Currently THREE.Spherical is disigned to work on referential where the poles (phi) are at the positive and negative y axis.
In architecture field the poles (phi) are at the positive and negative z axis.
How to use THREE.Spherical object in case z is the positive and negative pole axis?
Thanks for your help.
Related
Let's suppose:
1) XY plane is perpendicular to the surface of the street given by an area between the two green lines and a normal to this plane has the same "direction" as the yellow line,
2) XY has only three degrees of freedom - moving along X, Y, Z (Z is X x Y),
3) a limit for an X- along translation is when P1 is coincident to P2.
How does the θ change while translating by an [x,y,z] vector? How would it change if I added a rotation along any of the given axes and got rid of the point 3) (6 degrees of freedom)? Can you give me any hint where to look for an answer?
Hope everything is clear!
I am programming a raycaster in JavaScript. I am having trouble finding or implementing an algorithm for the raycast. Currently I am trying out the cast in the following style.
Theres an grid with equal block width and height. My player has a position within the grid, a direction as an angle 0 - 360 degrees where hes looking at. In the first step I need to figure out the distace from my Player to the Green dot at the Grid Intersection. I know that the Green Point is at the Intersection therefore i can figure out the length of the red line. Theres an right angle at the intersection. When I ve calculated the distance or x and y position of the green dot I have to do a simular thing in the 2nd step. The distance of the orange line is known, the position of the green dot and the angle is known. Again the right angle is on the intersecting border line.
I am not even sure if its possible this way, but maybe you got any other idea how I should then work it out. Thank you very much.
(Apologies for formatting below; I'm tapping this out on a phone)
From trigonometry, cos(aplha) = (length of red line)/(length of hypotenuse).
Therefore: length of hypotenuse = (length of red line)/cos(alpha).
You'd use sin for a vertical intersection.
A word of caution though: if you think about what would happen when the player is looking directly at a wall, all the lengths should be the same so that it's a constant height on screen, but they'll actually be different because the diagonals are different. You need to multiply by cos of the relative angle between the player's direction and the casting direction (so, if you've a 60 degree field of view then 0 at the centre of the display, off to +30 at one end and down to -30 at the other).
Also don't fall into the common trap of thinking that the angles you cast at should be evenly spaced. Think again of a person looking directly at a wall and use atan to get the proper relative angles.
The parametric equations of the ray read
X = x + t cos α, Y = y + t sin α
with t>0.
Assuming a unit grid (but you can rescale), and the angle in the first quadrant, the first intersections with the grid are
X = ceiling(x) => t = (X - x) / cos α => Y = y + (X - x) . tan α
and
Y = ceiling(y) => t = (Y - y) / sin α => X = x + (Y - y) . cot α
The smallest of the two t will tell you which of the horizontal and vertical is met first.
The next intersections are with X = ceiling(x) + i, and Y = ceiling(y) + j, hence the Y increase in steps tan α and the X in steps cot α.
For the other quadrants, the ceiling's are replaced by floor's.
For an ellipsoid of the form
with orientation vector and centre at point , how to find whether a point is inside the ellipsoid or not?
An additional note that the geometry actually is with a=b (spheroid) and therefore one axis is sufficient to define orientation
Note: I see a similar question asked in the forum. But, it is about an ellipsoid at origin and without any arbitrary orientation and here both arbitrary position and orientation are considered.
Find affine transform M that translates this ellipse in axis-oriented one (translation by -p and rotation to align orientation vector r and proper coordinate axis).
Then apply this transform to point p and check that p' lies inside axis-oriented ellipsoid, i.e.
x^2/a^2+ y^2/b^2+z^2/c^2 <= 1
Create a coordinate system E with the center at p and with the long axis of the ellipse aligned with r. Create a matrix that can transform global coordinates to the coordinate system E. Then put the transformed coordinates into the ellipse equation.
A center point p and an "orientation vector" r do not suffice to completely specify the position of the ellipsoid, there is one degree of freedom left. Your problem is indeterminate.
If your vector r is a unit vector from the origin to the pole, then the test for whether a point q is in (or on) the ellipse is:
v = q-p; // 3d vector difference
dot = v.r; // 3d dot product
f = dot*dot;
g = v.v - f; // 3d dot product and scalar subtraction
return f/(b*b) + g/(a*a) <= 1
Note that if the ellipse was aligned so that r was the z unit vector, then the test above translates into the usual test for inclusion of a point in an ellipse.
i have a coordinate system where the Y axis is UP. I need to convert it to a coordinate system where Z is UP. I have the rotations stored in quaternions, so my question is : if i have a quaternion X,Y,Z can i switch the Y with the Z and get the result that Z is actually UP?
Just swpping two axes in a quaternions? No this doesn't work because this flips the chirality. However if you flip the chirality and negate the quaternion's real part then you're back in the original chirality. In general form you can write this as
Q'(Q, i'j'k') = εi'j'k' Qw_w + Qi_i + Qj_j + Qk_k
where
is the totally antisymmetric tensor, known as the Levi-Cevita symbol.
This shouldn't be a surprise, as the i², j², k² rules of quaternions are defined also by the same totally antisymmetric tensor.
