# 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, nautical miles (nmi) or statute miles (mi).

## 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 using 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 43% · 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 43% · 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'
 $L$ ' =' ' 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)

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.

## 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 whether 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.

## Submarine Cables

Some flat earther claim submarine cables somehow prove the earth is flat. But if you look closer, they prove the earth can not be flat at all, on the contrary:

On the website Submarine Cable Map you can see all cables and get Infos about it. For example the cable between Sydney and Perth is 4850 km long:

 RFS Q1 2019 Cable Length 4850 km Owners Superloop, SingTel, Google, Indosat Ooredoo, Australia’s Academic and Research Network (AARNET) URL http://www.subpartners.net/indigo

The cable follows not the shortest line between this cities (3290 km) because it has to follow the shore line. So its length is 4850 km. The shortest line on the flat earth map is 8300 km between Sydney and Perth. So even the shortest connection on the flat earth is 1.7 times longer than the cable that follows the shore line.

## 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 distance. There are multiple methods to derive equations for the great circle distance. I am using vector algebra, which for me is the easiest method:

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

(4)
(5)
where'
 $\varphi_i$ ' =' 'Latitude. Negativ values for South. $\lambda_i$ ' =' 'Longitude. Negativ values for West.

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

(6)
(7)
width
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)
(9)
where'
 $\hat v_i$ ' =' 'Unit vector from center of the sphere to point $P_i$ $\varphi_i$ ' =' 'Latitude. Negativ values for South. $\lambda_i$ ' =' 'Longitude. Negativ values for West.

According to the vector scalar product, the cosine of the angle $\alpha$ (in radians) between the vectors is:

 (10)

And the great circle distance $L$ between the Points is:

(11)
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 $P_2$

Expressed in a single equation we get:

(12)
where'
 $L$ ' =' 'Great circle distance between $P_1$ and $P_2$ $R$ ' =' 'Radius of the sphere (earth) $\varphi_i$ ' =' 'Latitude. Negativ values for South. $\lambda_i$ ' =' 'Longitude. Negativ values for West.

We don't change the great circle distance if we rotate the system the amount $-\lambda_1$. So we can set $\lambda_1 = 0$ and replace $\lambda_2$ by $\lambda_2 - \lambda_1$ to simplify (12) further to:

(13)

## 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:

(14)
where'
 $r_i$ ' =' 'length of vector from north pole to point Pi $\varphi_i$ ' =' 'Latitude in degrees. Negativ values for south of the equator. $E$ ' =' '10,008 km = Distance from north pole to the equator of the flat earth

#### Explanation

If the latitude is 0, then r sould be the distance of the equator from the north pole E. If latitude is positive, then r should decrease by the amount $(\varphi / 90°) \cdot E$. So if latitude is 90°, the amount of decrease is Δ r = (90° / 90°) · E = E, i.e. at the north pole r = E − Δ r = 0. Accordingly if the latitude is negative, the distance from the equator should increase. This gives the equation for the distance of a latitude from the north pole:

 (15)

where the latitude $\varphi$ is given in degrees. After factoring out E we get the formula (14).

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

(16)
(17)
where'
 $x_i, y_i$ ' =' 'cartesian coordinates of point Pi $\lambda_i$ ' =' 'Longitude. Negativ values for West.

The distance between the points P1 and P2 is according to Pythagoras:

 (18)

Expressed in a single equation we get:

(19)
where'
 $L$ ' =' 'Shortest distance between $P_1$ and $P_2$ (straight line) $\varphi_i$ ' =' 'Latitude. Negativ values for South. $\lambda_i$ ' =' 'Longitude. Negativ values for West.

If we set $\lambda_1 = 0$ and $\Delta\lambda = \lambda_2 - \lambda_1$ we can (19) further simplify to:

(20)
with

Note: this is nothing else than the Law of cosines for a triangle with sides r1, r2 and angle Δλ for the unknown side L.

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?

4wabis (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

6wabis (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.

8wabis (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?

13wabis (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:

(21)
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.

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

16wabis (Walter Bislin, Author of this Page) 9/6/2018 | 16:29

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?

19wabis (Walter Bislin, Author of this Page) 1/13/2019 | 20:36

### 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:

(22)
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:

 (23)

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?

23Max in Perth 2/21/2019 | 11:29

Having experienced the flight from Perth to Johannesburg (and back) and having cycled across Australia from west to east, I can say from personal experience that the spherical distances are obviously correct. :)

24Kathy Evans 2/21/2019 | 22:02

Wondering where you got that artificially inflated number for the FE distance between Sydney and Perth? Logically, the FE distance will be actually less than the distance driving it in a car on the map, since it would be a straight line, and the highway isn't exactly a straight line.

