3D Printed Kinetic Sculpture

3D Printed Kinetic Sculpture from Yui Shiozawa on Vimeo.

Inspired by botanical phyllotaxis, this 3D printed sculpture reveals the abundance of mathematics hidden in nature. Each of 400 similar spikes creates a mesmerizing optical illusion when rotated by the golden angle at the coordinated imaging frequency. When it appears to be not spinning, it rotates by that angle at the imaging frequency. This is a pure videography; no computer graphic animation is involved.

I got the privilege to work on this project with one of my colleagues, Yui Shiozawa. I showed her some work by John Edmark and she agreed to do the math involved in the modeling, so the sculpture is her design. I built the stage to rotate the sculpture and the electronics to sync the LED flashing frequency with the stage rpm. Here’s some cool pictures of the process.

Sculpture getting 3D printed
Sculpture getting 3D printed
Yui is removing the graft from the sculpture and cleaning it up.
Yui is removing the graft from the sculpture and cleaning it up.
I hacked up a broken magnetic stirrer and used a hobby box for the stage.
I hacked up a broken magnetic stirrer and used a hobby box for the stage.
Great job.
Great job.
Really beautiful. Happy with how it turned out.
Really beautiful. Happy with how it turned out.

Special thank you to Nick Bontrager for helping us with the 3D printing process and letting us use his printers at TCU. He’s super great and if you ever get the opportunity to work with him you should take it.

Silver Nanoprisms

I made some silver nanoprisms following this very nice Mirkin paper. Took me a couple of tries to get the sodium borohydride reduction to go nicely. It’s possible to tune the thickness of the prisms by changing the concentration of the NaBH4 added, which changes the plasmon resonance. The leftmost yellow solution is a colloid of spherical particles, then it is the nanoprisms from thickest to thinnest.


Core-Shell Nanoparticles

I’ve been growing some core-shell nanoparticles for one of the projects I’m working on. I’m growing a thin Ag shell on these ceramic nanoparticles. The process is pretty interesting and the SEM images turned out nice so I thought I would share them.

Here are the bare ceramic nanoparticles. You can probably guess their crystal class based on the shape of these.
SEM of ceramic nanoparticles
Here is the seed layer of Ag. You can see beads of Ag forming on the surface of the nanoparticles.
Ag seeding of the NPs
And finally, condensation of the Ag layer. There are some obvious aggregation problems and the shells aren’t quite continuous. I’ve since learned how to deal with these problems and will save those pictures for a publication (hopefully). This picture turned out a bit fuzzy too. Still getting used to this new SEM.
Ag condensation

Magnetic Field Along the Axis of a Current Loop

The magnetic field of a current loop is one of the problems that is easily solved using Biot-Savart Law. This simple derivation is done in most college physics textbooks, but I had a difficult time locating it online, (although I’m sure it exists somewhere.)

Let start with stating the Biot-Savart Law.

B(r) = \frac{\mu_0}{4\pi} \oint \frac{I dl \times \hat{r}}{|r|^2}

Quick reminder of the terms used here. \mu_0 is the permeability of free space, which you should look up and use in the appropriate units if you will be doing calculations. I is the magnitude of current flowing through the loop. The little circle in the integral symbol means that it’s a closed loop integral, so we need to make sure that we integrate dl, the wire length differential, all the way around the loop so that it connects back to itself as a closed loop. r is the distance from the wire to the point of interest we would like to know the magnetic field. However, \hat{r} is the direction that r is pointing. Make sure you understand the difference between these!

This integral may look a bit daunting, but it simplifies very nicely. First, we will convert the cross product to it’s sine equivalent.

B(r) = \frac{\mu_0}{4\pi} \oint \frac{I dlsin(90^o)}{|r|^2}

I used 90 degrees as the angle for the sine function because the direction of dl is always perpindicular to the direction of \hat{r} as we integrate around the loop. That’s a little hard to visualize so maybe stare at the diagram for a bit until you are convinced. We are left with,

B(r) = \frac{\mu_0}{4\pi} \oint \frac{I dl}{|r|^2}

Now we want to focus on the z-axis component of the B-field. We only worry about that part because the other components of the B-field will cancel out with each other due to the symmetry of the loop.

B_z(r) = \frac{\mu_0}{4\pi} \oint \frac{I dlsin(\theta)}{|r|^2}

From trig we can see the sin(\theta) can be replaced in terms of r and R.

B_z(r) = \frac{\mu_0}{4\pi} \oint \frac{I dl}{|r|^2}\frac{R}{r}

We need to get rid of this r and get it into terms of something we are more comfortable with. From pythagorean theorem, we see that r is just,

r = \sqrt{R^2 + z^2}


B_z(z) = \frac{\mu_0}{4\pi} \oint \frac{IR dl}{(R^2+z^2)^{3/2}}

All that is left is to finish the integral dl, which is a circle. So, it’s just the circumference of the circle, 2 \pi R!

B_z(z) = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}

Nice! Notice if we set z=0, then we get the B-field at the center of the loop.

B(0) = \frac{\mu_0 I }{2R}