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

**pseudo**efficiency 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.