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Distances on Globe and Flat Earth

Tuesday, May 23, 2017 - 21:27 | Author: wabis | Topics: FlatEarth, Geometry, Mathematic, Interactive | Comments(14)
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 enter Latitude/Longitude in degrees with decimal point (e.g. 54.5°) or degrees, minutes and seconds (e.g. 54° 30min). Latitudes on the Southern Hemisphere and Longitudes westerly of the 0-Meridian must be entered as negative values. See Input Format for Latitude and Longitude for some input examples.

Perth-Sydney Johannesburg-Perth London-Moskau

The distance between 2 points P1 and P2 is computed on a Globe width radius R and on Flat Earth with distance from north pole to equator EQ. If you want to use other planet sizes, enter your values into the corresponding fields, else leave them as is. You can reset the fields to their default values by clicking into the fields and pressing the ESC key.

You can choose the units you want the distances to be calculated as km, sm or mile. sm is nautical miles as used in aviatic, mile is statute miles.

Input Format for Latitude and Longitude

The 3 fields for Latitude and Longitude accept all common input formats. You can enter the latitude or longitude uning only the degrees field when entering any common format or enter it in degrees, minutes and seconds in the corresponding fields. You can set N,S,E or W before or after the number in the dregrees field or use negative numers for S and W.

All in degrees: 51.507351° | 120 58 57W | 120°58'57"W | W69 01 29.74 | 41 25N | N41 25.117 | 32 S
Enter the value into the Degree-Field and leave the Minute and Second field 0.
Degrees and Minutes: 51° 30.44106 min
Enter the whole degrees into the Degree-Field and the minutes into the Minute field and leave the Second field 0.
Degrees, Minutes and Seconds: 51° 30 min 26.4636 sec
Enter each part into the corresponding field.

If you have a negative Latitude or Longitude with the zero degrees then enter 0 for degrees and a negative minute: 0° −7 min 39.929 sec. If Degree and Minute are zero, enter the negative sign into the Second field: 0° 0 min −12.34 sec.

Use the button →DMS to convert from Degree-Format to Degree-Minute-Second-Format. Use the button →Deg to convert from Degree-Minute-Second-Format to Degree-Format. This is useful to copy the position to google earth as one value.

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!

Check it Yourself

Google Maps gives a Distance from Perph to Sydney along a road of 3934 km. You can drive this route with your car and check the distance with your car's odometer. The distance is much shorter than what the Flat Earth model predicts.

Route Perth Sydney on Google Maps

To check wether the latitude and longitudes are correct on this page, you can buy a cheap Sextant e.g. on Amazon and measure the coordinates yourself using Celestial navigation technique.

Used Formulas for Globe Earth

Note: To prove that the following formulas are used in the calculations, you can insepct the Source Code of this Page.

Illustration of great-circle distance

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

Illistration Flat Earth Distance

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 }

Comments

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)
https://scholar.google.com/citations?user=YRLOHxoAAAAJ&hl=en

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

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.

3Ray Brown 11/14/2017 | 01:32

Could you provide some examples of how the fields should be filled out? Also I’m trying to find out if 3 suffeiciently spaced towers are placed on the same latitude will the middle tower be out of line when viewing the horizon outer towers lined up on a globe?Will your flat/globe simulator show this?

4wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 11/14/2017 | 23:13

Ray, see the new paragraph Input Format for Latitude and Longitude.

The projection of the line of sight onto the globe surface is always a so called Great Circle Arc. That is the shortest connection of 2 points along the surface of a globe. The Latitude lines are not Great Circles except at the equator. Following a certain Latitude line from one point to another on the same Latitude is not the shortest path between these points. You can immediately see that for Latitudes very near a pole. You would move almost on a circle from one point to the other.

So if you have 3 sufficiently spaced big enough towers placed on the same latitude and look from the first to the last tower, the middle tower would not lie between these towers, exept on the equator.

5fred 11/18/2017 | 17:10

Hi
I clicked the Perth-Sydney blue button and in the LGlobe distance it has 3 290,578 721 - I cannot tell what this number is? It' can't be 3...million... it can't be 3290 point 578 etc...
Also, the Radius for both R and EQ don't change at all... any FE who see's this will instantly dismiss this for these reasons... I think its a shame for the hard work you have put into it all to fail on these trivial points, which FE's will latch onto and use as an examples of Globe retardness, as they would call it!

hope this helps
Fred K

6wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 11/18/2017 | 20:31

fred, How can I make this easier? I have improved and simplified the description. Hope this helps.

Globe = 3290.578721 km is the distance between Perth and Sydney in km on the globe, as indicated. Check this with google earth, it matches the real world. The value labeled Flat Earth is the corresponding distance computed for the flat earth, assuming a flat earth size where the distance from the north pole to the equator is the value in the field EQ. If you don't agree with this value, enter your value and the distances are computed using your values.

R is, as described below the form, the radius of the globe used to compute the distance on the globe. Of course R and EQ do not change, because the size of the earth does not change. This fields are only for information. If you want to compute the distances using more accurate values for the radius of the earth or for the size of the flat earth, you can enter your values into this fields. That's all.

I am aware that there are different formats in different countries for numbers. My website is international, so I choose the recommendet format by ISO, which recommend spaces for grouping big numbers. Dots are not recommendet because dots are used in some countries as decimal points and this can be confusing.

