Was the moonwalk fake? No, not the Apollo landings. I am talking about Michael Jackson’s moonwalk. You got to admit, he had a big impact on a lot of stuff and this is my way to give him respect – physics.
I am sure you know about the moonwalk. Maybe you can even do the dance move yourself, but how does it work? First, here is a clip of MJ doing his stuff.
As a side note, I can’t remember where I saw it but there was a great discussion of the history of the moonwalk. If I recall correctly, some were saying Michael didn’t create this move. One thing is for sure, he made it popular. Now for the physics.
The key concept here is friction. Friction is actually uber-complicated, but a simple model works for many cases. Static friction is a force exerted on an object when it is in contact with some surface but those two surfaces do not move relative to each other. Kinetic friction is a force exerted on an object when the two surfaces are moving. Suppose I have a block at rest on a table and I pull it with a slowly increasing force. This is what it would look like:
Two key things from this graph. As you pull on the stationary block, the block doesn’t move. If I pull with 1 Newton, and it doesn’t move then the frictional force is 1 Newton. If I then pull with 2 Newtons and it still doesn’t move, the frictional force is 2 Newtons. The static frictional force does what it can to make the thing not move – but not more than it can. This leads to the static friction model of:
In this model, the force is less than or equal to the product of some coefficient (that depends on the two types of surfaces) and the normal force (how hard the two surfaces are pushed together). The direction of this frictional force is parallel to the surface in the direction that prevents the object from sliding.
The other key feature in the graph is the small jump down when the thing starts to slide. This is because the coefficient of kinetic friction is typically smaller than that for static friction. Also, if the object is sliding, the frictional force is constant.
Back to Michael and the moonwalk. The key here is: how do you make one foot slide and the other not slide? If both feet are stationary, then this is dealing with static friction. I could make the frictional forces on these two feet different by changing my center of mass. Here is a free body diagram:
Since he is not accelerating up and down, the following must be true:
These are the forces in the y-direction. They must all add up to zero so that:
There is another condition that must be satisfied. Since he is not rotating, the total torque about any point must also add up to zero. If you want more info on torque, check out this post (Note: link doesn’t work). But for this post I will just say that torque is like the ‘rotational force’. It depends on the point about which you want to rotate and is essentially the force applied times the perpendicular distance to the point of rotation. For the free body diagram of Michael, I have chosen one of his feet to be the point about which he is not rotating (I could chose any point). This makes 3 of the forces have zero torque (N2, F2 and F1 have zero torque because the perpendicular distance to point O is zero). Here I labeled the other important distances:
The only two forces that exert torque about O are the weight and the N1 force. They have opposite directions of torque because they would cause rotation in different directions. This along with the previous equation gives:
Eliminating mg, and solving for N1, I get: (I know the indices for the forces and distances don’t match)
If his center of mass is in the middle, then r2 – r1 = r1 and the two normal forces would be equal (as you would expect). If the center of mass is more towards the foot on the right, then r2 – r1 is less than r1 and N1 will be larger than N2. This will make the frictional force on the foot to the right greater and the other foot slide.
Well, what if r1 is greater than r2? One of two things would happen. Either he would fall over, or there would have to be a force pulling the foot on the left down. This is similar to Michael Jackson’s trick in “Smooth Criminal”.
Ok. So that is how Michael gets one foot moving. How does he keep one foot sliding and the other not sliding? It is really the same thing as above except that he can increase the force on the moving foot a little bit more since it is sliding. Sounds easy, but Michael could really make it look cool.
Finally, I just want to show another demo that is essentially the same idea.
Oh, it has a built in timer—a leaking bottle of water. Here is the shot from the show.
And here is my rough sketch.
Oh snap. That pay phone is broken. Wait! What’s a pay phone?
MacGyver fixes the missing headset by using two speakers from the car. One speaker is used for the speaker and the other speaker is used for the microphone.
It’s actually pretty cool, but most speakers can be used a microphone. The normal speaker is basically just three parts:
A coil of wire.
Some type of surface area to push air.
The coil of wire is connected to the speaker surface. When current is run through the wire, the coil makes a magnetic field. This magnetic field from the coil interacts with the other magnet to either push or pull the surface of the speaker. This in turn pushes the air into compressions—and it is these compressions in the air that make the sound.
