# Distances on Globe and Flat Earth

Tuesday, May 23, 2017 - 21:27 | Author: wabis | Topics: FlatEarth, Geometry, Mathematic, Interactive
Flat Earth and Globe Earth are two models for the earth that are very different in geometry. In order to decide which model matches reality, we can calculate the shortest distance between two points for both models and check which distance corresponds to reality.

I derive the formulas to compute the distance between two points for both models. The computations can be carried out using a calculation form. An example is then used to decide which model corresponds to reality.

## Calculation Form

You can specify latitude/longitude in degrees with decimal point (e.g. 54,5° or minutes and seconds (e.g. 54° 30min). Use →DMS to covert to format dd°mm'ss" and →Deg to convert back to Degrees.

The distance L between points P1 and P2 is computed on a Globe width radius R and on Flat Earth with distance from north pole to equator EQ.

sm is nautical miles as used in aviatic, mile is statute miles.

## Which Model fits Reality?

Flat Earth and Globe Earth are two models for the earth that are very different in geometry. Lets assume we don't know yet the right shape. On both models exists a way to define the shortest distance between two points. A point is defined on both models by a degree of latitude and longitude. The point with the same latitude and longitude correspond on both models to the same location, e.g. a certain city. We can derive for both models a formula to compute the distances. So we can make a corresponding prediction for both models. In order to decide which model is the correct one, we can calculate the shortest distance between the same two points for both models and check which prediction matches reality. If we find two points that give very different predictions for the two models, it is easy to figure out the right model.

On the globe earth the shortest distance is the so-called Great-circle distance, on the flat earth this is a straight line.

The real distance has to be gained by experiment, that means measuring the real world, not by distance indications on maps, since from this distance the correct model should be deduced. We can get the real distance between two locations by measuring the flight time of a direct flight between two locations and multiplying it by the average flight speed. Time and speed of a commercial airplane over a sufficient great distance give a good estimate for the real distance.

In order to obtain a clear result, we use distances which are as different as possible in both models for the same two locations. These are mainly distances in the southern hemisphere.

I choose the distance between Perth and Sydney Australia. Because they lie on the same continent this distance could also be obtained by driving with a car between this cities. The following dataset is used for both models:

Dataset
Perth (PER) Latitide = -31.950527, Longitude = 115.860457
Sydney (SYD) Latitude = -33.868820, Longitude = 151.209296
Flight-Time SYD/PER TSP 5 h 05 min = 5,08 h, from expedia.com for Qantas A330-200
Flight-Time PER/SYD TPS 4 h 10 min = 4,17 h, from expedia.com for Qantas A330-200
Ground Speed SYD/PER VSP 427 kts = 791 km/h, from flightradar24.com for A330-200
Ground Speed PER/SYD VPS 535 kts = 991 km/h, from flightradar24.com for A330-200
Taxi-Time TTaxi 20 min = 0,33 h, from www.rita.dot.gov, United States Department of Transportation
Hold-Time THold 10 min = 0,17 h, reserve time for holding patterns and vectors
Taxi+Hold TTH 30 min = 0,5 h
Climb/Descent-Time TCD 46% · 1,5 h = 0,645 h, from www.cfidarren.com, Statistical Summary of Commercial Jet Airplane Accidents

See Umrechnen von Fluggeschwindigkeiten how speeds are converted.

The Taxi and Hold times must be subtracted from the Flight-Times to get the time relevant for the distance traveled. During the Climb and Descent time the speed is increasing/decreasing. So I take the mean speed for this period. That is half of the cruise speed. The climb/descent time is for all flights the same and I found a value of 45% · 1,5 h = 0,645 h here which matches my experience with flight simulator times.

The formula to compute the distance traveled is then:

(1)
 L = T_\mathrm{CD} \cdot 0{,}5 \cdot V + (T - T_\mathrm{TH} - T_\mathrm{CD}) \cdot V
where'
 L ' =' ' Distance traveled T ' =' ' Total Trip time T_\mathrm{CD} ' =' ' Climb + Descent time T_\mathrm{TH} ' =' ' Taxi + Hold time V ' =' ' Cruise ground speed

Lets compute the distance from Sydney to Perth:

(2)
 L_\mathrm{SP} = 0{,}645\ \mathrm{h} \cdot 0{,}5 \cdot 791\ \mathrm{km}/\mathrm{h} + ( 5{,}08\ \mathrm{h} - 0{,}5\ \mathrm{h} - 0{,}645\ \mathrm{h} ) \cdot 791\ \mathrm{km}/\mathrm{h} = 3368\ \mathrm{km}

Now lets compute the distance from Perth to Sydney:

(3)
 L_\mathrm{PS} = 0{,}645\ \mathrm{h} \cdot 0{,}5 \cdot 991\ \mathrm{km}/\mathrm{h} + ( 4{,}17\ \mathrm{h} - 0{,}5\ \mathrm{h} - 0{,}645\ \mathrm{h} ) \cdot 991\ \mathrm{km}/\mathrm{h} = 3317\ \mathrm{km}

The difference is about 1,5%. The real distance is some km shorter because the airports are not aligned with the flight track so the airplane diverges after start and before landing some km from the direct route.

