Pump Forces, and Pressures

Threads about how water guns work and other miscellaneous water gun technology threads.
Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Pump Forces, and Pressures

Post by TheEngineer » Sat Jan 13, 2007 2:51 am

Ok, so I am running into some problems with my work.

I am trying to figure out the pump diameter I would need to get a 31.4 cubic inch Pressure Chamber to 55psi in 1 pump. Im looking into a 2 stage pump. First one to up the pressure, then the final(longest part of the stroke) to finish it off.

Now the thing is, this has to have a realistic force exerted on the pump. somewhere along the lines of 5-10 pounds of force for both stages.

This is where I run into a problem. I don't know of any formulas or equations that I can use to help me with this. Even an equation to get how many pumps to use to get to a certain psi would be great. I think I could go from there.

I saw on this forum that a guy made some graphs showing pumps to psi, and pumps to force on the pump. But he never showed how he got those graphs.

Im currently running everything off of MS Excel. Which is a great program for this kind of stuff.

And I would like to say, you guys do some pretty cool things on this site. I love home modding forums and this is one of the best.

Thanks in advance.

User avatar
Posts: 6449
Joined: Sat Mar 22, 2003 1:00 pm

Post by SSCBen » Sat Jan 13, 2007 3:57 am

Welcome to Super Soaker Central! If you like the site now, just wait until the update is completed (scheduled for March 1st). There'll be a lot more of the stuff you like there with a more in-depth physics section, one that would actually have the answer to your question. ;)

I don't really get what you mean by the two-stage pump thing. Could you explain in more detail? Do you mean a pump that pumps on both strokes? Or do you mean a pump that has one larger pump and a smaller pump in it? Both have been done before.

Once you let me know which type of pump you are using, I'll explain everything step by step. That is relatively easy. However, I'm not sure that any pump made to get 55 PSI in one pump with over 30 cubic inches of space will be easy to pump... will do that math as well.

For the first thing, you should use Boyle's law. Check out Wikipedia for more information on that equation. It's fairly straightforward.

Hope you enjoy your time here. I really love it when people come here with a problem to solve because that's something I try to do myself.


Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Sat Jan 13, 2007 4:18 am

Yeah the pump in a pump. (wierd edit for el oh el). I know its been done. But thats what I want to do. That way I can get to the pressure I want with little force and less pumping.

I was hoping for a very large diameter 1st stage.

My knowledge so far has taken my nowhere with boyles law. This math is kicking my brains butt.

Im just missing like one equation and its the one I need. :)

So let me narrow this down. I need a formula that uses: Pump force vs Pressure chamber psi, Pump Volume vs Pressure Chamber volume, and i think that ought to do it. Or just a formula comparing the relationship between pump strokes to psi.
Last edited by TheEngineer on Sat Jan 13, 2007 4:41 am, edited 1 time in total.

User avatar
Posts: 6449
Joined: Sat Mar 22, 2003 1:00 pm

Post by SSCBen » Sun Jan 14, 2007 9:56 pm

I've done the math for a single stage pump and have it written on a paper somewhere. I'll post it tonight when I have more time. Then I can start working on the two stage pump. Right now I have to do something. Thought I should let you know.


Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Sun Jan 14, 2007 10:39 pm

Right on. If you post the math for the single stage, I would be more than happy to help with the double stage.

You guys rock.

User avatar
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

Post by Silence » Mon Jan 15, 2007 1:43 am

Hmm, I must have missed this thread. Welcome to SSC, and hope you learn what you're looking for!

I daresay you could use maybe 15 pounds of force if necessary. However, it does depend on the application and how much you want to exert while pumping. Also, the length of the pump is an important thing to consider. If you're willing to let the back of the pump extend up to a foot behind the main gun grip, you could very well have an 18" pump! Try it with a yardstick. ;)

31.4 cubic inches * (15 PSI/55 PSI) = 8.6 cubic inches
Probably not put on paper/digital paper correctly, but the point is that there should only be 8.56 cubic inches of space left in the pressure chamber (PC) for the air. I'm presuming we're pumping water into the pressure chamber.

