The Physics of FitFlix

Ooooh, you knew this post was inevitable. Any science blog has to have one of these posts.

In our age of technology and rising rates of obesity, the question is often pondered as to how can we get people moving when they can just sit on the couch and watch TV? It’s a rather trying question that I’m sure doctors, fitness specialists, and game developers have been working on for a while. Go on the app store and you can find many Couch to 5k applications. I actually use one myself called Zombies, Run! I like Zombies, Run! because it sorta makes the running worth it. There’s a podcast playing behind you and that makes the run a little more bearable than hearing the weezing of your scrawny pale body trying to move more than 2mph. So now there’s this post flying around on the internet about what if you had a bicycle that was hooked up to your computer and you could use the internet to maybe watch Netflix if you could pedal the bicycle to power your machine. Seems fairly plausible, right? Let’s find out.

So how are we going to evaluate whether or not this is possible? Well, if you go through my past of science posts, you’ll see that there’s a common principle between them all (all two of em): Conservation of Energy. It feels bad that I keep coming back to this well but it’s one of the guiding principles when it comes to problems like this. So what are we looking at?

The first thing we want to do is establish how much energy would be necessary to power your computer so that you may watch Netflix (or your service of choice). And this is hard to pin down. Well, let’s start with the basics. We want to know how much energy your computer and modem use. According to MichaelBlueJay a desktop computer uses “about 65 to 250 Watts” and then we need to add another “20-40 Watts” for the (LCD) monitor. Griffith University seems to agree with the range somewhat, shifting the lower bound by 20 Watts. These are rather large ranges, and rightly so. The factors that go into how much power your computer consumes are the components and peripherals. So let’s just use the worst-case numbers. If you can do it for the worst-case, you can probably do it for the best-case. So we’re dealing with 290 Watts (250 Watts for the desktop computer, and 40 Watts for the monitor). Now we need the modem because that’s how your computer will be connected to the internet. And this is where I run into trouble. Because I usually don’t go beyond page 2 of Google when looking for sources. MakeItCheaper sounds like some random blog and it says 2-20 Watts (but the average is around 6 Watts). So I’ll continue with my worst-case scenario calculation but we may want to come back later and run the numbers again. So, the final total: 310 Watts.

A Watt is a Joule per second. Which means that every second we must supply 310 Joules to the bicycle. We also have to assume that every Joule we supply to the bicycle is sent to the computer and modem without any loss due to resistance in the wire… but hey, we’re working with idealized systems here. So how do we determine how much energy we’re supplying to the bicycle. If your exercise bicycle is anything like mine, it only has one wheel, or at least one main wheel. And if we just look at that one wheel, we have our problem solved. There’s an equation we know for the energy of a rotating object: Hyperphysics to the rescue.


Normally, if we were considering a rolling wheel we’d have to consider its rotational kinetic energy AND its translational kinetic energy. But exercise bicycles are notoriously stationary, so we only need to worry about rotational kinetic energy. Now we get to the I and the ω. What do these two symbols mean?

I denotes the object’s moment of inertia. You can think of it like resistance to rolling. It’s the rotational analogue to mass, resistance to external forces. The moment of inertia of an object depends on some factors like mass distribution and how the object is rotating. It is for this reason that the moment of inertia is better described as a tensor, but I won’t get into that. Just know that it’s easier to rotate some objects (like a wheel) about certain axes compared to others axes. The symbol ω represents the angular velocity, or how fast the object is rotating.

We’ll begin with the angular velocity, mostly because I want to make an assumption on it. I want to assume a constant angular velocity comparable to what a user may do. The question you want to ask yourself now is: why? Well I do this for two reasons. The first is that it simplifies the number of variables that we’re playing with from three to two (moment of inertia will have two components, plus the angular velocity). And the second reason is because I want to simulate real life as much as I can. And I find it highly unlikely that the rider will go race pace while they watch their show. So I plan to use a nice, leisurely pace for my calculations. On the exercise bicycle that we have here, I can do about 60-70 RPM comfortably. Let’s split the difference and say that when you’re biking you’ll do an average of 65 RPM for the duration of your show. Convert that that RPS (rotations per second) and we get 1.0833 RPS. Not quite done yet though, as we need to convert that rotations per second into radians per second. That’s not difficult though. 1 rotation per second is 2π radians per second. So at the end of it all, our angular velocity is going to be: 6.807 radians per second.

