Zigunov's labbook

Sound Visualization: Nonlinearities – Part III

Posted on May 12, 2020May 12, 2020 by Fernando Zigunov

As discussed in Parts I and II, we established that we can use a Schlieren or a shadowgraph apparatus to visualize sound waves. The shadowgraph is not as interesting an instrument, due to its stronger high-pass behavior. Nevertheless, both instruments are viable as long as one makes it sensitive enough to see the waves.

When reading the previous posts, the keen reader might be weary of my isothermal assumption when applying the ideal gas law. Shouldn’t temperature also be a spatial wave? The consequence, obviously, would be that the waves become sharper with increasing loudness and easier to visualize as the non-linearity kicks in. The reader would be rather right, the non-linearity effect can be significant in some cases. But for reasonable sound visualization, the pressure fluctuation amplitude has to be above about 1 order of magnitude of the ambient pressure in order for this effect to change the wave enough. Let’s look into the math:

For a compressible flow, assuming isentropic sound waves, we have:

$\displaystyle \frac{\rho}{\rho_a} = \bigg(\frac{P}{P_a} \bigg)^{1/\gamma}$

Where $\gamma$ is the specific heat ratio for the gas under analysis. A pressure wave of the form $P=P_a + P_0 e^{i(\omega t-kx)}$ then becomes a density wave of the form:

$\displaystyle \rho=\rho_a \bigg[1+\frac{P_0}{P_a} e^{i(\omega t - kx)} \bigg]^{1/\gamma}$

Taking the $x$ derivative:

$\displaystyle \frac{\partial \rho}{\partial x} = -\frac{i\rho_a P_0/P_a e^{i\omega t} (k/\gamma) \big[1+(P_0/P_a) e^{i(\omega t-kx)}\big]}{P_o/P_a e^{i\omega t} + e^{i kx}}$

Staring at this expression for long enough reveals we can separate the non-linearity term from the remaining terms:

$\displaystyle \frac{\partial \rho}{\partial x} \frac{\gamma}{\rho_a k} = -\frac{i P_0/P_a e^{i\omega t} \big[1+(P_0/P_a) e^{i(\omega t-kx)}\big]}{P_o/P_a e^{i\omega t} + e^{i kx}}=N(P_0)$

The non-linear function $N(P_0)$ represents the wave steepening as the amplitude becomes sufficiently high. Looking at the phasor in the denominator ( $P_o/P_a e^{i\omega t} + e^{i kx}$ ) we see the derivative operator split the time fluctuations from the spatial component of the fluctuation. For our detectability purposes, what really matters is the minimum and maximum value of $N(P_0)$ to get a sense of a peak-to-peak amplitude. In this case, the phasor minimum amplitude happens at:

$\displaystyle \min |N(P_0)|=\frac{P_0/P_a \big[1-P_0/P_a \big]^{1/\gamma}}{1+P_0/P_a}$

Similarly, the maximum happens when the phasor at the denominator minimum point:

$\displaystyle \max |N(P_0)|=\frac{P_0/P_a \big[1+P_0/P_a \big]^{1/\gamma}}{1-P_0/P_a}$

Note this value can even be zero, at about 194dB SPL, in which case these equations blow up to infinity. Nevertheless, the difference gives us the peak-to-peak amplitude of the non-linear derivative of density:

$\displaystyle N_{pk-pk}=\frac{P_0}{P_a} \frac{\big[1+P_0/P_a \big]^{1/\gamma} - \big[1-P_0/P_a \big]^{1/\gamma}}{1-(P_0/P_a)^2}$

Finding the minimum SPL as a function of wavelength is not as simple, since the non-linear function makes it difficult to isolate $P_0$ . However, we can do the opposite – isolate the frequency and plot the points with the axes flipped. To do so, let’s isolate the wavenumber:

$\displaystyle k=\frac{2}{\rho_a} \frac{\partial \rho}{\partial x}\bigg|_{\min} \frac{\gamma}{N_{pk-pk}}$

The factor of 2 comes from the fact that the calculations performed in the previous section considered the amplitude of the sine wave, not its peak-to-peak value. Now we can replace the equations for the Schlieren sensitivity to obtain:

$\displaystyle F=\frac{c}{\pi \rho_a } \frac{\gamma}{N_{pk-pk}} \bigg[\frac{1}{G(\lambda)} \frac{n_0}{L} \frac{a}{f_2} \frac{N_{lv}}{2^{n_{bits}-1}} \bigg]$ (Schlieren)

For the shadowgraph, a second derivative needs to be taken. But as shown in the chart below, the non-linearity only becomes relevant above about 180dB. In the case you get to these SPL values, you probably have a lot more to worry on the compressibility side of things. But it comes as good news that we can use a simple relation (like the one derived in Part I) to draw insight about sound vis.