I'm adapting my answer from this post since the one here was the older and likely more generic one.
It's probably best to consider this in the context of how you convert angle and axis to a quaternion. In Wikipedia you can read that you describe a rotation by an angle θ around an axis with unit direction vector (x,y,z) using
q = cos(θ/2) + sin(θ/2)(xi + yj + zk)
Your post only tells us Y ↦ Z, i.e. the old Y direction is the new Z direction. What about the other directions? You probably want to keep X ↦ X, but that still leaves us with two alternatives.
Either you use Z ↦ Y. In that case you change between left-handed and right-handed coordinate system, and the conversion is essentially a reflection.
Or you use Z ↦ −Y, then it's just a 90° rotation about the X axis. The handedness of the coordinate system remains the same.
Change of chirality
Considering the first case first. What does changing the coordinate system do to your angle and axis? Well, the axis coordinates experience the same coordinate swapping as your points, and the angle changes its sign. So you have
cos(−θ/2) + sin(−θ/2)(xi + zj + yk)
Compared to the above, the real part does not change (since cos(x)=cos(−x)) but the imaginary parts change their sign, in addition to the change in order. Generalizing from this, a quaternion a + bi + cj + dk describing a rotation in the old coordinate system would be turned into a − bi − dj − ck in the new coordinate system. Or into −a + bi + dj + ck which is a different description of the same rotation (since it changes θ by 360° but θ/2 by 180°).
Preserved chirality
Compared to this, the second case of Z ↦ −Y maintains the sign of θ, so you only have to adjust the axis. The new Z coordinate is the old Y coordinate, and the new Y coordinate is the negated old Z coordinate. So a + bi + cj + dk gets converted to a + bi − dj + ck (or its negative). Note that this is just a multiplication of the quaternion by i or −i, depending on which side you multiply it. If you want to write this as a conjugation, you have θ=±45° so you get square roots in the quaternion that expresses the change of coordinate system.
Try :
Quaternion rotation = new Quaternion(X,Z,Y, -W); //i had to swap Z and Y due to
No, you cannot exchange y and z -- it will turn into a Left-Handed Coordinate system, if it was Right-Handed (and vice-versa).
You can, however, do the following substitution:
newX = oldZ
newY = oldX
newZ = oldY
I suspect that what you really want is a simple rotation about the x axis. If that's why you want to switch y and z, then you should instead apply a rotation of -90 degrees about the +x axis (assuming you have a Right-Handed coordinate system).
I'm using Electro in Lua for some 3D simulations, and I'm running in to something of a mathematical/algorithmic/physics snag.
I'm trying to figure out how I would find the "spin" of a sphere of a sphere that is spinning on some axis. By "spin" I mean a vector along the axis that the sphere is spinning on with a magnitude relative to the speed at which it is spinning. The reason I need this information is to be able to slow down the spin of the sphere by applying reverse torque to the sphere until it stops spinning.
The only information I have access to is the X, Y, and Z unit vectors relative to the sphere. That is, each frame, I can call three different functions, each of which returns a unit vector pointing in the direction of the sphere model's local X, Y and Z axes, respectively. I can keep track of how each of these change by essentially keeping the "previous" value of each vector and comparing it to the "new" value each frame. The question, then, is how would I use this information to determine the sphere's spin? I'm stumped.
Any help would be great. Thanks!
My first answer was wrong. This is my edited answer.
Your unit vectors X,Y,Z can be put together to form a 3x3 matrix:
A = [[x1 y1 z1],
[x2 y2 z2],
[x3 y3 z3]]
Since X,Y,Z change with time, A also changes with time.
A is a rotation matrix!
After all, if you let i=(1,0,0) be the unit vector along the x-axis, then
A i = X so A rotates i into X. Similarly, it rotates the y-axis into Y and the
z-axis into Z.
A is called the direction cosine matrix (DCM).
So using the DCM to Euler axis formula
Compute
theta = arccos((A_11 + A_22 + A_33 - 1)/2)
theta is the Euler angle of rotation.
The magnitude of the angular velocity, |w|, equals
w = d(theta)/dt ~= (theta(t+dt)-theta(t)) / dt
The axis of rotation is given by e = (e1,e2,e3) where
e1 = (A_32 - A_23)/(2 sin(theta))
e2 = (A_13 - A_31)/(2 sin(theta))
e3 = (A_21 - A_12)/(2 sin(theta))
I applaud ~unutbu's, answer, but I think there's a simpler approach that will suffice for this problem.
Take the X unit vector at three successive frames, and compare them to get two deltas:
deltaX1 = X2 - X1
deltaX2 = X3 - X2
(These are vector equations. X1 is a vector, the X vector at time 1, not a number.)
Now take the cross-product of the deltas and you'll get a vector in the direction of the rotation vector.
Now for the magnitude. The angle between the two deltas is the angle swept out in one time interval, so use the dot product:
dx1 = deltaX1/|deltaX1|
dx2 = deltax2/|deltaX2|
costheta = dx1.dx2
theta = acos(costheta)
w = theta/dt
For the sake of precision you should choose the unit vector (X, Y or Z) that changes the most.