Lines don't actually have normals. You were right with the cross prod idea, but you need a plane for that. Is it reasonable to define a plane instead of a line for this problem?

EDIT: Not sure if this is what you are asking but you can draw a circle in 2D space with the following:

	// Calculate how much we have to increment the angle by each frame	float fAngInc = 360 / a_iTess;	float fCurAng = 0.0f;	// Set the generic attributes	for (int count = 0; count < a_iTess + 1; count++)	{		// Position		m_pvertsCircle[count].m_v3Translation.x = a_v3Pos.x + (float)sin((double)DEG2RAD(fCurAng)) * a_fRad;		m_pvertsCircle[count].m_v3Translation.y = a_v3Pos.y + (float)cos((double)DEG2RAD(fCurAng)) * a_fRad;			// Increment the angle for the next round		fCurAng += fAngInc;	}

Matt

[Edited by - matthughson on December 3, 2005 11:30:30 PM]

Quote:Original post by bkenwright
Now I want to get a normal which is perpendicular to this line...how would I go about doing this?

There's an infinite number of vectors (normals) perpendicular to a line. I presume you are okay with any one of them.

You were thinking about using the cross product, which was the right idea. You have a vector in the direction d, d = p2 - p1, of the line. If you had another vector v, not parallel to d, the cross product u of d and v, u = cross(d, v), would give a vector perpendicular to d (as well as v).

Thus, all you have to do is pick any vector v not parallel to d and compute the cross product. For example, you can pick one of the axis vectors (1,0,0), (0,1,0), or (0,0,1).

For reasons of numerical robustness you want to try to pick the vector v such that it is as perpendicular to d as possible (or as little parallel as possible, if you prefer).

One way to pick between (1,0,0), (0,1,0), and (0,0,1) is to look at the component of d that has the largest magnitude, here largest absolute value. Let's say that the x component has the largest absolute value. This means that d points more in the direction of the x-axis than along the y or z axis. Thus, don't cross d with (1,0,0) because they are the most parallel. Similarly, don't cross d with the axis with which it is secondmost most parallel. Instead cross d with the axis it is the least parallel with, which must be the axis corresponding to the component of d with the least magnitude!

Thus, your code would look something like this (untested):


If you want the result normalized, you would have to normalize the vector returned by this function.

Quote:Original post by jyk
We can now adapt Matt's code to suit your needs ('hope you don't mind, Matt...):

Heh, not at all, glad it could be used [wink]

@OP: If this stuff still isn't helping, maybe draw what you want the final result to look like and post it here.

Matt