position = centre;
normal = normal_Surface;
Vector3 binormal = Vector3.Cross(normal, tangent); //up=tangent?
//=============================================
//------- [Calculate Boundary Planes] ---------
//=============================================
float d = Vector3.Dot(centre, tangent);
plane_Left = new Vector4(tangent, width * 0.5f - d); // Left Plane (As Vector4)
plane_Right = new Vector4(-tangent, width * 0.5f + d); // Right Plane
d = Vector3.Dot(centre, binormal);
plane_Bottom = new Vector4(binormal, height * 0.5f - d); // Bottom
plane_Top = new Vector4(binormal, height * 0.5f + d); // Top
d = Vector3.Dot(centre, normal);
plane_Front = new Vector4(-normal, depth + d); // Front
plane_Back = new Vector4(normal, depth - d); // Back
How to construct an OBB/AABB from the following 6 planes?
Author
I have the following 6 planes:
The problem is that if I want to convert the combination of the planes into a bounding volume for testing against triangles how would I find the rotation needed for a OBB, or would an AABB suffice.
Both the OBB and AABB are constructed via a central point (which I can take from the centre paramter in above code) and Extents (width/2.0f, height/2.0f/, depth2.0f).
For the AABB it should be easy enough to use the provided width, height and depth to create an AABB.
However, in the case of the OBB it also requires a rotation matrix.
How do I obtain the rotation matrix using the above case?
The OBB will be used to determine which triangles intersect it for displaying a decal on a curved surface. If it is not necessary to use an OBB in this case please let me know.
Just some ideas:
Assuming the only transformations are rotation/translation then each plane pair (opposite sides) is the transformed version of either x/y, x/z, y/z planes.
You could get each corner point of the box by solving for the intersection of 3 planes (left, bottom, back intersection would give that corners point). Once you have those you could use 3 of the co planar ones (say 3 points on the left plane) along with that planes normal to work out a basis matrix.
This makes some assumptions that might not be true. Lets say where you say "left plane" you know that that plane will be the z/y plane with minimum x value. You can use the point, bottom, left, near to bottom, left, far to determine the direction of the z axis in your transformed space. Do the same thing to get the y axiz (you already know the x axis, it is the plane normal).
Those 3 vectors now should form an orthogonal basis (providing no scaling was used) which is pretty much a rotation matrix.
If you don't know which plane represents which side of your box then there are any number of rotation matrices that could be the right one. Can't think of how to get that down to a unique solution.
Assuming the only transformations are rotation/translation then each plane pair (opposite sides) is the transformed version of either x/y, x/z, y/z planes.
You could get each corner point of the box by solving for the intersection of 3 planes (left, bottom, back intersection would give that corners point). Once you have those you could use 3 of the co planar ones (say 3 points on the left plane) along with that planes normal to work out a basis matrix.
This makes some assumptions that might not be true. Lets say where you say "left plane" you know that that plane will be the z/y plane with minimum x value. You can use the point, bottom, left, near to bottom, left, far to determine the direction of the z axis in your transformed space. Do the same thing to get the y axiz (you already know the x axis, it is the plane normal).
Those 3 vectors now should form an orthogonal basis (providing no scaling was used) which is pretty much a rotation matrix.
If you don't know which plane represents which side of your box then there are any number of rotation matrices that could be the right one. Can't think of how to get that down to a unique solution.
Quote:How do I obtain the rotation matrix using the above case?You already have it (the basis vectors are the normal, binormal, and tangent vectors in your example code).
To compute an AABB from an OBB, you can simply compute the projection of the OBB onto the cardinal axes. (Don't have time to post the details at the moment, but post back if you have questions about this.)
Author
I thought I might be making this too hard.
So is the rotation matrix as follows :
Matrix rotation = (tangent, binormal, normal)
Or is local space different to tangent space for rotation?
I'm not sure I am going to need an AABB but can't I just ignore the rotation matrix for an AABB, or is the projection method the only option?
Thanks for your help.
