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#40388 - 09/27/11 11:49 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul


the 100 kg mass would not have a velocity !!!
and it would not have momentum !!!

in fact it has been decelerated from +40 m/s to 0m/s


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.

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#40389 - 09/27/11 11:50 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
let me rephrase that just for you , so that you can comprehend it.
a = 67 m/s - 40 m/s / 1 second = 27 m/s/s



Where did the 40m/s come from? Each mass has -40m/s before entering the turn, doesn't it?

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#40390 - 09/27/11 11:53 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
that makes absolutely no sense.

Correct.

Quote:

twice the velocity !

no matter what the momentum is of the 2 bodies as long as one of them is more massive than the other.


"much more massive" is not the same as "more massive"
">>" is not the same as ">"

Again, it's an approximation. It's never correct. It gets closer and closer to correct the more massive the body is. So we better not use it.

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#40391 - 09/28/11 12:07 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Quote:
">>" is not the same as ">"


so its like >^2?
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#40392 - 09/28/11 12:10 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
because I used a minus sign.

Quote:
a = 67 m/s - 40 m/s / 1 second = 27 m/s/s


put a -40 m/s in there and you get 107 m/s/s



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

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#40393 - 09/28/11 12:15 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
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.

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#40394 - 09/28/11 03:26 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
because I used a minus sign.

a = 67 m/s - 40 m/s / 1 second = 27 m/s/s

put a -40 m/s in there and you get 107 m/s/s


Yes. 107m/s^2 is the acceleration, and is also consistent with the formula you used:

Quote:

a = (sf - si)/t


sf = 67m/s
si = -40m/s
a = (67m/s - -40m/s) / 1s = 107m/s^2

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#40395 - 09/28/11 03:38 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
ok , lets do the rest from outside then.

OK. I probably should have mentioned that sooner.



Quote:

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


Again, you've ignored the other 49 masses that are about to hit the turn. They're more than enough to stop the pipe then reverse its direction.



Quote:
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. !!!

I think we should leave that for another time. The d has a common meaning that's universally used across math and physics, so they don't need to say it. It's much more standard than symbols for velocity, etc. But we better not go into what it actually means. Easier to just not use that formula.


Quote:



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

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?


Yes to all the above. Except I don't trust the vi values you're using. But that's an issue with the accelerator not the turn.

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#40406 - 09/28/11 01:09 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
Again, you've ignored the other 49 masses that are about to hit the turn. They're more than enough to stop the pipe then reverse its direction.


it wouldnt be 49 it would be 50

I havent ignored the other 50 masses that are being accelerated towards the turn.

during the 51 st second they are whats providing the force that pushes the pipe.

you have agreed that the formula is correct other than the initial velocity values.

so I will plug in that formula into my program.

Quote:
They're more than enough to stop the pipe then reverse its direction.


m2 vi +12.84147 m/s

m2 v = +11.52999459 m/s

this single mass managed to slow the pipes velocity by
1.31147541 m/s in 1 second

however the pipes mass becomes less each second
and
the force driving the pipe in the (+) direction
becomes stronger each second.

for the next 50 seconds the force driving the pipe will increase by a minimum of 80N each second and more if the speed of the exiting mass has a velocity
greater than +40 m/s.

and the mass that the increasing force has to push becomes less and less each second.

anyway , I'll plug the formula in today.

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

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#40408 - 09/28/11 02:13 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
TheFallibleFiend Offline
Megastar

Registered: 06/08/05
Posts: 1940
Loc: http://thefalliblefiend.blogsp...
Originally Posted By: paul


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. !!!



The d's don't have values, per se, because in that usage, d is an operator (roughly, like + or -). Actually, they are PART of an operator. It's first semester calculus.

It's a *unary* operator, so it doesn't sit between two operands. Instead, it sits before the operand.

d()/dt is an operator that means the "differential" of something "with respect to" t. For example, d(p)/dt means the change in momentum given an "infinitesimally" small change in time.

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#40411 - 09/28/11 04:04 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: TheFallibleFiend]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
thanks TFF

I didnt think it had any value.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#40412 - 09/28/11 04:05 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
OK , I got it working.

and WOW !!!

here are the results after 101 seconds.

hold on to your hat !!!!

the force that accelerated each of the 100 kg masses
was +80N x 1000 meters = +4000N

the force that is required to decelerate each of the 100 kg masses increases every second.

so the pipe feels +4000N plus the extra for the additional
+velocity that the decelerating mass has gained through the turn.

this pushes the pipe faster and faster each second.

by the time (101 seconds) the 1st mass reaches the place it started from
the pipe has a force that causes it to accelerate in the (+) direction of +15987.072 N

and the pipe has moved a distance of +9604.559 meters.

by the time (202 seconds) the last accelerated mass reaches the place where it started from the pipe has a force that causes it to accelerate in the (+) direction of +1302187.380N

and the pipe has moved a distance of +2863843.953 meters
or +2863.843 km

its needs some improvement because it takes 6.8 minutes to reach light speed.

