Center of Pressure on a Plane

y_{cp} = y_{c} + \frac{I _{x c} sin θ}{A y _{c}}

Location of the hydrostatic resultant, measured along the plane from the free-surface intersection. The extra I_xc sinθ /(A yc) term is nonnegative, so the center of pressure lies below the centroid unless the plane is horizontal (sinθ = 0). Deeply submerged, CP approaches the centroid.

$y_{cp}$ (m): Center-of-pressure coordinate along the plane
$y_{c}$ (m): Centroid coordinate along the plane
$I_{x c}$ (m⁴): Centroidal second moment about an axis parallel to the surface
$θ$ (deg): Angle of the plane from the horizontal
$A$ (m²): Area

Step-by-step solved example

Vertical rectangle 1.2 m wide × 2.0 m tall, top edge at the free surface. Locate y_cp from the surface.

y_cp sits below the centroid because pressure increases with depth.

1. Centroid and I
yc = 1.00 m, A = 2.40 m², θ = 90°, I_xc = bh³/12 = 1.2×8/12 = 0.800 m⁴.
2. CP
y_cp = 1.00 + 0.800/(2.40×1.00) = 1.00 + 0.333 = 1.33 m (2/3 of the height).
$y_{cp} = 1.33 m$

Answer: 1.33 m below the surface

10 practice questions

0/10 correct

1.On a vertical plane the CP is

2.For a horizontal plane, sinθ = 0 so y_cp

3.I_xc is

4.The extra term I_xc sinθ /(A yc) is

5.Vertical rectangle with top at the free surface: CP is at

6.As depth → ∞, I/(A yc) → 0 so CP

7.The y in the CP formula is measured

8.The resultant magnitude is still

9.A = 4.0 m², yc = 3.0 m, I_xc = 2.0 m⁴, θ = 90°. y_cp =

10.An inclined gate: pressure uses vertical depth, while y_cp is