• # Dietary Intake

Mon 11 March 2013 - 14:10:00 CDT

I was reviewing customer requirements on a recent project, and we came across something unintentionally comical. There was a requirement for illumination intensity which is given as power per unit area, in this case microwatts per square centimeter (uW/cm^2). The problem was that the m was missing leaving uW/c^2.

6 ounces of cake over 10 minutes is roughly 0.28g/s.

Now c is not a unit, but in the context of light, it represents the speed of light with units of meters per second. Just for fun, I plugged this into Wolfram Alpha as “500uW/c^2”. This reduced to having units of kilograms per second which the tool interpreted as “recommended dietary intake”!

This instance was an obvious typo, but it does highlight the need to be careful when authoring and reviewing requirements. Missing a digit, for example, could have far-reaching consequences for your project. And completely wrong units can cause your Mars probe to crash. Better to catch these things early.

And while we‘re on the subject, what do you use for requirements management? I’m guessing Word/Excel is a popular answer. Leave a comment below. This is a topic I will be covering in future posts.

• # User Error

Mon 04 March 2013 - 12:48:00 CST

I’ve been keeping this link around for a while because the first one on the list actually happened to me in the Detroit airport. Well, something very similar. I had a flight leaving soon, wanted coffee, but can’t drink it—especially quickly—when it’s too hot. So I asked the barista to put some ice in the cup to which she replied, “Do you want the ice at the bottom or the top?”

I had no idea how to respond to this except to think to myself, “If you can keep the ice at the bottom, that would be a neat trick.”

At any rate, it is a short, interesting article that reveals a bit about what all scientists, engineers, and product designers are up against sometimes. Remember these when you are:

• planning user research.
• building requirements for your new product.

And contact us if you need another set of eyes on your requirements, need people to help with your STEM initiative, or if you have an idea for something you would like to learn from this blog.

• # Shoreline Waves

Tue 12 February 2013 - 19:05:00 CST

In a previous post I mentioned that an optical plane wave—or any plane wave for that matter—when viewed in the frequency domain has a spatial frequency that depends on its arrival angle. Oftentimes, the data you see in the frequency domain is referred to as a spectrum, and plane waves arriving from different angles form an Angular Plane Wave Spectrum.

Waves near Jacksonville Beach, FL.

Looking at unit conversion, going from the time domain to its corresponding frequency domain converts from seconds to 1/seconds (or cycles per second or Hertz). Space, millimeters, goes to 1/mm (or sometimes line pairs per mm). But how do we go from an angle in radians to 1/mm?

First, consider a shoreline with waves coming in. The waves are comprised of a series of peaks and valleys with some spacing—the wave period—and they are traveling along a line that intersects the shoreline at some angle. If you were to freeze time and walk down the shoreline, you’ll also see peaks and valleys, and the spacing here is what defines the spatial frequency of the wave in the plane of the shoreline. (In optical terms, the “shoreline” could be any plane of interest in the system: an aperture, a lens, a sensor, etc.)

A traveling wave. Note how when the wave is sliced at an angle, the cross section is also a sine wave but with a lower frequency.

The simplest case, of course, contradicts what I just said. If the waves are coming in directly at the shore, as you walk along, the amplitude does not have peaks and valleys. It is constant, or has zero frequency (or is DC if you are an electrical engineer). As the angle of incidence increases, peaks and valleys appear, but they are spaced very far apart from each other. That is they have a low frequency, and the frequency increases as the angle increases. (In the limit, the angle is 90 degrees, the wave is traveling parallel to the shore, and the apparent frequency is equal to the wave frequency.)

Peaks and valleys of a plane wave crossing the “shore” at an angle. The double arrow denotes the period of the traveling wave. The spacing between two peaks along the shore gives the wave period—one over the frequency—in that plane.

Finally, a little trigonometry will give us the frequency of the wave in the angular plane wave spectrum as a function of its angle and period. Look at the same figure with a superimposed triangle. The wave (with period lambda) arrives at an angle, theta, relative to the dashed line which is perpendicular to the plane of interest (i.e. theta is relative to the plane normal). The spacing between peaks in the plane of interest is the hypotenuse of the triangle, and the length of this is the inverse of the spatial frequency, f, in the plane.

The same figure with a superimposed triangle. (Get that trig identity table out.)

So f is given by:

$$f=\frac{\sin(\theta)}{\lambda}$$

This has units of 1/length which is expected for spatial frequency, and the sine function takes care of the angle units.

• # Coins, Math, Things in the Universe

Thu 02 August 2012 - 12:52:00 CDT

A few weeks ago the big news about the Higgs boson came out, and I read an interesting article that talked about the experimental process. Now, I am not a particle physicist. If I’m any kind of physicist, I am a laser physicist—and yes, I know, photons are particles, but I deal more with the wave model. Anyway, more than the particles, what caught my attention right at the beginning of the article was the discussion about probabilities.

A lot of the discussion about Higgs also talks about matter vs. anti-matter and why we exist at all. If a particular particle has an even chance of decaying into, say, a proton and an anti-proton (and some other stuff), and when a proton and anti-proton meet they annihilate, why do we have protons?

This raises the probability question that goes back to the apocryphal fair coin: if you flip a fair coin ten times, what is the chance of getting five heads and five tails (the zero-sum, total annihilation, proton-free nothingness)? As it happens, the chance is a bit less than one in four. How do we calculate this?

Forget about ten flips for now. Putting on my computer scientist hat, ten flips with two possible outcomes each gives a ten-bit representation with 2^10=1024 combinations. Let’s go with four flips, sixteen possible combinations. These are easily written as the four-bit binary values with a round zero as a head and, for lack of a good analogy, a one for a tail:

Of these 16, six have two heads and two tails. 37.5%.

Remember when I said to forget about ten flips? Now forget about that, but go a step further. How do we calculate the probability of total annihilation for M heads in N flips?

First, what is the chance of M heads followed by (N - M) tails? This is just the chance of a head multiplied by itself M times, and that multiplied by the chance of a tail (N - M) times. For a fair coin, head and tail both appear with a probability of ½, so the overall chance is (½)^N. (For the four flips in the table, only the fourth row out of sixteen rows is two heads followed by two tails.)

But we don’t care about the order, so how many different ways can we distribute those M heads in N flips? This is where the binomial coefficient (or sometimes the “choose” notation) comes in. You have N slots. How many ways can you choose M of them?

If you have two slots and want to choose one, you have two possibilities. Three slots, three possibilities. With three slots and picking two, it’s also three possibilities because it is like picking one and then taking the two that you didn’t pick. In general? N-choose-M is written as:

$$\binom {N}{M} = \frac{N!}{M!\left(N-M\right)!}$$

So if you take the probability of getting two heads followed by two tails, 1/16, and multiply that by all the different ways you can pick two slots to put the heads in, 4-choose-2 = 6, you get 6/16=37.5% which is what we simply counted earlier.

Ten flips?

$$\left(\frac{1}{2}\right)^{10} \cdot \left( \frac{10!}{5! \cdot 5!} \right) = 24.6\%$$

Ten billion flips? A hundred billion? 1E1000? Exercise left for the reader.

So based on this math, the chance of getting an even number of protons and anti-protons becomes vanishingly small over eons of decay events. When you factor in that the probabilities of the results are actually unequal—the coin isn’t even fair—the overall imbalance becomes larger still. When you factor in that the universe is expanding, and the chance of x/anti-x annihilation decreases due to just having a lot more space, even more imbalance.

And this is—at least a part of—why there is something.