For the sake of this calculation, I’m going to imagine the wheel of the bicycle being a thin hoop (not terribly unrealistic I hope). The moment of inertia of a thin hoop (again from Hyperphysics) rotating about its center axis is:


With M being the mass of the hoop and R being its radius. This is where we have room to play with numbers. I have an infinite number of moments of inertia to choose from, and I have to narrow it down to one realistic one we can use as a baseline. But since we’re going with the whole “bicycle” motif, let’s use a standard bike wheel size. Let’s use the racing wheel sizes here. According the wikipedia, a standard size for the rim of the wheel is 622 mm. But that’s a diameter, not a radius, so chop that number and half and we get a 311 mm radius for our wheel. That’s .311 meters. I convert everything to these base units because I want the kinetic energy equation to yield Joules without any conversion. Now we’re pretty much done. I’ll solve for the mass of the wheel, and we’ll just guess as to whether or not you believe you could push that wheel that fast.

So, what are we doing? We’re going to solve our kinetic energy equation for the mass of the wheel.


We know that we must supply 310 Joules per second, and we’re going to use a leisurely pace of 1.0833 rotations per second as our angular velocity. We’re going to be using a wheel of radius 0.311 meters. So let’s plug in our numbers and see what we get.


That 69.1719 is the mass in kilograms. To put that into perspective, that’s about the mass of a healthy adult human male. That’s far more massive than any wheel you’ve ever used in a bicycle, so the chances are that you being able to rotate that wheel fast enough are probably slim.

But that was a worst-case scenario. Let’s try a best-case scenario! 80 Watts for the desktop (using Griffith University source numbers) and 20 Watts for the monitor. Oh, and let’s use 6 Watts for the modem (the average power for modems cited above). This gives us 106 Watts of power that we must provide, or 106 Joules per second.


23.6523 kilograms isn’t that bad. That’s close to 50 pounds. That’s still quite massive though. That weight is comparable to… maybe a large dog? My chalkboard weighs 60 pounds so slightly less than that. Anyway I still think that’s unreasonable. That’s bench press material… for me (I am not a strong person).

We see that we lose a lot of energy by treating the computer and the monitor separately. What if we instead used a laptop? Using the MichaelBlueJay numbers, a laptop uses 15 Watts in its best-case scenario. Tack on another 6 Watts for the modem and we get 21 Watts. Plugging in those numbers, we get:


4.686-ish kilograms, about 10 pounds. Which isn’t bad at all. 10 pounds is about how much my cat weighs. So I think this one is reasonable. I’d say that the best-case scenario for the laptop works fine. But I don’t like judging things on their best-case scenario. How about we use the worst-case scenario for a laptop. How about we use that 60 Watts for the laptop plus the 6 Watts for the modem.


That’s about thirty pounds or a little less than the weight of a 32-pack of water bottles. Not sure if that’s reasonable. But I think somewhere in the middle we can get a feasible case for FitFlix working on your laptop. Especially since when I was going 70 RPM the machine told me that I was producing some 80 Watts.

So where do we stand? FitFlix doesn’t work for desktop computers. It’s plausible that it could work for laptops. What about TVs? Simple answer to that: do TVs or more specifically: do Smart TVs (requiring that internet connection) consume comparable power to a laptop? Well according to RTINGS it’s 35 Watts for my 36″ LED TV. If it were instead an LCD TV, MichaelBlueJay would say around 100 Watts. So the answer really depends on your type of TV. But for my TV at least, Fitflix would work out.

Of course, we’re all missing the obvious solutions of using a phone or table, which use vastly less power. So you could definitely power your phone and Fitflix. Depending on your TV you could watch TV and Fitflix. And it’s plausible that you could use your laptop and Fitflix. However, I find it unlikely that you could desktop computer and Fitflix.

That’s just my take on it. If I’ve made any mistakes in the physics or the math or you have some problem with my method, or maybe you have a better method, please let me know. I’m more than welcome to criticism. That’s all for tonight though. Thanks for reading.

Artemis Hunt


The Physics of FitFlix

The Physics of Geizeer

Okidokes. So as many of my readers should know by now, I have a B.S. in Physics. I also happen to be very skeptical when it comes to sensationalist claims. Usually these take the form of some ‘Green initiative’. The goal being to appeal using virtue signal (saving the planet) while also claiming to be cheaper (economic incentive). Solar roadways were a bust. The Fontus water bottle was also a bust. But they still gathered a lot of traction due to being green and claiming to be cheaper and better for the world. And this hearkens back to the post I made about the lottery… what? I didn’t make a post about it? Maybe it was only a Facebook rant. Anyway there was a post a while ago saying that we could solve poverty by splitting the lottery jackpot among all United States citizens. They claimed we could give everyone about $4 million dollars!


