I have to find the axis and angle of rotation of a camera with an UP and Direction vector(They both are perpendicular to each other). I have the initial and final positions of the UP and direction vectors of the camera that is rotated. I want to find the axis and angle of the rotation for the camera. I am using C# for my project. I am new to this 3D rotation. So pardon my questions if you find them silly.
From the direction (forward) vector f and up vector u you can get the side vector s by performing a vector cross product (s = f x u). All three vectors are now orthogonal. You should also make them orthonormal by normalizing each one of them. Taken together, these vectors form an orthonormal basis.
You now have two such basis: the one from your initial camera orientation and the one from your final camera orientation. Both basis can be represented as a rotation matrix. A rotation matrix is simply a 3x3 matrix where the 3 rows are respectively:
The forward vector
The up vector
The side vector
For example, the matrix:
[[1 0 0]
[0 1 0]
[0 0 1]]
could be your initial camera orientation at start-up with its forward vector, up vector and side vector pointing towards the positive x axis, y axis and z axis, respectively.
You can now convert these two basis (M1 and M2) to two unit quaternions (Q1 and Q2) using this algorithm which takes care about potential problems like divides by zero.
At this point, you have two unit quaternions representing your initial and final camera orientation. You must now find the quaternion qT that transforms Q1 into Q2, that is:
q2 = qT * q1
q2 * q1^-1 = qT * (q1 * q1^-1) = qT
=> qT = q2 * q1^-1
Knowing that the inverse of a unit quaternion is equal to its conjugate:
q1^-1 = q1* iif ||q1|| = 1
qT = q2 * q1^-1 = q2 * q1*
There is a single step left: extracting the axis and angle from quaternion qT:
angle = 2 * acos(qw)
x = qx / sqrt(1-qw*qw)
y = qy / sqrt(1-qw*qw)
z = qz / sqrt(1-qw*qw)
The angle is, of course, given in radian. Beware of the divide by zero when calculating x, y and z. This situation would happen when there is no rotation or a very small one, so you should test if angle > epsilon where you would choose epsilon to be quite small an angle (say 1/10 of a degree) and not calculate the vector if that is the case.
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circle in three dimensional coordiantes [closed]
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I want to draw a circle in three dimensional coordiantes, i'm given a vector, the angle where vector's intersects with the circle is 90 degrees, the intersection point is the centre of the circle. The radius can be parametrized. EDIT: I am programming a server plugin for minecraft. At this point I have made a sword that can be thrown. I want to add some decor. I want that after the sword there was a trace in the form of a circle. But I don't understand how to draw a circle in 3D coordinates so that the angle of intersection of the sword throw vector with the center of the circle is 90 degrees. The radius can be arbitrary, and the vector can enter the center of the circle. I thought I could just rotate the throw vector on 3 axes and get a circle, but nothing worked. I need an equation with which I can draw a given circle.
You have center C, normal vector N, radius R. Seems you want to get points at the circumference.
At first get some base vector in the circle plane.
Possible way:
Reveal normal component with the largest magnitude and with the second magnitude. For example, abs(N.X) is the largest, abs(N.Z) has the second magnitude, and abs(N.Y) is the smallest. Make the smallest component zero, exchange two larger ones, and negate the largest. For this example base vector will be:
A = (N.Z, 0, -N.X)
It is perpendicular to normal, hence lies in the circle plane.
Then get the next basis vector using vector product (B will be perpendicular both to A and to N, it lies in the plane too)
B = N x A
Now normalize vectors A and B (make them unit length)
A = A / len(A)
B = B / len(B)
and you can get any point at the circumeference with parametric equation where t changes in the range 0..2*Pi
P(t) = C + R * A * Cos(t) + R * B * Sin(t)
or in components:
P.X = C.X + R * A.X * Cos(t) + R * B.X * Sin(t)
and so on
I am currently using Atan2 to calculate the player heading angle.
however after some trial and error I discovered that the in-game angle's are rather different to that of a "normal" lay out :
ReturnedAngle = Math.Atan2(Y2 - Y1, X2 - X1); /// ArcTan2 the difference in our Y axis is always passed first followed by X
ReturnedAngle = (180 / Math.PI) * ReturnedAngle; /// Converting our radians to Degrees the convervion ends at 358 not the full 360 degrees.
