In my project previously i used 2d Map. as per costumer requirement they want 3d map. But 3rd party provide only X and Y. In that project some Objects are static so I placed manually. But some Objects are dynamically changed the position. That object need X , Y , Z How to internally convert X Y to X Y Z
Some Static Objects Coordinates : in 2D X: 865 Y: 950
In 3d X:-54.2,Y:11.6,Z:17.5
in 2D X: 813 Y: 945
In 3d X:-54.4,Y:12.4,Z:24.1
2D Map
Find the relation between xy coordinates in 2D and xz coordinates in 3D. When you get it, you’ll solve a half of the question.
Next, to find y in 3D. I would use THREE.Raycaster() with THREE.Vector(0, -1 ,0) direction and its origin is higher than the highest point on the 3D map.
Related
A similar question was asked before, unfortunately I cannot comment Samgaks answer so I open up a new post with this one. Here is the link to the old question:
How to calculate ray in real-world coordinate system from image using projection matrix?
My goal is to map from image coordinates to world coordinates. In fact I am trying to do this with the Camera Intrinsics Parameters of the HoloLens Camera.
Of course this mapping will only give me a ray connecting the Camera Optical Centre and all points, which can lie on that ray. For the mapping from image coordinates to world coordinates we can use the inverse camera matrix which is:
K^-1 = [1/fx 0 -cx/fx; 0 1/fy -cy/fy; 0 0 1]
Pcam = K^-1 * Ppix;
Pcam_x = P_pix_x/fx - cx/fx;
Pcam_y = P_pix_y/fy - cy/fy;
Pcam_z = 1
Orientation of Camera Coordinate System and Image Plane
In this specific case the image plane is probably at Z = -1 (However, I am a bit uncertain about this). The Section Pixel to Application-specified Coordinate System on page HoloLens CameraProjectionTransform describes how to go form pixel coordinates to world coordinates. To what I understand two signs in the K^-1 are flipped s.t. we calculate the coordinates as follows:
Pcam_x = (Ppix_x/fx) - (cx*(-1)/fx) = P_pix_x/fx + cx/fx;
Pcam_y = (Ppix_y/fy) - (cy*(-1)/fy) = P_pix_y/fy + cy/fy;
Pcam_z = -1
Pcam = (Pcam_x, Pcam_y, -1)
CameraOpticalCentre = (0,0,0)
Ray = Pcam - CameraOpticalCentre
I do not understand how to create the Camera Intrinsics for the case of the image plane being at a negative Z-coordinate. And I would like to have a mathematical explanation or intuitive understanding of why we have the sign flip (P_pix_x/fx + cx/fx instead of P_pix_x/fx - cx/fx).
Edit: I read in another post that the thirst column of the camera matrix has to be negated for the case that the camera is facing down the negative z-direction. This would explain the sign flip. However, why do we need to change the sign of the third column. I would like to have a intuitive understanding of this.
Here the link to the post Negation of third column
Thanks a lot in advance,
Lisa
why do we need to change the sign of the third column
To understand why we need to negate the third column of K (i.e. negate the principal points of the intrinsic matrix) let's first understand how to get the pixel coordinates of a 3D point already in the camera coordinates frame. After that, it is easier to understand why -z requires negating things.
let's imagine a Camera c, and one point B in the space (w.r.t. the camera coordinate frame), let's put the camera sensor (i.e. image) at E' as in the image below. Therefore f (in red) will be the focal length and ? (in blue) will be the x coordinate in pixels of B (from the center of the image). To simplify things let's place B at the corner of the field of view (i.e. in the corner of the image)
We need to calculate the coordinates of B projected into the sensor d (which is the same as the 2d image). Because the triangles AEB and AE'B' are similar triangles then ?/f = X/Z therefore ? = X*f/Z. X*f is the first operation of the K matrix is. We can multiply K*B (with B as a column vector) to check.
This will give us coordinates in pixels w.r.t. the center of the image. Let's imagine the image is size 480x480. Therefore B' will look like this in the image below. Keep in mind that in image coordinates, the y-axis increases going down and the x-axis increases going right.
In images, the pixel at coordinates 0,0 is in the top left corner, therefore we need to add half of the width of the image to the point we have. then px = X*f/Z + cx. Where cx is the principal point in the x-axis, usually W/2. px = X*f/Z + cx is exactly as doing K * B / Z. So X*f/Z was -240, if we add cx (W/2 = 480/2 = 240) and therefore X*f/Z + cx = 0, same with the Y. The final pixel coordinates in the image are 0,0 (i.e. top left corner)
Now in the case where we use z as negative, when we divide X and Y by Z, because Z is negative, it will change the sign of X and Y, therefore it will be projected to B'' at the opposite quadrant as in the image below.
Now the second image will instead be:
Because of this, instead of adding the principal point, we need to subtract it. That is the same as negating the last column of K.
