## Saturday, February 4, 2017

## Tuesday, September 1, 2015

### Sun Tracker

Here is a demonstration that probably won't work, but it was fun to make. It is an image of the path of the corner of my building throughout the year. Each black curve is the path through the day (call it an equi-day curve) and each red curve is the path of the shadow at the same time every day (on your watch) throughout the year (call it an equi-time curve). The green dot should be the current position of the end of a shadow of a vertical pole of height 1 m, and the black line from the origin to the green dot is the shadow. The darker black curve and darker red curve are the current equi-day and equi-time curves (related to the analemma). Note that if you are viewing this when UTC-5 (Eastern Standard Time) is dark, there shouldn't be any figures. Also, you really have to be patient/fight with the CDF plugin to get it to work. But it did work on a random windows machine... Good luck.

## Tuesday, October 7, 2014

### Rolling and Slipping Sphere

The above demonstration illustrates the dynamics of a rigid sphere on a hard frictional surface with friction coefficient $\mu$. I started thinking about this problem after a discussion with students about the following problem. Imagine a spinning sphere (where the rotation axis of the sphere is in the plane of the surface) is dropped from a short distance onto the surface with no initial center of mass velocity. How long does it take to roll without slipping? I worked on this problem before and had a nice demonstration of it to aid my thinking, but I thought the generalization would be very difficult.

Later in the day, my friends Rich C and Howard A and I talked about what would happen when a ball that was spinning at a $45^\circ$ angle with respect to the perpendicular was dropped on the surface. Eventually with enough thinking, we determined the initial velocity of the contact point and I convinced myself that the ball would, seemingly paradoxically, move in a straight line.

Over the next few days, I convinced myself and eventually worked out the following:

- The force and torque on the sphere is set by the initial velocity of the center of mass and the initial angular velocity and
*remains constant relative to the space coordinates while the sphere is slipping.* - Since the force and torque are in the plane (the black and red arrows in the plane at the origin in the demonstration), the initial angular velocity component perpendicular to the surface never changes. A ball that is perfectly spinning like a top won't move.
- The velocity of the contact point is composed of two velocities: the center of mass velocity and the velocity of the contact point relative to the body. These two vectors change according to the force and torque (perpendicular to each other), but they seemingly collude to eventually produce a zero velocity of the contact point relative to the surface (shrinking blue vector) at the point of rolling without slipping. This seems like magic to see happening, but Rich had a good explanation of why this would happen. More on that later.
- Since the force and torque are constant from the very beginning of the problem, both the center of mass velocity and angular momentum vector change according to KINEMATICS. That's all! The velocity changes linearly (2D kinematics) and the angular velocity changes linearly (2D kinematics). The only complication is integrating the Euler angles forward. Originally I had thought to express all force and torque laws in terms of the Euler angles. This lead to 6 terrible coupled nonlinear second-order differential equations that gave me stability problems due to the coordinate singularities. I eventually realized that (just like the Hamiltonian approach) I could solve for the velocity dynamics first, then extract the Euler angles later.
- The time-till-rolling can be calculated by evaluating the cross product of the center of mass velocity (a linear function of time) and the relative velocity with respect to the body (a linear function of time). There is no real reason to do the cross product, but when they are antiparallel, the velocity of the contact point relative to the surface is zero. The cross product is just a nice way to get a scalar from vectors and to do the derivation without dealing with magnitudes of vectors (which are ugly).
- The torque and force laws must be done in the space frame since the center of mass is accelerating and therefore a non inertial frame.

$$t_{\textrm{roll}} = \frac{-v_x \omega_x - v_y \omega_y}{\mu \left( \beta f_y v_x - \beta f_x v_y + f_x \omega_x + f_y \omega_y \right)}$$

After the slipping phase, the sphere rolls without slipping and travels in a straight line.

