So I am currently trying to create a function that will take two 3D points A and B, and provide me with the quaternion representing the rotation required of point A to be "looking at" point B (such that point A's local Z axis passes through point B, if you will).
I originally found this post, the top answer of which seemed to provide me with a good starting point. I went on to implement the following code; instead of assuming a default (0, 0, -1) orientation, as the original answer suggests, I try to extract a unit vector representing the actual orientation of the camera.
void Camera::LookAt(sf::Vector3<float> Target)
{
///Derived from pseudocode found here:
///https://stackoverflow.com/questions/13014973/quaternion-rotate-to
//Get the normalized vector from the camera position to Target
sf::Vector3<float> VectorTo(Target.x - m_Position.x,
Target.y - m_Position.y,
Target.z - m_Position.z);
//Get the length of VectorTo
float VectorLength = sqrt(VectorTo.x*VectorTo.x +
VectorTo.y*VectorTo.y +
VectorTo.z*VectorTo.z);
//Normalize VectorTo
VectorTo.x /= VectorLength;
VectorTo.y /= VectorLength;
VectorTo.z /= VectorLength;
//Straight-ahead vector
sf::Vector3<float> LocalVector = m_Orientation.MultVect(sf::Vector3<float>(0, 0, -1));
//Get the cross product as the axis of rotation
sf::Vector3<float> Axis(VectorTo.y*LocalVector.z - VectorTo.z*LocalVector.y,
VectorTo.z*LocalVector.x - VectorTo.x*LocalVector.z,
VectorTo.x*LocalVector.y - VectorTo.y*LocalVector.x);
//Get the dot product to find the angle
float Angle = acos(VectorTo.x*LocalVector.x +
VectorTo.y*LocalVector.y +
VectorTo.z*LocalVector.z);
//Determine whether or not the angle is positive
//Get the cross product of the axis and the local vector
sf::Vector3<float> ThirdVect(Axis.y*LocalVector.z - Axis.z*LocalVector.y,
Axis.z*LocalVector.x - Axis.x*LocalVector.z,
Axis.x*LocalVector.y - Axis.y*LocalVector.x);
//If the dot product of that and the local vector is negative, so is the angle
if (ThirdVect.x*VectorTo.x + ThirdVect.y*VectorTo.y + ThirdVect.z*VectorTo.z < 0)
{
Angle = -Angle;
}
//Finally, create a quaternion
Quaternion AxisAngle;
AxisAngle.FromAxisAngle(Angle, Axis.x, Axis.y, Axis.z);
//And multiply it into the current orientation
m_Orientation = AxisAngle * m_Orientation;
}
This almost works. What happens is that the camera seems to rotate half the distance towards the Target point. If I attempt the rotation again, it performs half the remaining rotation, ad infinitum, such that if I hold down the "Look-At-Button", the camera's orientation gets closer and closer to looking directly at the target, but is also constantly slowing down in its rotation, such that it never quite gets there.
Note that I don't want to resort to gluLookAt(), as I will also eventually need this code to point objects other than the camera at one another, and my objects already use quaternions for their orientations. For example, I might want to create an eyeball that tracks the position of something moving around in front of it, or a projectile that updates its orientation to seek out its target.
Normalize Axis vector before passing it to FromAxisAngle.
Why are you using a quaternion? You're just making things more complex and requiring more computation in this instance. To set up a matrix:-
calculate vector from observer to observed (which you're doing already)
normalise it (again, doing it already) = at
cross product this with the observer's up direction = right
normalise right
cross product at and right to get up
and you're done. The right, up and at vectors are the first, second and third row (or column, depending on how you set things up) of your matrix. The final row/column is the objects position.
But it looks like you want to transform an existing matrix to this new matrix over several frames. SLERPs do this to matricies as well as quaternions (which isn't surprising when you look into the maths). For the transformation, store the initial and target matricies and then SLERP between them, changing the amount to SLERP by each frame (e.g. 0, 0.25, 0.5, 0.75, 1.0 - although a non-linear progression would look nicer).
