SDK Question Reference Frames and Atmosphere

[Calsir]

Hi, I have a couple of question regarding planetary atmosphere and the ecliptic reference frame:

a) is the atmosphere "solidly" rotating with the planet? That is, is

[math]\mathbf{v}_{air} = \omega (R_{P} + h) \cos \lambda \hspace{5pt}\mathbf{x} [/math] where h is the altitude, R_P the radius of the planet, [math]\lambda[/math] the latitude and [math]\mathbf{x}[/math] the vector pointing east?

b) I cannot find how the ecliptic reference frame is defined for a given planet: I understand that x points toward the vernal equinox. Also, since orbiter uses a left-handed system, I assume that y points towards the ecliptic North pole and is the cross product of the two (pointing east, while looking at the vernal equinox from earth, from the northern hemisphere). Is that correct?

I was lying, the questions are three . Is there a function to invert/transpose a matrix in the sdk? I can find ways to multiply matrices and vectors in the api, but no transposition or inversion is allowed. Transposition is sufficient because I need to invert rotation matrices.

Thank you.

 

[n72.75]

I was lying, the questions are three . Is there a function to invert/transpose a matrix in the sdk? I can find ways to multiply matrices and vectors in the api, but no transposition or inversion is allowed. Transposition is sufficient because I need to invert rotation matrices.

I seem to remember a hedder file which included matrix opperations i don't remember where you can get it though. I think it was called Matrix.h or something. Does anyone remember?

-Matt

 

[tblaxland]

a) is the atmosphere "solidly" rotating with the planet? That is, is

[math]\mathbf{v}_{air} = \omega (R_{P} + h) \cos \lambda \hspace{5pt}\mathbf{x} [/math] where h is the altitude, R_P the radius of the planet, [math]\lambda[/math] the latitude and [math]\mathbf{x}[/math] the vector pointing east?

AFAIK, yes, except that the manual states that TAS diverges from GS at altitude. I'm not quite sure what the cause of that is.

b) I cannot find how the ecliptic reference frame is defined for a given planet

The ecliptic reference plane is not defined for a "given planet". It is defined once only as a "global" reference frame.

I understand that x points toward the vernal equinox. Also, since orbiter uses a left-handed system, I assume that y points towards the ecliptic North pole and is the cross product of the two (pointing east, while looking at the vernal equinox from earth, from the northern hemisphere). Is that correct?

To clarify: the +X axis is aligned with the ascending node of the mean ecliptic plane through Earth's mean equatorial plane at the epoch J2000 (the use of "mean" there is because the short term small variations are smoothed out in determining that orientation). The +Y axis is aligned with the ecliptic north pole and the +Z axis is obtained from crossp(+X,+Y).

I was lying, the questions are three . Is there a function to invert/transpose a matrix in the sdk? I can find ways to multiply matrices and vectors in the api, but no transposition or inversion is allowed. Transposition is sufficient because I need to invert rotation matrices.

tmul(...) will transpose and multiply in the one operation. I've not found the need to do a transposition by itself.

 

[Calsir]

AFAIK, yes, except that the manual states that TAS diverges from GS at altitude. I'm not quite sure what the cause of that is.

If we assume a horizontal velocity for the vessel, its GS will be equal to its TAS multiplied by Rp / (Rp + h), since the aircraft is further from the centre of the planet than the ground is. Basically it is like to measure the track of the same angular velocity on two different circles.

The ecliptic reference plane is not defined for a "given planet". It is defined once only as a "global" reference frame.

To clarify: the +X axis is aligned with the ascending node of the mean ecliptic plane through Earth's mean equatorial plane at the epoch J2000 (the use of "mean" there is because the short term small variations are smoothed out in determining that orientation). The +Y axis is aligned with the ecliptic north pole and the +Z axis is obtained from crossp(+X,+Y).

Sorry for the bad wording in both instances, and thank you for the clarification.

tmul(...) will transpose and multiply in the one operation. I've not found the need to do a transposition by itself.

In fact, I found out that I did not need it. I can simply use the coordinate transformation functions .

 

[tblaxland]

If we assume a horizontal velocity for the vessel, its GS will be equal to its TAS multiplied by Rp / (Rp + h), since the aircraft is further from the centre of the planet than the ground is. Basically it is like to measure the track of the same angular velocity on two different circles.

Of course... :facepalm:

 

[Hielor]

If we assume a horizontal velocity for the vessel, its GS will be equal to its TAS multiplied by Rp / (Rp + h), since the aircraft is further from the centre of the planet than the ground is. Basically it is like to measure the track of the same angular velocity on two different circles.

I think that equation is backwards--the TAS is lower than the GS at altitude, since the air around you is moving faster. GS would be TAS multiplied by (Rp + h) / Rp.

