Seeing spin like an artist

In the last note, Towards the road less traveled with spin, we were looking at the asymmetric top spinning about its z-axis, and we have analyzed the path taken by the top, using the stereographic projection. The result turned out to be really boring. A straight half-line, not even worse of looking at. But who or what is the guilty party? The object of the painting, the spinning top, or the painter?

On the net you can find 3 Handy Tips For Learning to See Like an Artist. The second of these tips concerns us:

2. Turn Your Subject Upside Down

We tend to have the most difficulty drawing what we see when we are drawing things we are very familiar with, like facial features, body parts etc.

Because we know we are drawing an eye, for example, instead of relying on what we are seeing, our brain takes over and starts saying “this is what an eye should look like”.

But if we turn the reference (and our drawing) upside down, the features appear much less familiar, and we essentially trick our brain into staying quiet so that we can get on with drawing what we see.

Obviously if you’re drawing from life, you can’t turn your model upside down, but even just turning your drawing around in 90 degree steps can help you see areas where your drawing isn’t quite accurate. You can tilt your head 90 degrees too, to get a fresh look at the model.

And that is what we are going to. But why 90 degrees? We will first try 30, the 60 degrees and see what comes out? And we will also turn around, for even better experience.

You will see that the road becomes more interesting, we will experience something similar to the passengers of the Serra Verde Express

Here are the details.

Suppose we tilt our head 30 degrees left. The effect is the same as if the whole universe, including our spinning top, would tilt right. It is rather difficult to tilt the whole universe without disturbing it. But we can do it in our mind. We do not want to touch the top while it spins. If we do, it will change its rotation, probably it would start to nutate. Like on this movie at 1:18

If we want to have our top rotating about its z-axis, but tilted, we should first tilt it, and only then set in motion respecting its orientation. It is safer to tilt our head or, more scientifically, to tilt the laboratory frame.

Suppose we want to tilt the laboratory frame by 30 degrees about its t-axis. In Towards the road less traveled with spin) we have found that rotation about the z-axis is described by the matrix (see

(1)   \begin{equation*}U(t)=\begin{bmatrix}\cos t+i\sin t&0\\0&\cos t -i\sin t\end{bmatrix}.\end{equation*}

Using the same method we can find that rotation by the angle \phi about y-axis is described by the matrix

(2)   \begin{equation*}V(\phi)=\begin{bmatrix} \cos \left(\frac{\phi}{2}\right) & -\sin \left(\frac{\phi}{2}\right) \\ \sin \left(\frac{\phi}{2}\right) & \cos \left(\frac{\phi}{2}\right)\end{bmatrix}.\end{equation*}

Rotation of the laboratory frame is described by acting from the left. Therefore rotation of the top observed from the tilted reference frame is described by the product V(\phi)U(t). That way we will obtain a trajectory. The axes of the tilted top will never coincide with the axes of the laboratory frame. Therefore the stereographically projected trajectory will never escape to infinity. We can plot it all. But one trajectory would be boring. It is better to tilt the laboratory frame (all the time by 30 degrees) using 50 different tilting axes (all in the xy plane. The following picture is then produced:

If we add trajectories from 60 degree tilts and look from the top, we get something that is not much worse than Serra Verde Express:

We can also try this tip: By simply squinting your eyes as you look at the subject, you can eliminate a lot of the detail, making it much easier to see simple shapes and values.

In the following posts we will be squinting our top.

Towards the road less traveled with spin

We will start with a simple road. Like that in Nebraska in the last post,

except that we take a nice blue sky with some puffy clouds:

Instead of T-handle (as in Taming the T-handle ) we take the nice, asymmetric, but as much symmetric as possible, spinning top, with principal moments of inertia I_1=1,I_2=2,I_3=3.
The x-axis, with the smallest moment of inertia, is along the two bronze spheres. The y-axis, with the middle moment of inertia, is along blue-red line. The z-axis, whose moment of inertia is the sum of the other two, is vertical.

The vertical z-axis is the natural axis to try to spin the thing. Imagine our top is floating in space, in zero gravity. We take the z-axis between our fingers, and spin the device. If our hand is not shaking too much, our top will nicely spin about the z-axis. This is the most stable axis for spinning.

