Waterzooka LR© - Piston Water Gun with Hydraulics

[Ben]

I just thought of something that completely disproves your theory SilentGuy.

What changes the force at each stage? The work equation, dependent on distance. What changes the force in your model? The diameter of the pipe. Notice that the pipe's diameter is different than the distance. The pistons are necessary to move a distance to change the force. A change in pipe diameter alone not only does not reapply the force, it also lacks the change in distance necessary. And that is why the seperators are necessary to change the force.



By your theory, all of these are the same. However, they are not in reality. Only B and D are the same.

If A were like B, A would be unable to shoot. Think about what you are proposing here. The final piston in B would have so little force that it wouldn't be able to move. A lacks the distance necessary to change the force.

A change in diameter does not change the force. It needs the distance and the intermediary to change, not only a change in diameter.

Force also is not a property of matter. Force is the ability to accelerate mass. I'll even quote your favorite, Wikipedia here: "In physics, force is an influence that may cause a body to accelerate. It may be experienced as a lift, a push, or a pull." A change in diameter alone does no push, pull, or lift. A piston however does do the push, pull, or lift. That is why the force changes with a piston and it does not change when a piston is not there. The piston itself imparts the force.

Hopefully this explanation will clear everything up because I don't know what else could be said.

[Silence]

I'm dropping the case for now because I need to do some experiments on my own - to either prove or disprove the theory that force is more important than pressure. There's no point going theoretical there, I just have the urge to test a 3" cannon and a 4" cannon. McMaster-Carr and other places show statistics for nozzles based on pressure, not force, so I want to look into this again.

My argument here has evolved into something that's essentially saying pressure is more important, and that you need a smaller pump tube to increase pressure and increase range. Pressure would not be affected by dividers, tube size, or anything. But since it's the force vs. pressure debate again, I want to do some testing before I continue. Last time you and joannaardway had the debate, I was convinced force was more important, but I really need to see for myself.

Anyway, back to what you've just said.

As far as I'm concerned, you don't need a piston to change the velocity of a fluid in a new tube. I guess this is why I brought up the flow equations earlier. If you have a certain flow throughout the system, then the velocity changes inversely as the area. You don't need a piston to change the velocity of the fluid.

In each of the example water guns you diagrammed, the flow is the same - because the same amount of force is applied to the same amount of area in the pump, etc. If flow is constant throughout the system, then
V(pump) * A(pump) = V(nozzle) * A(nozzle)
I don't think we even need to get into force, for simplicity's sake. If you pump with a certain velocity, and if the ratio of pump area to nozzle area is the same for all the guns, then the velocity of the stream coming from the nozzle should be the same for all the guns.

Thus, all the guns you showed should have the same forces and velocities at the nozzle if everything is the same at the pump.

Pressure can also be used in my argument, because like the flow, it is constant throughout the system. If all the guns have the same pump area and if you apply the same pump force, then there's equal pressure in all the systems. That also means there's equal pressure at the nozzle. If the nozzle areas are the same, then there's equal force at the nozzles. Clearly, the presence of a divider will do little to hinder equalization of pressures throughout the system.

The problem is that this theory challenges your "force > pressure" argument. It's saying that it doesn't matter how much force is anywhere inside the gun - just the pressure created by that force, because pressure and a nozzle area turns into forces at the nozzle. Thus, it's challenging classic guns like SuperCannon II.

Some research has told me that statistics are found for nozzles according to area. Also, SuperCannon II had higher pressures than normal, and unlike many other guns, it had high bores and flow throughout. Other things like the constant pressure of CPS are interesting. I'm fairly sure the rubber exerts more force the more you stretch it, but over a greater area.

It would be helpful if you could reiterate the force vs. pressure argument , but I am really inclined towards testing two high-flow but differently-sized water cannons.

[Ben]

The reason why McMaster-Carr and such don't display information for things such as force is because many things don't create a force to push the water out, rather, they simply create a flow. I think most things there are rated in GPM. Flow is what matters, and creating that flow takes a force. Mathematically, calculating flow is difficult given thing such as pressure, so force is the easiest, most accessible model and it has proved to be accurate. Force can create pressure as we easily know. However, high force can create low pressure and have high flow. All three are related and important. Pressure is not a "deciding factor" in the power of water guns, but it is nonetheless important.

I'll agree with you if you are saying that there is more to powerful water guns than force. Flow created is what matters most, but we're not complicating things with fluid dynamics. Flow is most closely related to the force applied.

