Quote from: TinselKoala on August 14, 2012, 05:12:14 PM
Thank you for explaining, Larry. Sorry about that crossed post above.



So... work, or energy, is force x distance. In LarryC's example, in the first case on the left, we are applying a certain force (the pressure of the 100 lb weight) and we are using that force to move a column of water the length of the piston stroke. Right?

And in the third case on the right, we are applying that same force, and we are using it to move the (essentially) same column of water..... the length of the piston stroke.

We have attained the same pressures in each case... and the work input / work output _in each case_ balances.

But the work input in the case on the left is greater than the work input in the case on the right.

The system on the right neither takes in... NOR RETURNS... the same amount of work as the system on the left in LarryC's illustration, even though both systems are operating at the same "head" pressure. And it will feel different "by hand" because the system on the right only needs a small displacement to get to the full pressure, while the one on the left needs a "softer, longer" push to get to the full pressure.


Congratulations, you have just invented the lever.

Congratulations, you have just invented the lever.

Are you trying to mislead or just don't understand levers?

A level that would move 3.5' with a 100# effort force to create a 7 Foot head would have a longer distance from effort force to fulcrum than a second level that would only move 1.16' and would require a much larger weight to create the same 7 Foot head.

Also, you ignore the fact that after lift you would need to raise a 100# weight 3.5' with the 1 U and 1.16' with the 3 U. Why?

Regards, Larry
 

I have put users MileHigh, Seamus and Microcontroller on Moderation as they have trolling this
thread too much  !

Regards, Stefan.

Quote from: Larry C .. Post #1776 Pg 92

Simple explanation of energy advantage.

Shown is 3 systems, each create a 7 Foot Water Head. Each system uses a 100# weight as force, but each applies that weight over different distances.

Half filled U shaped column reduces the distance to apply force. Each additional U shaped columns further reduces the distance to apply the same force.

It doesn't account for air compression, but a water, mercury wouldn't have compression and mercury weight advantage over water is 13.534 to 1.0. Thinking Travismobile.

Also, if the piston was removed after stroke, the 3 U version would re-balance first, the 2 U second and the 1 U last.

Regards, Larry

Larry .. Perhaps I don't understand 'head' like you do ?!

Do you mind explaining to me why you say all 3 examples in your drawing have the same 7 foot head ? - & pressure ?

http://en.wikipedia.org/wiki/Hydraulic_head

Quote from: fletcher on August 14, 2012, 07:01:07 PM

Larry .. Perhaps I don't understand 'head' like you do ?!

Do you mind explaining to me why you say all 3 examples in your drawing have the same 7 foot head ? - & pressure ?

http://en.wikipedia.org/wiki/Hydraulic_head

Hi Fletcher,

I just picked 7 Foot off the top of my head. Could have been better and I may change and re-post.

Anyway 7' = 84", 84" X .036127 (Cu In weight of water) = 3.034668 PSI.

The SI of the water columns is such that the weight of the 7' water column would be a little less than 100#. Not important to me, so I'll let you figure that one out. Height * PI() * Radius * Radius * .036127.

The reason they all have the same head is because in 2 U and 3 U, the last water column is pressing up on the next air column and the next air column is pressing down on the next water column, etc. etc. They appear to the piston as a single column of water pushing up.

Regards, Larry
PS: This is not the Travis system, but it is a piece of the system broken out in an attempt to help increase the understanding.

@Harti, Thanks.

Yes, Stefan, I must protest as well.
Sorry... but I don't see any trolls. I see some people with genuine legitimate doubts and questions that aren't being sufficiently answered.

Simple questions like.... What was the reason that Mark Dansie wasn't completely satisfied on his first visit?  And my questions are the same as some of theirs too: what is the ratio of work out, to work in, and how is it known that the three layer system is clearly overunity itself?


Meanwhile.... I am once again operating completely without funding, scrounging stuff that I happen to have lying about.

In the LarryC case it's pretty clear that each individual system can only return the work that's put into it. The only question is whether or not the work in the third system on the right is the same as the work in the first system on the left.

Since the input force--- the weight of the 100 pound weight pressing on the piston... is the same in both cases (RIGHT?) but the travel is less in the one on the right, we know at least that the INPUT WORK isn't the same.
RIGHT? The force is the same and is constant in both cases: 100 pounds downward. (More correctly measured in Newtons but we'll let that slide for the moment.)
And the distance travelled is different. Therefore the product Force x Distance = Work is different, with the first system on the left making the greater product, since the distance through which the force has acted is greater. RIGHT?

SO the only question is whether the output work is different or the same. Since the system on the RIGHT cannot push up a water column to the same height, when released, as the system on the LEFT.... it should be clear that the work output available from the system on the right is less.

We don't even need any assumptions about pressures. All we need is to note that the input FORCE is constant and equal in both cases, the input DISPLACEMENT is less for the one on the right, therefore the input WORK is less for the system on the right.
Since the OUTPUT-- the return height, or the height of an additional column as in my thought experiment, is clearly LESS for the system on the right.... well... DO THE MATH (tm RA).

Now, the issue of the pressures. Is it a confirmed fact that the "heads" are the same in both cases? I don't know the answer to this myself... but I am preparing to measure it.

https://www.youtube.com/watch?v=c8WYI7QCj0k

Now... Koalas really don't like to get their paws wet. We get all the moisture we need from the leaves we chew, or otherwise ingest, and the beers that we drink. Getting wet makes us cross and jumpy. Especially when there  is no reward in the offing. However, even without funding enough to buy lunch, and working with garbage materials, we are still proud of our workmanship.

May I please have some suggestions as to how to measure the "head pressure" in this system? I could put a basic improvised manometer on the bottom of the first receiving cylinder... but this whole system is a manometer anyway, so I'm not sure of the probative value of that, although I myself would accept its indications. Bear in mind that I am not about to go out and buy a sensitive pressure gauge. I do have on hand a very precise and accurate electronic force gauge for mechanical linear force, a Mark 10 Series 5, model M5-05, that cost "somebody I know" about a thousand dollars.... but I dare not get it wet !!