For a microphone, the reverse happens. Compressions in the air push the surface. This moves the coil closer (or farther) from the magnet. This motion changes the magnetic flux (via Faraday’s Law) which induces a current in the coil. This current is then recorded as an audio signal.
Don’t believe me? You should try it. Oh, make sure you use a speaker like this:
OK, there are some weird things in these old phone head sets. I think they have to use super low resistance microphones and speakers since the power comes over the phone line. But still, this is very plausible.
Break a chain with handcuffs
Oh, and a steel rebar thing. MacGyver loops the handcuffs around the chain and then puts the rebar through the cuffs and twists.
With a longer metal bar, you can get a high torque on the chain. This should break the chain—at least as long as the hand cuffs are stronger than the chain.
I told my students I would solve this problem for them. It’s a real life problem too.
Is it possible to use a Neodymium magnet and a coil of wire to get an LED to light up?
That’s the real version. But because I was afraid students would be overwhelmed, I added the following:
The magnet has a maximum magnetic field of 0.2 Tesla.
The LED requires 1.5 Volts and 10 mA to light.
I like this problem because I don’t know the answer. Also, the answer is useful. If you want to do a physics demo showing the voltage induced by a changing magnetic field—what better way than with a hand-held magnet and a small light?
But will it even work? Let’s get started. Here is a diagram.
By changing the magnetic flux through the coil, this will create a curly electric field and an electromotive force (a change in electric potential). The magnetic flux is defined as:
Where N is the number of turns in the loop, A is the area of the loop and is the angle between the magnetic field and a vector perpendicular to the area. In the diagram above, since the magnetic field is perpendicular to the coil.
A change in flux produces a voltage according to Faraday’s Law:
Note: yes, I’m different. I think that the number of loops (N) is part of the magnetic flux and that the minus sign in Faraday’s Law doesn’t really mean anything.
Putting the flux into Faraday’s Law, I get (assuming ):
Now for some estimates. I could just estimate everything and then calculate the voltage—but instead I’m going to estimate everything except the number of turns. I can then solve for N and see if it’s reasonable.
Here’s what I have.
B = 0.2 T
A: radius = 0.01 m
Time interval = 1 second
Voltage = 1.5 V
Solving for N:
This is the perfect case to use python for your calculator. You can put your estimates as variables so that you can easily change things up. Here is my code. I get the following output.
Umm….yeah. That’s 23 thousand turns. I’m not going to do that. Even if decreased the time to 0.1 seconds, I would still need 2000 turns. Arg.
Oh, what if I just make a HUGE loop? Nope. That wouldn’t work. In my estimation for the change in flux, I assumed a constant magnetic field—this is obviously not true, but good enough for a small loop. With a big loop, you would have some of the magnetic field creating a negative flux. It would just make things worse.
What if I put the magnet on a spinning stick (run by a motor)?
I’ll be honest. This connection between the electric potential (change in electric potential) and the electric field can get sort of crazy. But let’s just start with a problem and then solve it in more ways than you wanted.
Here is the problem.
Let’s start with the energy to bring an electron to point B. The energy needed would be equal to the change in electric potential energy which is equal to:
That means I just need to calculate the change in electric potential from infinity to point B. Yes, you could also calculate the work needed to move the charge—I’ll do that also.
Since I am dealing with two point charges, I can use the following expression for the potential due to a point charge (with respect to infinity):
Where k is the Coulomb constant ( and r is the distance from the point charge to the final location. Since there are two point charges, the total potential will just be the sum of the two potentials. Let me call the positive charge “1” and the negative charge “2”. That means the total potential will be:
From the original problem, and . The distance will be 6 mm and the distance will be 4 mm (need to convert these to meters).
Putting this all together, I get the following. I will do my calculations in python. Here is the code.
Running gives the following output.
Boom. There is your first answer. What about point A instead of B? Well, in this case, I just have different distances. The distance for both AND are the same. Since they have the same distances but equal and opposite charges, the two potentials will be opposite. When added together, the total potential is zero volts. Yes, the energy needed to put a point charge at A from infinity is zero Joules.
What? Yes. How about this? Suppose you take the electron from infinity on the positive y-axis. As you move down the axis to point A, the electric field is in the x-direction. That means the electric force is in the negative x-direction. You would have to push it in the positive x-direction and move in the y-direction. But that requires ZERO work since the force and displacement are perpendicular.
