# Distances on Globe and Flat Earth

Tuesday, May 23, 2017 - 21:27 | Author: wabis | Topics: FlatEarth, Calculator, Geometry, Mathematics
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.

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.

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

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?

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

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.

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.

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?

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)

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.

15Ronald 9/6/2018 | 14:43

Your calculation for the AE map is not correct. Just like the globe you can not just put a ruler on it and measure. The AE projection is exactly like the globe, north south is nautical miles, east west is hour arc-minutes. You have to use degrees to calculate any east west distance. They both use 360 degrees along the equator where each degree is 60 nautical miles. Remember the globe was not made to be used as a map and is not one. It was made an indexing system for navigation. The only place the globe is proportional to size is along the equator.

Regards Ronald

Roland, you do confuse real shapes with maps. The real model/map of the earth is spherical. All flat maps of the globe are distorted projections, as we know. We only have flat map projections because globes are not so hany. We know that distances are only approximately correct on maps representing small portions of the globe.

Navigation systems on the other hand don't use 2D maps/representations of the earth but spherical 3D systems and spherical trigonometry. Each location given with with latitude/longitude coordinates is internally a 3D point on a sphere.

A map for a flat earth however is always flat without distortions and navigation systems could use plane trigonometry. Each location is a 2D point on a plane.

Flat Earther claim the earth is really a flat disc, not only a projection/map of the globe to a flat disk. The globe model is not a projection of the earth, it is the right 3D representation of the earth, because the earth is a globe.

Roland: Just like the globe you can not just put a ruler on it and measure.

Wrong. You can measure distances on the globe model using a string and a ruler and they correspond to scaled real distances on earth. Take the string and span it between two locations A and B (both given with latitude/longitude) along the surface of the globe so it follows a great circle and then measure the length of the string with a ruler. After scaling up the measured length corresponds to the real distance on the globe earth, anywhere.

We have to apply the same rules on the flat earth: Take the string and span it between the two locations A and B (given with the same latitude/longitude) on the flat earth and then measure the length of the string with a ruler. After scaling up the measured length would corresponds to the real distance on the flat earth, anywhere.

This is exactly what I have implemented in my App. Only the model that gives the right distances, veryfied on reality, is the right model. And that is the globe model.

Roland: Remember the globe was not made to be used as a map and is not one. It was made an indexing system for navigation.

The globe model was not made as a map or indexing system. The globe model is a geometrically accurate representation of the spherical earth as it is measured by geodetic survey. So the natural way to express positions is to use spherical coordinates, latitude and longitude and use spherical trigonometry to compute tracks and distances between points.

The AE map is a plane projection of the globe. Flat earth is not a projection, but is claimed to be a real physical thing. Flat earther state unmistakable the earth is flat, not is a projection of the globe. That's a huge difference. Real distances are measured on the real thing or on a scaled model with the same shape as the real thing. You have to use the corresponding model of the real thing to measure or compute tracks and distances.

You must not use the globe model and spherical trigonometry to compute distances for a real flat earth. You must use the flat earth model and plane trigonometry to compute distances for the flat earth. Likewise you must not use a flat earth model and plane trigonometry to compute distances for the globe earth.

Think about: Although flat earth uses latitude and longitude as a coordinate system like the globe, that does not imply that they represent the same geometrical points. One system represents points on a 3D sphere the other points on a 2D plane - big difference!

Other difference: On a flat earth each line with constant latitude is a circle around the north pole. This includes latitude -90° latitude, which is the border of the flat earth. On a globe the line with latitude 0° (equator) is the straightest possible path around the globe along a great circle. Lines of latitude greater than 0° curve around the north pole on the surface of the sphere. Lines of latitude less than 0° curve around the south pole. Latitude -90° is a single point, the south pole.

Do you claim, that the border of the flat earth with latitude -90° is a single point as the globe coordinate system tells?

17Hegel 10/27/2018 | 21:22

Would you be so kind as to calculate the distance between Santiago Chile and Wellington New Zealand on both the Flat Earth and the Ball Earth. In addition, would be possible to calculate the distance into Nautical miles? I would be extremely thankful for your effort. It is truly important. Many thanks!!

18Nicolas 1/13/2019 | 02:49

It seems to be the absolute proof of spherical Earth, although it doesn't show if the Earth is convex or concave. One thing left: it is very problematic to check this proof practically, even if you are rich. Although I don't like conspiracy theories. And if to make the prooving task more complicated and to take off tight accordance of selestial coordinates and earth coordinates, and if to have only information about distances by time of flights between separate points, if goes about the most part of southern hemisphere. For fun. Else one crazy idea, but maybe not so crazy, how do you think?

### Hollow Earth or Globe

Nicolas: It seems to be the absolute proof of spherical Earth, although it doesn't show if the Earth is convex or concave. One thing left: it is very problematic to check this proof practically, even if you are rich.

The geometry, that is the location and distances and their relationships work for both, the globe and a hollow earth. But the physics is not the same and tells without any doubt that we live on a rotating globe. Devices like navigation systems depend on this physical facts.

Let me explain: On a rotating globe there must be a centrifugal acceleration perpendicular to the rotation axes away from the axes. The magnitude and direction wrt. the surface would depend on the latitude of the observer. It has to be added vectorial with the raw gravitational acceleration according Newtons laws to get the effective gravitational acceleration we feel and measure. If the globe or the hollow earth would not rotate, then there would not be such centrifugal accelerations. If the earth would be a rotating hollow earth, then the centrifugal accelerations would point towards the ground in an angle depending on the latitude of the observer.