I don’t know whether @Livesosa or Philipe Andolini came up with this. However, the math is completely wrong. Everyone would get about $4. Enough to buy everyone a small pocket calculator to check their math. The worst part is that I tried to justify this by remembering that we in the states (and eventually the entire world) use short form when it comes to large numbers. What this means, for those of you not in the know, is that when we say ‘a billion’ what is understood is that we are referring to 1000 millions. In long form, ‘a billion’ would refer to a million millions. The problem is that the order of magnitude is off by a thousand, so I could not justify the calculation. In other words, people are just dumb.

Anyway, so I came across this new type of air conditioner. It’s called Geizeer and the hyperlink leads to their kickstarter. I plan to take on their claims and apply physics to evaluate their truth values. By the end of this post, we might be able to determine whether or not donating to it is a waste of money. So what are the claims?


The first claim is that you will be able to cool your home by 3° C for a (well-insulated) 12 square meter room. So first, we need to think about how large 12 square meters is. And this is where I often run into issues of convention. By my interpretation, 12 square meters would be the area covered by a rectangle that has sides of length three meters and four meters. So to put it into terms we Americans are familiar with, it’s about a 10′ x 13′ rectangle. Which is probably close to the size of my apartment’s floor now that I look around. So it’s not what I would call a small area. Because you want the room to be cool, not just a plane, let’s tack on a third dimension. Realistically you’re going to have an expanding ‘sphere of cooling’ because of the way temperature diffusion works. So our third dimension should be about the height of a room. The height of a room varies based on… well, the room really. So I’m going to stand up in my room and just guesstimate the height. [Will the real Artemis Hunt please stand up] Okay, looks like it’s probably close to nine feet tall. So that’s about 2.75 meters. So the dimensions of the final room which we will be considering for the sake of this thought experiment and calculation is 3m x 4m x 2.75m.

It also claims to cool the air by 3° C for this area. How large of a temperature change is that? Close to about 6° F. This is not a significant change. Room temperature is usually defined to be about 20° C or 21° C. This is close to about 68° F or 70° F. Just for the sake of consistency and to not have to do a ton of range calculations, let’s just stick to 21° C and 70° F being room temperature. The effective change in temperature would bring the room down to 64° F. But you’re likely not using air conditioner if the room is at room temperature. No, you’re probably using it to keep the house AT 70°. Which means your room is around 76° F.

All of this temperature stuff was just to give you an idea of the scale of changes going on. The calculations I’ll be performing are to lower a room from 24° C to 21° C. The first thing we need to do is figure out what equations we’re going to use. Perhaps you’re familiar with an old post of mine in which we did similar calculations, applying it to cartoon physics. We’re going to be using the same equations we used from that post, just with different things. So what equation do we need? Good old rapper name mcDeltaT


We want to figure out how much heat is necessary to change the temperature of a cube of air from one temperature to another. Now you’ll note that the equation refers to heat added. Initially you might be inclined to think that the equation is not applicable. After all, it says heat ADDED. However we want to cool the air! Surely we wouldn’t want to add heat to cool the air, right? Well, fret not, because the equation works both ways. Remember that Δ generally refers to the difference of a final condition and an initial condition. And I worded it that way intentionally; it has to be final condition minus initial condition. Subtraction is not commutative. The order in which you subtract things matters. And if you don’t believe me you can test it yourself. Subtract three from four and four from three and tell me whether or not you get the same answer both times.

Why am I emphasizing that the difference must be final condition minus initial condition? Well, let’s do a small thought experiment to demonstrate the concept. If the final temperature is higher than the initial temperature, then the  ΔT will have a positive value. If the final temperature is lower than the initial temperature, then the  ΔT will have a negative value. Mass m will obviously always be positive and specific heat c is defined to be positive. Just as a reminder, specific heat is the a measure of how much energy must be dumped into a mass to raise it by one degree Celsius.

So the only thing that can change the sign of Q, our heat added, is the sign of  ΔT. If the temperature increases,  ΔT is positive meaning heat was added to the system. If  ΔT is negative, that means heat was removed from the system. So the equation will still work just fine, we will just have a minus sign to denote that heat was removed from the system.