7Steven Elliott 12/17/2017 | 17:05

Thanks for making this page. It's really handy for making the case that the flat Earth map is terribly distorted in the Southern hemisphere. Here are my thoughts on how some enhancements you could make:

  • When on land include the actual driving distance, possibly only for the Perth-Sydney button, that you can get from Google maps or otherwise. The reason being is is that it suggests that anyone can verify the distance simply by looking at their odometer which might persuade conspiracy minded people who think that surveying is suspect. For example, in the case of Perth-Sydney Google says the drive is 3934 km, which is a bit more than the globe Earth distance of 3290 km, which makes sense, but much less than the flat Earth distance of 8301 km, which is impossible.
  • Replace "EQ" with "Pole to Equator". A bit longer, but it makes it clear.
  • Include a brief explanation that anyone can independently verify longitude and latitude with a sextant, which has been used by navigators for hundreds of years, so it can't be that the longitude and latitude for cities is hugely incorrect. Also, it's possible to buy a plastic sextant on Amazon for $45, so it's not that sextants are exotic and inaccessible equipment.
  • Make it clear that the code is open source. This both addresses the objection that you have some profit motive, and that the code is doing something sneaky.
  • Parse "NSEW" suffixes and interpret as positive or negative. I think you could do this with a regex, or stepping through the string backwards without too much trouble. I could maybe take a shot at it.
  • I have to agree with fred that the number format is confusing. I know that you're trying to make your site as international as possible, but I still think it's confusing to some readers. So for the example fred gave "3 290,578 721" I'd recommend "3290.578". Three decimal places should be plenty since that gets it down to a meter. No need to break up big numbers. A number in such form should be ready to enter into a spreadsheet, software or calculator.

Thanks again for making a page that promotes science and reason.

8wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 12/18/2017 | 02:40

Steven, thank you for the suggestions, I have incorporated them all. I even decided to change the number format on all english pages to US format instead of ISO, because the ISO format seems not to become common soon in english regions.

9Steven Elliott 12/18/2017 | 15:56

Wow, you did that quickly. And it looks great.

I'll see if I can get you some more hits by referencing this page in comments on YouTube.

Thanks again.

10David Cunningham 1/8/2018 | 03:44

Pick a long straight road, something you can drive down on cruise control with one of the metre wheel contraptions to verify and then input the start and end coordinates.
For myself i chose highway 10 in saudi arabia since the full road is 162 miles and the vast majority is straight (although i chose a point several miles in to start due to a couple of bends).
Starting at 24.137401,49.063824
Ending at 24.137528,51.312851
The calculated globe distance is 141.804772 miles.
Flat earth calculated distance is 178.615313 miles.
The distance had to be less than 162 miles since i wasnt doing the complete highway due to the bend and what do you know the globe distance is less.
Just doing small straight roads in should come out with durther mistakes in flat earth model especially south of the equator.
Well done for making this tool

11Falcon Crest 5/22/2018 | 02:05

Great Site. I've just discovered it today and it's worthwhile taken both assumptions on how best to translate Global Coords into a feasible Flat Earth Model,

1) by keeping the official coords at the expense of a sizable distorted land
2) by adjusting the coords and keeping the official mass of land

I have done both using the Law of Cosines, and assuming the Earth Radius as the one from North Pole to the Equator Line, giving 111.32 km for each Degree of Latitude wherever I needed to locate a certain place of interest.

As I know, the official coords are fully wrong along official distances as they both fail the summation of angles within Triangles equaling 180º when calculating 3 different places at once. It´s quite visible whenever there is Sea in the way.

A new Flat Earth map model needs to be developed sidelining the Gleeson's and Hammond's maps with their distorted masses of land for the Southern Hemisphere.

All in all, leaving the Maths apart with all those confusing Greek letters, good job.

12Ray Brown 5/26/2018 | 21:49

What accounts for the difference in distance of a line from 0.0 Longitude along the Equator to 180.0 longitude?

13wabiswalter@bislins.ch (Walter Bislin, Author of this Page) 5/27/2018 | 15:32

Ray, distances on the Gleason Map that go over the north pole are always correct. That is the purpose of this map in the first place. Because the shortest distance between 0 , 0 and 0 , 180 goes over the north pole on the FE map, 20,015 km is the same for Globe and FE.

If you measure the distance between 0 , 0 and 0 , 180 on a path along the equator on the FE map, you get another answer:

(16)
d = EQ \cdot \pi = 10{,}008\ \mathrm{km} \cdot \pi = 31{,}440\ \mathrm{km}
where'
d ' =' 'distance from 0 deg longitude to 180 deg longitude along the equator on FE
EQ ' =' 'the distance from the north pole to the equator

On the Globe the distance will always be the same 10,008 km no matter which path you take, because it will always be half around the Globe on a great circle.

14Martin 6/23/2018 | 16:32

Sorry to join the party so late but I found formula to calculate shortest distance on Flat Earth between two points with known longitude and latitude without a need to convert them into Cartesian coordinates.

alfa=abs(long_1 - long_2)/2.
L=sqrt((cos(alfa)*EQ*(lat_1 - lat_2)/90)^2 + (sin(alfa)*EQ*(2 - (lat_1 + lat_2)/90))^2)

where longitude and latitude are in degrees and follow same convention as described on this page.

Note: it give identical result as equation on this page.

This was done by calculating diagonal of equilateral trapeze created by bases of two equilateral triangles based on a two circle with common center (North Pole) each containing one point of given coordinates.

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