31.4 cubic inches - 8.6 cubic inches = 22.8 cubic inches displaced by the water. In other words, the volume of the pump has to be 22.8 cubic inches. The way you set up the pump, in terms of (length1 * area1 + length2 * area2), where the 1 and 2 signify each stage of the pump, is up to you. You could draw the line for the diameter change at the point where the pressure is 30 PSI, although you do start at 15 PSI.

10 pounds of force / (25pi/256 square inches) = 33 PSI
The pounds are the force you're exerting on the pump, you can change that to something else if you wish. The area, in square inches, is the area of the pump tube. 1/2" Schedule 40 PVC actually has an inner diameter, or ID, of 5/8", so its area is 25pi/64 square inches. McMaster-Carr has a chart of PVC IDs for different sizes of PVC; I can get the measurements if you wish.

There's a problem though. That 33 PSI isn't anywhere near the goal of 60 PSI, which I know is attainable with homemade water guns. So either (1) my math is wrong, and somebody should correct me, or (2) the human body can exert way more force than that. Or both.

At any rate, disregarding the pressure/math problem, this last calculation gives you the max pressure you can reach with each size of PVC, given a certain max amount of force. So couple that information with a chart/graph of the PC volume displaced vs. pressure, and that should tell you what lengths you'll need. But Ben has probably already created an insane formula for this. :)

I presume you got your information from joannaardway's posts in the dual-stage pump thread. You might also want to check out the two-stroke (stroke as opposed to stage) pump concept, which pumps water as you move the pump rod in each direction. You could use the weaker outstroke with the two-stroke concept to get the initial pumping done, then face the increased pressure as you do the instroke.

Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Mon Jan 15, 2007 1:58 am

Well the thing is. Im not exactly sure how you got the distinction between area/length and psi. Maybe Im missing something. Or maybe its just been a long day. And depending on the length of pump needed, I may end up halving it and two stroking it.

Im going to be pumping pure air.

User avatar
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

Post by Silence » Mon Jan 15, 2007 3:47 am

Pure air...that's what I suspected, but I was thinking in a water gun sense.

I connected the area/length and the PSI because of the water displacement inside the PC. The area determines the amount of displacement and the pressure increase as you move the plunger along the length of the pump tube. You must then use the area again to connect the force you're exerting and the building pressure.

I did not fully connect the two parts of the problem, but it sounds like Ben has the answers to that.

Now, you say you're pumping air instead of water. I assume this is for something other than a water gun, but I might be wrong. Anyway, it's a similar thing, but different in some ways.

31.4 cubic inches * (55 PSI /15 PSI) = 115 cubic inches' worth of STP air
somewhat goes the other way around. Earlier, I figured out how much space you'd have to push 31.4 cubic inches' worth of air into to get 55 PSI. With this, I'm figuring how much air you'd have fit into 31.4 cubic inches to get 55 PSI. The air gets displaced with in the first case, but extra air is pumped in for this case.

115 cubic inches - 31.4 cubic inches = 84 cubic inches' worth of air needs to be pumped in. The pump's volume in this case has to be 84 cubic inches. Which is enormous, and it could prove to be a problem later on.

10 pounds / (25pi/256 square inches) = 33 PSI, which is the same figure as in my previous post. But let's assume you can actually exert 20 pounds of force for the final few pumps. That would double the maximum pressure, to a workable 66 PSI. Or going backwards, we can do:
55 PSI * (25pi/256 square inches) = 17 pounds of force needed to pump at the end. Not bad at all, I would think.

Now, the main problem here would be the amount of air you'd have to pump in. If you were pumping water, you're only trying to fill 8.6 cubic inches of space to 55 PSI. With this, you're trying to fill 31.4 cubic inches of space to 55 PSI.

