Re: Konstantin Tsiolokvsky was wrong?!

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"Greg Neill" <gneillREM@xxxxxxxxxxxxxxxxxx> wrote in message
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It can never be relatively constant,

Why? If the fuel is fed at a constant rate and he
burn continues at a constant rate, and the pressure in
the combustion chamber is constant,

Bingo!
There is your problem,
A pressure chamber can not remain constant,
when it is not closed.
Again you seem to not even know basics
about an open or closed "system".

Hey James, when you're driving your car down the
highway (assuming that you have a car), how do
maintain a constant speed?

I hit the cruise control,
and sadly, my fuel does not stay constant

No, it wouldn't if there are hills and variable
winds. Fortunately we don't have that problem
with the empty space of our gedanken experiment.

You still burn fuel, and that is not constant
so the mass is changing so even with a constant
fuel burn, you will not have a constant speed.

You fail to understand the equation. It is specifically
taking into account the changing mass; it's the dm and
M - dm bits.

No it does not,
It is using the "beginning mass" and does not even have a time
operation included.

The time is implicit. You could do short burns for an
indefinite period, and as long as the exhaust velocity
was the same for each and the prescribed amount of fuel
mass was used, then the final speed would be as given.

Thrust is not velocity.

I did not say thrust is velocity ***,
I said the thrusting force must have a velocity

Force doesn't have a velocity. Different units.
Thrust is in Newtons (a force). Velocity is in
meters per second.

and such a force can't be constant if the mass is
changing over time.

Sure it can. It just means that the mass it's
pushing is less, so that the net effect is a
higher acceleration. F = M*A.

You are ignoring mass loss over time when you
use that equation.

No, it's specifically part of the derivation. The
conservation of momentum equation for any instant
is

M(t)*dv = Ve*dm

Where M(t) is the mass at the given instant, dv is
the velocity change produced by the bit of mass dm
leaving in the exhaust at speed Ve.



James has not plugged any numbers into the trivial
rocket equation, or else he wouldn't be making
such bone-headed statements.

The "trivial" rocket equatrion is missing
factors of reality all over the place.

Such as? The derivation was given to you. What
specifically is missing?

Why would I use such a silly equation
that is even ignoring time?

Time is implicit. It doesn't matter over what
period of time or how many individual engine burns
it occurs.

You work it out. You've been spoon fed the equation,
now plug in some numbers. You're obviously typing your
posts from a computer, use it for something other than
typing lame usenet posts -- write a program to plot
the terminal velocity of rockets with various starting
masses and fuel capacities. Use the Saturn V exhaust
velocity information.

You are a freaking moron,
I will use data that has been fixed by an equation
to make an equation?

The data given by the provided link was the empirically
measured static test exhaust velocities.

What a dipwad, Is that how you do your science?
Take data that has already been screwed by an equation
to show the data itself?

Clearly you didn't even check the data given, so you're
talking throuigh your hat as usual. What a maroon.

LOL
You are a fool Greg.

James has lost another argument, so he's slipped
into insult mode. Right on schedule, too.


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