### Result

The shortest distances computed between Perth and Sydney for the two models, according to the Calculation Form, are:

Globe Earth Model Flat Earth Model Flight Distance SYD/PER Flight Distance PER/SYD
3291 km 8301 km 3368 km (2) 3317 km (3)

The result matches without any doubt the Globe Earth Model.
The Flat Earth Model is wrong by a factor of 2,5!

## Used Formulas for Globe Earth

On a Globe the shortest distance between 2 points is a great circle.

The points are given in polar coordinates latitude \varphi and longitude \lambda:

(4)
P_1 = ( \varphi_{1,deg}, \lambda_{1,deg} )
(5)
P_2 = ( \varphi_{2,deg}, \lambda_{2,deg} )
where'
 \varphi_{i,deg} ' =' 'Latitude in degrees Nord. Negativ values for South. \lambda_{i,deg} ' =' 'Longitude in degrees East. Negativ values for West.

Converting into cartesian vector format using the center of the sphere as (0,0,0):

(6)
\vec P_1 = R \cdot \hat v_1
(7)
\vec P_2 = R \cdot \hat v_2
width
| \hat v_1 | = | \hat v_2 | = 1
where'
 \hat v_i ' =' 'Unit vector from center of the sphere to point P_i | \hat v_i | ' =' 'Lengt of vector \hat v_i R ' =' 'Radius of the sphere

The unit vectors to the 2 points are:

(8)
\hat v_1 = \left[ \matrix{ v_{ 1,x } \\ v_{ 1,y } \\ v_{ 1,z } } \right] = \left[ \matrix{ \cos\left( \lambda_1 \right) \cdot \cos\left( \varphi_1 \right) \\ \sin\left( \lambda_1 \right) \cdot \cos\left( \varphi_1 \right) \\ \sin\left( \varphi_1 \right) } \right]
(9)
\hat v_2 = \left[ \matrix{ v_{ 2,x } \\ v_{ 2,y } \\ v_{ 2,z } } \right] = \left[ \matrix{ \cos\left( \lambda_2 \right) \cdot \cos\left( \varphi_2 \right) \\ \sin\left( \lambda_2 \right) \cdot \cos\left( \varphi_2 \right) \\ \sin\left( \varphi_2 \right) } \right]
where'
 \hat v_i ' =' 'Unit vector from center of the sphere to point P_i \varphi_i ' =' '\mathrm{rad}(\varphi_{i,deg}) = Latitude in radian. Negativ values for South. \lambda_i ' =' '\mathrm{rad}(\lambda_{i,deg}) = Longitude in radian East. Negativ values for West.

The cosine of the angle \alpha between the vectors is:

 (10) \cos\left( \alpha \right) = \hat v_1 \cdot \hat v_2 = \left[ \matrix{ v_{ 1,x } \\ v_{ 1,y } \\ v_{ 1,z } } \right] \cdot \left[ \matrix{ v_{ 2,x } \\ v_{ 2,y } \\ v_{ 2,z } } \right] = v_{ 1,x } \cdot v_{ 2,x } + v_{ 1,y } \cdot v_{ 2,y } + v_{ 1,z } \cdot v_{ 2,z }

And the great circle distance L between the Points is:

(11)
L = R \cdot \alpha = R \cdot \arccos \, \left( v_{ 1,x } \cdot v_{ 2,x } + v_{ 1,y } \cdot v_{ 2,y } + v_{ 1,z } \cdot v_{ 2,z } \right)
where'
 L ' =' 'Great circle distance between P_1 and P_2 R ' =' 'Radius of the sphere (earth) v_{1,x} ' =' 'X component of the unit vector \hat v_1 from center to P_1 v_{2,x} ' =' 'X component of the unit vector \hat v_2 from center to P_2

## Used Formulas for Flat Earth

I first convert the points polar coordinates to cartesian coordinates. Then the points are 2 vectors in cartesian coordinates. Subtracting one vector from the other and computing the lengt of the resulting vector gives the distance between the 2 points on a flat earth, which is a straight line.

The length ri of the vector i from the north pole to a point Pi(φ,λ) is:

(12)
r_i = \left( 1 - { \varphi_i \over \pi / 2 } \right) \cdot E
where'
 r_i ' =' 'length of vector from north pole to point Pi \varphi_i ' =' '\mathrm{rad}(\varphi_{i,deg}) = Latitude in radian. Negativ values for South. E ' =' 'Distance from north pole to the equator of the flat earth

Having the length of the vector together with its longitude the cartesian coordinates of Pi can be computed:

(13)
x_i = r_i \cdot \cos( \lambda_i )
(14)
y_i = r_i \cdot \sin( \lambda_i )
where'
 x_i, y_i ' =' 'cartesian coordinates of point Pi \lambda_i ' =' '\mathrm{rad}(\lambda_{i,deg}) = Longitude in radian East. Negativ values for West.

The distance between the points P1 and P2 is:

 (15) L = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }

1T Mark Hightower 9/6/2017 | 04:37

Got referred here by Wolfie6020
I just recently derived equations for flat and sphere and programmed them into Excel.
I look forward to checking my results with your online calculator. This is great. Thank you.
Mark
T Mark Hightower
San Jose, CA
NASA Ames Research Center Engineer (retired)

2T Mark Hightower 9/6/2017 | 10:00

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