semifluid.com http://semifluid.com Intermediate in flow properties between solids and liquids; highly viscous. Mon, 07 May 2012 13:42:03 +0000 en hourly 1 http://wordpress.org/?v=3.3.2 QR code contact information stereogram http://semifluid.com/2012/03/14/qr-code-contact-information-stereogram/ http://semifluid.com/2012/03/14/qr-code-contact-information-stereogram/#comments Thu, 15 Mar 2012 02:17:03 +0000 Steve http://semifluid.com/?p=1235 I felt that my resume/curriculum vitae needed a quick way for people to get in touch with me, so I generated a QR code to encode my contact information. The QR code is scannable using a barcode reader app on most modern smartphones and will automatically pop up with my email and website contact information. I wanted the code to stand out, so, rather than embedding a logo/image into the center, I decided to go the psychophysical route and create a stereogram!

To view, focus eyes behind the image (diverge) or in front (converge, cross-eyed) so that the two images combine into one.

The stereogram was created by shifting a surface of pixels to the left in one image and right in the other (see an example in Wikipedia’s Random Dot Stereogram entry). When the two images are perceptually fused, you perceive the shift as a change in depth of the surface (more info in Wikipedia’s Autostereogram entry). The shift needed to be small enough that the codes would still be scannable (see Wikipedia again), but provide a convincing amount of depth.

I’ve also attached an animated version (also scannable!) using the left and right images as frames (See Lee (1970) Binocular stereopsis without spatial disparity for additional discussion):

]]> http://semifluid.com/2012/03/14/qr-code-contact-information-stereogram/feed/ 0 Switching hosts http://semifluid.com/2012/01/04/switching-hosts/ http://semifluid.com/2012/01/04/switching-hosts/#comments Wed, 04 Jan 2012 18:10:20 +0000 Steve http://semifluid.com/?p=638 I’ve switched hosting providers, so I am currently transitioning my blog from the old provider to the new one. There are a number of broken links and a number of issues that I am trying to resolve (specifically with my old folder organization structure for JAL code), but hopefully this will all be cleaned up soon!

426 - Upgrade Required

Update 01/09/2012: Still slowly transitioning the old posts to WordPress. Unfortunately, there isn’t an easy way (that I am aware of) to convert my old static-page format into the dynamic WordPress format, meaning that I have to upload each image individually and fix the code in the posts. So, unfortunately, it’s taking longer than I would prefer. In addition, I have been transitioning from YouTube to locally hosted videos, but unfortunately, that has been a battle as well. So, things are in a bit of disarray at this point, but will get better soon (I promise!).

Update 02/29/2012: Ok, done! (I hope…)

200 - OK

]]> http://semifluid.com/2012/01/04/switching-hosts/feed/ 0 Repel DEET Numbers Game http://semifluid.com/2011/08/17/repel-deet-numbers-game/ http://semifluid.com/2011/08/17/repel-deet-numbers-game/#comments Wed, 17 Aug 2011 17:45:35 +0000 Steve http://www.semifluid.com/?p=608 Repel produces a number of DEET products intended to scare away those little flying buggers that seem to be everywhere on these warm summer nights. While walking through my local Target, I noticed that there were varying concentrations available for purchase at my local store (please excuse the hand-held photos):

23% DEET 40% DEET 100% DEET

After looking a bit closer at the labels though, I noticed something peculiar in the “Active Ingredient” list…

Although the 23% and 40% concentrations included the proper percentages of DEET in their formulations, the 100% DEET repellent had only 98.11% DEET.  Here are crops from the above photos:

23% DEET Crop
 40% DEET Crop
 100% DEET Crop

All of the other variants include the labeled percentages of DEET, but Repel decided that the 100% variant would not. Instead, it contains 100% DEET* (where the asterisk notes that the percentage is now the “Relative % DEET”).  Hmm… so what exactly does “Relative % DEET” mean?  I could not find any information on the packaging (old or new) nor could I find information on the internet, so who knows?

I called Repel (1-888-880-1181), which is a subsidiary of United Industries Corporation, to find out why there were incongruities with the labeling.  The young woman I spoke with told me that although the 23% and 40% formulations include the percentage stated on the label, the 100% formulation (that only has 98.11% DEET) was marketed as such due to industry nomenclature.  She could not tell me what “Relative % DEET” meant.