So now we need to figure out what and m are for the air. We know ΔT already (24° C – 21° C = 3° C). According to this source, the specific heat of at around room temperature is about 1.005 kJ/kg*K. So we’ll roll with that. What about the mass of the air? This has to be horrendously tiny. Well… the density of air depends on on its temperature and pressure. We’ll be assuming standard atmospheric pressure because that’s what most rooms you might be using this in will be at. We’ll also use the density at room temperature because it’s reasonably close to that. We’ll be reducing the temperature by about 12.5 percent. Also since we’re dealing with a closed system, we shouldn’t expect the volume (and by extension, mass and density) to change. So from the same page, we see that the density of air at room temperature and standard atmospheric pressure to be 1.205 kg/m^3. Now if we include the temperature change, we can see how much energy we need to remove from the system to produce the temperature change.


Holy smokes, are there really 39 kg of air in my room? Probably not. Myself and my furniture displace a lot of the air so realistically the number is not quite as large. But 39 kg? That’s amazing! I wasn’t expecting such a large number, huh. Anyway, let’s carry on. So that’s the mass of the air in my room in kilograms. Let’s complete the equation to see how much energy we must add to the system to cool it by 3 degrees Celsius.


You may have noticed that I put the units of temperature change in Kelvin, rather than Celsius. Surely that would be breaking the rules? No, it’s not. Well, not in the end. The Kelvin scale is a rescaling of the Celsius scale. The only difference is where the zero is placed. So a change in a degree Kelvin is a change in a degree Celsius. At the end of the day, we see that we need to add -119.891 kJ of energy to the air in the room to produce this temperature change. If it makes you feel better, you may say the same thing by saying we need to remove 119.891 kJ of energy from the air in the room to produce this temperature change.

Now we get into one of the most important laws of physics. Its use has guided our hand in thermodynamics through the centuries. It is practically one of our ten commandments. What I am referring to is the Law of Conservation of Energy. We want to remove energy from the air in the room. But we’re not going to dump it out of the room. Remember that qualifier in the claim. The room has to be well-insulated. This means that energy doesn’t enter the room, and energy doesn’t leave the room. The room is a self-contained environment. What does this mean for the energy? Well it can’t leave the room, so it has to go somewhere else in the room. The natural flow of heat is from higher temperature regions to low temperature regions. Now the Geizeer actually provided such a low temperature region. There’s some kind of gel pack in its center which has been frozen in the refrigerator. Its temperature would probably be comparable to that of something in your freezer. Heat flows from higher temperature regions to lower temperature regions. So the heat in the warm air will flow to this cool gel pack. As the energy leaves the air and is absorbed by the gel pack, the air cools down (and the gel pack heats up).

Unfortunately it becomes very difficult to evaluate the energy necessary to change the temperature of ice packs without knowing what they’re made of. If it’s ice, we’re good to go. I can easily evaluate the physical properties of ice. But it’s some kind of weird gel that I can’t get the material properties of, then my journey stops here. So let’s just look at the ice situation. I would imagine that if ice were better, it would be used instead of some gel. Therefore ice is probably the worse of the two materials to use for the ice pack. So if our calculations work for ice, then we can be reasonably sure that they would work for the gel. However if the calculations don’t work for ice, we may be stuck with no way to decisively conclude anything.

So we’re going to use that heat added equation. But we’re going to rearrange it to solve for mass. We know the specific heat of ice. By this source it’s… temperature dependent. Ouch. Well let’s at least figure out how cold it would be at the outset. If we assume that the inside of the freezer is maintained at a temperature, we can assume that when we remove the ice pack that it will be at that temperature. So what temperature would that be? Well, according to the USDA it should be at 0° F. Which is about -17.8° C. So when we remove the ice pack from the freezer it will be about -17.8° C as well. As the ice pack increases in temperature, its specific heat changes. So let’s use an average specific heat over the range of ice temperatures. The average specific heat from -20° C to 0° C is 1.9984 kJ/kg*K. So that is the value that I will use. Since the density also varies by temperature and we calculated an average specific heat, we should probably do an average density as well. In that case, the average density of our ice will be 918.28 kg/m^3

Now we need to figure out the mass of the ice. Granted, the entire block is not made of ice but we will assume it is for the sake of simplicity. But how are we going to do that? This is where I start doing some pixel math. Let’s use this image


Assuming cubical symmetry we can probably figure everything out. So the box itself is 175 pixels tall. The kickstarter page lists the height of the box to be 134 mm. The length of the box is stated to be 144 mm. So let’s check the length to see if we get a comparable number. Doing another measurement I see that the box is 175 px by 175 px. So they have it as a square. What does this mean for our calculations? It means that the picture is likely not drawn to scale. If it is drawn to scale, then it means the resolution of each pixel probably has an uncertainty of 5 mm. But I’m too far in to pull out now, so forge ahead we must. Since it’s a square, let’s just take the average of the two lengths, or 139 mm and say that 175 pixels is 139 mm. Now let’s get the dimensions of the ice pack. And this is where we run into a bit of a problem. It’s not a simple object like a cube. So to get the volume of the ice pack, I’ll assume it is two separate objects. A rectangular prism on top of a rectangular prism.

