# Math: GPS Flight Data and the Shape of the Earth

On this page are all equations listed and explained as used in the Blog post GPS Flight Data and the Shape of the Earth.

## Calculating East/West Stretch Factor k

East/west directions on the Flat Earth are stretched by a factor k with respect to the Globe as follows, see GPS Latitude/Longitude on Flat Earth and Globe:

(1)
where'
 $k$ ' =' 'how much distances in east/west direction are stretched on the Flat Earth compared to the Globe, depending on latitude $U_\mathrm{fe}$ ' =' 'length of latitude line at latitude $\varphi$ on Flat Earth $U_\mathrm{globe}$ ' =' 'length of latitude line at latitude $\varphi$ on the Globe $\varphi$ ' =' 'latitude in radian $R$ ' =' 'radius of the Globe Earth

#### JavaScript Code

var kFact = {
Latitude: -66.5,
k: 0,

Update: function() {
if (this.Latitude == 90) {
this.k = 0;
} else if (this.Latitude == -90) {
this.k = Infinity;
} else {
var phi = rad( this.Latitude )
this.k = (Pi2 - phi) / cos(phi);
}
ControlPanels.Update( 'KFactPanel' );
}
};



## Calculating Attitude Angle and dPitch

Using recorded GPS flight data, it can be shown that Airplanes drop their Nose as they fly over the curvature of the earth. Below is the math used to calculate the attiude and pitch angles:

To calculate the Attitude angles, we can use the scalar product between 2 vectors. One vector will be the vertical at the current data point $\hat z$ the other vector will be the vector pointing to the next data point $\vec v_1$ or previous data point $\vec v_2$. The vertical is simply the unit vector pointing from the center of the earth, which is the origin of the ECEF coordinate system, to the location of a data point. My App uses the reference ellipsoid rather than a perfect sphere to calculate the verical, but this does not matter. The result will be the same.

If we have the coordinates of 2 data points in a cartesian coordinate system, like the ECEF coordinate system of GPS, we can calculate a vector from one point to the other point by vector subtraction:

(2)
where'
 $\vec v_1$ ' =' 'vector from current data point P2 to next data point P3 $\vec v_2$ ' =' 'vector from current data point P2 to previous data point P1

The scalar or dot product between 2 vectors $\vec u$ and $\vec v$ is defined as:

(3)
where'
 $\vec u \cdot \vec v$ ' =' 'scalar or dot product, which is not a vector but a scalar $\vec u, \vec v$ ' =' 'arbitrary vectors of any length $| \vec u |$ ' =' 'length of vector $\vec u$ $| \vec v |$ ' =' 'length of vector $\vec v$ $\varphi$ ' =' 'angle between the vectors

The length of a vector is calculated using Pythagoras as follows:

 (4)

It is very simple to calculate the scalar product in cartesian coordinates:

 (5)

Using the scalar product it is now easy to calculate the zenith angle $\varphi$ between the vertical $\vec z_i$ at the current data point i and the vector $\vec v_j$ to the next data point.

 (6)

Solving for $\varphi_i$ yields:

(7)
where'
 $\varphi_i$ ' =' 'zenith angle, angle between $\vec z_i$ and $\vec v_i$ $\vec z_i$ ' =' 'vector perpendicular to surface at current data point i $\vec v_j$ ' =' 'vector from current data point to next/previous data point $\vec z_i \cdot \vec v_j$ ' =' 'scalar product of the vectors z and v $\hat z_i$ ' =' '$\vec z_i / | \vec z_i |$ = unit vector in direction of the vector z $\hat v_j$ ' =' '$\vec v_j / | \vec v_J |$ = unit vector in direction of the vector v

From the zenith angles $\varphi_i$ we can calculate the attitude angles as follows:

(8)
(9)
where'
 $\alpha$ ' =' 'attitude angle
(10)

Note: I could choose any plane as the reference plane to calculate the dPitch angles, even the flat earth plane. It can be shown that the choice of the reference planes does not change the sum of all dPitch angles between start point and reference point, as long as they lie in a straight line or on a great circle. The choice of the horizontal to the reference ellipsoid as the reference plane at each point gives the correct result even if the points don't lie inline.

## Calculating Acceleration on the Flat Earth

Because distances in east/west direction on the Flat Earth are always stretched by a latitude dependen factor k with respect to the Globe, considerably in the southern hemisphere, the east/west distances and speeds derived from GPS measurements are way greater on the Flat Earth than in reality. If an airplane changes direction e.g. from south heading to west heading, the distances increase considerably. Because the time passed between 2 GPS datapoints remains constant, this means the airplane speed increases accordingly.

The corresponding acceleration to increase or decrease the speed is only one factor contributing to the G-Force load. Because the airplane has a much greater speed in the turns, the accelerations in curves get dangerously big on the Flat Earth.

The accelerations can be derived from the equations of motion.

First we write the equation for position as a function of time. With respect to a perfect circle on the Globe the positions on the Flat Earth are stretched in the east/west direction by the factor k. This factor is the ratio between east/west direction distances on the Flat Earth to the east/west direction distances at the same latitude on the globe, see Calculating East/West Stretch Factor k.