Some comments about the simulation:

- $\mu$ is the frictional coefficient.
- $\beta = \frac{m r^2}{I}$ is the ratio of mass times the radius squared of the sphere and the moment of inertia.
- $mag$ and $\theta \phi$ are the magnitude and direction of the initial angular velocity (black vector attached to the sphere's center). Horizontal is $\theta$ and vertical is $\phi$ of the spherical coordinates of the angular velocity vector.
- $v_0$ is the initial velocity of the center of mass. Centered in the square is (0,0).
- The track of the sphere over the surface is plotted in red while it is slipping and blue when it is rolling without slipping.
- The simulation time is displayed in the graphic. You can speed up and slow down the simulation by pressing the double up and double down arrows next to the time slider.

Further comment about the center of mass and the velocity of the point relative to the body colluding to cancel:

Rich suggested that from the ball's perspective, the dynamics should be the same as a ball being held by pincers (or pads or parallel compacting planes) on top and on bottom. Since in this instance, the relative velocity of the contact point(s) is the same as a plane (or two compacting parallel planes), the direction of each force will remain the same throughout the dynamics. There is no way for the pincers or planes to do anything other than slow the ball down in the direction that it sees. The advantage of the pads is that now there is no center of mass velocity and it the angular velocity slowly adjusts so that there is no motion at the contact points and thus the ball ends up spinning around the line connecting the contact points.

Rich suggested that from the ball's perspective, the dynamics should be the same as a ball being held by pincers (or pads or parallel compacting planes) on top and on bottom. Since in this instance, the relative velocity of the contact point(s) is the same as a plane (or two compacting parallel planes), the direction of each force will remain the same throughout the dynamics. There is no way for the pincers or planes to do anything other than slow the ball down in the direction that it sees. The advantage of the pads is that now there is no center of mass velocity and it the angular velocity slowly adjusts so that there is no motion at the contact points and thus the ball ends up spinning around the line connecting the contact points.

## Monday, September 29, 2014

### 3 Ball Stress Calculator

Some people may have to make sure to allow/unblock ads to see the cdfs. Additionally you may need to download the Wolfram CDF player.

This calculates the stress applied in a 3 point ball loading configuration according to the formula:

$$\sigma = \frac{-3}{4 \pi} P \frac{\left( X - Y \right)}{b^2}$$

where

$$X = (1+\nu) \ln \left( \frac{r_2}{r_3}\right)^2 + \frac{1-\nu}{2} \left( \frac{r_2}{r_3}\right)^2,$$

$$Y = (1+\nu) \left(1+\ln \left( \frac{r_1}{r_3}\right)^2 \right) + (1-\nu) \left( \frac{r_1}{r_3}\right)^2,$$

$P$ is the applied load in Newtons,

$r_1$ is the radius of the support circle in mm,

$r_2$ is the radius of the loaded area in mm,

$r_3$ is radius of the specimen in mm,

$b$ is the sample thickness in mm,

and $\nu$ is Poisson's ratio.

The radius of the support circle is, as far as I can tell, the radius of the circle produced by the points of the 3 balls supporting the sample. The radius of the loaded area is the radius of the piston on top pressing down on the sample. The radius of the specimen is the radius of the circular sample. From what I can tell, $r_2 < r_1 < r_3$ by definition.

This calculates the stress applied in a 3 point ball loading configuration according to the formula:

$$\sigma = \frac{-3}{4 \pi} P \frac{\left( X - Y \right)}{b^2}$$

where

$$X = (1+\nu) \ln \left( \frac{r_2}{r_3}\right)^2 + \frac{1-\nu}{2} \left( \frac{r_2}{r_3}\right)^2,$$

$$Y = (1+\nu) \left(1+\ln \left( \frac{r_1}{r_3}\right)^2 \right) + (1-\nu) \left( \frac{r_1}{r_3}\right)^2,$$

$P$ is the applied load in Newtons,

$r_1$ is the radius of the support circle in mm,

$r_2$ is the radius of the loaded area in mm,

$r_3$ is radius of the specimen in mm,

$b$ is the sample thickness in mm,

and $\nu$ is Poisson's ratio.

