“Ever present,

never twice the same;

Ever changing, never less than whole”

~Robert Irwin

If for an instant he dozed off, the image of Fantômas took shape in his mind, but

never twice the same: sometimes he saw a colossal figure with bestial face and muscular shoulders; sometimes a wan, thin creature, with strange and piercing eyes; sometimes a vague form, a phantom—Fantômas!

Marcel Allain, Pierre Souvestre,Fantomas

In the last post, Taming the T-handle continued, we have learned how to model free rotation of an asymmetric rigid body in that very-very special case when the ratio of the angular momentum squared to the kinetic energy doubled is exactly – the middle moment of inertia. In that case only one flip of the axis happens. For an infinite time, in the past, and in the future, the T-handle rotates along the axis, or rather: very close to this axis. In its whole infinitely long life only once it flips, in our case near I say “near”, since what is, say , when compared with infinity?!

Of course in practice we will never be able to achieve this particular ratio of angular momentum and kinetic energy **exactly**! But who cares? In practice there are no perfect straight lines and there are no perfect spheres. So what? When the ratio is “slightly” different, when the parameter that we used in all previous posts, is “slightly” different from , then the motion is quasi-periodic, though I am not sure if it is indeed quasi-periodic according to the mathematical definition of quasiperiodicity. From Wikipedia:

In mathematics, a quasiperiodic function is a function that has a certain similarity to a periodic function. A function is quasiperiodic with quasiperiod if where is a “simpler” function than . What it means to be “simpler” is vague.

In our case there are flips occurring periodically, the pattern repeats itself is a very similar way (though not exactly the same) again and again, the period can be calculated using elliptic -function. We have discussed it before, though we did not stress this phenomenon of “quasi” periodicity.

In Taming the T-handle continued we have derived several important formulas. We have

(1)

where

(2)

(3)

and (for ) is given by:

is periodic, with period But is not periodic at all. The functions and in are periodic, but is a sum of two terms. The first term, linear in , would make periodic. with period , but the second term spoils any periodicity whatsoever. That is why the flips, though governed by the periodicity from the elliptic functions never repeat **exactly** the same way.

### Answers to Reader’s questions

In a comment to the last past Bjab formulated the following hypothesis:

So now there is a hypothesis that:

velocity of the end of the leg of T-handle is always perpendicular to the plane of T-handle.

Always means: when m=1

Proving (or disproving) this hypothesis is a good exercise. My T-handle is defined by the following Mathematica code:

In fact I have two slightly shifted T-handles, of different colors, so that it looks like one but painted top red, bottom yellow. The shift is in -direction, so that my T-handle is all in the plane The leg is along the axis, from -0.9 to 0.2. We can assume that the leg end is at The point moves with time along the trajectory:

The plane of the T-handle is also rotated by the matrix therefore the angle between the velocity vector and the plane is the same as the angle between and the plane. But

Thus has components . This vector is orthogonal to but is at the angle to

Therefore it seems that the hypothesis, though fruitful and interesting, should not be considered as proven.

**Update**: Why ? We know that Thus We want to find the angle, so we are interested in normalized vector. Therefore we can skip the common factor. Thus we see that **the angle is constant in time – which is an interesting fact**. In particular for we have

The square root of the sum of squares is Therefore the normalized first component is We have

“but is at the angle to .”Not possible.

Indeed. I have made a mistake with the components. Fixed. They are Coeeficients in and in front of the same function, are and Their squares add to 1. Therefore of the angle with is . Therefore the angle is

Impossible? Why?

Better, but rather not possible.

“Better, but rather not possible”.

Yes, I know, Trump expects Russia to ‘return’ Crimea to Ukraine.

Why coeeficients in and , in front of the same function, are and ?

It follows from (10) in the previous post when For other values of we will get different and (though the sum of their squares is always 1).

Oh, in your former post we find (10) and (11)

so is 3 times smaller, isn’t it?

Yes, indeed, I have forgotten about the proportionalities of and

Anyway, the end of the leg is in (0,-1,0) when t = – infinity (I suppose)

therefore

= infinty

therefore

then

therefore

we can tell that is perpendicular to

No. It is not perpendicular. After the correction with 1/3 factor, the angle is constant in time and exactly .

Think it over please.

when T-handle has its leg in (0,-1,0)

Zero vector is orthogonal to every vector. And you can’t normalize it. Not very useful.

I do not know, we have a problem, Houston. I have to think.

Where is the limit of this sequence?

The limit is when all mistakes are fixed, and there is only a finite number of them. I think we are rather close.

So it would be good if we know the orientation of T-handle when t = 0.

We know. The angle is constant in time and equal to . I have updated the end of the post.

eguals to what when t=0 ?

is the identity matrix. But not

Not possible.

I think that the problem is with using formulas where start of time (t=0) is set in different way.

Is eqal to or rather equal to ?

OK. You are right. I checked. In my scheme

” think that the problem is with using formulas where start of time (t=0) is set in different way.”

The argument at the end of the post does not depend on the value of $Q(0).$ And I do not see where this argument can go wrong. But I will triple check. There is something fishy …

I rechecked. Can’t find a fault.

so how much is ?

It looks like

for your Q(0) data.

But for t=0

indeed.

But what has to do with ?

On a picture of T-handle in this your post we have the state of T-handle which is (and ) and

Something is fishy.

“…and

Which and what it has to do with

What we need is derivative of and this is .

Can we agree on this?

BTW

What means “The direction of the velocity vector will be the same.” in your post?

It was “fake news” inserted into an otherwise good post. Removed.

Anyway, the direction of movement of the leg on your animations definitely does not look like pi/6.

In the animation I have an extra rotation by the matrix

so that my T-handle rotates about x axis rather thatn z-axis.

OK. I will make a separate animation showing just the normalized velocity and the moving e_1(t) vector, both on the sphere.

Update. I did it. Numerically it comes out as constant pi/3. Need to check again the calculations in the note and fix them.

Update. Done. Everything was OK, only in the last sentence I have exchanged the two components.

Let’s talk about formula:

It is true for every point of the top. There is some corelation between t and .

So for some t should be identity matrix.

For which t is identity matrix?

We will know then how T-handle is position at that time and where its leg is at that time.

is a particular solution of a differential equation. On animation I am selecting it so that the conserved angular is along the x axis. There is no reason for one particular solution to be the identity matrix at some . But we can always choose the unique solution with the property that

It is sufficient to define

Thanks for all corrections!->

->

->

Thanks. Fixed. I am truly sorry for my sins. But even if I try, more sins I do not remember.