But we seriously need to realize how much air you'd need to pump. 84 cubic inches' worth of air. If you had a pump tube with an area of 1", so maybe 1.25" in diameter, you'd need it to be 84 inches or 7 feet long in order for it to have this capacity. And you would need a ton of force. But I guess that's why you decided to use a dual-stage pump in the first place, so you could use a large pump diameter.


I'm going to try to come up with a reasonable pump setup. It might even require 3 strokes. I'll use numerical calculations, although an algebraic formula would probably be better.

Stage 1 - 2" PVC.
20 pounds / pi square inches [the cross-sectional area] = 6.4 PSI

Just figured something out, after I saw that you could only get negative pressure with a 2" pump. You can actually add 15 PSI, or one atmosphere, to the PSI that you can pump up to. Why? Because the atmosphere is also pushing the pump in because there's 15 PSI, or one atmosphere, acting on all side of the pump. So even with the 10 pound, 1/2" pump calculation, you're getting an additional 15 PSI, for a total of 48 PSI. Which isn't too shabby.

6.4 PSI + 15 PSI = 21.4 PSI. That's the maximum pressure you can pump up to with this stage of the pump. We'll use this pressure to identify how much air we can displace with this stage and how long we can make it.

31.4 cubic inches * (21.4 PSI / 15 PSI) = 45 cubic inches' worth of STP air
This calculation is the same as the first calculation in this post--determining how much air is in the PC. The difference is that I replaced the goal of 55 PSI with the current 21.4 PSI, simply because we have only reached 21.4 PSI at this point.

45 cubic inches - 31.4 cubic inches = 14 cubic inches
The 45 cubic inches was really the volume's worth of air packed into the PC. If you eliminate the 31.4 cubic inches that was in there since the beginning, you'll see that you only added 14 cubic inches. That's the capacity of this stage of the pump.

14 cubic inches / pi square inches [the pump's area] = 4.5 inches
Given that the pump's capacity is 14 cubic inches, and the area (the base of a cylinder, if you will) is pi square inches, the length of the stage (the height of the cylinder) is 4.5 inches.

So we know how short we can make a 2" wide stage, so that we pump the maximum amount of air in a short distance. All while keeping the force below 20 pounds.

Rinse and repeat, but for a different diameter for the pump tube. However, we now assume that the pressure inside the PC starts at 21.4 cubic inches.

Stage 2 - 1" PVC
20 pounds / (pi/4) square inches [the cross-sectional area] = 25 PSI

25 PSI + 15 PSI = 40 PSI
This is your atmospheric pressure bonus kicking in. Now, with a maximum of 40 PSI, we're nearing the goal of 55 PSI. At least compared to the previous maximum of 21.4 PSI with the 2" tube.

31.4 cubic inches * (40 PSI / 15 PSI) = 84 cubic inches' worth of air

The PC can now take a total of 84 cubic inches' worth of air until it reaches the maximum pressure we can have during this stage.

84 cubic inches - 45 cubic inches = 39 cubic inches
Subtracting the 45 cubic inches of air that were inside from the second stage, we find that this stage of the pump can add up to 39 cubic inches' worth of air to the PC.

39 cubic inches / (pi/4) square inches [the pump's area] = 50 inches
Given that the pump's capacity is 39 cubic inches, and the area is pi/4 square inches, the length of the stage is a whopping 50 inches.

To be continued... I'll finish the calculations tomorrow. You'll probably want to have 1.5" inch PVC as another stage in between, as the current pump is already 64 inches long - the sum of each stage's length - and not quite at maximum capacity. You'll still need to add the 1/2" PVC stage, and some stages in between (1.5", etc.) or replacing some of these might be better. So experiment with the different combinations.

No matter what, this thing will be complex, probably using at least 4 stages. If you get it working, that's great. Note that 1/2" PVC doesn't fit in 3/4" PVC, which doesn't fit in 1" PVC--so it'll have to go directly from 1/2" to 1". If a system with more than two stages can be designed, and if you manage to keep the pump short, then congratulations! Also, if you can reliably exert 25 pounds of force in a pumping motion, that would be really sweet. 20 pounds is just an estimation.

