That’s a Big Transistor

Here's some fun for a Friday.

A few years back a bunch of my coworkers and I got to discussing the space program over lunch. Someone asked why it is that we continue to launch devices into orbit by strapping a big old tank full of liquid oxygen to the device and then set it on fire. Why haven't we developed better technology using magnets or something?

I did some poking around the web and wrote up a little summary of the analysis, which I present here for your amusement.


Suppose you've got a 1000 kilogram object that you'd like to be an arbitrary height above the surface of the earth. In order to get an object to an arbitrary height, we need to accelerate it to the escape velocity of the earth, which is 11 kilometres per second. How much energy does that take? The kinetic energy of a projectile is half the mass times the square of the velocity:

Kinetic Energy = 0.5 x 1000 kg x (11000 m/s)2 = about 60 gigajoules

Sixty billion Joules of energy sounds like rather a lot, but really it isn't. The electrical power for your house is measured in Joules, but because Joules are so tiny, the bill comes in kilowatt hours. One Watt is a power consumption of one Joule of energy per second, so a kilowatt hour is a power consumption of 1000 Joules per second for 3600 seconds = 3.6 million Joules. That costs about ten cents. So that sixty billion Joules would only cost about $1700 -- near what it costs to keep 20 one hundred Watt light bulbs burning for a solid year. And power costs less if you buy it in bulk.

$1700 to put a tonne in orbit is incredibly cheap. Why doesn't NASA use electricity instead of chemical power? Turning the electricity into acceleration quickly is harder than turning it into heat and light slowly, but surely we can figure something out.

One way to do it would be to build a coil gun. Coil guns are very easy to build -- in fact, I've built one myself. A coil gun works like this: you have a metal projectile sitting in a tube. One end of the tube has a coil of wire wrapped around it. When an electric current is applied to the wire, it turns the coil into an electromagnet, which pulls the projectile towards it.

Of course, the current has to shut off before the projectile passes entirely through the coil, otherwise the electromagnet will be pulling the object back towards it, slowing it down.  The projectile sails on through the coil, out into space.

You can build multi-stage coil guns. The first coil gets the projectile moving. As the projectile leaves the first coil, that triggers a switch that turns on the next coil, and so on. Each coil makes the projectile a little bit faster.

So a coil gun stage has three basic parts: the coil itself, something which stores the electricity -- usually a capacitor of some kind -- and a switch to connect the power source to the coil at exactly the right moment.

I once had an old television set that didn't work anymore. Before I took it to the dump I ransacked it for capacitors. I had everything I needed around the house -- a few capacitors, a diode, some old telephone wire, a piece of spare kite spar tubing, and a light switch. Run wall power through the diode to charge the capacitors, run power from the capacitors to the coil, interrupted by a switch. Put a nail in the tube, hit the switch, and the nail goes flying. Cool!  (Note: large capacitors can kill or wound you when charged.  Handle with care.)

The problem is that none of these pieces scale up.

Suppose we've got a really huge multi-stage coil gun. Just to keep the math easy, let's say we have ten thousand coils, each a metre long with which we're going to accelerate our projectile to 10000m/s by applying a uniform acceleration of 5000m/s2. That acceleration would squash humans like bugs, but we could use this thing to move large quantities of equipment into orbit. When the on switch is hit, a mere two seconds later the projectile will be heading into space at 10000m/s.

Consider only the last coil of the ten thousand. The projectile is going to be moving at almost 10km/s, so it will only be in the coil for about 100 microseconds. We can assume that this is about how long the last coil is going to be energized.

It is vitally important that the pulse of electricity delivered to the coil be short, for three reasons. First, as I mentioned before, the magnet had better be off by the time the projectile reaches the far side, otherwise the magnet will be slowing the projectile down. Second, the farther the projectile is from the coil when the coil is energized, the weaker the magnetic pull will be; you don't want to waste power by turning the magnet on while the object is too far away to get much oomph from it. And most important, the magnetic field strength of an electromagnet is proportional to the electric current. Current is electrons moved per second, so if you want high current you have to either make the number of electrons moved larger, or the amount of time you spend moving them smaller, or preferably both.

For all these reasons we can assume that the pulse is going to be extremely short, on the order of 100 microseconds.

