The plane is given by three points(P1, P2, P3). It's 3D. How to get the the steepest slope direction vector on this plane? It looks like a high school geometry problem. But I forget those things. thanks!

Hi Patrick ,

Create two vectors, Vector 1 = P2-P1 Vector 2 = P3-P1 The cross product of those two vectors will give you vector which represents the normal direction ( steepest gradient).

The above answer is for a single plane with its point representations. When an entire flow field is taken into consideration, the implementation differs depending upon the problem taken into consideration.

cheers karthik

That's normal direction. What I mean is the steepest slope alone the plane. Thank you anyway.

If h is the height of the plane above a reference plane, the steepest slope is in the direction of grad(h).

Unless you specify some reference point, line, or plane, the expression 'steepest slope' has no meaning. Let's assume that your reference plane is the x-y plane at z=0, so you are interested in the steepest descent to zero altitude (z). The steepest descent from any point in space to this plane is simply described by a unit vector in z-direction (either positive or negative, depending on the position of that point). Now, all you need to do is to project that vector onto your plane. In general: Given a reference plane, find the normal vector on that reference plane, and project it onto your plane.

Hi Patrick,

Find the gradient of the flow paramater ( eg velocity ) to determine its steepest slope. Consider the stencil arrangement of 9 cells in 2D and 27 in 3D around the partcular node of a cell in which the steepest slope has to be calculated.

cheers Karthik

"Given a reference plane, find the normal vector on that reference plane, and project it onto your plane." But how to project? In which direction? That is the problem.

Now I derived the expression my self. Suppose the unit normal vector n(A,B,C). Here A,B,C is the vector component. (I use y not z as the elevation direction, sorry) Then the steepest slope in the plane is: (A*B/sqrt(A^2+C^2), -sqrt(A^2+C^2), C*B/sqrt(A^2+C^). (check the signs, not so sure)

For example: A plane parallel to the z axis, and has a angle of 45 degree with XZ plane. The unit normal vector(1/sqrt(2), 1/sqrt(2), 0), then the steepest slope direction (-1/sqrt(2), 1/sqrt(2),0).

Thank you all guys!

"But how to project? In which direction? That is the problem."

The projection of a vector onto a plane is uniquely defined. There is no ambiguity about "directions". And that's exactly what you're doing with your formula.