Maximizing a Nozzle's Stream Velocity

[Silence]

Thanks, just saw your edit. The strange thing is I came across the "Ask a Scientist" pages (didn't really read it, just noted the pink background ) while constructing my reply.

And before you ask, I'm replying today because the first reply took only half an hour - I think.

Originally Posted by Ben

The pressure and velocity are proportional. They both are dependent on the area. But, that's not the entire picture of flow because it's missing the area of the cross section.

Pressure and flow and proportional, I'm thinking. Flow and velocity are proportional if you make the area a constant - after all, it is the area of the nozzle, and the nozzle is a control.

Originally Posted by Ben

And yes the thread does cover flow as well as velocity. To calculate the time it takes to empty a syringe, I'm pretty sure the best way would be to calculate the flow first. And that seems to be what they were doing. It's nearly exactly the same situation Supercannon II is in.

Sorry, I didn't read the entire thread. Even by the end of the first page, they had moved on from flow again - they were more concerned with the momentum and force of and exerted by the hammer.

If they have flow/velocity, I'll check the thread again tomorrow.

Originally Posted by Ben

What I am saying essentially is that the flow coefficient equation only works over a valve or a system. It does not actually calculate flow created. It calculates flow lost. Consider a point in the pressure chamber and one right before the nozzle. There is no or little pressure differential. Yet there can be flow given that the valve was open. That is where this equation falls apart. Yes, there can be flow there, but there also can not be flow there. A force model makes more sense because a closed valve exerts back a force, equalizing all forces. But in your model, with or without the valve, there shouldn't be any flow using points inside of the water gun.

I only brought in the flow coefficient as I was searching various books and the web for different ways to relate the pressure and flow - I wasn't quite sure what I was looking for at the time. Then I realized the flow coefficient was also extremely important, even though nozzle size - a constant - was also a factor.

I don't get what you're saying in the latter half of the paragraph. Perhaps a drawing would help . (Just noting, if the valve is closed, there is an area for the pressure to apply force. I'm not sure how that fits in with your argument though.)

Originally Posted by Ben

And anyone using these equations to calculate flow created from pressure is using them incorrectly. Again, I looked in my book for the flow coefficient equations. This book uses different equations (more of them and more advanced versions), but the same coefficient. I believe that the version you posted is a simplified version that puts more constants in the coefficient. I know it's the same coefficient because it's titled the same.

Yes, this is indeed a very simplified version - or there are multiple equations that involve flow coefficient. I think it's mainly the simplification though.

Some of the equations involve things like density and other pressures, but most of those are either constant or nearly constant here. The fact that the fluid - a liquid - is incompressible (for all intents and purposes) already rules out density. But I see what you're saying.

Feel free to introduce more complex versions of the formulas, I don't mind. Just keep in mind we're only aiming for a rough model.

Originally Posted by Ben

The book dedicates a grand total of 9 pages to the use of flow coefficients out of about 680 pages. The one example problem they give with it already has a flow going and the coefficient is used solely to calculate flow lost. Cv is first mentioned on page 451. In contrast, on page 114, they derive Newton's second law, the one concerning force, for fluid flow. This is the first chapter on fluid flow and it's on the second page of that chapter. Which one is more important? You would say Cv, but it is a minor part of fluid mechanics in the eyes of Fox & McDonald.

Once again, I used Cv to connect flow and pressure while also adding in an important variable.

Originally Posted by Ben

No, I was explaining that a pressure differential powers all air pressure water guns. You replied with a statement saying that because the SS 300 patent says it uses a pressure differential to shoot the water, it has to operate on pressure. What I said was that the pressure differential also powers the force model. So, anything stating that it works by a pressure differential is not saying that it operates on pressure.

Beyond the fact that pressure * area = force, the pressure difference really doesn't support your argument. A massive piston inside a cannon does not have a pressure gradient - it's equalized, even when the valve is closed.

The pressure differential they speak of is between the internals and the atmosphere. The piston does not separate the internals and the atmosphere.

Let me quote the same passage from the patent, but also what follows:

Such pressure activated water guns work upon the principles of pressure differentials between the water held within the toy and the atmosphere. The water within the toy is subjected to a pressure higher than that of the ambient air. As a result, when the water within the toy is given an avenue of escape, the water will stream out under the pressure.

I'm reading through the rest of the application right now. They have a lot of information - they mention 15-30 gauge psi and some other stuff. I'm digging deeper into the goldmine of information now.

The problem with simple two-stroke squirting systems is that the amount of liquid that can be expelled is limited to a single volume of the compressible area; also, the pressure of the liquid exiting the device is dependent directly upon the force being applied during the time of expulsion. Consequently, when water is squirted in this manner, only a small volume is released with each pumping action. When attempts are made to increase the amount of water propelled by increasing the volume [read: the pump's cross-sectional area] of the compressible area, the pumping action cannot displace the water at a high pressure, resulting in expulsion of water at low pressures.

They're implying that low pressure is bad for range. We know the force is the same - you apply the same force to both pumps.

The paragraph right after is also great. Just great!

Water guns have advantageously involved squirting large volumes of water at high pressures. Generally, the higher the pressure, the longer the distance the water can be propelled, thus increasing the range and power of the water gun [sic].

Caught the last sentence? The next part of the paragraph focuses on turbulence and Cv.

The present invention water gun uses a two-stroke pump to store and pressurize large amounts of water, but relies upon low pressure and wide avenue of release and a wide nozzle to achieve squirts of large volumes of water which are generally non-turbulent squirts as they exit the nozzle.

Amazing, wouldn't you say? It's all in the pressure. Not once has the patent mentioned force (except outside of the context; such as force water into the pump tube, force open the valve, etc.), at least as far as I can tell. They haven't even mentioned the area of the PCs. Mentioning volume just was a claim as to one of the features of the gun - the same paragraph goes on to describe other usability benefits. It also mentions 20 psig as an example pressure, with a range of 10-30 psig.

There is a fair amount of redundancy. Later D'Andrade and Johnson repeat:

The more water or air is added, the higher the low pressure and the farther and longer the invention may propel water...

There's more, too, but I think you get the point.

Originally Posted by Ben

Well, I remember calculating this sort of stuff before. I used to be a "pressure diehard" myself, but I then saw that reality does not match up with that theory. It wasn't until I took AP Physics that I realized that FORCE is what causes things to move. A flow is movement, so force matters. And it matches up with reality very well.

I remember a thread at WWN where ZOCCOZ asked whether more pressure chambers will be benefitial in water gun designs given the same pressure. I said no because the pressure was the same and I felt that was all that mattered. Now, I would not dare say such a thing because it simply does not match reality or the actual physics.

Yes, force causes things to move. But the equations show that it's the pressure gradient that causes flow.

