There’s really not too much to explain here. MacGyver rolls a guy up in a carpet and uses that to keep him in place. Seems like this should work.
Reminder—even though there’s not any great science to talk about here, I love these kinds of hacks.
Pick a lock with an antenna
MacGyver pulls a metal antenna off a motorcycle and then uses it to pick the lock on the same motorcycle. Oh, this is a police motorcycle—he uses it to turn on the siren.
Is this possible? Possible, yes. Normally you need two things to pick a lock—something to move the lock pins and something to provide torque. Theoretically, you could do it with one—but it would be tough.
Radiator fire extinguisher
There’s a car on fire and they need to get something out of it. MacGyver improvises a DIY fire extinguisher by using the cooling system of another car. He grabs a pipe and then pokes a hole in the radiator. When the engine is revved up, the coolant shoots out through the pipe and onto the burning car.
Again, the theory here is solid. You could probably even get the coolant to shoot fairly far. However, it would take a significant amount of coolant to put out a fire.
Statue battering ram
This one is good. MacGyver and Jack take a marble statue and support it from some rope and stuff such that it can swing. Then they swing this statue into a door. Boom—battering ram.
Hacking airplane wifi
OK, not hacking to get free wifi but using the wifi to control the whole plane. That’s what Riley does as she is stuck on a plane with a virus. Is this even possible?
This is not a MacGyver-hack. But he is talking about radioactive stuff. Here is my super short explanation.
An atom is made of a positive nucleus plus some electrons. Normally, there will be the same number of electrons as protons in the nucleus. This makes the overall charge of the atom zero. But wait! There are also neutrons in the nucleus. Let’s just look at an example.
Strontium-90 is an atom with 38 protons and 38 electrons. It also has 52 neutrons in the nucleus—oh, and it’s radioactive. Strontium-88 also has 38 protons and electrons, but it only has 50 neutrons. Since these two atoms have the same number of protons, but different neutrons—they are isotopes of each other.
Sr-90 is radioactive. That means its nucleus is unstable. It goes through radioactive decay and produces Yttiruim-90 which has 39 protons. So, one of the neutrons in Sr-90 turns into a proton and also produces an electron (so that charge is conserved).
The half-life of Sr-90 is 28.8 years. Since the decay process is random, there are more decays when you have more atoms. This means that the decay rate depends on the number of atoms. So, you can’t say how long it will take for all the material to go through radioactive decay. Instead, we say how long it will take for half of it to decay—that is the half life.
So, does that mean that all of the material will decay in 2 times 28.8 years? Nope. It means that after 28.8 years, you will have half as much stuff. After another 28.8 years you will again have half of what you started with. Every 28.8 years, you will have half as much.
What about Plutonium? There are several different isotopes. Plutonium-238 has a half-life of 87.7 years. Plutonium-239 has a half life of 24,000 years. The shorter the half-life, the more “radioactive” it is. But for longer half-life stuff, it’s still going to stick around for a long time.
DIY Geiger Counter
A Geiger counter is a device to measure radioactivity. It consists of an outer conducting tube with a conducting wire running down the middle. The wire and tube are at different electric potentials—usually a fairly high voltage.
When a charged particle (the result of a radioactive decay) enters the tube, it interacts with the gas molecules to ionize them—knock out an electron. This free electron then accelerates in the high potential and collides with other gas molecules to produce even more free charges. More and more charges are produced—it’s called an electron avalanche. This avalanche is then detected as an electron current—usually to produce an audible click. That’s the clicking sound you hear.
Here is MacGyver’s build.
He uses a motorcycle muffler as the tube—you can’t see the wire in the middle, but it’s there. The battery provides the voltage and then there is an audio speaker to convert the electron avalanche into sound.
Oh, I forgot—here is a video demo.
Bonus. Here is my sketch for the DIY Geiger counter.
DIY Ice Climbing Gear
There’s really not much to explain here. MacGyver uses some cut up chain link fence and some rebar to make spike-shoes and a type of ice pick. He uses these to climb up an old brick wall—sticking the metal into the old mortar. It seems like it would work.
MacGyver builds one of those rolling things that mechanics use to get under cars (I forget the name). With that, he disables the trucks (all but one). Really, once he is under the truck he could do a number of things. He could cut the fuel line or cut an electrical line—let’s just say he does it.
In order to convince the mob that the key guy they are trying to catch isn’t worth it. They make it look like he is super radioactive. The mob guys have a Geiger counter—so MacGyver needs to remotely set that off.
The electron gun would shoot electrons into the Geiger counter and set off an electron avalanche in the same way that a radioactive source would. Unfortunately, it was too complicated of a build.
What about a super high amplitude electromagnetic wave? The idea is that when the electric field part of the wave hits the detector, it will increase the field inside and make it more sensitive to ionizing radiation. This would make it seem like something is more radioactive than it is. Yes, I know this is a stretch—but it’s based on some real idea.