Don't forget that you're converting a quaternion back into a matrix in order to pass it to the rendering pipeline (unless there's some new features in the shaders to handle quaternions natively). So any efficencies due to quaternion use has to take into account the conversion process as well.
Related
I'm currently in the process of finishing the implementation for a camera that functions in the same way as the camera in Maya. The part I'm stuck in the tumble functionality.
The problem is the following: the tumble feature works fine so long as the position of the camera is not parallel with the up vector (currently defined to be (0, 1, 0)). As soon as the camera becomes parallel with this vector (so it is looking straight up or down), the camera locks in place and will only rotate around the up vector instead of continuing to roll.
This question has already been asked here, unfortunately there is no actual solution to the problem. For reference, I also tried updating the up vector as I rotated the camera, but the resulting behaviour is not what I require (the view rolls as a result of the new orientation).
Here's the code for my camera:
using namespace glm;
// point is the position of the cursor in screen coordinates from GLFW
float deltaX = point.x - mImpl->lastPos.x;
float deltaY = point.y - mImpl->lastPos.y;
// Transform from screen coordinates into camera coordinates
Vector4 tumbleVector = Vector4(-deltaX, deltaY, 0, 0);
Matrix4 cameraMatrix = lookAt(mImpl->eye, mImpl->centre, mImpl->up);
Vector4 transformedTumble = inverse(cameraMatrix) * tumbleVector;
// Now compute the two vectors to determine the angle and axis of rotation.
Vector p1 = normalize(mImpl->eye - mImpl->centre);
Vector p2 = normalize((mImpl->eye + Vector(transformedTumble)) - mImpl->centre);
// Get the angle and axis
float theta = 0.1f * acos(dot(p1, p2));
Vector axis = cross(p1, p2);
// Rotate the eye.
mImpl->eye = Vector(rotate(Matrix4(1.0f), theta, axis) * Vector4(mImpl->eye, 0));
The vector library I'm using is GLM. Here's a quick reference on the custom types used here:
typedef glm::vec3 Vector;
typedef glm::vec4 Vector4;
typedef glm::mat4 Matrix4;
typedef glm::vec2 Point2;
mImpl is a PIMPL that contains the following members:
Vector eye, centre, up;
Point2 lastPoint;
Here is what I think. It has something to do with the gimbal lock, that occurs with euler angles (and thus spherical coordinates).
If you exceed your minimal(0, -zoom,0) or maxima(0, zoom,0) you have to toggle a boolean. This boolean will tell you if you must treat deltaY positive or not.
It could also just be caused by a singularity, therefore just limit your polar angle values between 89.99° and -89.99°.
Your problem could be solved like this.
So if your camera is exactly above (0, zoom,0) or beneath (0, -zoom,0) of your object, than the camera only rolls.
(I am also assuming your object is at (0,0,0) and the up-vector is set to (0,1,0).)
There might be some mathematical trick to resolve this, I would do it with linear algebra though.
You need to introduce a new right-vector. If you make a cross product, you will get the camera-vector. Camera-vector = up-vector x camera-vector. Imagine these vectors start at (0,0,0), then easily, to get your camera position just do this subtraction (0,0,0)-(camera-vector).
So if you get some deltaX, you rotate towards the right-vector(around the up-vector) and update it.
Any influence of deltaX should not change your up-vector.
If you get some deltaY you rotate towards the up-vector(around the right-vector) and update it. (This has no influence on the right-vector).
https://en.wikipedia.org/wiki/Rotation_matrix at Rotation matrix from axis and angle you can find a important formula.
You say u is your vector you want to rotate around and theta is the amount you want to pivot. The size of theta is proportional to deltaX/Y.
For example: We got an input from deltaX, so we rotate around the up-vector.
up-vector:= (0,1,0)
right-vector:= (0,0,-1)
cam-vector:= (0,1,0)
theta:=-1*30° // -1 due to the positive mathematical direction of rotation
R={[cos(-30°),0,-sin(-30°)],[0,1,0],[sin(-30°),0,cos(-30°)]}
new-cam-vector=R*cam-vector // normal matrix multiplication
One thing is left to be done: Update the right-vector.
right-vector=camera-vector x up-vector .