The corresponding solution of Euler’s equation is \vec{\omega}=(0,0,\omega_3), where \omega_3 is a constant. The solution of the attitude matrix equation is Q(t)=\exp(W(t\vec{\omega})).
The square of angular momentum vector is I_3^2\omega_3^2, the doubled kinetic energy is I_3\omega_3^2, the parameter d that we were using in the previous notes is 1/I_3. The parameter m is just zero, m=0. In what follows for simplicity we will take \omega_3=1.

We will start with describing the spin history in \mathbf{R}^3 using stereographic projection described in Chiromancy in the rotation group.

In fact, we already did it in Dzhanibekov effect – Part 1, and Dzhanibekov effect – Part 2, but that was at the very beginning of this series, and we did not know yet what we were doing! So, now we do it again, in a more “legal” way.

Rotation about an axis \mathbf{k} by an angle \theta can be described by (see Eq. (4) in Putting a spin on mistakes) an \mathrm{SU}(2) matrix:

(1)   \begin{equation*}U(t)=\exp(i\frac{t}{2}\vec{k}\cdot \vec{s})=\cos\frac{t}{2}I+i\sin\frac{t}{2}(\vec{k}\cdot\vec{s}).\end{equation*}

In fact in Putting a spin on mistakes I have made a mistake that I corrected only at this moment, when checking it again: I have forgotten the imaginary i in the formula!

We will be rotating about the z-axis, so we take \mathbf{k}=(0,0,1). Then \vec{k}\cdot\vec{s}=s_3. Therefore

(2)   \begin{equation*}U(t)=\begin{bmatrix}\cos t/2+i\sin t/2&0\\0&\cos t/2 -i\sin t/2\end{bmatrix}.\end{equation*}

Comparing with Eq. (1) in Chiromancy in the rotation group, we get

(3)   \begin{equation*} \begin{align} W(t)&=\cos t/2,\\ X(t)&=0,\\ Y(t)&=0,\\ Z(t)&=\sin t/2. \end{align} \end{equation*}

The stereographic projection (Eq. (4) in Chiromancy in the rotation group ) is

(4)   \begin{equation*} \begin{align} x(t)&=0,\\ y(t)&=0,\\ z(t)&=\frac{\sin t/2}{1-\cos t/2}. \end{align} \end{equation*}

We can use trigonometric formulas to simplify:

(5)   \begin{equation*}z(t)=\cot t/4.\end{equation*}

Here is the plot of \cot t/4 from 0 to 2\pi:

At t=0 we have z=\infty. That is OK, because at t=0 the frame of the body coincides with the laboratory frame. The rotation Q(t) is the identity, its stereographic image is at infinity.
At t=2\pi we get x=y=z=0. That is the point representing the matrix U=-I. It also describes the identity rotation in \mathbf{R}^3. We get a straight path, the positive part of the z-axis. It is like the road in Nebraska at the top.


In the next posts we will learn how to travel more dangerous, less traveled roads.

Chiromancy in the rotation group

When we travel by a car, we decide which path to take. Choosing the right path may be crucial. Sometimes we choose a path that is straight, but nevertheless is scary. Like this road in Nebraska:

Sometimes the path is just “unusual”:

And sometimes we will think twice before choosing it:

We are interested in a particular kind of traveling: we want to travel in the rotation group. Think of an airplane. When it flies, it of course moves from a location to a location. But that is not what is of interest for us now. During the flight the plane turns and tilts, and all its turns and tilts can be (and often are) recorded. There are essentially three parameters that need to be recorded:

Usually they are referred to as roll, pitch and yaw. When you throw a stone in the air, and if you endow it with three axes, its roll, pitch, and yaw will also change with time. As there are three parameters, they can be represented by a point moving in a three-dimensional space. Instead of roll, pitch and yaw we could choose Euler angles. But, in fact, we will choose still another way of representing a history of tilts and turns of a rigid object as a path. We will use stereographic projection from the 3-dimensional sphere to the 3-dimensional Euclidean space.

You may ask: why to do it? What is the point? What can we learn from it? And my answer is: for fun.
One can also ask: what is the point with inspecting your hand? And yet there is the whole art of chiromancy, or palmistry. From Wikipedia:

Chiromancy consists of the practice of evaluating a person’s character or future life by “reading” the palm of that person’s hand. Various “lines” (“heart line”, “life line”, etc.) and “mounts” (or bumps) (chirognomy) purportedly suggest interpretations by their relative sizes, qualities, and intersections. In some traditions, readers also examine characteristics of the fingers, fingernails, fingerprints, and palmar skin patterns (dermatoglyphics), skin texture and color, shape of the palm, and flexibility of the hand.