Current is voltage divided by resistance, power is voltage times current, and the heat produced by resistance is a function of current.  (That's why we have high-voltage power lines to deliver power: we step the power up to high voltage so that it moves lots of power without creating too much heat. We then step it back down to more useful low voltage/high current at local transformer stations.) The resistance in copper wire is going to be non-trivial, and worse, the magnetic fields set up in the electromagnet are going to themselves push against the electrons moving through the coil -- we're going to get inductance on the line. The closer together the coils, the stronger the electromagnet will be, but the more impedance will be produced. In order to get enough current, we're going to need huge voltages. And no matter how you slice it, the coils are going to generate heat, lots of it. But that's OK. We can cool the coils, and make sure that we don't pump so much power into each coil that the copper wire melts.

But there is another place where heat is produced. 100% of the current for each coil has to pass through a switch. A conventional switch, where you have two pieces of metal with a gap between them, and you remove the gap, isn't going to work for the kind of power we're talking about. We're talking about currents way larger than arc welders use, so basically the switch will weld itself shut. And as the switch closes, the high voltages will cause current to leak across the channel when the switch is half closed, thereby increasing the amount of time that electrons are flowing. That will weaken the magnetic field. Mechanical switches just aren't going to cut it.

Fortunately we have a very fast, very precisely controllable switching technology: transistors. A transistor is a chunk of silicon which has been carefully constructed so that it can act as either an electrical insulator or an electrical conductor. They are incredibly high-speed switches -- that's why we make computers out of them. Let's just build ten thousand transistors, one for each coil switch. The switch for a coil will be triggered by the projectile interrupting a laser beam crossing the previous coil.

Hold on, a minute though. Transistors are semiconductors -- they do not transmit electricity perfectly. They're maybe 90% efficient. About 10% of the power transferred through the switch is going to be turned into heat inside the switch. How much heat can a transistor take before its wrecked? Anyone who has overclocked a Pentium knows what I'm talking about! Transistors get real hot real fast when you try to push a lot of power through them.

How much power?

Power = mass x acceleration x distance / time

By the time we hit the last coil, the 1000kg projectile will be moving at about 9999.5m/s. We need to accelerate it up to 10000m/s, we have 100 microseconds in which to do so, and one metre. The acceleration is 5000m/s2. If we work it out, the total power required by the last stage is fifty billion Watts. The five million Joules cost about a dime; it's getting them where they need to be in 0.0001 seconds that's the hard part.

We're going to turn 10% of that power directly into heat in the switch, so that's five billion Watts of heat to dissipate. Imagine the heat of ten million 500 Watt spotlights concentrated in a small area, and all turned on for 100 microseconds.  It'll be hot.

Note that we're assuming that 100% of the force generated by the magnetic field is translated into motive force on the projectile. In reality, only about 25% of the magnetic force actually turns into acceleration.  We'd need to up the power by at least a factor of four, so really we've got more like 20 billion Watts of heat to dissipate. But let's ignore that for now.

Let's assume that the transistor is extremely thin and therefore the transistor is equally hot everywhere. An extremely thin transistor is also a good idea because a thin transistor has large surface area. The larger the surface area an object has, the faster you can cool it. Let's wrap our transistor in a huge copper heat sink.

Slabs of copper do not diffuse heat infinitely quickly. The temperature of the surface of the heat sink that is touching the transistor will rise based on many factors -- higher power consumption, smaller surface area and longer application of heat by the transistor all cause a larger temperature rise at the interface between the transistor and the sink. The heat capacity and thermal diffusion of copper are well known, and from these facts we can work out that:

temperature rise at surface in = 0.6 x ( Watts / cm2) x seconds

Clearly if the heat sink is not sucking heat out fast enough, the heat sink is going to itself get hot enough to wreck the transistor. Let's suppose that the transistor can take a rise of 100 degrees Celcius before it is destroyed. We have 5 billion Watts of heat, and 100 microseconds to get rid of as much of it as possible. How much area do we need to ensure that the surface of the heat sink rises only 100 degrees?

Solve the above equation for area and you get 300000 square centimeters. That's a square transistor 5.5 metres on a side, about the size of a typical kitchen floor.

Transistor-grade wafer-thin silicon costs about $2 for a square centimeter, so the transistor for the last stage alone will cost us about $600000 dollars. And that's for the stage that only adds 0.01% of the total oomph, for a device that can only launch a one-tonne projectile, and we've neglected inefficiency in the coil. Clearly to build the whole device would cost multiple billions of dollars for the on switch alone. That's an expensive on switch!

