# Centrifugal and Gravitational Acceleration in an Aircraft

Wednesday, January 25, 2017 - 00:08 | Author: wabis | Topics: FlatEarth, Calculator, Knowlegde, Aviatic, Mathematics In a fast-flying aircraft heading east near the equator, an additional centrifugal acceleration to earths centrifugal acceleration acts away from the earth so that a part of the earth's attraction force is canceled out and therefore one is slightly lighter in the aircraft than on the ground. This is called the Eötvös effect .

In this article, this effect can be calculated with a calculation form for any location and flight direction and altitude. The calculation is verified by means of a real flight. In addition, all the formulas used are listed and explained. Deutsche Version: Zentrifugal- und Gravitationsbeschleunigung in einem Flugzeug

## Calculation Form

See Legend for a description of the fields. Set GS = 0 if you want to compute the weight of the test mass for some not moving location. If you set GS = 0, the centrifugal acceleration is calculated with the formulas for a stationary object rotating with the earth! Use the Lat and Alt to specify this fixed location then. The results of this calculation are shown in the right column. JavaScript for the Calculation Form: Centrifugal and Gravitational Acceleration in an Aircraft

## Experiments with a Test Mass

In the following videos Wolfie (a pilot) weighs a test mass on different locations on the earth and in airplanes flying in different directions. The test mass gains and looses weight according to the calculations on this page:

## Other Influences

The following table compares the magnitudes of the most varied influences on the gravitational acceleration. The first 5 influences are taken into account in the calculations obove.

Value m/s2 Value / 9.806 Description
9.806 100 % Gravitation: Mean effective gravitational acceleration on earth
0.099 1 % Aircraft: Centrifugal acceleration in an aircraft at a speed of about 500 kt with a strong tailwind of about 100 kt, direction of flight is east on the equator at an altitude of 12.5 km
0.034 0.35 % Max Centrifugal: Amount of centrifugal acceleration at the equator as a part of the effective gravitational acceleration
0.026 0.264 % Latitude: Maximum variation of the mean gravitational acceleration depending on latitude: plus at the poles, minus at the equator
0.003 0.03 % Altitude: Reduction of effective gravitational acceleration per 1,000 m altitude
0.002 0.0204 % Geoid: Maximum Geoid variation
0.001 0.0102 % Range of the maximum Geoid variations on Geoid-Maps 
0.0005 0.0050 % Maximum Range of the Geoid variations on most places on earth
+30 mGal correspond to −100 m = −330 ft altitude. 100 mGal = 0.001 m/s2.
0.00014 0.0014 % Air Density: Apparent increase of gravitation per 1,000 m altitude due to air density decrease (buoyancy) for an iron weight with a density of 7,874 kg/m3
0.000062 0.00063 % Coriolis: Maximum influence of the Corriolis-Effect at the poles on an airspeed of 480 kt, if the airplane follows a great circle on the surface of the earth. On the equator the influence is 0 .
0.00000163 0.0000166 % Moon+Sun: Maximum influence of gravitation of the moon and sun combined 
0.00000113 0.0000115 % Moon: Maximum influence of gravitation of the moon
0.00000050 0.0000051 % Sun: Maximum influence of gravitation of the sun

The calculations are based on an ellipsoid. The real shape of the earth varies from place to place. Therefore, gravitation also varies accordingly. The variation, however, is less than ±0.01% of gravitation at most of the places on earth.

If the deviation of the gravitation of the geoid from the ellipsoid is known, you can consider this in the calculation form by adding a correction value to Alt: +30 mGal−100 m = −330 ft

 Influence of the Coriolis-Effect:

(1)
where'
 $c$ ' =' ' Increase of the effective gravitational acceleration due to the Coriolis-Effect $g$ ' =' ' effective gravitational acceleration $c_\mathrm{H}$ ' =' ' coriolis acceleration, acting always horizontally $v$ ' =' ' speed of the airplane (Ground Speed) $\omega$ ' =' ' $2\,\pi / T$ = angular velocity of the earth $T$ ' =' ' 24 · 3600 s = one day period $\varphi$ ' =' ' degree of latitude

Note: the Coriolis effect increases the effective gravitationals acceleration slightly. It actually works horizontally (cH) but, together with the gravitational acceleration, causes a slight deviation of the same and a minimal increase by c.

The influence of the tides is negligible in the calculations. The influence of the buoyancy of the atmosphere and the coriolis effect are very tiny and can be neglected too.

## Legend

Reset resets all fields ​to their initial values. To reset only a specific field to its initial value, place the cursor in the field and press the ESC key. Clear sets all fields to 0.

