Water gun power

Threads about how water guns work and other miscellaneous water gun technology threads.
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cantab
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Water gun power

Post by cantab » Wed Jun 25, 2008 10:46 pm

I've been thinking about how to measure power of a water gun quantitatively. isoaker use range * output to get a relative power, based on what seems like some dubious assumptions. So I thought I'd see if I could come up with something more rigourous.

Thinking about the basic definition of power

Power = Work done / time taken

Let's work per unit time, so take time taken as 1 second

Now I am going to relate the work done to the kinetic energy of the stream of water, as it exits the nozzle.

W = 0.5 m v^2

Velocity is tricky to measure, but I realise we can calculate it, by dividing output by nozzle area

v = Q / A

Now the mass of water m, will also depend on the output

m = p Q t

p is the density of water, t is the time taken, again we'll use one second

Thus we get an equation for power

P = ( 0.5 p Q^3 ) / A ^ 2

Checking dimensions

p is kg m^-3, and is 1000 for water
Q is m^3 s^-1
A is m^2

P = kg m^-3 m^9 s^-3 m^-4 = kg m^2 s^-3
which is what we expect, Watts, http://en.wikipedia.org/wiki/Watt#Definition

Notice that we do not use range at all! This seems surprising, but when I think about it, range is determined not just by the power of a water gun, but by how that power is used.

This formula should remove the oddity of a bucket throw getting high power - as while the output is high, so is the 'nozzle'.

We could then perhaps define pseudoefficiency as range / power. A better designed gun will get more range using less power. (I use the term pseudoefficiency as I'm not defining this rigourously. A rigourous efficiency definition would require dividing two powers. Maybe I could figure out drag from velocity and range, and go from there somehow, thinking about how rapidly the kinetic energy is lost to in the air).


I'd be interested to see how this calculation compares with the one currently used by isoaker. However, that would require accurate nozzle diameter measurements. I'm not sure how one measures small internal diameters precisely. Photography is one possibility, seeing how thick a wire/pin can be stuck in the nozzle is another.

Hopefully I've shed some light on the matter, and not confused you too much.

EDIT: Changed symbol for output to Q as was suggested below.
Last edited by cantab on Thu Jun 26, 2008 10:10 am, edited 1 time in total.

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Silence
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Re: Water gun power

Post by Silence » Wed Jun 25, 2008 11:23 pm

"Power" as used in the general sense, and which iSoaker was referring to, is more a vague term for how strong a stream is. You may find that, while

Code: Select all

P = ( 0.5 p o^3 ) / A ^ 2
is more scientifically correct, it is actually less useful metric than

Code: Select all

P = range * output
Range and output are two very important characteristics that are easy to determine and appreciate. In contrast, the actual equation involves the density of water rho (which is hard to remember in Imperial units) and the exact area of the nozzle. Neither is as useful as range.

That is, it's a lot more rewarding to brag about range than about the density of your water. Or the area of the orifice, if you will.

That said, good analysis! That equation combines with some basic equations could help somebody who's really trying to model the range of a water gun. My only recommendation is to make rho a constant and unroll it from the main equation (essentially, combine it and the 0.5 coefficient). Heuristics are always better than variables for rules of thumb.

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SSCBen
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Re: Water gun power

Post by SSCBen » Thu Jun 26, 2008 12:37 am

Interesting analysis cantab. Thinking about it this might describe stream "concentration" more than power though (roughly output over area). That could be a measure of power I suppose. There doesn't have to be one measure of water gun power after all.
I'm not sure how one measures small internal diameters precisely. Photography is one possibility, seeing how thick a wire/pin can be stuck in the nozzle is another.
When I forget what size hole I drilled in something I stick progressively larger drill bits into the hole until I find one that fits. A similar technique could be used for small nozzles, but you will need a good selection of drill bits.

I think a weighted version of iSoaker's formula would be best for the reasons SilentGuy mentioned. We need to give range more weight so we could either raise range to a power greater than 1 or we could raise output to a power less than 1.

Thinking about it, fluid mechanics and electricity have many similarities. Perhaps something like P = I^2*R could be adapted for water guns. I (current) is analogous to water flow (Q) and resistance could be analogous to nozzle area. This doesn't include range still but P = Q^2*A would be another interesting formula.