Comparison of minimum sensitivity for the linear (black) relationship derived in Part I and the compressible, non-linear relation derived here. Parameters are the same: $f_2=2$ m, $a=1$ mm, $L=0.1$ m and $N_{lv}=5$

Comparison of minimum sensitivity for the linear (black) relationship derived in Part I and the compressible, non-linear relation derived here. Parameters are the same: $f_2=2$ m, $a=1$ mm, $L=0.1$ m and $N_{lv}=5$

Schlieren vs Shadowgraph for Sound Visualization – Part II

Posted on May 5, 2020May 5, 2020 by Fernando Zigunov

This continues our saga we started at Part I (spoiler alert: you’re probably better off with a Schlieren). Thanks to a good discussion with my graduate school friend Serdar Seckin, I got curious about applying the same sensitivity criterion to a shadowgraph system. Turns out, Settles’ book also has the equations for contrast in the case of a shadowgraph (equation 6.10). It is even simpler, since the contrast is a purely geometrical relation:

$\displaystyle C_x=\frac{\Delta E_x}{E}=\frac{\partial \epsilon}{\partial x} g$

Where $g$ is the distance between the “screen” (or camera focal plane, in case of a focused shadowgraph) and the Schlieren object. Since the ray deflection occurs in the two directions perpedicular to the optical axis ( $x$ and $y$ ), we get a contrast in the $x$ direction ( $C_x$ ) and another in the $y$ direction ( $C_y$ ). The actual contrast seen in the image ends up being the sum of the two:

$\displaystyle C = C_x + C_y=g \bigg(\frac{\partial \epsilon}{\partial x} + \frac{\partial \epsilon}{\partial y}\bigg)$

Since the ray deflection $\epsilon$ is dependent on the Schlieren object (as discussed in Part I), we can plug its definition to get:

$\displaystyle C = \frac{L g}{n_0} \nabla_{xy}^2 n$

Which is the “Laplacian” form of the Shadowgraph sensitivity. Laplacian between quotes, since it does not depend on the $z$ derivative. We can plug in the Gladstone-Dale relation and the ideal gas law to isolate the pressure Laplacian:

$\displaystyle \nabla_{xy}^2 P = \frac{C R T}{G(\lambda)} \frac{n_o}{L g}$

Replacing the traveling wave form $P=P_0 \exp(i[\omega t - \kappa_x x - \kappa_y y])$ , now obviously considering the two-dimensional wavenumber in the directions perpendicular to the optical axis:

$\displaystyle P_0 = -\frac{\big(\nabla_{xy}^2 P \big)\exp(-i[\omega t - \kappa_x x - \kappa_y y])}{\kappa_x^2 + \kappa_y^2} = -\frac{\big(\nabla_{xy}^2 P \big) P \exp(-i[\omega t - \kappa_x x - \kappa_y y])}{\kappa_{xy}^2}$

Where $\kappa_{xy}^2$ is the squared wavenumber vector magnitude, projected in the $x,y$ plane. Applying the same criterion of $N_{lv}$ levels for minimum contrast, we get the following relation for the minimum detectable pressure wave:

$\displaystyle P_{min} = \frac{RT}{G(\lambda)} \frac{n_0}{Lg} \frac{N_{lv}}{2^{n_{bits}-1}} \frac{1}{\kappa_{xy}^2}$

Which we can easily convert to minimum $SPL$ based on the wave frequency $F$ :

$\displaystyle \boxed{SPL_{min} = 20\log_{10} \bigg(\frac{RTc^2}{4\pi^2 P_{ref}G(\lambda)} \frac{n_0}{Lg} \frac{1}{F^2} N_{lv}\bigg)- 6.02(n_{bits}-1)}$

design consequences

The boxed formula for shadowgraph is very similar to the one for Schlieren found in Part 1. By far, the most important point is the dependence on $1/F^2$ [Shadowgraph] instead of $1/F$ [Schlieren]. This is obvious, since Schlieren detects first derivatives of pressure, whereas Shadowgraph detects second derivatives, popping out an extra $F$ factor.

From a detector apparatus perspective, however, this is not of the most desirable features: The shadowgraph acts as a second-order high-pass spatial filter, highlighting high frequencies and killing low frequencies. The Schlieren has a weaker, first-order high pass filtering behavior. But this means they have a crossing point, a frequency $F$ where the sensitivity is the same for both. The apparatus parameters are different, however we can do the exercise for the rather realistic apparatus given in Part 1. If we make $g=0.5$ m (i.e., the focal plane is 0.5 meters from the Schlieren object), we get the following curve for a 12-bit camera:

Minimum detectable sound levels – Schlieren versus Shadowgraph. Schlieren parameters: $f_2=2$ m, $a=1$ mm. Shadowgraph has $g=0.5$ m. Both consider $N_{lv}=5$ and $L=0.1$ m.