It appears as though Repel has realized “Relative % DEET” could not be easily quantified or marketed without the asterisk, so they have removed the percentage sign from newer packaging, labeling the product as simply “Repel 100″.  However, many online stores are still marketing the Repel as containing 100% DEET:

Walmart.com: Repel 100 Percent Deet Pump, 4 oz Amazon.com: Repel 100 1 oz Insect Repellent Pump Spray 100% DEET 402000

Although this is a minor example, some companies allow their marketing departments a bit too much freedom, so it is important to be a savvy consumer (e.g., see: Lawsuit: Taco Bell Ground Beef Is Really Just “Meat Filling”).  So, when looking at varying concentrations, caveat emptor.

]]> http://semifluid.com/2011/08/17/repel-deet-numbers-game/feed/ 4 Calculating the COM of a Spherical Plot http://semifluid.com/2010/06/30/calculating-the-com-of-a-spherical-plot/ http://semifluid.com/2010/06/30/calculating-the-com-of-a-spherical-plot/#comments Wed, 30 Jun 2010 21:15:39 +0000 Steve http://www.semifluid.com/?p=319 What is everyone excited for??? More math! Ok, maybe not, but I’ll come back to the electronics projects later (including investigating the Parallax RFID reader, Arduino FIO, and heck, maybe even some more LED fun). In the meantime, let’s get back to math.

What do the volume calculations look like, and how would one go about calculating the center of mass/geometric centroid of a spherical plot? More after the break…

So, using our good buddy, Wolfram Mathworld, we know that the spherical coordinate volume element, basically the volume we will be using when integrating the solid, is equal to:

dV = \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta

Note that we are going to use the ISO 31-11 convention for our coordinate system, meaning that \theta is the “longitude” and ranges from 0 to 2\pi, \phi is the “latitude” and ranges from 0 to \pi, and \rho is the radius at any given point.

Integrating this around a sphere gives us the volume of the sphere:

V = \int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta

V = \int_0^{2\pi} \! \int_o^{\pi} \! \frac{1}{3} r^3 \sin(\phi) \, d\phi \, d\theta

V = \int_0^{2\pi} \! \frac{2}{3} r^3 \, d\theta

V = \frac{4}{3} \pi r^3

To convert from spherical to Cartesian coordinates, we have to simply use the following:

x = \rho\sin(\phi)\cos(\theta)

y = \rho\sin(\phi)\sin(\theta)

z = \rho\cos(\phi)

So, to solve for the x, y, and z centers of mass, we will need to convert from spherical cooridnates to Cartesian coordinates. We will do this while calculating the function centroid for each axis. The basic formula is:

COM = \frac{\int \! x f(x) \, dx}{\int \! f(x) \, dx}

Our f(x) \, dx in this case is the dV from above and so the basic forms for the COM for each of the three axes is:

COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! x \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}

COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! y \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}

COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! z \, dV}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} dV}

So, substituting our Cartesian coordinate conversions and our dV from above:

COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\sin(\phi)\cos(\theta) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}

COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\sin(\phi)\sin(\theta) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}

COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho\cos(\phi) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta}{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} d\rho \, d\phi \, d\theta}

Simplified:

COM_{x} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi)^2 \cos(\theta) \, d\rho \, d\phi \, d\theta}{V}

COM_{y} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi)^2 \sin(\theta) \, d\rho \, d\phi \, d\theta}{V}

COM_{z} = \frac{\int_0^{2\pi} \! \int_o^{\pi} \! \int_0^{r} \! \rho^3 \sin(\phi) \cos(\phi) \, d\rho \, d\phi \, d\theta}{V}

Solving these for a constant (i.e., r) leads to:

COM_{x} = 0

COM_{y} = 0

COM_{z} = 0

No big surprise, because, as we would assume, the center of mass for a sphere should be at the origin, (0,0,0).

Now, instead of a perfect circle, let’s try a different shape, say defined by:

r(\phi,\theta) = 1+\sin(\frac{\theta}{2})\sin(\phi)

A shape that looks somewhat like a peach and leads to COM values of:

COM_{x} = \frac{-\frac{784}{225}-\frac{41\pi^2}{128}}{5\pi} \approx -0.423

COM_{y} = 0

COM_{z} = 0

Or a spiraling shell:

r(\phi,\theta) = 1+\theta

COM_{x} = \frac{6\pi^2(-3+6\pi+4\pi^2)}{-1+(1+2\pi)^4} \approx 1.165

COM_{y} = \frac{6\pi^2(1+\pi)(-5+2\pi+2\pi^2)}{-1+(1+2\pi)^4} \approx -1.833

COM_{z} = 0

Or perhaps something really out there:

r(\phi,\theta) = 1+\sin(3\theta)\sin(4\phi)+\sin(\theta)\sin(\frac{\phi}{3})

