Quote:
Good God. This is hopeless. All the time I have been assuming all velocities are measured from a single observer, "stationary in space".

Now it turns out you've been measuring some of them from different reference frames - sometimes the tube, sometimes the other observer.

It's much easier if you use just one observer! Then we get -80m/s for the mass you described.


Good God. This is hopeless. All the time I have been assuming all velocities are measured from a single observer, "stationary in space".

Now it turns out you've been measuring some of them from different reference frames - sometimes the tube, sometimes the other observer.

really , I didnt know that I thought we were inside the pipe system.

ok , lets do the rest from outside then.

Then we get -80m/s for the mass you described.

hows that?

what Im showing so far is that after 50 seconds

and this is from an observer outside during the 50 seconds.
Im looking at the pipe and the pipe is moving to my left.
after the 50 seconds has passed

the pipe has moved +321.036744117737 meters to my left
the mass has moved -678.963255882263 meters to my right

the pipe has a velocity of +12.84147 m/s
the mass has a velocity of -27.15853 m/s

the pipe has a momentum of +77048.82 Ns
the mass has a momentum of -2715.853 Ns

the pipe has a mass of 6000 kg
the mass has a mass of 100 kg

so the 100 kg mass enters the turn
at a velocity of -27.15853 m/s

now you only have -2715.853 Ns vs +77048.82 Ns
thats a difference of
+74332.967 Ns momentum

LOL

that was the sound of me sawing a leg off of your new table.

is this the equation your talking about?


if so I suppose that the "d" in the formula has no real value because so far I havent been able to find just what is used to put where the "d" 's are. !!!

and it only takes into account 1 mass and 1 velocity

normaly when a symbol represents an element that has value the person describing the usage of the symbols in the formula will tell you what the different symbols are supposed to represent.

is this true in the above formula?

so that where the formula uses symbolism such as
dt or dp
it really just means
t or p

?????

or are these the 2 formulas your talking about?





theres no unexplained "d"'s in it and it really needs no explanation as to how your supposed to use it.

not that they have any descriptive instructions on using them
as wiki has in their informative articles related to physics.

I of course would guess that
u1 = m1 velocity initial
u2 = m2 velocity initial

because a guess is all I have to work with using this
non informative article.

ok , I got them to work the results are below.

using initial velocities of

m1 vi -27.15853 m/s
m2 vi +12.84147 m/s

m1 mass 100 kg
m2 mass 6000 kg

results

m1 v = +51.52999459 m/s
m2 v = +11.52999459 m/s

that is using these formulas




I plugged the formulas into an excel file
I havent programmed them into my program yet.

is that about what you get?

do you think that this would be a good way to determine the
velocity of the 100 kg mass as it leaves the turn and the velocity of the pipe?

if so then we can get on with this and cover the next 50 seconds.

thats where the pipe gets really fast.





3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.