 

[tblaxland]

I think that equation is backwards--the TAS is lower than the GS at altitude, since the air around you is moving faster. GS would be TAS multiplied by (Rp + h) / Rp.

Doesn't it depend on your heading? Heading east at altitude you will be going with the wind. Heading west you will be going against it.

 

[Hielor]

Doesn't it depend on your heading? Heading east at altitude you will be going with the wind. Heading west you will be going against it.

No, because when you're in the air, you're moving with the air. Wind direction doesn't matter.

 

[tblaxland]

Wind direction doesn't matter.

Wind direction matters to the ground speed.

 

[Hielor]

Wind direction matters to the ground speed.

Then it's convenient that Orbiter has no wind...

---------- Post added 11-12-09 at 12:10 AM ---------- Previous post was 11-11-09 at 11:51 PM ----------

More specifically, if you think about it in the rotating frame of reference, whether your airspeed is in the direction of rotation or not makes no difference. Think about standing on the edge of a record, with your friend standing halfway between you and the center. (the friend is on the "ground", and your motion with respect to the record represents your TAS, so your motion with respect to your friend represents your groundspeed) If you walk spinward with velocity x, your speed relative to your friend will be the same as if you were walking antispinward with velocity x at the same radius.

 

[Notebook]

Maybe somene should do a Michelson-Moreley experiment, and see if they detect wind...

N.

 

[tblaxland]

More specifically, if you think about it in the rotating frame of reference, whether your airspeed is in the direction of rotation or not makes no difference. Think about standing on the edge of a record, with your friend standing halfway between you and the center. (the friend is on the "ground", and your motion with respect to the record represents your TAS, so your motion with respect to your friend represents your groundspeed) If you walk spinward with velocity x, your speed relative to your friend will be the same as if you were walking antispinward with velocity x at the same radius.

OK, I can see that. So I return to my original question: why does the manual state that TAS diverges from GS at altitude? EDIT: Don't worry, I see it now. In the rotating frame of reference, your apparent velocity (observed from a point closer to the centre) will be higher when closer to the observer than when further from the observer, for the same linear velocity relative to the rotaing frame. Rotating reference frames always mess with my head...

 

[Calsir]

I think that equation is backwards--the TAS is lower than the GS at altitude, since the air around you is moving faster. GS would be TAS multiplied by (Rp + h) / Rp.

Doesn't it depend on your heading? Heading east at altitude you will be going with the wind. Heading west you will be going against it.

The airspeed (TAS) is greater than the ground speed, let me explain with a Gedankenexperiment:

An object is moving at altitude A over the equator. Earth has radius Re. Earth rotation rate is [math]\Omega[/math].

1) this object is not moving with respect to a non-rotating reference frame dragged by earth (for instance, its right ascension stays constant). Airspeed and ground speed will be
[math]TAS = \Omega (R_e + A)[/math]
[math]GS = \Omega (R_e)[/math]
towards West since the only contribution comes from earth rotation. Therefore [math]TAS > GS[/math].

2) the object moves matching Earth rotation (like when it is on the ground):
[math]GS = TAS = 0[/math] since the relative angular velocity is zero.

3) the object is moving in the same direction of Earth rotation (East) at [math]v = 2\Omega(R_e +A)[/math].
again, we will have
[math]TAS = \Omega (R_e + A)[/math]
[math]GS = \Omega (R_e)[/math],
this time towards East.

Earth rotation does not matter in this case, it only matters with the introduction of coriolis and centrifugal effects, which don't play any role in this scenario. Therefore, heading does not come into play either.

 

[Hielor]

An object is moving at altitude A over the equator. Earth has radius Re. Earth rotation rate is [math]\Omega[/math].

1) this object is not moving with respect to a non-rotating reference frame dragged by earth (for instance, its right ascension stays constant). Airspeed and ground speed will be
[math]TAS = \Omega (R_e + A)[/math]
[math]GS = \Omega (R_e)[/math]
towards West since the only contribution comes from earth rotation. Therefore [math]TAS > GS[/math].

2) the object moves matching Earth rotation (like when it is on the ground):
[math]GS = TAS = 0[/math] since the relative angular velocity is zero.

3) the object is moving in the same direction of Earth rotation (East) at [math]v = 2\Omega(R_e +A)[/math].
again, we will have
[math]TAS = \Omega (R_e + A)[/math]
[math]GS = \Omega (R_e)[/math],
this time towards East.

Earth rotation does not matter in this case, it only matters with the introduction of coriolis and centrifugal effects, which don't play any role in this scenario. Therefore, heading does not come into play either.

Okay, you win

I also tested it in Orbiter, and TAS was indeed higher than GS at altitude.