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.
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.
PerthSydney JohannesburgPerth LondonMoskau
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.
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 socalled Greatcircle 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 
FlightTime SYD/PER T_{SP}  5 h 05 min = 5,08 h, from expedia.com for Qantas A330200 
FlightTime PER/SYD T_{PS}  4 h 10 min = 4,17 h, from expedia.com for Qantas A330200 
Ground Speed SYD/PER V_{SP}  427 kts = 791 km/h, from flightradar24.com for A330200 
Ground Speed PER/SYD V_{PS}  535 kts = 991 km/h, from flightradar24.com for A330200 
TaxiTime T_{Taxi}  20 min = 0,33 h, from www.rita.dot.gov, United States Department of Transportation 
HoldTime T_{Hold}  10 min = 0,17 h, reserve time for holding patterns and vectors 
Taxi+Hold T_{TH}  30 min = 0,5 h 
Climb/DescentTime T_{CD}  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 FlightTimes 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) 
 
where^{'} 

Lets compute the distance from Sydney to Perth:
(2) 

Now lets compute the distance from Perth to Sydney:
(3) 

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.
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!
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^{'} 

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^{'} 

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^{'} 

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^{'} 

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 r_{i} of the vector i from the north pole to a point P_{i}(φ,λ) is:
(12) 
r_i = \left( 1  { \varphi_i \over \pi / 2 } \right) \cdot E
 
where^{'} 

Having the length of the vector together with its longitude the cartesian coordinates of P_{i} can be computed:
(13) 
x_i = r_i \cdot \cos( \lambda_i )
 
(14) 
y_i = r_i \cdot \sin( \lambda_i )
 
where^{'} 

The distance between the points P_{1} and P_{2} is:
(15) 
L = \sqrt{ (x_2  x_1)^2 + (y_2  y_1)^2 }

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)
https://scholar.google.com/
I got exactly the same answers with my spreadsheet as with your online calculator. I did it for San Jose CA to New York City NY, Perth Australia to Sydney Australia, and for San Francisco (SFO) to Dubai (DXB). Also I had already done a similar analysis as you did for Perth to Sydney, although I did not go into great detail on the flight time calculation like you did. I just cited what Google maps gave as a flight time. I had posted this on webpilot71 Youtube channel. Here it is. So I think a key is to calculate distances between points on both the spherical earth map and the flat earth map and compare them as well as against actual measured distances on the earth. So I took Perth Australia and Sydney Australia. For spherical earth map shortest distance (great circle distance on spherical surface) is about 2050 miles. For flat earth map shortest distance (straight line between two points) is about 5160 miles. Google maps shows driving distance of about 2440 miles. Now all we need it to get someone in Australia to drive from city to city and see what distance they get with their car's odometer. The google map also shows a flying time between the two cities of 5 h 5 min. This means that if the flat earth distance is correct the plane would need to fly around 1000 mph. For flat earth map I am assuming Azimuthal Equidistant map with distance from north pole to equator of about 6225 miles, the same distance from north pole to equator on the the spherical earth map (i.e. great circle distance). At least from what I am finding out so far, the cutting edge flat earth researchers are not yet willing to commit to a flat earth map, which makes it impossible to do comparisons between spherical and flat models, obviously.