Super Soaker Central forums

Silence · 06-26-2006, 02:59 PM

All of a sudden, I have a nasty feeling that this whole thing might be impossible after all--and you might get the feeling too when you see the next question (though you'll hopefully have a good reply

). Why can't a person pump up a gun in such a way that there is low surface area and resistance, and then turn the gun so that there is great surface area and power? Surely something must prevent this from being the case...

I've been wrestling with the possibilities, such as different amounts of water being pumped and the water level changing at different rates, but I can't seem to come to any solid conclusion. Sorry if I can't explain the problem very well, but it's the best I can do--basically, you have an easy time pumping, but you get more force out of it. Strange...

joannaardway · 06-26-2006, 03:58 PM

That's more or less where I started out at with this concept. Having people firing explanations at you starts to blur the lines.

I think I'm returning to my first idea - I can see holes in my theory now.

The force theory is flawed - the chamber ID has no effect - the nozzle ID is the important point.

Ben · 06-26-2006, 04:22 PM

Chamber ID does matter, as does nozzle ID. However, nozzle ID limits while chamber ID multiplies.

This works on the same exact principle that the pumps do. Do you know why large ID pumps are harder to pump? Same exact reason - they take more force to move. A certain PC has a certain amount of pressure. The same pressure pushing on a larger area will inheritly have a higher force by definition.

Quote:

Why can't a person pump up a gun in such a way that there is low surface area and resistance, and then turn the gun so that there is great surface area and power? Surely something must prevent this from being the case...

If I understand this correctly, I may have already answered your question in my previous paragraph. Though, explain it again if I haven't.

joannaardway · 06-26-2006, 04:31 PM

Your previous paragraph contains the death toll of the theory. You are maintaining that greater chamber ID results in more output (thus more force acceleraring the water) on a nozzle of a given ID.

Thus, following that theory, a greater chamber ID must result in a greater force on a pump of a given ID.

Are you maintaining that a 2" chamber is easier to pump than a 3" chamber? That's what the theory says.

The sideways chamber and pump counter theories must at least show error with the principle.

Ben · 06-26-2006, 04:34 PM

Quote:

Are you maintaining that a 2" chamber is easier to pump than a 3" chamber? That's what the theory says.

Exactly. That's Hydraulics 101.

Check out any hydraulics equations on the subject. Big Bee himself told me via email back in 2004 that they use basic hydraulics equations to determine which pump ID to use.

Edit: Just was doing some thinking, and I'd have to say that the 2" chamber would be easier to pump than the 3" chamber given the volume of both chambes is the same. Obviously, how fast one chamber gains pressure also will have an effect on the difficulty.

Silence · 06-26-2006, 06:48 PM

Quote:

Originally Posted by joannaardway

Your previous paragraph contains the death toll of the theory. You are maintaining that greater chamber ID results in more output (thus more force acceleraring the water) on a nozzle of a given ID.

Thus, following that theory, a greater chamber ID must result in a greater force on a pump of a given ID.

Are you maintaining that a 2" chamber is easier to pump than a 3" chamber? That's what the theory says.

The sideways chamber and pump counter theories must at least show error with the principle.

My thoughts exactly. In fact, in the compact APH thread, I said that I chose 3" chambers for my first homemade because they would supply more force, even if they were harder to pump up.

Now, let me explain my problem again--this time with the aid of diagrams.

The spinning feature clearly won't be there in any soaker, but I added it for demonstrative purposes. You pump up the gun in the first position--which should be easy due to minimal surface area. In the second position, there's a lot of surface area, so you get a ton of force when you shoot. Is this "phenomenon" possible, and if not, what am I missing?

Ben · 06-26-2006, 08:11 PM

Yes, you will get more force with the chamber rotated. Interesting idea actually.

Silence · 06-26-2006, 08:33 PM

My question is why? How can you get extra force after putting in only a certain amount? Some unknown factor must make this impossible...

That last sentence makes me sound like some popular physicist (whose name I can't remember) quoted in The Physics of Star Trek who said that a certain phenomenon couldn't happen, because "something must prevent it from such an absurd fate!" Needless to say, the phenomenon (possibly dealing with the death of stars, though I can't remember) was perfectly legitimate, and the criticized theory proved true. Let's hope it's the case with this...

Ben · 06-26-2006, 08:39 PM

With more surface area, it's like pushing with both hands on a heavy door instead of one. Same force from a bunch of different points. That's what pressure is.