GS: Speed of the aircraft with respect to the ground surface (ground speed).

Lat: Degree of Latitude of the aircraft position. The equator is 0, the North Pole 90 and the South Pole -90 degrees.

Latcal: Latitude at which the scale has been calibrated with the test mass Wcal.

Wcal: Weight of the test mass.

Course: heading in degrees North. North is 0, east is 90, south 180 and west is 270 degrees.

Alt: Altitude of the airplane above sea level.

Altcal: Altitude above sea level, on which the scale has been calibrated with the test mass Wcal.

Model: The earth is not a perfect sphere and can be approximated by an Ellipsoid . Model determines with which model the calculations are performed. For Sphere, the average radius R of the earth is used, see (2).

grel: Relative effective acceleration. This is the effective acceleration gh in the aircraft relative to the effective acceleration go on the earth, see (23). If the aircraft is on the ground, grel = 1. Values less than 1 mean that due to the centrifugal acceleration, one is correspondingly lighter in the aircraft than on the ground.

Wo: Weight of the test mass displayed on the scale at the position of the aircraft at sea level.

go: Effective acceleration on the earth's surface, see (5). This corresponds to the vectorial sum of gravitational acceleration goG and centrifugal acceleration aoZ.

goG: Gravitational acceleration on the earth's surface, see (9). Depending on the mass M of the earth and the distance R from the center of mass.

aoZ: Centrifugal acceleration on the earth's surface, see (7). This is dependent on the rotational speed or rotation period T and the effective radius for the latitude Lat. The effective radius is the perpendicular distance of the point P from the earth axis. On the poles this radius is 0, on the equator it is equal to the radius of the earth at that location.

R: Radius of the earth at latitude Lat. This is the distance of the point P from the center of the earth, see (3). In the sphere model, this radius is the same size everywhere.

ρ1: Radius of the red cutting ellipse at point P, i.e. the geodesic in east/west direction, see (41).

veq: Tangential velocity at the equator, veq = ω · R. Depends on the rotational speed ω or the rotational period T and the radius R on the equator, see (2).

go,cal: Effective gravitational acceleration at the point where the scale has been calibrated, ie at the latitude Latcal at altitude Altcal.

Wh: The weight of the test mass as displayed on the scale in the aircraft at the cruise altitude Alt.

gh: Effective acceleration in the aircraft, see (11). This is the vectorial sum of gravitational acceleration ghG and centrifugal acceleration ahZ in the aircraft.

ghG: Gravitational acceleration in the aircraft, see (17). Depending on the mass M of the earth and the distance from the mass center, i.e. the earth radius at latitude Lat and the altitude Alt.

ahZ: Centrifugal acceleration in the aircraft, see (21). This is dependent on the tangential velocity v of the aircraft on the geodesic with respect to space in altitude Alt and the radius ρh of the geosdesic at the position of the aircraft.

ρh: Radius of the geodesic, see (25). In the sphere model, this is the radius R of the earth plus the altitude Alt of the aircraft. In the ellipsoid model, ρh is also dependent on the direction of flight Course and is a value between ρ1 and ρ2 plus the altitude Alt of the aircraft, see (24).

ρ2: Radius of the blue cutting ellipse at point P, i.e. the geodesic in north/south direction, see (27).

vrot: Tangential speed of the point Ph due to the earth's rotation, see (16). Is dependent on latitude Lat and the altitude Alt.

v: Combined speed of the aircraft and the rotation of the earth at latitude Lat, see (13). This is the tangential velocity of the aircraft on the geodesic with respect to a non-rotating earth. Together with the radius ρh of the geosdesic, this rate determines the centrifugal acceleration ahZ acting in the aircraft.

## G-Display on an Aircraft

The default values in the Calculation Form are from a real flight with a Bombardier Global Express . The aircraft had about 100 kt tailwind and flew at a speed of about 500 kt True Airspeed (TAS) east (kt or kn = knot = Nautical miles per hour). The velocity causing the centrifugal acceleration is composed of the tangential velocity due to the rotation of the earth vrot at a latitude of −35° and the airspeed with respect to the earth surface of 600 kt (true airspeed TAS + tailwind), see (13).

Even under these extreme conditions, the centrifugal acceleration ahZ is so small that the relative effective acceleration $g_\mathrm{rel}$ barely shows a value below 1 on the G-display: in particular 0.9907796, which is rounded to 0.99.

This means that a man who has a weight of 100 kg on the ground at position P weighs only 99 kg in the aircraft.