With COM values of:

COM_{x} = 0

COM_{y} = \frac{32383290752127}{62456794017280} \approx 0.518

COM_{z} = -\frac{359881023}{2416474112} \approx -0.149

]]> http://semifluid.com/2010/06/30/calculating-the-com-of-a-spherical-plot/feed/ 0 Calculating the COM of a 3D Surface of Revolution http://semifluid.com/2010/06/29/calculating-the-com-of-a-3d-surface-of-revolution/ http://semifluid.com/2010/06/29/calculating-the-com-of-a-3d-surface-of-revolution/#comments Tue, 29 Jun 2010 20:10:53 +0000 Steve http://www.semifluid.com/?p=242 My current research at Rutgers focuses on identifying the cognitive mechanisms that are employed when perceiving the shape and structure of 3D objects. One of the little problems that I had to overcome was finding out how to calculate the center of mass/gravity of a 3D surface of revolution analytically. Specifically, I needed to calculate the COM for conical frustums with and without an attached part (which was also analytically generated).

Unfortunately, it wasn’t as easy to derive as I assumed, but the steps toward that end are pretty straightforward.  After the break you can see the general equation I ultimately derived for calculating the COM in 3D-space.

A conical frustum will be defined with attributes R_{1} (the top radius), R_{2} (the bottom radius), and h (the height of the truncated cone).

The radius of the frustum at any given height z is:

r_{fr}(z)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2}

The area of this slice at any given height z is:

A(z) = \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r \, dr \, d\theta = \pi r_{fr}(z)^2

Then the volume can be calculated by integrating across all heights:

V = \int_0^{h} \! A(z) , dz

Combine these three equations together and we have:

V = \int_0^{h} \! \pi (\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2})^2 \, dz = \frac{1}{3} h \pi (R_{1}^2 + R_{1} R_{2} + R_{2}^2)

Ok, so now we have the radius, area, and volume of the conical frustum. Using this information, we can now calculate the center of mass.

COM_{x} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r^2 \cos(\theta) A(z) \, dr \, d\theta \, dz = 0

COM_{y} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z)} \! r^2 \sin(\theta) A(z) \, dr \, d\theta \, dz = 0

COM_{z} = \frac{1}{V} \int_0^h \! z A(z) \, dz

Doing a little substitution with COM_{z}:

COM_{z} = \frac{1}{V} \int_0^{h} \! z \pi (\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2})^2 \, dz

= \frac{1}{V} \frac{h^2 \pi (R_{1}^2 + 2 R_{1} R_{2} + 3 R_{2}^2)}{12}

= \frac{h (R_{1}^2 + 2 R_{1} R_{2} + 3 R_{2}^2)}{4(R_{1}^2 + R_{1} R_{2} + R_{2}^2)}

Note that because the conical frustum is radially symmetric around the z-axis, the COM for both x and y will be 0. This obviously will not be the case for shapes with more complex surfaces of revolution, but is a nice little proof.

Complex, you want complex??

So, we will add a simple part onto the shape (like the one in the image above), defined using both a radial function, f_{rad}(\theta), and a vertical function, f_{vert}(z). The functions themselves can be any analytical functions, as long as they are continuous. And frankly, it doesn’t matter for this proof, so we’ll leave them undefined for the time being.

Remember how simple the radius at any given height was? (Just to remind you: r_{fr}(z)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2}). Now we need to add our radial and vertical functions, making the function dependent upon not only height, but also angle:

r_{fr}(z,\theta)=\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(theta)

Given that the radius is now a bit more complex, the area of this slice at any given height z is now:

A(z) = \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta

Then the volume is now a bit more complicated:

V = \int_0^{h} \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta \, dz

And now we can calculate the COMs:

COM_{x} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r^2 \cos(\theta) (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz

COM_{y} = \frac{1}{V} \int_0^h \! \int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r^2 \sin(\theta) (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz

COM_{z} = \frac{1}{V} \int_0^h \! z (\int_0^{2\pi} \! \int_0^{r_{fr}(z,\theta)} \! r \, dr \, d\theta) \, dz

The good news, all you really need to define is r_{fr}(z,theta) and h. The bad news, pentuple integrals. Substituting all of the necessary equations, we get these monstrosities:

COM_{x} = \frac{\int_0^h \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r^2 \cos(\theta) (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}

COM_{y} = \frac{\int_0^h \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r^2 \sin(\theta) (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dr \, d\theta \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}

COM_{z} = \frac{\int_0^h \! z (\int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta) \, dz}{\int_0^{h} \! \int_0^{2\pi} \! \int_0^{\frac{h-z}{h} R_{1} + \frac{z}{h} R_{2} + f_{vert}(z) f_{rad}(\theta)} \! r \, dr \, d\theta \, dz}

Not very pretty, ehh? Well, in the meantime, I’ll leave the final calculations for you to work on. Suffice to say, it is a pain in the butt.