To give some background to this question, I'm creating a game that needs to know whether the 'Orbit' of an object is within tolerance to another Orbit. To show this, I plot a Torus-shape with a given radius (the tolerance) using the Target Orbit, and now I need to check if the ellipse is within that torus.
I'm getting lost in the equations on Math/Stack exchange so asking for a more specific solution. For clarification, here's an image of the game with the Torus and an Orbit (the red line). Quite simply, I want to check if that red orbit is within that Torus shape.
What I believe I need to do, is plot four points in World-Space on one of those orbits (easy enough to do). I then need to calculate the shortest distance between that point, and the other orbits' ellipse. This is the difficult part. There are several examples out there of finding the shortest distance of a point to an ellipse, but all are 2D and quite difficult to follow.
If that distance is then less than the tolerance for all four points, then in think that equates to the orbit being inside the target torus.
For simplicity, the origin of all of these orbits is always at the world Origin (0, 0, 0) - and my coordinate system is Z-Up. Each orbit has a series of parameters that defines it (Orbital Elements).
Here simple approach:
Sample each orbit to set of N points.
Let points from first orbit be A and from second orbit B.
const int N=36;
float A[N][3],B[N][3];
find 2 closest points
so d=|A[i]-B[i]| is minimal. If d is less or equal to your margin/treshold then orbits are too close to each other.
speed vs. accuracy
Unless you are using some advanced method for #2 then its computation will be O(N^2) which is a bit scary. The bigger the N the better accuracy of result but a lot more time to compute. There are ways how to remedy both. For example:
first sample with small N
when found the closest points sample both orbits again
but only near those points in question (with higher N).
you can recursively increase accuracy by looping #2 until you have desired precision
test d if ellipses are too close to each other
I think I may have a new solution.
Plot the four points on the current orbit (the ellipse).
Project those points onto the plane of the target orbit (the torus).
Using the Target Orbit inclination as the normal of a plane, calculate the angle between each (normalized) point and the argument of periapse
on the target orbit.
Use this angle as the mean anomaly, and compute the equivalent eccentric anomaly.
Use those eccentric anomalies to plot the four points on the target orbit - which should be the nearest points to the other orbit.
Check the distance between those points.
The difficulty here comes from computing the angle and converting it to the anomaly on the other orbit. This should be more accurate and faster than a recursive function though. Will update when I've tried this.
EDIT:
Yep, this works!
// The Four Locations we will use for the checks
TArray<FVector> CurrentOrbit_CheckPositions;
TArray<FVector> TargetOrbit_ProjectedPositions;
CurrentOrbit_CheckPositions.SetNum(4);
TargetOrbit_ProjectedPositions.SetNum(4);
// We first work out the plane of the target orbit.
const FVector Target_LANVector = FVector::ForwardVector.RotateAngleAxis(TargetOrbit.LongitudeAscendingNode, FVector::UpVector); // Vector pointing to Longitude of Ascending Node
const FVector Target_INCVector = FVector::UpVector.RotateAngleAxis(TargetOrbit.Inclination, Target_LANVector); // Vector pointing up the inclination axis (orbit normal)
const FVector Target_AOPVector = Target_LANVector.RotateAngleAxis(TargetOrbit.ArgumentOfPeriapsis, Target_INCVector); // Vector pointing towards the periapse (closest approach)
// Geometric plane of the orbit, using the inclination vector as the normal.
const FPlane ProjectionPlane = FPlane(Target_INCVector, 0.f); // Plane of the orbit. We only need the 'normal', and the plane origin is the Earths core (periapse focal point)
// Plot four points on the current orbit, using an equally-divided eccentric anomaly.