We are going to learn about the lines in the rotation group. Some will be short, some will be long. We will learn how read from them about the body that recorded these lines during its flight.

From the internet we can learn:

Which Hand to Read
Before reading your palm, you should choose the right hand to read. There are different schools of thought on this matter. Some people think the right for female and left for male. As a matter of fact, both of your hands play great importance in hand reading. But one is dominant and the other is passive. The left hand usually represents what you were born with physically and materially and the right hand represents what you become after grown up. So, the right hand is dominant in palm reading and the left for supplement.

The same is with rotation groups. Which one to choose? We have rotations represented naturally as 3\times 3 orthogonal matrices, but we can also represent rotations by quaternions or by unitary matrices from the group \mathrm{SU}(2). Both play a great importance. We will choose unit quaternions or, equivalently, \mathrm{SU}(2), they seem to be primary!

Let us recall from Putting a spin on mistakes that every 2\times 2 unitary matrix U of determinant one is necessarily of the form

(1)   \begin{equation*}U=\begin{bmatrix}W+iZ&iX-Y\\iX+Y&W-iZ}\end{bmatrix}.\end{equation*}

and it determines the rotation matrix R(U):

(2)   \begin{equation*} R(U)=\begin{bmatrix} W^2+X^2-Y^2-Z^2&2(XY-WZ)&2(WY+XZ)\\ 2(WZ+XY&W^2-X^2+Y^2-Z^2&2(YZ-WX)\\ 2(XZ-WY)&2(WX+YZ)&W^2-X^2-Y^2+Z^2 \end{bmatrix}, \end{equation*}

where W,X,Y,Z are real numbers representing a point on the three-dimensional unit sphere S^3 in the 4-dimenional Euclidean space \mathbf{R}^4, that is

(3)   \begin{equation*}W^2+X^2+Y^2+Z^2=1.\end{equation*}

The three-sphere S^3 is “homogeneous”, the same as the well-known two-sphere, like a table tennis ball.

And yet I want to paint one particular point on our three-sphere, namely the point represented by numbers W=1,X=Y=Z=0. It follows from Eq. (2) that the rotation matrix corresponding to this particular point is the identity matrix:

    \[I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.\]

I am not going to hide the fact that also the opposite point, W=-1,X=Y=Z=0, on our 3D ball also produces the identity rotation matrix, that is, in ordinary language: no rotation at all.

We are going now to go back to the stereographic projection that has been introduced already in Dzhanibekov effect – Part 2.

The idea is as in the picture below:

except that our sphere is 3-dimensional, and our plane onto which we are projecting is also 3-dimensional. Our North Pole is now the point W=1,X=Y=Z=0, our South Pole the point W=-1,X=Y=Z=0, and our “equator”, that is the set of all points on S^3 for which W=0 is now a 2-sphere rather than a circle like on the picture above. The projection formulas read (see Dzhanibekov effect – Part 2):

(4)   \begin{eqnarray*} x=\frac{X}{1-W},\\ y=\frac{Y}{1-W},\\ z=\frac{X}{1-W}. \end{eqnarray*}

The inverse transformation is given by

(5)    \begin{eqnarray*} W&=&\frac{r^2-1}{r^2+1},\\ X&=&\frac{2x}{r^2+1},\\ Y&=&\frac{2y}{r^2+1},\\ Z&=&\frac{2z}{r^2+1}, \end{eqnarray*}

where r^2=x^2+y^2+z^2.

All the southern hemisphere (points on S^3 with W<0 are projected inside the unit ball in \mathbf{R}^3! The “equator”, where the three-sphere intersects the three-plane, is projected onto the unit sphere in \mathbf{R}^3. The North Pole has no image, it is the only point on S^3 that has no image. In fact, it has an image, but at “infinity”. We will not need it. The South Pole, which, as we know, represents the trivial rotation, is mapped into the origin of \mathbf{R}^3, the point with x=y=z=0.
We will continue this course of rotational palmistry in the next posts.