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Please help me with the following. I want to know if there is an equation/set of equations to find out the projected area of a (3-D) cube when it is oriented at different angles of attack to the fluid flow, rendering the velocity vector of the impinging fluid on the cube surface to change with each angle of attack. In particular, what is the projected area when the cube is oriented at such an angle that the flow velocity vector passes through one of the vertex and the geometric centre of the cube? Thanks.

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Which hydrodynamic problem are You going to calculate? There are few which can be solved with such a precision to need exact projected area. In practice one regularly just takes the diameter of "equivalent sphere". – Georg Jun 6 '11 at 13:32
up vote 3 down vote accepted

The projected area is $\sqrt{3} \ l^2$

Imagine the long diagonal of a cube. This diagonal is the same direction as the velocity vector you're interested in.

The length of the diagonal is $\sqrt{3}l$, with $l$ the side length of the cube. This come from the Pythagorean Theorem.

If we project this diagonal onto a side, we get simply $l$. Therefore, the projection operation at this angle divides by $\sqrt 3$.

The cube has three faces exposed to a fluid flowing coming in from the direction of the long diagonal, each with area $l^2$. Each of these is projected onto the plane perpendicular to the fluid flow, so the projected area is $(3/\sqrt{3})\ l^2 = \sqrt{3} \ l^2$.

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General Case $$l^2 (|\hat{n_1} \cdot \hat{u}|+|\hat{n_2} \cdot \hat{u}|+|\hat{n_3} \cdot \hat{u}|)$$ where $\hat{u}$ is the unit vector of the fluid flow and $\hat{n_i}$ is the normal vector to one of the cubes three orthogonal sides. – David May 22 '11 at 1:32
@David: you might want to make that an answer. (If you also explain where it comes from, it'll be an even better answer.) – David Z May 22 '11 at 2:41

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