So we have 240 - 240 = 0 (where the second 240 is the principal point in x, cx) and the same for Y. The pixel coordinates are 0,0 as in the example when z was positive. If we do not negate the last column we will end up with 480,480 instead of 0,0.
Hope this helped a little bit
I have a THREE.Plane plane which is intersected by a number of THREE.Line3 lines[].
Using only this information, how can I acquire a 2D coordinate set of points?
Edit for better understanding the problem:
The 2D coordinate is related to the plane, so imagine the 3D plane becomes a Cartesian plane drawn on a blackboard. It is pretty much a 3D drawing of a 2D plane. What I want to find is the X, Y values of points previously projected onto this Cartesian plane. But they are 3D, just like the 3D plane.
You don't have enough information. In this answer I'll explain why, and provide more information to achieve what you want, should you be able to provide the necessary information
First, let's create a plane. Like you, I'm uing Plane.setFromNormalAndCoplanarPoint. I'm considering the co-planar point as the origin ((0, 0)) of the plane's Cartesian space.
let normal = new Vector3(Math.random(), Math.random(), Math.random()).normalize()
let origin = new Vector3(Math.random(), Math.random(), Math.random()).normalize().setLength(10)
let plane = new Plane.setFromNormalAndCoplanarPoint(normal, origin)
Now, we create a random 3D point, and project it onto the plane.
let point1 = new Vector3(Math.random(), Math.random(), Math.random()).normalize()
let projectedPoint1 = new Vector3()
plane.projectPoint(point1, projectedPoint1)
The projectedPoint1 variable is now co-planar with your plane. But this plane is infinite, with no discrete X/Y axes. So currently we can only get the distance from the origin to the projected point.
let distance = origin.distanceTo(projectedPoint1)
In order to turn this into a Cartesian coordinate, you need to define at least one axis. To make this truly random, let's compute a random +Y axis:
let tempY = new Vector3(Math.random(), Math.random(), Math.random())
let pY = new Vector3()
plane.projectPoint(tempY, pY)
pY.normalize()
Now that we have +Y, let's get +X:
let pX = new Vector3().crossVectors(pY, normal)
pX.normalize()
Now, we can project the plane-projected point onto the axis vectors to get the Cartesian coordinates.
let x = projectedPoint1.clone().projectOnVector(pX).distanceTo(origin)
if(!projectedPoint1.clone().projectOnVector(pX).normalize().equals(pX)){
x = -x
}
let y = projectedPoint1.clone().projectOnVector(pY).distanceTo(origin)
if(!projectedPoint1.clone().projectOnVector(pY).normalize().equals(pY)){
y = -y
}
Note that in order to get negative values, I check a normalized copy of the axis-projected vector against the normalized axis vector. If they match, the value is positive. If they don't match, the value is negative.
Also, all the clone-ing I did above was to be explicit with the steps. This is not an efficient way to perform this operation, but I'll leave optimization up to you.
EDIT: My logic for determining the sign of the value was flawed. I've corrected the logic to normalize the projected point and check against the normalized axis vector.
What we have is angle(0-360), our small object is in the center and we have width and height of each object. I have tried to do dividing widths and heights without success. Task is to place small object in the bottom of the big object, and when big object get rotated we need small object be on the bottom again, but that bottom could be up or left or right, that why i guessed we need the angle.
So basically we need to create circle movement of small object. Radius of that circle will be big object height/2. But how to calculate X and Y locations from the center to place the small object?
Represantation in images:
Here we have the default state with angle 0
Here we have angle 47
And here we have angle 227
Lets say you want to calculate your new coordinates r pixels away from the point (X,Y) for an angle a. And if your new coordinates would be (x1, y1),
x1 = X + r * COS(a)
y1 = Y + r * SIN(a)
Here is some more info and techniques
I'm trying to draw an ellipsoid in 3d space out of individual blocks.
I have no problem with 2D ellipses, but as far as 3D goes I'm having some trouble. I'm using Bresenham's circle algorithm to draw 2D ellipses. What I'm trying to do is draw 2D ellipses in layers with an increasing (starting from the bottom going up, using symmetry for the other half) radius on both the X radius and Y radius.
It all sounds like it would work, but when I go to implement it, I can't figure out how to alter the x radius and y radius to make the curve of the ellipsoid.
Your 2D slices should all have the same orientation and aspect ratio.
If your ellipsoid is axis-aligned, they should also have the same center.
Your slices should scale proportionally to:
scale = sqrt(1 - ((center-z)/half_vsize)^2)
where:
z = height of the current slice
center = height of the largest slice
half_vsize = half the vertical size of the ellipsoid
If (x0, y0) is the x- and y-width of the largest slice, (x, y) = (scale*x0, scale*y0) is the x- and y-width of the slice at height z.
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.