# 3-D Earth Geometry

An interesting topic in 3-dimensional geometry is **Earth geometry**. The Earth is very close to a sphere (ball) shape, with an average radius of `6371\ "km"`. (It's actually a bit flat at the poles, but only by a small amount).

Earth geometry is a special case of **spherical geometry**. When we measure distances that a boat or aircraft travels between any 2 places on the Earth, we do not use straight line distances, since we need to go around the curve of the Earth from one place to another. (Think about the **direct** or straight-line distance between London and Sydney, through the Earth. That's going to be a lot less than the distance a plane flies around the surface of the Earth.)

Let's start with an example. What distance does a plane fly between Beijing, China and Perth, Western Australia?

To figure this out, we need to understand **latitude** and **longitude **first.

## Latitude and Longitude

First, we represent the Earth by a sphere:

We slice the Earth through the equator and remove the top. We show the Earth's axis which (by convention) points to North at the top.

## Parallels of Latitude

We want to be able to indicate how far North or South of the equator we are for any point on the Earth's surface.

For example, we wish to connect all the points that are 30^{o} N of the equator. We do that by drawing a line from the centre of the Earth to the surface of the Earth, 30^{o} up from the horizontal. We then draw a circle around the Earth parallel to the equator through the point where the 30^{o} line meets the surface of the Earth. We can do the same thing for below the equator, in a southerly direction.

The angle from the horizontal through the equator determines the name given to each parallel of latitude. Five parallels of latitude are shown in the diagram; 60^{o} N, 30^{o} N, 0^{o} N (the equator), 30^{o} S and 60^{o} S.

Each 1^{o} of latitude on the Earth represents **60 nautical miles**, or around 110.9 km.

Of course, 90^{o} N represents the North pole and 90^{o} S is the south pole.

Parallels of latitude give an indication of where a place (or a boat or a plane) are positioned in a **northerly** or **southerly** direction from 0^{o} N (the equator).

Continues below ⇩

## Lines of Longitude

What about the East and West directions for a point on the Earth?

We choose some point on the equator and draw a line from the centre of the Earth (point C) to the chosen point on the equator. We then draw a circle around the whole Earth so that it goes through the point on the equator and the North and South poles. (Historically, the position for 0^{o} was chosen so that the circle we have drawn goes through Greenwich in England).

The line we drew above is called a **line of longitude**. At the 'front' of our diagram, the line is called 0^{o} E and at the back, the line is called 180^{o} E, for reasons we'll see in a moment.

[The line of longitude that goes through 0^{o} has a special name: **the prime meridian.**]

Lines of longitude have the same radius as the radius of the Earth. They are examples of **great circles. ** (A great circle has the same radius and same centre as the earth itself).

Let's now define some more lines of longitude. We move around the Earth in an Easterly direction by 90^{o} and draw a great circle through the poles as before. The front line of longitude is labeled 90^{o} E and the one opposite to it at the back of our diagram is labeled 90^{o} W.

We could then draw the lines of longitude for every angle between 0^{o} E and 180^{o} E (covering Europe through India, Asia and the Pacific Ocean) and then through 0^{o} W and 180^{o} W (through the Atlantic Ocean, the Americas and the Eastern Pacific Ocean).

**Note: **0^{o} E = 0^{o} W = 0^{o} and 180^{o} E = 180^{o} W = 180^{o}.

Lines of longitude give an indication of where a place (or a boat or a plane) are positioned in an **easterly** or **westerly** direction from 0^{o} E.

For lines of longitude, we cannot say that 1^{o} represents any particular distance, because the physical distance changes between each line of longitude as we move from the equator to the poles. At the equator, 1^{o} represents around 60 nautical miles or 111 km. but at the poles, it represents 0 km.

## Finer Divisions of Longitude and Latitude

Since 1^{o} of latitude represents 60 nautical miles on the Earth's surface, we need finer divisions of 1^{o} to accurately determine places on the Earth's surface.

As we learned earlier in the Angles section, we can divide 1^{o} into smaller divisions called **minutes** (where 1^{o} = 60 minutes) or **seconds** (where 60 seconds = 1 minute).

For example, London is 51^{o}32' North of the equator. This means 51 degrees plus 32/60 of a degree North.

Historically, it was the Babylonians who divided units of time and angles into 60 equal parts.

### Examples of Latitudes and Longitudes for Cities

Any place on the Earth can be uniquely described by its latitude and longitude.