Minimum detectable sound levels – Schlieren versus Shadowgraph. Schlieren parameters: $f_2=2$ m, $a=1$ mm. Shadowgraph has $g=0.5$ m. Both consider $N_{lv}=5$ and $L=0.1$ m.

If the news weren’t all that great for the Schlieren apparatus, the Shadowgraph is actually worse: The crossing point for this system occurs at $F=200$ kHz. There might be applications where these ultrasound frequencies are actually of interest, but I believe in most fluid dynamics problems the Schlieren apparatus is the to-go tool. Remembering that, even though I show SPL values above 170dB in this chart, the calculations considered a constant-temperature ideal gas, which is not the case anymore at those non-linear sound pressure levels.

With this, I’m declaring Schlieren a winner for sound-vis.

Seeing Sound Waves with your Schlieren apparatus – Part I

Posted on May 1, 2020May 5, 2020 by Fernando Zigunov

Tell me: Would you be excited to see sound? Well, I sure damn was when I first attempted to see the sound produced by an ultrasonic transducer! And still am, to be honest! So let’s learn how to do it with a Schlieren apparatus and the sensitivity considerations necessary in the design of a Schlieren optical system. For this, I’ll use my handy Schlieren and Shadowgraph techniques by no one else than the great Prof. Gary Settles. My PhD advisor, Prof. Alvi, was his PhD student – so I should get this right!

I’m a practical engineer, but I’ll derive the formulas in this section. If you want, you can skip ahead to the next section where I apply them and produce nice design charts.

the math

If you would like more details on the physics, please refer to Settles. This is more of a derivation section. We’ll start with the Gladstone-Dale law, which relates the index of refraction of a gas to its density:

$\displaystyle n-1 = G(\lambda) \rho$

Where $n$ is the index of refraction, $G(\lambda)=2.2244\times10^{-4}\big[1+(6.37132\times 10^{-8}/\lambda)^2\big] m^3/Kg$ (for air) is the wavelength-dependent Gladstone-Dale constant and $\rho$ is the density of the gas.

Assuming a parallel light beam, the beam deflection $\epsilon$ of a uniform Schlieren object of length $L$ is given by:

$\displaystyle \epsilon = \frac{L}{n_0} \frac{\partial n}{\partial x}$

Where $x$ is the direction of deflection. This deflection is magnified in the focal point of the optical apparatus mirror/lens close to the camera/screen; where the knife-edge is placed, according to:

$\displaystyle \Delta a=f_2 \epsilon$

Where $\Delta a$ is the change in the focal point location in the direction perpendicular to the knife edge and $f_2$ is the focal length of the second optical element (lens/mirror). Then, the contrast change $C$ observed in the camera/screen is given by Equation 3.6. in Settles’ book:

$\displaystyle C=\frac{\Delta E}{E} = \frac{f_2 \epsilon}{a}$

We can use the ideal gas law to combine the expressions as a function of $\partial P/\partial x$ :

$\displaystyle \frac{\partial P}{\partial x} = \frac{RT}{G(\lambda)} \frac{a}{f_2} \frac{n_0}{L} C$

We can then assume we are looking for sinusoidal sound waves of the form $P(x,t)=P_0 \exp(i[\omega t - \kappa x])$ :

$\displaystyle \frac{\partial P}{\partial x} = i\kappa P_0 \exp(i[\omega t - \kappa x])$

OK. Now let’s stop here for a while. I think it goes without saying you’ll probably need a high speed camera to see the sound waves, as they aren’t among the mundane things you can use your regular DSLR to observe. Given that, we can consider whichever camera you’ll use has some bit depth $n_{bits}$ . When setting up your Schlieren apparatus, you’ll probably use half of the dynamic range of the camera for the “background” frame. This means, your intensity $E$ is about half of the dynamic range of the camera. In bit count, you’ll have $\displaystyle E=N_{BG}=2^{n_{bits}-1}$ as the background, undisturbed luminance level. Now, let’s assume you need at least $\Delta E = N_{lv}$ levels to distinguish the signal from the noise. Maybe you have some de-noising method, like POD, that improves your SNR and your sensitivity. How loud do the sound waves need to be in order to produce a $N_{lv}$ signal given you have a $N_{BG}$ background luminance?