This doesn't seem to violate any physics rules I know about. Then again, it might be a good idea to post about this at some physics forum and see what they think... I'll start looking at DMOZ to see what I can find.

Silence · 06-26-2006, 09:39 PM

Yeah, I suppose that counts--so it's like 10 hands pushing at the same time, as opposed to each one pushing in turn. Needless to say, shot time will be less--but that's a given anyway. Part of my question had to do with pumping, though--if this is the case, why wouldn't it take longer to pump (unless it does take longer after all, when I'll happily drop my case)? Well, thanks for the help, and I'll ask my dad about it, too.

joannaardway · 06-27-2006, 05:48 AM

Ok. The idea about the pump must be wrong:

Force = pressure * area.

Thus, for a 1/2" pump with an area of approximately .2 square inches, a given chamber pressure must result in a given force on the pump.

Correct?

Assuming 50 psi to make the figures neat:

50 psi*.2 si = 10 pounds of force regardless of chamber area.

Regardless of how hard the chamber is "pushing" the water in it, the pump force is the same.

Thus, the principle applied to the nozzle must be dependant on pressure, not force.

insanitys_engineer · 06-27-2006, 08:01 AM

Be glad i deleted all the rambling that was here. It was ALLL WRONG!

this is right however...
Pressure is not constant: it is only accurate for the air in the tank. Elsewhere in the guns pressure can be a lot bigger, and for a 4" dia tank and with a 1/5" nozzle all at 4 bar you can get a 16 bar pressure jet.
Tank orientation does matter, and will be significant for a long horizontal tank.

Silence · 06-27-2006, 03:42 PM

Yeah, I realized that a fixed tank would work just fine--although you wouldn't want to shoot while pumping! Also, if you use a cylinder or something for the PC, then not all the water will be shot out. You could put the opening to the PC in a "corner", and while that would be complicated, it would work...

insanitys_engineer · 06-27-2006, 05:20 PM

If you can bend a pipe or drill an angled hole you could angle the tank. Also, you won't be firing dead level so the horizontal tank should drain well if the hole is at the back bottom edge. I still cling to the idea of a wooden tank, and so the angling of the whole shape to how you need it.

NB.
Pressure differs throughout the soaker thats the mistake i made. I assumed it didn't, it was all 50 psi, But of course its not. I was stupid earlier.

joannaardway · 06-27-2006, 05:51 PM

Pressure must be constant throughout the soaker.

Any pressure "peaks" will almost instantly "smooth" so that the soaker is isobaric.

If you can find a way of getting a 16 bar jet from a soaker pressurized to 4 bar, you've just solved all the world's energy problems - energy from nothing.

ENERGY IS CONSERVED! (except in energy-mass conversion, but unless your soaker contains a nuclear reactor, we're fine there)

Energy = force * distance.

If we suddenly assume that the mystical sideways chamber idea is correct and force on the pump is less then...

Assuming the same fill, distance must be the same for the same number of pumps?

If we have a 50 newton average over 10 metres, then 500 joules have been stored.

If we have a 100 newton force for the "other chamber", then 1000 joules have been stored.

How come the other chamber has more energy stored? This would make it less efficent for use? How exactly can this even vaguely make sense?

How much physics do I need to shoot across these forums before I can sink this theory for good?

Should I bring out mass-flow computions:

flow = flow co-effiecent * sqrt((pressure in chamber squared - air pressure squared)/temperature)

The flow co-efficent is dependant on NOZZLE area, not chamber area.

And does force even come into that calculation? - NO!

I know this theory is completely flawed - and I will drag out every equation I can to prove it.

EDIT/Sanity revert: This is a post that I would slay other members on sight for, but I can get very, very persistant and aggresive when other people apparantly refuse to at least consider my theory. Sorry, but I've expained my reasons for this before - thus I am leaving the post as it began.