I'll give you one example that I think may illustrate what appears to be a discrepancy in the force model. Water gun A uses 500 pounds of force with a 6 inch piston. Water gun B uses 500 pounds of force with a 3 inch piston. Which will perform better? I am inclined to think B. I think pressure might come into play when comparing the same forces. Of course, each system involves the same amount of energy, so there really should not be too much difference.

In each of the example water guns you diagrammed, the flow is the same

The flow is the same in each respective water gun. However, the ones with the higher force will push the water out more easily, creating more flow for that individual system as a whole. It's not very hard to understand. It's like how larger diameter regular pistons are harder to pump. Once the force you apply is increased (through growth, constant use, weight lifting, steroids, whatever works for you), the water guns become easier to pump. It's exactly the same.

Other things like the constant pressure of CPS are interesting. I'm fairly sure the rubber exerts more force the more you stretch it, but over a greater area.

This is something I have wondered about myself. The tubing definitely gets harder to stretch as it inflates. However, the performance and difficulty to pump remains the same. I believe it has something to do with how the tubing inflates, but I can not be sure. I assume the way the tubing inflates makes most of the forces cancel out.

You also are continually referring to "force at the nozzle." There is no force at the nozzle. Force is APPLIED. Force is a push, pull, or lift as Wikipedia said. You seem to think force is a property of matter, but it is not. The force pushes the water out. It creates the flow. The nozzle affects the flow, not the force applied.

If you want to say that the larger your nozzle, the more flow, you can easily state that. But the nozzle DOES NOT AFFECT THE FORCE. It affects the flow, not the force. That is why the statistics for nozzles that matter such as GPM at a certain pressure are affected by the area of the nozzle.

Now, by your theory, all of the water guns I posted in my previous post are the same, correct? You think that any changes in diameter are the same as if they were changes with pistons, so I thought I would challenge that with examples of where that goes wrong.

Edit:

After some thinking, I have a better understanding of what matters right now.

Force matters the most still. However, since force is the ability to accelerate mass, more mass will accelerate less. This is why smaller diameter pumps are easier to move and why larger diameter pumps are harder to move. For each force, there must be an ideal range of diameters to accelerate the most mass at the highest velocity.

A rather simple concept that explains a lot. For this reason, between equal forces, the higher pressure one will perform better. This is part of what SilentGuy has stated that I agreed with, that smaller diameter pumps are easier to move. The full extent of the benefit of higher pressure is limited. Considerably higher force given low pressure is definitely better than lower force with higher pressure.

[Silence]

Originally Posted by Ben

The reason why McMaster-Carr and such don't display information for things such as force is because many things don't create a force to push the water out, rather, they simply create a flow. I think most things there are rated in GPM. Flow is what matters, and creating that flow takes a force. Mathematically, calculating flow is difficult given thing such as pressure, so force is the easiest, most accessible model and it has proved to be accurate. Force can create pressure as we easily know. However, high force can create low pressure and have high flow. All three are related and important. Pressure is not a "deciding factor" in the power of water guns, but it is nonetheless important.

That explains a lot, while inevitably raising some other questions . Thanks. And yes, I was referring to the flow. It's clear that flow and pressure aren't directly proportional.

From what I've seen, including a number of my dad's physics and fluid dynamics books, flow is mainly affected by pressure.
Q = A * v, where Q is the flow
Cv = Q / sqrt(dP), where Cv is the flow coefficient and dP is delta pressure, or the pressure differential
Cv * sqrt(dP) = A * v
So I don't really know where force fits in, unless it does make it easier to pump, as you say. I don't know. Gotta do those tests...

Originally Posted by Ben

This is something I have wondered about myself. The tubing definitely gets harder to stretch as it inflates. However, the performance and difficulty to pump remains the same. I believe it has something to do with how the tubing inflates, but I can not be sure. I assume the way the tubing inflates makes most of the forces cancel out.

Yup, that's what it sounds like. Just like a spring, the more you deflect rubber, the more force it applies - Hooke's law, if you will. Except that a piston powered by a spring won't increase in area like LRT will.

But that seems to imply, once again, that pressure is more important than force. If our theory about the rubber holds true, then performance stays the same even as force increases. The performance would be constant, therefore, because it is dependent on the constant pressure.

Originally Posted by Ben

You also are continually referring to "force at the nozzle." There is no force at the nozzle. Force is APPLIED. Force is a push, pull, or lift as Wikipedia said. You seem to think force is a property of matter, but it is not. The force pushes the water out. It creates the flow. The nozzle affects the flow, not the force applied.

If you want to say that the larger your nozzle, the more flow, you can easily state that. But the nozzle DOES NOT AFFECT THE FORCE. It affects the flow, not the force. That is why the statistics for nozzles that matter such as GPM at a certain pressure are affected by the area of the nozzle.