Let’s isolate $P_0$ :

$\displaystyle P_0 = \frac{\partial P}{\partial x} \frac{1}{\kappa} \underbrace{\big\{-i\exp(-i[\omega t - \kappa x])\big\}}_{\text{included in } N_{lv}} = RT \frac{1}{\kappa} \frac{1}{G(\lambda)} \frac{a}{f_2} \frac{n_0}{L} C$

Using our bit depth, we get the minimum pressure disturbance to be:

$\displaystyle P_{min} = RT \frac{1}{\kappa} \frac{1}{G(\lambda)} \frac{a}{f_2} \frac{n_0}{L} \frac{N_{lv}}{2^{n_{bits}-1}}$

Converting to sound pressure level ( $SPL$ ):

$\displaystyle SPL_{min} = 20 \log_{10} \bigg(\frac{P_{min}}{P_{ref}} \bigg) = 20 \log_{10}\bigg({\underbrace{\frac{RT}{P_{ref}}}_{1/\rho_{ref}} \frac{1}{\kappa} \frac{1}{G(\lambda)} \frac{a}{f_2} \frac{n_0}{L} \frac{N_{lv}}{2^{n_{bits}-1}}}\bigg)$

$\displaystyle SPL_{min} = 20 \log_{10}\bigg(\frac{RT}{P_{ref}}\frac{1}{\kappa} \frac{1}{G(\lambda)} \frac{a}{f_2} \frac{n_0}{L} N_{lv}\bigg) - 20 (n_{bits}-1) \log_{10}(2)$

Evaluating the $\log_{10}(2)$ and replacing $\kappa = 2\pi F/c$ :

$\displaystyle \boxed{SPL_{min} = 20 \log_{10}\bigg(\frac{cRT}{2\pi P_{ref}} \frac{1}{G(\lambda)} \frac{1}{F} \frac{a}{f_2} \frac{n_0}{L} N_{lv}\bigg) - 6.02(n_{bits}-1)}$

A capital $F$ is used for frequency to eliminate confusion with $f$ , which in this context is the mirror focal length. If you know the background level, $N_{BG}$ , then you might want to expose it in the formula instead of the bit depth of the camera pixels:

$\displaystyle SPL_{min} = 20 \log_{10}\bigg(\frac{cRT}{2\pi P_{ref}} \frac{1}{G(\lambda)} \frac{1}{F} \frac{a}{f_2} \frac{n_0}{L} \bigg) + 20 \log_{10}\bigg(\frac{N_{lv}}{N_{BG}}\bigg)$

usage in design

The boxed formula is extremely useful for sound visualization design, as it allows us to define whether we can even see the sound waves we want to. First, a few remarks: I separated the bit depth because the $6 n_{bits}$ appears in so many places. In this case, we have it with a negative sign, which is awesome, since in the end we want to minimize $SPL_{min}$ to maximize sensitivity. The first fraction inside the $\log()$ is a constant, dependent on the gas and the reference pressure. For air at 25ºC, $cRT/2\pi P_{ref} = 2.335\times 10^{11} m^4/Kg\cdot s$ .

Note that the minimum SPL is inversely proportional to the frequency $F$ . This means, higher frequencies are easier to see. Ultrasonic frequencies, in particular, are somewhat straightforward to see, as we will discuss. Let’s plot $SPL_{min}$ as a function of frequency for a typical large-scale Schlieren apparatus:

Sensitivity of an air-based Schlieren aparatus with $f_2=2$ m, $a=1$ mm, $L=0.1$ m and $N_{lv}=5$

Sensitivity of an air-based Schlieren aparatus with $f_2=2$ m, $a=1$ mm, $L=0.1$ m and $N_{lv}=5$

The black line represents the bit depth of most contemporary scientific cameras. As you can see, the news aren’t that great: For low frequencies, we have to scream really loud to see anything – and this is after blowing up the image such that we can only see 5 levels! For example, for a frequency of 5kHz we have about 128dB of minimum sound pressure level to be perceived by our camera. It’s not impossible, and with the advent of new data processing techniques like POD it is well feasible. But wouldn’t it be great to have 4 more bits per pixel (red dashed line)? That would bring the minimum SPL to 104dB, which would enable the visualization of a lot of phenomena (and even a rock concert!).

Well, I hope this helps you design your own Schlieren apparatus. Or maybe you lose hope altogether and quit – but at least this saved you time! If you want to design your own Schlieren setup to visualize sound waves, you can download the code that generated the chart above here. Anyhow, here’s a little video of how sound waves look like for a really loud (~140dB, ~5kHz) supersonic jet:

The anti-gravity piddler: A demonstration of aliasing

Posted on April 25, 2020May 1, 2020 by Fernando Zigunov

So you’ve probably already seen demos on Youtube showing this really weird “camera effect” where they stick a hose to a subwoofer and get the water to look like it’s being sucked back to the hose, seemingly against gravity.

I personally love this effect. In the case of the subwoofer, the effect is due to what is technically called “aliasing”. Aliasing is an effect important to all sorts of fields, from data analysis to telecommunications to image processing. In technical jargon, you get aliasing when you don’t satisfy the Nyquist criterion when sampling your signal. This might not be accessible to everyone, so I’ll explain it differently.

