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Author Topic: ROC  (Read 4050 times)

Offline webby1

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Re: ROC
« Reply #15 on: December 09, 2017, 02:06:56 AM »
This one is interesting.


I supplied the heights and angles and stuff,, there are 2 1kg weights,, one moves up as the other moves down.

Free Energy | searching for free energy and discussing free energy

Re: ROC
« Reply #15 on: December 09, 2017, 02:06:56 AM »

Offline sm0ky2

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Re: ROC
« Reply #16 on: December 09, 2017, 07:24:26 AM »
Changing the roc by doing work may change the rate
But the change still happens, at some rate.
We can draw a relationship between the change in roc
and the work we did by changing it.


For example:


A diver can pull the co2 cord on his vest from the deep sea
He will go up, at an roc. Based on his change in bouyancy,
and his change in water pressure.
 If we then do work with the buoyant force, the rate will
change based on the work we do.
But wether we do work or not, the diver will reach the top.
So, how much “energy” is contained in the divers roc?
Does this apply to Any roc?
And how does this affect our assessment of energy based on
the rate?

Offline webby1

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Re: ROC
« Reply #17 on: December 09, 2017, 08:27:32 AM »
That is why I am playing with mechanical stuff,, well in sim anyway.


Most of the stuff I am doing up in sim are copies of what I have actually built over the years,, it is way faster to make the changes in the sim and see if the sim shows any interesting trends or tendencies.
The program is somewhat limited in what it can do,, but since these are from things I have built I can get an idea of what the changes would "feel" like.


If you were to use a device or 3 such that the roc that is desired from the source,, the first device,, must go through the second device to get to the third device and the second device itself is a net zero device relative.  IOW, one mass falls and lifts another mass but to do so it does it via a device I put in the way,, and the rate that one falls is not the same as the other lifts,, or vice-versa, and it is the second device that allows that difference to happen.


Simple enough and it should leave nothing over, nothing extra, but this is not always the case and it is this that I am looking into to see exactly what it is within the relationships that creates the odd condition.  If I can get to the root of it then I would thin anything that has a ROC could also be setup to have a difference in the ROC as a source.


All things seem to have a ROC since that rate is a relative function of observation between the points of change.




Offline sm0ky2

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Re: ROC
« Reply #18 on: December 09, 2017, 01:46:18 PM »
I would recommend studying the mechanisms of 1600-1800’s clockworks
A lot of those machines do exactly what you are talking about.

Offline webby1

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Re: ROC
« Reply #19 on: December 10, 2017, 05:34:47 PM »
I was not expecting the sim to show this,, then I did not expect it to rotate past 90 degrees without error either,,,


ETA:  I am going to run a few variations to make sure the sim is not having an error that is not showing up,,,

Free Energy | searching for free energy and discussing free energy

Re: ROC
« Reply #19 on: December 10, 2017, 05:34:47 PM »
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Offline webby1

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Re: ROC
« Reply #20 on: December 10, 2017, 08:27:22 PM »
I relocated my input and output a little bit so I could run it backwards from the end of the other run.

Offline Low-Q

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Re: ROC
« Reply #21 on: December 12, 2017, 08:28:56 PM »
Time rate of cha4nge,, that is kind of superfluous since rate of change includes time,, rate,,  :)

What if I have two quantities of work that are identical, they will be performed within the identical time period and against each other, but the rate of change during that time period is not the same between each work quantity.

If both work and timeframe is identical, one or the other cannot do this in different timeframes. You just said they are performed within the same time period.


Two identical work is identical per definition. If not, they are per definition not identical.


Vidar

Free Energy | searching for free energy and discussing free energy

Re: ROC
« Reply #21 on: December 12, 2017, 08:28:56 PM »
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Offline webby1

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Re: ROC
« Reply #22 on: December 12, 2017, 09:01:44 PM »
The time from start to end is the same,, so the total time that the work is preformed for the total work performed is the same but the "path" the work takes is not.


The "job" moves a force of 10N 2m,, 20J, so if the power component is not constant then the rate of change between the input and output is not the same as in.


If I move 10N 1m in 1 second and then take 5 seconds to move it the rest of the way,, or if I use 1N for 1.5m and then use 37N for .5m.


