I have found in the book "Unmanned Rotorcraft Systems" following expressions (eqn. 2.25 and 2.26).

enter image description here


$\lambda$ = geodetic longitude

$\phi$ = geodetic latitude

But I don't understand how this expression can be derived or where it originates from. In the book they only say it can be derived by using spherical triangles, but do not show how.

I have found a similar equation in the derivation of the transport rate between ECEF-Frame and NED-Frame in this paper: enter image description here But unfortunately it is also not explained, how he got to after the second "="


  • $\begingroup$ Would you add a comment about what $N_E$ and $M_E$ stands for? But I suspect this is simple trogonometry... $\endgroup$
    – U_flow
    Commented Apr 6 at 10:35
  • $\begingroup$ $N_E $ is the radius of the curvature in the prime vertical of the meridional plane. $ M_E $ is the radius of the curvature in the meridional plane. See also the first link , p.25 eqn 2.2 - 2.7 (newbooks-services.de/MediaFiles/Texts/4/…) $\endgroup$ Commented Apr 6 at 14:03

1 Answer 1


The relationship between peripheral speed and angular velocity in 3d is given by

$\vec{v}=\vec{\omega} \times\vec{r}$

therefore in the 2D case this is

$v = r \cdot \dot{\alpha}$

with $v$ being the velocity, $r$ the distance between the center of rotation and the object and $\dot{\alpha}$ being the angular velocity. Solving this for the angular speed we arrive at:

$\dot{\alpha} = \frac{v}{r}$

Given these relationships, we can easily see that in the velocity north case, the angular velocity is simply $\dot{\phi}$, the velocity $v_{nv}$ and the radius the (mean) radius of the earth plus the flight height $r=N_E+h$. Inserting this into the equation above exactly yields the formula you asked about.

For the east case, a $cos(\phi)$ term is inserted into the sum for the radius (resulting in $r=(M_E+h)\cdot cos(\phi)$, as the radius decreases with increasing latitude.

  • $\begingroup$ @Christian Mändel I hope this answers your question sufficiently well. If not, I would have to draw a sketch, but it should be obvious that this is simply the equations for peripheral speed ("Umfangsgeschwindigkeit" in German) applied to the earth. $\endgroup$
    – U_flow
    Commented Apr 8 at 19:49
  • 1
    $\begingroup$ Thank you ! I was having trouble imagining it. You made it very clear. Cheers! $\endgroup$ Commented Apr 12 at 12:17
  • $\begingroup$ @ChristianMändle thanks. If that explains the question, please upvote and accept... $\endgroup$
    – U_flow
    Commented Apr 12 at 15:29

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