const float ECCAngle = PI / 2.f;
for (int32 i = 0; i < 4; i++)
{
// Plot the point, then project it onto the plane
CurrentOrbit_CheckPositions[i] = PosFromEccAnomaly(i * ECCAngle, CurrentOrbit);
CurrentOrbit_CheckPositions[i] = FVector::PointPlaneProject(CurrentOrbit_CheckPositions[i], ProjectionPlane);
// TODO: Distance from the plane is the 'Depth'. If the Depth is > Acceptance Radius, we are outside the torus and can early-out here
// Normalize the point to find it's direction in world-space (origin in our case is always 0,0,0)
const FVector PositionDirectionWS = CurrentOrbit_CheckPositions[i].GetSafeNormal();
// Using the Inclination as the comparison plane - find the angle between the direction of this vector, and the Argument of Periapse vector of the Target orbit
// TODO: we can probably compute this angle once, using the Periapse vectors from each orbit, and just multiply it by the Index 'I'
float Angle = FMath::Acos(FVector::DotProduct(PositionDirectionWS, Target_AOPVector));
// Compute the 'Sign' of the Angle (-180.f - 180.f), using the Cross Product
const FVector Cross = FVector::CrossProduct(PositionDirectionWS, Target_AOPVector);
if (FVector::DotProduct(Cross, Target_INCVector) > 0)
{
Angle = -Angle;
}
// Using the angle directly will give us the position at th eccentric anomaly. We want to take advantage of the Mean Anomaly, and use it as the ecc anomaly
// We can use this to plot a point on the target orbit, as if it was the eccentric anomaly.
Angle = Angle - TargetOrbit.Eccentricity * FMathD::Sin(Angle);
TargetOrbit_ProjectedPositions[i] = PosFromEccAnomaly(Angle, TargetOrbit);}
I hope the comments describe how this works. Finally solved after several months of head-scratching. Thanks all!
I have to possitions, p1 and p2, p2 is attached to p1, not only to p1's position but also to it's rotation, so q1 is a quaternion which represents p1's rotation.
If q1 rotates, then p1's position must also rotate around p1 accordingly.
I only need to calculate p2's position, not it's rotation, I worked the rotation out already.
So basically is a spaceship docked to a station, I need to move and rotate the station around with the ship docked to it.
How do I do it?
the code i wrote for it works as long as the station is not rotated during the time of docking:
bool docked[100];
Quaternion quatTarget[100];
double distance_dock[100];
vector3 docking_position(int ship, int station)
{
if (!docked[ship])
{
docked[ship] = true;
distance_dock[ship] = distances(position[ship], position[station]);
vector3 direcc = normalized(position[station] - position[ship]);
quatTarget[ship] = vecToVecRotation(direcc, { 0, 0, 1 });
QuaternionNormalize(&quatTarget[ship], &quatTarget[ship]);
}
Quaternion orientation = total_rotation[station] * quatTarget[ship];
Matrix docking_place;
MatrixRotationQuaternion(&docking_place, &orientation);
vector3 axis_z = { docking_place(0, 2), docking_place(1, 2), docking_place(2, 2) };
return position[station] + -axis_z * distance_dock[ship];
}
What I do here is take an orientation quaternion from the ship to the station at the time of docking and then traslate the ship "distance_dock" units along the negative z axis of the orientation, so the ship will always move accordingly, but somehow if I dock the ship when the station is already rotated then I get the initial docking position wrong, though it still rotates perfectly along with the station.
If I understand you correctly, you have two objects that have a rigid transformation between them. The problem is that you want to calculate the pose (position + orientation) of one, given the pose of the other.
Let's say you have three frames; the Station frame "S", the Vehicle frame "V" and the Global frame "G" (I assume your graphics environment has a global 3D Cartesian frame).
The transformation between frames S and V is fully known (translation and orientation) and constant, and is denoted S_p_SV (the position of the Vehicle w.r.t the Station, expressed in the Station frame) and SV_q (the quaternion orientation of the Vehicle, expressed in the Station frame).
This will be confusing if you have not had experience in rigid-body mechanics, in which case you should read some introductory notes/slideshows on "Rigid-Body Mechanics" which are plentiful on Google results.
I have written the expression in LATEX but unfortunately StackOverflow does not support it, so I have attached it as an image. The original LATEX can be found here.
In my notation below, for example on the first line Sp_SV , is the position of the Vehicle w.r.t. the Station, expressed in the Station frame (of rotation).