Originally Posted by Ben

Also, the marketing department definitely came up with the name CPS. It is not a statement as to how the system works, rather, it's calculated marketing that serves to people's minds. Most people assume more pressure is good, and the smarter people know that pressure drops. Obviously, "constant pressure" works for people. Product names definitely do not always state how the system works.

I'd be lured by something that claimed Constant Force, if it were true - as implied - that constant force meant constant range. But no matter. (If people believed in pressure more than force, it's because earlier guns capitalized on and advertised pressure as opposed to force - because pressure is important and force isn't.)

Well then, let me quote the patent I found that suggested bladders for water guns. The first phrase is in the "Background of the Invention."

The object of the present invention is to provide a liquid projecting device which maintains liquid in a pressurized reservoir at a relatively constant pressure independent of reservoir volume enabling uniform fluid projection characteristics throughout operation.

I'll more investigation for the weekend. Patent searching and quoting has already made this response take much more time than the earlier one today.

Perhaps you're right as to how the constant pressure is obtained. We'll see.

Originally Posted by Ben

It approximates the flow loss through a valve. They use it in nozzles because they slow the flow. I definitely do not think it's inaccurate for nozzles because that's what it's meant for. The Cv equation is meant to be used to calculate flow loss through a valve. Nozzles being a valve, that is a legitimate use for them. But, it only works given a flow before. The equation is not used to calculate flow created by a pressure. None of the examples I see in the book use it to calculate flow from pressure. You can't get more authoritative than that. You're not arguing with me. You're arguing with established physics.

I might understand your argument if you used some other equation. This one most definitely is not meant to be used as you are using it. It is meant to calculate flow loss through a valve. I am completely confident that the flow coefficient equations are the wrong model.

I've no time to dig up more links. All I can say is look back through this thread and through all the links I've already given to show how people use flow to calculate pressure or pressure to calculate flow.

The SS 300 patent might not have specifically stated "flow coefficient," but its "large avenue of escape" or whatever screamed "flow coefficient" to me.

Originally Posted by Ben

Not a single scientific critique? I don't know if you are reading my posts after reading that. EVERYTHING I mention is scientific. Especially the stuff I read from my book. How can you get more scientific than by reading a book on the subject? (Don't answer that, it's rhetorical.)

You still have not scientifically critiqued my physics. Yes, you've stated "you're not using it in the right way," but that's not even the physics itself. And I've shown how plenty of others have used this equations that way.

Also, I don't see how flow coefficient can't be involved in calculating flow. It's whole point is to determine efficiency of flow.

Originally Posted by Ben

As for the "problem" with derivation I've found, I've responded to everything you have said that is a problem with force model. The book's derivation is one you can not take issue with. Again, I know that my original model was wrong, so there's nothing you can say about that. I'm going with the actual stuff in the book.

Fine:
Why can mass flow rate be substituted for mass?

If you can answer that, then you have an argument.

By the way, I'd still like to see the pictures of the pages in your physics book. Until then, even your qualifications I can't believe. Perhaps many of my sources are on the Internet, but at least I've linked to them.

Originally Posted by Ben

Check out the link in my last post too. I edited it in about the same time as your posting.

Thanks. I think I mentioned it at the top of this post.


EDIT: Saw your new edit.

Originally Posted by Ben

Also, consider a situation I raised when joanna said that pressure matters more than force. Put a ball valve after the PCs, but before the firing valve, in a standard APH design. Pressurize the PCs, and close the ball valve. Now, according to the pressure model, because the pressure differential is the same between the first ball valve and the nozzle, the flow should be the same (for a shorter period, of course). I do not believe that you will state that this situation will perform the same if the ball valve were not there. Again, pressure differentials matter because they create a force to push, but simply having a pressure by itself without something pushing (applying a force), is not going to create flow. The water is slightly compressed for our purposes because the volume does not change noticable. Thus, negligible expansion occurs and little flow will.

I don't get it. Could you do a diagram? Thanks.


EDIT again: I had a bunch of other pages open in tabs, forgot to post them.

Took a look at a list of flow calculators at eFlow - not a bad site.

The second site on the list has what looks to be a neat nozzle flow calculator. You have to pay to calculate, but looking at the fields and the formulas, the pressure differential is important. Force is mentioned nowhere. The calculator also uses one of the versions of the flow coefficient equations with extra variables.

The tenth site's description reads:

Originally Posted by Gossman Consulting, Inc.

Flow calculator for pipes to determine if flow is turbulent

Messing around with it, sounds interesting. Obviously doesn't take into account the joints and all. I'm still thinking that the nozzle generates more turbulence, though.

[Ben]

Beyond the fact that pressure * area = force, the pressure difference really doesn't support your argument. A massive piston inside a cannon does not have a pressure gradient - it's equalized, even when the valve is closed.

The pressure differential they speak of is between the internals and the atmosphere. The piston does not separate the internals and the atmosphere.

Again, you don't understand some basic fundamentals of physics. First off, there is no pressure gradient aside from the water decompressing once it exists the nozzle. There's a pressure difference, but inside of the water gun there is no gradient.

Second off, the pressure differential is what powers all water guns, even in a force model. There would be no force without a differential. I have no idea why you think the SS 300 patent mentioning a pressure differential breaks anything I've said.

Anyway, I'm looking through the quote you posted and they don't phase me one bit. You say catch the last sentence, and it says "Generally, the higher the pressure, the longer the distance the water can be propelled, thus increasing the range and power of the water gun." The word generally exists there. I would say that sentence under the same contexts. You know, higher pressure is benefitial in the force model.

And when they're saying "When attempts are made to increase the amount of water propelled by increasing the volume of the compressible area, the pumping action cannot displace the water at a high pressure, resulting in expulsion of water at low pressures." I don't think they're speaking about the pump's cross sectional area. What increases the amount of water propelled is more PC volume. More pump volume can make it pump faster, but it does not increase the volume. Also, they never said low pressure was bad.

Also, while you love to say "high flow coefficient!" and such, you know the SS 300's flow coefficient is not extremely high. It's probably not even high at all. It's about as big as the nozzle if you look through the patent images. I don't think you will state that it's increasing the performance on the level necessary. And wide avenue of release could easily mean wider pressure chambers... and that seems to be what they did in this design. They also could easily mean wider streams, another thing they did in the patent. In fact, "wide avenue of release" sounds more like a wide nozzle than anything else. When you match reality with what they are saying, you get a better picture.

Once again, I used Cv to connect flow and pressure while also adding in an important variable.

Again, you're not responding to anything I say. Why is Cv stuff only on 9 out of over 680 pages of my book? Why itn't it some sort of "fundamental" thing they use? Why are you defending an equation that obviously is not used to calculate flow created by a pressure.

I've no time to dig up more links. All I can say is look back through this thread and through all the links I've already given to show how people use flow to calculate pressure or pressure to calculate flow.

