| This week:
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
TOP |
Tuesday,
May 17, 2005
(1100 Rome time)
If you do everything else just right, you get older. When you get older you
get jet lag that doesn't quit. Get your traveling done when you're younger
and recover from time zone changes more quickly. Or travel by boat...
I keep getting mail on the DREAD system physics. DREAD is a centrifugal gun
that sends pellets out at a fairly high rate of speed; as one correspondent
observes, it's an area denial system intended to produce a beaten area, not
a pinpoint accuracy system to be used against small individual targets. The
advantages are in the logistics. You carry fuel, not explosive propellants.
The confusions seem to come with regard to the "recoil" of such a system,
and there is considerable confusion of angular vs. linear accelerations.
First, consider the system as a whole, and put it into free space. We're not
concerned about angular momentum on the first pass: what we see is a system
in space that spews out mass. For the moment we'll assume the mass all comes
out along a single vector which we can call "aftward". Any system that sends
out mass in a single direction is a rocket, and subject to the rocket
equation; since the mass comes out in chunks rather than particles it won't
be an efficient rocket and you won't get the constant e in the equation, but
we'll ignore that for the first cut. (The classic rocket equation assumes
small continuous particles and thus is concerned only with the ratio of the
starting mass to the mass at velocity, and the exhaust velocity; we're
violating that assumption big time here, but we're ignoring the violation.)
The point is that we don't need to know how the reaction mass -- the
particles, balls, whatever that is going overboard -- were accelerated to
velocity. All we need to know is the starting mass of the system, the
mass after the ammunition is exhausted, and the exhaust velocity, and we can
calculate the delta-v (change in velocity) of the system as a whole assuming
it is in free space and not subject to friction (wheels on pavement,
atmospheric drag). For refinement we can do the calculation for each
projectile and sum over the whole mass one projectile at a time, thus
correcting for the size of the projectiles (once again, the rocket equation
assumes continuous flow of negligible sized particles; this where the
constant e appears in the rocket equation, and why rocket science uses
natural logs.) That's the linear acceleration effect. Over
time it can be quite large, but in practice the accelerations due to the
rocket effect will be overcome by the effects of friction, wheels or treads
on the ground, and the motive system of the vehicle which isn't, after all,
in free space. They won't be negligible, though, particularly if the exhaust
velocity of the projectiles is high and the "fuel" weight is an appreciable
fraction of the total system weight, so there will be fuel consumption
required to keep the vehicle moving toward the target. Now we
have the angular momentum, which is more complicated. Assume that the
projectiles start at rest with respect to the system, travel up a tube, out
the moment arm of the thrower, and are released. Assume the vehicle is
traveling directly toward the target. Think about the point in the travel of
the throwing arm at which the projectile is released (this confuses people
who haven't studied the problem, or who haven't modeled the system. It may
help to think of a man standing in an L5 colony with artificial gravity and
what happens if he simply drops a baseball.) And that's the
best I am going to do jetlagged in Rome with other things to do. I leave the
rest as an exercise for the reader. ===================
Last night we successfully navigated the Roman Metro system, and today we
managed a bus to the Vatican, but we were tired enough coming back that we
took a taxi. The bus system is 1 Euro each for 75 minutes, although I have
never seen anyone actually check a ticket. The taxi is about 5 Euros. It's
odd to be in Italy without Lira. My Roman friends tell me they accepted far
too low a value for the Lira when they joined the EU, and in the year after
they lost the Lira and went to prices in Euros the actual price of nearly
everything doubled. That's serious inflation, but it seems stable now.
Tuesday TOP
Current Mail
|