Some examples (moving from West to East):

Los Angeles: 34

^{o}3' N, 118^{o}14' WNew York: 40

^{o}45' N, 73^{o}59' WLondon: 51

^{o}32' N, 0^{o}5' W. (London is just slightly west of Greenwich.)New Delhi: 28

^{o}35' N, 77^{o}12' ESingapore: 1

^{o}21' N, 103^{o}49' E (Singapore is almost on the equator).Sydney: 33

^{o}52' S, 151^{o}12' E

## What is the Distance Between 2 Places on the Earth?

We need one more key bit of information to solve the problem posed before (distance from Beijing to Perth). What is the distance around the curve of the Earth along a great circle between any 2 points?

Let's call the latitude of the first place *φ*_{1} and the longitude of the first place *λ*_{1}. Similarly, the latitude of the second place is *φ*_{2} and the longitude of the second place is *λ*_{2}.

We need to know the **central angle** (the angle from the centre of the Earth to the 2 places of interest). We'll see how to do that in 2 ways:

- Step by step, using Pythagoras' Theorem and sine ratio
- More directly, using Haversine's formula

### a. Using Pythagoras' Theorem and sine ratio

Given the Earth's center is at (0,0,0), we can find the *x*-, *y*-, and *z*-coordinates of a point on its surface given the latitude *P* and longitude *Q*, as follows:

x=RcosPcosQ

y=RcosPsinQ

z=RsinP

With those coordinates for the 2 points on the Earth's surface, we can find the direct (straight line) distance between two points in 3-D space (which we found on the page, 3-Dimensional Space) using:

`"distance"\ AB = ` `sqrt ((x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2)`

With that distance, we can then find the central angle (i.e. the angle at the center between the 2 points on the Earth's surface), using trigonometry.

Let's use the above to find the distance along the great circle from Beijing to Perth.

For Beijing, the latitude is *P*_{1} = 39^{o}54' N and the longitude is *Q*_{1} = 116^{o}24' E.

For Perth, the latitude is *P*_{2} = 31^{o}57' S and the longitude is *Q*_{2} = 115^{o}52' E.

We need to express these angles in **radians** (if you are rusty, see Radians):

Beijing: *P*_{1} = 39^{o}54' N = 0.696386; *Q*_{1} = 116^{o}24' E = 2.031563.

Perth: *P*_{2} = 31^{o}57' S = −0.557633; *Q*_{2} = 115^{o}52' E = 2.022255.

[Note that the latitude for Perth is negative, since it is South of the equator.]

We now find the *x*-, *y*-, and *z*-coordinates for Beijing and Perth, given that the radius of the Earth is 6371 km:

#### Beijing

*x*_{1} = 6371 cos 0.6964 cos 2.0316 = −2173.2

*y*_{1} = 6371 cos 0.696386 sin 2.0316 = 4377.9

*z*_{1} = 6371 sin 0.6964 = 4086.7

#### Perth

*x*_{2} = 6371 cos −0.5576 cos 2.0223 = −2358.5

*y*_{2} = 6371 cos −0.5576 sin 2.0223 = 4864.3

*z*_{2} = 6371 sin −0.5576 = −3371.4

Now for the straight line distance between the 2 cities (directly, through the Earth), using the 3-D distance formula:

`sqrt ((-2358.5- {:-2173.2:})^2+ (4864.3-4377.9)^2+ (-3371.4-4086.7)^2)` `=7476.2` km

Next, we find the central angle.

In the case of Beijing to Perth, the central angle looks like the following, with O at the centre of the earth:

We draw the triangle joining Beijing, Perth and the Earth's center, and construct an altitude for our triangle, *OS*. Since the direct distance from Beijing to Perth is 7476 km, the distance from point *S* to Beijing is `7476/2=3738` km, as shown:

Earth's central angle for Beijing-Perth

Since `sin theta = 3738.11/6371` and the central angle is twice *θ*, we have:

Central angle = `2arcsin(3738.11/6371)` `= 2 xx 0.6270249` `= 1.254049`

(Radians, of course, and full calculator accuracy was used throughout, but not shown.)

Now, to find the **distance travelled**, we need to use our formula for **arc length** that we learned before (see Arc Length). If *r* is the radius of the great circle and θ is the angle subtended at the centre (in radians), the arc length *s* is given by:

s=rθ

Now, the average radius of the Earth is `6371\ "km"`, and the angle we just found is `1.254049` radians, so the flying distance from Beijing to Perth is given by:

s= 6371 × 1.254049 = 7989 km.