The SS 300 patent might not have specifically stated "flow coefficient," but its "large avenue of escape" or whatever screamed "flow coefficient" to me.

Well, I'm using textbook examples, while you're using forum links. You seem to think forum links are a more authoritative statement. That and I think "wide avenue of release" is actually referring to a wider nozzle. It makes much more sense overall. And, you know that the Cv is not very good for this water gun because it's ID is about as big as the nozzle.

You still have not scientifically critiqued my physics. Yes, you've stated "you're not using it in the right way," but that's not even the physics itself. And I've shown how plenty of others have used this equations that way.

Also, I don't see how flow coefficient can't be involved in calculating flow. It's whole point is to determine efficiency of flow.

I suppose you can say that I haven't scientifically criticized anything, but I could say the same about you then. What would be "scientifically criticizing?" And why are my TEXTBOOK examples given less weight than you ONLINE FORUM examples?

You're right. The Cv equation is about efficiency. It's not used at all to calculate flow created. It's used to minimize flow lost typically with valves with lower Cv valves.

Fine:
Why can mass flow rate be substituted for mass?

If you can answer that, then you have an argument.

By the way, I'd still like to see the pictures of the pages in your physics book. Until then, even your qualifications I can't believe. Perhaps many of my sources are on the Internet, but at least I've linked to them.

Okay, I can see now that you aren't reading my posts closely at all.

THE EQUATION I POSTED BEFORE IS COMPLETELY WRONG. Got that? I hope you do because I've said it several times. What's wrong with the book's equation? I'll get you a picture today, but you seem to deny anything you can't see yet. I've tried to find the same equation online, but it appears that it's far too advanced for most people online.

I don't get it. Could you do a diagram? Thanks.

I find this unusual because I swear you understood it before when I mentioned it to joanna. I'll get you the diagram later today.

Took a look at a list of flow calculators at eFlow - not a bad site.

The second site on the list has what looks to be a neat nozzle flow calculator. You have to pay to calculate, but looking at the fields and the formulas, the pressure differential is important. Force is mentioned nowhere. The calculator also uses one of the versions of the flow coefficient equations with extra variables.

Again, I don't know how many times I have to say it, but none of these calculators are used to calculate how much flow is created. That's because the Cv stuff is not supposed to be used to do that. Typically people know how much flow they have because they use pumps with measured flow values already. That's why places like McMaster-Carr put those values there.

Also, get to reading the .gov link I posted. I'm assuming that because you didn't reply to it, you haven't looked at it in much detail.

[joannaardway]

I'm very sorry if I cover points that have already been discussed, but the volume of stuff was too much to take in all at once. I have however been thinking about this for ages.

As far as I can tell, Ben's theory of force is dependent on a few things, and pretty high up the list is the fact that water is incompressible.
The literal definition of this is: An incompressible substance is one which can never be under pressure.

However, defining incompressible as having no (notable) volume change when it is under pressure is unfortunately a common fallacy.
Only an ideal gas follows a perfect pressure/volume inverse relationship, so trying to apply this to a fluid is a mistake.

It is perfectly possible for a substance to have it's pressure double when it's volume has only been reduced by a quarter - or even less in the case of liquids.
In fact all real gases display slightly similiar effects. For example, when it's volume is halved, the pressure in air doesn't quite double.

Now, before you get high and mighty on me, I have to state a fact - there are no incompressible substances in the universe (Even solids are compressible, but as they keep their own shape, then the result can be viewed as a force*).
Any incompressible substance, by definition, must have a infinite speed of sound within it (check me up on this by all means, but it is true.)
Now, as nothing can exceed the light barrier, no incompressible substances can ever exist.

*I'll come back to this later.

As the speed of sound in water is known, and it is certainly not infinite, we can conclude that water must be able to carry a pressure. And indeed, water's volume does change when under pressure. But it doesn't follow the same inverse relation ship of gases.

I assure you that all the facts I have given are entirely correct, and please feel free to check up on them - You will find them true.

Now, this means that the equal and opposite force on the air in a PC is coming from PRESSURE in the water, rather than a force.

Put a piston in between, and the pressure is converted to a force (as the solid keeps it's shape), and then back to a pressure.

Now, I promised to come back to the thing about things keeping their shape. Only things that (vaguely) keeps their shape will - although it carries a pressure - act as if it is carrying a force. As a thought experiment to prove this, I suggest the following:

Imagine a pipe containing water, trapped between two pistons of equal size. Push on one piston, and an equal force will arise at the other end. (And the water's shape has remained a perfect cylinder through out)
But if the pistons are of different sizes (and the pipe changes diameter in the middle somewhere), the fluid is changing shape (because the two bits of pipe are of different diameters), so it no longer acts as if it were carrying a constant force. (Because it is carrying pressure, and changing shape).

Now, in both cases, imagine that the pipe's contents were frozen, so that they kept their shape (the fact that ice is less dense than water doesn't affect this experiment).
In this case, the forces on both ends are the same regardless of whether the pistons are of different sizes - because the ice keeps it's shape at all times.

Only when a substance keeps its shape can the force argument be applied. (Again, check up my core facts) As water doesn't, the force argument must be mistaken in this case. Not to say that the evidence is wrong, but the wrong conclusion has been drawn from it, as has happened many times in science (Pholostogen theory comes to mind).

I have to note the following flaws:
-The theory cannot account for all water guns (only "special" ones like the Supercannons, and the Waterzooka seem to be usable [And I still disagree with the force that can be practically applied]), and it's effects are not noticeable on all water guns (I pose the 3xA and external pressure chamber problem. No effect on range or stream velocity at all. Even from many times the force oddly enough.)
-Linked to this, it's been based on only a couple of soakers - not the widest list of sources ever. (Was the data ever made available to view in full?)
-The theory has tried to automatically build in how stream velocity affects range, however, this should be removed and handled seperately. To solve the issue we need to look at the relationship to stream velocity, then handle stream velocity to range seperately.

That's all I've got to say for now. I don't expect it will do anything to change anyone's views, but I think it's totally conclusive. None of the facts are false or inaccurate, and I have been very careful about my logic organising it into an argument.

So, by all means, try to find a problem with it - but I am certain that there isn't, which dispels force theory.

[Silence]

Originally Posted by Ben

Again, you don't understand some basic fundamentals of physics. First off, there is no pressure gradient aside from the water decompressing once it exists the nozzle. There's a pressure difference, but inside of the water gun there is no gradient.

Second off, the pressure differential is what powers all water guns, even in a force model. There would be no force without a differential...

Hence my claim that a pressure gradient between the internals and the atmosphere is not what creates the force in the chamber.

The differential is not what creates the force in the chamber. What creates the force in the chamber is the air pressure applied over area. The differential does not do a thing. In your model, the pressure is the same on both sides of the piston, and there is equal force in opposite directions, and zero net force.

