The following calculations show us how to convert between the head and pressure of a known mass in both metric and Imperial units.
Head (h) is the height of a column of the matter of unit - cross-section - area that generates pressure 'p' at its base.
Pressure (p) is the pressure at the base of a column of matter of height 'h'
The following variables and units are used here:
head (h) {m or ft}
mass-density (ρ): kg/m³ or lb/ft³
mass-pressure (p): kg/m² or lb/ft²
force-pressure (p): N/m² or lbf/ft²
mass-pressure: p = h.ρ {m . kg/m³ = kg/m² or ft . lb/ft³ = lb/ft²}
force-pressure: p = h.ρ.g {m . kg/m³ . m/s² = kg.m/s² / m² = N/m² or ft . lb/ft³ . ft/s² = lb.ft/s² / ft² (lbf/ft²)}
head =10m; density = 1000kg/m³; gravitational acceleration = 9.80663m/s²
mass pressure
metric; p = 10 x 1000 = 10000 kg/m²
Imperial; p = 32.80839895 x 62.42796058 = 2048.1614363435 lb/ft²
force pressure
metric; p = 10 x 1000 x 9.80663 = 98066.3N/m²
Imperial; p = 32.80839895 x 62.42796058 x 32.17398294 = 65897.51111 lb.ft/s²/ft² (lbf/ft²)
mass-pressure: h = p/ρ {m . kg/m³ = kg/m² or ft . lb/ft³ = lb/ft²}
force-pressure: h = p / g.ρ {m . kg/m³ . m/s² = kg.m/s² / m² = N/m² or ft . lb/ft³ . ft/s² = lb.ft/s² (lbf)}
mass-pressure = 10332.2956kg/m²; density = 1.2928kg/m³; gravitational acceleration = 9.80663m/s²
mass pressure
metric; h = 10332.2956 ÷ 1.2928 = 7992.184097m
Imperial; h = 2116.21674 ÷ 0.08070687 = 26221.01985ft
force pressure
metric; h = 101325 ÷ 1.2928 ÷ 9.80663 = 7992.184097m
Imperial; h = 68087.11176 ÷ 32.17398294 ÷ 0.08070687 = 26221.01985ft
You will find further reading on this subject in reference publications(3 & 12)