06-26-2006, 02:59 PM	#31
Silence Administrator Join Date: Apr 2006 Location: Virginia Posts: 3,293 UserID: 576	All of a sudden, I have a nasty feeling that this whole thing might be impossible after all--and you might get the feeling too when you see the next question (though you'll hopefully have a good reply ). Why can't a person pump up a gun in such a way that there is low surface area and resistance, and then turn the gun so that there is great surface area and power? Surely something must prevent this from being the case... I've been wrestling with the possibilities, such as different amounts of water being pumped and the water level changing at different rates, but I can't seem to come to any solid conclusion. Sorry if I can't explain the problem very well, but it's the best I can do--basically, you have an easy time pumping, but you get more force out of it. Strange... __________________ Forum rules

06-26-2006, 03:58 PM	#32
joannaardway Do not adjust your TV set Join Date: Feb 2006 Location: SW Hertfordshire, England Posts: 875 UserID: 549	That's more or less where I started out at with this concept. Having people firing explanations at you starts to blur the lines. I think I'm returning to my first idea - I can see holes in my theory now. The force theory is flawed - the chamber ID has no effect - the nozzle ID is the important point. __________________ "Over the hills and far away, she prays he will return one day. As sure as the rivers reach the seas, back in his arms again she'll be." - Over the Hills and far away, Gary Moore "So many people have come and gone, their faces fade as the years go by. Yet I still recall as I wander on, as clear as the sun in the summer sky" - More than a feeling, Boston

06-26-2006, 04:31 PM	#34
joannaardway Do not adjust your TV set Join Date: Feb 2006 Location: SW Hertfordshire, England Posts: 875 UserID: 549	Your previous paragraph contains the death toll of the theory. You are maintaining that greater chamber ID results in more output (thus more force acceleraring the water) on a nozzle of a given ID. Thus, following that theory, a greater chamber ID must result in a greater force on a pump of a given ID. Are you maintaining that a 2" chamber is easier to pump than a 3" chamber? That's what the theory says. The sideways chamber and pump counter theories must at least show error with the principle. __________________ "Over the hills and far away, she prays he will return one day. As sure as the rivers reach the seas, back in his arms again she'll be." - Over the Hills and far away, Gary Moore "So many people have come and gone, their faces fade as the years go by. Yet I still recall as I wander on, as clear as the sun in the summer sky" - More than a feeling, Boston

06-26-2006, 08:11 PM	#37
Ben Founder Join Date: Mar 2003 Location: College Park, Maryland Posts: 6,054 UserID: 1	Yes, you will get more force with the chamber rotated. Interesting idea actually. __________________ email: / Forum rules / Read this page before emailing me. Do not send me a PM or email with a water gun question if someone else could answer the question. Post at the forums. You will get a response from me along with others' views or ideas.

06-26-2006, 08:33 PM	#38
Silence Administrator Join Date: Apr 2006 Location: Virginia Posts: 3,293 UserID: 576	My question is why? How can you get extra force after putting in only a certain amount? Some unknown factor must make this impossible... That last sentence makes me sound like some popular physicist (whose name I can't remember) quoted in The Physics of Star Trek who said that a certain phenomenon couldn't happen, because "something must prevent it from such an absurd fate!" Needless to say, the phenomenon (possibly dealing with the death of stars, though I can't remember) was perfectly legitimate, and the criticized theory proved true. Let's hope it's the case with this... __________________ Forum rules

06-26-2006, 08:39 PM	#39
Ben Founder Join Date: Mar 2003 Location: College Park, Maryland Posts: 6,054 UserID: 1	With more surface area, it's like pushing with both hands on a heavy door instead of one. Same force from a bunch of different points. That's what pressure is. This doesn't seem to violate any physics rules I know about. Then again, it might be a good idea to post about this at some physics forum and see what they think... I'll start looking at DMOZ to see what I can find. __________________ email: / Forum rules / Read this page before emailing me. Do not send me a PM or email with a water gun question if someone else could answer the question. Post at the forums. You will get a response from me along with others' views or ideas.

06-26-2006, 09:39 PM	#40
Silence Administrator Join Date: Apr 2006 Location: Virginia Posts: 3,293 UserID: 576	Yeah, I suppose that counts--so it's like 10 hands pushing at the same time, as opposed to each one pushing in turn. Needless to say, shot time will be less--but that's a given anyway. Part of my question had to do with pumping, though--if this is the case, why wouldn't it take longer to pump (unless it does take longer after all, when I'll happily drop my case)? Well, thanks for the help, and I'll ask my dad about it, too. __________________ Forum rules

