Air-powered CPS without a regulator

[Ben]

It's not ID - it's surface area. Surface area is controlled by by the internal diameter in shapes where the cross-sections are circles. So technically speaking, a larger internal diameter means greater force with the same pressure.

There's plenty of information about this on the internet, but Wikipedia appears to be a good start.

[Silence]

This sounds great! However, I am quite sure that it would work with a narrow shape--as long as the areas have similar differences at different water levels, you should be able to provide constant pressure. Naturally, the water would fill up a smaller pressure chamber more quickly--but more PCs would work, and it would be large enough as it is.

@ joannaardway and other confused people: This is actually a superb idea of Ben's, once you understand the way it works.

pressure = force / area
force = pressure * area

That should be simple enough. Clearly, force varies directly with both pressure and area. Assuming force is equal to the constant of 100 pounds, and also given that the starting pressure is 100 PSI, we find the following:

100 = 100 * sq. inches
sq. inches = 1

So you shoot a bit, and the pressure drops to 50 PSI. We still want the force to be 100 pounds and stay constant, so a little math can find what must be the correct area for the force to stay constant, even as the pressure drops.

100 = 50 * sq. inches
sq. inches = 2

I have not taken physics yet, so I really wouldn't know the formulas for pressure when the air has been displaced by a specific amount of volume; however, Ben does, and his calculations are on the sheet he scanned. I had original supposed that the correct curve would be a quadratic (in order for the area to increase in a linear fashion), as in the picture, but I'm assuming the pressure does not increase in a linear fashion and that the shape must be similar to what Ben drew. Nevertheless, I am excited, because I would never have thought that a drop in pressure could be accounted for in this way.

[Ben]

The equation I used to calculate the pressure if the air was in a specific volume was Boyle's Law, which actually was an equation they teach you in chemistry class.. We actually never discussed pressure once during my Physics class, and only after the AP exam did I ask my teacher a few things about it. I did however use my calculus knowledge to find the volume of a function rotated around the x-axis... you would not know how to do that sadly.

The way you described for approximating the ideal pressure chamber curve is simple and I like it. You'd still need calculus knowledge to fill in the gaps however. I think I'll try that and see what I get.

What I tried initially was too complicated - I tried to find a curve where the derivative of the force equation always equaled zero (no change in force). Needless to say, I have no idea how to do that at the moment without using far more advanced calculus and probably differential equations. I opted instead for trial-and-error.

Just another place where physics, calculus, and random thinking make better water guns.

[Silence]

What my weak Calculus knowledge (I just finished Algebra II) tells me is that only a horizontal line will produce a derivative of zero. Of course, any linear line will produce a constant derivative. And this happens to be where I think we differ in thinking, although I can't be sure yet...

When I suggested a quadratic curve, I was thinking that it would be quadratic from both the front and the side--somewhat like a dome. With that assumption, points at different levels will form an arithmetic sequence, so the derivative might be what you're looking for. If you thought the sides of the PC were flat, then a triangle from the side (not a cone, because it's only in one dimension) would also produce the same results. By the way, such a triangle shape would also be narrow enough to be made enormous.

I do like this, too--we find our own ways to bypass or face challenges, rather than using the exact methods the teachers tell us to. I guess I forgot about Boyle's Law, though--that's what I was looking for, but I couldn't identify it. I guess it proves that we are looking for an area that increases at a steady rate, as opposed to whatever rate I thought you might have been referring to. However, I'm now convinced that my shapes might be the right ones--and since I've got all the formulas now, I can help test the various shapes.

It's been quite a while since I last used TI-BASIC, but here goes:

NOTE: due to some text limitations in the post editor and my being too lazy to use OOo to write out the best version, there are some text imperfections; for example, "->" is the store symbol and "L ___" signifies a custom list (the lowercase "L" looks like "l", so I just did this). Also, the underscores signify the spaces.

prgmCAPSHAPE

clrhome

disp "P_=_F_/_A","A_=_F_/_P","PRESS_[ENTER].

pause

input "FORCE=",F

input "MAX._PRESSURE=",M

0 -> N

for(P,M,15,(M-15)/8

N + 1 -> N

P -> L P(N)

end

F / L P -> L A

disp "AREA=

output (8,1,"[ENTER]_TO_EXIT

pause L A

clrhome

It isn't that long, but I still haven't tried this yet--feel free to do so now if you wish, since it might be a while before I do it myself. I could probably do a really fancy thing on the computer, but that seems like even more work.

[Ben]

Oh, believe me, I know what you're saying about the quadratic shape. That was the first shape I tried. It's force is not constant. I later tried the inverted shape because I felt that the air would have less space and therefore more pressure, yet have a larger ID due to the way the curve bends outward. And consequently I got a relatively constant force equation.

