RE: [Orekit Users] transformation

Indeed, I just tried the Veis1950 and I get an inclination angle of 9.9e-5°, so it is better than the EME2000. For the earth frame I use ITRF_2005.

Unfortunately, it creates another problem. The computation which previously took about 20s to finish is now taking forever, I did not let it go till the end, but it is more something like 20s per iteration now. Is it something know about the conversion between frames that it takes longer with this frame? Or is it something else?

Vincent

From: pascal parraud [mailto:pascal.parraud@c-s.fr]
Sent: 19 October 2011 18:24
To: orekit-users@orekit.org
Subject: Re: [Orekit Users] transformation

You can try the Veis1950, instead of EME2000, it is the best (quasi)inertial Z aligned with earth frame (ITRF93 for example).
It is supposed to be perfectly Z aligned but, in practice, it does not fit exactly (a quick test shows that a 7000.0 m/s velocity in the ITRF93 equatorial plane gives a 0.01 m/s Z component in the Veis1950).

Pascal

Le 19/10/2011 16:37, Vincent Rivola a écrit :

Dear Pascal,

Thank you for your reply. It confirms what I was thinking but unfortunately I do not know why they are not z aligned.

Do you think I could you other frames than earth and EME2000 which would be z aligned to avoid this problem. If yes, which one?

The important part for me is to be able to have some kind of ENU (East North Up) frame in which I can easily define the velocity vector of my body, and another frame which must be inertial so I can use it to create the orbit element (which only takes inertial frame as input).

Do you have other frames to suggest me which would help me to do this?

Regards,

Vincent

From: pascal parraud [mailto:pascal.parraud@c-s.fr]
Sent: 19 October 2011 16:28
To: orekit-users@orekit.org
Subject: Re: [Orekit Users] transformation

Hello Vincent,

At first sight, and if I understand well the way you do, the non zero inclination angle comes from the transform between the earth reference frame, underlying frame of the TopocentricFrame, and the inertial EME2000 frame. As these two frames are not perfectly "Z aligned", defining the velocity in the equatorial plane of the earth frame doesn't provide a zero inclination orbit in EME2000.

Does it help you ?

Best regards,

Pascal

Le 19/10/2011 10:51, Vincent Rivola a écrit :

Dear Pascal,

Thank you for your reply. I am now facing another problem.
Let me describe my code, I have some information about a body (longitude, latitude, altitude, inclination angle of the orbit, speed) as some point and I want to use orekit to propagate the trajectory.
First step I create a topocentric frame based on the body location:

GeodeticPoint station = new GeodeticPoint(latitudeRad, longitudeRad, altitude);
AbsoluteDate date = getInitialDate(); //defined by the user
TopocentricFrame frame = new TopocentricFrame(earth, station, "initial"); //earth is a OneAxisEllipsoid

Then I create a transformation to go from my topocentric frame to an EME2000 inertial frame:

Transform t=null;
try {
t = frame.getTransformTo(inertialFrame, date); //inertial frame is my EME2000 frame
} catch (OrekitException e) {
         // TODO Auto-generated catch block
      e.printStackTrace();
      }

Then I compute the speed of rotation of the earth at body position which gives me the relativeVelocity vector:

PVCoordinates pv2000 = t.transformPVCoordinates(new PVCoordinates(Vector3D.ZERO, Vector3D.ZERO));
Vector3D relativeVelocity = t.getInverse().transformPVCoordinates(new PVCoordinates(pv2000.getPosition(), Vector3D.ZERO)).getVelocity();

Then I define the coordinates of my body speed in the topocentric frame, but also including the relative velocity due to earth rotation:

Vector3D velocityVector = MakeVelocityENU(flightPathAngleRad, A, velocity, relativeVelocity); //MakeVelocity returns a Vector3D with the coordinates of the body velocity in topocentric frame