First, let’s look at the bottom rectangular prism. Again, we’ll be assuming cubical symmetry, so its length will also be its width. And that length happens to be 120 pixels. Using our scale of 175 pixels being 139 mm, we conclude that 120 pixels is 95.314 mm. Now let’s get the height of the rectangular prism. Counting pixels, we see that it’s 44 pixels tall, or 34.949 mm. So the dimensions of the rectangular prism below are 95.314 mm x 95.314 mm x 34.949 mm. Now let’s attempt to approximate the cube on top. Now I’m going to display to you the box that I will be approximating our smaller rectangular prism with.


Yes, some of the sloping sides are not included but the middle part isn’t entirely filled so this eyeballing should be good enough to give us some ballpark results. The length is 45 pixels and the height is 25 pixels. Wow, I free-handed that box, and got such nice numbers? Let’s convert those pixels to real-world measurements. Converting the pixel measurements, we get a length of 35.743 mm and a height of 19.857 mm. So the dimensions of the smaller prism are 35.743 mm x 35.743 mm x 19.857 mm.

Now we have the dimensions of the boxes. And it is here that we will convert our measurements to meters. Why? Because we have the density of ice in kg/m^3 and I don’t want to deal with converting cubic millimeters to cubic meters. Much better to start with cubic meters. So let’s do that.

Big Prism: 95.314 mm x 95.314 mm x 34.949 mm = 0.095314 m x 0.095314 m x 0.034949 m
Lil Prism: 35.743 mm x 35.743 mm x 19.857 mm = 0.035743 m x 0.035743 m x 0.019857 m

(Oh wow, I never noticed that the font wasn’t monospaced. I’m too used to programming)

We’re almost ready to use our heat equation. Now we just need to get the volume of our sum prisms. This happens to be 0.000342872 m^3. To get the mass of this ice, we multiply the volume we found ( 0.000342872 m^3 ) by the density we settled on earlier (918.28 kg/m^3) and what we get is a block of ice with a mass of 0.3149 kg. Before we go any further, let’s do a little bit of a check. Does this seem reasonable? Well, that’s about .7 lbs. It weighs about as much as a baseball. That seems about right for an ice pack. So our guess doesn’t seem unreasonable.

Alright! So we have our mass (0.3149 kg), we have our specific heat (1.9984 kJ/kg*K), all we need now is our change in temperature! If we permit the ice block to come into thermal equilibrium with the room, we will see a change in temperature of 39° C. Final temperature (21° C) minus initial temperature  (-18° C). I personally do not think that this would be the actual change in temperature, but we may as well do the calculation to see what would happen as time went to infinity. We do need to note a few things though. First, ice won’t exist above 0° C. Water will. So we can only do the heat equation for the energy necessary to raise the temperature of ice from -18° C to 0° C and the energy necessary to raise the temperature of water from 0° C to 21° C. We’ll have to use a different equation for the melting bit but that’s simple enough. It’s the mass of the object multiplied by its latent heat of fusion. The latent heat of fusion of ice/water being 334 kJ/kg. So here’s the process step by step.

  1. Calculate the energy required to raise the temperature of ice to 0° C
  2. Calculate the energy required to melt all of that ice into water
  3. Calculate the energy required to raise the temperature of water to 21° C

Simple enough, right? Let’s get at it. First, we need the energy required to raise the temperature of ice to 0° C from -18° C.


11.3273 kJ. Alright, now let’s figure out how much energy we need to melt all of the ice


105.177 kJ. Excellent, moving right along. Now the ice has been fully melted into water at 0° C. We didn’t add any mass, we didn’t take any away. it was only a phase change. The only thing that really changed was the density of water (it increased). The specific heat of water however, is very different from that of ice. The specific heat of water is 4.186 kj/kg*K. That’s one you memorize quite quickly whenever you do enough of these types of problems. Here’s the energy calculation for the water.


