$P=ρgh$ - intuitive understanding of the equation? I've come across this equation recently which relates pressure with the product of density, gravitational acceleration and height difference in a medium. 
I understand that


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*$P = ρgh$ expands to $\frac{m}{V}g(h_2-h_1)$


Therefore the $V$ (in $\mathrm{m}^3$) in the denominator gets "reduced" by $h$ (in $\mathrm m$) and becomes a surface area $A$ (in $\mathrm m^2$), breaking down to the definition of pressure:


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*$P = \dfrac{F}{A}.$


So, it's clear that mathematically it works, of course... but in my head, not so much. Could someone give me some intuitive way how to think about it?
How do density, gravitational acceleration and height of a column of a medium "give" pressure?
$$\frac{mg}{V}h\to\frac{F}{V}h\to\frac{Fh}{V}\to\frac{W}{V}$$
I tried doing this up here and I ended up with energy density, it seems... I'm at a loss how to think about this, as you can see. Hah, I've just checked Wikipedia, it seems that pressure and energy density indeed share the same units and at times could be considered synonyms. Well, that's neat, but still doesn't help me.
Much obliged!
 A: This tells you how much pressure there is in a fluid of density $\rho$ at a depth of $h$, on the Earth's surface. The reason it works is because the fluid at depth h has to hold up the fluid above it. If you look at a column of fluid with cross section area A stretching from h up to 0, the mass of the fluid is $\rho Ah$, its weight is $\rho A g h$, and this is the force holding it up. Dividing by the area gives the pressure at depth h.
A: One possible example way to make this more intuitive is estimating the weight of the atmosphere. 
We know that under 10 m of water the pressure in our lungs roughly doubles.
So a 10 m water column ($\rho_{water}=1 g/cm^3, \rho_{air} =$  dependent on altitude) is equal in weight to the atmosphere above it. 
To estimate the total weight of the atmosphere (if you e.g. want to compare the observed annual  rise in $CO_2$ of 2 ppm with the total worldwide emissions), you just multiply the surface of the earth with 10m and use the density of water.
So the reason we substituted mass with density is that the area of the water /atmospheric column just doesn't matter.  
A: Your question is "How do density, grav. acceleration and height of a column of a medium "give" pressure?"
It is a one-to-one mapping between pressure exerted by a column of water and its height. 
Building on the earlier Ron Maimon's answer, the intuition here is that,there is a one-to-one mapping between pressure and height. For example, in case of water, where the density is always 1 and g, of course is constant, the pressure of 14.7 PSI is "equivalent" to 33.4 feet of head. Like wise, for every value of pressure in PSI, there is one and only equivalent feet of head. Hence pressure and head are in that sense equivalent ways of talking about the same thing.
A: If you use something like Moon's units (dm, kg, ds), then  $\rho=1\mbox{ kg/dm}^3$, and $g=0,980665\mbox{ dm/ds}^2$.  So a dm of water, corresponds to a kilogram per sq decimetre, and since a kilogram * a leo (an acceleration unit = 10 m/s²), is a dekanewton.
So a deciametre of water = 1 kgf/dm² = 1 pieze (kilopascal).  Note this is the same measure that watch-makers give for 'our watch is tested to 300 ft of presure.  The torr uses the same idea but with mercury, which is 13.6 times denser (ie 1 cm Hg = 13.6 cm Water.  
Wilberforce Mann suggested these units in the LinnBase system, under the names Linn (dm), Arr (dm²), Soll (dm³), Capp (L), and Soll (kg), along with Monn (currency).  Using Temm (time), Dynn (force), etc, one can replicate Moon's system into something really interesting.
Atmosphere is about 100 of these units, so 100 dm = 10 m,  100 kgf/dm² = 1 kgf/cm², and 100 pieze = 1 bar.  
Pendlebury designed a whole system (TGM = timm grafut masz) based on this particular relation, using the hour, divided into powers of 12 to derive the time unit.  It has a good deal of acceptance at the duodecimal society, but one can construct versions for other number systems too.  
You can include the other 'rubbery constant' of the joule constant into this mix, so that Joule's mechanical equivalent of heat (778 ft/defF) becomes converted at 1 degF = 778 feet, so converted.  The unit is more or less something like a 'second F', where 60" = 1', 60'=1 deg.  Raising something by 1 dm against gravity, is then a second of Heat, where 100 K is 116.3 degrees.  