25wabis (Walter Bislin, Author of this Page) 2/22/2019 | 01:43

Kathy, that is the crux of this whole page! On the flat earth all distances east west on the southern hemisphere, whether straight or along a road, are way too long compared with reality. Straight lines on the flat earth map are shorter than roads on the flat earth map, sure. But they both are way too long compared with reality. That's proof that the earth can't be flat. The distances of the flat earth do not match reality. But the distances calculated on a globe match reality exactly. So the earth must be and is a Globe.

See Used Formulas for Flat Earth how a distance on the flat earth is calculated from latitide and longitude of 2 locations.

The distances on an AE map (flat earth map) are only correct in the north-south direction. That was the whole purpose to create an AE map. It is a special kind of projection of the Globe.

26Jason Barnes 2/23/2019 | 07:45

Great site, Walter. I only became aware of the FE a couple of months ago and was kind of intrigued by the idea, along the lines of, how could anyone think that. Your site addresses my biggest dilemma: distance in the real world vs globe model vs flat earth model. In the real world it takes about the same time to drive or fly across both the United States and Australia. From that I conclude they must be about the same width. On a globe model they measure about the same distance across, but on the flat earth model Australia is almost twice the width of United States. I could shrink the width of Australia on a flat earth to make it fit real earth (I can't expand United States), but if I do that I still have the problem of the distance between Johannesburg and Perth. I can't shrink that without making the Atlantic Ocean ridiculously out of proportion. Therefore I have to conclude that the globe model more closely represents reality. I have not yet found a flat earth model that addresses this problem. There are many flat maps that represent the globe world in different projections and they have distance problems, but we accept that and use them for the information they contain. If the flat earth model is a map representation then I could accept it with it's limitations. However, if it is a representation of reality then it seems to have serious problems. So, thanks for the time and effort you put into this site. Helped me alot.

27Globetardos Extremos 3/6/2019 | 07:30

(1) You cannot have a pressurized atmosphere without a container.
(2) Water can never conform to the exterior of an object.
(3) A body of water will always find its own level.
(4) Even considering refraction & atmospheric obstruction, we can see buildings & mountains too far away.

28wabis (Walter Bislin, Author of this Page) 3/6/2019 | 17:27

### Atmospheric Pressure Gradient, no Vacuum

Globetardos Extremos: You cannot have a pressurized atmosphere without a container

But you can have a pressure gradient in the atmosphere due to gravity without a container, and that can be measured with a barometer. But there is no boundary where the pressure suddenly changes like with a container. The pressure decreses more and more with distance from earth's center of mass until there are only a few atoms left per m3 and you can no longer use a barometer to measure the pressure.

But there is everywhere some gas in space. Never heard about solar wind? How can there be a wind without gas? Do you not know how stars and planets are formed? From gas clouds attracted and condensed by gravity until they form a ball of gas whith enough pressure and temperature to start fusion. You can see gas clouds in space with a telescopt all over the place. For example here. And even the moon is technically still inside the very diluded gas that belongs to the atmosphere of the earth.

### The Shape of level Water

Globetardos Extremos: Water can never conform to the exterior of an object. A body of water will always find its own level.

The surface of water is dictated by all acceleration potentials, e.g. the sum of gravitational potential and centifugal acceleration potential of an object like the earth. The water surface must lie on a layer with constant acceleration potential everywehere.

On a globe the equipotential layers have the shape of an ellipsoid. In a rotating container the sum of gravity and centrifugal acceleration builds equipotential layers with the shape of parabolas, so the surface of the water builds a parabola in a rotating container as explained in the following video:

Explaining The Spinning Water Demo; Youtube

If there where any difference in the potential at some place on the surface of water, this would result in a local difference in acceleration between neighboring molecules of water that would distributes the water to neighboring reagions until all water is on the same level again. This level is not flat on a planet.

So water would only be flat in a hypothetical gravitational field without any centrifugal accelerations, that builds flat parallel equipotential layers. Such fields do not exis naturally in the universe. All graviational fields are spherical or sums of overlapping spherical fields and other accelerations. Only locally on a very big sphere the field is approximately flat. This is the reason why water in a container is almost perfectly flat to the atomic level. But big lakes and oceans are not flat.

The curvature of a big lake like the lake Ponchartrain is measured with precision survey GPS devices in more than 6500 data points. The points, which are raw 3D coordinates (not assuming a globe) lie on an arc for each lane. From this points the calculated radius of the arcs matches exactly the official radius of the earth at this locations. If we overlay all data points with Soundly's image, both show exact the same curvature. So Soundly's images are not due to atmospheric effects, because GPS is not influenced by the atmosphere.