(11)
where'
 $\vec{s}(t)$ ' =' ' position as a function of time $r$ ' =' 'radius of curvature in north/south direction $v$ ' =' 'tangential speed in north/south direction $k$ ' =' 'stretching factor in east/west direction from Globe to Flat Earth model

From this equation we can derive the velocity and acceleration using Calculs:

(12)
(13)
where'
 $\vec{v}(t)$ ' =' ' velocity as a function of time $\vec{a}(t)$ ' =' ' acceleration as a function of time $r$ ' =' 'radius of curvature in north/south direction $v$ ' =' 'tangential speed in north/south direction $k$ ' =' 'stretching factor in east/west direction from Globe to Flat Earth model

$\vec{a}(t)$ is the centripetal acceleration. The passengers feel the reaction force, the centirfugal acceleration, which is the vector with the same magnitude pointing in the opposite direction.

(14)
where'
 $\vec{c}(t)$ ' =' 'centrifugal acceleration

The passenger also feel the gravitational acceleration $\vec g$ towards the ground $-g \cdot \hat z$. So the total acceleration vector felt by the passenger is:

 (15)

The G-Force load a passenger feels is then the vector sum of all acceleration diveded by the gravitational accelation g:

 (16)

At the Antarctis trip the airplane flew at south latitudes of around −66.7°. At this latitude the east west distances on the Flat Earth are more than 6.9 times longer than on the Globe! This means that on a Flat Earth the airplane would have flown seemingly 6.9 times faster in east/west directions than in north/south directions.

Airplanes fly standard curves of 1/4 circle in 1 minute. On the Antarctic trip, according to the recorded GPS data, the airplane flew circles at a speed of about 605 km/h. So on the flat earth the north/south part of the circle would be flown at 605 km/h while the east/west part of the circle would be flown at 6.9 · 605 km/h = 4175 km/h.

This is the predicted acceleration. See Calculating Speed and Acceleration from GPS data for the accelerations displayed in the GPS Flight Data Visualisation App. The Globe acceleration and the Flat Earth acceleration are calculated the same way, using the corresponding positions mapped on the Globe or Flat Earth respectively.

## Calculating G-Force Load on the Flat Earth

The equation for the G-Force load plotted in the graph below is:

(17)

To calculate the G-Force load for the Globe simply set k = 1. The Geogebra file to produce the plot below can be downloaded here:  Acceleration Airplane Antarctica Flat Earth.zip .

The green line of the plot shows the G-Forces felt by a passenger on the Flat Earth on a circular track at a north/south speed of 167.55 m/s = 610.4 km/h at a standard turn rate of 4 minutes for a full cicle, which results on the Globe in a circle with a radius of 6400 m. On a Flat Earth the circle is stretched to an ellipse width the semi major axes 6.69 times the height.

Acceleration on a Flat Earth in Antarctica

As we can see, at 0° and 180° of a circular Globe track, which is at the east and west point of the circle/ellipse, the acceleration a passenger would feel on the Flat Earth is more than 3 g. At the north and south point the accelerations are the same in both models. On the Flat Earth the speed of the airplane in north/south direction is 610 km/h and in the east/west direction 4080 km/h, which is more than Mach 3.5, 3.5 times the speed of sound! No airlines can fly such maneuvers, which would not make any sense anyway.

The claim of Flat Earthers, that GPS works the same on the Flat Earth and Globe is proven false. The tracks recorded by GPS show speeds and accelerations on the Flat Earth model, which no airplane and passenger ever have experienced and make no sense. GPS data proves, that the earth is not flat.

## Calculating Speed and Acceleration from GPS data

In the paragraph above I have calculated the predicted accelerations. The effective accelerations from the GPS data is calculated as shown below. This calculation is the same for Flat Earth and Globe. The difference is only, that for the Flat Earth the data mapped onto a Flat Earth is used, while for the Globe the data mapped onto the Reference Ellipsoid is used.

How to get cartesian Flat Earth coordinates from the GPS data, which is stored as cartesian ECEF Globe coordinates, is described at Globe and Flat Earth Transformations and Mappings.

Note: this is not a coordinate system transformation but a mapping projection, which does not retain the original vectors. That's why distances, angles, speeds and accelerations on the Flat Earth are not the same as on the Globe.

Velocity is the change of position over the change in time between 2 points:

 (18)

So I have to calculate the distance to the next point and divide is by the difference of the time stamps between this 2 points.

The speed is simply the length of the velocity vector:

 (19)

This is the speed displayed at Speed, using the data mapped onto the Globe model, and SpeedFE, using the data mapped onto the Flat Earth model.

The acceleration is the change in velocity over the change in time:

 (20)

The forces the passengers feel is the reaction force to this acceleration, so we have to take the vector pointing in the opposite direction of $\vec a_i$. Then we have to add the gravitational acceleration acting in the $-\vec z_i$ direction to get the total acceleration $\vec u$ the passengers feel:

 (21)

The G-Force load a passenger feels is then the magnitude of this vector $\vec u$ divided by the gravitational accelation g = 9.806 m/s2:

 (22)

This is the value that is displayed at Accel and AccelFE.

If the time between the previous and current point $\Delta t_{i-1}$ is not the same as the time between next and current point $\Delta t_i$, then the calculation is not accurate. If the difference between $\Delta t_{i-1}$ and $\Delta t_i$ is more than 10%, then the App places a ? after the value. If the difference is more than 10% the text unreliable is displayed instead.

The Globe acceleration and the Flat Earth acceleration are calculated the same way, using the corresponding positions mapped on the Globe or Flat Earth respectively.

### Resources

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