Because the aircraft flies towards the east, that is, with the earth's rotation, the rotation and flight speed add up, resulting in a maximum centrifugal acceleration which makes the man lighter.

When the aircraft flies towards the west, i.e. contrary to the earth's rotation, the two speeds partly cancel out, and the man then weighs in the airplane roughly the same as on the ground. Check this by entering 270 at Course.

This effect is actually measurable: see Experiments with a Test Mass.

## Model of the earth, reference ellipsoid

In the following sections, all the formulas used in the Calculation form are listed and explained:

The earth is not a perfect sphere. It is slightly flattened at the poles and the diameter at the equator is 42.8 km greater than at the poles. Gravitation on the surface is also not uniform, but varies by mass distribution on the surface and inside the earth.

For the calculations on this page I use a Reference ellipsoid . This is a rotation ellipse with the earth axis as the rotation axis and the origin at the center of the earth. This ellipsoid has the following parameters :

(2)

$a = 6{,}378{,}137\ \mathrm{m}$

Semi major axes, i.e. radius on the equator

$b = 6{,}356{,}752{.}3142\ \mathrm{m}$

Semi minor axes, i.e. radius on the poles

$R = 6{,}371{,}008{.}8\ \mathrm{m}$

Average radius of the sphere earth

$g_\mathrm{e} = 9{.}7803253359\ \mathrm{m}/\mathrm{s}^{2}$

Effective acceleration on the equator

$g_\mathrm{p} = 9{.}8321849378\ \mathrm{m}/\mathrm{s}^{2}$

Effective acceleration on the poles

$G \, M = 3{.}986004418 \times 10^{14}\ \mathrm{m}^{3}/\mathrm{s}^{2}$

Geocentric gravitational constant 

$T = 86{,}164{.}098903691\ \mathrm{s}$

Rotation period with respect to space (sidereal day) 

$\omega = 7{.}292115 \times 10^{-5}\ \mathrm{rad}/\mathrm{s}$

Angular speed

where'
 $G$ ' =' 'Gravitational constant $M$ ' =' 'Mass of the earth $\omega$ ' =' '2 π / T

The product G·M can be determined more precisely than the individual factors .

## Effective acceleration on the earth

### Position on the Reference ellipsoid

In order to calculate the accelerations at a specific point P on the reference ellipsoid, we must calculate the position of the latitude of degree φ . Along the latitude through this position the accelerations are constant in magnitude.

The following formula can be used to calculate a point P on the surface of the earth with h = 0 or to compute a point Ph that is at a distance h from this point above the surface . The connecting line of these two points is perpendicular to the ellipsoid.

(3)
(4)
with
and
where'
 $P_\mathrm{x}$ ' =' 'Position of the point P measured from the earth axis $P_\mathrm{z}$ ' =' 'Position of the point P measured from the equatorial plane $h$ ' =' 'Altitude measured perpendicular to the surface of the ellipsoid $\varphi$ ' =' 'Degree of latitude in radian = degree · π / 180 $a$ ' =' 'Semi major axis of the ellipse (radius at the equator), see (2) $b$ ' =' 'Semi minor axes of the ellipse (radius to the poles), see (2)

The origin of the coordinate system is at the center of the ellipsoid.

### Effective acceleration

In geosciences, the effective (gravitational) acceleration of a celestial body is composed of its gravitational acceleration (gravitation) due to the mass of the body and the centrifugal acceleration in the reference frame, which rotates with the body  .

The effective acceleration on the earth's surface $g_\mathrm{o}$ is needed as a reference for the calculation of the relative effective acceleration $g_\mathrm{rel}$ in the aircraft. The index o stands for the altitude 0, that is, the sea level.

The earth is slightly flattened on the poles due to its rotation. It has the form of a rotational ellipsoid. For such an ellipsoid, according to WGS84 the effective acceleration can be calculated as follows :

(5)
where'
 $g_\mathrm{o}$ ' =' 'Effective acceleration at latitude φ $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180 $a$ ' =' 'Semi major axis of the ellipse (radius at the equator), see (2) $b$ ' =' 'Semi minor axes of the ellipse (radius to the poles), see (2) $g_\mathrm{e}$ ' =' 'Gravitational acceleration on the equator, see (2) $g_\mathrm{p}$ ' =' 'Gravitational acceleration at the poles, see (2)

The effective acceleration acts perpendicular to the surface of the ellipsoid at point P. It can be specified in vector form as follows:

(6)

### Centrifugal acceleration

The centrifugal acceleration on the earth's surface aoZ acts perpendicular to the rotation axis, depends on the degree of latitude φ and always acts outwards, i.e. its Z component is 0. The magnitude of the centrifugal acceleration is therefore equal to its X component.