Equipotential levels on earth have the shape of a Geoid due to small variation in the gravity field, which is approximately an Ellipsoid.

So sea level has exactly the shape of the Geoid. Geodetic survey has measured the gravity field to such a precision, using Gravimeters and satellites, that we can measure the Geoid to sub cm accuracy. GPS navigation devices must know the exact shape of the Geoid to correctly calculate the elevation of the observer above mean sea level everywhere on earth. Elevation is defined as the vertical distance from the Geoid surface.

### Refraction and Equation for Hidden Height

Globetardos Extremos: Even considering refraction & atmospheric obstruction, we can see buildings & mountains too far away.

What do you mean by atmospheric obstrucion? Fog?

I never ever saw any image which does not exactly show what the globe model with refraction predicts, not even from Flat Earthers. Flat Earther commonly use the wrong equations for hidden height: 8 incher per mile squared. This equation calculates the drop from the tangent at the observer with distance. Or they do not take the accurate observer height into account (which is very sensitive to how much is hidden) and ignore Refraction.

See here for the the Exact Equations for the Hidden Height (Advanced Earth Curvature Calculator).

Flat Earther like to look directly through layers of atmosphere directly above water, where refraction is considerably higher than standard refraction some meters above the surface everywhere. See all the laser test over water at night. In these conditions the earth seems to be flat, because the laser can be seen over any distance. On day light you would recognize the strong distortion directly over water.

No surveyor does measure under those conditions, knowing about how refraction works. See Refraction Coefficient as a Function of Altitude.

The globe model predicts that directly over cold water or ice refraction is so strong (k > 1) that the light gets bent along the surface, making a laser visible at any distance at the ground. All this laser tests from flat earther are worthless to measure the shape of the earth. They only prove that refraction is strong over the surface of water.

See Simulation of Atmospheric Refraction with strong refraction over water compared with flat earth with no refraction.

29Ben 3/23/2019 | 15:39

One flight that is impossible on flat earth is Qantas flight QF27. These are direct non-stop flights between Sydney, Australia and Santiago, Chile.

Globe distance = 7050 miles
Flat Earth distance = 15,959 miles

Simply the airliner (a Boeing 747-400) would run out of fuel on the flat earth. And would have to fly faster than twice the speed of sound to make the schedule on time.

QF28 is the return journey, so its not wind assistance. And these flights occur 3 or 4 tines a week, every week, regardless of the wind speed and direction.

It amuses me that Flat Earthers can not agree between themselves what FE model is correct and they're unable to provide a model with no flaws or controversy. While everyone is in agreement with the spherical globe model and all the physics, maths and navigation works.

30Algersis 3/29/2019 | 19:38

A question from Italy

I put Lat and Long from the airport of Sidney to the airport of Santiago of Chile, picked up from Google Earth.

Sydney           - La   33° 56' 28.81" S - Lo 151° 10' 23.63" O
Santiago         - La   33° 23' 08.70" S - Lo  70° 48' 09.37" O


The result of calculated distance is about 7,224 Km, for Lo 151° and 70°.
The result of calculated distance is about 11,341 Km, for Lo 151° and -70°.
But the distance must be about 7,040 miles, (about 11,327 Km)

What is right and what is wrong?

31wabis (Walter Bislin, Author of this Page) 3/29/2019 | 22:37

The only accepted characters are E for east and W for west, N for north and S for south. There is no recognition of characters used in other languages than english. So the O is simply ignored, which is in your case wrong. So 70° O should be entered as 70° E or -70°.

My coordinates from https://www.latlong.net/ are:

Sydney:   Lat = 33° 51' 54.5148" S   Long = 151° 12' 35.6400" E
Santiago: Lat = 33° 27' 33.2244" S   Long =  70° 38' 43.2528" W


Which leads to a globe distance of 11,347 km = 7050.9 mile. Of course the distances depend on the coordinates you enter. The cities are some km in diameter. Another thing is, that this App calculates the distances for a perfect sphere. But the earth is an ellipsoid. So software that uses the official WGS84 ellipsoid model, like google earth and GPS, get slightly different values. I think 11,347 km or 11,327 km is close enough to show the difference between globe and flat earth. The difference is less than 0.2%.

32Shawn 5/8/2019 | 01:42

The country dimenstions are already measured. Whether flat or globe Earth, the distance between two points in Australia does not change.

If you really want to prove something, calculate the distance between Chile and one end of Australia. They would be closer on a globe Earth, and farther on a doughnut shaped or flatish Earth.