I dont think that would be a proper acceleration for bags of salt water.

so we might want to slow it down a bit.

but to mars at 56,000,000 km its a slow 6.8 minute cruise.

it might be best to experiment with this type of thing without people in it first.

because if something breaks they might find themselves in another galaxy the next day.

I let it run for 1 hour

and the pipe traveled a distance of

5.02576160643331E+73 meters

my program breaks it down into 1 million km units also

5.02576160643331E+64 million km







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

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#40420 - 09/29/11 12:56 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Quote:

the force that accelerated each of the 100 kg masses
was +80N x 1000 meters = +4000N

No. That quantity is 80,000Nm. Nm is not force. Actually it's energy. Using it as a force means everything else in your post will be meaningless so I didn't read it. Also the multiplication is done wrong.

Can you show your calculations step-by-step? Without missing any steps?



Edited by kallog (09/29/11 12:57 AM)

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#40421 - 09/29/11 01:04 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
LOL

your right , I did mess that up.

it is +80N applied for 50 seconds.

+80N each second for 50 seconds over a distance of 1000 meters

+80N x 50 seconds = +4000N

Quote:
Can you show your calculations step-by-step? Without missing any steps?


I think it would be best if you did your own calculations and if our calculations are different then we can discuss them.

you said you have vb 6

I could just rename the .prj + .frm + .vbw files to .txt and you could just remove the .txt from them and run it yourself.

open up your notepad , paste the below in it.
save it as (form1.frm.txt) then rename it to
form1.frm and see if you can run it.
-----do not copy this line-----------------------

VERSION 5.00
Begin VB.Form Form1
Caption = "Form1"
ClientHeight = 3195
ClientLeft = 60
ClientTop = 345
ClientWidth = 4680
LinkTopic = "Form1"
ScaleHeight = 3195
ScaleWidth = 4680
StartUpPosition = 3 'Windows Default
Begin VB.CommandButton Command2
Caption = "timer = 1 second per interval value"
Height = 375
Left = 1200
TabIndex = 2
Top = 2160
Width = 3015
End
Begin VB.Timer Timer1
Left = 360
Top = 2160
End
Begin VB.CommandButton Command1
Caption = "Button"
Height = 375
Left = 1560
TabIndex = 1
Top = 480
Width = 1215
End
Begin VB.TextBox Text1
Height = 375
Left = 1560
TabIndex = 0
Text = "0"
Top = 1080
Width = 1215
End
End
Attribute VB_Name = "Form1"
Attribute VB_GlobalNameSpace = False
Attribute VB_Creatable = False
Attribute VB_PredeclaredId = True
Attribute VB_Exposed = False
Private Sub Command1_Click()
Text1.Text = 80 * 50

End Sub

Private Sub Command2_Click()
Timer1.Interval = 1
End Sub

Private Sub Timer1_Timer()
Text1.Text = Val(Text1.Text) + 80

End Sub




-----------do not copy this line---------------

if you can run the above then this might work.





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

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#40424 - 09/29/11 04:54 AM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
kallog Offline
Megastar

Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
LOL
+80N x 50 seconds = +4000N


4000Ns, not 4000N. This is the impulse applied to the pipe, not the force. It's also the momentum added to it. The force is 80N because you defined the force to be 80N.


Quote:

I think it would be best if you did your own calculations and if our calculations are different then we can discuss them.

No. That will lead to another confusing mess. You can just show each step of the machine, and what happens to the relevant variables (velocity/etc).

You can probably get your program to show the results after each change.

I already wrote a program for the complete 1-mass system. You didn't bother to even read the code or paste it into your own program, or run it, or anything. So I don't want to waste time doing that again.

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#40429 - 09/29/11 02:50 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
4000Ns, not 4000N. This is the impulse applied to the pipe, not the force. It's also the momentum added to it. The force is 80N because you defined the force to be 80N.


q: what is the amount of force that you would need to apply to a 100kg mass in order to accelerate it at a acceleration rate of .8 m/s/s for a distance of 1000 meters over a time period of 50 seconds.

answer: 4000N

80N x 50 seconds = 4000N

1 newton = the force required to accelerate a 1 kg mass at a
acceleration of 1 m/s/s.

or you can say

1 newton = the force required to accelerate a 1 kg mass at a
acceleration rate of 1 m/s each second.

ie ... in 1 second the 1kg mass will accelerate to 1 m/s/s if you apply a force of 1N to it for a time of 1 second


so 50 seconds x 80N = 4000N

http://en.wikipedia.org/wiki/Newton_(unit)

Quote:
it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared


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

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#40430 - 09/29/11 03:04 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
TheFallibleFiend Offline
Megastar

Registered: 06/08/05
Posts: 1940
Loc: http://thefalliblefiend.blogsp...
Originally Posted By: paul


80N x 50 seconds = 4000N



No.
80N x 50 = 4000N

80N x 50 seconds = 4000Ns

We don't throw away the seconds. Force x Time yields a Force x Time, not a Force alone.

