Is the triangle's shape and size invariant with time? What would you consider representative of it? Its geometrical center? The center of mass of the three objects?

Thanks Kai for answering the question.

The triangles are allowed to change their size with time and the shape becomes distorted if we consider a regular shape at the initial time. Regarding the geometrical quantity that we may consider in characterizing the rotation of this fictive triangle, I have no idea yet. Could you please suggest any idea that could lead to the computation of the angular velocity pseudovector? Thanks!

Would you mind sharing something the context of your question?

Anyway, in rigid but finite size object mechanics, we know that we may at all times decompose its general movement into a translation (represented by the center of mass movement) and a rotation (about the center of mass). So here my guess would be to do the same thing. At any time you can compute the location of the center of mass and its velocity. Also, you can compute the total angular momentum (peudo-) vector (defining the axus of rotation) and the moment of inertia with respect to the rotation axis. From Angular momentum and moment of inertia you compute the angular velocity.

What do you think?

Thanks Kai! Unfortunately data are not available for the center of mass of the 3 particles, I mean of course the instantaneous velocity components. Actually it is not a problem since we do not need this in order to compute the angular momentum of our 3 particles. But how can I find out the axis of rotation based on the computation of the angular momentum? Provided that I know that I think That is it easy to derive the angular velocity pseudovector knowing the moment of inertia. Thanks a lot! you gave me a great idea.

Do you think that I could compute directly omega based on the formula below? I is the moment of inertia tensor of the 3 particles computed with respect to the 3 axis of reference at the center of gravity.

Well, I am interested in the study of particle triplet dispersion in turbulent flow. We have a set of more than 1,330 million triplets (group of 3 particles) released from a localized source initially at a scale comparable to the Kolmogorov dissipative scale eta. We aim to characterize the geometrical evolution and size distortion and also look to the PDFs. We found that is also nice to see what does the rotation of these triangles look like, that is why I tried to find a good formulation of my problem. I think that you gave me the right answer. Thanks!

The data base in fact allows us to have access to more than 27 million million triplets! and thus we could obtain unrepresented statistics but it is really time consuming if calculations are made on a single computer. Do you think that we could collaborate together in that direction? Thanks!

The unit vector along the pseudovector of total angular momentum IS (assumed to be by me) the unit vector of the (momentary) axis of rotation.

Converning center of mass: if you have the individual masses and the individual locations, then you can compute the center of mass.

Edit1: Please don't take this to be the truth yet. I'll give it yet another thought, the same goes for your equation, which is of the kind I envisaged. But I am not totally sure yet.

Edit2: If your liquid is incompressible then you may take the , masses to be the same for the three "particles"? Do these really follow the liquid movement or do they experience friction and inertia om their own right...?

Great thank you. Yes there is no a problem with the center of gravity of the bunch. So we need to see just the moment of inertia with respect to this center of rotation. Thanks!

best

Kai. I have just got a powerful answer in StackExchange; Physics (the first one). It works well for 2D case thus I guess the same for 3D. Thanks!

http://physics.stackexchange.com/questions/109978/how-can-i-compute-the-angular-velocity-of-a-triangle-formed-by-three-particles-k?noredirect=1#comment224759_109978

Abdallah, the answer you got there looks nice and quite general, I agree. And you don't need to use discrete difference evaluations, since you have the velocities. On the other hand, i am not sure whether that answer is sufficiently general and if i were you I'd compare the result with the one obtained by computing the total angular momentum about the center of mass.

The reason for my slight scepticism is the following. Consider the very special example where the three particles move along circular trajectories of constant radius and equal angzular velocity about the center of mass. Then imho you obtain a time independent "^n" vector. Therefore "omega" as defined there would be zero (zero derivative of "^n") despite the fact that this movement describes a rotation. in any case, please cross check my argument, I only gave it 2 minutes!

And one other thing might be interesting, it came to my mind while riding my bike homewards on Friday: whatever the correct method shall be (in the kind of framework we've now been setting up), the method somehow "constructs" an instantaneous, quasi-rigid body (the triangle). Depending on the properties of the liquid I expect there to be length scales (i.e. distances between the particles) below which this idea is resonable and above which it stops to make sence since there may be e.g. be vortices between the particles, so that as a matter of fact their motons become less and less correlated, while the computation shall always give a resulting rotation of some kind. I would expect that such length scales are known in the literature on fluid dynamics and Navier Stokes equations, which I am not quite familiar with. But maybe an evaluation of such motions as you're trying to do could help to evaluate such quantities from time series data? If you had data for more than three particles you could evaluate the motion for each triple (for any n-tuple, in fact) and see to what extent the different resulting motions correlate and when they stop to do so.