Here's a scan of 6 shapes I've already tested: http://www.sscentral.org/images/chamber-force-2.jpg

Please note the orientation of the equations - the first equation appears quadratic, but it actually is the graph of a root function. That is due to the way the equations are rotated, around the x-axis. The quadratic shape you mention SilentGuy is the root function on this paper.

I've also tested a few other shapes since then, but they really haven't broken any new ground. I'm not sure what other types of functions I could try without trying a piece-wise function.

[joannaardway]

I always work with the ID at the point of pressure drop - in other words, the nozzle diameter.

Pressure is almost always constant throughout the fluid, so surely ID in the chamber is irrespective.

In assuming what you are, you are effectively saying that the chamber is not isobaric with the remainder of the pipework.

Air pressure can force as much as it likes over the entire surface area, but only a certain amount is applied at the pressure difference.

Ok, assuming that everything I've been taught so far is wrong:

Is this a matter of applying differentiation and integration to a function of force? That's about the only sense I could see here.

[Ben]

@joanna: I don't think that you understand in the least what we are talking about, so I'll explain again.

We're talking about the force of the air pressure, which is a function of the pressure and the surface area the pressure is pushing on. The pressure is constant throughout the fluid, in this case air. The water is not pressurized very much at all, rather, it exerts a force on the pressurized air because water resists pressurizationb. So technically, yes, the chamber is not "isobaric" with the rest of the piping. The pressure is not the same, but the force is. Force and pressure are two different things, as are mass and inertia.

No differentiation or integration is performed on the force. We're not talking about a power or work function here, though I suppose you could derive one from what I have described. The only integration I perform is to find the volume of the air chamber at specific points, and that's the only calculus used.

Read the Wikipedia article and I think you'll figure out what I have been trying to explain. They explain the force of pressure just about as well as it gets.

@everyone: I've found a small error in my force calculation which creates a slight, but distinct difference in the final force function. In my previous generalized force function, there was still force when the pressure was equalized, which is incorrect. We have to go by gauge pressure, that is, pressure minus atmospheric pressure. So F = A*(P-14.7).

My new corrected generalized force equation:



E - where in the function the pressure chamber ends
Po - pressure at h=E, in our case atmospheric pressure
r(h) - function describing the radius of the chamber are height h

z is only necessacy for the integration and is not defined.

This generalized function will make creating the force functions far easier.

[Silence]

Well, theory often doesn't keep up with real life due to unknown factors. At any rate, I'm assuming your root function was such in both the y and z coordinates, and that the same goes for everything else. If not, what results did the triangle have?

Anyway, I trust that your calculations are correct, so the least we can do is support the shape that does work. Also, do you agree that the shapes can just be made narrower in order to make them more "streamlined," or will that have major negative effects?

EDIT: Didn't see the last two posts because I picked up on this post from earlier, but that shouldn't matter that much. The conversation went a bit over my head there, but I really don't think it needs to be that complicated. Note the picture I posted: because the chamber is relatively full, the pressure is high; but the surface area of the water where it touches the air is low (as opposed to greater/constant surface area, as found in standard air pressure PCs), so the force is relatively average. When the water level drops, the pressure will also drop--but since the surface area increases greatly, the force still stays somewhat average. Naturally, all this assumes that force=pressure*area.

What Ben is trying to do is find the PC shape that matches the pressure and area at different water levels in such a way that the force remains constant. Because the two are inversely varied, there are a variety of ways to find the best shape--and the various ways are what Ben and I had been discussing recently in this thread. I just find it revolutionary that PC shape has such a great effect--for example, an APH with a 1/2" pump and 3" PCs might have 9/4 as much power as an APH with a 1/2" pump and 2" PCs due to the extra area. In fact, I just realized why Duxburian says large PCs provide more power than small ones!

Anyway, if you're confused, Ben's probably right in stating that the Wikipedia article will help (who knows, he might even edit it to include this stuff!). I haven't read through the whole thing though, but it is absolutely necessary to understand the two equations I quoted earlier: P=F/A (or more specifically, F=PA) and Boyle's Law, or PV=k.

[Ben]

Well, theory often doesn't keep up with real life due to unknown factors. At any rate, I'm assuming your root function was such in both the y and z coordinates, and that the same goes for everything else. If not, what results did the triangle have?

I'm not quite sure what you mean by the y and z coordinates. It's a 2D function rotated around the x-axis. Could you explain in more detail?

As for the results of different shapes, I think I'll be working on redoing those functions for a large variety of shapes. The curved inward (convex?) shape provides relatively constant force, and is the best we've seen so far.

The image you posted also will not provide constant force. It will provide slightly more constant force, but not close to constant force. The chamber has to be curved inward to provide close to constant force.

[Silence]

I apologize--I was just too lazy to explain or draw what I was trying to describe in reasonable detail. I did it just now, and it barely took me five minutes!