]]> http://semifluid.com/2010/06/29/calculating-the-com-of-a-3d-surface-of-revolution/feed/ 0 8×8 RGB LED Display http://semifluid.com/2010/06/28/8x8-rgb-led-display/ http://semifluid.com/2010/06/28/8x8-rgb-led-display/#comments Tue, 29 Jun 2010 02:21:23 +0000 Steve http://www.semifluid.com/?p=213 So, this project was sidelined until I had to make another BatchPCB purchase.  Thankfully it wasn’t too long until I had the opportunity to work on it again! The current setup is basically 4 of the original 4 RGB LED Controller boards and 12 of the updated DR1r3 boards. All 16 are wired in parallel and being controlled by my desktop machine. You can see an extended version of this RGB test sequence after the break and I’m also including the (uncommented, sorry!) Processing 1.1 code that I used to control the boards.

Here’s an extended cut of the video display (which is displaying an RGB Test Sequence).  Note that the current maximum throughput is approximately 12-13 frames per second due to the RS-232 baud rate bottleneck, but I’m looking for ways to speed up the data transfer without requiring a faster oscillator.:

The Processing 1.1 code:
VideoDisplay (includes the .pde file and the video used)

]]> http://semifluid.com/2010/06/28/8x8-rgb-led-display/feed/ 12 4 RGB LED Controller Update http://semifluid.com/2010/02/16/4-rgb-led-controller-update/ http://semifluid.com/2010/02/16/4-rgb-led-controller-update/#comments Tue, 16 Feb 2010 17:44:18 +0000 Steve http://www.semifluid.com/?p=191 I recently began working on a consulting project that required the creation of some PCBs.  Since I have had such great success with BatchPCB.com in the past, I decided to use them again to fab the custom PCBs.  The BatchPCB purchasing system adds a few static fees (set-up, handling, and shipping), so I felt that this was as good a time as any to make some additional of my PIC16F628 4 RGB LED PWM Controller boards with a couple of modifications.

As I noted in the previous post, there was an error on the first revision of the board and a pull-up resistor on RA5 (pin 4 in the schematic above) was necessary.  I added the MCLR resistor to the board along with a couple of other modifications:

  • Removed the extraneous capacitors, we only need one.
  • Added a breakout for the one remaining I/O pin.
  • Added a small perfboard to the PCB with +5V and ground lines.
  • Relocated the resistors to make them much easier to solder.  I may use a SIL resistor array in the future.

I kept the LED locations exactly the same because, hey, if I put all of the time and effort into the 4 boards that I previously ordered, then I might as well keep the same form factor.  My scheme for the short-term is to create a 4×4 array of the 4 RGB LED Controller boards, which will give me a 20cmx20cm 8×8 RGB LED display.

Here are some pictures of the boards:

Here is the updated schematic and board (note, you can open the BRD and SCH files in Eagle Layout Editor):

And finally, here is an updated firmware that improves the PWM performance:

I will make sure to post when I have the full array put together (I currently only have 12 of the 16 boards I need for the 4×4 array).

]]> http://semifluid.com/2010/02/16/4-rgb-led-controller-update/feed/ 12 Life of a Graduate Student http://semifluid.com/2010/02/15/life-of-a-graduate-student/ http://semifluid.com/2010/02/15/life-of-a-graduate-student/#comments Tue, 16 Feb 2010 01:07:07 +0000 Steve http://www.semifluid.com/?p=186 I have been neglecting semifluid.com for sometime now, but fear not! I will post new projects shortly as I try to dig through the comments.  My time has been split between coursework (which is diminishing), research, and teaching.  As I send revisions of papers back and forth with my advisor, I will devote time to updating the site.  In the meantime, here are 3 videos demonstrating aftereffects that I created for my sensation and perception lab:

]]> http://semifluid.com/2010/02/15/life-of-a-graduate-student/feed/ 0