Of course, we could solve these problems by inventing new magic materials. If we had cheap superconductors that conducted electricity and heat with 100% efficiency at reasonable temperatures, supported large currents and fast switching, then sure, we could build mass drivers that put objects into orbit at low cost. Tragically, we do not have magic materials.

And of course I haven't even mentioned the difficulties of generating and storing enough power to light Seattle and then discharging it all in two seconds. Generating large quantities of power spread out over lots of time and space is easy. Concentrating that power into a very tiny time and space is hard. Fifty gigawatts for the last stage alone -- we'd need the Mr. Fusion from Back To The Future!  (Actually, we'd need around 40 of them, as Mr. Fusion generated the mere 1.21 gigawatts required to power the flux capacitor.) I've also glossed over many serious difficulties in the implementation of the coils; dealing with self-inductance, magnetic eddys and other issues is non-trivial.

Until there are radical new advances in material science and power management we're going to be stuck with strapping big tanks of liquid oxygen onto the sides of projectiles if we want to get them into space.


Further reading:

I stole this argument from this excellent, more technical description of the switching problem:

This site is a good one if you're interested in exotic orbital technologies:

This guy builds

coil guns with switches rated to 14000 amps at 1300 Volts. I would imagine that these are not cheap.

This page has some great explanations of the

magnetic and electric physics involved in coil guns.


Comments (24)

  1. Jonny D says:

    well there is on other huge factor that has been ignored.

    basically there will be inductance in the object traveling through the coil. so whatever you are trying to send up to space will heat up.

    Sure you could try to use no-magnetic materials, but then you can’t accelerate it. So. you are bound to melt anything that you try to send up. and further more, the magnetic fields will rip it to pieces…

    this really might be the most entertaining way to quite literally burn money.

  2. G. Man says:

    Awesome! Man what the heck did I read before blogs?

  3. Robert Hahn says:

    Fascinating writeup, Eric. You got me wondering though – if 1000kg is going to be a problem, what if we scale the payload *down*?

    A lot of progress is being made in the robotics field. I would suggest the following: You want an object weighing 1000kg in orbit. Design the object into 500 2kg objects with enough smarts (and a bit of power) to track and locate their ‘buddy parts’ and self-assemble.

    Naturally, I would expect that a design like that may result in an object that would weigh more than 1000kg, because there will be overhead, but how much really? Artificial intelligence code optimized for self-assembly won’t be needed after – so the RAM can be flashed with the mission specs to make that object useful. 500 small powersources might be a nice way to design this object because it would be more fault tolerant. Or maybe the power source for each object can be detached and collected in one place so that orbital changes can be made as the mass of the object increases. Or, have the object self-assemble on the moon, then blast off from there into a stable earth orbit.

    it’s easy for me to say this stuff – I’m not doing the math 😉

    good article though. really fun!

  4. "Unfortunately, electrical resistance is equal to current divided by voltage"

    Err… That’s not how Ohm saw it. Ohm’s law states that V = IR. If you rearrange that to express R in terms of voltage and current you get:

    R = V/I

    So that would be voltage divided by current, not vice versa. So given what you go on to say:

    "That’s why we have high-voltage power lines to deliver power"

    given that resistance is apparently voltage divided by current, wouldn’t a high voltage system be worse because there is more resistance?

    Actually, no – stepping the voltage up or down doesn’t in fact change the resistance of electrical cables at all. The resistance remains constant. (Well, it changes according to temperature, but it’s not directly dependent on the voltage or current.)

    So in the equation that relates current, voltage, and resistance, resistance is effectively a constant when dealing with power cables. All you do by upping the voltage is to reduce the current – the resistance remains the same.

    So why do we use high voltage power cables? The answer lies in *power*. The goal is to minimize the loss of power. Here’s the equation that lets us calculate the power disippated:

    Power = VI

    The the amount of power dissipated by an elecric power cabls is equal to current multiplied by the voltage drop across that cable. (Note that this is *not* the voltage of the whole power supply. It’s just the amount by which the voltage drops across the cable – this is the power loss in the cable we are remembering.)

    So how do we work out what the voltage drop is actually going to be across a given cable? Well for that we use Ohm’s law, which remember is:

    V = IR

    So the voltage drop will be equal to the cable’s resistance multiplied by the flowing current.

    By substituting in this equation for V into the equation for power loss, we can work out the power loss as follows:

    Power = V * I

    = (I*R) * I

    = I*I*R

    So the power loss in an electric cable is equal to the square of the current multiplied by its resistance.