Because there is no gradient inside the gun's chamber, the gradient can't be the cause of force in the chamber. The chamber is equalized because the piston can move.

Originally Posted by Ben

...I have no idea why you think the SS 300 patent mentioning a pressure differential breaks anything I've said.

Anyway, I'm looking through the quote you posted and they don't phase me one bit. You say catch the last sentence, and it says "Generally, the higher the pressure, the longer the distance the water can be propelled, thus increasing the range and power of the water gun." The word generally exists there. I would say that sentence under the same contexts. You know, higher pressure is benefitial in the force model.

So, why do they mention pressure everywhere, but force nowhere? Surely they'd mention force if it were important? Surely they'd mention surface area if that was the reason for more PCs, not volume?

Originally Posted by Ben

And when they're saying "When attempts are made to increase the amount of water propelled by increasing the volume of the compressible area, the pumping action cannot displace the water at a high pressure, resulting in expulsion of water at low pressures." I don't think they're speaking about the pump's cross sectional area. What increases the amount of water propelled is more PC volume. More pump volume can make it pump faster, but it does not increase the volume. Also, they never said low pressure was bad.

I forgot to specify they were talking about piston-powered guns. No PCs. And how else would pressure be reduced (with the force applied to the pump the same) unless the area is increased? That also increases the volume.

This is just like the Waterzooka case. The LR got more range than the original Waterzooka because it had a smaller pump. The ratio of pressures and pump areas was nearly the same. The range was the same as a regular Waterzooka with a narrow pump.

"They never said low pressure was bad." Did you even read the next thing I posted? The paragraph after the pumper one, and the next paragraph I quoted, read:

Water guns have advantageously involved squirting large volumes of water at high pressures. Generally, the higher the pressure, the longer the distance the water can be propelled, thus increasing the range and power of the water gun.

Your point?

Originally Posted by Ben

Also, while you love to say "high flow coefficient!" and such, you know the SS 300's flow coefficient is not extremely high. It's probably not even high at all. It's about as big as the nozzle if you look through the patent images. I don't think you will state that it's increasing the performance on the level necessary. And wide avenue of release could easily mean wider pressure chambers... and that seems to be what they did in this design. They also could easily mean wider streams, another thing they did in the patent. In fact, "wide avenue of release" sounds more like a wide nozzle than anything else. When you match reality with what they are saying, you get a better picture.

Nope, somewhere they mentioned a wide nozzle and a wide avenue to the nozzle. I'll take your word for it on the actual dimensions though, as I'm feeling too lazy to look at the patent again.

However, a lower flow coefficient than a homemade gun does in part explain the lower range. But if you look at the internals of a stock gun, there are tiny vinyl tubes everywhere. They also mentioned the use of a high flow valve as opposed to the use of a low-flow valve.

Originally Posted by Ben

Again, you're not responding to anything I say. Why is Cv stuff only on 9 out of over 680 pages of my book? Why itn't it some sort of "fundamental" thing they use? Why are you defending an equation that obviously is not used to calculate flow created by a pressure.

I can not answer your questions as to why they barely mention Cv. However, I'll still say that you haven't actually critiqued the physics. If Cv is efficiency, then certainly it should affect the flow. And the pressure-flow correlation perfectly models the figures shown in the nozzle advertising PDF (I've mentioned it many times now).

Originally Posted by Ben

Well, I'm using textbook examples, while you're using forum links. You seem to think forum links are a more authoritative statement.

I've used a huge variety of sources. All you've referred to is an alleged textbook derivation (and yes, I'm still skeptical), as it is in my eyes, and a patent that ended up proving my own point.

Originally Posted by Ben

That and I think "wide avenue of release" is actually referring to a wider nozzle. It makes much more sense overall. And, you know that the Cv is not very good for this water gun because it's ID is about as big as the nozzle.

I've mentioned both the fact that their wide avenue of release is better than what you'd find on modern water guns. They also claimed the importance of using a ball valve for flow. And once again, if there is low Cv, that would explain the diminished range compared to an APH.

Originally Posted by Ben

I suppose you can say that I haven't scientifically criticized anything, but I could say the same about you then. What would be "scientifically criticizing?" And why are my TEXTBOOK examples given less weight than you ONLINE FORUM examples?

A scientific critique would be a point that actually disputes the validity of my physics. Find an error in the physics, an inconsistency in the math. You can't, because there isn't one.

I found an error in the derivation you posted. You claim you've scrapped some equation, but I don't know what that one is (is it the derivation), and I don't know what equation you're using now.

The error I found was a fundamental physics error. A real problem. Not a battle of qualifications, credentials, authority, or anything.

I'm not claiming to have any more sources than I have. Maybe I don't have what you call the authoritative fluid dynamics book, but I've seen it before, and it isn't that great. Check the reviews at Amazon - not too shiny. But I haven't felt a need to bring up issues like that, because I have the content down.

And needless to say, my only sources aren't from online forums. Let's see:
- A patent you directed me to
- A CPS water gun patent
- An information sheet for industrial nozzles
- A PDF guide to flow at an industrial engineering site
- Several calculators at university websites
- Physics forums in which respected members have connected flow and pressure
- Several books that my dad has bothered to keep

I'll add more to the list as I recall them. I know there are more. These are just the more qualified sources, by the way. And once again, this should not be a battle of qualifications.

Originally Posted by Ben

You're right. The Cv equation is about efficiency. It's not used at all to calculate flow created. It's used to minimize flow lost typically with valves with lower Cv valves.

Depends on how you look at it. And looking at the actual simplified Cv equation, it can be used to solve for flow given pressure.

Don't tell me that's not how people use it. You know very well that these equations are free to be manipulated as long as you do it right. And if I need proof, the calculator I linked to had a different version solving for flow.

Originally Posted by Ben

THE EQUATION I POSTED BEFORE IS COMPLETELY WRONG. Got that? I hope you do because I've said it several times. What's wrong with the book's equation? I'll get you a picture today, but you seem to deny anything you can't see yet. I've tried to find the same equation online, but it appears that it's far too advanced for most people online.

Which equation is this? I really must not have been reading your posts properly, as I don't know which equation you're talking about.

Is it the derivation's equation, or a different one? If you've scrapped the derivation, then what equation are you using? I'm guessing the one you scrapped is different though, as I think you're still arguing for the book's derivation.

And yes, I do deny this, partly because I haven't seen it yet. I've shown a problem with the science, and if qualifications are as important as you claim, then this doesn't mean a thing as I haven't seen it yet.

Originally Posted by Ben

I find this unusual because I swear you understood it before when I mentioned it to joanna. I'll get you the diagram later today.

All right, thanks.