ReturnedAngle = Math.Round(ReturnedAngle + 360, MidpointRounding.AwayFromZero) % 360; /// MOD and round our angle.
Above is the C# code I am using to calc the heading angle. My questions is how would I go about converting this angle from the "normal" angle system to the in-game one.
I think this is your situation. You have a right-hand coordinate system, but you are measuring a clock-wise angle, which is inconsistent.
In any case, draw a small positive angle from 360 (red below) to form a right triangle (purple below) with positive sides.
To measure the angle θ of the triangle, measure the short side Δx and the long side Δy and compute.
var θ = Math.Atan2(Δx, Δy);
This would work for any positive or negative values for the two sides. For example, if the angle goes above 90° then Δy would flip signs, as your target point is going to be below the origin. But the beauty of Atan2() is that you don't need to worry about these cases as it works on all four quadrants if you make it work for a small positive angle.
In reverse you have
var Δx = R*Math.Sin(Θ);
var Δy = R*Math.Cos(Θ);
where R is the distance between the target and the reference point.
Math.Atan2(Y2 - Y1, X2 - X1) computes the angle anticlockwise from the x axis. Math.Atan2(X2 - X1, Y2 - Y1) computes the angle clockwise from the y axis, which is what you want.
The 'clockwise from north' convention is used in navigation and mapping. Over the years I've found that it easiest to think in terms of vectors having components north, east. This means that atan2 is called the same way, that is, to get the direction of q from p:
dirn = atan2( q[1]-p[1], q[0]-p[0]);
If you are thinking of p and q as x,y vectors this gives you the angle anti-clockwise from the x axis. If you are thinking of p and q as n,e vectors it gives you the angle clockwise from north.
It also means that the formulae for a rotation matrix is the same. To rotate through an angle a, you use the matrix
R = ( cos(a) -sin(a) )
( sin(a) cos(a) )
Again, if you are thinking of the vectors as being x,y then applying R rotates through and angle a, anti-clockwise from the axis, while if you think of vectors as being n,e applying R rotates through an angle a, clockwise from north.
I am attempting to fuse skeletons from two separate Kinects.
From first Kinect I have skeleton1, from it I choose 4 points corresponding with 4 of it's joints. With those 4 points I can construct a plane on which they all are. The plane coefficients Ax + By + Cz = D are known. As I understand the planes Quaternion would be Q = (D; A; B; C)
From the second Kinect I have the same data, but in second Kinects coordinate system.
How can I rotate the plane from second Kinect so that it would have the same orientation as the plane from the first Kinect?
Unfortunately, quaternions don't work that way. They do not represent a plane, and cannot be created just by re-arranging the components of a plane equation (plane equation has two degrees of freedom, while quaternion is a non-parametric representation of rotation/orientation).
However, if you are looking for a rotation between two normal vectors (one for each plane), which will transform one plane to another, it is a relatively simple task:
if the two normal vectors are the same, there is no rotation needed and the result is identity,
if the two vectors are parallel (their dotproduct is zero) but not identical, the transformation between them is a rotation of 180 degrees around an arbitrary axis perpendicular to them,
in all other cases, you can compute the axis of rotation using a crossproduct (and normalization), and the angle using arc cosine of their dotproduct.
If you still need to create a quaternion out of them, you can do it with a bit of trigonometry.
I have read some of the duplicate answers about angle between two vectors, but I'm still stuck with my problem. I have two vectors and I want that the angle between them to always be 90 degrees. To achieve that I need to find the angle between them, so that I can subtract or add the correct amount of degrees so that the angle between them always is 90 degrees.
The picture illustrates a sprite and two vectors. How do I find the angle A between them two? I have tried to use this code to get the angle between two vectors, but I must have missed something out, because I don't get the correct results:
public float GetAngleBetween (Vector2 A, Vector2 B)
{
float DotProd = Vector2.Dot (A, B);
float Length = A.Length () * B.Length ();
return (float)MathHelper.ToDegrees ((float)Math.Acos (DotProd/Length));
}
Any input is welcome and thank you in advance for any answers.
The actual angle in radians is
Math.ACos(Vector2.Dot(a, b));
Make sure that a and b are normalized vectors or the results can get pretty weird.
I think you may be looking for the Vector2.Dot method which is used to calculate the product of two vectors, and can be used for angle calculations.
