Area of a Plane below a Plane

Question

this is the first question I've asked on StackOverflow, so hopefully I'll do it properly.

I'm trying to simulate paddle physics, i.e. the force generated by moving a paddle through water. The most important data for this calculation is the submerged surface area of the paddle.

How can I calculate the surface area of a plane beneath (intersecting) another plane?

• The y of the "water" plane is always 0
• The coordinates (x,y,z) of the "paddle" plane's corners are always known
• The "paddle" plane may be at any depth, i.e. 0, 1, 2, 3 or 4 corners may be beneath the "water" plane (of course the 0 & 4 cases are trivial)
• I have access to many functions for operations on the coordinates (multiply, divide, distance, etc.)

Thank you so much.

Answer 1

You must find the points $P$ and $Q$, on $BC$ and $AD$ respectively, having $y=0$. To find the distance of $P$, for instance, form $B$ and $C$, remember that $BP/CP=|B_y/C_y|$. Once you know $CP$ and $DQ$ it is easy to find the area of trapezium $CDQP$. The cases when 1 or 3 corners are underwater can be treated in an analogous way.

Answer 2

If only one corner is dept, for example D: $$\frac 1 2|\frac{y_D}{y_D-y_A}\vec{DA}\times \frac{y_D}{y_D-y_C}\vec{DC}|$$ If D and C are dept, F.E. D is deeper: $$\frac 1 2|\frac{y_D-y_C}{y_D-y_A}\vec{DA}\times \vec{DC}|+|\frac{y_C}{y_D-y_A}\vec{DA}\times \vec{DC}|$$ This is for the case if your paddle is some kind of parallelogram