Oh. You want to get to A from a point of infinity on the positive x-axis? OK. That still works. Remember that for electric potential, the path doesn’t matter—only the change in position (path independent). I can take whatever path I like. I’m going to move in a circle from positive infinity x to positive infinity y. The electric field is zero out there, so that requires zero work. Now I’m at positive infinity y—and I just did that problem. Zero work.
Another way—by calculating the work
Remember that work-energy principle? It says this:
And the work can be defined as the following (if the force and displacement are constant):
Oh, and the force will be the opposite of the electric force where:
So, as you push a charge towards point B (point A is boring—for now) the electric field changes. That means we have a problem. We can’t use the above formula to calculate the work—unless we cheat. Let’s cheat.
Instead of calculating the total work to move the charge to point B, I’m just going to move it a tiny bit. During this tiny move, the electric field (and thus the force) will be approximately constant. Then I can do another tiny move. Keep repeating this until I get to point B. Here is a diagram.
If this distance is short () then the force is approximately constant. That means the tiny amount of work (which I will call ) would be equal to:
OK, just to be clear. This is the force needed to PUSH the electron (with a charge e)—it’s not the electric force on the electron (which is in the opposite direction). Also, the angle between F and the displacement is zero. That means the cosine term is just one. I wrote the force and displacement as scalars because it’s just the magnitude that matters.
Now we are ready for some stuff. Here are the steps I am going to use.
Start at some position far away (but not actually infinity because that would be crazy). It just needs to be far enough away such that the electric force is negligible.
Calculate the total electric field and the force needed to push the electron at this point.
Move some short distance towards point B.
Over this distance, assume the force is constant and calculate the small work done—add this to the total work.
Repeat until you get to point B.
Before making this one program, I’m going to just make a program to plot the electric field from some value up to point B. Here is the plot from that program. (here is the code)
Note that I started from just 5 cm away from the origin—which is TOTALLY not infinity. However, it makes the graph look nice. But still, this is good because it looks like the calculation is working. Now I can use this same calculation go find the work needed to move the electron. Here is the code.
And the output:
Notice that gives a close, but wrong answer (compared to my previous calculation). Why is it wrong? Is it because I started at y = 0.5 meters (I just realized I’ve been using the variable y instead of x—but it should be fine). Or is it wrong because my step size is too big?
The answer can be found by just changing up some stuff. If you move the starting point to 1 meter, you get about the same answer. However, if you change dy to 0.0001, you get the following output.
That works. Oh, I added some more stuff to the output.
One more thing (and then I will look at the electric field in another post). What if I use a different path to get to point B? Instead of coming along the x-axis (which I previously called “y”), I come parallel to the axis a distance of 2 mm above it. Once I get right over point B, I turn down.
This introduces some “special” problems.
I can break this path into two straight pieces (path 1 is parallel to x-axis and path 2 is parallel to y-axis).
Along path 1, the force needed to push the electron is NOT parallel to the path. So, the angle is not zero in . This means I’m going to have to calculate the actual vector value of the electric field at every step along this path.
The same is true along path 2.
But in the end, I should get the same work required—right?
OK, hold on because this is going to get a little more complicated. Let me just include one sketch and then I will share the code for this new path. Here is how to calculate the electric field and work for a particular step in path 1.
Here’s what needs to happen to calculate the electric field (and force) for each step:
Find the vector from the positive charge to step location.
Use this vector to find a unit vector (to give the electric field a direction).
Use that vector to also find the magnitude of the electric field.
Calculate the electric field due to the positive charge (as a vector).
Repeat this for the negative charge.
Add the two vector values for the electric field to get the total electric field.
Multiply by the charge to get the force (which would be in the opposite direction).
Now, to calculate the work done during each small step, I could use the angle between the force and displacement. But I don’t know that. Instead, I can use the vector definition of work:
Yes, that is the dot product. Fortunately, the dot product is already built into VPython (Glowscript). So, once I get a vector value for the force and the displacement I can just use the “dot()” function.
Don’t break into people’s houses. Oh sure, there are lots of ways to get into someone’s house. Think of a locked door as a social norm. People agree to not go past that locked door (or window)—even though they probably could.
In this case, MacGyver gets in through a window. In some cases, it’s possible to use the friction between your hands and the glass to shake the window up and down. This can slowly force the window lock into the unlocked position.