(7)
where'
 $\vec a_\mathrm{oZ}$ ' =' 'Centrifugal acceleration at sea level $\omega$ ' =' 'Angular speed, see (2) $P_\mathrm{x}$ ' =' 'Distance of the point P from the axis of rotation, see (3) $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180

The angular velocity ω indicates how fast the earth rotates about its axis. It can be calculated from the sidereal rotation period T as follows:

 (8)

The Sidereal day is the duration of a full revolution of the earth around itself against the fixed starry sky. The mean sidereal day on the earth is almost 4 minutes shorter than the sloar day of 24 hours, see T in (2).

### Gravitational acceleration

The gravitational acceleration at point P can be determined vectorially:

(9)
(10)
where'
 $\vec g_\mathrm{oG}$ ' =' 'Gravitational acceleration at sea level $g_\mathrm{oG}$ ' =' 'Magnitude of gravitational acceleration at sea level $\vec g_\mathrm{o}$ ' =' 'Effective acceleration at sea level, see (5) $\vec a_\mathrm{oZ}$ ' =' 'Centrifugal acceleration at sea level, see (7)

## Effective acceleration in the aircraft

Airplanes fly the shortest possible connection between two points on the earth's surface. Such a connection line is the Geodesic . On a sphere, each geodesic is located on a Great circle .

If the earth is approximated by an Ellipsoid , geodesics generally do not form closed curves, see picture. To simplify the calculations on this page, I approximate the geodesics by a cutting ellipse optained by an intersection plane through the ellipsoid. These ellipses pass through the point P and contain the perpendicular through the point P. These ellipses are rotated by a certain angle α from the longitude through P. For these ellipses the radius of curvature ρ relevant to the centrifugal acceleration can then be calculated at point P.

The effective acceleration in the aircraft is composed of gravitational and centrifugal acceleration. Please note the following:

The centrifugal acceleration in the aircraft does not act perpendicular to the axis of rotation of the earth but upwards perpendicular from the surface of the ellipsoid because the aircraft moves on a geodesic (blue) with respect to a non-rotating ellipsoid. This has a different curvature ρ than the longitude or latitude of the ellipsoid at the point P and the center of the osculating circle of the geodesic lies not on the axis of rotation of the earth but on the violet axis with angle φ in the image.

The Inertial Reference System measures the accelerations in 3 mutually independent orthogonal axes. The Z direction points upwards in respect to the aircraft. In cruise, all forces on the aircraft are balanced so that the effective acceleration acts along the negative direction of the Z axis.

The effective acceleration is obtained by vectorial addition of the gravitational and centrifugal acceleration:

(11)
(12)
where'
 $\vec g_\mathrm{h}$ ' =' 'Effective acceleration in the aircraft as a vector $g_\mathrm{h}$ ' =' 'Magnitude of the effective acceleration acting along the direction of the negative Z axis of the aircraft $\vec g_\mathrm{hG}$ ' =' 'Gravitational acceleration in the aircraft, see (17) $\vec a_\mathrm{hZ}$ ' =' 'Centrifugal acceleration in the aircraft, see (21)

### Position of the aircraft

In order to calculate the accelerations in the aircraft, we must calculate its position Ph from the latitude φ and the altitude h over sea level.

The position Ph of the aircraft can be calculated by the formula (3).

### Absolute velocity of the aircraft

To calculate the centrifugal acceleration ahZ in the aircraft, we need its absolute velocity (speed) v on the geodesic. This is the speed with respect to a non-rotating ellipsoid. It is vectorially composed of the tangential velocity vrot of the point Ph due to the earth's rotation and the velocity vgs and the course α of the aircraft with respect to the surface (ground speed GS).

For the calculations, a flat 2-dimensional coordinate system is used at the points P and Ph, respectively, with the Y coordinate facing north and the X coordinate facing east.

(13)
(14)
where'
 $\vec v, v$ ' =' 'Absolute velocity (speed) on the non-rotating geodesic $\vec v_\mathrm{rot}$ ' =' 'Tangential velocity of the point Ph, see (16) $\vec v_\mathrm{rel}$ ' =' 'Velocity of the aircraft with respect to the surface corrected for altitude h, see(15)

The speed vrel at altitude is somewhat higher than the ground speed vgs because of the earth curvature.