Alright, now for the final step, let’s add all of these energies together. That’s 11.3273 kJ + 105.177 kJ + 27.6816 kJ = 144.1859 kJ. That’s how much energy that must be dumped into the ice pack to produce that type of temperature change. Now the question is… do we have enough? Remember how much energy the air needed to lose to drop 3° C? It was 119.891 kJ. We have 144 kJ of energy. So we do have enough… if we raise the temperature of the ice pack from -18° C to water at 21° C. Let’s figure out how high the temperature of the ice pack has to become to cool the room exactly. To do that, we first subtract the energy necessary to raise the ice to melting temperature and the energy necessary to melt the ice from the amount we predicted would be necessary to lower the air’s temperature. So 119.891 kJ – 11.3273 kJ – 105.177 kJ = 3.3867 kJ. Now we solve for change in temperature using 3.3867 as our heat added value. We get this:


The water would have to be raised to a little over 2.5° C for this to work.

So what does this all mean? It means that in this worst case check, for which the ice pack is literally made of ice, that there is a place to put the energy we removed from the air to cool the room. In fact, we could even cool it a little bit more. Especially when you remember that we likely overestimated the mass of the air in the room because your room probably has furniture in it – which take of space that air could be taking up. Of course… that presents its own slew of problems which we will not be addressing. This blog post is long enough already and I still have another claim to address! So this part of the Geizeer gets a pass by my calculations.


In fact, using it for the duration of 24 hours will have an economic consumption of less than 1 cent.

Let’s check that one out. So the device facilitates the transfer of energy through the use of a fan. This fan is powered by a 3.7 V battery with a lifetime of 7 hours. This means that you must recharge the battery at least twice if you want to run the device for 24 hours. So it’s theoretically impossible to have it running all day however that doesn’t mean we can’t evaluate the claim of energy cost. The fact that I have no idea how much energy is required to charge a battery would mean we can’t evaluate the claim of energy cost. Maybe we can get around that though.

What do we know? We know the energy necessary to to cool the air as claimed. We know that each of these ice packs supposedly works for four hours. This allows us to figure out how much energy is being dumped into the ice pack(s) per day. This next calculation works regardless of whether or not the ice pack is ice or some kind of gel. The reason being that we’re calculating how much energy is being dumped into it per day, not whether or not there’s enough energy for the temperature change to occur. We calculated earlier that we need to remove 119.891 kJ of energy from the air. This is the work necessary to cool the air. We do this over four hours. This means we’re operating at about 8.3258 Watts (8.3258 Joules per second). Why have I done this conversion? Because the cost of power in the United States is rated by kW*hr. The average being about 12 cents per kiloWatt hour. So if we want to figure out how much this power would cost, we multiply by the kiloWatts we’re using and by 24 hours because we’re doing this by the day. We’re using 0.0083258 kW and we’re doing it for 24 hours, so the cost of the energy transfer alone is…


The claim is false. If you ran it all day it would cost you a little over 2 cents. However that’s just the cost to move the energy from one place to another. It does not include the cost necessary to operate your freezer to freeze the ice packs repeatedly. It doesn’t include the cost to operate the fan. Realistically, it will cost you far more than 2 cents per day. But not all of that cost is going directly to the Geizeer’s operation. Your freezer was likely already running all day long anyway, so the addition of the ice pack probably doesn’t add an effective cost to your home.

Now we come up to the big question. Is it cheaper to run a Geizeer instead of your usual air conditioning unit? That’s a hard claim for me to work with and let me begin by stating that I AM NOT AN EXPERT. If your home is controlled by some central air system like mine is, it’s working for far more than just one room. It’s working for your kitchen, your bedroom, your dining room, your living room, and your guest room. That’s a whole lot more air than this little apartment room that I calculated the numbers for. Sure, it’s 2.5 cents for one small room, but when you add up all of the rooms, do the numbers still work out? I can’t say for sure. I did look up a site to figure out how to compare the cost. But the problem is it just depends on the size of the house and the type of air conditioner far too much. But their example is well over even $1 which I don’t see a number of these Geizeers in your home ever reaching. one thing to note, is that you need to buy the Geizeers, while you likely just pay for the operation of your current air conditioner. So if you buy a Geizeer, you’ll likely recoup the cost of the Geizeer at a rate of one Geizeer per month. So at the end of the day, looks like the Geizeer passes my test. Despite one of its claims being false, it does seem to be effectively cheaper to operate compared to an air conditioner. It’s also portable, while your window AC or your house AC probably isn’t. (Shower thought: Do people even use window ACs anymore? I haven’t seen them in forever… but I do currently live in Alaska)

So the Geizeer gets a pass! Thumbsup.jpg

Here is the link to their Kickstarter again.

