Spherical Geometry Lecture Notes

This document contains my lecture notes for Spherical Geometry. The goal of this document is to be used as a reference, so I will keep it pretty compact.

Coordinate systems

There are multiple coordinates systems that are used in this course, these are:

• Cartesian $(x, y, z)$
• Polar $(r, \theta)$
• Cylindrical $(r, \theta, z)$
• Spherical $(r, \theta, \phi)$

Polar coordinates

Polar coordinates is a 2D coordinate system, expressed in $(r, \theta)$.

The cylindral and spherical coordinate systems are based on the polar coordinate system. We can convert between the cartesian and polar coordinate systems with the following equations.

Cartesian to polar

If we want to convert from $(x, y) \rightarrow (r, \theta)$: $$ \begin{align} \begin{cases} r &= \sqrt{x_p^2 + y_p^2} \\ \theta &= \textrm{atan}\left(\frac{y_p}{x_p}\right) \end{cases} \end{align} $$

Note about $\textrm{atan}$

We also need to keep track of the quadrant in which $P$ falls.

Sign $x$	Sign $y$	Quadrant	$\theta$
$+$	$+$	Q1	$\theta' = \theta$
$-$	$+$	Q2	$\theta' = \frac{\pi}{2} + \theta$
$-$	$-$	Q3	$\theta' = \pi + \theta$
$+$	$-$	Q4	$\theta' = \frac{3\pi}{2} + \theta$

Polar to cartesian

If we want to convert from $(r, \theta) \rightarrow (x, y)$: $$ \begin{align} \begin{cases} x &= r\cdot\cos\theta \\ y &= r\cdot\sin\theta \end{cases} \end{align} $$

Cylindrical coordinates

Cylindrical coordinates are expressed in $(r, \theta, z)$, which are illustrated below.

We can convert between the cartesian and cylindrical coordinate systems with the following equations.

Cartesian to cylindrical

If we want to convert from $(x, y, z) \rightarrow (r, \theta, z)$: $$ \begin{align} \begin{cases} r &= \sqrt{x_p^2 + y_p^2} \\ \theta &= \textrm{atan}\left(\frac{y_p}{x_p}\right) \\ z &= z_p \end{cases} \end{align} $$ Note that $\textrm{atan}$ is used, so the result needs to be modified based on the quadrant.

Cylindrical to cartesian

If we want to convert from $(r, \theta, z) \rightarrow (x, y, z)$: $$ \begin{align} \begin{cases} x &= r\cdot\cos\theta \\ y &= r\cdot\sin\theta \\ z &= z_p \end{cases} \end{align} $$

Spherical coordinates

Spherical coordinates are expressed in $(r, \theta, \phi)$, which are illustrated below:

We can convert between the cartesian and spherical coordinate systems with the following equations.

Cartesian to spherical

If we want to convert from $(x, y, z) \rightarrow (r, \theta, \phi)$: $$ \begin{align} \begin{cases} r &= \sqrt{x^2 + y^2 + z^2} \\ \theta &= \textrm{atan2}(y, x) \\ \phi &= \textrm{acos}\left(\frac{z}{r}\right) \end{cases} \end{align} $$ The function $\textrm{atan2}$ automatically adjusts for quadrants.

Spherical to cartesian

If we want to convert from $(r, \theta, \phi) \rightarrow (x, y, z)$: $$ \begin{align} \begin{cases} x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \\ z &= r\cos\phi \end{cases} \end{align} $$

Geographical coordinates

Geographical coordinates are expressed in latitude and longitude, as illustrated below:

These coordinates are very similar to $\theta$ and $\phi$ in the spherical coordinate system. However, notice that $\phi$ is measured with the x-axis in the geographical coordinate system, and with the z-axis (up) in the spherical coordinate system.

DMS format

Geographical coordinates are often expressed in the day, minute, second (DMS) format. An example of this are the coordinates of Amsterdam, which are 52° 22' 45.0804'' N, 4° 53' 57.9516'' E. To convert these into degrees, we divide the minutes by 60, and the seconds by 3600. $$52 + \frac{22}{60} + \frac{45.0804}{60^2} = 52.379189$$ If we have a coordinate with 5 W, this is the same as -5 E. But if it happens to be -5 W, then the minus sign indicates that it is west, so it sould not be written as 5 E, but as 5 W. However, in my opinion this is just bad notation, and 5 E should have been used in the first place.

