the observations when the expedition returned. This article by Robert N. Bergantino addresses this question in detail.

Calculating latitude from measurements of the sun was relatively simple. Longitude was more difficult to obtain, especially without an accurate way to measure time. The expedition had a chronometer, the most expensive piece of equipment on the expedition that was accurate enough to determine the relative time between observations, but not accurate enough to determine Greenwich time. A more accurate chronometer would have been too expensive and too fragile to carry on the voyage.

For more on calculating longitude, see Dava Sobel’s book Longitude.

Related posts

[1] At least 17 years ago. I don’t keep a log of what I read, but I mentioned Undaunted Courage in a blog post from 2008.

[2] More formally known as Marcq Saint-Hilaire’s intercept method.

Zero knowledge proof of compositeness

Posted on by John

A zero knowledge proof (ZKP) answers a question without revealing anything more than answer. For example, a digital signature proves your possession of a private key without revealing that key.

Here’s another example, one that’s more concrete than a digital signature. Suppose you have a deck of 52 cards, 13 of each of spades, hearts, diamonds, and clubs. If I draw a spade from the deck, I can prove that I drew a spade without showing which card I drew. If I show you that all the hearts, diamonds, and clubs are still in the deck, then you know that the missing card must be a spade.

Composite numbers

You can think of Fermat’s primality test as a zero knowledge proof. For example, I can convince you that the following number is composite without telling you what its factors are.

n = 244948974278317817239218684105179099697841253232749877148554952030873515325678914498692765804485233435199358326742674280590888061039570247306980857239550402418179621896817000856571932268313970451989041

Fermat’s little theorem says that if n is a prime and b is not a multiple of n, then

bn−1 = 1 (mod n).

A number b such that bn−1 ≠ 1 (mod n) is a proof that n is not prime, i.e. n is composite. So, for example, b = 2 is a proof that n above is composite. This can be verified very quickly using Python:

    >>> pow(2, n-1, n)
    10282 ... 4299

I tried the smallest possible base [1] and it worked. In general you may have to try a few bases. And for a few rare numbers (Carmichael numbers) you won’t be able to find a base. But if you do find a base b such that bn−1 is not congruent to 1 mod n, you know with certainty that n is composite.

Prime numbers

The converse of Fermat’s little theorem is false. It can be used to prove a number is not prime, but it cannot prove that a number is prime. But it can be used to show that a number is probably prime. (There’s some subtlety as to what it means for a number to probably be prime. See here.)

Fermat’s little theorem can give you a zero knowledge proof that a number is composite. Can it give you a zero knowledge proof that a number is prime? There are a couple oddities in this question.

First, what would it mean to have a zero knowledge proof that a number is prime? What knowledge are you keeping secret? When you prove that a number is composite, the prime factors are secret (or even unknown), but what’s the secret when you say a number is prime? Strictly speaking a ZKP doesn’t have to keep anything secret, but in practice it always does.

Second, what about the probability of error? Zero knowledge proofs do not have to be infallible. A ZKP can have some negligible probability of error, and usually do.

It’s not part of the definition, but practical ZKPs must be easier to verify than the direct approach to what they prove. So you could have something like a primality certificate that takes far less computation to verify than the computation needed to determine from scratch that a number is prime.

Proving other things

You could think of non-constructive proofs as ZKPs. For example, you could think of the intermediate value theorem as a ZKP: it proves that a function has a zero in an interval without giving you any information about where that zero may be located.

What makes ZKPs interesting in application is that they can prove things of more general interest than mathematical statements [2]. For example, cryptocurrencies can provide ZKPs that accounting constraints hold without revealing the inputs or outputs of the transaction. You could prove that nobody tried to spend a negative amount and that the sum of the inputs equals the sum of the outputs.

Related posts

[1] You could try b = 1, but then bn−1 is always 1. This example shows that the existence of a base for which bn−1 = 1 (mod n) doesn’t prove anything.

[2] You might object that accounting rules are mathematical statements, and of course they are. But they’re of little interest to mathematicians and of great interest to the parties in a transaction.

Monero subaddresses

Posted on by John

Monero has a way of generating new addresses analogous to the way HD wallets generate new addresses for Bitcoin. In both cases, the recipient’s software can generate new addresses to receive payments that others cannot link back to the recipient.

Monero users have two public/private keys pairs: one for viewing and one for spending. Let Ks and ks be the public and private spending keys, and let Kv and kv be the public and private viewing keys. Then the user’s ith subaddress is given by

\begin{align*} K^s_i &= K^s + H(k^v, i) G \\ K^v_i &= k^v K^s_i \end{align*}

Here G is a generator for the elliptic curve Ed25519 and H is a hash function. The hash function output and kv are integers; the public keys, denoted by capital Ks with subscripts and superscripts, are points on Ed25519. The corresponding private keys are

\begin{align*} k^s_i &= k^s + H(k^v, i) \\ k^v_i &= k^v + k^s_i \end{align*}

As with hierarchical wallets, the user scans the blockchain to see which of his addresses have received funds.

