6 replies to this topic

[~Helgon]

Hey, i have a little problem. I want my look vector to follow the up and downs of the terrain.

 

Can someone give me a hint how to accomplish it?

 

As in the picture for example. If i move onto the incline i want that the look vector moves the same pitch amount up. I've read that it can be done with a binormal vector but I'm not sure where to start or if it is even right (I've already tried so much :/)

 

Maybe someone can give me a hint. Would really appreciate it

 

Regards

 

[Morphex]

Well, if you have the terrain normal its trivial isnt it? Since you can just Dot Product it to get the inclination of the slope. After than just calculate the Look Vector.

[dAND3h]

Equally, you can rotate "the object you are moving" 's UP vector until it points in the same direction as the terrains normal. Then do a cross product on the new UP vector and the right vector, which will give you the look vector.

Edited by dAND3h, 21 December 2012 - 12:21 PM.

[haegarr]

A variation of the standard way to compute a look-at matrix works here as well: If m denotes the movement direction (e.g. projected onto the x-z plane) and n the terrain normal at the current position, then
r := m cross n
denotes the side vector perpendicular to both m and n. Then
f := n cross r
denotes the forward vector that can be used as look-along vector as requested. It usually need to be normalized, though.

If, on the other hand, the normal isn't available because you're working with a height map, then this topic may help you. (Please notice that my first posts therein haven't considered triangulation correctly, but down from post #15 inclusive things are done well). After computing the height at the current position and the height a bit in direction of the movement vector, then the normalized difference vector is the result.

BTW: To be pedantic: A bi-normal is a construct that can be computed at locations on a line. In case of a surface the correct term is bi-tangent.

Edited by haegarr, 21 December 2012 - 12:51 PM.

[JohnnyCode]

the way I would do it is I would track last position of the object, and subtract last position from current position. You then get a vector that is a derivate of path that object follows. This vector then would be new lookat vector...Also I would never set up vector other than 0,1,0!

[Steve_Segreto]

Here's a link that can help you find the height at a particular point - the same concept can be used to find the normal of the terrain at a particular (non-grid) location.

 

http://softwright.com/faq/support/Terrain%20Elevation%20Interpolation.html

 

The concept is called bi-linear interpolation.

[haegarr]

Not using the pitch but the method similar to "look-at" has a reason: It avoids to concatenate a rotation but instead is in itself a full-blown solution, i.e. it computes the end result instead of a delta that has to be applied. You can find parallels in some dozen threads about the look-at computation here in GDnet everywhere.

 

Looking at the original code snippet

 

XMVECTOR incline = XMVector3Dot(mCam.GetLookXM(), mTerrain.GetNormal(camPos.x, camPos.z));
...

then mCam.GetLookXM() gives the vector m, and mTerrain.GetNormal(camPos.x, camPos.z) gives the vector n. You can compute all the mRight, mUp, and mLook from them.

 

 

Of course, also computing the matrix directly as above suffers from abrupt changes when the camera crosses edges of adjacent triangles with noticeably different normals. If you want smoothness in those situations, a possibility may be what I've described in the previous post: Assume that the height field is spaced 10 by 10 length units. Assume further that you use a forward vector with a length of 2 length units (i.e. a scaled mCam.getLookXM() vector). Then compute the position (inclusive height) of the camera as well as of the point at the end of the forward vector but again projected onto the terrain. Then use the difference vector as new look-along vector. As long as the camera is more or less in the middle of a triangle, the result will be the same as with the other methods of computation. But when the end of the forward vector falls beyond an edge of the triangle (i.e. in the given example it is closer than 2 length units to the edge), then the other triangle influences the result, and that the more the camera gets closer to the edge. Just after crossing the edge, the new triangle has 100% influence on the result. If you are heading the camera to look back than the just left triangles gets most of its influence back.