Ben, is there a function that graphs a variable pump area and the infinitesimal lengths of each area for a given force? Ie, using calculus? Maybe some appropriate stage sizes could be decided upon using such a function. The function I'm thinking of would be area vs. length, or length vs. area, with the total area below the function as the total volume pumped.

Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Mon Jan 15, 2007 3:59 am

Wow, thats alot to ponder.

Ill get to working on things tomorrow. Throw something together in excel.

Your right though, this is not for a water gun... Yet.

Im trying to work on a small air powered cannon, and you guys are the only people close enough to what I wanted to do, that use math instead of just guessing.

However, after reading through the threads and whatnot. Im planning on using this information and try and make a water gun that will go for distance, with minimal pumping. I will use the built up air pressure, and shove the water out of the nozzle. Im not sure how well it will work. But It may be something different.

User avatar
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

Post by Silence » Mon Jan 15, 2007 4:49 am

Here's the link I wanted to get: http://www.mcmaster.com/param/html//plasticschedule/
Basically tells you the ID and OD of every type of PVC fitting you could want. McMaster-Carr also sells an incredible assortment of O-rings for the seals.

Yeah, my explanations tend to get very long, very fast. Quite confusing, like with my argument with the Waterzooka LR. But if you have any questions, just ask. ;)

Yup, the air-pumping system is used in just about everything except water guns. Pneumatic cannons, Nerf, water balloon launchers...everything. You should also glance at pump designs intended for air-pumping use, that essentially have the intake check valve built into the plunger. Pretty neat.

Building a water gun using something like this would be very difficult. The best water guns typically need a high PC capacity--meaning either higher pressure or greater volume, neither of which are single-pump friendly. Air cannons can use very small PCs as long as the valve opens quick enough. In contrast, water guns need a ton of force and a large nozzle diameter to reduce the effects of drag on the stream.

EDIT: A few more random thoughts (I'm usually not so inspired). The Splat Cannon, a water gun by Water Warriors/Buzz Bee Toys, is a shotgun that has its tiny, spring PC charged with a single pull of the lever. You could create an effective, short-ranged water gun using this system. Or just use a pumper, which has no pressure chamber--you fire as you pump.

The other, more important thing is the use of latex rubber tubing (LRT), which you can also get from McMaster-Carr. It reaches maximum pressure/force nearly instantly, and that force is constant--it's applied by a rubber tube. Although the rubber thins out and applies less pressure as the tube expands, the area increases so the force ends up staying constant.

User avatar
Posts: 6449
Joined: Sat Mar 22, 2003 1:00 pm

Post by SSCBen » Mon Jan 15, 2007 2:32 pm

The math Andrew and I did was for pumping water. That math is very different from pumping air.

Looks like SilentGuy has you on the right track. I didn't check it all, but it seems to be what I would have done. One thing though. You don't add atmospheric pressure to the internal pressure of the pressure chamber (you did this somewhere). What matters is the pressure differential between the air and the pressure chamber. ;)
Ben, is there a function that graphs a variable pump area and the infinitesimal lengths of each area for a given force? Ie, using calculus?
Not really quite sure what you mean.

You could just use an equation like

pump volume = Area * Length

and simplify. Though I'm not quite sure that's what you mean. I'll reread everything shortly, but if you could explain again I'm sure I could be more helpful.
Last edited by SSCBen on Mon Jan 15, 2007 2:37 pm, edited 1 time in total.

User avatar
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

Post by Silence » Mon Jan 15, 2007 3:06 pm

Ben wrote:You don't add atmospheric pressure to the internal pressure of the pressure chamber (you did this somewhere). What matters is the pressure differential between the air and the pressure chamber.
Well, at least you figured out what I meant. I guess you could also subtract 15 PSI from inside the chamber, but by the time I realized we're using the pressure differential, I didn't feel like doing all the calculations again. :p
Ben wrote:Not really quite sure what you mean.
Actually, never mind that. I was thinking about a function that essentially plotted how long each diameter stage would need to be, but as an actual curve of area vs. length instead of a piecewise function. The area of the graph below the curve would be the total volume you're pumping.

