Mass Drivers Cost Far Too Much -- by Keith H. Lofstrom

Posted December 22, 1995 on sci.space.tech newsgroup.

There are frequent proposals to build electromagnetic launchers based on
coils, heaps of power switching electronics, and so forth, attempting to
beat the high cost of rockets.  Proponents point to benchtop experiments
that push sub-kilogram masses to tens of meters per second, and extrapolate
to launch systems that can push multi-ton payloads into orbit.

It isn't that easy.

Power electronics cost can be measured in fractions of a dollar per watt
handled.  Perhaps you can posit some super-duper new semiconductor
technology that somehow gets around power handling limits, but in
real life pushing large amounts of power through an electronic switching
device deposits heat in it.   If you find a way around this, there is an
existing multi-billion dollar electronic power market just waiting for
you to take over.  Miracles being unlikely, assume you are stuck with
what exists now.

Electronic systems are not 100% efficient - 90% is a more likely number.
If you are putting 10 watts into pushing the payload, about 1 watt will
be wasted as heat - mostly in the switching device.  The system scales
to the amount of waste heat it has to absorb.

Magic materials aside, what is available for switching power in
semiconductors is silicon, and a number of exotic 3/5 compounds.
Silicon is a mediocre conductor of heat, but the other materials are
worse.  A "nearly magic" device would consist of a thin silicon die
firmly welded to a chunk of copper.  The thicker the silicon, the 
longer it takes to get heat out.  Let's assume a very thin but practical
silicon die, perhaps 20 microns thick.

Silicon has a specific heat of 1.6 joules/cm3-C, and a thermal conductivity
of 1.5 watt/cm-C.  These two numbers result in a "thermal diffusion
constant" of 1.1 seconds per cm2.  A 1 cm thick slab of silicon will
reach thermal equilibrium in around a second, a 1 mm thick slab in
around 10msec, a 20 micron thick slab in about 40 nanoseconds.  We
can treat time periods longer than equilibrium time as "steady state".

Mass drivers and such pulse their power through switches.  How long are
the pulses?  Well, a 1 meter payload nearing the exit end of an orbital
launch mass driver will be moving at about 10,000 meters per second, so
we can assume pulses of around 100 microseconds.    Much longer than
40 nanoseconds - effectively, our silicon is in steady state.

The copper slab we are welded to doesn't reach equilibrium.  Copper
has a specific heat of 0.8 joules/cm3-C and a thermal conductivity
of 4 watt/cm-C, so it diffuses heat about 5 times faster than silicon.
However, the block of copper is where the heat from the pulse ends up,
so it can be considered to be infinitely thick, and we are instead
worried about heat diffusion into copper.  The temperature rise at the 
surface of the copper is about 0.6 degrees Celsius times the power
in watts/cm2 and the square root of the time in seconds.  Thus, for
a dissipation of 1000 watts per square cm, and a pulse time of 100
microseconds, we get a temperature rise of 6 degrees Celsius.

We can probably stand a 100 degree instantaneous rise on the silicon
surface - higher than that, and things will rapidly degrade.  The silicon
device can thus take a pulse proportional to 170 watts/cm2, divided
by the square root of the pulse length.  A cm2 of masked, processed
silicon costs about 2 dollars.  Thus, for one dollar we can dissipate
around 100 watts * seconds^(1/2).  A 100 microsecond pulse allows us to
dissipate 10,000 watts for a dollar.  Assuming 10 percent efficiency,
we can deliver 100,000 watts into accelerating a payload, per dollar,
at this pulsewidth.

(Note:  this is *incredibly* optimistic - typically, power electronics
costs more like $1/watt.  I have a catalog with a microscope illuminator
power supply for $1500 that delivers 12 watts!)

Let's assume our mass driver is built up out of sections of length
LS, an exit velocity of VE, and we drive payloads of mass M with an
acceleration of A.   The total length of the mass driver is 
LT = VE^2 / (2 * A), and thus there are NT = LT / LS sections.  Number
the sections N, from 0 to NT.  The payload mass is M.

The velocity at the Nth section is 

      VN = sqrt( 2 * A * LS * N )

The time pulsewidth at the Nth section is

      TN = LS / VN = sqrt ( LS / ( 2 * A * N ) )

The power at the Nth section is

      PN = Force * Velocity
         = M * A * VN
         = M * sqrt( 2 * A^3 * LS * N )

The power dissipation cost of the Nth section is

      CN = PN * sqrt( TN ) / ( 100 / 10% )

         = M * 2^(1/4) * A^(5/4) * LS^(3/4) * N^(1/4) / 1000
         (M in kg , A in m/s2 , LS in m, results in dollars )

The total power dissipation cost is approximately

         = M * 2^(1/4) * A^(5/4) * LS^(3/4) * NT^(5/4) / 1250
         (M in kg , A in m/s2 , LS in m, results in dollars )
          
        ********************************

         = M * sqrt(  VE^5 / LS ) / 2500

        ********************************

      ( VE in meters/second,  LS in meters, M in kg, results in dollars )

Note that the cost is independent of total length or acceleration.  A
longer system uses less power but more sections.  

A benchtop device with an exit velocity of 100 meters/second,
a section length of 0.1 meters, and a payload of 0.1 Kg should have
a silicon cost of about 13 dollars.

Assuming a 1 meter section, a 1000 Kg payload, an exit velocity of 10,000
meters per second, we get a resulting cost of about 4 billion dollars.
This is just silicon cost;  everything else is assumed free, including
a power delivery system capable of producing many gigawatts.

Power source cost?  The peak power at the muzzle is

        Pmax = M * V^3 / 2 * LT.  For a 100 Km long accelerator, 1000 Kg,
and 10Km/sec, that's 5 GW worth of power.  A shorter accelerator is 
correspondingly higher power.  That's a *lot* of Die Hards.

-------------------------------------------------------------------------

A more conservative power cost estimate is based on just the peak watts,
not the pulse width.  If one assumes a constant 200 controlled watts per
dollar, the equation turns into

    Cost = M * VE^3 /( 600 * LS )

and our two examples become $830 and $830 billion dollars respectively.

-------------------------------------------------------------------------

Conclusion about mass driver costs:

The electronic cost of an electronically switched mass driver goes up as
something between V^2.5 and V^3, and is proportional to the mass.  A
high velocity mass driver is far more expensive than a benchtop toy,
and real driver costs will be far higher than these incredibly optimistic
estimates, due to the cost of coils, structure, heat sinking, and assembly.
Low rep-rate pulsed systems are far more costly than steady state systems.