The radius of the support circle is, as far as I can tell, the radius of the circle produced by the points of the 3 balls supporting the sample. The radius of the loaded area is the radius of the piston on top pressing down on the sample. The radius of the specimen is the radius of the circular sample. From what I can tell, $r_2 < r_1 < r_3$ by definition.

## Thursday, October 24, 2013

### Tetrahedral Structures ala Bit-Player

The wonderful blog entry found here by Brian Hayes at Bit-Player was inspirational and led me to an area that I had long thought barren! I recently got my son some of the Geomag sticks and ball bearings and playing with them I got the impression that they were fun, but the models that I could make were limited. Well I was SO wrong.
In his blog entry Brian Hayes introduced me to the tetrahelix, but also issued an irresistible challenge concerning a bridge that he built. He said
"The result (below) is no longer a helix at all but a weird sort of bridge with an arched spine and two zigzag rails. If you extend the arc further, will the two ends meet to form a closed loop? I don’t have enough Geomags to answer that question."
I just couldn't resist. I modelled it on Mathematica and was eventually led to a notation for a chain of attached tetrahedrons. I started by labelling his original figure

I started with arbitrarily numbering the vertices of the tetrahedron on the end. From this, I could specify the next tet by writing down the vertex that was reflected through the plane of the other three. For example, in his picture above, 1 was reflected to 1'. I continued this forward to get the sequence:

I started with arbitrarily numbering the vertices of the tetrahedron on the end. From this, I could specify the next tet by writing down the vertex that was reflected through the plane of the other three. For example, in his picture above, 1 was reflected to 1'. I continued this forward to get the sequence:

{1,4,3,2,4,1,3,4,2,1,4,3,2,4,1,3,4,2,1,4,3,2,4}

The great thing is this repeats! The unit is {1,4,3,2,4,1,3,4,2}. The natural questions now just flood in with this notation. For example, I can do random walks using this sequence. Also, I believe the fact that this structure curves is due to the over abundance of the number 4. For example, the unit {1,2,3,4} leads to the tetrahelix (as do all of the permutations of this unit) which makes me believe that there is definitely some deeper math going on here. For example I believe an even permutation leads to a right-handed helix and an odd permutation leads to a left-handed helix. Anyway, I found with 95 steps using the above unit, the ring comes close, but does not close as you can see below.

This made me wonder if any units can actually close after repeating. I think not, but I haven't checked yet. I imagine that if you are dealing with units, an integral multiple of the unit would have to match up with the full circle. But then, you could meander away and then come back, or you could have a non-repeating number that closed on itself.

Anyway here is a short cdf where you can explore units up to 6.

This made me wonder if any units can actually close after repeating. I think not, but I haven't checked yet. I imagine that if you are dealing with units, an integral multiple of the unit would have to match up with the full circle. But then, you could meander away and then come back, or you could have a non-repeating number that closed on itself.

Anyway here is a short cdf where you can explore units up to 6.

I have not worried about intersection, and if you relaxed the requirement that consecutive numbers aren't equal (this just flips a tet back to where it started) you can make any tetrahedral structure, you just label the tetrahedra sequentially. When you get to the end of a limb, you reflect the sequence till you get back to the main line. This is not the best way of doing things, but it seems to be complete. Anyway, many more things to explore! I think I have to invest in a few more Geomag sets.

## Tuesday, March 26, 2013

### Effective Mass Surface of an Ellipsoid

Here are the ellipsoidal effective mass surfaces.

$E = E_0 \pm \frac{\hbar^2 k_{r}^2}{2 m_e} f(\theta, \phi)$

$E = E_0 \pm \frac{\hbar^2 k_{r}^2}{2 m_e} f(\theta, \phi)$

## Friday, January 18, 2013

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