06-27-2006, 05:48 AM	#41
joannaardway Do not adjust your TV set Join Date: Feb 2006 Location: SW Hertfordshire, England Posts: 875 UserID: 549	Ok. The idea about the pump must be wrong: Force = pressure * area. Thus, for a 1/2" pump with an area of approximately .2 square inches, a given chamber pressure must result in a given force on the pump. Correct? Assuming 50 psi to make the figures neat: 50 psi*.2 si = 10 pounds of force regardless of chamber area. Regardless of how hard the chamber is "pushing" the water in it, the pump force is the same. Thus, the principle applied to the nozzle must be dependant on pressure, not force. __________________ "Over the hills and far away, she prays he will return one day. As sure as the rivers reach the seas, back in his arms again she'll be." - Over the Hills and far away, Gary Moore "So many people have come and gone, their faces fade as the years go by. Yet I still recall as I wander on, as clear as the sun in the summer sky" - More than a feeling, Boston

06-27-2006, 08:01 AM	#42
insanitys_engineer Member Join Date: Jun 2006 Location: England, kent, south of London Posts: 32 UserID: 679	Be glad i deleted all the rambling that was here. It was ALLL WRONG! this is right however... Pressure is not constant: it is only accurate for the air in the tank. Elsewhere in the guns pressure can be a lot bigger, and for a 4" dia tank and with a 1/5" nozzle all at 4 bar you can get a 16 bar pressure jet. Tank orientation does matter, and will be significant for a long horizontal tank. __________________ Sanity is just an excuse for not having an imagination Water water everywhere, and not a drop to drink. Last edited by insanitys_engineer : 06-27-2006 at 09:52 AM.

06-27-2006, 03:42 PM	#43
Silence Administrator Join Date: Apr 2006 Location: Virginia Posts: 3,293 UserID: 576	Yeah, I realized that a fixed tank would work just fine--although you wouldn't want to shoot while pumping! Also, if you use a cylinder or something for the PC, then not all the water will be shot out. You could put the opening to the PC in a "corner", and while that would be complicated, it would work... __________________ Forum rules

06-27-2006, 05:20 PM	#44
insanitys_engineer Member Join Date: Jun 2006 Location: England, kent, south of London Posts: 32 UserID: 679	If you can bend a pipe or drill an angled hole you could angle the tank. Also, you won't be firing dead level so the horizontal tank should drain well if the hole is at the back bottom edge. I still cling to the idea of a wooden tank, and so the angling of the whole shape to how you need it. NB. Pressure differs throughout the soaker thats the mistake i made. I assumed it didn't, it was all 50 psi, But of course its not. I was stupid earlier. __________________ Sanity is just an excuse for not having an imagination Water water everywhere, and not a drop to drink. Last edited by insanitys_engineer : 06-27-2006 at 05:25 PM.

06-27-2006, 05:51 PM	#45
joannaardway Do not adjust your TV set Join Date: Feb 2006 Location: SW Hertfordshire, England Posts: 875 UserID: 549	Pressure must be constant throughout the soaker. Any pressure "peaks" will almost instantly "smooth" so that the soaker is isobaric. If you can find a way of getting a 16 bar jet from a soaker pressurized to 4 bar, you've just solved all the world's energy problems - energy from nothing. ENERGY IS CONSERVED! (except in energy-mass conversion, but unless your soaker contains a nuclear reactor, we're fine there) Energy = force * distance. If we suddenly assume that the mystical sideways chamber idea is correct and force on the pump is less then... Assuming the same fill, distance must be the same for the same number of pumps? If we have a 50 newton average over 10 metres, then 500 joules have been stored. If we have a 100 newton force for the "other chamber", then 1000 joules have been stored. How come the other chamber has more energy stored? This would make it less efficent for use? How exactly can this even vaguely make sense? How much physics do I need to shoot across these forums before I can sink this theory for good? Should I bring out mass-flow computions: flow = flow co-effiecent * sqrt((pressure in chamber squared - air pressure squared)/temperature) The flow co-efficent is dependant on NOZZLE area, not chamber area. And does force even come into that calculation? - NO! I know this theory is completely flawed - and I will drag out every equation I can to prove it. EDIT/Sanity revert: This is a post that I would slay other members on sight for, but I can get very, very persistant and aggresive when other people apparantly refuse to at least consider my theory. Sorry, but I've expained my reasons for this before - thus I am leaving the post as it began. __________________ "Over the hills and far away, she prays he will return one day. As sure as the rivers reach the seas, back in his arms again she'll be." - Over the Hills and far away, Gary Moore "So many people have come and gone, their faces fade as the years go by. Yet I still recall as I wander on, as clear as the sun in the summer sky" - More than a feeling, Boston Last edited by joannaardway : 06-27-2006 at 05:56 PM.

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