In the case of the hose stuck to the subwoofer, the speaker shakes the hose back and forth with a single frequency (a single tone) and generates a snaking/spiraling effect on the water stream. If the water stream is slow enough (that is, if its Reynolds number is low enough for the flow to be laminar), then no weird stuff (non-linear effects) occur, and we get a simple, single-toned spatial wave in the water jet. That being the case, the cycles repeat very nicely, becoming indistinguishable from each other. If you can fulfill this criterion, then aliasing also occurs in a nice manner, that is, if you happen to fail to satisfy the Nyquist criterion, you don’t get a jumbled mess but a nicely backward or forward motion that looks like it’s in slow motion.

It is a simple thing to do, but there’s some beautiful fluid dynamics on it. Generating repeatable patterns and laminar flows is not that simple, especially when you are engineering a device. If you attempt to reproduce the video linked, you’ll find yourself suffering through a parametric search of flow rate/shake amplitude until you get the right combination that displays a nice effect.

Here, I’ll discuss a different device, though – I’ll talk about the piddlers of Dr. Edgerton, that inspired awe in many people around the world – including myself. I have never personally seen one. But I understood what it was and that the effect was not just a camera artifact, but something that could be seen with the naked eye because they use a stroboscopic light to show the effect to the viewer. I have not – as of 2018 – found any instructions on how to make these. And since it turns out it’s quite simple, I think it should be popularized. Here’s my take:

the piddler design considerations no one talks about in practice

The water piddler is a device that generates a jet of water droplets. Water jets are naturally prone to breaking down into droplets – if you’re a man you know it! But on a more scientific tone: Water jets are subject to a fluid dynamic instability called the Rayleigh-Plateau instability. This document here is an incredible source that enables the prediction of what are the conditions for this unstable behavior without hassling you with all the complicated math behind. The Rayleigh-Plateau instability looks like shown below:

Naturally occurring Rayleigh-Plateau instability in my kitchen

It’s beautiful – but it is also not really a single frequency. There seems to be some level of repeatability to it, but not enough to make the strobe light trick work. The reason for this non-repeatability is the following curve:

Dispersion relationship for the Rayleigh-Plateau instability

This is the dispersion relationship, extracted from equation (23) of Breslouer’s work. It corresponds to an axisymmetric disturbance in the radius of the jet – $R(z,t)=R_0 + \varepsilon e^{\omega t + ikz}$ , where $\varepsilon$ is a small disturbance in the radius $R$ and $z$ is the streamwise coordinate. $k$ gives us a wavenumber of the ripple in the jet, and $\omega$ gives us a frequency of this disturbance. The dispersion relation normalizes the wavenumber $k$ by the original radius of the stream, $R_0$ . When $\omega >0$ , we get an exponentially growing disturbance in the radius, which eventually makes the jet break down into droplets. So the black curve in the chart shows that any disturbance between $0<k R_0<1$ will grow, but higher frequency disturbances will not – they simply oscillate. Disturbances closer to the peak at $kR_0 = 0.697$ will grow faster, which is an important design guideline when we want to break down the jet into a stream of droplets.

The problem, though, is that any other frequencies around the peak also grow. The peak is somewhat smooth, so there will be a lot of space for non-uniformity, especially when the disturbances themselves start at different amplitudes.

So what would be a good design procedure? Well, first, we need to make sure the jet will be laminar. One way to guarantee that is to make the Reynolds number of the nozzle that makes the stream lower than 2000. That guarantees the pipe flow is laminar, which in turn makes the stream laminar. Of course, this is a little limiting to us because we can only work with small jet diameters. You can try to push this harder, since the flow inside the stream tends to relaminarize as the stream exits the nozzle due to the removal of the no-slip condition generated by the nozzle wall.

The other constraint has to do with reasonable frequencies for strobing. You don’t want to use too low of a strobe frequency, because that is rather unbearable to watch. Strobe frequencies must be above 30Hz to be reasonably acceptable, but they only become imperceptible to the human eye about 60Hz. We get a design chart, then:

Design chart for f=60Hz, water/air, 25ºC

The chart shows the growth rates (real part of $\omega$ ) for combinations of realistic jet diameters and velocities, which are the actual design variables. The line of constant Reynolds number looks like a 1/x curve in this space. The white line shows the upper limit for laminar pipe flow. You want to be under the white line, as well as in the growing region, which is about the region enclosed by the dashed line. For higher frequencies, the slope of the $\approx 45^\circ$ black boundary decreases, meaning you need smaller diameters to make the strobe light work. For lower frequencies, the slope increases, improving the available parameter space, but too low frequencies will be uncomfortable to watch. In case you want to develop your own piddler, a Matlab implementation to generate the colorful chart above is here.