(15)
where'
 $\vec v_\mathrm{rel}$ ' =' 'Velocity of the aircraft with respect to the surface corrected for altitude $v_\mathrm{gs}$ ' =' 'Speed with respect to the surface of the earth in m/s = knots · 0.5144 m/s/kt $\rho$ ' =' ' Radius of curvature of the geosdesic on the earth's surface, see (24) $h$ ' =' 'Altitude above sea level in m = feets · 0.3048 m/ft $\alpha$ ' =' 'Azimut: direction of flight path with respect to north in radian = degrees · π / 180°; east is 90°

The tangential velocity of the point Ph has only a component in the X direction and is dependent on the degree of latitude φ :

(16)
where'
 $\vec v_\mathrm{rot}$ ' =' 'Peripheral speed at latitude φ and altitude h $\omega$ ' =' 'Angular velocity of the earth's rotation, see (2) $P_{\mathrm{h},\mathrm{x}}$ ' =' 'Distance of the point Ph from the rotational axis, see (3) $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180°

The trajectory of the aircraft is a geodesic which is located on an extended ellipsoid which has the local perpendicular distance h to the ellipsoid of the earth's surface. The position Ph is calculated using formula (3), by using Alt for the altitude h.

### Gravitational acceleration in the aircraft

To calculate the effective acceleration at the point Ph at a distance h from the reference ellipsoid, there is a formula according to WGS84 . The corresponding acceleration acts perpendicular to the surface of the ellipsoid and is composed of gravitational acceleration and centrifugal acceleration due to earths rotation.

However, the aircraft is not connected to the surface and is therefore not exposed to the centrifugal acceleration of the surface. The Centrifugal acceleration in the aircraft must be determined in a different way.

The gravitational acceleration $g_\mathrm{hG}$ in the aircraft is obtained by subtracting the centrifugal acceleration $a_\mathrm{ohZ}$ due to earths rotation from the effective acceleration $g_\mathrm{oh}$ at altitude h calculated according to WGS84.

(17)
where'
 $\vec g_\mathrm{hG}$ ' =' 'Gravitational acceleration in the aircraft $\vec g_\mathrm{oh}$ ' =' 'Effective acceleration according to WGS84 for altitude h, see (18) $\vec a_\mathrm{ohZ}$ ' =' 'Centrifugal acceleration with respect to the rotational axis of the earth at the Point Ph, see (20)

The effective acceleration according to WGS84 for the altitude h can be calculated as follows :

(18)
with
and
where'
 $g_\mathrm{oh}$ ' =' 'Effective acceleration according to WGS84 at a distance h from the ellipsoid $g_\mathrm{o}$ ' =' 'Gravitational acceleration according to WGS84 at point P on the ellipsoid, see (5) $a$ ' =' 'Semi major axis of the ellipse (radius at the equator), see (2) $\omega$ ' =' 'Angular velocity of the earth's rotation, see (2) $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180 $h$ ' =' 'Altitude above the surface of the sea in m = feets · 0.3048 m/ft

The effective acceleration according to WGS84 acts perpendicular to the surface of the ellipsoid. The vector representation is therefore:

 (19)

The centrifugal acceleration due to the rotation of the earth at point Ph is:

(20)
where'
 $\vec a_\mathrm{ohZ}$ ' =' 'Centrifugal acceleration due to earths rotation in altitude h $\omega$ ' =' 'Angular speed, see (2) $P_{\mathrm{h},\mathrm{x}}$ ' =' 'Distance of the point Ph from the rotational axis, see (3) $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180

### Centrifugal acceleration in the aircraft

In contrast to the centrifugal acceleration of the earth's surface, the centrifugal acceleration in the aircraft acts away from the geodesic curvature, i.e. perpendicular to the surface of the ellipsoid at the point Ph. Its magnitude can be calculated from the absolute velocity of the aircraft v on its non-rotating trajectory (geodesic) and the radius of curvature ρh of the geodesic at altitude h.

(21)
where'
 $\vec a_\mathrm{hZ}$ ' =' 'Centrifugal acceleration in the aircraft $v$ ' =' 'Absolute speed on the non-rotating trajectory, see (14) $\rho_\mathrm{h}$ ' =' 'Radius of curvature of the geodesic of the aircraft, see (25) $\theta$ ' =' 'Heading of $\vec v$ measured from the north direction in radian, see (22) $\varphi$ ' =' 'Degree of latitude in radian = degrees · π / 180°

The angle $\theta$ of the geodesic with respect to the non-rotating ellipsoid can be calculated from the total velocity $\vec v$, see (13). This velocity takes the direction of flight α and the rotation of the earth into account. Note that $\vec v$ does not have the same heading as the Ground Speed $\vec v_\mathrm{gs}$ because of the rotation of the Earth.

