edit n/a anymore.

Thanks for the feedback argona369. Sorry if im missing your point here but, isn't that just normal hydraulics? I realise everything balances out in your example. My idea is trying to manipulate the fluid required to push a piston using a quadrant shape, without changing the surface area of the pistons. They can be the same size even.

ok what about the quadrant? anyone have any thoughts on this?

bump

its simple physics compared to most of the stuff on here someone please debunk it in a way i can understand.

I haven't got the time to build something like this, but im sure the difference in displaced volume will be obvious. The question is weather the force quadrant piston will remain the same barring friction. 

I have recieved a helpful email from Rocha Raymundo from pxd.com explaining why it won't work, but i still can't get my head round why! I emailed him back to say why i don't understand and he tried explaining it again but i still can't completely grasp it.

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You already understand that the ratio of the area of the pistons determine their mechanical advantage so I wont go there. But there is the concept of static symmetry that you need to think about. your piston in the first diagram has a hole for the block to slide into, as pressure is applied to the work piston the block is thought to keep the area of the piston constant while using less fluid to accomplish the work of lifting the piston well in that respect it does use less fluid but the block then behaves like a piston and since it is not moving back with the work piston it is robbing the are from the piston and thus make it use more pressure so here is an equation for you

 

let W = Work piston area

let B = Block cross-sectional area

let Wd = Work piston diameter

let Bd = Block diameter

let D = Drive piston area

 

B = (Bd * .5)^2 * 3.14

 

Assuming the block is cylindrical this equation calculates the area, this is important because we need to know how much force it will see as it is proportional to the cross-section of the hole in the Work piston (just like the hole the work piston slides through)

 

W = ((Wd * .5)^2 * 3.14) ? B

 

If you take the cross-sectional area of the block away from the piston (or take the area of the hole the block slides in) you will have the area of the work piston

 

M.A. = W/D

 

If you notice when you change the Bd the M.A. will change with it and this is where your system equalizes

 

Here is another way of looking at it. Most pistons are round but what happens when you have a piston with a square notch in its side? You simply calculate the properties of the piston with notch in it, the wall of the cylinder still has to fill the notch or fluid would be everywhere. But it will behave exactly the same way as a piston with no notch.

 

Now on to your quadrant based system.

 

It is simple really why the system balances. The quadrant is like the notch before mentioned but instead of being and integral part of the wall it is ?floating? but still anchored, so the pressure enacted on it is trying to push the quadrant out of the piston but being held there by the linkage. The piston is forced to move around the quadrant and as expected less water has to be pumped into it to get the piston to move but equally more pressure is required to do it also. And pressure times volume is still constant.

 

I guess one more way to look at it is if you allow the block and the piston to move together it will produce the mechanical advantage that you are looking for. But the moment you stop the block from moving the piston area is suddenly reduced and the M. A. is reduced as well.

 

There is one thing that I left out

 

Work is defined as Force times distance

 

Power is defined as Work times time.

 

Energy is defined as power divided by time

 

So you balance it like this        energy = (area of piston minus block cross-sectional area times pressure times distance traveled)/time of stroke = (area of pump piston times pressure times distance)/time of stroke

 

Or        Volume = area of piston minus block cross-section area times stroke = area of pump piston times stroke

 

Hope this help the light come on and you can move on to your next educational project.

 
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Hi thanks very much for looking at this! Im glad you found the moon idea it is freaking crazy, thanks for going into detail about it i didnt realise you could calculate the potential energy of the moon! haha, quite a few kW hours though! I really appreciate you going into detail about these ideas you are the only person on the internet who has done.
 
About the quadrant idea. You say that while less water is required, more pressure is required to push the piston upwards because of the quadrant shape, even though same amount of surface area on the piston is exposed to the water. Regardless of what shape is sliding through the hole because it is not an itegral part of the wall it will not work, because there is water pressure on the quadrant aswell.
 
Just to clarify am i not right in saying that because there is a device holding the quadrant place which is anchored to the wall , there is the same amount of pressure holding it in place so any force on the quadrant is cancelled out? Like me pushing on a building there will be an eqaul and opposite force pushing back. The piston moves up but the quadrant just rotates agreed?
 
Phil

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Hey not a problem, I generally don?t fallow rabbit trails but every once in a while I take a challenge, and I saw that people were only giving you a blanket answer which would annoy the piss out of me. But anyway if I understand your new question right: when you anchor a ?piston? it simply behaves like a vessel forcing the pressure to look elsewhere to escape, so look at it like this your quadrant is a fancy piston (it pivots but in reality simply slides through the work piston) this fancy piston will have a certain amount of force on it relative to the cross--section of the quadrant, and the work piston has pressure on it relative to the cross-section, this force that is exerted does not change whether the piston moves or not, if you add the force of the quadrant to the force of the piston you will get the total force of the system and if you allowed both to move together doing the same work they would both contribute to the work, however if you restrict the movement of one you take away its contribution of force making only one of the pistons do the work.

 

So in answer of your question as I understand it? no you are not correct in thinking that the force is canceled out. The flaw in this thinking is in: work is force x distance if you take either of the two out of the equation you get zero work. So instead of looking at is as the force is canceled out just think about it as the force being contained.

 

If you were to make a cannon the internal pressure on the walls is trying to force it apart. So let?s put some real numbers to it:

 

The cannon has 3? boar

The powder in the cannon develops 25000psi

The powder chamber is 6? long in the breach

 

The total area that the pressure is exposed to is 7.06 + 56.5 + 7.06  = 70.62 square inches

 

The total force on the walls and projectile is then 25000 x 70.62 = 1,765,500 pounds

 

The question at this point is: is the force on the projectile 1,765,500 pounds?

 

Actually the force on the projectile is only equal to the area behind the projectile or: 176,500 lbs

 

The quadrant would behave the same as the walls of the cannon so if the total area of the piston + quadrant is 10 and exposed to 100psi the total force would be 1000lbs if the quadrant then had an area of 2 the force of the piston would only be 800lbs so although the work piston would move 20% faster it would also take 20% more force to do the same amount of work thus the pressure supplied would have to be 120 psi to do the same work.

 

Of course this is only if I am understanding your question right.

 

Cheers

Well I don't really see how a cannon is identical to a cylinder with two
differently shaped pistons in it...

But if I had to come up with an explanation on why it won't work,
I'd say: the "bent" piston is nothing but a bent version of a straight one,
and if you calculate the total displacement of water by said pistons
and the amount is equal, then it won't do anything.

I am still more interested in why you think it will/ work.