Re: Konstantin Tsiolokvsky was wrong?!

"Spaceman" <Realspace@xxxxxxxxxxx> wrote in message news:wZWdnWpiEL9ckZ_ZRVn-sw@xxxxxxxxxxxxxx

<matt271829-news@xxxxxxxxxxx> wrote in message
news:1140984124.925164.265460@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
No, the answer is not calculated "at" or "for" any particular time. The
answer encompasses the *whole* time that the motor burns, as well as
the fact that the fuel mass is constantly decreasing and the rocket
speed is constantly increasing over that time. It may sound like
alchemy, but that really is the way it works.

But Like I have said,
At half burn,
The end result would be less,

Of course, you've only used half the fuel.
Plug in the numbers and see what the results are.

You are using the beginning time
and calling it the end result.

No. The time is implicitly the time during which
the burn took place, and it doesn't matter how
long a time or how short a time that is.

you are giving the end result as the starting
result.

No. You don't know how to read or apply an equation.




It does not do such if it is using the 10:1 that is only
present at time 0 and lacks a time function itself.

As I said, the answer takes into account the fact that the ratio
changes continuously with time.

It does not actually.
It is only using the 10:1 ratio.


Is e changing over time?

No, e is presumed to be constant over time. This is an assumption that
is made in the statement of the problem. As such it is not a debatable
point - it just *is* constant because it is stated to be so.

Supposedly and I guess logically, No,
Is mass changing over time,
Yes. and logically yes.

Yes, the total mass of the rocket decreases over time as the fuel is
expelled.

Yet, the end result you get is coming only from
the initial starting factors?

Initial and end conditions determine what happened.

It is ignoring the in between,

The in-between was taken care of by the integration of
the equation. That's the genious behind the Calculus.

and since you got such data from Earth based
experiment, it is also ignoring the "air pressure"
as a factor in the difference between Earth
based and "space based" equations.

The equation did not reference any specific data, so
how the information is obtained is irrelevant to the
equation. If you desire, you can await actual firings
of rockets and use data gatherered from them to plug
into the equation.




So..
Where is the "timing" function for the mass change
in that equation?

I assume you are referring to the equation:

Max. rocket speed = e * Log(1 + m1 / m2)

Of course I am.


There is no "timing function" for mass change here because it is
irrelevant to the answer. You don't need to know how quickly the fuel
is burnt, only that all the fuel eventually shoots out of the back of
the rocket at speed e. The fuel could burn out in one millisecond or
burn over one million years, and the final speed of the rocket would be
the same. According to my calculations anyway.

And that is where it is wrong.
In your case, it is happening "instantly"
and the rocket reaches ful speed in "no time".

No, the time is imlicit and can be long, short, or
medium in length. Seconds or years. It makes no
difference. If you could build a rocket engine that
would burn all of its fuel in a single instant and
eject it all at once at the same speed, then it would
have the same effect as squirting it out slowly over
months (assuming a rocket without other external
influences, of course).

It is missing reality completely.

No, that's your personal bailywick.

and the equation was also based on a rockets
exhaust speed with a pressure against it. (Earth based pressure)

Can you point out the term for the ambient pressure in the
derivation? No, no mention of back pressure was made in the
derivation. James is wrong again.

and of course, no such pressure is found in space.

And no pressure is mentioned in the equation.


It is all way to "predicted" without any actual physical measurements
occuring.

What measurements are required to derive an equation?


.

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