Originally Posted by Ben

Again, I don't know how many times I have to say it, but none of these calculators are used to calculate how much flow is created. That's because the Cv stuff is not supposed to be used to do that. Typically people know how much flow they have because they use pumps with measured flow values already. That's why places like McMaster-Carr put those values there.

Take a look at the calculator I pointed out - in the second link, I think. The only button, the "calculate" one, would calculate the mass flow rate and volumetric flow rate if you pay them. Their equation for flow is the only one that mentions pressure (and pressure difference is one of the fields) - and it gets that flow by using the flow coefficient equation.

Just because you use flow and pressure to calculate Cv doesn't mean you can't use Cv and pressure to calculate flow.

What use is Cv if it's not being used to calculate something? I thought it was used to calculate flow in valves with known Cv. And in any case, my physics are right, and I'm plugging in the equations correctly - you've brought up no problems there - so I presume I'm doing things right.

Originally Posted by Ben

Also, get to reading the .gov link I posted. I'm assuming that because you didn't reply to it, you haven't looked at it in much detail.

Right indeed. I'm taking a look at it right now.

*reads*

I'll admit, it took a little bit, but I found what I was searching for.

Originally Posted by Tim Mooney

Atmospheric pressure below the bottom hole, acting over the area of the
bottom hole; this is your only upward force and it is equal to P_o * A, where
P_o is the pressure outside the bottle, and A is the area of the hole.

I'd explain how you can use either force or pressure for the fixed area of the opening, but then you'd get annoyed by the whole "there is no 'force at the nozzle' " argument. And that's essentially what they're talking about - "force at the nozzle."

But we don't like that, do we?


EDIT: Forgot to reply to Jo. Dunno why, sorry.

Interesting arguments, a new face definitely changes things. All we need is Drenchenator or Waterzooka to really get things going.

Anyway, I'll be thinking about all the stuff you said, as the information there is completely new (as far as I can remember). Be sure to read this thread though before it gets even longer.

EDIT again: Good point about the velocity vs. range - forgot to thank you for that. Focusing on velocity now for a fixed nozzle should definitely help.

[Ben]

As far as I can tell, Ben's theory of force is dependent on a few things, and pretty high up the list is the fact that water is incompressible.
The literal definition of this is: An incompressible substance is one which can never be under pressure.

However, defining incompressible as having no (notable) volume change when it is under pressure is unfortunately a common fallacy.

Again, people do not read my posts in their entirety and ignore key parts.

I NEVER once said that water was incompressible. Go through every post I ever made and you'll see that. Because I know people will get "technical" on me, I only say that water is incompressible for our purposes. It will still carry a pressure (that is, a force per unit area). That's why the hydraulics work. By compressable for our purposes, I mean experiences a change in volume.

I never once said water was incompressible. The only thing I say is that the water is experiencing a force (a push) from something such as air pressure, a hand piston, a spring, rubber tubing, or anything. This matches reality perfectly. It is the only thing I have noticed is consistent and the physics is on my side.

Imagine a pipe containing water, trapped between two pistons of equal size. Push on one piston, and an equal force will arise at the other end. (And the water's shape has remained a perfect cylinder through out)
But if the pistons are of different sizes (and the pipe changes diameter in the middle somewhere), the fluid is changing shape (because the two bits of pipe are of different diameters), so it no longer acts as if it were carrying a constant force. (Because it is carrying pressure, and changing shape).

Now, in both cases, imagine that the pipe's contents were frozen, so that they kept their shape (the fact that ice is less dense than water doesn't affect this experiment).
In this case, the forces on both ends are the same regardless of whether the pistons are of different sizes - because the ice keeps it's shape at all times.

Only when a substance keeps its shape can the force argument be applied. (Again, check up my core facts) As water doesn't, the force argument must be mistaken in this case. Not to say that the evidence is wrong, but the wrong conclusion has been drawn from it, as has happened many times in science (Pholostogen theory comes to mind).

Sounds to me like you're describing a hydraulic system at first. And we know how to calculate the force on the other end. I don't know how making something solid somehow disproves anything I'm saying though. Sounds like it'll just do nothing in that case. I don't get this argument. Could you explain in more detail?

You can apply a force to a fluid to create movement (flow). Again, it's the second thing my book mentions in the chapter on fluid flow. It works even when no pressure is created, like in a piston water gun with a nozzle the same diameter as the pipe. While the water will experience some pressure for a short time before the other water can move, the entire thing moves together after that and it's the force of the push that is powering the water gun.

-The theory cannot account for all water guns (only "special" ones like the Supercannons, and the Waterzooka seem to be usable [And I still disagree with the force that can be practically applied]), and it's effects are not noticeable on all water guns (I pose the 3xA and external pressure chamber problem. No effect on range or stream velocity at all. Even from many times the force oddly enough.)

This I find laughable. The force theory explains how every water gun operates. No matter what it is, force is involved in all water guns. The force of a piston being pushed by one's hands. The force of air pressure. The force of rubber tubing. The force of springs. The force of magnets. It all comes down to the applied force. And I've found that force required for a certain range matches up remarkable well. Just because we don't know how to calculate the force for certain systems (rubber CPS mainly), does not mean that the force model is incorrect. Huge logical fallacy.

Also, due to the lack of flow, your "3xA" will perform poorly regardless of how much force of pressure is applied. That's what the flow coefficient equations are actually for. Lots of lost flow due to inefficiencies. And I am not sure that you will say that more pressure created more performance as well. From what I remember, you weren't getting good performance even at 100 PSI.

-Linked to this, it's been based on only a couple of soakers - not the widest list of sources ever. (Was the data ever made available to view in full?)

And I suppose the pressure theory is based upon data from hundreds of water guns? I think data from 10 - 15 or so water guns is better than data from none.

And yes, I have used several water guns, though exclusively homemade, in my models. Due to poor weather and lack of time, I have not finished a definitive equation desribing the relationship between force and range for efficient systems.

As for whether or not the data was public, actually, it is for the most part. Aside from the few tests I have made and never told anyone about (I can't think of any because I usually am excited to post about them, but I suppose a few aren't posted), the remainder of the data I used comes from others water guns and their reported performance. Most of that comes from this forum. A few come from email and other websites.

-The theory has tried to automatically build in how stream velocity affects range, however, this should be removed and handled seperately. To solve the issue we need to look at the relationship to stream velocity, then handle stream velocity to range seperately.

SilentGuy's theory or the force model? My model doesn't really say much about stream velocity. It mainly is a simplified model to describe how much force is necessary to get a certain range.

Hence my claim that a pressure gradient between the internals and the atmosphere is not what creates the force in the chamber.

The differential is not what creates the force in the chamber. What creates the force in the chamber is the air pressure applied over area. The differential does not do a thing. In your model, the pressure is the same on both sides of the piston, and there is equal force in opposite directions, and zero net force.