Q: What is 2 bananas times 2 bananas?
Hint: NOT 4 bananas.

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#40431 - 09/29/11 03:18 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: TheFallibleFiend]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
the reason I used 50 seconds is because kallog cant seem to understand things that Im trying to communicate to him.

and here is what he commented on.

Quote:
the force that accelerated each of the 100 kg masses
was +80N x 1000 meters = +4000N


if you want to accelerate a 100kg mass from 0m to 1000m
in 50 seconds , at an acceleration rate of .8 m/s/s
you are required to apply a 80N force for the 1000 meter
distance thus 80N x 1000m = 4000N

proof = 4000N / 80N = 50 seconds
proof = .8 m/s/s = 80N / 100kg
proof = 80N = 100kg x .8 m/s/s

meaning that the total force over the 50 second time period that the 80N was applied for over the 1000 meter distance would be 4000N

Quote:
that accelerated

past tense = it has already happened.
80N was applied for 1000 meters.

what the meaning of the sentence means is that
the total force that I paid for to accelerate the 100 kg mass was 80N x 1000 meters = 4000N


that is how I would interpret that sentence.

he commented on my use of 1000 meters as if he didnt understand what it was representing.

if it cost you 1 dollar a second to walk and you walked a distance of 1000 meters in 50 seconds

would it cost you 50 dollars or 50 dollar seconds?

1 dollar x 50 seconds = 50 dollars
it certainly doesnt equal 50 dollars a second
or 50 dollar seconds
it equals 50 dollars


obviously he didnt rembember what we have been discussing.

I changed the 1000 meters to 50 seconds and I used the word seconds to describe what the (50) represented!!
ie.. so that he would possibly be able to understand what the
50 REPRESENTED
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#40432 - 09/29/11 04:25 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: kallog]
paul Offline
Megastar

Registered: 03/21/06
Posts: 4136
Quote:
You can probably get your program to show the results after each change.


it already does.

it shows all changes to all possible elements that I can think of.

mass , velocity , force , acceleration , time , distance

it adds 80N each second to accelerate each mass to the total force that accelerates the pipe.

it subtracts 100kg from the total mass of the pipe each second because each launched mass is 100kg.

it has separate calculations for every second.
it has a separate calculation for the turn.
(it just got faster because the 1 second turn wont require 1 full second)but I havent included that yet.

it replaces the initial velocities of the pipe and the 100kg mass with the resultant velocities of the pipe and 100kg mass from the turn calculation.

it adds the force required to decelerate the 100kg mass to the total force applied to the pipe each second.

so that when an element changes such as velocity or mass
the program determines the force required to apply to either accelerate or decelerate the pipe and the 100kg mass.

its pretty accurate other than the 1 second turn time that is pre-programmed into it.
and that turn time will be lower every time a mass passes through it.

but that will change too.

I think that if I use the initial velocity of the mass and the final velocity of the mass as it passes through the turn
I can get the time required for the mass to pass using
the following formulas

va = .5(vi+vf)
where
vi = initial velocity
vf = final velocity
va = average velocity
( = bracket
) = bracket
. = .
= = equal to
+ = plus

this is the average velocity of the mass through the turn
meaning that the result from the above formula would be the average velocity of the mass passing through the turn.

where average means average.

then
t = vi / vf
where
t = time
vi = initial velocity
vf = final velocity
/ = divided by
= = equal to

so the only way to keep track of the actual time would be to keep a separate time value.

the program will track only the programming time.
meaning that the program follows a set interval when dealing with time , that is the program will execute code at a given time that represent actual time.


so every time a mass passes through the turn the program will
add a partial time to the time value vs adding a full second to the time value

but the program itself will only use full seconds to execute the code.






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

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#40433 - 09/29/11 05:04 PM Re: Record Cold in New Zealand 1903 -26.6 C [Re: paul]
TheFallibleFiend Offline
Megastar

Registered: 06/08/05
Posts: 1940
Loc: http://thefalliblefiend.blogsp...
Paul,
I think you would enjoy taking a pre-calculus class. You would obviously do your homework which is the main problem that most people have with it - and I bet you could find some acceptable classes offered online.

It probably wouldn't be any more effort than you are investing now.

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