A user may choose to give a different subaddress for each transaction for added security, or to group transactions for accounting purposes.

Note that in addition to subaddresses, Monero uses stealth addresses. An important difference between subaddresses and stealth addresses is that recipients generate subaddresses, and senders generate stealth addresses. Someone could send you money to the same subaddress twice, failing to create a new stealth address. This is not possible if you give the sender a different subaddress each time.

Related posts

A triangle whose interior angles sum to zero

Posted on by John

Spherical geometry

In spherical geometry, the interior angles of a triangle add up to more than π. And in fact you can determine the area of a spherical triangle by how much the angle sum exceeds π. On a sphere of radius 1, the area equals the triangle excess

Area = E = interior angle sum − π.

Small triangles have interior angle sum near π. But you could, for example, have a triangle with three right angles: put a vertex on the north pole and two vertices on the equator 90° longitude apart.

Hyperbolic geometry

In hyperbolic geometry, the sum of the interior angles of a triangle is always less than π. In a space with curvature −1, the area equals the triangle defect, the difference between π and the angle sum.

Area = D = π − interior angle sum.

Again small triangles have an interior angle sum near π. Both spherical and hyperbolic geometry are locally Euclidean.

The interior angle sum can be any value less than π, and so as the angle sum goes to 0, the triangle defect, and hence the area, goes to π. Since the minimum angle sum is 0, the maximum area of a triangle is π.

The figure below has interior angle sum 0 and area π in hyperbolic geometry.

Strictly speaking this is an improper triangle because the three hyperbolic lines (i.e. half circles) don’t intersect within the hyperbolic plane per se but at ideal points on the real axis. But you could come as close to this triangle as you like, staying within the hyperbolic plane.

Note that the radii of the (Euclidean) half circles doesn’t change the area. Any three semicircles that intersect on the real line as above make a triangle with the same area. Note also that the triangle has infinite perimeter but finite area.

Related posts

A circle in the hyperbolic plane

Posted on by John

Let ℍ be the upper half plane, the set of complex real numbers with positive imaginary part. When we measure distances the way we’ve discussed in the last couple posts, the geometry of ℍ is hyperbolic.

What is a circle of radius r in ℍ? The same as a circle in any geometry: it’s the set of points a fixed distance r from a center. But when you draw a circle using one metric, it may look very different when viewed from the perspective of another metric.

Suppose we put on glasses that gave us a hyperbolic perspective on ℍ, draw a circle of radius r centered at i, then take off the hyperbolic glasses and put on Euclidean glasses. What would our drawing look like?

In the previous post we gave several equivalent expressions for the hyperbolic metric. We’ll use the first one here.

d(z_1, z_2) = 2\, \text{arcsinh}\left( \frac{|z_1 - z_2|}{2\, \sqrt{\Im z_1\, \Im z_2}} \right)

Here the Fraktur letter ℑ stands for imaginary part. So the set of points in a circle of radius r centered at i is

\{ x + iy \mid d(x + iy, i) = r \}

Divide the expression for d(x + iy, i) by 2, apply sinh, and square. This gives us

\sinh^2\left(\frac{r}{2}\right) = \frac{x^2 + (y-1)^2}{4y}

which is an equation for a Euclidean circle. If we multiply both sides by 4y and complete the square, we find that the center of the circle is (0, cosh(r)) and the radius is sinh(r).

Summary so far

So to recap, if we put on our hyperbolic glasses and draw a circle, then switch out these glasses for Euclidean glasses, the figure we drew again looks like a circle.

To put it another way, a hyperbolic viewer and a Euclidean viewer would agree that a circle has been draw. However, the two viewers would disagree where the center of the circle is located, and they would disagree on the radius.

Both would agree that the center is on the imaginary axis, but the hyperbolic viewer would say the imaginary part of the center is 1 and the Euclidean viewer would say it’s cosh(r). The hyperbolic observer would say the circle has radius r, but the Euclidean observer would say it has radius sinh(r).

Small circles

For small r, the hyperbolic and Euclidean viewpoints nearly agree because

cosh(r) = 1 + O(r²)

and

sinh(r) = r + O(r³)

Big circles

Note that if you asked a Euclidean observer to draw a circle of radius 100, centered at (0, 1), he would say that the circle will extend outside of the half plane. A hyperbolic observer would disagree. From his perspective, the real axis is infinitely far away and so he can draw a circle of any radius centered at any point and stay within the half plane.

Moving circles

Now what if we looked at circles centered somewhere else? The hyperbolic metric is invariant under Möbius transformations, and so in particular it is invariant under

z ↦ x0 + y0 z.

This takes a circle with hyperbolic center i to a circle centered at x0 + i y0 without changing the hyperbolic radius. The Euclidean center moves from cosh(r) to y0 cosh(r) and the radius changes from sinh(r) to y0 sinh(r).