Okay, I'll stop misusing terminology. I guess I was suggesting that it's the pressure that counts. Sorry for the confusion and annoyance.

Once again, I'm just going to drop this argument until I can do some testing. Even at this point, nobody has done a controlled pressure vs. force experiment.

EDIT: I love the vB smilies.

[Ben]

From what I've seen, including a number of my dad's physics and fluid dynamics books, flow is mainly affected by pressure.
Q = A * v, where Q is the flow
Cv = Q / sqrt(dP), where Cv is the flow coefficient and dP is delta pressure, or the pressure differential
Cv * sqrt(dP) = A * v
So I don't really know where force fits in, unless it does make it easier to pump, as you say. I don't know. Gotta do those tests...

This is the calculus stuff I was talking about previously. Too advanced for most people. You can use force to get the pressure and then take the derivative of that to get dP. It's not too advanced, but few know how to do it. That's where the force could fit in in this equation. Physics is mainly fitting loads of equations together and converting things from one thing to another as this example shows. Just because it doesn't have a place to plug something in doesn't mean it can't work like that. You yourself show that by substituting in for Q.

Now, I do not know whether or not this equation is the right one to use for the flow created by an applied force. Most of the fluid dynamics things I've seen aside from the basic physics book my school has uses partial differential equations, something I know very little about. Consequently, I have not looked further than the first few chapters of the fluid mechanics/dynamics book I have and the chapter on it in my college introductory physics book for AP Physics.

But that seems to imply, once again, that pressure is more important than force. If our theory about the rubber holds true, then performance stays the same even as force increases. The performance would be constant, therefore, because it is dependent on the constant pressure.

I don't think you read my post from this. The force can be constant. The long tubular shape makes every force applied except for the very end of the tube cancel out. This is hard to visualize when explaining through words, but the final force is the vector sum of all of the forces. An animation might help though, so if someone understands what I am saying, please animate it!

The rubber applies force at angles perpendicular to the rubber itself. That is, it pushes inward from where the rubber is. Most of the tube can be said to be a cylinder aside from the end. The force of the cylinder part is not pushing the water out. Instead it is pushing the water in, and when pushing from all sides it cancels out. The cylinder part does nothing to push the water in the direction that matters. I find this is very similar to much of the electric field work we have done in AP Physics in that the system can be simplified greatly. The end part that is pushing horizontally is the only part that does not cancel out. And that part is constantly there except when the bladder is very low. Correlates perfectly.

The entire tube collapses, reducing the length of the cylinder part, but keeping the end part that is pushing. This is why the force and pressure is constant from what I can tell. This explanation makes perfect sense and I believe it is everything that is going on. It also explains why BBT's diaphragm system is less constant than regular CPS (due to the fact that not everything cancels out).

Again, if this explanation is inadequate, I will try to explain differently.

[isoaker_com]

@Ben: something seems slightly amiss with your otherwise decent explanation of force. Let's put it this way; if you create a piston system where the piston pushes against a liquid perpendicular to where the nozzle is, you can still squeeze out water despite having applied a force perpendicular to the direction of final flow. What happens is that force being applied by any source in any direction to a liquid ends up increasing the pressure experienced on the liquid. When the nozzle is closed, all the force vectors cancel out and there is no motion. However, open the nozzle and you end up with a small area (the cross-sectional area of the nozzle) where there is no force vector pushing back, thus water flows forth out the nozzle at the pressure experienced by the water.

[Ben]

Hmm... I think you're confused a bit by my explanation. I never said the flow had to be parallel to the flow from the nozzle. I wrote concerning flow from the rubber tubing. And that flow doesn't have to be parallel to the bladder (to get to the path to the nozzle). As long as everything doesn't cancel out, it flows. For example, sometimes I reuse rubber tubes, installing them in backwards from their original use, and find that the side the was used more often is the one that inflates the easiest. This can make the tube push the water in a different direction than the exit. While I suppose it's not completely efficient, it nonetheless pushes the water out. What I explained was a simplified system.

Anyway, I drew a picture and hopefully I can get it scanned in soon. My scanner isn't working right.

[isoaker_com]

Perhaps I am slightly confused by your explanation. The sentence that got me was the one saying the the rubber tubing pushing from all sides of the cylindrical part ends up cancelling each other out. I'd argue that the force pushing from the sides increases inner pressure and still does contribute to the available force to push water out of the nozzle.

[Ben]

Force is the ability to accelerate mass. By that definition, the cylindrical parts are not applying a net force because they are not accelerating mass. That is what it comes down to. They are definitely storing energy and creating pressure, but they are not contributing to the applied net force until the "end" (for lack of a better term) is at that part. That is when their stored energy is expended.