Artemis Hunt


The Physics of Geizeer

Where’s the Physics? – Episode 1

The image in question


I’d like to begin a series where I pick apart physics in certain media. I found this gif on Facebook, and I feel like it’s a pretty good place to start. The image is from a cartoon called Totally Spies and I remember that it used to air on Cartoon Network. I don’t know the details too much except that it’s supposed to be some female spy show. You could read the Wikipedia article I linked for details.

So before we go into what’s wrong with this picture, let’s go into what’s right. The resolution is pretty poor, but you can see that the middle of the burned door is yellow while the edges are fading shades of red. Because the door is presumably melting, it means that the door is getting rather hot. Lasers are just light beams and the reason they can melt things is because they deposit energy into what they strike. When the door receives that energy, it heats up at the area of contact. Heat then flows from warmer areas to colder areas via thermodynamics. Got a warm spot in the center, cooler areas nearby, the heat will flow. So why do the colours matter? Because hotter materials are different colors. Red hot is colder than yellow hot which is colder than white hot. The middle is the warmest, so it should be the colour that corresponds to highest temperatures. There’s also a small case to be made with how the door melts in that there are two “large” areas with let’s say a “medium” region and “smaller” regions that melt. Two lasers striking the door at different points, those spots should be the hottest, and should melt first. The region between them should reach melting temperature shortly after the striking regions do, so it melts next.

However that’s where the physics stops working. There are two major things that I would like to focus on in the way of bad physics. The laser and the melting door. I’ll start with the melting the door.

When I gave the show props for the colours of the melting door being okay, perhaps I should’ve given them half credit. So remember that when things get warm they change colors? Now obviously, the metal must be yellow-hot before melting. That much is apparent from how the metal was yellow hot before it started melting. But if you watch the melting animation, you’ll see that the edges of the yellow-hot region remain red-hot BUT they start to melt! Bad physics!

The reason this is important is because we view the door as melting from the ‘out-side’ but the laser strikes it on the ‘in-side’! So if anything, the side of the door facing into the room should melt first. That’s why I didn’t dock it points for the melting in the first section. Because we only see the door melting from the outside, we can’t tell too well the order in which it melts.

Now when I looked at the door, I had wondered as to whether or not it may have been easier to simply melt the glass. It’s probably some kind of laminated glass from what I’ve read on sound booth creation. The problem with glass is that it’s hard to really pin down its melting temperature without knowing what type of glass it is. For the sake of argument, let’s just take a high end estimate of 1700 degrees celsius. Many other sources talk about when glass becomes malleable which would be around 1300 degrees celsius but I’m going with melting because it’s easier to equate melting for steel.
(I suppose technically I should go with vaporization because the door doesn’t seem to show where there melt-steel is…) Anyway, the melting point of steel is also tricky because it also depends on the steel’s composition. But according to this source the melting point of steel could be as high as 1500 degrees celsius.  This seemed odd to me because I always thought that glassblowing came before metalworking. The Iron Age, at the very least, came around 1200 B.C. The earliest glassblowing was apparently 1500 B.C. So glassblowing came before iron, which I would infer means that the temperatures necessary to melt glass were easier to achieve than that of iron. So the glass we’re talking about for windows is probably not the same type of glass that was used in early glassblowing.

Why am I talking about all of this melting business? Well, I’d like to talk about how much energy is being delivered to the materials (since it’s easy to talk about lasers in terms of Watts). There’s a material property called specific heat. Specific heat is how much energy is required to raise the temperature of a material. The Watt is a unit of power, and it is defined as a Joule per second. The specific heat of water is 4186 Joules per kilogram-kelvin. You have a 5kW laser, you can raise the temperature of one kilogram of water one degree celsius every second (you actually do a little more but I’m rounding here). Let’s look at the specific heats of glass and steel. DISCLAIMER: To be perfectly honest, I don’t really know if the glass we want is technically on the glass table. I’m going to assume it is, but I don’t know. I could be wrong. So we’re going to use the pyrex glass (753 J/kg*k) and the carbon steel value (120 J/kg*k). Obviously there are different masses of glass, so let’s compare how much glass that we need to melt and how much steel that we need to melt. When comparing densities, I will be using pyrex glass’ density because I’m using pyrex glass’ specific heat. The density of pyrex glass is 2.23 g/cm^3 (though looking at that page, we also see that the melting point is about 1500 degrees celsius, which helps narrow down our calculations). The density of carbon steel is 7.83 g/cm^3. So if we need to melt some mass M of steel to make an exit, we would need to melt about 1/3 M of glass to make an exit. Let’s do some math. We want to know how much energy would we would have to deliver to each material to melt some volume of it, let’s say 1000 cm^3. And we’ll be raising the temperature from room temperature to 1500 degrees celsius. So 1500 °C – 20 °C  (room temperature) = a change in 1480 degrees celsius.