Here’s how he could build it.
Start with a radio – it has to be a transmitting radio. Maybe a CB, maybe a portable radio?
Boost the power to the radio. I think plausibly connect the power source from the radio to the 12 volt car battery?
Get a bent piece of metal to put around the radio antenna to make a focusing dish. Alternatively, you could find some pre-existing dish- aim the dish-radio at the Geiger counters. boom.
Sorry for the delay in science notes. There were things that got in the way.
Block and Tackle
I don’t really like calling it that. A better term is a compound pulley. But the key to all of the simple machines is force vs. distance. If you increase the distance over which you apply a force, you can get a larger force output of the machine over a shorter distance. That’s exactly what happens with a “block and tackle”.
But basically it’s alcohol oxidase with some other stuff added. When mixed with alcohol, this makes aceteldehyde—the main thing that causes hangovers. MacGyver could get alcohol oxidase from alcohol test kits.
Note: I’m not a biochemist, I got this info from my brother (a biochemist).
Don’t make this stuff. But it is possible. Actually, I’m not even going to include the link.
Using a hammock as both a ladder and a body sling? Yup, that’s good. Using a chain as a DIY car boot. That works too. Not much to explain, but I think both of those hacks are great.
In order to knock out everyone in the room (including himself), MacGyver throws some chloroform in a container into a fire. It explodes and everyone gets knocked out. Sure, this would be tough to do in real life—but it’s a least plausible.
DIY Arc Lamp
Wait. There wasn’t an arc lamp in this episode was there? Nope. It get cut out of the beginning. However, I made an arc lamp anyway as part of my DIY videos. It’s awesome.
Here are some things I need to share regarding this meeting. Overall, it was great to see everyone. As usual, the conversations were the best. I regret that there were some people I did not get to meet or talk to—maybe next time we will meet up.
Blog vs. WIRED
One question that came up multiple times was about this blog vs. my posts at WIRED. How do I decide where a post goes? OK, here is my explanation.
In the beginning, there was a blog. A blog was super informal and free form and alive with comments. It was like the 60s and I was a hippie. A physics hippie. I don’t know if this early blogging era was like the 60’s or maybe the wild-wild west. Well, the comments eventually turned into the wild west with a shoot out at the O.K. Corral.
When I moved to WIRED, everything was the same except it was at a different site. But I’ve been there for a LONG time (9 years?) and things evolve. My posts at WIRED are more edited and geared for a specific audience. That’s not bad, it’s just different. I don’t think I can just write whatever I want like I did in the old days. No more random posts that just talk about my cat (I don’t even have a cat).
So, that’s where this blog comes in. It’s a place where I can post whatever I want and no one can stop me. These are the kinds of things you will find here.
Random posts (like this one) that are just an outlet for me to write stuff and tell stories.
Explicit educational material. If a post needs too many equations, I would rather put it here. Many WIRED readers (while very education) don’t really get into all the equations. Also, since WIRED is paywalled it makes it more difficult for educators to access the stuff (in the off chance that they might find it useful).
MacGyver science notes. Oh sure, I post some MacGyver stuff on WIRED—but I really don’t think they want to see 50 posts on different episodes. So, those are here.
I think that pretty much covers it, so I don’t even need this last point.
In the end, I apologize for the confusion with the two blogs. Oh, actually there are three. I recently wrote a post on OneZero Medium (analysis of a car crash from Stranger Things) also. Not sure how much I will write there—but it’s still me.
What’s up with all the drone videos?
Yes, I have a drone. I love my drone. I can only hope my drone loves me as much as I love it. Honestly, I am honored that you even noticed my drone videos.
Oh, wait. You haven’t seen them? I can fix that.
In case you are curious. This is a DJI Spark. Great drone.
Here are some more short comments.
Meeting with Bruce Sherwood and Ruth Chabay was great. I wish I had a picture with both of them (I did get one with Bruce though).
Bruce made this epic comment in regards to numerical vs. analytical calculations. People claim that analytical solutions are better because you can solve a problem in terms of known functions like sine and cosine. But how do you find the value of the cosine function? YUP – numerically or in a table or in an infinite series. So, in a way all solutions are numerical. Win for numerical.
The other deep thought by Bruce was a discussion on his AJP on energy. Read that paper. This sums it up. You can not find the work done by friction. Friction is crazy hard. I think I might write a WIRED post on this.
Eric Ayars had an excellent presentation on chaotic systems. One system was a bouncing ball on a moving floor. I wonder if there is a case where the ball just stops—this could happen if the relative collision speed of the ball and floor is zero.
The 30 demos in 60 minutes was pretty good. I love these things. Even though I’ve seen many of these demos before, I always find something new. Here is their site http://30demosin60minutes.com/
I went hiking. It was super hot, but I had a great time.
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.