    Notice that the voltage doesn’t figure anywhere in this equation. Only the current. Of course voltage and current are related, in that in order to send a given amount of power down a cable, Voltage*Current must be equal to the power. But you are at liberty to trade one off against the other – if you need a kilowatt, you could have 1V at 1000 amps, or 1000V at 1 amp.

    And bearing in mind that the equation for power loss in a cable only features current, and not voltage, one of these looks more attractive than the other. With 1000V at 1 amp, you’ve got a power loss in the cable of I*I*R, thats 1*1*R, or R. I.e. the power loss will be whatever the resistance is. But with a volate of 100V and 10A, the power loss is now 10*10*R, i.e. 100 times greater!

    The resistance remains constant throughout these calculations. It’s just not right to say that resistance drops as you increase voltage. It doesn’t. The power loss through heat dissipation in the cable drops, but that’s because this loss is a function of current and not voltage.

    (Of course the fact that cable heats up as it dissipates heat and that changes the temperature complicates matters a little. R isn’t really a constant. But even taking this into account, high voltage still ends up looking attractive.)

  5. Eric Lippert says:

    You’re absolutely right — I totally screwed up that explanation. I’m not sure what I was thinking when I wrote that. I’ll make the offending paragraph more clear. Thanks!

  6. Enigma2e says:

    Back in high school, my senior year, i did this as a science fair project. I actually went through several designs before i settled on one. I still havent tested the latest incarnation for one reason alone. The switching transistors i use are 7 bucks a-piece and rated at 70 Amps each. However, last time i ‘tested’ it with only 3 of the 6 stages operational. I ended up with just a metal backing of the transistor + a heatsink. the ceramic and chip had literally disintegrated themselves. And the only thing i was using, as i couldnt find any caps, was a 12v deep cycle marine battery. I have pictures of my experiements here hosted on my {slow} DSL server. Username is "coilgun", Password is "coilgun", if prompted.

  7. Scott says:

    There’s also a problem that has nothing to do with electronics. Inertia. You’re taking an object at rest (an object with alot of mass) and accelerating it at an incredible rate. You’re going to need a large platform capable of holding its own against the (for lack of a better term) "back energy" in order to achieve the maximum forward energy.

    The platform size is already going to be required to be large and strong enough to support the weight of 10k meters of coils, the electronics and cabling, potentially a lift system for maintenence personnel.

    It could be an interesting sight though. Imagine the cabling running from the ground to the upper levels of the coil gun, like you would see on a cellular tower. Now imagine firing the gun, the inertia actually compressing the structure toward the ground and those cables going slack. Then going taut again when the projectile is released (hopefully!)

  8. Eric Lippert says:

    Since it would have to be kilometres long, you’d want to build it up the side of a mountain, going from horizontal to vertical gradually. Of course, then there are all the problems associated with making an object with lots of inertia curve!

    I think one of Heinlein’s novels featured a coil gun up the side of Pike’s Peak for launching stuff into space.

  9. Peter Torr says:

    Eddy’s in the current again! Somebody get him out!!

    Plus strapping stuff to objects and blowing it up in a sem-controlled fashion is much cooler than having silly electrons do all the work… imagine watching "Armageddon" or "Apollo 13" without all the cool take-off effects.

    On second thoughts, you don’t have to imagine anything at all — just watch "The Core" and realise how crap the world would be without rocket take-off sequences in blockbuster Hollywood movies about deep-sea oil drillers saving the world from ulitamte destruction.

    What would Jerry Bruckheimer do?!?

  10. Patrick says:

    "Until there are radical new advances in material science and power management we’re going to be stuck with strapping big tanks of liquid oxygen onto the sides of projectiles if we want to get them into space."

    The conclusion does not follow from the argument. You’ve demonstrated that coil guns would require radical material advances (radical, but hardly magical).

    You haven’t demonstrated that all alternatives are infeasible. There may yet be a non-liquid oxygen solution with current materials and power management technology. Perhaps solid fuel rockets, or propulsion by ground-based lasers. Who knows.

  11. Zac says:

    Eric, your Mr. Fusion arithmetic is bogus. If you listen closely, the flux capacitor clearly required 1.21 "jigga"-watts. Clearly Doc had some sort of communication back from the future related to the power of Jay-Z, but we don’t really know what those units scale to.