Here's a diagram detailing what I meant by 2D/shape only in the y-coordinates (with x pointing up, as in your sketches):


And here's what I meant by 3D/shape in both the y- and z-coordinates:


As you can see, the perspective and quality isn't perfect; but it was done quickly in Paint, so what can I say? Note that I just chose random functions for each one: the first is a 3D-function in the shape you were describing as the best one, and the second is basically a triangle.

If we're talking about three dimensions, the area of the root function in 3D and the triangle in 2D increases by a constant. At any rate, I can now see why the shape in which the walls curve inward (concave for me, convex for you) is the best--and I really answered it in my last post. Because Boyle's Law involves an inverse variation, the area should increase at a quicker and quicker rate as the PC is emptied, because the pressure actually starts dropping more quickly (and why you really only see dropoff at the end, though that's also due to the smaller area at the end in spherical PCs). Thus, the area has to increase much more rapidly than I had expected. Thanks for sticking to your original conclusion until I had cleared this up...

[Ben]

Good to see that we're on the same page here now! Now, it's time for someone other than SilentGuy, joanna, and myself to be the ones discussing.

I just came back from my nightly run, and I thought of something I had not considered: angled shots. The way I designed this was so that the water line was parrallel to the ground. Thus, at an angle, the water line with shift and the force will not be constant. I could redo my math and redesign the chamber to work at a 45 degree angle, but that would be a little harder and unnecessary. An angled joint before the valve will fix this problem. We're not going for perfect laminar flow obviously, so this isn't a problem.

I also admittedly think backwards when it comes to concave or convex. I meant curved inwards, which is concave. My mistake.

What should we call this pressure chamber design? I'm thinking something like the "concave cone" design. What do you think?

[Silence]

Yes, it is nice to have all the various confusing points and delusions sorted out. For a name, I was thinking of something along the lines of "unregulated CAP" (or "UCAP")--it seems to allow for more shape flexibility than you'd get with "curved cone," which nevertheless is more appropriate. However, the choice is up to you--you're the one doing all the math and theory for this idea. Chances are, though, that if a firm such as Buzz Bee starts using this, they'll make up a new name for it. Technically, the name for their PreCharger series is "Aqua Master Pre Charge System"--but we should have a technical name ourselves, I suppose.

Are you suggesting that we use a weight and a joint or some other method to "hang" the PC and have it at least somewhat vertical all the time? That seems quite unreliable, and I wouldn't trust it. In fact, I'm going radical here--what about putting the reservoir in the handheld part and the PC in the backpack? Yes, that might be a bit excessive, but it's still something. Anyway, this could be a problem--even the slightest of tilts could slightly alter the stream, and if a cone is turned only 60 degrees, you'll have much greater dropoff than normal. I'll keep on looking for a solution (I do my best thinking when I'm in bed at night and before I fall asleep, but I don't have a chance to write the stuff down--and I forget the next day ), but you shouldn't give up hope at all--you're incredibly close.

Yes, more opinions would be greatly appreciated. I'm hoping enough people understand this stuff, especially since we have a ton of engineers here--and I'd just like to know how well the community is taking this. EDIT: And Ben, since you know where I live, you probably know there's a strong university nearby--maybe my dad can have somebody machine a concept version, just for testing. Naturally, though, I can't guarantee this at all, and I'll need a completely finalized design anyway. Having such a model in front of us would be cool as it is, even with the tilting problem unsolved.

[Ben]

No, I simply suggested use of a 45 degree bend before the nozzle and valve.

[Silence]

Oh, I see--so would the nozzle be angled upward at 45 degrees by default? That might work, but average use would probably be closer to 0 degrees than to 45 anyway--so this might be more harmful than good. I've been focusing so much on the benefits of the water level and surface angle that I've forgotten the disadvantages.

Having a "rotating" PC, even with a hose connecting to the nozzle, would just add more complexity (especially if compensating for sideways tilting) that wouldn't be present with a CPS soaker--and I don't know about you, but I've been considering CPS performance to be the goal--as well as increased simplicity. If the design could ensure a vertical PC, then a hose could be used to aim and shoot. I'm still thinking, and after all this, it wouldn't be right to abandon the project...

[joannaardway]

I know exactly what you are saying.

However, it's not making any sense.

A few errors I can see: The water and air must be at the same pressure.

The water and air are making contact with each other over the same area - nothing else is possible. If the pressures were different, the forces would be different, and thus, the water would be moving in one direction or the other...

Which of course is what happens when you fire...

You lose pressure from the water, so the forces are inequal, thus forcing the water down the chamber and out of the nozzle...

Assuming the pressure loss is consistent regardless of chamber ID, (Why shouldn't it be?) the resultant force will be greater when you fire.

A greater resulant on the same mass means greater acceleration...

Right, filling in a few gaps off the screen - it all makes sense now. I need to see the reason on screen, rather than just have the result stated at me.

Damn it - I hate disproving myself.