Equal things that don’t look equal

Posted on by John

The previous post described a metric for the Poincaré upper half plane. The development is geometrical rather than analytical. There are also analytical formulas for the metric, at least four that I’ve seen.

\begin{align*} d(z_1, z_2) &= 2\, \text{arcsinh}\left( \frac{|z_1 - z_2|}{2\, \sqrt{\Im z_1\, \Im z_2}} \right) \\ &= \text{arccosh}\left(1 + \frac{|z_1 - z_2|^2}{ 2\, \Im z_1 \, \Im z_2} \right) \\ &= 2 \, \text{arctanh}\left| \frac{z_1 - z_2}{z_1 - \overline{z}_2}\right| \\ &= 2\, \log\left( \frac{|z_2 - z_1| + |z_2 - \overline{z}_1|}{2\, \sqrt{\Im z_1\, \Im z_2}} \right) \\ \end{align*}

It’s not at all obvious that the four equations are equivalent, or that any of them matches the expression in the previous post.

There are equations for expressing arcsinh, arccosh, and arctanh in terms of logarithms and square roots. See the bottom of this post. You could use these identities to show that the metric expressions are equal, but I don’t know of a cleaner way to do this than lots of tedious algebra.

Before diving into the calculations, you might want some assurance that you’re trying to prove the right thing. Here’s some Python code that generates random pairs of complex numbers and shows that the four expressions give he same distance.

import numpy as np

def d1(z1, z2):
    return 2*np.arcsinh( abs(z1 - z2) / (2*(z1.imag * z2.imag)**0.5) )
def d2(z1, z2):
    return np.arccosh(1 + abs(z1 - z2)**2 / (2*z1.imag * z2.imag) )
def d3(z1, z2):
    return 2*np.arctanh( abs( (z1 - z2)/(z1 - np.conjugate(z2)) ) )
def d4(z1, z2):
    return 2*np.log( (abs(z2 - z1) + abs(z2 - np.conjugate(z1)))/(2*np.sqrt(z1.imag * z2.imag)) )

np.random.seed(20251127)
for n in range(100):
    z1 = np.random.random() + 1j*np.random.random()
    z2 = np.random.random() + 1j*np.random.random()
    assert( abs(d1(z1, z2) - d2(z1, z2)) < 1e-13 )
    assert( abs(d2(z1, z2) - d3(z1, z2)) < 1e-13 )
    assert( abs(d3(z1, z2) - d4(z1, z2)) < 1e-13 )

Perhaps you’re convinced that the four expressions are equal, but why should any of them be equivalent to the definition in the previous post?

The previous post pointed out that the metric is invariant under Möbius transformations. We can apply such a transformation to move any pair of complex numbers to the imaginary axis. There you can see that the cross ratio reduces to the ratio of the two numbers.

More generally, if two complex numbers have the same real part, the distance between them is the log of the ratio of their imaginary parts. That is, if

\begin{align*} z_1 &= x + iy_1 \\ z_2 &= x + iy_2 \end{align*}
then

d(z_1, z_2) = \log \frac{y_2}{y_1}

if x, y1, and y2 are real and y2 > y1 > 0.

Here’s a little Python code that empirically shows that this gives the same distance as one of the expressions above.

def d5(z1, z2):
    assert(z1.real == z1.real)
    return abs( np.log( z1.imag / z2.imag ) )

for n in range(100):
    x = np.random.random()
    z1 = x + 1j*np.random.random()
    z2 = x + 1j*np.random.random()
    assert( abs(d1(z1, z2) - d5(z1, z2)) < 1e-13 )

So now we have five expressions for the metric, all of which look different. You could slug out a proof that they’re equivalent, or get a CAS like Mathematica to show they’re equivalent, but it would be more interesting to find an elegant equivalence proof.

Update: Although the four expressions at the top of the post are analytically equal, they are not all equally accurate for numerical evaluation. I did a little testing and found the arctanh method to be the least accurate and the rest roughly equally accurate.

Hyperbolic metric

Posted on by John

One common model of the hyperbolic plane is the Poincaré upper half plane ℍ. This is the set of points in the complex plane with positive imaginary part. Straight lines are either vertical, a set of points with constant imaginary part, or arcs of circles centered on the real axis. The real axis is not part of ℍ. From the perspective of hyperbolic geometry these are ideal parts, infinitely far away, and not part of the plane itself.

We can define a metric on ℍ as follows. To find the distance between two points u and v, draw a line between the two points, and let a and b be the ideal points at the end of the line. By a line we mean a line as defined in the geometry of ℍ, what we would see from our Euclidean perspective as a half circle or a vertical line. Then the distance between u and v is defined as the absolute value of the log of the cross ratio (u, v; a, b).

d(u, v) = |\log (u, v; a, b) | = \left| \log \frac{|a - u|\,|b - v|}{|a - v|\,|b - u|} \right|
Cross ratios are unchanged by Möbius transformations, and so Möbius transformations are isometries.

Another common model of hyperbolic geometry is the Poincaré disk. We can use the same metric on the Poincaré disk because the Möbius transformation

z \mapsto \frac{z - i}{z + i}

maps the upper half plane to the unit disk. This is very similar to how the Smith chart is created by mapping a grid in the right half plane to the unit disk.

Update: See the next post for four analytic expressions for the metric, direct formulas involving u and v but not the ideal points a and b.