Water gun power
 DX
 Posts: 1780
 Joined: Wed Feb 04, 2004 1:00 pm
Re: Water gun power
Hehe, the relativity of water guns strikes yet again. *cue Imperial March*
Mess With the Best, Get Soaked Like the Rest!
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2007 Red Sox  World Series Champions!
 Drenchenator
 Posts: 807
 Joined: Fri Jun 18, 2004 12:00 pm
Re: Water gun power
But there should be an overall measure of how powerful something is. Power is a measure of work per time, so the the higher the number the more work it can do in less time. The overall "goodness" of a gun might be impossible to measure, but we should be able to find out how much work a gun can do in a certain amount of time.I think what this whole discussion has shown is that the qualities of a water gun can't be reduced to one single number, no matter how much one might want to. Different people want different things. Some favour range, others output, others will want a balance of both, still others care about shot time, or want a weapon that's small and light, and so on.
The Drenchenator, also known as Lt. Col. Drench.
 SSCBen
 Posts: 6449
 Joined: Sat Mar 22, 2003 1:00 pm
Re: Water gun power
If you want to get into literal power from a physics standpoint, you could figure out how much work the system is doing as a function of time and take the derivative. That shouldn't be that hard with basic thermodynamics. You didn't take that class though Drenchenator so for the sake of enlightenment I'd suggest taking a look at that Wikipedia page.
I might figure out some actual power numbers to compare them against what's been posted.
I might figure out some actual power numbers to compare them against what's been posted.
 Drenchenator
 Posts: 807
 Joined: Fri Jun 18, 2004 12:00 pm
Re: Water gun power
I actually based the equation I came up with from that one in a similar method to what Ben said. Since taking the derivative is basically a funky form of division (think about it), I divided by time on both sides to get power from work and flow rate from volume. Not to hard to understand. I also used the pressurevolumework equation to get the energy put into an air pressure gun in this article. There's no time component, so it's pretty much impossible to take the derivative to get the power. If you knew how the pressure dropped with respect to time, it could be done though.
The Drenchenator, also known as Lt. Col. Drench.
 Silence
 Posts: 3825
 Joined: Sun Apr 09, 2006 9:01 pm
Re: Water gun power
Well, that's dependent on the type of pressure chamber, but it could be done. Tie flow to the change in volume and use Boyle's Law to get to the change in pressure. Unfortunately, things get weird with precharging and "constant" pressure, plus the volume of the chamber makes a difference, too.Drenchenator wrote:If you knew how the pressure dropped with respect to time, it could be done though.
 SSCBen
 Posts: 6449
 Joined: Sat Mar 22, 2003 1:00 pm
Re: Water gun power
It'd really only be reasonable to figure out literal power in theoretical approximations of systems, and the most reasonable system would be "constant " pressure. You'd still have to do 3D integrals but if the pressure is constant with respect to time the math would be much simpler. It'd probably be interested to do for some example CAP systems to see what actual power numbers will come out.
Needless to say figuring out the pressure with respect to time would nearly always involved inaccurate approximations, and the limits of the integral would change with time as well so it gets rather messy and inaccurate.
If anyone doesn't understand that, that's okay, take Calculus 3 and you'll follow. A year ago I wouldn't understand what I said here.
I guess this is why simpler measures of water gun power are necessary.
Needless to say figuring out the pressure with respect to time would nearly always involved inaccurate approximations, and the limits of the integral would change with time as well so it gets rather messy and inaccurate.
If anyone doesn't understand that, that's okay, take Calculus 3 and you'll follow. A year ago I wouldn't understand what I said here.
I guess this is why simpler measures of water gun power are necessary.
 Hunt_and_Annoy
 Posts: 404
 Joined: Sat Feb 12, 2005 1:00 pm
Re: Water gun power
cantab wrote: Athlete peak power ~ 1000 Watts. If you could harness your full power for a water gun (by connecting a pump to a bike or rowing machine I guess), think what you might be able to get.
It should be noted that peak power is different from sustained power. Even Lance Armstrong couldn't sustain more than 500 watts for very long, and the peak power of a "very fit" regular human is quite a bit lower than that of superathletes.
 cantab
 Posts: 1492
 Joined: Fri Oct 19, 2007 1:35 pm
Re: Water gun power
True. Peak power for a normal person is likely to be lower, and sustained power lower still. Even so, mine are about 400 and 200 Watts respectively (measured on a concept 2 rowing machine). And I'm hardly the most powerful person around.
 Hunt_and_Annoy
 Posts: 404
 Joined: Sat Feb 12, 2005 1:00 pm
Re: Water gun power
True. But carrying around any device which would allow one to use that power would be very difficult anyway, so moot point.
 cantab
 Posts: 1492
 Joined: Fri Oct 19, 2007 1:35 pm
Re: Water gun power
Haha. And now you have set the new challenge  build a portable, lightweight, device that allows a person to generate a lot of power to power a water pump.
