## Quaternion derivativegreenspun.com : LUSENET : quaternions : One Thread |

A quaternion q = [ x, y, z, w] where [x, y, z] = sin(a/2)U and w = cos(a/2); a is the angle of rotation around the axis of rotation U. If a rigid body is rotating with angular velocity Omega=[wx ,wy ,wz] around a fixed point O in 3 dimensional space. What is the derivative of q (i.e dq/qt)?

-- Zahi Ahmad Khalil (zkhalil2002@yahoo.com), October 21, 2002

Hello Zahi:I wrote about how to approach Newton's second law in a noninertial, rotating reference frame (URL at end - but now that I look at it again, I worry about the multiplication steps, arg). The short path to the acceleration is to multiply together the following:

(d/dt, Omega) (d/dt, Omega) (0, R) = (d/dt, Omega) (-Omega.R, dR/dt + OmegaxR) = (-d Omega/dt . R - 2 Omega.dR/dt, -Omega.R Omega + d^2 R/dt^2 + d Omega/dt xR + 2 OmegaxdR/dt + OmegaxOmegaxR).

A few things to notice. The scalar is not zero. Still I don't get what that means, sorry. We see some familiar things in the 3-vector, such as: translational acceleration (d^2 R/dt^2), coriolis (2 OmegaxdR/dt), azimuthal (d Omega/dt xR), and centrifugal (OmegaxOmegaxR). Any label for the -Omega.R Omega? Will have to think about that term again...

Your question starts from a different place. Instead of (0, R), you begin from (cos a/2, U sin a/2) - where I am using my convention of the scalar first, followed by the 3-vector which is capitalized. The equation you might be looking at is this:

(d/dt, Omega) (d/dt, Omega) (cos a/2, U sin a/2)

From my broader work with quaternions, that looks like a relativistic noninertial rotating frame simply because there are no zeroes anywhere which is the sign of relativity at work. Yes, I could multiply it out, but to spin at a relativistic speed this late at night is too much!

doug

Starting to doubt the contents here. Anyone want to check it? http://world.std.com/%7Esweetser/quaternions/classical/2ndLaw/2ndLaw.html

-- Doug Sweetser (sweetser@theworld.com), November 06, 2002.

Hello:

This bulletin board understands HTML. I got the space between hello and this paragraph using two break tags. White spaces are not something built into the HTML syntax, but breaks are easy to use.

(Just used a pair of breaks again). If the first term of a second derivative of a spatial vectoriszero, then the force is conservative. That implies many things, such as the force in question can be described by a scalar potential, the energy used is independent of the path, the curl of the 3-vector is zero, and a line integral of a closed loop is zero.

Since the scalaris notzero, the system is not conservative. That is my take on it.

doug

-- Douglas Sweetser (sweetser@theworld.com), January 18, 2003.

Take components of q, then figure out each component as a function of time. Then relate that to the angular velocity. (You'll have to get a general form of U, and I can't see if you'll need partial derivitives)Tom Willis

Advert: www.durham.ac.uk/thomas.willis ... Please visit! :-)

-- Tom Willis (thomas.willis@durham.ac.uk), November 06, 2002.

Doug, considering the first case, looking at a point p =[0,R] fixed in a body rotating with angular velocity W we have: [d/dt, W][d/dt, W][0, R]=[d/dt, W][-W.R, dR/dt+WxR], so far so good! =[-dW/dt.R-2W.dR/dt-W.(WxR),...Scalar Part -W(W.R)+d^2R/dt^2+dW/dtxR+WxdR/dt+WxdR/dt+Wx(WxR)...Vector Part]. If W is the only motion, then dR/dt is perpendicular to W, also WxR is perp to W so W.dR/dt and W.(WxR) are zero in the Scalar part. The first and last terms in the Vector part -W(W.R)+Wx(WxR)=W^2R. Gathering: The second derivative as seen from the fixed frame is: [-dW.R, d2^R/dt^2+2WxdR/dt+W^2R+dW/dtxR] The vector part is the usual second derivative+Coriolis+centrifugal+ azimuthal. We have to find out why the scalar is not zero! Any progress.

-- John H. DuHart III (jduhart@nyc.rr.com), January 17, 2003.