So is the rotation matrix as follows :
Matrix rotation = (tangent, binormal, normal)
Or is local space different to tangent space for rotation?
Quote:
To compute an AABB from an OBB, you can simply compute the projection of the OBB onto the cardinal axes.
I'm not sure I am going to need an AABB but can't I just ignore the rotation matrix for an AABB, or is the projection method the only option?
Thanks for your help.
Quote:I thought I might be making this too hard.Yes, that should be right, assuming you've placed the right vectors in the right columns/rows.
So is the rotation matrix as follows :
Matrix rotation = (tangent, binormal, normal)
(There is the question of whether it should be 'binormal' or 'bitangent', but that's another issue.)
Quote:I'm not sure I am going to need an AABB but can't I just ignore the rotation matrix for an AABB, or is the projection method the only option?If you simply ignore the rotation, the AABB may not fully enclose the clipping volume (draw some example diagrams if you're unsure why this is the case).
AFAIK at least, the projection method is the fastest and most straightforward way to compute an AABB from an OBB.
Author
Quote:
Yes, that should be right, assuming you've placed the right vectors in the right columns/rows.
Well the matrices in XNA are row-major matrices.
So therefore:
rotation.M11 = tangent.X; rotation.M12 = tangent.Y; rotation.M13 = tangent.Z; rotation.M21 = binormal.X; rotation.M22 = binormal.Y; rotation.M23 = binormal.Z; rotation.M31 = normal.X; rotation.M32 = normal.Y; rotation.M33 = normal.Z;Where a 4x4 row-major matrix =
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
Is this correct or should I arrange them in columns?
EDIT: If I wanted to get the yaw, pitch and roll from this rotation matrix, how would that be found?
It looks like you have some of your terminology confused.
You certainly can extract them if needed, but it's kind of a pain (you have to get all of the conventions right - including rotation axis order - or the results will likely be incorrect).
Also note that Euler-angle triples can alias one another (that is, two sets of Euler angles can produce the same orientation), so the Euler angles you get out won't always be the same as the angles you put in.
Quote:Well the matrices in XNA are row-major matrices.That doesn't have anything to do with whether the matrices are row major or column major. It does however show that your matrices are set up to work with row vectors rather than column vectors.
So therefore:rotation.M11 = tangent.X; rotation.M12 = tangent.Y; rotation.M13 = tangent.Z; rotation.M21 = binormal.X; rotation.M22 = binormal.Y; rotation.M23 = binormal.Z; rotation.M31 = normal.X; rotation.M32 = normal.Y; rotation.M33 = normal.Z;
Quote:Where a 4x4 row-major matrix =And that doesn't relate in any way to majorness or vector notation convention :) (AFAIK at least, matrices in linear algebra are always indexed by row and then column.)
11 12 13 14
21 22 23 24
31 32 33 34
41 42 43 44
Quote:Is this correct or should I arrange them in columns?That depends on the conventions of the math library you're using. I haven't used XNA, but the DirectX math library uses row-major storage and row vectors, so if XNA uses the same conventions as DirectX, then you're good.
Quote:If I wanted to get the yaw, pitch and roll from this rotation matrix, how would that be found?First, the usual question: what do you need the Euler angles for?
You certainly can extract them if needed, but it's kind of a pain (you have to get all of the conventions right - including rotation axis order - or the results will likely be incorrect).
Also note that Euler-angle triples can alias one another (that is, two sets of Euler angles can produce the same orientation), so the Euler angles you get out won't always be the same as the angles you put in.
Author
Quote:
First, the usual question: what do you need the Euler angles for?
I don't sorry, it seems I just needed to use a Quaternion.CreateFromRotationMatrix() method which I overlooked by mistake.
If I want to compute a tight AABB from a rotation matrix and a central point, how would I go about doing that correctly?
