×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Rigid Body Dynamics: The situation

Rigid Body Dynamics: The situation

Rigid Body Dynamics: The situation

(OP)
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.
Replies continue below

Recommended for you

RE: Rigid Body Dynamics: The situation

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

RE: Rigid Body Dynamics: The situation

(OP)
Greg,

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

Cheers.

RE: Rigid Body Dynamics: The situation

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

RE: Rigid Body Dynamics: The situation

(OP)
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.



RE: Rigid Body Dynamics: The situation

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

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login



News


Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close