Howto calculate density in the IPS system

(1)[tab]P = [ρ]gh

so for water in the IPS system:
[tab][ρ] = 62.4 lbm/ft[sup]3[/sup] / (1,728 in[sup]3[/sup]/ft[sup]3[/sup]) / 386.16 lbm/slinch = 0.000094 slinch[sup][†][/sup]/in[sup]3[/sup]
where:
(2)[tab]F=ma[tab]so[tab]F/m=32.18 ft/s[sup]2[/sup]*12 in/ft = 386.16 lb/slinch

which gives, for a one inch water column in an inertial frame on the earth's surface:

[tab]0.001122 slinch/in[sup]3[/sup] * 386.16 in/s[sup]2[/sup] = 0.0361 lb/in[sup]2[/sup]

which is about what you would expect.

[†][tab]"slinch" is a slang word for the mass unit in the IPS system since there is no formal word for mass in this system.

Note:[tab]People brought up in the English system confuse force with mass, while many people brought up in the metric system confuse mass with force, eg, torque wrenches calibrated in kg-m.
Nobody said the Imperial (IPS) units were easy.