[Thanks to reader Paul Holland for the above suggested procedure.]

### b. Haversine's Formula

Haversine's formula for the central angle of a sphere is more direct than the procedure we used above. It is given by (angles in radians):

Central angle = `2\ arcsin sqrt( sin^2 ((varphi_2-varphi_1)/2) +cos\ varphi_1\ cos\ varphi_2\ sin^2 ((lambda_2-lambda_1)/2) )`

where

*φ*_{1} and *φ*_{2} are the latitudes of the first and second place respectively (we used *P*_{1} and *P*_{2} before); and

*λ*_{1} and *λ*_{2} are the longitude of the first and second place respectively (we used *Q*_{1} and *Q*_{2} before);

(See Haversine Formula for a derivation of the above formula.)

Using our results from above, we have:

Beijing: *φ*_{1} = 39^{o}54' N = 0.696386; *λ*_{1} = 116^{o}24' E = 2.031563.

Perth: *φ*_{2} = 31^{o}57' S = −0.557633; *λ*_{2} = 115^{o}52' E = 2.022255.

Applying the Haversine formula gives us:

Central angle =

`2 arcsin sqrt( sin^2 ((-1.254)/2) +cos (0.696) cos (-0.557) sin^2 ((-0.009)/2) )`

= 1.254049

(radians once again, and full calculator accuracy was used throughout, but not shown)

Once again, using the arc length formula

s=rθ,

we one again obtain:

s= 6371 × 1.254049 = 7989 km.

### Exercises

1. Find the shortest distance that an aircraft flies from Athens (38^{o} N, 24^{o} E) to Hong Kong (22^{o} N, 114^{o} E).

Answer

Expressing latitude and longitude in radians:

Athens: *P*_{1} = 38^{o} N = 0.6632251;

*Q*_{1} = 24^{o} E = 0.418879.

Hong Kong: *P*_{2} = 22^{o} N = 0.3839724;

*Q*_{2} = 114^{o} E = 1.989675

We now find the *x*-, *y*-, and *z*-coordinates for Athens and Hong Kong, given that the radius of the Earth is 6371 km:

#### Athens

*x*_{1} = 6371 cos 0.6632 cos 0.4189 = 4586.4

*y*_{1} = 6371 cos 0.6632 sin 0.4189 = 2042.1

*z*_{1} = 6371 sin 0.6632 = 3922.3

#### Hong Kong

*x*_{2} = 6371 cos 0.3840 cos 1.9897 = -2402.7

*y*_{2} = 6371 cos 0.3840 sin 1.9897 = 5396.3

*z*_{2} = 6371 sin 0.3840 = 2386.8

Now for the straight line distance between the 2 cities (directly, through the Earth), using the 3-D distance formula:

`sqrt ((-2402.7- 4586.4)^2+ (5396.3-2042.1)^2+ (2386.8-3922.3)^2)` `= 7902.9" km"`

Next, we find the central angle.

`7902.9/2=3851.5`,

and since `sin theta = 3851.5/6371` and the central angle is twice *θ*, we have:

Central angle = `2arcsin(3851.5/6371)` `= 2 xx 0.64918 ` `= 1.29836`

(Radians, of course, and full calculator accuracy was used throughout, but not shown.)

We use

s=rθ

giving

s= 6371 × 1.29836 = 8272 km.

[Thanks to reader Paul Holland for the above example.]

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2. Find the distance that an aircraft must fly from London to Los Angeles if it flies directly, using Haversine's Formula.

Answer

"If it flies directly" means that it will fly a great circle route.

Here's the route (London is near the bottom of the image, Los Angeles near the top):

Image from Google Earth.

We learned above that the latitude and longitude for London and Los Angeles are as follows:

London: φ_{1} = 51^{o} 32' N = 0.899426,

λ_{1} = 0^{o} 5' W = −0.001454

Los Angeles: φ_{2} = 34^{o} 3' N = 0.594285,

λ_{2} = 118^{o} 14' W = −2.063561

Applying the formula for the central angle gives:

Central angle =

`2 arcsin sqrt( sin^2 ((-0.305)/2) +cos (0.899) cos (0.594) sin^2 ((-2.062)/2) )`

= 1.374296 (full calculator accuracy was used throughout)

So the distance between London and Los Angeles is:

s=rθ= 6371 × 1.374296 = 8756 km

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