This is fairly complicated, but I am completely confident that my expantion is exactly what is occuring.

[isoaker_com]

I fear this has strayed a little far from the original topic of this thread. I still feel that either I'm really misreading your explanation or that something is still amiss in it. If the rubber tubing walls help add to the pressure experienced by the water, they must be contributing to the flow out the nozzle; their contribution may not be as great as the horizontal contraction length, but there still must be some contribution. If one puts two pistons squeezing towards each other, water will move sideways to get out of the way despite the fact that the force from the pistons are initially only towards each other.

[Silence]

iSoaker, looking back at the last few posts, I see what you're saying.

Pressure is what drives a hydraulic system. When a piston applies force perpendicular to the nozzle, it's creating pressure that acts in all directions. The walls react back in all directions, except if there's an opening.

Force in one place just can't explain how water will act in another place. Flow and pressure are constants, and they do explain how water will act in another place.

EDIT: Remember the tests I did with the weighing scale and the sand? They showed that the force used to accelerate the stream is proportional to the area of the stream. When you fire SuperCannon II with its 1250 pounds of force, that's only in the large chamber, which has a 4" (not exact) diameter. That's at 100 PSI.

The pressure in the stream is also 100 PSI. But the area of the stream is much lower - 1/64th that of the piston. So only about 19.5 pounds of force are being used to accelerate the water (I got that number from a ratio of the two areas and forces).

[Ben]

I fear this has strayed a little far from the original topic of this thread. I still feel that either I'm really misreading your explanation or that something is still amiss in it. If the rubber tubing walls help add to the pressure experienced by the water, they must be contributing to the flow out the nozzle; their contribution may not be as great as the horizontal contraction length, but there still must be some contribution. If one puts two pistons squeezing towards each other, water will move sideways to get out of the way despite the fact that the force from the pistons are initially only towards each other.

The rubber tubing adds to the pressure, yes. But that does not necessarily mean it adds to the flow.

I understand that my explanation is a simplification of things. Where there is no movement, there is no force. I do not think anyone will deny that only part that is moving in the CPS chamber is the "end" of the chamber, and thus it is the only one creating force, which translates into a flow. The force is constant. The entire system is a tension system and can be thought of as millions of rubber bands. One end is contracting, while the other end stays open as a path for the water to travel through.

If one puts two pistons squeezing towards each other, water will move sideways to get out of the way despite the fact that the force from the pistons are initially only towards each other.

Yes, but in a CPS system there is another direction the rubber can move that avoids this conflict. I find that the rubber always does what is the easiest. One end collapsing is indeed easier to do than all ends collapsing in simultanously.

And again, to reiterate, what matters for flow and force is movement. Only one part of the CPS chamber is moving. That part is the one that is imparting the force and consequently creating the flow.

The pressure in the stream is also 100 PSI. But the area of the stream is much lower - 1/64th that of the piston. So only about 19.5 pounds of force are being used to accelerate the water (I got that number from a ratio of the two areas and forces).

Let me say this again for the millionth time: THERE IS NO FORCE AT THE NOZZLE. No ratio can be made to calculate force at the nozzle. Force is the ability to accelerate mass. A change in diameter does not change the force applied where it matters unless the change uses a piston or something similar to apply the force again.

Just because you seem to like throwing equations around, here's the derivation to flow from force:

F = m * a

m = p * Q
p is density. For water, p is about 1 at most temperatures (let's not get too technical though).

F = p * Q * a
Drop p because it equals one.

Q = A * v

F = A * v * a

a = dv/dt

F = A * v * dv/dt

To use this, you'll have to do at least some nasty differential equation work that I don't care for. But, there's the relationship between applied force and flow. There are absolutely no errors in the math, unlike your application of an equation to calculate flow lost over an orifice. This in fact uses the most basic equations of physics to get the job done.

Interestingly enough, what this does show is that the velocity of the fluid (not the flow rate) is proportional to the pressure. F / A = P, so P = v * dv/dt. Very interesting. I thought they would at least be close since both rely on the area.

Anyway, I'm making a reply in your new thread right now.

[Silence]

Originally Posted by Ben

Where there is no movement, there is no force.

I disagree. The classic textbook scenario: you push on the wall. Force is being exerted, but no work is being accomplished.

Originally Posted by Ben

Let me say this again for the millionth time: THERE IS NO FORCE AT THE NOZZLE. No ratio can be made to calculate force at the nozzle. Force is the ability to accelerate mass.

I said force is being used to accelerate the stream. The mass of the stream depends on the area, and the force also depends on the area.

Look at my critique of the derivation in the other thread.