Of course you'd likely end up having to stay put while using it  but you'd probably be sucking water from a lake or river anyway.
You may as well just carry an engine mind  but where's the fun in that?
In any case, since you're by the water, there's a much simpler way of soaking someone, and more effective than any other  push them in!
Of course you'd likely end up having to stay put while using it  but you'd probably be sucking water from a lake or river anyway.
You may as well just carry an engine mind  but where's the fun in that?
In any case, since you're by the water, there's a much simpler way of soaking someone, and more effective than any other  push them in!
 Hunt_and_Annoy
 Posts: 404
 Joined: Sat Feb 12, 2005 1:00 pm
Re: Water gun power
Stationary water guns? God, soon the helicopters outfitted with WBL's that I dreamed up years ago will be reality...
BTW, you had 140 posts and your userid is 1400, just thought it was kind cool.
BTW, you had 140 posts and your userid is 1400, just thought it was kind cool.
 cantab
 Posts: 1492
 Joined: Fri Oct 19, 2007 1:35 pm
Re: Water gun power
Helicopters with WBL's? Pah! Just get one of those firefighting planes to airstrike the opposition!

 Posts: 13
 Joined: Mon Jul 07, 2008 5:54 pm
Re: Water gun power
Hmmm, sounds like a good addition to the car project
When I die, I want to go peacefully in my sleep, like my grandfather did. not screaming, like the passengers in his car.
 SSCBen
 Posts: 6449
 Joined: Sat Mar 22, 2003 1:00 pm
Re: Water gun power
I just had a "doh" moment about actual water gun power. Finding the average power (yes, true power) of a water is pretty easy if we assume constant pressure or linearly dropping pressure. All you have to do is find the total amount of work necessary to charge the water gun (W = the integral of P*dV), realize that's the amount of energy stored, and then divide by the shot time to get average power. It's that simple.
For some reason before I assumed finding the total work would be much more difficult than it actually is. I don't know what I was thinking but it's probably safe to assume I wasn't thinking, especially considering I've done the math before.
I'd love to make a derivation of a formula for air pressure and constant pressure but I need to get back to studying for a thermodynamics midterm.
Edit: For the constant pressure case power is simple.
power = (pressure*volume)/(shot time)
With SI units you have (Pa)*(m^3)/(s) which is the same as Watts.
With truly constant pressure the average power is the same as the instantaneous power.
Edit #2: Some approximate average power figures for the largest nozzles of various CPS water guns are below:
Supercannon II  ~7800 Watts (I need to find the exact numbers and do a nonconstant pressure calculation)
CPS 2000  247.5 Watts
CPS 2500  103.4 Watts
Monster XL  101.0 Watts (assuming pressure of 23 psi)
Arctic Blast  90.6 Watts (assuming pressure of 23 psi)
CPS 3000  67.6 Watts
CPS 1500  47.7 Watts
CPS 1000  32.9 Watts
SC 600  8.0 Watts
Some data is assumed, some came from memory, some from iSoaker.com, and some came from a website that had pictures of Super Soaker boxes that listed pressures.
Edit #3: I just realized that this method is the same as what Drenchenator used before, just taken from a different angle and more general. Drenchenator's numbers are eerily similar. Remember that flow rate = volume/(shot time) so what he had is the same in the constant flow case.
So... for CPS... P = p*Q
Edit #4: I took a few minutes to derive the average power of a more general system. If we assume the pressure drops linearly with volume (as it does in most cases) then we can simply use the % pressure drop figures I had described before in the LPD thread.
P = p0*V*(10.5*b)/d = p0*Q*(10.5*b)
where
P is power
p0 is the initial pressure
V is the total volume of water being shot
b is the fraction that describes pressure drop (i.e. 25% drop is 0.25)
d is the shot duration
Q is average flow
As you can see this is a simple variation of the formula I derived earlier. If you make b=0 so the pressure is truly constant the equation is the same.
Edit #4: I think it should go without saying that total energy stored = p0*V*(10.5*b).
For some reason before I assumed finding the total work would be much more difficult than it actually is. I don't know what I was thinking but it's probably safe to assume I wasn't thinking, especially considering I've done the math before.
I'd love to make a derivation of a formula for air pressure and constant pressure but I need to get back to studying for a thermodynamics midterm.
Edit: For the constant pressure case power is simple.
power = (pressure*volume)/(shot time)
With SI units you have (Pa)*(m^3)/(s) which is the same as Watts.
With truly constant pressure the average power is the same as the instantaneous power.
Edit #2: Some approximate average power figures for the largest nozzles of various CPS water guns are below:
Supercannon II  ~7800 Watts (I need to find the exact numbers and do a nonconstant pressure calculation)
CPS 2000  247.5 Watts
CPS 2500  103.4 Watts
Monster XL  101.0 Watts (assuming pressure of 23 psi)
Arctic Blast  90.6 Watts (assuming pressure of 23 psi)
CPS 3000  67.6 Watts
CPS 1500  47.7 Watts
CPS 1000  32.9 Watts
SC 600  8.0 Watts
Some data is assumed, some came from memory, some from iSoaker.com, and some came from a website that had pictures of Super Soaker boxes that listed pressures.