These are just as examples and whenever you have a non-equal lever you have these kind of interactions,, or say a compound lever,, lots of mechanical devices provide for a nice scenic route that no one seems to bother looking at on the journey from here to there :)

Offline sm0ky2

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Re: ROC
« Reply #23 on: December 13, 2017, 01:17:36 AM »
Like a ball rolling up and down a series of hills
Only to end up a few inches lower than it started.

Offline sm0ky2

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Re: ROC
« Reply #24 on: December 13, 2017, 01:21:21 AM »
Here’s an interesting thing I found
If you take a steel ball and drop if height x
Measure momentum impact force.


Now take the same ball and roll it down a long spiral
With a small angle of incline
And measure its final momentum impact force


Here’s a hint: calculate the gravitational constant of the angle of incline
to figure how how long of a spiral you need to make
For the ball to be going fast enough for E=mgh to break down.




Free Energy | searching for free energy and discussing free energy

Re: ROC
« Reply #24 on: December 13, 2017, 01:21:21 AM »
Sponsored links:




Offline webby1

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Re: ROC
« Reply #25 on: December 13, 2017, 01:35:50 AM »
I saw a youtube thing on that,, those kind of things are interesting.


The sim I am actually playing with at the moment has a 1kg mass falling down but it ends up at a higher elevation than when it started,, no it is not magically lifting itself, there are other things happening at the same time, like another 1kg mass falling as well as a counter balance thing and that allows me to extract work from the falling mass that is lifting itself.

Offline webby1

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Re: ROC
« Reply #26 on: December 13, 2017, 07:15:48 AM »
Here is another file but this time I made a small change to the sim and I identified in the spreadsheet what I call the wobble.


I am trying to make sure that the cross-over point in a lever arm to an arc which creates a condition of 0 change distance but an infinite force and throws sims way off base.  I tried this another way but the forces blew the sim apart in a few ways which may of been fun to watch but did not provide any answers as to whether or not there is the cross-over condition.


If you are a Bessler fan,, this one is NOT what he would suggest :)  This one is taking most of the output gain in the wobble section which is a pulse rather than a constant.


The sim itself actually takes a bit of adjusting to get the mass velocity, resistance from the damper and stuff all working at the correct time of interaction to create the wobble AND leave the system in a state to continue in the direction of rotation,, small changes make a difference.

Offline telecom

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Re: ROC
« Reply #27 on: December 13, 2017, 10:54:50 AM »
Here’s an interesting thing I found
If you take a steel ball and drop if height x
Measure momentum impact force.


Now take the same ball and roll it down a long spiral
With a small angle of incline
And measure its final momentum impact force


Here’s a hint: calculate the gravitational constant of the angle of incline
to figure how how long of a spiral you need to make
For the ball to be going fast enough for E=mgh to break down.

Do you mean the law of energy conservation fails?
Being superseded by the conservation of the momentum?

Offline Low-Q

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Re: ROC
« Reply #28 on: December 13, 2017, 04:29:46 PM »
Here’s an interesting thing I found
If you take a steel ball and drop if height x
Measure momentum impact force.


Now take the same ball and roll it down a long spiral
With a small angle of incline
And measure its final momentum impact force


Here’s a hint: calculate the gravitational constant of the angle of incline
to figure how how long of a spiral you need to make
For the ball to be going fast enough for E=mgh to break down.
If you take away friction, both scenarios privide the same kinetic energy at the bottom.


Say the spiral is 1m high and its track is 10°.
Then the length of the track is approx 5.7587704m.
The acceleration of the ball at this angle is sin(10)×9.81ms^2=1.7035ms^2.
The final velocity is 9.81ms.
This result is valid if the ball does not roll, but slides along the frictionless track. If it rolls, the spinning mass gains rotational momentun, and the velocity will be less than 9.81ms at the bottom.


If you drop the ball from 1m hight, its kinetic energy is the same as the ball sliding or rolling along the spiral track.


Vidar

Offline telecom

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Re: ROC
« Reply #29 on: Today at 12:54:04 AM »
in terms of the ball going down the spiral. it will be rolling on a side due to a
centrifudal force, extracting considerable pressure.
Will this decrease a final energy?

Free Energy | searching for free energy and discussing free energy

Re: ROC
« Reply #29 on: Today at 12:54:04 AM »

 

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