(22)
where'
 $\theta$ ' =' 'Heading $\vec v$ measured from the north direction in radian $v_\mathrm{y}$ ' =' 'Y component of the total velocity $\vec v$ $v$ ' =' 'Magnitude of $\vec v$, siehe (13)

If $v = 0$ then $\theta = 0$ can be set because the centrifugal acceleration in the aircraft is in this case 0 anyway.

## Relative Effective Acceleration in the aircraft

On the G-display, the Z-component of the effective acceleration in the aircraft is displayed with respect to the effective acceleration $g_\mathrm{o}$ of the earth's surface under the aircraft:

(23)
where'
 $g_\mathrm{rel}$ ' =' 'Effective acceleration in the aircraft with respect to the effective acceleration on the ground under the aircraft $g_\mathrm{h}$ ' =' 'Effective acceleration in the aircraft, see (12) $g_\mathrm{o}$ ' =' 'Effective acceleration on the ground under the aircraft, see (5)

If the aircraft is on the ground, $g_\mathrm{o}$ and $g_\mathrm{h}$ are equal, so $g_\mathrm{rel} = 1{.}00$ is displayed.

Notice: May be the aircraft uses a mean effective acceleration of go = 9.80665 m/s2 instead of the above computed real effective acceleration at the point P.

## Calculating the curvature radius of the geodesic

In order to be able to calculate the centrifugal acceleration ahZ (21) in the aircraft on its trajectory around the earth, we need the radius of curvature ρh of the trajectory at point Ph. We obtain the radius of curvature by calculating the radius of curvature ρ of the geodesic on the earth ellipsoid and adding h.

The calculation of ρ takes place via the green hatched cutting ellipse. The rotation axis of the green cutting ellipse is the connecting line PQ and is rotated by the angle γ with respect to north direction. This rotation axis is in general not identical with the major axis of the ellipse!

I could not find any formulas for calculating the parameters of the green cutting ellipse. However, I can estimate the radius of curvature ρ (perhaps the formula shown here is even correct) by calculating the maximum radius of curvature ρ1 of the red cutting ellipse and the minimum radius of curvature ρ2 of the blue cutting ellipse and then interpolating the angle γ with a cosine function. The radius ρ depends on γ and has a value between the minimum and maximum radius:

An equation for the Directional Ellipsoid Radius can be found in the Wikipedia. Although the presented equation below is not correct, it is a good approximation. If I find time, I will incoorporate the equation from Wikipedia.

(24)
(25)
where'
 $\rho$ ' =' 'Radius of curvature of the green-dotted ellipse (geodesic) at the point P $\rho_\mathrm{h}$ ' =' 'Radius of curvature of the geodesic at height h $\gamma$ ' =' 'Heading of the aircraft (either heading α or angle θ of the geodesic, see (22)) $\rho_1$ ' =' 'Radius of curvature of the red ellipse at point P, see (41) $\rho_2$ ' =' 'Radius of curvature of the blue ellipse at point P, see (27)

Note that in the conversion of the ground speed to the speed at altitude (15) the geodesic in the coordinate system rotating with the earth must be used. These geodesic correspond to the cutting ellipse rotated by the angle γ = α, where α is the heading (azimuth) with respect to the rotating earth.

For the centrifugal acceleration ahZ, the geodesic must be described in the non-rotating coordinate system since the aircraft is detached from the earth. The rotation angle γ for the corresponding cutting ellipse can be calculated using the absolute velocity $\vec v$, see (22).

### Calculating the blue cutting ellipse

For the calculation of the radii of curvature ρ1 and ρ2, I need the lengths of the semi major axes of the red and blue ellipse.

The blue ellipse is the north-south section through the ellipsoid. The two semi axes correspond to the maximum and minimum radius of the earth:

 (26)

### Calculating the radius of curvature of the blue ellipse

The smaller radius of curvature ρ2 at the point P can be calculated with the data of the blue ellipse (see figure under Calculating the curvature radius of the geodesic), :

(27)
where'
 $\rho_2$ ' =' 'Smaller radius of curvature at point P at azimuth α = 0° or 180° $a_\mathrm{S}$ ' =' 'Semi major axes of the ellipse $b_\mathrm{S}$ ' =' 'Semi minor axes of the ellipse $P_\mathrm{x}$ ' =' 'X position of the point P, see (3) $P_\mathrm{z}$ ' =' 'Z position of the point P, see (3)

### Calculating the red cutting ellipse

#### Intersect point Q

In order to obtain the radii of the red cutting ellipse (see figure under Calculating the curvature radius of the geodesic), we must calculate the intersection point Q of the straight line PQ with the blue cutting ellipse. For this, we introduce the equations for the ellipse and the line:

 (28) Ellispe equation (29) Line equation with

The intersections of the ellipse with the straight line are obtained by inserting the straight line equation in the ellipse equation.