Edit #3: I just realized that this method is the same as what Drenchenator used before, just taken from a different angle and more general. Drenchenator's numbers are eerily similar. Remember that flow rate = volume/(shot time) so what he had is the same in the constant flow case.
So... for CPS... P = p*Q
Edit #4: I took a few minutes to derive the average power of a more general system. If we assume the pressure drops linearly with volume (as it does in most cases) then we can simply use the % pressure drop figures I had described before in the LPD thread.
P = p0*V*(10.5*b)/d = p0*Q*(10.5*b)
where
P is power
p0 is the initial pressure
V is the total volume of water being shot
b is the fraction that describes pressure drop (i.e. 25% drop is 0.25)
d is the shot duration
Q is average flow
As you can see this is a simple variation of the formula I derived earlier. If you make b=0 so the pressure is truly constant the equation is the same.
Edit #4: I think it should go without saying that total energy stored = p0*V*(10.5*b).
Last edited by SSCBen on Sun Oct 19, 2008 3:20 pm, edited 1 time in total.
 cantab
 Posts: 1492
 Joined: Fri Oct 19, 2007 1:35 pm
Re: Water gun power
I have yet another measure of power. Well, I thought I did, but actually it reduces to isoaker's!
Basically, my original method used the ACTUAL velocity of the stream.
If, however, I instead use the THEORETICAL MINIMUM velocity, vmin, according to simple projectile motion, then I get what I shall call 'Effective Power, Pe'
For the 45 degree angle, range
R = vmin^2/g
Then power
P = 0.5 p Q R g
Where p is the density of water, Q is the output, R is the range, and g is the acceleration due to gravity.
************************************
So now we have three dimensionally correct measures:
Ben's new method, based on the release of the stored energy (as pressure). I'll call this internal power, Pi
My original method, based on the kinetic energy of the stream exiting the nozzle. I'll call this stream power, Ps.
My new method, based on projectile motion. I'll call this effective power, Pe.
Pi > Ps > Pe
This is ALWAYS the case or energy is added from nowhere and you have a perpetual energy machine.
From these, we can calculate three measures of efficiency:
Internal efficiency, Ei = Ps/Pi
This measures energy losses within the gun
Stream efficiency, Es = Pe/Ps
This measures energy losses in the air, eg from atmospheric drag.
Overall efficiency, Eo = Pe/Pi
This measures all losses.
Basically, my original method used the ACTUAL velocity of the stream.
If, however, I instead use the THEORETICAL MINIMUM velocity, vmin, according to simple projectile motion, then I get what I shall call 'Effective Power, Pe'
For the 45 degree angle, range
R = vmin^2/g
Then power
P = 0.5 p Q R g
Where p is the density of water, Q is the output, R is the range, and g is the acceleration due to gravity.
************************************
So now we have three dimensionally correct measures:
Ben's new method, based on the release of the stored energy (as pressure). I'll call this internal power, Pi
My original method, based on the kinetic energy of the stream exiting the nozzle. I'll call this stream power, Ps.
My new method, based on projectile motion. I'll call this effective power, Pe.
Pi > Ps > Pe
This is ALWAYS the case or energy is added from nowhere and you have a perpetual energy machine.
From these, we can calculate three measures of efficiency:
Internal efficiency, Ei = Ps/Pi
This measures energy losses within the gun
Stream efficiency, Es = Pe/Ps
This measures energy losses in the air, eg from atmospheric drag.
Overall efficiency, Eo = Pe/Pi
This measures all losses.
I work on Windows. My toolbox is Linux.
Arsenal:
Super Soaker: XP215, 2xXP220, Liquidator, Aquashock Secret Strike M(odded), Arctic Blast M, CPS1200, CPS2100, SC Power Pak, 3l aquapack, 1.5l aquapack
Water Warriors: Jet, Sting Ray M, Shark, Argon M, Tiger Shark, PulseMaster
Others: Waterbolt, The Blaster, Storm 500, Shield Blaster 2000, generic PR gun, generic backpack piston pumper (broken), 3l garden sprayer M, 10l water carrier:
Arsenal:
Super Soaker: XP215, 2xXP220, Liquidator, Aquashock Secret Strike M(odded), Arctic Blast M, CPS1200, CPS2100, SC Power Pak, 3l aquapack, 1.5l aquapack
Water Warriors: Jet, Sting Ray M, Shark, Argon M, Tiger Shark, PulseMaster
Others: Waterbolt, The Blaster, Storm 500, Shield Blaster 2000, generic PR gun, generic backpack piston pumper (broken), 3l garden sprayer M, 10l water carrier: