What does 1714 mean in hydraulics? After reading this answer: http://www.answers.com/Q/What_does_1714_mean_in_hydraulics, I still do not have a good idea of what 1714 represents.  In fact, in the equations I was working with I saw a constant 0.0005834 and it took me until just now to realize that it is a another way to say 1/1714.
So what is this magical 1714, what does it represent, why is it needed? 
In short I am looking for "1714 represents ... " in plain English that a 9th grader can understand, and yet be correct and true to its purpose and place in life to satisfy a seasoned hydraulic engineer.
 A: 1 horse power = 33000 foot-pounds per minute (by definition)
1 US gallon = 231 cubic inch (by definition)
1 psi = 1 pound per square inch (by definition)
In the equation 
$$HP = k \Delta P F$$
where $F$ = flow rate in gallons per minute, $\Delta P$ is pressure difference in psi, and $HP$ is power in HP, you need a conversion factor. Doing everything in inches:
$$\frac{HP}{33000*12 *inch-pounds/min} = k\frac{\Delta P}{pounds/in^2}\frac{flow}{231 inch^3/min}$$
from which it follows that
$$k = \frac{231}{33000*12}\approx\frac{1}{1714}$$
In other words, "in plain English that a 9th grader can understand, and yet be correct and true to its purpose and place in life to satisfy seasoned physicists":

1714 represents the numerical scale factor needed to obtain pump power in HP given pressure in units of psi and flow rate in gallons per minute. It is not an exact number - only approximate."


Note - I had to hold my nose a bit while writing this answer, as working with non-SI units does not come naturally to me. But I think that it's a fair question - and NASA put a man on the moon with this system of units. No SI-based operation ever put a man on the moon. So here's to you, NASA!
A: 
So what is this magical 1714, what does it represent, why is it needed?

Some people use archaic units. That's all it means.
Suppose you measured pressure in pascals, or newtons per square meter, flow rate in cubic meters per second, and power in watts. With these units, power = pressure * flow rate. There is no scale factor.
A similar thing happens to Newton's second law. The sane thing to do is to express force in newtons, mass in kilograms, and acceleration in meters/second2. This results in the very nice form of Newton's second law, $F=ma$. Those who one insist on using customary units, with force in pounds force, mass in pounds mass, acceleration in feet/second2 have to deal with the uglier $F=kma$, where $k=1/32.174049$. That 1/32.174049 is just a consequence of using inconsistent units. The exact same applies to your magical 1714.
A: It's just an artifact of using different units for the same types of quantities in the same equation.
Suppose we have an ideal pump that puts a force $F$ on a fluid to have it move at a steady velocity $v$. The power required to do this is
$$ P = Fv. $$
Since pressure $p$ is force per unit area, then a flow with cross sectional area $A$ has $F = Ap$. At the same time, the volumetric flow rate is $q = Av$, so $v = q/A$. Thus we have
$$ P = pq. $$
This formula is right without modification, but of course you should be consistent when you plug in dimensional quantities. The units involved are
\begin{align}
P & \sim \mathrm{(mass)} \cdot \mathrm{(length)}^2 \cdot \mathrm{(time)}^{-3} \\
p & \sim \mathrm{(mass)} \cdot \mathrm{(length)}^{-1} \cdot \mathrm{(time)}^{-2} \\
q & \sim \mathrm{(length)}^3 \cdot \mathrm{(time)}^{-1}
\end{align}
If you choose a single base unit each for mass, length, and time, and you measure $P$, $p$, and $q$ according to the above products of those base units with no additional factors, then there will be no $1714$. This is the case, for example, in SI units, where $P$ is measured in watts, $p$ in pascals, and $q$ in cubic meters per second.
Unfortunately, hydraulic engineers once used (still use? apparently?) hodgepodge units. In this case we can write
$$ \frac{P}{1\ \mathrm{horspower}} = k \left(\frac{p}{1\ \mathrm{lb./in.^2}}\right) \left(\frac{q}{1\ \mathrm{gal./min.}}\right), $$
where
$$ k = \frac{(1\ \mathrm{lb./in.^2})(1\ \mathrm{gal./min.})}{1\ \mathrm{horsepower}}. $$
Now
\begin{align}
1\ \mathrm{horsepower} & \approx 745.7\ \mathrm{W} = 745.7\ \mathrm{kg\,m^2/s^3}, \\
1\ \mathrm{lb./in.^2} & \approx \frac{4.448\ \mathrm{N}}{(2.54\times10^{-2}\ \mathrm{m})^2} \approx 6.895\times10^3\ \mathrm{kg/m\,s^2}, \text{ and} \\
1\ \mathrm{gal./min.} & \approx \frac{3.785\ \mathrm{m^3}}{60\ \mathrm{s}} \approx 6.309\times10^{-5}\ \mathrm{m^3/s}, \\
\end{align}
so
$$ k \approx \frac{(6.895\times10^3\ \mathrm{kg/m\,s^2})(6.309\times10^{-5}\ \mathrm{m^3/s})}{745.7\ \mathrm{kg\,m^2/s^3}} \approx 5.834\times10^{-4} \approx \frac{1}{1714}. $$
That is, our equation $P = pq$ is equivalent to "power in horsepower divided by pressure in psi and flow rate in gpm is 1/1714."