33Silvio Zonzini 8/31/2019 | 15:04

Fantastic ! Good work !

34MartinITALY 9/26/2019 | 06:13

I'm a flat earther and the reason why you see Australia too wide in the flat map is because you need to know how to measure in the Gleason Map.

Your Google indicates 3.934 km from Sydney to Perth.

in the Gleason Map, the distance from these two cities is 36 degrees and the map says very well, each degree is 60 nautical miles.

36x60 = 2160 NM

2160 NM = 4000.3 Km

35wabis (Walter Bislin, Author of this Page) 9/27/2019 | 21:04

MartinITALY, I know how to measure the Gleason map. And you are totally wrong:

1. Google Earth gives a distance between Sydney and Perth of 3290 km, like my calculations above, not 3934 km.
2. The 60 nautical miles per degree is only valid for degrees of latitudes. You may not apply this for longitudes!

Think about: if you are e.g. 1° = 60 nmi from the north pole, 360° longitude is a circle with circumference 2·π·60 nmi = 377 nmi. That is 1.05 nmi per degree longitude, not 60 nmi (nautical miles)! You get only correct results from 60 nmi per degree longitude at the equator. For x angles of longitude at all other latitudes you have to use d = x · 60 nmi/° · cos(latitude). So for Sydney/Perth latitude −33°, longitude difference 35.3° you have to calculate:

d = 35.3 · 60 nmi · cos( −33° ) = 1786 nmi = 3305 km

which is much closer to the real 3290 km than your 4000 km. The small difference between 3290 km and 3305 km is due to the fact that Sydney and Perth are not exactly at the same latitude. Where does the cosine come from? Because the earth is a globe and on the globe all longitude lines converge on the north and on the south pole. To take also different latitudes on the Gleason map into account, you have to use much more complicated spherical trigonometry (see Used Formulas for Globe Earth), because the earth is spherical!

The Gleason map is a projection of the globe onto a flat plane. No projection keeps all distances of the real world correct. The Gleason map keeps only the radial distances correct. We know for sure that Australia has not the shape as shown on the Gleason map. On the Gleason map Australia is 2.5 times too wide. So the Gleason map can not be a scale model of the real world. No 2D map can be a scale model of the real world, because you cannot map a 3D shape like a globe onto a 2D plane without distortions. If the earth were flat, this would be no problem at all.

If the earth were flat in reality then the Gleason map were not a projection but a scale model of the earth. In this case you could measure any distance on the map using a ruler and multiply the measured distance by the scale of the map to get the real distance.

If you do that with the Gleason map e.g. for the distance between Sydney and Perth, you get 8301 km, while the real value is only 3291 km. Of course, if you know the mapping algorithm from the globe earth onto the Gleason map, you can get the real distance from the map by applying the inverse mapping. But you never can directly measure the distance and simply multiply it by the scale of the map to get the real distance.

But because the earth is a sphere, you can take any globe model and measure any distance between 2 locations with a string, measure the length of the string and multiply it with the scale of the globe and you will get the correct real distance. This proves that the earth is a globe in reality and the globe model is the correct scale model of the earth.

Try to calculate the distance between Sydney and San Francisco from the Gleason map using your method. You can't do that without applying spherical trigonometry to get the real shortest distance airplanes fly, so called great circle (GC) routes. Hint: the GC distance between Sydney and San Francisco is 11,942 km, which you can measure from a globe model directly as explained. If you do that on the Gleason map, you will get 14,615 km, which is demonstrably wrong.

Here is how you can calculate the real distance from the Gleason map using spherical trigonometry: Measure latitude and longitude of 2 locations on the map and use the equations shown at Used Formulas for Globe Earth. This works because this way you apply the inverse projection of the Gleason map onto the globe.

36Pako 9/27/2019 | 07:52

I have calculated for both cases, and once I take the angle that separates both cities as 35.35º I get for a Flat Earth to a NP a distance of 8,308.89 km. While if I do it for a Global Earth to the SP there is a distance of 3,863.9 km.

Whoever made and assign the coordenates for longitude and latitude was quite aware of this crucial problem and must have decided to shift the coords in order to fit the purpose of a fake Globe. I have no doubt about it.
Deception talks Money so the man who does it gets its money right up there.

37wabis (Walter Bislin, Author of this Page) 9/27/2019 | 21:04

Pako: Whoever made and assign the coordenates for longitude and latitude was quite aware of this crucial problem and must have decided to shift the coords in order to fit the purpose of a fake Globe. I have no doubt about it.

