Rigid Body Dynamics: The situation

[themaniac]

Rigid Body Dynamics:
The situation is an aircraft performs a manouver which involves rotation about all 3 axis. The result of this manouver is that any point offset from the center of gravity (CoG) experiences an acceleration that is the combination of the linear accelerations of the CoG and a component due to the rotations about the CoG.

If the aircraft is defined using a standard right-hand coordinate system the equations can be located in any number of dynamics texts and is usually given by:

[a] = [a]CoG + [w] X ([w] X [r]) + [alpha]X[r]

[a]- Acceleration at offset point
[a]CoG - Acceleration at center of gravity
[w] - Angular velocity about center of gravity
[r] - Distance offset from center of gravity
[alpha] - Angular acceleration about center of gravity
X - Cross product

If however the aircraft was to be defined in a left-hand coordinate system would the above equation still apply.

Now, I know that this is the type of question we were all asked in the first year of our engineering studies but that was years ago and i'm just a little rusty on these things.
Any assistance would be appreciated.

 

[GregLocock]

So long as all the axes were consistent then the equation would hold. That is rotation around z axis would be positive if it was from x to y, etc.

There is no inherent physical basis for choosing rh or lh coordinate systems , for this particular job, so far as I know.



Cheers

Greg Locock

 

[themaniac]

Greg,

That is the same conclusion I had come to. I just really wanted comfirmation before i moved on.

Cheers.

 

[astroclone]

Would the Acceleration equation be:

[a] = [a]CoG + [w] X ([w] X [r]) + [alpha] X [r] + 2 * [w] X [rdot]
In place of:
[a] = [a]CoG + [w] X ([w] X [r]) + [alpha]X[r]

T

 

[themaniac]

astroclone,

From my understanding the equation you have posted is correct. The difference is that yours allows for the radius between the center of gravity and the point of interest to change (thus the 2 * [w] X [rdot] term). This of course goes to zero if the distance is fixed, reducing it to the equation i originally posted.
Naturally feel free to correct me if i'm wrong.

 

[astroclone]

Yeah, you're right. I just love the part where you are figuring something out and get to cancel a term like that. Makes me all warm and fuzzy.
T