Distance between two points

If we want to calculate the distance between two points, there are three different cases that occur. An important formula to use is the arc of a circle, which is $L = \theta\cdot r$, where $\theta$ must be in radians. Other useful conversions to know is that $1^\circ = 60\textrm{nm}$ (nautical miles), and $1\textrm{nm} = 1.852 \textrm{km}$.

Equal longitude (GRC)

In the case where the longitude is equal, we can subtract both latitudes, and multiply them by the radius of the earth, which is $R=6371$.

If we have for example point A at 36 N, 22 E, and point B at 29 N, 22 E, the difference in latitude is $36 \textrm{N} - 29 \textrm{N}=7^\circ$. The distance between A and B is $7^\circ \cdot 60\textrm{nm} \cdot 1.852 \textrm{km} = 777.84 \textrm{km}$. Because the latitude lines are all great circles (GRC), this distance is a GRC distance, which is the shortest.

Equal latitude (parallel)

In the case where the latitude's are equal, we need a different method because the longitude lines are not great circles. In this case the angle is $$ \theta = |\Delta L|\cos\phi. $$

Note that $\phi$ is in radians.

If we have for example point A at 12 N, 53 E, and point B at 12 N, 21 E, the difference in longitude is $\Delta L = 53\textrm{E} - 21\textrm{E} = 32^\circ$. Now we can find the angle, which is: $$ \theta = 53\cdot\cos\left(12\cdot\frac{\pi}{180}\right) = 31.3007 $$ Thus the distance from A to B is $31.3007 \cdot 60 \cdot 1.852 = 3478.14$. This is not a GRC distance, but a parallel distance (line is parallel to longitude lines). This also means that this is not the shortest distance.

No equal latitude and longitude (GRC)

To calculate the distance between two point if the latitude and longitude are not equal, we can use the dot product formula, which is $$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|\cdot|\mathbf{b}|\cdot\cos\theta. $$ We can use this to find the angle between two vectors, if we rewrite it as $$ \theta = \textrm{acos}\left(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\cdot|\mathbf{b}|}\right), $$ which can then be used to find the distance in km. Note that this is quite some work, because the coordinates need to be converted from $\textrm{DMS} \rightarrow (r,\theta,\phi) \rightarrow (x,y,z)$. This again is a GRC distance, thus the shortest.

Loxodrome (rhumb line)

A path along the rhumb line, is not the shortest, but has a constant azimuth. That means the angle $\beta$ with the meridian is always the same. A rhumb line will converge into the poles.

Determine direction for loxodrome

We can calculate the length and angle of the path if we construct a small triangle.

Based on this triangle, we can set up the following equation to determine the direction:

$$ \tan k = \frac{R'\ \text{d}\lambda}{R\ \text{d}\varphi} $$

Because $R' = R\cdot\cos\varphi$ (from the parallel distance equation), we can substitute:

$$ \tan k = \frac{R \cos\varphi\ \text{d}\lambda}{R\ \text{d}\varphi} = \frac{\cos\varphi\ \text{d}\lambda}{\text{d}\varphi} $$

Now we want to rearrange the equation, such that $\lambda$ is on the left, and $\varphi$ is on the right, and integrate it.

$$ \int_{\lambda_P}^{\lambda_Q}\text{d}\lambda = \tan k \cdot \int_{\varphi_P}^{\varphi_Q} \frac{1}{\cos \varphi}\ \text{d}\varphi $$

Integrating the left-side, and converting it from radians to degrees gives:

$$ \underbrace{\frac{180}{\pi}\left(\lambda_Q-\lambda_P\right)}_{\Delta L_{PQ}} = \tan k \cdot \frac{180}{\pi}\int_{\varphi_P}^{\varphi_Q}\frac{1}{\cos\varphi}\ \text{d}\varphi $$

Substituting $\Delta L_{PQ}$, and integrating the right side gives:

$$ \Delta L_{PQ} = \tan k \cdot \underbrace{ \frac{180}{\pi}\left( \ln \left( \frac{\tan\left(45^\circ + \frac{\varphi_Q}{2}\right)}{\tan\left(45^\circ + \frac{\varphi_P}{2} \right)} \right) \right) }_{\Delta VB} $$ Which we can now simply write as: $$ \tan k = \frac{\Delta L_{PQ}}{\Delta VB}. $$

Determine the distance for loxodrome

The distance is a lot simpler, but again starts with a small triangle.