It is actually rather remarkable that the parameter space looks like this, because feasible diameter/frequency combinations actually will break down into droplets if excited with 60Hz – the line frequency in the US. Say, for example, for 3mm jet diameter and 1m/s speed, we have a high growth rate and the piddler will produce a nice effect. At 6mm, 0.5m/s, we still have laminar flow but the instabilities won’t grow at 60Hz (lower frequency instabilities grow instead). Thus, you’ll not get a good piddler out of that combination. You might be able, for example, to push the bounds a bit (which I did) and make the jet diameter 4.75mm and the jet speed about 1.2m/s. In that case, the Reynolds number is about 3200, which still makes a reasonably repeatable piddler pattern.

Another thing you can attempt (I did) is to try to use a more viscous fluid. More viscous fluids will increase the viable diameter/velocity combinations where the Reynolds number is still low by pushing the white line up and to the right. For example, propyleneglycol allows us to approximately double the diameter of the pipe. The problem, obviously, is that it’s incredibly messy!

now to the real world

The design map is a good guideline to start this, but there are a few tricks that no one really describes in the internet. I’ll save you weeks of suffering: The best way to generate the disturbance is to use a coffee machine pump. Yes, a coffee machine pump! It accomplishes the two tasks for this device: Recirculating the fluid and generating a strong enough, single-frequency disturbance such that you don’t really need to trust the Rayleigh-Plateau instability alone to generate the droplets.

This is the basic implementation of the piddler I built. The coffee machine pump is a ULKA-branded one (like this one). I believe it doesn’t really matter which model you use, since mine had too much flow rate and I had to reduce the flow with the flow control valve indicated in the schematic. These pumps are vibratory pumps. They function by vibrating a magnetic piston back and forth with a big electromagnet. The piston pumps a small slug of fluid, that goes through a check valve inside the pump. When the piston returns, the suction shuts off the check valve, preventing back-flow. Suction fills the piston cavity of new fluid, and the cycle repeats.

Since the piston is pumping the fluid in tiny slugs at the correct frequency (assuming you have a viable Rayleigh-Plateau instability on your design), an acoustic wave will go through the water until the nozzle, generating the intended velocity disturbance in the mean flow. It will not be a choppy flow, but an oscillating one, due to the really strong surface tension forces in water. The figure in the left shows that there’s a full wave of instability before the stream breaks down into droplets.

Now that we discussed the design, let’s go for a little demo. In this video, I’ll also go through the z-Transform, which is a cool mathematical tool for modeling discrete-time control systems. I used this piddler as a prop for the lecture!

Smoke rings to the tune of AC/DC

Posted on April 23, 2020May 12, 2020 by Fernando Zigunov

So I’ve been spending quite a bit of time thinking about vortex rings. Probably more than I should! I decided I wanted something that shot vortex rings filled up with smoke, but in a way that can last for very long periods of time. I came up with this idea that if I had an ultrasonic mist maker, I would be able to generate virtually endless fog that I could use for this. As a tease, this is what I came up with:

Working on a room humidifier that shoots smoke rings to the tune of AC/DC – what do you think? from functionalprint

Video also available here. Smoke in the Water here.

So hopefully you also found this cool! To be honest, I don’t know how this isn’t a thing yet – this could very well be a product. (i.e., it still performs the function as a room humidifier, but vortex rings are just cooler!). Ok, so how did I do it?

What’s a vortex ring?

A vortex ring is a (rather important) fluid mechanical structure. It is present in pretty much all realistic flows and plays an important role in turbulence. Generating one is simple: When fluid is squeezed through a round nozzle, it forms a temporary jet that “curls” around the edges of the nozzle. Since the nozzle is round in shape, the curling happens all around, like a ring of spinning fluid. If the “squeezing” stops, the curling continues, though, through inertia. One thing we learn in fluid mechanics is that a vortex (this curled fluid structure) induces a velocity everywhere in the flow field – i.e., it tries to spin everything around it. If the nozzle blows upward, the left-hand side vortex core induces an upward speed on the right-hand side. The same happens from the right-hand side vortex core, it also induces an upward speed on the left-hand side. It actually happens all around the circle, meaning the vortex ends up propelling itself upward.

If the flow of the vortex ring is somewhat laminar and we seed it with smoke, we can see the vortex ring propelling itself as a traveling ring (as in the video) because it persists for quite a long time. Eventually, it becomes unstable and stretches until it twists and crosses itself, rapidly breaking down to tiny vortices and spreading itself in a turbulent cloud.

How do I make one?