It'd be worthwhile to plug a bunch of numbers into a spreadsheet and see how closely they follow intuition. I think I'll try that...

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Drenchenator
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Re: Water gun power

Post by Drenchenator » Thu Jun 26, 2008 1:26 am

Nice work and brilliant derivation. I was hoping that someone read all of our physics articles, and this just verifies you know your stuff!

This should calculate the actual power of water leaving the nozzle, which could be useful if you really want to see how powerful a certain nozzle is. However, since we are often concerned with how powerful a stream is, a typical measure of power doesn't always account for what we are considering. The product of range and output comes up with the wrong units for power, but is good enough if we only compare water guns.

Also, we like to use Q for output instead of o. It's less confusing and goes along with the standards already setup for fluid mechanics.
This formula should remove the oddity of a bucket throw getting high power - as while the output is high, so is the 'nozzle'.
I've used that example in the past to illustrate that output isn't really that much to brag about. But since your equation accounts for the nozzle area, it creates a very fair way to measure the overall power of a nozzle. I think this will be its best use.

Edit: Another way to get power (just based on some quick dimensional analysis) is the product of the pressure and flow rate (power = pressure x output). The units match up ( N x m / s = (N / m^2) x (m^3 / s) ).
Last edited by Drenchenator on Thu Jun 26, 2008 1:35 am, edited 1 time in total.
The Drenchenator, also known as Lt. Col. Drench.

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cantab
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Re: Water gun power

Post by cantab » Thu Jun 26, 2008 4:14 pm

rho (which is hard to remember in Imperial units)
And easy to remember in SI :-P
It's probably constant enough for our purposes. However, if we assume power depends primarily on the gun and not the liquid it fires, we predict that if Q scales with p^-3. More loosely, a more dense liquid will give less output. But since mostly we're just shooting water that's not very relevant.

Rough-and-ready calculations (my nozzle areas are subject to bad errors as I only have a normal ruler to measure them). I also provide some other things for comparison.

Arctic Blast
Main nozzle, diameter 1.5 mm, area 1.77 x 10^-6 m^2
Output 33 mL s^-1 (from isoaker.com), 3.3 x 10^-5 cumecs (m^3 s^-1)
Power = 5.7 Watts

Flood nozzle, diameter 16.0 mm, area 2.01 x 10^-4 m^2
Output 560 mL s^-1, 0.00056 cumecs
Power = 2.2 Watts

I suggest that for a given firing chamber, the power from the different nozzles is going to be fairly similar, within errors and drag considerations. This hypothesis remains to be proven.

CPS 2000
Nozzle diameter half inch or 12.7 mm (from a bit of googling), area 1.27 x 10^-4 m^2
Output 849 mL s^-1, 0.000849 cumecs
Power = 19 Watts

Unsurprisingly, this is rather more than the Arctic Blast. Though not by more than maybe one order of magnitude.

Low-energy lightbulb ~ 10-20 Watts

1 horsepower = 746 Watts

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.

This equation does have some practical uses. If you are planning on making a motorised gun, you can decide the output and nozzle diameter you want, and get how much power it needs. Of course you'll then need to take into account the gun's not 100% efficient, but at least it would stop you trying to obtain CPS 2000 power with a 1 watt motor, or conversely using a lawnmower engine to power an XP-like stream.
Also you might be able to figure out a reasonable human pumping power, and use that to figure out the pump time/shot time ratio for an air pressure gun, or get a power limit on a piston pressure gun.

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Silence
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Re: Water gun power

Post by Silence » Thu Jun 26, 2008 4:30 pm

Not bad...as long as the end result does show a positive correlation towards water guns that are better (according to common sense), this could be a useful metric. I still favor basic range-based metrics, but this may work fine too. The only problem is it varies unpredictably (well...as estimates go) with nozzles. For example, most people would say the Flash Flood nozzle is more powerful than the regular one on an Arctic Blast.

I'm not sure how easy it is to connect stream power to the power of a lightbulb, vehicle, or athlete.