Because there is no gradient inside the gun's chamber, the gradient can't be the cause of force in the chamber. The chamber is equalized because the piston can move.

Can you explain the entire "pressure gradient" thing to me? I am not finding that term used in the links you have posted (unless I'm missing something). Air pressure water guns do not have a pressure gradient anyway, unless this is related to time. The pressure is constant throughout the system unless things such as valves change it.

So, why do they mention pressure everywhere, but force nowhere? Surely they'd mention force if it were important? Surely they'd mention surface area if that was the reason for more PCs, not volume?

They don't need to mention force. I think it's implied in the entire patent and the fact that the SS 300 has wider PCs. And I don't think you can find anything that says the pressure combined with some Cv value alone controls the power of a water gun.

Anyway, can't the same be said about your ignoring my FLUID MECHANICS book postings? Surely they would give more than 9 pages to Cv values if they were important.

I forgot to specify they were talking about piston-powered guns. No PCs. And how else would pressure be reduced (with the force applied to the pump the same) unless the area is increased? That also increases the volume.

This is just like the Waterzooka case. The LR got more range than the original Waterzooka because it had a smaller pump. The ratio of pressures and pump areas was nearly the same. The range was the same as a regular Waterzooka with a narrow pump.

"They never said low pressure was bad." Did you even read the next thing I posted?

I don't think you understand. No matter what, all water guns can be simplified into a piston system of some sort. Even if pressure is what matters, a force still is applied that creates that pressure. All water guns are the same essentially. And I think something that is an actual water gun is better than what you were posting, which were very far from water guns.

Hey, if you believe the LR system only worked through higher pressures, go ahead and make one. I find it really annoying that you insist things are the same between each one when the guy reported they were not. You do know that 5 more feet of range is significant, right? If you are so sure, go do some statistical tests on it to statistically state that both perform the same. I did something like that before to statistically state that the SS 300 shoots up to 50 feet for a statistics project in school. Not hard. And it is very definitive.

Yes, I read what you posted and I responded DIRECTLY to it:

Anyway, I'm looking through the quote you posted and they don't phase me one bit. You say catch the last sentence, and it says "Generally, the higher the pressure, the longer the distance the water can be propelled, thus increasing the range and power of the water gun." The word generally exists there. I would say that sentence under the same contexts. You know, higher pressure is benefitial in the force model.

How about you start reading my posts as opposed to skimming them? Sure, I do skim to a certain extent, but I let people know I did or that I was pressed for time.

Nope, somewhere they mentioned a wide nozzle and a wide avenue to the nozzle. I'll take your word for it on the actual dimensions though, as I'm feeling too lazy to look at the patent again.

However, a lower flow coefficient than a homemade gun does in part explain the lower range. But if you look at the internals of a stock gun, there are tiny vinyl tubes everywhere. They also mentioned the use of a high flow valve as opposed to the use of a low-flow valve.

I suppose you would also call the SS 300's valve "high flow" despite the fact that it's not. You should realize that they are writing at a time when no water guns had internals as large as the SS 300's. High flow then is very low flow now. The SS 300's internals are about the same size as the Max-D 6000's from what I'm looking at. Not high flow.

I can not answer your questions as to why they barely mention Cv. However, I'll still say that you haven't actually critiqued the physics. If Cv is efficiency, then certainly it should affect the flow. And the pressure-flow correlation perfectly models the figures shown in the nozzle advertising PDF (I've mentioned it many times now).

I have critiqued the physics plenty. I don't know how stating that this equation is not used for that purpose is not critiquing the physics. If you want me to find an error in the math, there it is. That's like using the wrong formula to calulate anything. That's still errors in math.

And I have definitely stated that Cv values affect the flow. Read the after the last quote of my last post on page one. You seem to be missing a large percentage of what I have said, so I find myself restating things. The Cv values do not determine the flow from the applied force however. They reduce it, but they do not determine the initial, unrestricted flow.

I also have taken no issue with the nozzle PDF stuff because that's what that equation is meant to be used for. In fact, I EXPLICITELY said that in the post I just mentioned.

A scientific critique would be a point that actually disputes the validity of my physics. Find an error in the physics, an inconsistency in the math. You can't, because there isn't one.

I found an error in the derivation you posted. You claim you've scrapped some equation, but I don't know what that one is (is it the derivation), and I don't know what equation you're using now.

The error I found was a fundamental physics error. A real problem. Not a battle of qualifications, credentials, authority, or anything.

It's becoming more obvious to me that you skim my posts after this.

Again, THE EQUATION I DERIVED IS WRONG. How many times must I say that? The mass flow rate must be changed to be used. What I am speaking about is the one in my book. I have been speaking about this for several posts, but you seem to have missed my first mention of the book's equation.

Depends on how you look at it. And looking at the actual simplified Cv equation, it can be used to solve for flow given pressure.

Don't tell me that's not how people use it. You know very well that these equations are free to be manipulated as long as you do it right. And if I need proof, the calculator I linked to had a different version solving for flow.

Sure, you can manipulate equations all you want. But manipulating one that's meant to calculate flow lost is a mistake. Again, flow is proportional to velocity, and pressure drop is proportional to velocity. That's why this equation works to calculate flow drop. This is applying an equation meant for a certain situation for another.

A coefficient most often is a measured value. Here it can contain a ton of things specific to the valve. Note that the cross sectional area of the valve is not included. There's a ton of things not included that you know are necessary for flow. These things are then included in the coefficient because it is specific to the valve. This equation does not calculate most everything necessary for flow.

I am completely confident that this equation is not meant to calculate flow created. I would highly suggest finding another one if you want me to believe your pressure theory.

Which equation is this? I really must not have been reading your posts properly, as I don't know which equation you're talking about.

Is it the derivation's equation, or a different one? If you've scrapped the derivation, then what equation are you using? I'm guessing the one you scrapped is different though, as I think you're still arguing for the book's derivation.

And yes, I do deny this, partly because I haven't seen it yet. I've shown a problem with the science, and if qualifications are as important as you claim, then this doesn't mean a thing as I haven't seen it yet.

Don't worry. I post the equation with other things below.

Take a look at the calculator I pointed out - in the second link, I think. The only button, the "calculate" one, would calculate the mass flow rate and volumetric flow rate if you pay them. Their equation for flow is the only one that mentions pressure (and pressure difference is one of the fields) - and it gets that flow by using the flow coefficient equation.

Just because you use flow and pressure to calculate Cv doesn't mean you can't use Cv and pressure to calculate flow.

What use is Cv if it's not being used to calculate something? I thought it was used to calculate flow in valves with known Cv. And in any case, my physics are right, and I'm plugging in the equations correctly - you've brought up no problems there - so I presume I'm doing things right.

I'm not finding what you're talking about, so link directly to it.

Cv is used to calculate flow lost through valves. It stands for coefficient of velocity. It represents how easily things can flow through a valve by comparing the lost velocity. There are more coefficients in my book and a larger K variable respresenting something involving all of them.

I'd explain how you can use either force or pressure for the fixed area of the opening, but then you'd get annoyed by the whole "there is no 'force at the nozzle' " argument. And that's essentially what they're talking about - "force at the nozzle."

What they're talking about is an actual force (a push) at the nozzle. Before you thought "force" was magically reduced at the nozzle only with the explanation that the diameter change changed the force somehow. I don't see how this disproves anything I'm saying. Could you explain?

And why don't you say anything about how they used force to determine flow? I thought you said that isn't what is happening? You expect me to respond to links you post, but you don't even make a real response to the ones I post!

Anyway, I started looking for the correlation between how easily something can flow, pressure, and range. I've compiled the following information from memory. Everything is self explanatory except for the last value, which is the square root of pressure multiplied by the internal diameter. Essentially, it is an approximation of the flow. The flow and range should be proportional for the most part. Let's see how it works out. Again, I'm going into this with no biases. If anyone thinks there are legitimate problems with my statistics, please tell me. The question mark ones are approximations and are correct to the best of my knowledge. A few of these are from memory and are a little sketchy, but they sound correct at the very least.

CPS 1000 - 23 PSI - 1/2 inch - 40 feet - 2.4
CPS 1500 - 22 PSI - 1/2 inch - 40 feet - 2.3
Supercannon II - 40 PSI - 1 1/2 inch - 65 feet - 9.5
SuperCAP - 40 PSI - 1 inch - 63 feet - 6.3
SS 300 - 40 PSI - 1/4? inch - 47 feet - 1.6
SuperCPS - 30 PSI? - 5/8 inch - 62 feet - 3.4
APH - 60 PSI - 3/4 inch - 55 feet - 5.8
CPS 2000 - 28 PSI - 3/4 inch - 53 feet - 4.0
CPS 2500 - 23 PSI - 3/4 inch - 45 feet - 3.6

I'm open to more additions.

Graph range vs. the coefficient. You'll find a very weak positive correlation. In fact, I only say a very weak positive correlation because it does appear to be increasing despite the fact that each point is spread out. And I'm sure that adding more water guns would show exactly how weak the relationship is. Certainly, more flow is definitely beneficial, but it does not determine how a water gun will perform.

This weekend is rainy, but I will test Supercannon II with a long 1/2 inch "barrel" soon. I expect the flow to definitely be reduced and the range to be decreased due to the lost from from such a quick reduction, but I do not think the reduction will be on the order of 10 or more feet as SilentGuy seems to imply. I know this because I actually have done tests with barrels before to see if they increased range.



Here's one example I brought up with you SilentGuy earlier that you did not understand. What I was saying was that inside of the water gun, there is no pressure differential. Yet, there is flow. This again shows that the Cv equation does not work in all situations. If this explanation is not adequate, I am not sure what is because this is very simple.



Here's the example I brought up a long time ago with joanna.

The water gun is pressurized. There are two ball valves. Initially valve 1 is open, while valve 2 is closed. After pressurization, valve 1 is closed, isolating the pressurized water. Valve 2 is opened. What happens?

Nothing happens. I don't think you will say anything will happen aside from a slow flow of water out.

But, there is a pressure differential! Why doesn't this perform the same (although with a shorter burst) as it would had this water not been isolated? It's because the air pressure is PUSHING the water out. It is applying a force. There is no applied force when the water shot is isolated. Building this device will prove that the pressure theory is fundamentally flawed.



Here is another device that when built will prove that the pressure theory is flawed. This is a pressure multiplier. With this, one could easily achieve 200 - 300 PSI if not more with very high flow (two inches would be great) from a lower pressure that is more easy to obtain with an air compressor. If you believe in your theory SilentGuy, build this device without reluctance. I have no doubts as to how it will perform. If you want to prove anything, this would be how to do it.



This is a simple to understand situation that shows the flaws in the pressure theory. This drawing shows a piston water gun with an open end. When the piston moves, the water flows. But, there is absolutely no pressure differential. Surely, if you want to get super technical (and I know you people love to), technically a small amount of pressure is created for a miniscule fraction of a second when the piston first starts moving as the atoms bump into each other. But, that's not what's powering this water gun! The force of the push is. The water is simply being pushed out! That's how every water gun system works except for the weird capacitor bank ones.



Here's the scan from my book. Look at the equation for force. Seems to me that the equation can very easily be used to calculate flow from an applied force, given that you know differential equations very well (and preferably partial differential equations for the more advanced stuff). The dV with a line through it represents the the volume differential, which includes the cross sectional area (for those who want to say it doesn't include cross sectional area).

If people don't understand that force is what powers water guns, I invite them to test them out for themselves. I came to this conclusion based upon what I saw in class and reality.

[joannaardway]

The second part of my statement about the hydraulics and freezing it is meant to state that something that changes shape cannot carry a force.

As we know that the water is definately changing shape within the system, it cannot therefore be carrying any force from the chamber to the nozzle (which your force theory implies), and it can only carry pressure (by virtue of water pressure).

I am not denying your data - I just cannot agree with the theory that comes with it. I think some factors must have been seriously overlooked.

I'll go off and create a pressure theory from known equations, and see how that matches reality. A theory created from data normally overlooks something.

As an example: There is a strong correlation between the number of ice creams sold on one day, and the length of the daylight hours. From that, you could conclude that longer days result in more ice cream sales.

But if you think more deeply, then the truth is that summer days (cause) are usually longer (effect) and hotter (effect), and more ice creams are sold on hot days (effect).

I postulate you are looking at two effects and linking them - so a cause must be found.
You believe force is a cause of the results. I believe it to be an effect of something else (pressure, and perhaps some other unforeseen factor)

[Silence]

I've been browsing the forums, and I keep seeing references to one of your CPHs that hits over 60 feet. I'll keep looking.


Water is essentially incompressible for our purposes - I'll agree with that. That said, there can still be pressure.

Originally Posted by Ben

You can apply a force to a fluid to create movement (flow). Again, it's the second thing my book mentions in the chapter on fluid flow. It works even when no pressure is created, like in a piston water gun with a nozzle the same diameter as the pipe. While the water will experience some pressure for a short time before the other water can move, the entire thing moves together after that and it's the force of the push that is powering the water gun.

There's still pressure built up, even if nothing gets compressed. The force you apply to the pump divided by the area of the pump results in gauge pressure. That pressure is what forces water out.

Originally Posted by Ben

This I find laughable. The force theory explains how every water gun operates. No matter what it is, force is involved in all water guns. The force of a piston being pushed by one's hands. The force of air pressure. The force of rubber tubing. The force of springs. The force of magnets. It all comes down to the applied force. And I've found that force required for a certain range matches up remarkable well. Just because we don't know how to calculate the force for certain systems (rubber CPS mainly), does not mean that the force model is incorrect. Huge logical fallacy.

My model also seems to describe flow and range very well. The true problem at this point is the fact that we have no tests with controls and significant results.

A 4" water cannon vs. a 2" cannon, operating at given pressures, and using the same nozzle, would be an effective test. That's what I'm aiming for, although part availability might be better for a 2" and a 3" cannon.

Looking at the patents, CPS water guns capitalize on the constant pressure. Once again, I haven't seen a single reference to force.

Originally Posted by Ben

Also, due to the lack of flow, your "3xA" will perform poorly regardless of how much force of pressure is applied. That's what the flow coefficient equations are actually for. Lots of lost flow due to inefficiencies. And I am not sure that you will say that more pressure created more performance as well. From what I remember, you weren't getting good performance even at 100 PSI.

The flow coefficient really is the killer here.

The reason 100 PSI doesn't help too much is because of the square root. That's only 85 PSIg (unless it was 100 PSIg), and the square root of 85 is little over 9. The square root of, say, 25 PSIg (for SuperCannon II or the SS 300) is 5. I'd say SuperCannon II has a flow coefficient much greater than 9/5 times the 3XA's flow coefficient, yielding greater velocity for a given nozzle.

Originally Posted by Jo

-The theory has tried to automatically build in how stream velocity affects range, however, this should be removed and handled seperately. To solve the issue we need to look at the relationship to stream velocity, then handle stream velocity to range seperately.

Originally Posted by Ben

SilentGuy's theory or the force model? My model doesn't really say much about stream velocity. It mainly is a simplified model to describe how much force is necessary to get a certain range.

Probably my theory. I used range and velocity pretty much interchangeably, at least in the first post. Thanks for the tip, joannaardway!

Originally Posted by Ben

Can you explain the entire "pressure gradient" thing to me? I am not finding that term used in the links you have posted (unless I'm missing something). Air pressure water guns do not have a pressure gradient anyway, unless this is related to time. The pressure is constant throughout the system unless things such as valves change it.

I referred to "a pressure gradient between the internals and the atmosphere is not what creates the force in the chamber." Essentially, the difference in pressure between the nozzle inlet and the aperture.

However, you're right in that there's no actual gradient within the water gun. That's why a pressure difference itself can't create the force acting on the piston.

Originally Posted by Ben

They don't need to mention force. I think it's implied in the entire patent and the fact that the SS 300 has wider PCs.

Yes, they need to mention force if that is what determines range. But they clearly say that low pressures don't allow the streams to obtain the same range. I can't see the basis of your argument.

Originally Posted by Ben

And I don't think you can find anything that says the pressure combined with some Cv value alone controls the power of a water gun.

Correct, there are no sources that specifically state that. But none even imply the use of force in determining range. The calculator I showed you used pressure to determine flow through a nozzle, however, and people in the forums used pressure and flow coefficients for flow.

And once again, that's ignoring the actual physics.

Originally Posted by Ben

Anyway, can't the same be said about your ignoring my FLUID MECHANICS book postings? Surely they would give more than 9 pages to Cv values if they were important.

Well then, I guess Cv isn't used that much. Happy?

But seriously, I haven't heard much about that book except the number of pages that include Cv. I haven't heard what it says about Cv, what it says about force (unless it's the derivation, which I've already addressed), what it says about pressure, etc.

Point taken, though.

Originally Posted by Ben

I don't think you understand. No matter what, all water guns can be simplified into a piston system of some sort. Even if pressure is what matters, a force still is applied that creates that pressure. All water guns are the same essentially. And I think something that is an actual water gun is better than what you were posting, which were very far from water guns.

Yes, force needs to be applied to create that pressure. However, the area it's applied onto is what determines how much pressure you get from a certain amount of force.

You may have a ton of force acting on SuperCannon II's piston, but that does not necessarily mean high pressure. The piston has a lot of surface area (which explains the high force), so the pressure is not exceptional.

I posted a description of a patent for a piston pumper. Such guns are definitely water guns - what do you think the Waterzooka LR is?

Originally Posted by Ben

Hey, if you believe the LR system only worked through higher pressures, go ahead and make one. I find it really annoying that you insist things are the same between each one when the guy reported they were not. You do know that 5 more feet of range is significant, right? If you are so sure, go do some statistical tests on it to statistically state that both perform the same. I did something like that before to statistically state that the SS 300 shoots up to 50 feet for a statistics project in school. Not hard. And it is very definitive.

5 feet? I was under the impression that the LR got 57 feet, and that the narrow Waterzooka got 55 feet to the puddle and 60 feet to the last drop. Depends on how you look at it, but they're pretty much even. Consider that pumper range varies from shot to shot and that the analysis of where the last puddle is varies.

However, I agree I need to get to testing. I'm going to do it with water cannons though, for more consistent performance and for more power.

Originally Posted by Ben

How about you start reading my posts as opposed to skimming them? Sure, I do skim to a certain extent, but I let people know I did or that I was pressed for time.

My comment was in response to your statement that "they never said low pressure was bad." What I asked was, "Did you even read the next thing I posted?" A rhetorical question, in other words. I know you responded. However, you asked why low pressure was bad even though the very next statement I quoted said it was bad for range.

Originally Posted by Ben

I suppose you would also call the SS 300's valve "high flow" despite the fact that it's not. You should realize that they are writing at a time when no water guns had internals as large as the SS 300's. High flow then is very low flow now. The SS 300's internals are about the same size as the Max-D 6000's from what I'm looking at. Not high flow.

The "high" flow was the reason the gun by far outranged everything else that came before it. I mentioned the fact that homemade have even better range partly due to better flow coefficient.

I understand that guns that came after the SS 300 often were closer in terms of range than pre-SS 300 guns were. The Max-D 6000, for example, has decent range.

Originally Posted by Ben

I have critiqued the physics plenty. I don't know how stating that this equation is not used for that purpose is not critiquing the physics. If you want me to find an error in the math, there it is. That's like using the wrong formula to calulate anything. That's still errors in math.

That's not errors in the math. Tell me why so many sources are using equations like this to calculate flow as I am doing.

However, there is no problem with the physics. I am using the pressure gradient across the nozzle in an equation where you are supposed to use the pressure gradient across the valve. I am using the flow coefficient, k, of the nozzle where you are supposed to use the flow coefficient of the valve. I am calculating the flow of the nozzle using an equation that calculates flow of a valve. I am using that flow and I am using the nozzle area in another equation to calculate the stream velocity.

My physics is perfect. Tell me why, scientifically, I cannot use these formulas in this manner.