Say I were calculating the pressure under 1 foot of water, in psi.

I'm thinking the pressure is force per unit area, and should be equal to the force exerted by the column of water standing over 1 square inch. So

$$F = ma~~~~~~~~~~~~~~~~~~$$ $$~~~~~~~~~~~~~~~~~= volume\times density\times g$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 1 inch \times 1 inch \times height(in inches)\times density \times g$$ $$=h\rho g~~~~~~~~~~~~$$

The Pressure under 1 foot of water is $12 inch^3 \times 0.036 pounds/in^3\times 386 inches/s^2$ which gives me $166.75~\text{pound-force per square inch}$. However, various internet references tell me that 1 foot of water exerts $0.43 \text{ psi}$. Eg. Wolfram Alpha

They seem to be leaving out acceleration due to gravity entirely. Why?


3 Answers 3


As John Rennie explained, in the American Engineering System, force is expressed in lb$_f$, mass in lb$_m$, and acceleration in ft/sec$^2$. This system is not coherent. Hence, a conversion factor other than one must be used in the equation for force; that is, $F=\frac{ma}{g_c}$, where $g_c=32.174 \frac{lbm \cdot ft}{sec^2 \cdot lbf}$ is a constant, known as the gravitational conversion constant.

Derivation of $g_c$

The principle of conservation of units is used to derive $g_c$. We wish to convert $F=ma$ from SI to AES.

First, find the dimension of the "hidden constant" in the given equation. In other words, $1=\frac{F}{ma}$ has dimension force divided by the product of mass and acceleration. Next, consider the units of the hidden constant -- the given units are $\frac{N \cdot s^2}{kg \cdot m}$ and the required units are $\frac{lbf \cdot sec^2}{lbm \cdot ft}$.

Now, convert the given units of the hidden constant to the required units. Thus, $$\left(\frac{1 N \cdot s^2}{kg \cdot m}\right) \left(\frac{0.45359 kg}{lbm}\right) \left(\frac{0.3048 m}{ft}\right)\left(\frac{lbf}{4.4482 N}\right)=\frac{1}{32.174}\frac{lbf \cdot sec^2}{lbm \cdot ft}$$

The hydrostatic pressure is calculated as follows:

$$p=D\frac{(\rho g_c)}{144g} \ ft \frac{ft^2 \ lbm \ ft \ sec^2 \ lbf}{in^2 \ ft^3 \ lbm \ ft \ sec^2}$$

where $p$ is hydrostatic pressure in psig, $D$ is depth (or height) of the fluid in ft, $\rho$ is the fluid density (lbm/ft^3). Note that the density of fresh water is 62.4 lbm/ft^3, $g_c$ is the gravitational constant of acceleration (32.2 ft/sec^2), and $g$ is units conversion to lbf (32.2 $\frac{lbm \ ft/sec^2}{lbf}$)

In AES the fluid density is often express in lbm/gal. There is 7.48 gal/ft^3, therefore $p=0.05194D\rho$.

It is often convenient to express the hydrostatic pressure as a fluid pressure gradient or hydrostatic pressure developed per unit height of fluid. $\nabla p =0.052\rho$, where $\nabla p$ is the hydrostatic pressure gradient (psig/ft). For fresh water the pressure gradient is: $$\nabla p=0.052(8.33)=0.433\ \text{psi/ft}$$


Sticking to SI units for now, the unit of force is the Newton while the unit of mass is the kilogram. To convert kilograms to Newtons you multiply by $g$.

So if we express a pressure in Newtons per square metre we multiply the mass by $g$. But if we were going to express the force in kilograms per square metre we wouldn't multiply by $g$. However in SI units we never express force in kilograms per square metre so there is never this ambiguity. We always multiply by $g$.

The trouble with Imperial units is that the pound is used both as a measurement of mass and a measurement of force, and too frequently it is not made clear which is being used. In your example the pressure is in pounds per square inch where the pound is the unit of mass, so you don't multiply by $g$. The PSI unit means the force that a one pound mass would exert in the gravity at the Earth's surface.

Sometimes the pound force, or lbf, is used as a unit of force, but this is even more confusing because in these units $g = 1$ i.e. one pound generates a force of 1 pound force. So if you express pressure in units of pound force per square inch you do have to multiply by $g$, but $g = 1$ so it makes no difference.

There's a good reason physicsts like using SI units, and it's not because we just want to be different.

  • 1
    $\begingroup$ Thank you, that helps, a bit. Damn imperial units. g=1 in what units? $\endgroup$
    – RaGe
    Commented Nov 5, 2014 at 18:13
  • $\begingroup$ g = 1 in what units? - good question, but I don't even want to think about it! :-) $\endgroup$ Commented Nov 5, 2014 at 18:23
  • 2
    $\begingroup$ This is why satellites don't get to Mars... $\endgroup$
    – ProfRob
    Commented Nov 5, 2014 at 22:01
  • $\begingroup$ Don't forget poundals and slugs... $\endgroup$
    – DJohnM
    Commented Feb 1, 2016 at 18:53
  • $\begingroup$ @JohnRennie, I agree. I'm a retired engineer in the U.S., and even I don't like imperial units. $\endgroup$ Commented May 6, 2016 at 2:26

Let's take a look at this in a very simple way. At 60 F, 1 ft^3 of water weighs 62.37 lbs. that weight is resting on the base of 144 in^2.

The force (weight) of the one foot high water on one square inch is:

             P = (62.37 lbs)/(144 in^2) = 0.433125 psi.
  • $\begingroup$ Thank you, but this doesn't help me much. You're saying 62.3 pounds of water exerts 62.3 pounds of force. My question really is, why isn't force = 62.3 x acceleration due to gravity (in the appropriate units, of course). Where does g go? $\endgroup$
    – RaGe
    Commented May 6, 2016 at 1:34

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