Can we measure such a change in the gravitational acceleration due to the rotation of the earth?

Lets calculate what we should expect on a rotating globe. The centrifugal acceleration (c) is dependent on the rotation period and the distance from the rotation axes, which is dependent on the latitude:

(17)
c = \omega^2 \cdot r = { \left( 2 \cdot \pi \over T \right) }^2 \cdot R \cdot \cos( \varphi )
where'
 c ' =' 'centrifugal acceleration 90° outward of the rotation axes of the earth \omega ' =' '2 \cdot \pi / T = angular velocity T ' =' '24 h = rotationa period of the earth r ' =' 'distance of the observer from the rotation axes R ' =' '6371 km = radius of the earth \varphi ' =' 'latitude of the observer

Because \cos(90°) = 0 at the poles the centrifugal acceleration is zero. It is maximal at the equator. Lets calculate the centirfugal acceleration at the equator:

 (18) c_\mathrm{eq} = { \left( 2 \cdot \pi \over 24 \cdot 60 \cdot 60\ \mathrm{s} \right) }^2 \cdot 6{,}371{,}000\ \mathrm{m} = 0{.}0337\ \mathrm{m}/\mathrm{s}^{2}

The raw gravitational acceleration according to Newton depends on the latitude too, because the earth is not perfectly spherical but approximately an ellipsoid so that the distance from the mass center of the earth varies with latitude, but lets take the mean gravitational acceleration for comparison: 9.81 m/s2. This means that the centrifugal acceleration is about 0.34% of the raw gravitational acceleration. You should weigh 0.34% less on the equator than on the poles due to the centrifugal acceleration.

Can this be measured? The answer is yes, even with a simple cheap precision scale! And what we measure is exactly what is predicted for the rotating globe earth. Precision scales have to be calibrated for each latitude with a known mass to compensate for the centrifugal acceleration at that latitude for this reason.

Wolfie6020 did such experiments to measure the centrifugal acceleration at different locations of the earth and in an aircraft flying in different directions, which causes changes in the centrifugal acceleration: more when flying east in the same direction as the earth rotates, less when fly west against the rotation of the earth, called the Eötvös effect:

The Eötvös effect observed in aircraft - how does it affect Gravity?

I have calculated what influences other than the rotation of the earth there could be: latitude, altitude, geoid variations, buoyancy, Coriolis effect, tides of moon and sun: see Centrifugal and Gravitational Acceleration in an Aircraft

Recap: Although the geometry would work the same way for a hollow earth and a globe earth, if we ignore the third dimension, pyhsics is different and thells us we live on a rotating globe. If we include the third dimension (up) in exact measurements of the earth, like geodetic surveyors do, we find clearly that the earth is a globe and can not be concave. Light does not bend in the atmosphere in such a way that the measurements could be all wrong. We know the physics of refraction very well and can measure it.

We can measure the centrifugal acceleration with simple cheap experiments as Wolfie showed. To find out whether there could be other physical effects that produce similar effects, read the linked page. Each possible effect is calculated and compared in magnitude with the centrifugal acceleration. And the finding is, that the centrifugal acceleration is by an order of magnitude the strongest effect of all possible other forces beside gravity.

As a sidenote, the ellipsoidal shape of the earth is such that the combination of the raw gravitational acceleration due to the mass of the earth and the centrifugal acceleration acts always perpendicular to the surface of the ellipsoid and not towards the center of the earth. Why is that the case? It's because the earth is not solid. If the effective gravitational acceleration would not be perpendicular to the surface, it would result in tangential force components that causes streams that shape the earth as long undtil there are no more tangential forces. This is the case, when the earth reaches a shape, which has the same equipotential at each point on the surface. This shape is an ellipsoid of rotation. It's the shape the water of the oceans forms. It's the shape of all rotating planets and suns.

So mean sea level has the shape of approximately an ellispoid. There are minor deviations due to some local inhomogeneities in the mass distribution, eg. due to mountains or different rock types in the crust. This deviations are accounted for in the so called Geoid and the GPS heights, which are calculated from the reference ellsipsoid, are correceted accordingly to get the real altitude wrt. mean sea level at each location of the earth.

20Nicolas 1/16/2019 | 22:29

Absolutely, if selestial sphere above us is real, and selestial bodies on it are the real physical bodies with definite places, that we see, with no doubt the Earth is a globe, of course. Then the meanings of atmospherical refraction are true, geographic coordinates are true, and many other things. If unreal, everything becomes very uncertain.

21Shaun 1/17/2019 | 01:55

Great information on this Article. There are a few sites that allows you to do distance calculations, like http://www.distancecalculator.co.za or http://www.DirectionsTo.com.au

22Nicolas 1/17/2019 | 20:46

Or else one model of Flat Earth:
Among different tryings to show spherical Earth terrains on a plain surface there is stereographic azimuthal projection. This is one of most ancient methods of mapping, and of mapping of whole known world, or Earth surface. Maybe most ancient one. The difference between usual azimuthal projection world map, that flatearthers believe the real world map, and stereographic azimuthal projection world map, that meridional distances are increased proportionally like distances along parallels. So this projection has no shape and angle distortions on short distances but only distortions of distances.
So my fantastic hypothesis claims, that this is the real world map of Flat Earth, but with some especial feature: this proportion between distances on globe and our map acts not only towards distances, but to all physical objects and velocities. This way everything, that moves along meridians, changes it's size, even light wavelengths and sizes of atoms. The question is: Is there any difference between observed geometry of that world and spherical Earth?

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