You have a complete wrong imagination of how latitudes and longitudes are defined and how real locations and distances are measured on earth. Nobody has faked anything. Everybody can measure latitude and longitude for any location himself simply by using GPS, or if you don't trust GPS, using a sextant and a clock, like ancient navigators did.

Measuring latitude and longitude of 2 locations (using which ever method you prefer) and applying spherical geometry you can calculate the shortest distance between the 2 locations for the globe. You can then check this distance with any method you want (car, airplane, ect.), it will match the globe calculation. Every single flight proves that every single day.

Nobody ever measured something else than what is told. The whole earth is measured with different methods (geodetic surveying, airplanes, GPS satellites) since centuries repeatedly. We now know the shape of the globe earth to cm accuracy. It's not that difficult to measure the curvature and radius of the earth. Everyone that has done the measurements confirms the earth is a globe the size as is told. I did it multiple times.

For example: I have 4046 GPS data points from a Sail Drone that circumnavigated Antarctica in 196 days. Every hour the GPS location was recorded. See Geo-Data Visualisation and Calculator App.

Using the flat earth model to calculate the distance between each point and summing all distances I get a track length around Antarctica of 77,000 km. That is 3.5 times the published distance and the distance calculted on the globe model of 22,000 km.

38Pako 10/2/2019 | 19:37

Wabis, you are not in charge of assigning coords like Latitude and Longitude for each location around the World and I doubt you have ever used a sextant and a clock like ancient navigators did supossing you had knowledge of the stars.

Therefore I doubt the Globe model is the right one. No matter the flights times and speeds you take to get a right picture of the correct distance we are looking for in correspondence with the real world.

As someone suggested earlier on in one of the posts (Number 11 from Falcon Crest), the South Hemisphere in the Gleason and Hammonds maps are obviously distorted and wrong because the authors took in good faith the present and official coords that were only fit for a Global map. If you do not understand this last statement, I cannot help you to reason.

If the land masses and distance triangulation do not correspond, it is a sign there is trickery on it. I´ve been calculating global data and flat earth data for quite some few years and as a land surveyor when I was employed and never having to compute for any sort of allowance in regards to curvature if Earth was a Globe. My experience is based mainly on railways and construction buildings.

So please do not come to us with that all technology is so reliable you will never get lost in the middle of the ocean in the South Hemisphere and wreck your ship if you are the captain and guide yourself with a sextant or hoping to get a signal from a remote crewless drone. Satellites do not exist, period, specially at the heights we are told to be in space. Learn some science and study the atmosphere.

Oh and by the way, the Bible is a good book where you can find some light to the truth they are trying to hide from us. Some few books that might help you to open your eyes while leaving apart the world of maths. Terra Firma by David Wardlow Scott; Zetetic Astronomy by Parallax; One hundred proofs Earth is not a Globe by William M Carpenter; Is the Bible from Heaven and the Earth a Globe? by Alex Gleason.

39wabis (Walter Bislin, Author of this Page) 10/2/2019 | 22:14

Pako, you are beyond help. Just look up Celestial navigation to find out how a sextant is used to find your location on the globe. And learn the difference between coordinate transformations and mapping so you can figure out how to read maps to accurately get the right distances from them. You know, all flat maps are distorting projections of the WGS84 Globe Model, all navigation, GPS and Google Earth is based on.

Pako: Learn some science and study the atmosphere.

Learn some science? Really? Look at all my other web pages. Do you still think I have to learn some science? I'm a scientist, physicist and engineer. How about you learn how geotetic surveyors measured the globe, before making a fool of yourself as a surveyor by claiming the earth is flat.

### I studied the Atmosphere and Refraction

Study the atmosphere? You mean refraction? I already did. Here is some work of mine:

Simulation of Atmospheric Refraction
The interactive Refraction Simulator on this page renders scenes as seen on atmospheric conditions you can provide. It is an accurate simulation of Atmospheric Refraction for both Globe Earth and Flat Earth.

The Rainy Lake Experiment
The Rainy Lake Experiment was designed to show, how we can figure out the shape of the earth, Flat or a Globe, by observing and measuring a clever arrangement of targets over a distance of 10 km, taking terrestrial refraction into account and using modern equipment.

Deriving Equations for Atmospheric Refraction
On this page I explain how terrestrial atmospheric refraction is calculated from basic physical laws and how the equations and constants used in refraction equations are derived. This equations are used by geodetic surveyors and implemented in modern theodolites to automatically compensate for refraction (and curvature).

### Geodetic Surveyor measure the Curvature of the Earth

And as you think geodetic surveyors do not measure the curvature of the earth:

Here Is The Curve!
Results obtained from the GNSS Survey of the Lake Pontchartrain Causeway, a mane-made concrete structure monumenting the curvature of the earth. Hangout Soundly and Jesse Kozlowski showing their data in my 3D App.