Detecting Metallic Ink
Here is another one that seems crazy, but it turns out to be not so crazy. MacGyver builds a detector to find some hidden cash. Yes, it’s indeed possible to detect the change in magnetic fields due to metallic ink in US currency.
But you have to be pretty close—and really it would only work to determine how many bills are in a container. However, there’s still a chance this could work.
In this case, MacGyver uses a hall effect sensor along with a speaker to create an audio-based system to search for the money. In the show it works like a metal detector—but it’s not a metal detector since the hall effect probe detects magnetic fields.
I’m not sure I should go over all the details of a hall effect sensor, instead I will just like to one of my WIRED posts on the subject.
But what about the speaker part of this build? Well, it is indeed true that you get a voltage signal out of a hall effect probe. If you run this into an audio amplifier, you probably won’t get any sound because you would need a changing magnetic field. But it seems likely that you could have the hall effect probe voltage control and audio tone.
Anyway, here is my very basic sketch for this detector.
Distraction with Streetcar Sparks
MacGyver grabs a chain and throws it onto the wire that the streetcar runs on. Sparks fly and cause a distraction.
The New Orleans streetcars are electric powered trains. They get power from two lines. There is a line above the car and the other is in the rails (at least I’m fairly sure that’s how it works). So, just touching a wire at the top with a conductor wouldn’t do anything. If you had a chain running from the top wire down to the ground, that would cause a short circuit and probably melt the chain. It would be bad.
Of course there is a way to get this to work. What if MacGyver throws the chain over the power line so that the chain hits both a power line AND a support pole? I imagine there is an insulator keeping the power line isolated from the ground, but getting that chain to make a connection would do the trick.
Infrared Chemical Tracker
MacGyver finds the following stuff:
Selenium powder (they make solar panels with this stuff)
Cadmium oxide – the stuff from the inner part of a battery
With this he is making a type of quantum dots.
Oh, I forgot to say something—quantum dot tracking dyes are real.
The idea of the quantum dot is that it is a very small particle that emits a particular frequency of light. If you “excite” it with an ultraviolet laser, it can emit infrared radiation that can be detected with a drone camera. Cool.
So, for MacGyver’s case—they skipped the whole UV light part. But still, this is another great example of something that seems crazy but is in fact based on some real science. Science is crazy.
MacGyver is a show. It’s fictional. It’s not real. Some of the things are BASED on real science (and some of them are legitimately real). But it’s still a show. It’s like Star Wars—but without the light sabers. Everyone knows there is no way you could even THINK of making a lightsaber with science, but we like them anyway.
So, even though some of the hacks in the show are only slightly plausible, there is still an element of truth in there somewhere. Honestly, I’m just happy that anyone even cares to make a show that even considers real science. Thanks Peter!
OK, now for this episode’s MacGyver hacks.
Tracking a vehicle with CO2 sensors
So, there’s this runaway robot car with guns and the Phoenix team has to find it. It’s got stealth technology, so they can’t find it from above. That leaves MacGyver, Riley, and another girl in a car to track it down.
The idea is to use the carbon dioxide emitted from the robot. Yes, it’s a hybrid vehicle. That means it has an internal combustion engine. These things take in gasoline and produce energy along with carbon dioxide and other stuff. Oh, it’s this same carbon dioxide that contributes to global warming and climate change—just to be clear.
The plan is to have two CO2 sensor sticking out of the car on tree branches (now you get the title). The sensors are connected to the dome light in the car so that they can tell which direction has a stronger CO2 and then they know which way to turn.
Here’s what it looked like.
Here is one of my diagrams (this went through quite a few iterations).
Here is one of my earlier diagrams—it was slightly more realistic using some MOSFETs for amplification and everything.
In the end, the CO2 level in the air from a vehicle is quite small. I think it would be seriously implausible to use two detectors to determine the direction to the robot. So, I will go ahead and give this is “real score” of maybe 1.5 out of 10. Here are some other hack scores—in case you are curious.
The basic idea is to beam high power radio waves at a vehicle to fry the electronics. In this case they say it just drowns out the network so that Brutus can’t communicate and it stops.
MacGyver builds this RF gun using a transmitter on the truck and a satellite TV dish. In order to get the power high enough, he uses the car battery.