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cantab
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Re: Water gun power

Post by cantab » Thu Jun 26, 2008 4:53 pm

Because the measured values are taken to powers, relative errors increase. To get an accurate power for the riot blast would require rather precise timing (counting frames in video would be the best affordable option). isoaker reckon it takes half a second; if the actual time is .4 seconds, the power goes up to 4.2 W. While for the main nozzle the diameter probably has an error of about +/- .25 mm or 16%, which means a 32% or +/0 1.8 W error.

That's the main drawback here, it's so sensitive to errors. For small nozzles nozzle size is a problem, while for short blasts (or trying to get at initial power when there's dropoff) timing is.
Video is good for precise timing. I also wonder if it would be good for directly observing velocity, which cuts the nozzle diameter measurement out altogether.


The other stuff was just to give an idea of perspective really. Like, the CPS 2000 seems phenomenal - but it's only about as powerful as a lightbulb.

As I said, it's just one characteristic of the stream. The relationship between this power and range depends on how the stream behaves in the air, which in turn is strongly influenced by the nozzle design. So if you have a gun with lower-than-expected range for its power (ie low pseudoefficiency), that suggests there's room for improvement in the nozzle. Whereas if pseudoefficiency is high, you probably should leave the nozzle alone, at least if range is your main concern. So it's relevant for deciding whether nozzle-related mods (or design tweaks for homemades) are worth doing, BEFORE you go and do it.

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SSCBen
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Re: Water gun power

Post by SSCBen » Thu Jun 26, 2008 5:09 pm

For the sake of comparison I took the info cantab put up, added some more stats to it, and added Supercannon II. Putting all that into a spreadsheet made it easy to see how it ranks things.

The only major difference between all of these methods as far as I can tell is whether it gives priority to high output or high range. The original method, the method I based on electricity, and Drenchenator's method all have output priority. cantab's method has range priority. I'd consider range priority to be ideal so cantab's method actually works despite it not taking into account the range. Interesting. I'm going to try a weighted original method right now...

Original method
1) 0.2761 m^4/s - Supercannon II (riot blast)
2) 0.0888 m^4/s - Supercannon II (stream nozzle)
3) 0.0135 m^4/s - CPS 2000
4) 0.0034 m^4/s - Arctic Blast (riot blast)
5) 0.0003 m^4/s - Artic Blast (stream nozzle)

cantab's method
1) 1979.2 Watts - Supercanonn II (stream nozzle)
2) 997.3 Watts - Supercannon II (riot blast)
3) 19.1 Watts - CPS 2000
4) 5.8 Watts - Arctic Blast (stream nozzle)
5) 2.2 Watts - Arctic Blast (riot blast)

Ben's method
1) 0.1735 m^4/s^2 - Supercannon II (riot blast)
2) 0.1257 m^4/s^2 - Supercannon II (stream nozzle)
3) 0.0057 m^4/s^2 - CPS 2000
4) 0.0016 m^4/s^2 - Arctic Blast (riot blast)
5) 0.0006 m^4/s^2 - Artic Blast (stream nozzle)

Drenchenator's method
1) 6246.0 Watts - Supercannon II (riot blast)
2) 2750.7 Watts - Supercannon II (stream nozzle)
3) 263.4 Watts - CPS 2000
4) 88.8 Watts - Arctic Blast (riot blast)
5) 5.2 Watts - Arctic Blast (stream nozzle)
Last edited by SSCBen on Thu Jun 26, 2008 7:22 pm, edited 1 time in total.

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cantab
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Re: Water gun power

Post by cantab » Thu Jun 26, 2008 6:51 pm

I knew Supercannon II was powerful, but I'm still astounded. TWO KILOWATTS! That's insane.

Another note: the dimensions for your method are the square of those for the original. You might consider taking the square root.

Graphs coming soon.

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SSCBen
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Re: Water gun power

Post by SSCBen » Thu Jun 26, 2008 7:21 pm

Actually, I did my method incorrectly. The equation should be P = Q^2/A because 1/A follows resistance, not A. I've updated the chart above. Nothing major changes.

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cantab
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Re: Water gun power

Post by cantab » Thu Jun 26, 2008 7:55 pm

Ah right. That becomes even more interesting, since if you multiply by the density, you'll get Newtons.

Actually, the isoaker method, if again you multiply by the density, gives Newton seconds, which is impulse or momentum.