 (30)

Multiplying and sorting the terms after λ yields a quadratic equation for λ:

 (31)

One of the two solutions is already known: it is the point P at which λ = 0. Substituting λ = 0 into (31) yields:

 (32)

Since P lies on the ellipse, this equation is true. Because the last term is 0, it drops from equation (31). We can divide the remaining equation on both sides by λ because λ for the point Q is not zero. Thus we obtain a linear equation for λ:

 (33)

Now we can simply solve for λ:

(34)
with
where'
 $\lambda$ ' =' 'Distance between the two points P and Q $a, b$ ' =' 'Semi axis of the blue ellipse, see (2) $P_\mathrm{x}, P_\mathrm{z}$ ' =' 'Coordinates of the point P, see (3) $s_\mathrm{x}, s_\mathrm{z}$ ' =' 'Direction vector of the connecting line PQ, see(29)

If we insert this λ into the line equation (29), the point Q = (x, z) is obtained:

 (35)

#### Calculating the axes of the red ellipse

In (29) we have chosen $\vec s = (s_\mathrm{x}, s_\mathrm{z})$ so that its length is 1. Therefore, the absolute value of λ is just the distance of the two points P and Q. The semi minor axis now is half of this distance:

(36)
where'
 $b_\mathrm{S}$ ' =' 'Semi minor axis of the red ellipse $\lambda$ ' =' 'Distance between P and Q , see (34)

For further calculations, we need the coordinates of the center O of the red ellipse:

 (37)

For the calculation of the semi major axis aS of the red ellipse, I use the following trick: If the ellipsoid together with the red ellipse is stretched in the Z direction so that the ellipsoid becomes a sphere with radius a and the red ellipse becomes the red dotted circle, the geometry in width does not change. The semi major axis aS of the red ellipse is therefore equal on the ellipsoid as on the sphere. The semi major axis on the sphere can easily be calculated.

To obtain a sphere, all coordinates in the Z direction must be multiplied by the factor a / b. The semi major axis aS of the red ellipse passes through the point O. Therefore, we must extend the coordinates of O accordingly:

 (38)

Next, we need the distance m between $O^{\,\prime}$ and the center M = (0,0) of the sphere:

 (39)

We now obtain the semi major axes of the red ellipse via Pythagoras:

(40)
where'
 $a_\mathrm{S}$ ' =' 'Semi major axis of the red ellipse $a$ ' =' 'Semi major axis of the blue ellipse or sphere, see (2) $m$ ' =' 'Distance of $O^{\,\prime}$ from the center of the sphere, see (39)

### Calculating the radius of curvature of the red ellipse

The desired radius of curvature of the red ellipse (see figure under Calculating the curvature radius of the geodesic) lies at the vertex of the semi minor axis and can therefore be calculated as follows :

(41)
where'
 $\rho_1$ ' =' 'Larger radius of curvature of the cutting ellipse at a rotational angle of γ = 90° or 270° $a_\mathrm{S}$ ' =' 'Semi major axis of the red ellipse, see (40) $b_\mathrm{S}$ ' =' 'Semi minor axis of the red ellipse, see (36)

## Sources

World Geodetic System 1984 ; Wikipedia
https://de.wikipedia.org/wiki/World%5FGeodetic%5FSystem%5F1984
Earth ; Wikipedia(en)
https://en.wikipedia.org/wiki/Earth
World Geodetic System ; Wikipedia
https://de.wikipedia.org/wiki/World%5FGeodetic%5FSystem
Referenzellipsoid ; Wikipedia
https://de.wikipedia.org/wiki/Referenzellipsoid
Schwerefeld ; Wikipedia
https://de.wikipedia.org/wiki/Schwerefeld
Gravitation und Schwere ; Das GOCE-Projektbüro Deutschland am Institut für Astronomische und Physikalische Geodäsie (IAPG) der TU München
http://www.goce-projektbuero.de/7863--~goce~Goce~Produkte~Level_2_Produkt~gravitation_und_schwere.html

Ellipse, Krümmungsradien ; Wikipedia

1number33 1/28/2018 | 13:36

First, I must congratulate you on producing such an excellent calculator. Its thoroughly designed and presented. I especially approve of the inclusion of the mathematical basis for the calculations although some of it is beyond my mathematical ability.

My first instinct when seeing a calculator like this is to apply a sanity test, ie put in some simple numbers and see if they produce a sensible result.

So I set the reference to ground level at the pole and set the plane to be stationary on the ground at the equator and Wcal=100. With a spherical model this produces a weight difference of -0.344% which is exactly what I calculated using the simple formula of w^2 x R/ g. Where w is the earths angular velocity, R is its radius and g = 9.81 m/s^2. So far so good.

I then changed the model to ellipsoid. Since I know that the eccentricity is about 0.34% and the gravitational force is inversely proportional to the square of the distance, I expect this to add approximately another -0.68% giving a total weight difference of about -1%. But your calculator only shows -0.53%. It has only added -0.18%. Theres a large error somewhere. So am I doing something stupid or is there problem with your ellipsoidal model?

I notice that when I switch between Spherical and Ellipsoidal models the value of R, which you say is the distance of the point P from the centre of the earth, only changes from 6371 to 6378 which is only 0.1%.

I also notice that if I change the position of the aircraft to the pole, then R still increases to 6378 for ellipsoidal, surely it should decrease.

2wabis (Walter Bislin, Author of this Page) 1/28/2018 | 16:36

@number33: Thank you for putting my calculations to the test. You made the following mistakes (I admit that on the first version of the article I made the same mistakes):

### Do not apply spherical formulas to an ellipsoid

number33: Since I know that the eccentricity is about 0.34% and the gravitational force is inversely proportional to the square of the distance,...

Here is the problem: Newton's formula g = GM / R2 is only valid for an exactly spherical object of mass M, or if the distance from the object is much greater than the size of the object. Newton's formula is not valid near any other shape!

You can see this on a simple example: Lets take 2 identical spheres of mass m with a distance of 2r between their centers. Lets assume the distance of the observer from the common center between the spheres is d on the extended line between the masses. The gravitational acceleration of the 2-mass system is the sum of the gravitational acceleration to each mass:

 (42)

If we compute the acceleration by simplifying the system to a single mass 2m acting on the center between the masses we would get:

 (43)

This is obviously not the same result. But if in (42) d >> r (d is much grater than r) we can ignore r and get Newtons formula (43) again.

### Why the earth has an ellipsoidal shape

Note that the effective acceleration (vectorial sum of gravitational and centrifugal acceleration) acts always exactly perpendicular to the surface of the ellipsoid. If that would not be the case, there would be acceleration components acting parallel to the surface. On a mostly liquid body that would produce surface streams or tectonic activities until the resulting shape is such that this components vanish. The resulting shape finally is the ellipsoidal shape we observe for each planet above a certain size. Note that the gravitational acceleration vector (not to confuse with the effective gravitational acceleration vector) on the surface of an ellispoid generally does not point to the geometric center.

So to compute the gravitational acceleration near the surface of an ellipsoid we must not use Newtons formula.

To get the formula for any shape, we have to divide the shape into infinitesimal tiny volume elements and compute the acceleration vectors for each element and integrate (sum) them. For an ellipsoid the resulting formula for the effective acceleration is shown at Effective acceleration. To get the gravitational acceleration you have to vectorial subtract the centrifugal acceleration from that.

number33: I notice that when I switch between Spherical and Ellipsoidal models the value of R, which you say is the distance of the point P from the centre of the earth, only changes from 6371 to 6378 which is only 0.1%.

On the ellipsoidal earth the equator radius is a = 6378.137 km and the pole radius is b = 6356.7523142 km. The difference is 21.3846858 km or about 0.34%.

number33: I also notice that if I change the position of the aircraft to the pole, then R still increases to 6378 for ellipsoidal, surely it should decrease.

Please check that again. When I set the position to the pole (latitude = 90°) I get R = 6356.752 km which is less than 6371 km. When I set the position to the equator (latitude = 0°) I get R = 6378.137 km which is more than 6371 km.

The ellipsoidal shape of the earth means not only a bulge at the equator but also a flattening at the poles compared to a corresponding sphere with a mean radius of R = 6371 km and the same volume and mass as the ellipsoid.

3number33 1/28/2018 | 19:21

Thank you for your explanation Walter, I see my error now in assuming that the gravitational attraction behaved as if it was concentrated at the centre of gravity. With hindsight this is obviously not true although I'm surprised that such a tiny eccentricity makes such a large difference in the weight calculation. Now I see why you have that complex formula (5) for gravitational acceleration. Now I just need to understand why the simple formula works ok for a perfect sphere. I'm retired now and my days of using integral calculus are long gone.

As for the value of R, I get the same value as you now. I must have set the reference latitude to 90 instead of the plane's latitude.