    Honestly, such a lack of rigour. Tsk tsk.

  12. Lance Fisher says:

    Like Patrick said coil guns aren’t the only way to convert electricity into kinetic energy in order to move an object to space. You could use a giant elevator, which also have their problems, but at least you can convert the energy slowly. Check this out:

  13. Eric Lippert says:

    As the article notes, a space elevator has to be made out of fictionite. Again, this is a technology which is predicated on radical advances in material science.

  14. Kalon Jelen says:

    Nice reference of Ghostbusters in the title, BTW. Great article too.

  15. Martin says:

    Has anyone worked out how many megawatts of power would be saved if the USA converted to the same standard as the rest of the world and used 230V as its voltage for domestic electrical supplies? 3kW electric heaters would only use 13A. Ohmic losses would be halved.

    Or you could halve the amount of copper used in the wiring…

  16. Steve says:

    I knew a 1.21 gigawatts reference would be in this article. Eric, you have a lot more time on your hands my friend.

  17. Martin,

    That’s one of those pie-in-the-sky savings though. The energy, (both human and electrical) that it would take to get the entire country to switch over would be immense, costly, and most likely tragic for the economy.

    Microsoft isn’t the only entity not interested in breaking backward compatibility just for the sake of new design or efficiency. 🙂

  18. Norman Diamond says:

    1/25/2005 1:20 AM Martin

    > if the USA converted to the same standard as

    > the rest of the world

    and had a mix of 100, 120, 220, and 240, matching the standard of the rest of the world?

    In a house in a medium-sized Indonesian city, when a refrigerator was moved from one room to another, they had to buy a transformer.

    > used 230V as its voltage for domestic

    > electrical supplies?

    Yes, that part would be beneficial, except for the minor issue that Carmen Crinoli pointed out. Maybe the whole world should move to a metric 1kV.

  19. Simon says:

    Excellent discussion. Got me completely in the mood to help my son with his physics homework.

    I’m glad someone remembered Eddy!

    I’m sure it won’t be long before one of the eggs at MIT or O&C discover a new form of energy that can aid propulsion.

    It just might take a few years to get past all the Health and Safety standards and application forms of our respective nations, before you can buy a domestic version of it in WalMart or Woolworths.

  20. Phylyp says:


    You missed your vocation – teaching. If you had been my physics teacher, I’d have been inspired.

    Reading through this, e = 1/2 * mv^2 seemed familiar. Then it hit me: I’d learned this so *many* times in drab proofs, this was a thrilling account of its actual usage!

  21. Eric Lippert says:

    Thanks, that’s a very nice thing to say.

    I’ve often considered whether I’d enjoy teaching, and the conclusion that I’ve reached is that trying to teach people who didn’t want to be there would be intensely frustrating for me. I don’t think I could be a high school teacher, for instance.

    I suppose I could teach college courses though. Perhaps after I retire from Microsoft I’ll spend my time teaching classes and writing books! 🙂

  22. Bramster says:

    Great discussion! And there are other considerations.

    Why not a mixture of technologies. It seems to me that it takes at least a couple of seconds for the shuttle to clear the launch tower. . . (I’ll try to time it the next launch). How much fuel is used to lift all that fuel past the end of the tower?

    How about an elevator with the mother of all counterweights for the first part of the journey up the inside of mountain, a coil gun for the rest of the time inside the mountain, and then conventional chemical rockets for the rest of the trip to orbit?

    Maybe even a super laser to pump energy into an ablative stage before the Goddards kick in.

    Recall that Armstrong & Aldrin’s Lunar Excursion Module had enough fuel to both slow from lunar orbit to a dead stop, and then return to the same orbit.

    Now, if Bill Gates really wanted to leave a legacy. . .

  23. There’s one other thing to consider: The sattelite you may want to launch has to include quite much fancy electronics to do it’s work, but if you have coils which induce a high current into the electronics (and you would need enormous shielding to prevent this), how do you manage to NOT fry the electronics? How could such a great shielding be made?

  24. Tanveer Badar says:

    And just not forget how particle colliders work these days. I am not talking about child stuff of betatron and synchronotron.

    Consider Stanford Linear Collider (SLC) and CERN’s proton-antiproton collider. They work exactly like having multiple electromagnets pulling the particles and imparting energy to them during the process.

    The only difference is that the payload is around 10^31 kg this time and travels an equivalent distance to a couple of earth’s circumference in case of CERN.

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