Learn: Centre of pressure — vertical plane
Water pressure increases with depth. On a submerged vertical plate, the lower part therefore pushes harder than the upper part, so the resultant force acts below the area centroid.
What the formula is saying
Start with centroid depth h. Add I/(A h), a positive correction representing the unequal distribution of hydrostatic pressure. I measures how the plate area is spread vertically about its horizontal centroidal axis.
Read the symbols in plain language
- h̄
- Centroid depth
Centroid depth. The second moment divided by area times centroid depth produces an additional depth.
mMetres measure length; 1 m = 1000 mm.
- IG
- Centroidal second moment
The area-weighted square of distance from the stated axis. It measures the spread of the cross-section, not its area or mass.
m⁴The fourth power of metres is used for a second moment of area; 1 m⁴ = 10¹² mm⁴.
- A
- Area
Area. The second moment divided by area times centroid depth produces an additional depth.
m²Square metres measure area; square the length conversion factor.
- hcp
- Result to find
Centre of pressure — vertical plane. Add the positive depth correction below the centroid measured from the same water surface.
m
Sort out the units first
h is measured vertically from the free water surface in m; A is the wetted plate area in m²; I is its area second moment in m⁴. I/(A h) has unit m, so it can be added to h.
Assumptions before calculating
Assume a plane vertical plate, hydrostatic liquid, a free surface at uniform atmospheric pressure, and pressure measured relative to that atmosphere. Use the centroid and second moment of the entire submerged area.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find hcp and explain the result in the stated output unit.
- h̄ · Centroid depth
- 3 m
- IG · Centroidal second moment
- 1.333 m⁴
- A · Area
- 4 m²
Calculate the pressure-distribution correction
The second moment divided by area times centroid depth produces an additional depth.
(1.333) ÷ ((4) × (3)) ≈ 0.1110833333 mLocate the centre of pressure
Add the positive depth correction below the centroid measured from the same water surface.
(3) + (0.1110833333) ≈ 3.111083333 m
Does this worked answer make sense?
The calculated depth must be greater than or equal to h. If the same small plate is moved much deeper, the correction I/(A h) decreases and pressure becomes relatively more uniform across it.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- h̄ · Centroid depth
- 2 m
- IG · Centroidal second moment
- 0.6666666667 m⁴
- A · Area
- 2 m²
Calculate the pressure-distribution correction
The second moment divided by area times centroid depth produces an additional depth.
(0.6666666667) ÷ ((2) × (2)) ≈ 0.1666666667 mLocate the centre of pressure
Add the positive depth correction below the centroid measured from the same water surface.
(2) + (0.1666666667) ≈ 2.166666667 m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- h̄ · Centroid depth
- 4 m
- IG · Centroidal second moment
- 2.25 m⁴
- A · Area
- 3 m²
Find: Learn: Centre of pressure — vertical plane
A hint, not the answer
Start with centroid depth h. Add I/(A h), a positive correction representing the unequal distribution of hydrostatic pressure. I measures how the plate area is spread vertically about its horizontal centroidal axis.
h is measured vertically from the free water surface in m; A is the wetted plate area in m²; I is its area second moment in m⁴. I/(A h) has unit m, so it can be added to h.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate the pressure-distribution correction
The second moment divided by area times centroid depth produces an additional depth.
(2.25) ÷ ((3) × (4)) = 0.1875 mLocate the centre of pressure
Add the positive depth correction below the centroid measured from the same water surface.
(4) + (0.1875) = 4.1875 m
Avoid the common trap
Do not use the depth of the top edge as h, or the mass moment of inertia as I. Do not subtract the correction: pressure is larger lower down, so the resultant lies deeper than the centroid.
When this method applies — and when it does not
This expression is for a vertical plane, not an arbitrary inclined or curved surface. Inclined plates need an inclination factor and curved surfaces need force-component analysis. It locates the force; it does not calculate plate stress or support reactions.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Centre of pressure — vertical plane. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.