Calculating Earth's Radius from Horizon Drop; The Maine Surveyor
Earth's radius can be calculated with reasonable accuracy from a very simple method: Measuring Horizon Drop. Ancient Mathematician Al-Biruni worked out an excellent equation for this.

Youtube channel Jesse Kozlowski, geodetic surveyor
Many videos about geodetic surveying and measuring the globe.

### Proofs that Satellites exist

Pako: Satellites do not exist, period, specially at the heights we are told to be in space.

Except you can see them and receive their signals yourself.

A telescope, a Nebula and plenty of Satellites
Watch the whole video for explanations of how the satellites were spotted by Wolfie.

GPS: An introduction to Satellite Navigation
Standford University - 13 October 2014

40MartinITALY 10/5/2019 | 00:56

Wabis, I don't trust your equations because a square has four corners of 90° and this multiplied is 360° in total. A circle also has 360° and we know a circle looks different from a square.

i've learned to fly Cessna with Gamin 420 and Boeing 737 with flight management system FMS and the plane goes on a flat plane locked to an altitude and there is no curvature calculation.

I designed my own ultra light plane, setting a center of gravity that I call center of weight, I calculated all airfoils dynamics and my plane took off at the sim and again it flies flat.

I made my own research at this simulators backend and they have the planet land either flat or cilinder shape.

You can set up the FMS or Garmin 420 with data from sky vector website and everything will match.

41MartinITALY 10/5/2019 | 01:13

Your simulation of atmospheric refraction is a digital interpretation of how things should be seen in real life with both models. You are basically trying to compensate the lack of curvature with an illusionary mirage. But in reality the thing is much more complex as you are ignoring important factors. There are distortion patterns in the lens hardware and limitations in the the human eyes field of view and depth which determine perspective and ocular resolution. You are ignoring the distortion generated at the atmospheric blocks also responsable to bend sun rays and you are ignoring the bounce of photons sun illumination the main enemy of lack of curvature mirages.

42MartinITALY 10/5/2019 | 01:54

I didn't take more than 5 minutes to accept the Flat Earth. Nobody is able to access into something like this with a rigid critical brain. The sub conscious is like a child but it can dominate the critical brain. we developed a very critical brain and it is not good. I wake up totally gasping for air and shaking like a foil similar to a convulsion. it is an ansia thing but I don't care anymore, my subconscious tells me is all fine and I continue sleeping with no problems at all. It is a flat earth after and I am created in perfection. The open globe reached its limits, we cannot proceed like that anymore,many pretend they do but they don't if you dig, and things will eventually restore into a calm flat earth with more subconscious around and accepting the creator as it is.

43wabis (Walter Bislin, Author of this Page) 10/5/2019 | 21:58

MartinITALY: Wabis, I don't trust your equations because a square has four corners of 90° and this multiplied is 360° in total. A circle also has 360° and we know a circle looks different from a square.

So you don't trust my equations because you are wrong.

A circle has no corners, so it has no sum of internal angles. The angle at the center of a figure like the circle does not contribute to the sum of the corner angles. The general formula to calculate the sum of the corner angles for any n-gon is 180° · (n2). So for a triangle you get 180°·1 = 180°, for a square 180°·2 = 360 for hexagon 180°·4 = 720° ect. If you increase the numer of corners to infinity, you get a circle with an infinite sum of internal angles, not 360°.

MartinITALY: i've learned to fly Cessna with Gamin 420 and Boeing 737 with flight management system FMS and the plane goes on a flat plane locked to an altitude and there is no curvature calculation.

Do you see or know what the FMS is calculating internally? No! You simply have to insert waypoints or latitude and longitude, which by the way are geodetic (spherical) coordinates. The rest is calculated automatically for you, eg. the great circle track (the shortest distance on the globe) and the distances between waypoints, using the World Geodetic System WGS84, the official globe model used by navigation devices, GPS, Google Earth, Surveyors, Maps ect. The inertial reference units even use cartesian earth centered earth fixed 3D vectors for all calculations like position, orientation and velocity of the aircraft, taking gravity, the rotation of the earth, the curvature of the earth (WGS84 model) and coriolis and centrifugal accelerations into account.

Inertial Navigation Systems (INS)
Very good introduction of how INS work.

MartinITALY: I made my own research at this simulators backend and they have the planet land either flat or cilinder shape.