OK, now for a homework question. Assuming the van they are in has a normal style car battery, how much current does the RF gun use so that it drains the battery in 5 minutes? Some estimations might be required.
Hotwire a car
I’m pretty sure I talked about this in a previous post. Modern cars are really tough to hotwire—good thing they found an older camero.
Final hack of the show. MacGyver uses his phone and a belt to pull and bend the vents on Brutus. He needs a space big enough to fit a USB stick through. It’s funny because MacGyver kills his own phone and not Jack’s.
It seems that most of the second semester algebra-based physics is magic. Since you need calculus to derive many of the expressions, the students just get them magically instead.
NOT TODAY. Well, I hope not. Today I am going to use python and the Biot-Savart Law to find the magnetic field due to a wire. Here is the expression I want to show:
Where I is the current in a wire and r is the distance from the wire. I guess I should start with the magnetic field due to a moving point charge.
Yes, that’s sort of a crazy equation. The weird part is the cross product. Here are some notes:
The “times” symbol is the cross product.
The cross product is an operation between two vectors that returns a vector as the resultant (unlike the dot-product that returns a scalar).
The resultant of this vector is perpendicular to both of the products—that makes this only work in 3D.
The magnitude of the resultant depends on the magnitude of the products and the sine of the angle between them.
OK, that’s enough of that. Fortunately, we don’t really need to compute cross products since it’s built into VPython (Glowscript). Let me do one more thing before calculating stuff. Suppose I have a charge q moving with a velocity v over some short length of wire, L. I can write qv as:
So, instead of dealing with qv, I can use IL. Note that L is a vector in the direction of motion for the current. Now my magnetic field looks like this:
I changed from L to dL since it has to be a short wire. So, dL is just a way to emphasize that the wire is super short.
Let’s do this. Here is my first calculation. Let’s say I have a super short wire (0.01 m) with a current of 0.1 Amps. What is the magnetic field a distance of 0.02 meters from the wire? I left off something important—but I will show you that in a second. Here is the code to calculate this magnetic field.
It looks like this (this is just an image—you need to go to the trinket site to actually run this code).
If you run this, you get an output of <-2.5e-7,0,0> T. I think that’s correct. But let’s make this better. Let’s make a visual representation of the magnetic field. Really, that is the power of VPython anyway. Here is the new code and this is what it looks like when you run it.
I rotated the camera angle a little bit so you could see the wire and the magnetic field. OK, now for MORE VECTORS. Here is the sloppy code.
Oh. I like that. It’s pretty. But you can see that the magnetic field makes a circular pattern around the wire. But what about a long wire? Here comes the part where we NEED python. I want to be able to represent a long wire as a series of a bunch of small wires. Then I can calculate the magnetic field due to each of the small wires and then add them up to get the total magnetic field.
In order to simulate a “long wire” I need to have the “observation” location in the center of the series of short wires. Maybe this diagram will help. Here is a side view of 8 small wires together along with the observation location.
Each of these parts of a wire will have a magnetic field at the “obs” location. So, here is how this will work.
Pick some distance from the wire (r) and create the observation location as a vector.
Take the wire and break it into pieces. The more pieces, the better the answer.
For each piece in the wire, calculate the vector r to the observation location.
Calculate the magnetic field due to this piece and add it to the total.
Let’s do this. Here is the code. Oh, I am going to use the same wire as before but I will make it 1 meter long. Also note—I’m not going to display an image of the magnetic field (or even the wire). I’m going to try to make this as simple as possible.
Using 8 pieces, I get the following output.
Where the theoretical B is the value calculated from the scalar equation up top (magnetic field due to a long wire). So, this is the scalar value (ignore the negative sign). Also, it looks quite a bit off—but there are a couple of points.
This calculation only uses 8 points.
There is a slight error. I put the first I*dL at the position x = -L/N. That assumes a super tiny dL—and that’s not true when N = 8.
The magnetic field due to a long wire equation (above) is for an infinite length wire.
So, with 50 pieces you get a pretty good agreement with the theory. I like that.
But wait! The theoretical value says the magnitude of the magnetic field decreases as 1/r. Does that work for this model too? Let’s test it. Here is the code.
Surprising that the two calculations don’t quite agree at very close distances. I suspect that is because I have an even number of wire pieces (50) which puts the observation location between two wires segments. Or something like that. But otherwise, this works.