The prefixed superscript indicates the rotation frame. For the quaternion G_Sq for example, this represents the orientation of the Station frame from the Ground Frame.
In terms of implementing this in C++, I am unsure of what library you are using for Quaternions, but you will need the following functions:
Convert Euler to Quaternions - If you are going to manually specify the rotation SVq (rotation of Vehicle w.r.t Station)
Convert Quaternion to DCM - For the first method in the LATEX
Quaternion Multiply - For the second method in the LATEX
Quaternion Conjugate - For the second method in the LATEX
I am trying to follow this course about computer graphics, but I'm stuck in the homework 1. I don't understand what's the role of the vector eye and up. The descripcion of the homework can be found in this link, there's also the skeleton of the first assignment.
So far I have the following code:
// Transform.cpp: implementation of the Transform class.
#include "Transform.h"
//Please implement the following functions:
// Helper rotation function.
mat3 Transform::rotate(const float degrees, const vec3& axis) {
// Please implement this.
float radians = degrees * M_PI / 180.0f;
mat3 r1(cos(radians));
mat3 r2(0, -axis.z, axis.y, axis.z, 0, -axis.x, -axis.y, axis.x, 0);
mat3 r3(axis.x*axis.x, axis.x*axis.y, axis.x*axis.z,
axis.x*axis.y, axis.y*axis.y, axis.y*axis.z,
axis.x*axis.z, axis.z*axis.y, axis.z*axis.z);
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
r2[i][j] = r2[i][j]*sin(radians);
r3[i][j] = r3[i][j]*(1-cos(radians));
}
}
return r1 + r2 + r3;
}
// Transforms the camera left around the "crystal ball" interface
void Transform::left(float degrees, vec3& eye, vec3& up) {
eye = eye * rotate(degrees, up);
}
// Transforms the camera up around the "crystal ball" interface
void Transform::up(float degrees, vec3& eye, vec3& up) {
vec3 newAxis = glm::cross(eye, up);
}
// Your implementation of the glm::lookAt matrix
mat4 Transform::lookAt(vec3 eye, vec3 up) {
return lookAtMatrix;
}
Transform::Transform()
{
}
Transform::~Transform()
{
}
for the left method it appears to be doing the right thing, which is, rotating the object around the y-axis (actually I'm not sure if the object is moving or what I'm moving is the camera, can someone clarify?).
for the up method I cannot make it work which will be rotating the object (or camera?) around the x-axis (at least that's what I think).
finally, I don't understand what should the lookAt method should do.
Can someone help me understand the actions to be performed?
Can someone explain what are the roles of vectors eye and up?
View transforms are often implemented using a "look-at" function. The idea being that you specify where the camera is, what direction it is looking in, and what direction represents "up" in your particular space, and you get a matrix back which represents that transform.
It looks like you're trying to implement some kind "rotating ball" navigation control. That's fairly simple - horizontal movement should rotate around some "Y" axis, and vertical movement should rotate around the "right" (or X) axis. Generally those rotations work around the current view axes, rather than globally, so that the movement is intuitive. I'm not sure exactly what you're looking for there.
A look-at function works as follows.
A 3x3 matrix representing a rotation can be viewed as being composed the 3 perpendicular unit axes of the space you are transforming into. So if you can supply those vectors, you can build the matrix.
The first axis is easy. A camera is typically oriented to look along "Z", so if you take the vector representing the direction of the thing being looked at from the camera's position, then normalise it, this is the Z axis.
Then you need to define a distinct 'up' vector - (0,1,0) is typical, but you will need to choose a different one in cases where the Z-axis is pointing in the same direction.
The cross product of this 'up' vector and the 'Z' axis gives the 'X' axis - this is because the cross product gives a perpendicular vector, and what constitutes horizontal will be perpendicular to both the 'forward' direction, and 'up'.
Then the cross product of the 'X' and 'Z' axes gives the 'Y' axis (which is not necessarily the same as the 'Y' axis - consider looking towards the ceiling or towards the floor).
These three axes, normalised, (x,y,z) directly form a rotation matrix.