You need a means of generating smoke. Smoke machines used in stages / parties is generally the easiest way to get started. You fill a bucket with smoke, have a hole about 1/4 of the bucket diameter on one end, and then tap the opposite end. This replicates the “squeezing” process described before. It is not really an ideal solution, though, because the smoke fluid has to be replenished quite often. Plus, routing the smoke from the machine to this device that produces the smoke rings is not really easy (the smoke condenses in the walls of a pipe and forms a lot of oil in it).

So this idea struck me. If I use an ultrasonic fog generator (like this one), then I can produce ungodly amounts of smoke from a relatively small water tank. This smoke can last for hours and be stored in the water tank to increase its density. This is what I came up with:

A speaker is connected to a little water bucket (an ex-ice cream container) through this funky-looking black 3d-printed part. It’s just a duct that adapts the size of the speaker to the size of the orifice in the bucket. The bucket has about 120mm height, and the water level is about 70-100mm. The ultrasonic transducer is simply submerged in the bucket, generating tiny water droplets (a mist). The mist will mostly stay in the container, since gravity makes the droplets rain back to the water eventually. The tank lid has a nozzle, which is the only exit available for the air and the mist, once it is pushed by the speaker’s membrane. Thus, the speaker acts as a piston, an electromechanical actuator, and displaces air inside the bucket. In the forward stroke, it squeezes the air out, forming a vortex ring. In the return stroke, it draws the air back in. The waveform has to be asymmetric, such that the suction power is less than the blowing power. Otherwise, the vortex rings are sucked back into the nozzle, and though they do propel, a lot of their niceness is destroyed.

The figure above shows the best waveform shape I found for driving the speaker. It is quite important to get the waveform right! Even more importantly, it is crucial to DC-couple the driver. If you AC couple this waveform, it will not work at the low frequencies (i.e. 1-2Hz).It’s easy to test with a function generator, since the waveform is already DC coupled. In the end application, however, I ended up building a Class-D amplifier, without the output filter stage. The speaker itself removes the high frequency content due to its inductance.

I would share my design (mechanical drawings, etc). But this is such a custom-built device to fit a random ice cream container I found that there’s no point in doing that. I’m sure if you are determined enough to make this, you’ll find your way! A few tips:

Fog height between water level and top of the tank is somewhat important. The particular fog machine I used generates 30-50mm of fog height above the water level. If the fog is not to the top of the tank, when the speaker pumps the fluid out there will be no fog carried with it, which will result in an un-seeded vortex ring, ruining the visual effect. I found that the fog doesn’t overflow through the nozzle even when the tank lid is closed, even with a high water level.
The displaced volume is important. The larger the speaker size (I used a 4″ speaker with a 20mm nozzle), the less it has to displace to produce a nice vortex. A ratio between speaker diameter and nozzle diameter of 5 seemed to work well to me.
Remember, velocity is dx/dt. This means when you increase the frequency of the signal, the velocity increases linearly (2x frequency, 2x velocity). This means that, as you increase the frequency of the signal, you don’t need as much amplitude to generate the same exit velocity. Since exit velocity roughly determines the vortex circulation and, therefore, the vortex Reynolds number, you want to keep that number the same in your experiment. Say, if you double the frequency keeping the amplitude of the voltage signal, you’ll get twice the exit velocity, which will make the vortices shoot twice as fast (i.e., they’ll go further) and with twice the Reynolds number (i.e., they will become turbulent and break down earlier). There’s a balance to strike here.

I hope this gives you some ideas!

Cutting mathematical sheets

Posted on April 10, 2020April 10, 2020 by Fernando Zigunov

Mathematics in the complex plane are sometimes surprisingly difficult to understand! Well, the complex numbers definitely earned their name! Maybe you’re also studying complex analysis, or have studied it in the past and didn’t quite understand it. The fact is, it requires a lot of imagination to see the concepts.

I sometimes like to compensate for my lack of imagination by writing an app that visualizes the things I wanna see. This one, I would like to share with others, so I wrote it in Javascript. You can have a look at it here.

It’s basically a way to visualize how a given variable z maps into a function f(z). If f(z) is multi-valued (like, for example if f(z)=sqrt(z)), then the complex map “compresses” the entire z plane in a fraction of the f(z) plane. In the example f(z)=sqrt(z), as you probably already learned, this fraction is 1/2 (since the exponent is 1/2). The function f(z)=sqrt(z) is then said to have two branches. Functions that have this behavior will have a branch point, which is this point where as you go along 360º in a small circle around it, the function f(z) does not make a 360º arc. The function f(z) then becomes discontinuous, “it branches”. The discontinuity is actually along a curve that starts at the branch point, and this curve is called “a branch cut”. The branch cut, however, can be any curve starting at any angle. It just needs to start at the branch point.

