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```From: higgins@mecheng.mcgill.ca (Andrew Higgins)
Newsgroups: sci.space.science
Subject: Re: Gun lanched space craft
Date: Wed, 09 Dec 1998 18:08:00 +0200

In article <19981207234614.12523.00001832@ng29.aol.com>,
rbynum3965@aol.com (RBynum3965) wrote:
>
> Why not use a catapult to launch satelittes? A 1 km arm with a 1/10 sec
> 90degree swing will accelerate a satellite somewhere in the ballpark of 15
> km/sec. Now  with enough styrofoam peanuts we might be able to send people up.
> Robert Bynum
>

This idea has been suggested before, notably by Robert Zubrin and David Baker:

Baker, D., Zubrin, R., "Lunar and Mars Mission Architecture
Utilizing Tether-Launched LLOX," AIAA Paper 90-2109, 26th
Joint Propulsion Conference, Orlando, FL, July 16-18, 1990.

Unfortunately, the problem is that the tension in the catapult arm exceeds
the tensile strength of the arm itself, and the arm will tear itself into
pieces.  Even for the very best materials, you cannot have a tip velocity
much in excess of 1 km/s, unless you happen to have some lengths of
buckytube robe hanging around.  [See detailed calculations below.]

Tapering the arm (which would more likely be a tether, rather than an
rigid catapult) can help overcome this material limit somewhat.  For
example, the Baker and Zubrin design I cited above used a Kevlar tether 7
km long, which tapered from 18 cm at the base to 10 cm at the tip, which
would spin at 2-3 rpm.  This is sufficient to launch a payload from the
lunar surface to low lunar orbit.

Another problem would be aerodynamic drag on the tether.  Baker and Zubrin
solve this problem by basing the launcher on the moon.

The material limitation is what leads to Derek Tidman's "Slingatron" idea,
mentioned earlier in this thread.  Rather than *pull* the payload in a
circular path by using tension in a sling, why not *push* the payload
along a circular track?  The track makes a "Hula-Hoop" motion synchronized
with the payload as it travels around the track, always pushing inwards.
By this technique, you can by-pass the material limit of sling-type
launchers.

This is the same mechanism by which you can accelerate a marble rolling
around the bottom of a bucket to very high speed by gyrating the bucket
with a small-amplitude circular motion.

Tidman's latest versions of the "Slingatron" dispenses with
electromagnetic levitation of the vehicle and instead relies on a
gasdynamic bearing or on evaporation/ablation of the underside of the
vehicle to maintain a nearly frictionless contact between vehicle and
track:

Tidman, D., "Slingatron Mass Launchers," Journal of Propulsion
and Power, Vol. 14, No. 4, July-August 1998, pp. 537-544.

The numbers on this concept appear quite promising, but just how low the
friction between vehicle and track can be kept is probably the biggest
unknown and is very difficult to estimate.

Appendix:  What is the fastest speed a sling-launcher can obtain?
=================================================================

Consider a sling-arm of length "R", cross-sectional area "A", and density
"rho".  Neglect the payload mass on the end of the tether (tether mass
ends up being much greater than the payload mass in most designs anyway).

The tension created by an increment of mass on a sling in constant angular
motion is given by:

T = dm V^2 / r

Where "dm" is the incremental mass, "V" the velocity of the mass, and "r"
the radius of its circular motion.  Thus, the total tension in the tether
is given by the integral over the tether length:

Ttotal = Integral[rho A (omega r)^2 dr/r, from 0 to R]

(where we've used rho*A*dr for the incremental mass dm)

where "omega" is the angular velocity (a constant) and "r" is the distance
along the tether.  Integrating, we get:

Ttotal = (rho A omega^2 R^2) / 2

Since omega*R is the tip velocity (Vtip) we can substitute:

Ttotal = (rho A Vtip^2)/2

Notice we could also have gotten this answer by dispensing with the
calculus and simply treating the tether as a point of mass rho*A*R
traveling at velocity Vtip/2 about a radius of R/2.

So, now that we know the total tension on the tether, we can solve for the
tip radius as a function of tension:

Vtip = Sqrt[2 Ttotal/(A rho)]

Total tension divided by area is tensile stress.  The best tether material
on the market today is Spectra 2000, with a tensile strength of 3.25 GPa
and a density of 0.97 g/cc.  Putting those values into the equation, we
get:

Vtip = 2.59 km/s.

So, the maximum velocity we can obtain is 2.59 km/s.  Any higher tip
velocity and the tether will tear itself to pieces.  Notice that this
result is independent of the tether length or angular velocity.

An interesting possibility is using buckytube cable, with a (theoretical)
tensile strength of 150 GPa and a density of 1.3 g/cc.  That would give a
tip velocity of 15 km/s!

Finally, tapering the tether can improve the situation somewhat, but this
is the basic limitation you have to deal with.
--
Andrew J. Higgins            Department of Mechanical Eng.
Shock Wave Physics Group     McGill University
higgins@mecheng.mcgill.ca    Montreal, Quebec
```

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