For example:
// the angle between the two vectors is less than 90 degrees.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) > 0
// the angle between the two vectors is more than 90 degrees.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) < 0
// the angle between the two vectors is 90 degrees; that is, the vectors are orthogonal.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == 0
// the angle between the two vectors is 0 degrees; that is, the vectors point in the same direction and are parallel.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == 1
// the angle between the two vectors is 180 degrees; that is, the vectors point in opposite directions and are parallel.
Vector2.Dot(vector1.Normalize(), vector2.Normalize()) == -1
Is this what you're looking for, or do you need the exact angle?
If I understand your question diagram and comments, the Dot product and Acos are not the only bits of info you need. You also need to account for when the sprite is not located at (0,0).
float angleInRadians = (float) Math.Acos(Vector2.Dot(Vector2.Normalize(vector1 - spritePosition), Vector2.Normalize(vector2 - spritePosition)));
int angleInDegrees = MathHelper.ToDegrees(angleInRadians);
I am learning XNA and in almost all of the educational kits found on http://creators.xna.com/en-US/. I always see a call to Normalize() on a vector. I understand that normalize basically converts the vector into unit length, so all it gives is direction.
Now my question is When to normalize and what exactly does it help me in. I am doing 2D programming so please explain in 2D concepts and not 3D.
EDIT: Here is code in the XNA kit, so why is the Normalize being called?
if (currentKeyboardState.IsKeyDown(Keys.Left)
|| currentGamePadState.DPad.Left == ButtonState.Pressed)
{
catMovement.X -= 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Right)
|| currentGamePadState.DPad.Right == ButtonState.Pressed)
{
catMovement.X += 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Up)
|| currentGamePadState.DPad.Up == ButtonState.Pressed)
{
catMovement.Y -= 1.0f;
}
if (currentKeyboardState.IsKeyDown(Keys.Down)
|| currentGamePadState.DPad.Down == ButtonState.Pressed)
{
catMovement.Y += 1.0f;
}
float smoothStop = 1;
if (catMovement != Vector2.Zero)
{
catMovement.Normalize();
}
catPosition += catMovement * 10 * smoothStop;
In your example, the keyboard presses give you movement in X or Y, or both. In the case of both X and Y, as when you press right and down at the same time, your movement is diagonal. But where movement just in X or Y alone gives you a vector of length 1, the diagonal vector is longer than one. That is, about 1.4 (the square root of 2).
Without normalizing the movement vector, then diagonal movement would be faster than just X or Y movement. With normalizing, the speed is the same in all 8 directions, which I guess is what the game calls for.
One common use case of vector normalization when you need to move something by a number of units in a direction. For example, if you have a game where an entity A moves towards an entity B at a speed of 5 units/second, you'll get the vector from A to B (which is B - A), you'll normalize it so you only keep the direction toward the entity B from A's viewpoint, and then you'll multiply it by 5 units/second. The resulting vector will be the velocity of A and you can then simply multiply it by the elapsed time to get the displacement by which you can move the object.
It depends on what you're using the vector for, but if you're only using the vectors to give a direction then a number of algorithms and formulas are just simpler if your vectors are of unit length.
For example, angles: the angle theta between two unit vectors u and v is given by the formula cos(theta) = u.v (where . is the dot product). For non-unit vectors, you have to compute and divide out the lengths: cos(theta) = (u.v) / (|u| |v|).
A slightly more complicated example: projection of one vector onto another. If v is a unit vector, then the orthogonal projection of u onto v is given by (u.v) v, while if v is a non-unit vector then the formula is (u.v / v.v) v.
In other words: normalize if you know that all you need is the direction and if you're not certain the vector is a unit vector already. It helps you because you're probably going to end up dividing your direction vectors by their length all the time, so you might as well just do it once when you create the vector and get it over with.
EDIT: I assume that the reason Normalize is being called in your example is so that the direction of velocity can be distinguished from the speed. In the final line of the code, 10 * smoothStop is the speed of the object, which is handy to know. And to recover velocity from speed, you need to multiply by a unit direction vector.
You normalize any vector by dividing each component by the magnitude of the vector, which is given by the square root of the sum of squares of components. Each component then has a magnitude that varies between [-1, 1], and the magnitude is equal to one. This is the definition of a unit vector.