It’s too bad I can’t embed trinket.io right into this blog. I guess I will have to upgrade my wordpress at some point.
Before I get to this science for this episode (there’s some great MacGyver hacks here), let me say something else about the show. The storyline for this episode was great. It had a nice plot, and I really enjoyed the MacGyver and Jack flashbacks. Now for some science.
The best Mac-hacks are real. This is real—very real. It is indeed true that this was an early idea for a phone. Here’s how it works.
You need a directed light. One way to do this is to get a mirror that reflects sunlight. A parabolic mirror works a little better, but still the mirror is a great idea.
Use your voice to shake this mirror. This changes the intensity of the reflected light to match the wave pattern of your voice.
Use some type of electrical photo device (solar panel, photo diode, photo resistor) to modulate an electric current to this same sound pattern from the light. Send this to a speaker.
For MacGyver’s build, he uses a microphone and connects it to a porch light. The idea is that the microphone will modulate the brightness of the bulb. For this to work, I think it has to be an LED bulb. An incandescent won’t work (I don’t think) since the hot bulb filament won’t change brightness quickly enough for sound frequencies.
There is also the problem of AC vs. DC. If MacGyver connects to the AC power line going to the bulb, this might not work. But still—it’s very plausible.
On the other end of the photophone, Riley uses a police car light as her transmitter. Again, if this is LED it should work (plus the car runs on DC, not AC). Finally, the only problem is aiming. In practice, you need your detector to pick up the changing brightness from that one light. Of course it’s daytime, so there are many “lights” outside. Putting a lens on the detector to aim it would help a bunch.
OK, now you want to build one of these yourself. You should. It’s actually not too terribly difficult. Let’s start with the simplest part—the receiver. The easiest way to get this to work is to connect a small solar panel to an amplified speaker.
Oh, do you know where I got that solar panel? Yes, it was from a garden light. You put these small lights outside and the solar panel charges a battery during the day and the light comes on at night. They were old and the battery was bad, so I took it apart.
Now for the transmitting side of the photophone. I tried to do this with a laser instead of a light (so that I could aim it). It mostly worked, but it’s a bit more tricky.
This is something I need to rebuild at some point in the future. Make it better. But still, this should be in my list of Top 10 MacGyver Hacks. I need to make that list.
Gum Wrapper and Battery to Start a Fire
Ok, actually this was to melt a wire. MacGyver takes a foil gum wrapper and connects it to two ends of a battery. The idea is that the foil will make a short circuit.
But there are two questions:
Is a gum wrapper an electrical conductor? I don’t think they are actually made from aluminum foil—but I suspect that many of these do in fact conduct.
Would it get hot? Even with a small battery, yes I think it would.
DIY Non-Contact Voltage Probe
MacGyver needs to find wires behind the wall. He puts together this awesome looking probe (or as Jack calls it—a doohicky)
Actually, this prop is great. Here’s why.
It looks cool and it’s clearly a combination of multiple items.
It’s not specific—it doesn’t show exactly what MacGyver uses. This is good because that way it could still be plausible.
Finally, it’s based on something real.
But how does it work? There are multiple ways to detect voltages without touching—I think the most common method measures a super tiny voltage that is created by nearby electric fields. The NCVP is essentially part of a capacitor. When in the presence of an electric field, there is a voltage across the capacitor and you detect this voltage. I need to build one of these—for fun. I’ve seen a very basic version somewhere.
Kitchen Chemistry to Detect Explosives
How do you detect explosives? MacGyver is correct that most explosives are based on nitrogen. If you measure the nitrogen, you can get an estimate of the type of explosive.
There are many things in the kitchen that can be used to detect chemicals. Here is one that you can do at home—it’s a chemical-based pH detector (to determine if something is acidic). The color of this cabbage juice will change color depending on the pH level of the material.
Here’s how to make it.
Laser-Based Wire Cutter
Here’s the problem. There are two bombs that need to be disarmed at the exact same time.
The idea is to use a laser that turns on two identical cutters at the same time. The first thing to use is a beam splitter. This takes a laser beam and breaks into two beams. I guess that’s fairly obvious from the name. Here is a video showing how that works.
For the cutter part, it uses a photocell as the “switch” to turn it on. Here is a rough diagram I created for this hack.
In the end, these two motors might not cut at the exact same speed. But it’s still a fairly fun MacGyver moment.