All modern flight simulator use the WGS84 globe earth model. This is the only way you can fly real aircraft flightplans accurately and import sceneries from Google Earth, because the real flight plans and Google Earth are based on WGS84 globe earth model.

"The Earth" Preview • Microsoft Flight Simulator 2020
Youtube video showing the globe earth from high altitude on the new FS2020 flight simulator from Microsoft.

Do flight simulators software and hardware simulate a sphere shaped earth model or a flat earth model?
Question and answers from simulator developers on Quora.

MartinITALY: You can set up the FMS or Garmin 420 with data from sky vector website and everything will match.

SkyVector News
Map projection uses WGS84 Ellipsoidal model of the earth. Flight planning uses T. Vincenty's Formulae.

Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid.

44wabis (Walter Bislin, Author of this Page) 10/5/2019 | 22:51

MartinITALY: Your simulation of atmospheric refraction is a digital interpretation of how things should be seen in real life with both models. You are basically trying to compensate the lack of curvature with an illusionary mirage. But in reality the thing is much more complex as you are ignoring important factors.

A simple 3D programm is not capable to simulate refraction physically correct, taking all properties of the atmosphere into account. But you can simply use a mean value in cases of low refraction of calm air, as it is some meters above the ground. Geodesy uses so called standard refraction values, which are implemented in my App. But I have also programmed a Simulation of Atmospheric Refraction using a Ray Tracer, that simultes complex refraction situations physically accurate.

We can measure refraction with different methods or dedicated devices directly and correct the optical measurements of theodolites accordingly. If we do so, the earth shows a mean curvature of 6371 km, always, everwhere.

MartinITALY: You are ignoring the distortion generated at the atmospheric blocks also responsable to bend sun rays and you are ignoring the bounce of photons sun illumination the main enemy of lack of curvature mirages.

And you are ignoring all facts, shown on this and other pages, that show the earth can't be flat but is a globe. Some aspects of reality require more than 5 minutes thinking to understand.

45MartinITALY 10/6/2019 | 02:49

The World Geodetic System WGS84 is just triangulation based on a predetermined grid.

In my experiment, I feed all airports data and set 3 floats as LAT , LON, and of course a predetermined RAD numeric size.

return rad * set(
-cos(lat) * cos(lon),
sin(lat),
cos(lat) * sin(lon)

v@P = spherToCart(
1.0



I get a sphere with all airport points, and that how GPS works in real life, but if change some parameters in the equation, if i eliminate the RAD function and change it to just a SET, then I get a flat rectangular map with all airports in what it looks in place.

I could continue tweaking functions and get a flat gleason map.

IMAGE:
46MartinITALY 10/6/2019 | 03:04

Your Simulation of Atmospheric Refraction as I said before, is only a projection that shows what you shouldn't suppose to see on a curve, but it turns out that is however visually there, the technical name is Mirage.

I will go straight to the point: ray tracing.

I will study this in depth and I will come back.

47wabis (Walter Bislin, Author of this Page) 12/18/2019 | 21:52

### World Geodetic System WGS84

MartinITALY: The World Geodetic System WGS84 is just triangulation based on a predetermined grid.

Nonsense! The WGS84 is not about triangulation based on a grid.

Wikipedia: The World Geodetic System (WGS) is a standard for use in cartography, geodesy, and satellite navigation including GPS. This standard includes the definition of the coordinate system's fundamental and derived constants, the ellipsoidal Earth Gravitational Model (EGM), a description of the associated World Magnetic Model (WMM), and a current list of local datum transformations.

### Projections versus Transformations

MartinITALY: In my experiment, I feed all airports data and set 3 floats as LAT , LON, and of course a predetermined RAD numeric size. I get a sphere with all airport points, and that how GPS works in real life, but if change some parameters in the equation, then I get a flat rectangular map with all airports in what it looks in place

Nope, that's not how GPS works in real life. Educate yourself:

GPS: An introduction to Satellite Navigation
Standford University - 13 October 2014

Now to your "experiment", which shows that you have no clue what you are doing:

You have applied a simple cylindrical projection transformation on the real vectors, given in geodetic (or spherical) coordinates, by simply using latitude and longitude as your x-y-plane coordinates and set the z-coordinate (elevation h) to 1, while any value for h would result in a plane.

This changes all vectors, so they don't have the same length and distances between them anymore and have nothing to do with the original vectors anymore. You can't even reverse this projection. But any transformation that preserves the lengths and distances between vectors can be reversed. Your transformed vectors of the airport locations have nothing to do with reality anymore! It can be shown that this is also the case for any Gleason map projection for the same reason. That's the reason why the distances on the Gleason flat earth map can never conform with realty.