The translation portion of the matrix is generally the position of the camera, transformed by the rotation's inverse (such that when transforming the camera position by the lookat matrix itself, it should end up back at the origin).
1) Your course is using the OpenGL library, and your homework assignment is to fill in the skeleton module "Transform.cpp".
2) The method you're asking about is "mat4 Transform::lookAt(vec3 eye, vec3 up)":
lookAt: Finally, you need to code in the transformation matrix, given
the eye and up vectors. You will likely need to refer to the class
notes to do this. It is likely to help to dene a uvw coordinate frame
(as 3 vectors), and to build up an auxiliary 4 4 matrix M which is
returned as the result of this function. Consult class notes and
lectures for this part.
3) A hint for what these two arguments "eye" and "up" mean should be in your class notes and lectures.
4) Another hint is to "define a uvw coordinate frame (as three vectors), and build up an auxiliary 4x4 matrix ... which is returned as a result...".
5) A final hint:
Q: What's the difference between an OpenGL mat3 and mat4?
A:
What extractly mat3(a mat4 matrix) statement in glsl do?
mat3(MVI) * normal
Returns the upper 3x3 matrix from the 4x4 matrix and multiplies the
normal by that. This matrix is called the 'normal matrix'. You use
this to bring your normals from world space to eye space.
The reason why the original matrix is 4x4 and not 3x3 is because 4x4
matrices let you do affine transformations and contain useful
information for perspective rendering. But to take a normal from world
space to eye space, you just need the 3x3 model view matrix.
I need to rotate the camera around a player from a third person view. (Nothing fancy).
Here it is how I try it:
// Forward, right, and position define the plane - they have x,y,z components.
void rotate ( float angle, Vector interestPoint )
{
Vector oldForward ( Forward );
forward = forward * cos(angle) + right * sin(angle);
forward.Normalize();
right = forward.CrossProduct ( up );
right.Normalize();
position = ( position + old_forward * position.Distance( interestPoint ) ) - (forward * position.Distance( interestPoint ) );
this->angle += angle;
}
The problem is that if, let's say just do turn left a lot, the distance between the object and the camera increases.
For a very simple orbit camera, like most 3rd person adventure games, you will need 4 things:
The position of the target
The distance from the target
The azimuthal angle
The polar angle
(If you want your camera to be always relative to the target in orientation, you need to provide the target's orientation as well, in this case I will simply use the world orientation)
See Spherical coordinate systems for a reference.
You should map your azimuthal angle on your horizontal control (and make it loop around when you reach 2 * PI) and your polar angle should be mapped on your vertical control (or inverted if the player selects that option and make it clamped between -PI and PI - watch out for calculations based on the world Up vector if you go parallel to it (-PI or PI)
The distance can be fixed or driven by a spline, for this case we will assume a fixed distance.
So, to compute your position you start with WorldForward, which is a unit vector pointing in the axis that you generally consider to be your forward, for example (1,0,0) (here, if we were building a relative camera, we would use our target's forward vector) and you invert it (* -1) to go "from the target" "to your camera".
(The following is untested pseudo code, but you should get the gist - also, keep note that it can be simplified, I just went for clarity)
Next step is to rotate this vector using our azimuth angle, which is the horizontal orientation component of your camera. Something like:
Vector toCamera = WorldForward * -1;
Matrix horizontalRotation = Matrix.CreateRotationZ(azimuth); // assuming Z is up
Vector horizontalRotationPosition = horizontalRotation.Transform(toCamera);
At this point, you have a camera that can rotate horizontally around your target, now to add the other axis, you simply transform again using the polar angle rotation:
Matrix verticalRotation = Matrix.CreateRotationY(polar); // assuming Y is right
Vector finalRotatedVector = verticalRotation.Transform(horizontalRotationPosition);
Now, what we have is a unit vector that points to the position where the camera should be, if you multiply it by the distance you want to keep from your target and add the position of your target, you should get your final position. Keep in mind that this unit vector, if negated, represents the forward vector of your camera.
Vector cameraPosition = targetPosition + finalRotatedVector * distanceFromTarget;