QMCDT (Hyperphysics image)

Let’s plug in our numbers. First for the steel.


So about 1.4 mega Joules or 1.4 MJ for short (Billie Jean, is not my lover….) The assumption of 1000 cm^3 of material getting melted went into that… is that reasonable? Well… that’s about 61 cubic inches. That door looks to be maybe 2 inches thick?  Look  at that huge gap, it’s probably a great deal more than 61 cubic inches. So our estimate is low if anything. But going through the mess of trying to figure out with scaling how much material is actually melted is probably more complex than what readers care about. Also I’m eyeballing everything because I don’t have photoshop to pixel-perfect this stuff. Now let’s see how much energy we need to melt the glass.


I’ve defined pg and ps to be the densities of glass and steel, respectively. We see immediately that you need about half the required energy to melt the glass! (Lord knows what would’ve been required to burn away the sound-absorbing material that surrounds the room…) And before you say that they would’ve had to climb up to that little window door – LOOK AT THE GIANT FREAKING WINDOW TO THE RIGHT OF THE DOOR. Also, it takes like 2 seconds (being generous) to melt the door. That means that the laser used is delivering about 700,000 Joules per second to the door. That’s .7 Mega Watts. This was delivered FROM the lipstick ehh… to be frank I have no idea what they’re called. Let’s say lipstick pencil. This .7 MW were delivered from the lipstick pencil. Which worries me because I want to know how this .7 MW laser was generated in such a small device, and why the device doesn’t heat up in the woman’s hand as she fires it. And why doesn’t it destroy the compact mirror that was used to reflect it? We’re also neglecting the most obvious of questions – why did she not just point the laser at the door and melt it like that? Reflecting lasers does not increase the power of lasers. There’s no additional energy being pumped into the laser. If anything, the laser is losing energy.

That’s all of the door stuff. Now let’s focus on the laser – mirror thing. It is its own boatload of problems but it should be much easier to deal with. Let’s suppose that, (for some reason or another) the lipstick pencil doesn’t melt as it is being used. The compact mirrors are lying on the floor. The lipstick pencil is CLEARLY pointed towards the mirrors. By the Law of Reflection (and now we have to assume that for some reason this .7 MW laser doesn’t destroy the compact mirror in a second) the reflected beam should’ve gone into the floor, not to the other mirror. But let’s say the light beam bends towards the other mirror. Then, in fractions of a second the light beam would STILL go into the floor. See the picture below. QMCDT

So how long would this take? Well, if each compact mirror is… say… about 7 cm tall (2.7 inches, 0.07 meters) and they’re placed… what looks to be a full tile apart. Those tiles look to about two female heads in length? What’s the average length of a female head? Well, a Google search led me here. Sure it’s circumferences, but we can work with that. About 53 cm to the circumference of a female head. These women appear to be teenagers from my point of view? But that’s fine, we’ll use the same value for an adult woman’s head. 53 cm divided by pi (which is pronounced ‘pee’ in Greek by the way) leads to about a 17 cm diameter head. We want two of those, so we’ll assume that the tiles are 34 cm (0.34 meters) apart.

(I keep putting things in meters because I can quickly recall the speed of light in meters per second)

Let’s do some math. Let’s say that the incoming angle is 1°. Completely unreasonable, but I’m using it to prove a point. Simple trigonometry shows that the first reflection, if the laser strikes the very top of the mirror, will travel 0.006 meters down. Okay, cool. So we can actually bounce a few times. We can bounce a lot of times, actually. How many bounces? We can bounce 11 times before hitting the floor. Almost 12 times but almost only counts in horseshoes and hand grenades. The total path length for those 11 reflections is 11 times is really close to 11 times .34 meters (this is actually somewhat expected due to the small angle approximation for tangent). So light would travel about 3.74 meters before it hit the ground. How long would that take? Well, c, the speed of light, is 3*10^8 meters per second. So, close to 1/1*10^8 of a second. The time of reflection in the gif is much longer than that, so bad physics.

So this has been a short excursion into the realm of cartoon physics. I hope you’ve learned a little something, as it has been a pleasure getting serious about silly little animations. That’ll be all for now.

Artemis Hunt

Where’s the Physics? – Episode 1