But I would say it's impossible. The area has to be in the domain instead of the volume, but then you've got some other problems. Don't worry about it.

EDIT: I really wanted a graph like that to prevent having to do so many calculations and guessing. But now I created a calculator program for that.

Some notes: if there's a space here, it's not in the program unless there's a _.
TI-83 wrote: :1 -> S: 15.4 -> Q
:LblA: ClrHome
: Disp "STAGE_
:O utput(1, 7, S
:Input "D (IN):", D
: pi(D/2)^2 -> A
:Input "MAX F (LB):", F
:F/A+15.4 -> P
:Input "PC V (C.IN):", V
:If S=1: V -> O
:VP/Q -> T
:T-O -> N
:N/A -> L
: Disp "MAX P (PSI):
:O utput(5, 12, P
: Disp "L (IN):
:O utput(6, 7, L
:Input "EXIT? (1/0):", X
:If X=1:Return
:N -> O
:P -> Q
:S+1 ->
:Goto A
Pretty messy, but it works. I wouldn't normally use a label, but since the label's at the top, I'm not losing anything.

D - diameter of the pump stage
A - area of the pump stage
F - force exerted on the pump
P - maximum pressure
Q - previous maximum pressure
V - volume of the PC
T - total volume's worth of STP air inside the PC
O - old volume's worth of air - what was already inside the PC
N - new volume's worth of air - what you're pumping in
L - length of the pump stage

EDIT again: Had to change the program I entered. ": Disp" and ": pi", without a space in between, produced " :D " and " :p ". Happened in numerous places, actually quite funny.
Last edited by Silence on Mon Jan 15, 2007 4:00 pm, edited 1 time in total.

Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Fri Jan 19, 2007 2:08 am

Well, Im running some numbers. And Ive found that anything over 40 psi creates this very steep PSI to Pump length ratio.

By that I mean. Anything above 40 psi. The pump diameter with enough max psi to get to said psi, would cause it to need a lot of length. To account for the volume.

Unless I work with extremely small PC volumes, the PSI above 40 would not be feasible.

I made a quick Excel Spread sheet of my multistage calculator. If you guys have any comments that I should change. They would be most welcome.

Multi-stage Calculator

Requires IE to run.

User avatar
Posts: 3825
Joined: Sun Apr 09, 2006 9:01 pm

Post by Silence » Fri Jan 19, 2007 2:19 am

I would need Office 2003 Web Components (I use OpenOffice.org) or OfficeWorks to run it...so I can't. I'm also fairly certain Ben uses Linux. Anyway, nice to know you got a similar calculation setup working, so good luck!

That's the problem with higher pressures--the diminishing returns. Note that if the application allowed for, say, 2 pumps, you could do a lot more. Also, you could just use a smaller tank...which is almost the same thing as reducing the pressure as far as air capacity is concerned.

I would go for a slightly lower pressure, and hone in on creating an efficient system that has a long barrel and a fast valve. Good luck!

Posts: 8
Joined: Sat Jan 13, 2007 1:18 am

Post by TheEngineer » Fri Jan 19, 2007 3:06 am

Yeah, what Im finding is that two pumps open up a whole new world.

Im working with this site

with mass of .032
barrel of .68
friction of 0
and a needed velocity of 180 to 200 miles per hour

From that I find the correct psi to volume.

Then plug that into my excel. So far, the smallest length of a pump I have gotten to is 45inches. Which is quite large.

I may make a javascript version later on. But right now Im just seeing if my calculations are even close to what they should be.

Next up, Ill be working with two pumps of the multistage.

QUESTION: Would I be able to use the same calculations, then just divide the final length by two?
Last edited by TheEngineer on Fri Jan 19, 2007 3:14 am, edited 1 time in total.