finding the rotation axis - math

I'm using eigen library to rotate a plane to be parallel to the ground plane.
The ground plane is defined using the normal vector (0,0,1)
The target plane is a set of 3D points and a normal
The rotation angle is known
the normal vector of the plane as well as every point on that plane has to be rotated to be parallel to the ground plane
I'd like to use affine transformation from
http://eigen.tuxfamily.org/api/TutorialGeometry.html
something like this
Transform t = AngleAxisf(a,axis);
axis in this case is a matrix representing an arbitary axis, along which the rotation takes place.
How to find this axis?
Many thanks

Making two planes parallel can be done by making their normals parallel, so you just need to find the axis to rotate the target plane normal about. This is just the axis that is perpendicular to both your ground plane normal and your target plane normal which can be found using the cross product. In your case, if your target plane has a normal of [x,y,z], then the rotation axis is [y,-x,0].

Related

How to orient a 3D object in relation to its direction and the ground?

In my game I have characters walking around a 3d terrain. The characters treat the terrain however as a 2d game map, so each character has a direction and a rotation on a 2d plane.
I want to rotate the characters as they're walking on the terrain, so that they are oriented to stand in relation to the terrain, rather then always be oriented as if they're walking on flat ground. This with keeping the original direction of the characters.
Basically I want
For each arbitrary x\z (width\depth) point on the game map I have
the (x,y,z) vector of the point on the terrain
The normal of the the specific terrain face related to the point
Using this, how do I set the rotation of the characters to achieve this?
Depending on which axis you would like to rotate the object the dot product of the faces normal with that axis will return you the cosine of the angle between the two vectors. By that angle you would have to rotate your object.

Find axis of rotation given rotation angle

I have a plane which is rotated 90 degrees around an unknown axis. I know a point and normal for the plane before and after the rotation. How can I find the axis of rotation?
I've done a sketch to illustrate - it's 2D but the problem is actually 3D.
I worked it out with some help from #davin.
Use the cross product to find the direction of the rotation axis. The two known points on the planes and the unknown point on the rotation axis make an isosceles triangle, so simple geometry finds the unknown point.
The axis of rotation is an eigenvector of the rotation matrix. Moreover, it has eigenvalue 1. Every rotation matrix has such an eigenvector. Then just apply the translation to the eigenvector (presuming you're rotating and then translating) to get the final axis of rotation.
Mathematically, you need to solve Rv = v, which is equivalent to finding the nullspace of R-I.

In a TBN Matrix are the normal, tangent, and bitangent vectors always perpendicular?

This is related to a problem described in another question (images there):
Opengl shader problems - weird light reflection artifacts
I have a .obj importer that creates a data structure and calculates the tangents and bitangents. Here is the data for the first triangle in my object:
My understanding of tangent space is that the normal points outward from the vertex, the tangent is perpendicular (orthogonal?) to the normal vector and points in the direction of positive S in the texture, and the bitangent is perpendicular to both. I'm not sure what you call it but I thought that these 3 vectors formed what would look like a rotated or transformed x,y,z axis. They wouldn't be 3 randomly oriented vectors, right?
Also my understanding: The normals in a normal map provide a new normal vector. But in tangent space texture maps there is no built in orientation between the rgb encoded normal and the per vertex normal. So you use a TBN matrix to bridge the gap and get them in the same space (or get the lighting in the right space).
But then I saw the object data... My structure has 270 vertices and all of them have a 0 for the Tangent Y. Is that correct for tangent data? Are these tangents in like a vertex normal space or something? Or do they just look completely wrong? Or am I confused about how this works and my data is right?
To get closer to solving my problem in the other question I need to make sure my data is right and my understanding on how tangent space lighting math works.
The tangent and bitangent vectors point in the direction of the S and T components of the texture coordinate (U and V for people not used to OpenGL terms). So the tangent vector points along S and the bitangent points along T.
So yes, these do not have to be orthogonal to either the normal or each other. They follow the direction of the texture mapping. Indeed, that's their purpose: to allow you to transform normals from model space into the texture's space. They define a mapping from model space into the space of the texture.
The tangent and bitangent will only be orthogonal to each other if the S and T components at that vertex are orthogonal. That is, if the texture mapping has no sheering. And while most texture mapping algorithms will try to minimize sheering, they can't eliminate it. So if you want an accurate matrix, you need a non-orthogonal tangent and bitangent.