To preserve reality, you must only apply transformations to vectors that preserve vector lengths and distances, like e.g. transformations from spherical coordinates into cartesian coordinates. In reality it is much more complicated, because the earth is a geoid, a slightly irregular ellipsoid with bumps. But the principle is the same as shown below. Such transformations preserve the real shape of the earth, which is a sphere, no matter what coordinate system you apply.

#### Transformation Spherical into Cartesian Coordinates

So lets look how a transformation from spherical into cartesian coordinates works:

(24)
where'
 $\vec P_{\mathrm{spher}}$ ' =' 'vector in spherical coordinates $\varphi$ ' =' 'latitude $\lambda$ ' =' 'longitude $R$ ' =' 'radius of the sphere $h$ ' =' 'elevation above the surface of the sphere $\vec P_{\mathrm{cart}}$ ' =' 'same vector in cartesian coordintes with the same origin $x,y,z$ ' =' 'cartesian coordinates of the vector

This transformation preserves the length of all vectors and the distances between them, as you can prove easily if you would know what you are talking about:

(25)
where'
 $l_{\mathrm{spher}}$ ' =' 'length of vecor from spherical coordinates $P_{ \mathrm{spher},3 }$ ' =' 'radial component of vector in spherical coordinates $R$ ' =' 'radius of the sphere $h$ ' =' 'elevation above the surface of the sphere

In any Cartesian coordinate system the length of a vector is according to Pythagoras:

 (26)

The expression in the square root can be simplified by first factoring out a cos(φ)2. The red terms are cos(x)2 + sin(x)2 = 1:

 (27)

So the whole expression under the square root is for any angle φ and λ always = 1! So we get for the length of the vector in the cartesian coordinate system:

 (28)

which is the same length as in the spherical coordinate system. Vector lengths are preserved by this transformation. It can be shown that distances between vectors are preserved too accordingly.

#### Projection Spherical into Cylindrical Coordinates

Now lets look at your cylindrical projection. First it is obvious that simply using the latitude and longitude angles as x and y does not represent any distances on the earth, angles are not distances. If we want to preserve at least the distances along the equator and along a meridian, we can multiply the angles in radian by R, the radius of the earth. We can even preserve the elevation h instead of setting it to 1, as you have done.

(29)
where'
 $\vec P_\mathrm{spher}$ ' =' 'vector in spherical coordinates $\varphi$ ' =' 'latitude in radian $\lambda$ ' =' 'longitude in radian $R$ ' =' 'radius of the sphere $h$ ' =' 'elevation above the surface of the sphere $\vec P_\mathrm{cyl}$ ' =' 'projected vector into cylindrical coordinates $x,y,h$ ' =' 'coordinates of the projected flat earth where 0-meridian at equator is the origin

Although, other than your projection, this projection can be reversed, because it retains the 3D information, but it does not preserve the original vector lengths and distances, as I show below. It changes the shape of the measured real earth. It has nothing in common with reality anymore.

 (30)

### Refraction Simulator is based on pysical laws

MartinITALY: Your Simulation of Atmospheric Refraction as I said before, is only a projection that shows what you shouldn't suppose to see on a curve, but it turns out that is however visually there, the technical name is Mirage.

Look at the simulator before juding it. It is based on basic physics of light propagation in the atmosphere, tested uncountable times and found to be in accordance to observation and experiments. This knowledge is applied in geodesy since centuries and in modern high precision measurements of large bodies like airplane wings.

You can provide the atmospheric conditions you have measured and my Simulation of Atmospheric Refraction calculates the image that will result on the globe and on the flat earth model, inclusive inferior and superior mirages, looming, sinking, towering ect. For the flat earth to appear curved as observed on clear days, according to the laws of physics the temperature gradient for such a refraction has to be at least -150°C/km and even getting much steeper with altitude, which is physically impossible. Such a gradient can only hold on very small layers directly above the ground, which produces extreme inferior mirages. Any blurry, unsteady images are a sign of non-standard refraction. Standard temperature gradient is -6.5°C/km. Under standard refraction the earth looks like a sphere with radius 7/6·R.

In The Rainy Lake Experiment we measured refraction. I then entered the measured refraction in my simulation for globe and flat earth and superimposed the images taken with a camera through an autolevel. The images match the globe model perfectly. The flat earth model is way off.

So as soon as you make real measurements you always get the earth is a globe, never flat! The earth only appears to be flat if you look with the unaided eye and ignore all images we have from high altitudes and space.

Next time you comment here and without providing measurements or experiments you have done, I will delete your comment. I don't accept baseless claims from you anymore.

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http://walter.bislins.ch/fe/Dist

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