And finally I use my transformation created earlier to go from the PVcoordinates in Topocentric frame to the PVcoordinates in intertial frame, and I create a new orbit:
pv2000 = t.transformPVCoordinates(new PVCoordinates(Vector3D.ZERO, velocityVector));
orbit = new CartesianOrbit(pv2000, inertialFrame, date, PlanetEarth.gravitation_coeff);

The problem I have is that the inclination of the orbit at the end of this process is equal to the inclination of the orbit the user is entering. For a very simple case with longitude and latitude (0,0) and an inclination orbit of 0° (so on the equator) if I print the result of "orbit.getI()" I get an inclination angle of 0.0023128007807688327° instead of 0°.
I also noticed that if I change the inital date I use for my computation, the inclination angle is changing.
Do you have any idea how I can solve this problem?

Best regards,

Vincent

On Mon, 2011-10-10 at 11:07 +0200, pascal parraud wrote:

Dear Vincent,

The altitude of a GeodeticPoint is defined with respect to some BodyShape.
If you usually construct the central body, namely the Earth in your case, as a OneAxisEllipsoid,
the only implementation of a BodyShape in Orekit, then the altitude is defined with respect to the underlying geoid.
By the way, the latitude is a geodetic one, it is not geocentric.

Best regards,

Pascal

Le 10/10/2011 10:19, Vincent Rivola a écrit :

Dear all,

I have a question about the altitude considered inside Orekit.

When I create a point with "new GeodeticPoint(lon, lat, alt)", what is the altitude to enter? Is it with respect to the sea, or with respect to the geoid? Or even something else?

While propagating the trajectory, I also use the function "getAltitude" and I have the same question about this. What type of altitude does it return?

Thank you for your help.

Vincent

-----Original Message-----

From: MAISONOBE Luc [mailto:luc.maisonobe@c-s.fr]

Sent: 09 October 2011 21:38

To: Martin

Cc: orekit-users@orekit.org

Subject: Re: Re: [Orekit Users] transformation

Martin <martin.spel@rtech.fr> a écrit :

Dear Luc, thank you very much for the idea.

It is giving the same results as something we were doing before, but this is

more elegant.

For anybody interested, the velocity vector in the ENU frame is build as:

  //velocity expressed in ENU referential

  double cosI = Math.cos(inclinationAngleRad);

  double sinI = Math.sin(inclinationAngleRad);

  double cosFPA = Math.cos(pFlightPathAngleRad);

  double sinFPA = Math.sin(pFlightPathAngleRad);

  double vE = pVelocity * cosI* cosFPA + relativeVelocity.getX();

  double vN = pVelocity * sinI* cosFPA;

  double vU = pVelocity * sinFPA;

The main issue we had is that this gives an orbit with an inclination in the

earth fixed frame. Plotting orbit->getI() gives a different

inclination due to

the east component being different in the inertial frame. The problem is that

in the input we have velocity and FPA being relative, and inclination in the

inertial. We now solved the problem by only allowing all parameters to be

either in the earth fixed frame or in the inertial frame. No more mixing!

One more question: defining an equatorial orbit going westward seems

to pose a

problem. I'm getting errors that the minimum step size is reached. Is this a

known issue?

Yes, perfectly equatorial orbits that are also retrograde cannot be

represented with the current implementation of EquinoctialOrbit. There

is an extension of EquinoctialOrbits that can handle retrograde

equatorial orbits, but it is not in Orekit yet (it will be in the

future). Meanwhile, if you use the current development version of

Orekit from the Git source repository, you can use the numerical

propagator with any orbit types, including Cartesian. Up to Orekit

5.x, numerical propagation always used EquinoctialOrbit internally, so

you hit the retrograde/equatorial problem.

The development version of Orekit depends on the development version

of Apache Commons Math, and will be released only when Apache Commons

Math 3.0 will be released, and this as not yet been decided.

So if you must have equatorial retrograde orbits, you must use both

development versions. If you can use only released versions, you will

not be able to handle equatorial retrograde orbits. I'm sorry about

this.

best regards

Luc

Kind regards,

Martin