Re: Cannon in space, how much more propellent efficent are they than missiles?

On Jan 13, 9:10 pm, Michael Price <nini_...@xxxxxxxxx> wrote:
On Jan 14, 12:53 am, IsaacKuo <mech...@xxxxxxxxx> wrote:

There's still a limited amount of it.

But that limited amount is approached asymptotically so you can add
energy by adding barrel length forever.  So the

It gets to very near the limit quickly, depending on the
expansion ratio.

The pressure never drops to zero no matter how much you
expand the volume because P = k/V (where k is a constant for each shot),

The pressure goes down faster than that.  You seem to be
assuming constant temperature, which is wrong.  As the
volume expands, the temperature goes down.

So how much of the kinetic energy comes from pressure and how much
from temperature?

Huh? The kinetic energy comes from heat, which is different
from temperature or pressure. Heat is measured in units
of energy.

Neither pressure nor temperature are anywhere near constant
when a piston is moving to rapidly change the volume of
working gas. That pressure changes should be intuitively
obvious. The change in temperature is not as familiar in every
day life, but it's just as real--it's called adiabatic heating or
cooling.

 Your math has gone wrong somewhere because it implies you
can add an unlimited amount of extra energy just by
making the barrel longer.
 And that's right, with an infinite barrel you could.

That violates conservation of energy.

  I mispoke.  You can add energy infinitely but you can't add infinite
energy.

Using your incorrect formula for pressure, the amount of energy
would have indeed been infinite. Your incorrect assumption of
constant temperature would actually be accurate if there were
some continuous addition of energy to keep the temperature
constant. Since this involves adding a potentially unlimited
amount of extra energy into the system, conservation of
energy is not violated.

The exhaust from the gun will never be at a lower velocity
than the projectile,

  But it will be lower than the exhaust from a rocket.

Nope. Almost all of the exhaust from a rocket will have a
much lower velocity--in the reference frame of the launching
platform. This is because you subtract the rocket's current
velocity from the nominal exhaust velocity.

since it can't push the projectile to go any faster than it's
going.  It will actually leave the gun at
an even faster velocity (typically MUCH faster).

  Where's your evidence for this?

It's common knowledge about real life gun muzzle blast.

This is wrong.  Up until a delta-v significantly larger than the
exhaust velocity, rockets efficiently put the kinetic energy
into the rocket, not the exhaust.

  The momentum is the same right?

Yes.

 So unless the fuel is heavier than
the rocket the kinetic energy of the fuel is a lot more than the
increase in KE of the rocket.

In an efficient rocket, the fuel IS heavier than the rocket.
The ideal efficiency results when the exhaust velocity is
tuned to result in a constant rearward speed. So, for
example, the mass of the fuel may be 4 times the mass
of the empty rocket. This will result in equal and opposite
momentum, and pumping four times as much kinetic
energy into the rocket as the exhaust. This implies an
efficiency of 80% (not counting the inefficiencies of the
reactions and general waste heat losses and such).

The guns in my game are only used for extremely short range
(there are no low acceleration targets in this game).  These guys
are in a desperate situation with very limited resources, so they
use what they have available even though it's not very good.

  If you're that close you're dead against anything with decent
missiles.

In my game there will be limited amounts of missile ammo.
Ships don't get "free reloads", those reloads have to be supplied
by explicit logistics. As such, a missile boat could easily find
itself
in a battle situation where it has little or no missile ammo left,
or its remaining missiles need to be targeted on a valuable enemy
target rather than tasked with defending the missile boat.

The advantages are that you use less propellant for a given
delta v

Wrong.

Prove it.

For an example, consider a 2:1 mass ratio. This mass ratio
implies an equal mass of propellant as payload. Using the
rocket equation, we see that the final velocity is 69% of
the exhaust velocity.

Now, let's assume a 100% efficient gun (for fairness).
With a 1:0 mass ratio, we'd end up with a final velocity
equal to 100% of the exhaust velocity (if you placed a
speck of dust in the gas stream). To get a final velocity
of 66.667%, you'd need to mix 2/3 fuel with 1/3 payload.
In other words, the mass ratio is 3:1.

The rocket gets slightly faster using half the fuel.

Things get even better looking for the rocket when
the mass ratio is higher than 2.78:1. In those cases,
the final velocity EXCEEDS the exhaust velocity. These
are velocities which guns can't reach at all, no matter
how high the mass ratio.

Isaac Kuo
.

Relevant Pages

  • Re: Obliterating the Rocket Equation with a Torusail
    ... >> some unexpected science breakthrough (ala Cold Fusion), ... >>sum of the kinetic energy and potential energy per kg ... >>two and take the square root to get the exhaust velocity ... >Trip dynamics are governed by the power of the rocket. ...
    (rec.arts.sf.science)
  • Re: Velocity and Kinetic Energy
    ... His initial velocity relative to the fixed point ... He energizes the rocket for a moment, ... 200 joules worth of velocity to himself. ... joules of energy to his velocity. ...
    (sci.physics)
  • Re: Cannon in space, how much more propellent efficent are they than missiles?
    ... propellant available which is probably the same as rocket fuel. ... If the desired "muzzle velocity" as any decent fraction of the ... if the muzzle velocity is a small fraction of the exhaust velocity. ... there's a particular amount of energy ...
    (rec.arts.sf.science)
  • Proof the Law of Conservation of Energy is Wrong
    ... immediately after its first burn ... In such a situation it is obvious that the first rocket will continue ... losing the velocity it had gained from the first burn. ... 'energy can neither be created or destroyed, ...
    (sci.physics.relativity)
  • =?ISO-8859-1?Q?Re=3A_Sensation_in_der_Physik_=3A_Nimtz_Experiment_wird_?= =?ISO-8859-
    ... So please state the front velocity. ... Antwort: Superluminal energy transport is exciting too i think. ... energy flux that you elect as being the flux buildung up the tunnel ...
    (de.sci.physik)