You are right Kai. In the example you provided the angular velocity vanishes based on the definition given in StackExchange. I didn't expect that the formula is false, however I think that the problem is in the definition of the vector to be considered. i.e. we need to chose a vector other than the normal to the triangle.

I tried to use the formula you suggested based on the angular momentum and the moment of inertia for 2D (the orientation vector is easy to compute, please see the file in attach here) and I found the good formula (but there is a multiplication by 2?). Thus the method you suggested would lead to the same result as StackExchange if we consider the center of mass of the rod (of triangle) as an origin.

Yes, what we are doing now is to see the evolution of the angular velocity pseudovector for a fixed radius of gyration. We expect to obtain a decay a la -2/3 following the Kolmogorov first similarity law which stipulates that the motion in the inertial range is universally determined by the considered scale. i.e. the angular velocity is just the inverse of the eddy-turn-over-time of the eddies at scale r. We indeed found this result in the case of a rod in turbulence in the inertial range. For large separations (when the distance between the particles becomes larger than the integral scale), there is no such a dependence and the physics is different.

In resume, we face a serious problem with the definition of the vector to characterize the rotation of our triangle. Once this vector is determined, we could then see what does the rotation looks like. Could you please let me know your thoughts on that? Thanks!

Hmm, this is not going to be a full answer....

You might want to ponder what moment of inertia to choose. Your particles do represent volume fractions of a (incompressible?) liquid, don't they. Corresponding masses they represent could therefore be equal. But their movement is in 3D and I would find it more natural to have them represent a 3D volume (sphere) of the liquid instead of a 2D volume (triangle or disk), especially if the rotation you consider is a most general one.

Under the conditions above this makes everything much more easy since the tensor of inertia would be diagonal and just scale with the size (diameter) of the sphere. This sphere could e.g. be chosen to just have the radius corresponding to the largest distance of any particle from the center of mass.

That IS somewhat arbitrary, but maybe you'll have to make a somewhat arbitrary choice, one that (hopefully) suits your goals best.

Edit: maybe you'll have to try and play with different definitions...

Thanks Kai. I really appreciate the idea of the sphere, it could tell us about the separation rate. However, your initial suggestion miraculously do work! I tried to compute the angular velocity based on your definition for your last example and found that the triangle is indeed rotating around z. Here are the formulas (please see below). I took the center of gravity as an origin. Thanks a lot!!

Abdallah, I might take some time later to look into it. I'm really glad if it worked. I'd just share a few further thoughts on the issue.

a) I would have expected the original idea would *have to* work (no miracle involved :-), unless there had been a serious error in consideration.

b) inverting the inertia tensor equation might numerically be unfavorable, though, (at times, i.e. when the three bodies are on one line or so, i.e. in some form of degenerate geometry -- I may be wrong, though) but this can probably be encountered with suitable inversion algorithms.

c) therefore I like the complementary idea of computing the quantities straightforwardly. (i) by using in-plane components only, you get the rotation about the normal to the triangle (around the center of mass) (ii) e.g. by using the "n cross n_dot" equation you get the aditional rotation of the normal (about an axis in the plane) . (iii) these two (pseudo-) vectors add to the total angular velocities, and since they're orthogonal you would easily compute the modulus as the sqare root of the sum of squares of the first two. (iv) in the end this should relate to my first approach straightforwardly and may serve as a cross check.

d) concerning additional information 1: if you look at the in plane movements (in the plane of the triangle) if really the movement was a quasi-rigid rotation about the center of mass (com), then the position vectors relative to the com and the velocity vectors sould be perpendicular (spherical motion of each particle in the plane). Otherwise you have fluxes in and out (while still in the instantanous triangle plane)

e) concerning additional information 2: similar arguments should hold for the rotation of the triangle plane (or its normal vector) it is just determined by the velocity components perpendicular to the plane. Again, it should be possible to determine deviations from a homogeneous rotation about the in plane axis and thus to infer on additional flows.

Good luck with your work. I'm glad I could help.

If the three particles are moving as a solid, then the instantaneous center of rotation is the intersection of the three perpendicuars to the three velocities. The velocity of each particle is the cross (vectorial) product of the radius vector (inknown) from the instantaneous center to the particles with the angular velicity (unknown). In this case you have 3 vectorial equations (9 scallar equatons) and 3 constaints (expressing that distances between partices are constants) in totat 12 equations with 12 unknowns tree radius vectors (9 scalars) of particles from the center and the angular velocity (3 scalars).

In the when three particles are free, their velocities should be divide in two parts: tranlation and rotation. You can find the traslation by calculating the velocity of their center of mass. The remaining veoctiy should be considered rotation relative to center of mass (its posirion you can find). Expressing each rotation velocity as cross product of the radious vector from the center with the angular velocity, you can find that angular veocity.