This currently works for non rotated OBBs but I get the feeling it is not correct as you mentioned a projection method being involved:
public static CD_AxisAlignedBoundingBox ComputeFromRotationMatrix(Matrix rotation, Vector3 centre) { CD_AxisAlignedBoundingBox aabb_New = new CD_AxisAlignedBoundingBox(centre, 0, 0, 0); // +ve halfwidth extents = radius aabb_New.Extents.X += Math.Abs(rotation.M11); aabb_New.Extents.X += Math.Abs(rotation.M12); aabb_New.Extents.X += Math.Abs(rotation.M13); aabb_New.Extents.Y += Math.Abs(rotation.M21); aabb_New.Extents.Y += Math.Abs(rotation.M22); aabb_New.Extents.Y += Math.Abs(rotation.M23); aabb_New.Extents.Z += Math.Abs(rotation.M31); aabb_New.Extents.Z += Math.Abs(rotation.M32); aabb_New.Extents.Z += Math.Abs(rotation.M33); return aabb_New; }Quote:If I want to compute a tight AABB from a rotation matrix and a central point, how would I go about doing that correctly?You would need a set of extents in addition to a center point and rotation matrix (which I think you already know).
All you really need is a function to project an OBB onto an axis. No guarantee I'll get this right, but:
float center = dot(box.center, axis);float radius = abs(dot(box.axis[0], axis)) * box.extents[0] + ... ditto for index '1' ... + ... ditto for index '2';min = center - radius;max = center + radius;
Author
Thanks jyk,
Is the following method what you meant:
I project the obb onto the world x,y,z axes and then obtain the tight fighting AABB from that.
Is this method sound, I have tried optimizing it as best as I could?
Is the following method what you meant:
public static CD_AxisAlignedBoundingBox ComputeFromOBB(CD_OrientedBoundingBox obb) { CD_AxisAlignedBoundingBox aabb_New = new CD_AxisAlignedBoundingBox(Vector3.Zero, 0, 0, 0); aabb_New.Centre.X += Vector3.Dot(obb.Position, GameConstants.WORLD_X_AXIS); aabb_New.Centre.Y += Vector3.Dot(obb.Position, GameConstants.WORLD_Y_AXIS); aabb_New.Centre.Z += Vector3.Dot(obb.Position, GameConstants.WORLD_Z_AXIS); // Compute the projection interval radius of obb onto world x axis float r_X = obb.Extents_HalfWidth.X * Math.Abs(Vector3.Dot(GameConstants.WORLD_X_AXIS, obb.Axis_X)) + obb.Extents_HalfWidth.Y * Math.Abs(Vector3.Dot(GameConstants.WORLD_X_AXIS, obb.Axis_Y)) + obb.Extents_HalfWidth.Z * Math.Abs(Vector3.Dot(GameConstants.WORLD_X_AXIS, obb.Axis_Z)); // Compute the projection interval radius of obb onto world y axis float r_Y = obb.Extents_HalfWidth.X * Math.Abs(Vector3.Dot(GameConstants.WORLD_Y_AXIS, obb.Axis_X)) + obb.Extents_HalfWidth.Y * Math.Abs(Vector3.Dot(GameConstants.WORLD_Y_AXIS, obb.Axis_Y)) + obb.Extents_HalfWidth.Z * Math.Abs(Vector3.Dot(GameConstants.WORLD_Y_AXIS, obb.Axis_Z)); // Compute the projection interval radius of obb onto world z axis float r_Z = obb.Extents_HalfWidth.X * Math.Abs(Vector3.Dot(GameConstants.WORLD_Z_AXIS, obb.Axis_X)) + obb.Extents_HalfWidth.Y * Math.Abs(Vector3.Dot(GameConstants.WORLD_Z_AXIS, obb.Axis_Y)) + obb.Extents_HalfWidth.Z * Math.Abs(Vector3.Dot(GameConstants.WORLD_Z_AXIS, obb.Axis_Z)); aabb_New.Extents = new Vector3(r_X, r_Y, r_Z); return aabb_New; }I project the obb onto the world x,y,z axes and then obtain the tight fighting AABB from that.
Is this method sound, I have tried optimizing it as best as I could?
This topic is closed to new replies.
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