Projection matrix point to sphere surface

I need to project a 3D object onto a sphere's surface (uhm.. like casting a shadow).
AFAIR this should be possible with a projection matrix.
If the "shadow receiver" was a plane, then my projection matrix would be a 3D to 2D-plane projection, but my receiver in this case is a 3D spherical surface.
So given sphere1(centerpoint,radius),sphere2(othercenter,otherradius) and an eyepoint how can I compute a matrix that projects all points from sphere2 onto sphere1 (like casting a shadow).
Do you mean that given a vertex v you want the following projection:
v'= centerpoint + (v - centerpoint) * (radius / |v - centerpoint|)
This is not possible with a projection matrix. You could easily do it in a shader though.
Matrixes are commonly used to represent linear operations, like projection onto a plane.
In your case, the resulting vertices aren't deduced from input using a linear function, so this projection is not possible using a matrix.
If the sphere1 is sphere((0,0,0),1), that is, the sphere of radius 1 centered at the origin, then you're in effect asking for a way to convert any location (x,y,z) in 3D to a corresponding location (x', y', z') on the unit sphere. This is equivalent to vector renormalization: (x',y',z') = (x,y,z)/sqrt(x^2+y^2+z^2).
If sphere1 is not the unit sphere, but is say sphere((a,b,c),R) you can do mostly the same thing:
(x',y',z') = R*(x-a,y-b,z-c) / sqrt((x-a)^2+(y-b)^2+(z-c)^2) + (a,b,c). This is equivalent to changing coordinates so the first sphere is the unit sphere, solving the problem, then changing coordinates back.
As people have pointed out, these functions are nonlinear, so the projection cannot be called a "matrix." But if you prefer for some reason to start with a projection matrix, you could project first from 3D to a plane, then from a plane to the sphere. I'm not sure if that would be any better though.
Finally, let me point out that linear maps don't produce division-by-zero errors, but if you look closely at the formulas above, you'll see that this map can. Geometrically, that's because it's hard to project the center point of a sphere to its boundary.

3d Parabolic Trajectory

I'm trying to figure out some calculations using arcs in 3d space but am a bit lost. Lets say that I want to animate an arc in 3d space to connect 2 x,y,z coordinates (both coordinates have a z value of 0, and are just points on a plane). I'm controlling the arc by sending it a starting x,y,z position, a rotation, a velocity, and a gravity value. If I know both the x,y,z coordinates that need to be connected, is there a way to calculate what the necessary rotation, velocity, and gravity values to connect it from the starting x,y,z coordinate to the ending one?
Thanks.
EDIT: Thanks tom10. To clarify, I'm making "arcs" by creating a parabola with particles. I'm trying to figure out how to ( by starting a parabola formed by a series particles with an beginning x,y,z,velocity,rotation,and gravity) determine where it will in end(the last x,y,z coordinates). So if it if these are the two coordinates that need to be connected:
x1=240;
y1=140;
z1=0;
x2=300;
y2=200;
z2=0;
how can the rotation, velocity, and gravity of this parabola be calculated using only these variables start the formation of the parabola:
x1=240;
y1=140;
z1=0;
rotation;
velocity;
gravity;
I am trying to keep the angle a constant value.
This link describes the ballistic trajectory to "hit a target at range x and altitude y when fired from (0,0) and with initial velocity v the required angle(s) of launch θ", which is what you want, right? To get your variables into the right form, set the rotation angle (in the x-y plane) so you're pointing in the right direction, that is atan(y/x), and from then on out, to match the usual terminology for 2D problem, rewrite your z to y, and the horizontal distance to the target (which is sqrt(xx + yy)) as x, and then you can directly use the formula in link.
Do the same as you'd do in 2D. You just have to convert your figures to an affine space by rotating the axis, so one of them becomes zero; then solve and undo the rotation.

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