Simulating earth rotation and (excess) Lenght of Day calculation

[MountFX]

Hello everybody ,

I'm new here and hope you can help me with this problem. I have to simulate the Earth rotation with eulers equations of motion (without external torques at first).

I have given:

Solution of eulers equation without external torques:

$\omega = (x, y, z)' \left[\frac{rad}{s}\right]$ (angle velocity vector of earth)

with:

$x = r_{earth} \cdot cos( C \cdot (t-t_0) )$
$y = r_{earth} \cdot sin( C \cdot (t-t_0) )$
$z = D~$

Meaning:

• C is a constant dependent on z (is also constant)
• D is constant in case of no external torques
• ω is a time dependent vector. I have many ω (e.g. every minute for a whole year). So every minute I have a new ω.
• ΩN is the nominal Earth rotation rate (which I am not sure how to calculate, I have taken 2*PI/(24*60*60)).
• T is the period of the day in s (24*60*60).

With every new ω (I think I have to take this ω but I am not sure!?) and formula:

$\Delta LOD = \frac{(\Omega^N - \omega) \cdot T}{\Omega^N}$

I have to calculate the LOD (so every minute a new one).

My problem is, that ω is a vector and not a scalar to calculate the LOD. In this special case (without torques), LOD should be constant I think.

Questions

• I think, without external torques the LOD is constant, right?
• Can I calculate ΩN or should I use 2∏/(60*60*60) or should I take this constant from the internet?
• How can I calculate the ω in LOD formula? I tried to calculate the distance (sqrt(x²+y²+z²)), but I think this is the wrong way to solve this.

Ideas

• I can look, when ω has turned around 2∏? That would be one day. Then calculate how long it takes for one sec and use this result in the LOD calculation for ω. But how can I realize that with the stuff I've given? Or is this idea stupid?

Please help me .

Best regards!

 

[MountFX]

Ok, found the mistake after controlling several times all equations. There was an error in my calculation.