Doug, considering the first case, looking at a point p =[0,R] fixed in a body rotating with angular velocity W we have:------------------- ---------------------------------------------------------------------- [d/dt, W][d/dt, W][0, R]=[d/dt, W][-W.R, dR/dt+WxR], so far so good!-- ---------------------------------------------------------------------- =[-dW/dt.R-2W.dR/dt-W.(WxR)...Scalar Part,---------------------------- -W(W.R)+d^2R/dt^2+dW/dtxR+WxdR/dt+WxdR/dt+Wx(WxR)...Vector Part].--- ---------------------------------------------------------------------- If W is the only motion, then dR/dt is perpendicular to W, also WxR--- is perp to W so W.dR/dt and W.(WxR) are zero in the Scalar part. The-- first and last terms in the Vector part -W(W.R)+Wx(WxR)=W^2R, so:----- ---------------------------------------------------------------------- [d/dt,W][d/dt,W][0,R][-dW/dt.R, d2^R/dt^2+2WxdR/dt+W^2R+dW/dtxR]------ ---------------------------------------------------------------------- The vector part is the usual second derivative+Coriolis+centrifugal+ azimuthal. We have to find out why the scalar, -dW/dt.R, is not zero! Any progress? PS: How do you insert spaces in this email?

-- John H. DuHart III (jduhart@nyc.rr.com), January 17, 2003.

Doug, we have a "MAJOR, IF NOT UNSURMOUNTABLE" problem with your question of what does the scalar part mean! Pure quaternions such as [0,R] are not closed under multiplication! ......Note that [a,B][0,R] = [-B.R, aR+BxR], a quaternion with a scalar part! ....If we apply your operator [d/dt,OMEGA] to [0,R] we get [d/dt,OMEGA][0,R] = [- OMEGA.R,dR/dt+OMEGAxR]. .....Since we know that the physical reality is described by OMEGAxR,(if R is a fixed point in the rotating body dR/dt in the body=0), and we should get a pure quaternion [0,OMEGAxR] for the rate of change of R in the fixed frame. ....After all, changes in vectors are themselves vectors!.......PS: Do you have "KILLING RULES" to take of this!

-- John H. DuHart III (jduhart@nyc.rr.com), January 18, 2003.

Yep, the scalar(Energy/Mass) might actually result from viewing Space/Time stuff happening in a non-inertial frame! Along these lines maybe dtau=Sqrt(dt^2+ds^2+dy^2+dz^2) has to be used instead of dt. Perhaps the scalar would become very small for Omega.R<

-- John H. DuHart III (jduhart@nyc.rr.com), January 18, 2003.

Yep, the scalar(Energy/Mass) might actually result from viewing Space/Time stuff happening in a non-inertial frame! Along these lines maybe dtau=Sqrt(dt^2-ds^2-dy^2-dz^2) has to be used instead of dt. Perhaps the scalar would become very small for Omega.R much much less than c!................................................ Thanks Doug, I haven't been in this kitchen since 1969!..........................PS: Tried the break for space and HTML is sensitive also to

-- John H. DuHart III (jduhart@nyc.rr.com), January 18, 2003.

dq/dt = (1/2)*Omega*q with Omega = [Omega,0]

-- Jim Appenzeller (superstringman@yahoo.com), January 31, 2003.

Two expressions, depending on the coordinate system used to define the angular velocity.1) Let Omega (capital 'O') be the angular velocity resolved along fixed axes. Then

dq/dt = 0.5*Omega*q.

2) Let omega (lower case 'o') be defined as follows:

omega == q\Omega*q

Thus omega is parallel to Omega, but is resolved in rotating coordinates. Obviously q*omega = Omega*q, so

dq/dt = 0.5*q*omega.

All products are quaternion products. The forward slash denotes left- multiplication by the inverse quaternion. Vectors Omega and omega are to be promoted to quaternions with zero scalar parts.

Depending on the application it may be more useful to resolve the angular velocity in rotating or non-rotating coordinates. In either case, however, the meaning of the vector is the same - it is the angular velocity of some object with respect to a particular frame of reference.

-- anonymously answered, May 29, 2003

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