Anyway, the graph. I've plotted all three new methods against the isoaker one. It's a log-log plot so we can see everything properly. Notes:

A reminder: from left to right, the data are for the Arctic Blast stream, AB riot, CPS 2000, Supercannon II stream, and SCII riot. Mine and Drenchenator's are on the left hand y axis, ben's is on the right

Drenchenator's method has the closest correlation with the isoaker one. Mine looks to have the least, with Ben's somewhere in the middle.
Gnumeric (my usually preferred spreadsheet program) won't do power regression, which is annoying, so I'll have to do it manually, that's coming soon.
Attachments
water_gun_power_graph1.png

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Silence
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Re: Water gun power

Post by Silence » Fri Jun 27, 2008 12:30 am

Hmm...all methods show a huge difference between the most powerful and the least powerful soakers. Then again, it's like that in most things - computer component benchmarks, for instance.

Here's the last version of my "battle practicality formula" from early 2007:

Code: Select all

BPF = D^2 * sqrt(t) * V / (2*m1 + m2)
BPF = battle practicality factor for a given gun and nozzle
D = distance in meters
t = shot time in seconds
V = volume of the reservoir and pressure chamber, in liters
m1 = gun mass in kilograms when full
m2 = backpack mass in kilograms when full
It kind of grew out of proportion as I added more elements. And still, nothing could be universal. Oh well.

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cantab
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Re: Water gun power

Post by cantab » Fri Jun 27, 2008 10:49 am

Hmmm, that's an interesting formula.
However, as reservoir size increases, it's cancelled by the increased mass (though backpacks are still better than onboard). So the formula fails for impossibly heavy weapons (ie you can't carry a 1000 litre backpack!)
I think you're right to care about range much more than shot time. But perhaps better than just plain shot time is shot time to pumps ratio. Some guns (like the water warriors aqua masters) have short shots but correspondingly few pumps. They lend themselves to different tactics (a tank empty on them can be little more than a tap shot on some other blasters), but shouldn't necessarily be downmarked for it. Which is better - 3 pumps for 1 second, or 20 pumps for 2 seconds? (that's a rhetorical question)

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Silence
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Re: Water gun power

Post by Silence » Fri Jun 27, 2008 2:24 pm

[quote="cantab"]However, as reservoir size increases, it's cancelled by the increased mass (though backpacks are still better than onboard).[/cantab]
Heh, that must have totally missed me. In light of that observation, volume and mass might as well be removed from the equation - as you could justify any volume/mass. The current ratio essentially just favors weight in the backpack over in the hands, so maybe I added volume partly to reduce that effect.

Since I'm just trying to model the effect (rather than the science), maybe a bell curve would be best. The maximum battle practicality would occur with volume/mass somewhere in the middle. Unfortunately that would make the equation more complex. Maybe subtracting a quadratic would be better. Then the problem would be that addition/subtraction is used instead of scalars - which means a water gun could have decent stats but zero volume (like a hose), or lots of volume and no range (like a big bucket), and still have a high rating.

Which is better: pumping with 10 pounds of force or pumping with 100? My point is, it's a trade-off once again. Or pumping 10 inches or pumping 100 inches? I'm thinking we should:
  • find the pressure of the water gun
  • find the pump area A = 15 pounds / pressure
And the area itself can be a factor. The problem with finding inches of pumping (which I naturally prefer to number of pumps) is it penalizes PC volume, just like mass penalizes reservoir (and PC) volume.

You could link A to the output, but it's probably better to just divide everything by Q. More output is bad when you could get the same performance with less output. That obviously depends on the game type, though. Your equations all reward output, but in my eyes, output alone only makes things harder because it hurts shot time and pumping time. The effects of output - like range - are a separate concern.

In a sense, because shot time t = volume V / output Q, shot time is a good substitute for both those values.

My current formula is:

Code: Select all

[b]D * D * sqrt(t) / Q[/b]
[b]D[/b] = distance (range) in meters
[b]t[/b] = shot time in seconds
[b]Q[/b] = output in liters/second
Q is the most likely factor to change for different scenarios. iSoaker does soakfests so he gives Q a power of 1. I don't need that output so I gave it -1.

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cantab
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Re: Water gun power

Post by cantab » Sat Jun 28, 2008 5:53 pm

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.

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