The mind-numbing part starts to happen when the argument of the multi-valued function (for example, sqrt(z) and log(z) are the most commonly seen functions in math) is another function. Then, the branch points lie at the roots of the argument. If there are multiple complex roots, each one of them will be a branch point, from which a branch cut has to be made. Determining one of the branch cut curves, however, determines all of the remainder branch cuts.

Let this settle a bit. Let’s say you place a cut in the target function f(z) around the branch point as a line at an angle ϑ. z in polar form can be written as z=R exp(iα). The reason why branches occur in a complex function is because f(R exp(iα))≠f(R exp(iα+2πi)) , even though z=R exp(iα)=R exp(iα+2πi) (i.e., as you go around a full rotation, the point in the complex z plane is the same. But f(R exp(iα)) is uniquely defined. This means that the function f(z) itself determines how much angular displacement one full rotation about a branch point in z incurs in the f(z) plane. Therefore, defining the branch cut line automatically defines the region of f(z) that is available in the mapping z->f(z), which is “the branch” of f(z).

The app I developed computes that, for simple functions like sqrt(z+5). However, it is not general. It assumes that the branch cut starts at (0,0) in the f(z) plane. If the branch cut has to start at a different coordinate (say, for f(z)=sqrt(z)+1 it would have to start at (1,0) ), then the app mapping does not work correctly. It, nevertheless, gives some insight (in my opinion). Especially for a student! If you wanna develop it further, let me know!!

An example

Consider the function f(z)=sqrt(z^2-4). This function has two branch points, i.e., where the argument of sqrt() is zero. These points are the roots of z^2-4, which are +2 and -2. The square root function maps the full z plane to a half plane, which is why in the app it appears like this:

On the right-half, we see the f(z) plane. The color represents the angle, i.e., arg(f(z)). We see the upper half plane in vivid colors, whereas the lower half is ‘dull’. The upper half is the current branch. The branch cut is defined, interactively, by the white solid line. The other branch cut line is, as discussed, automatically defined in the dashed line by the half-plane mapping the square root function gives. This upper-half plane branch cut, shown in the z plane, looks like two lines spanning out from the branch points (+2 and -2). They appear in the left-hand figure as a discontinuity in the color (which represents the angle of f(z)).

The cool thing is that, by dragging the little black dot in the right hand figure, we can move the branch cut interactively. Here’s a video of what happens if we go around the circle:

And another one going though other random functions:

I hope this helps you as much as it helped me understand these concepts! It’s actually quite cool once you visualize it!

Driving a hundred solenoid valves

Posted on January 26, 2020January 26, 2020 by Fernando Zigunov

As I discussed in this past post about MIDIJets, I was attempting to make a platform for surveying microjet actuator location and parameters in aerodynamic flows for my PhD research. But I think this is something that can be quite useful in many other contexts. After working with this for a couple months now and realizing how robust the driver I developed was (yes, I’m proud!), I decided to release this project as an open-source hardware. Maybe someone else might find this useful?!

With that said, the project files can be found at this GitHub page: https://github.com/3dfernando/Jexel-Driver . The files should be sufficient for you to both build your own board, program it with a PICKit4 (I’m pretty sure you should be fine with older PICKit versions) and communicate with the Serial port through a USB connection.

What can I do with it?

Now, let’s talk about the device’s uses. Being able to control many solenoids with a single board can be very useful. In my case, the application is aerodynamic research. We can activate or energize a boundary layer of a flow. But maybe the applications could transcend aerodynamic research? Imagine a haptic feedback glove that makes vibrating air jets on your fingers, how cool would that be? Or maybe an object manipulator by controlling where air is issuing from? I think there’s some other possibilities to be explored. If you would like to replicate this, let me know.

Visualizing the jets

Here’s some quick flow-vis showing the pulsating jets with a small phase delay of 60º. Just as a reminder, visualizing jets of 0.4mm diameter is not easy – so I apologize if the video looks noisy! There’s a dust particle floating in the air in some frames. That’s kinda distracting but is not part of the experiment!

Some caveats

Well, I’m a mechanical engineer, so board design is not really something I do professionally. Therefore, expect some issues or general weirdnesses with my design. If you’d like to replicate this, I used a Matrix Pneumatix DCX321.1E3.C224 solenoid. It is not a large valve. The right connector is on the project BOM. The issue is that this valve is a high voltage, low current valve (24V, 50mA). The driver shield I designed has those specs in mind. This means a different driver circuit would probably be needed for valves with different specs. Also, for higher currents, be mindful that the motherboard carries the current through it, possibly generating some noise if the driver current is too high (yes, I was not very smart in the board design!).

In conclusion

Well, I hope you found this mildly interesting. If you think you could use this project and you made something cool inspired by this, I would be pleased to know!