Re: Propellantless propulsion fun 3 (recirculative propelant)

"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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If you would think about it without ignoring
relative mass and kinetic energy differences,
You might understand that it does work.

While you're explaining the difference in motion that
a large mass in motion makes (we are eager to hear),
you might take a moment to explain what "relative mass"
is in your version of Newtonian physics.

Oh Greg,
I am not going to go over classical physic with you,
You are too brainwashed by your rubber rulers to
think about such classical thought without it interfering
with your religion.

The reason I ask is that, to the best of my knowledge,
"relative mass" is not a concept from classical Newtonian
physics. I am eager to learn of this. Since you seem
to be in the know here, I am asking enlightenment.

Why are you now limiting me to using Newton terms?
You really love to make diversions huh?

You keep harping on about classical physics and rubber
rulers, so I thought you didn't want to employ Relativity.
What's brought you around?


Force isn't transfered in the way you're implying. It's
not a quantity of stuff. The forces that are engendered
in an encounter depend upon the total momentum transferred
and the time over which the encounter takes place.

An object impacting a stiff surface generates higher forces
than one impacting a soft, maleable one -- the deceleration
of the impacting object is spread over more time for a
surface that "gives". Remember, Dv = a*Dt. The change in
velocity is equal to the acceleration multiplied time over
which the acceleration takes place. And the force that
results from the deceleration is given by F = m*a. So,
combining the two, Dv = F/m*Dt, or F = m*Dv/Dt.

So to stop a given mass m moving initially at velocity Dv
over a time period Dt with a constant deceleration, the
force required is F = m*Dv/Dt. A pliable surface will
extend the time Dt, making the force lower.

The force will be "lower" Greg?
Wow. you are getting a clue now!
With a lower force over time, yet a larger force
occuring now over the same time, who wins?

The accounting balances. A lower force over a longer
time trasnfers the same amount of momentum as a
larger force over a shorter time. The momentum
transfered, for a constant force, is F*dt.
.

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