Based on the diagram, we can find that: $$ \cos k = \frac{R\ \text{d}\varphi}{v_{\text{lox}}} \iff v_{\text{lox}} = \frac{R\ \text{d}\varphi}{\cos k}, $$ which gives us a simple formula for the distance. $$ v_{\text{lox}} = \frac{60\cdot\Delta\varphi}{\cos k} $$

Right angled spherical triangles

The illustration belows shows a right angled spherical triangle.

We can use the following identities to solve right angled spherical triangles. Another important theorem is that if $a \lt 90$, then $A \lt 90$, and this holds for the converse too! This also applies to $b$ and $c$. Each of the identities can be proved by using the fig 64. $$ \begin{align} \cos c &= \cos a \cos b \\ \sin A &= \frac{\sin a}{\sin c} \\ \sin B &= \frac{\sin b}{\sin c} \\ \cos A &= \frac{\tan b}{\tan c} \\ \cos B &= \frac{\tan a}{\tan c} \\ \tan A &= \frac{\tan a}{\sin b} \\ \tan B &= \frac{\tan b}{\sin a} \\ \cos A &= \sin B \cos c \\ \cos B &= \sin A \cos b \\ \cos c &= \cot A \cot B \end{align} $$

It is also important to know that the signs in $\cos c = \cos a \cos b$ must match. For this we can construct the following table.

$\sin a$	$\sin b$	$\sin c$
$a \gt 90$	$b \gt 90$	$c \lt 90$
$a \lt 90$	$b \gt 90$	$c \gt 90$
$a \gt 90$	$b \lt 90$	$c \gt 90$

Since $C = 90$, we don't know if it is acute or obtuse. For this we need to calculate $c$, and check if it is acute/obtuse.

Oblique angled spherical triangles

If the triangle is oblique, then we need to use some other rules to solve them. The first equation we can use is the first cosine rule. $$ \cos a = \cos b \cos c + \sin b \sin c \cos A $$ Which is also defined for the other angles: $$ \begin{align} \cos b &= \cos a \cos c + \sin a \sin c \cos B \\ \cos c &= \cos a \cos b + \sin a \sin b \cos C \end{align} $$ A second equation we can use is the sine rule: $$ \frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C} $$ It doesn't matter if the small letter is on top, or below, the identity still holds. However, if have $a, B, C$, then we can find $A$ with: $$ \cos A = -\cos B \cos C + \sin B \sin C \cos a $$ To make sure that we have the correct angle, we can use the following two rules to check:

1. Check the signs in the equation $\cos a = \cos b \cos c + \sin b \sin c \cos A$
2. $a \gt b - c$
3. $B + C - A \lt 180$

If we want to use these equation to find the GRC distance between two points, we use the cosine rule: $$ \cos c = \cos a \cos b + \sin a \sin b \cos C, $$ but we change it to use geographical coordinates instead: $$ \begin{align} \cos v &= \cos(90 - \varphi_A) \cos(90 - \varphi_B) + \sin(90 - \varphi_A) \sin(90 - \varphi_B)\cos\Delta L \\ \cos v &= \sin \varphi_A \sin \varphi_B + \cos\varphi_A \cos\varphi_B \cos\Delta L \end{align} $$

If for example, we want to find the GRC distance from Nagasaki ($32^\circ 34'N - 129^\circ 20'E$) to San Fransisco ($37^\circ 48'N - 122^\circ 20'W$).

First we find $\Delta L$, which is:

$$ \Delta L = |\lambda_B - \lambda_A| = \left|-\left(122 + \frac{20}{60} \right) - \left(129 + \frac{49}{60} \right)\right| = 252.15^\circ. $$

Because we want to find the shortest distance, and $\Delta L \gt 180^\circ$, we will use the other angle, thus $\Delta L = 360^\circ - 252.15^\circ = 107.85^\circ$.

Now we can calculate the distance between the two points, which is:

$$ v = \text{acos}\left( \sin 32.71 \sin 37.8 + \cos 32.71 \cos 37.8 \cos 107.85 \right) = 82.67^\circ $$

If we multiply the angle by $60$ nautical miles, we find that the distance is $82.67 \cdot 60 = 4960.5$ nautical miles.