Limit a ray to not go beyond a plane - roblox

As you can see from the image
I want the mouse hit to not go beyond the origin’s position. If anyone knows about this, I’d like to know.
My first idea was subtract the origin to mouse hit's position and get their magnitude then subtract the mouse hit with mouse hit's look vector(unit vector) multiplied by the magnitude, but this does not work.

From your image, it's a little unclear what defines the plane that is your limit? I'm going to assume that it is the plane defined by the face of a box, and the normal vector to that face.
You know the camera position, and the direction of your ray, so once you define your plane, you should be able to calculate the intersection of that plane with your ray.
See Ray-plane intersection by Scratchapixel. They have done an excellent job breaking down the equations that make this work. (Note, for this code I have renamed the l variables to r, as the variable l0 was somewhat easy to mistake for the constant 10)
In a LocalScript, try this...
local WorldWorkspace = game:GetService("Workspace")
local Players = game:GetService("Players")
local part = WorldWorkspace.Part
local camera = WorldWorkspace.CurrentCamera
local player = Players.LocalPlayer
local mouse = player:GetMouse()
--[[
GIVEN THE EQUATIONS...
(p - p0) ⋅ n = 0
where...
- p is the intersection point (Vector3)
- p0 is the origin of the plane (Vector3)
- n is the plane perpendicular to the origin point (Vector3)
(r0 + r * t) = p
where...
- r0 is the origin of the ray (Vector3)
- r is the ray direction (Vector3, normalized)
- t is the length of the ray (Vector3)
- p is the intersection point (Vector3)
THEREFORE...
r * t ⋅ n + (r0 - p0) ⋅ n = 0
AND SOLVING FOR t WE GET...
t = ((p0 - r0) ⋅ n) / (r ⋅ n))
]]
local function calculateIntersection(n : Vector3, p0 : Vector3, r : Vector3, r0 : Vector3) : (boolean, Vector3?)
-- returns false if the ray defined by r0 and r will never intersect with the plane defined by p0 and n
local denominator : Vector3 = n:Dot(r)
if denominator > 0 then
local p010 : Vector3 = p0 - r0
local t : Vector3 = p010:Dot(n) / denominator
local doesIntersect : boolean = t >= 0
local p : Vector3? = if doesIntersect then (r0 + (r * t)) else nil
return doesIntersect, p
end
return false, nil
end
local function getOrigin() : Vector3
-- calculate the origin of the intersection plane by simply using the rough bounding box.
local cubeCenter = part.CFrame.Position
local cubeSize = part.Size
-- return the back face of the cube for testing
return cubeCenter + Vector3.new(0, 0, cubeSize.Z * 0.5)
end
local function getPlane() : Vector3
local right = part.CFrame.RightVector
local up = part.CFrame.UpVector
local normalPlane = right:Cross(up).Unit
-- multiply by -1 to invert the outward direction of plane
return -1 * normalPlane
end
local function onClick()
-- NOTE - tweak the values of p0 and n to get the plane right
local p0 = getOrigin()
local n = getPlane()
local r = mouse.Origin.LookVector
local r0 = mouse.Origin.Position
local success, result = calculateIntersection(n, p0, r, r0)
if success then
-- draw a ball where the point intersects with the plane
-- NOTE - this will only draw on the local client, not the server
local ball = Instance.new("Part")
ball.Shape = Enum.PartType.Ball
ball.Size = Vector3.new(0.2, 0.2, 0.2)
ball.Position = result
ball.Anchored = true
ball.CanCollide = false
ball.Color = Color3.new(1, 0, 0)
ball.Parent = WorldWorkspace
end
end
-- listen for when the player clicks
mouse.Button1Down:Connect(onClick)
In my testing, I was able to lock collisions to the defined plane, but I needed to mess around with the getOrigin and getPlane functions to properly define where I wanted the plane to be.

Related

Calculate 3D cordinates from with camera matrix and know distance

I have been struggeling with this quiz question. This was part of FSG 2022 registration quiz and I can't figure out how to solve it
At first I thought that I can use extrinsic and intrinsic parameters to calculate 3D coordinates using equations described by Mathworks or in this article. Later I realized that the distance to the object is provided in camera frame, which means that this could be treat as a depth camera and convert depth info into 3d space as described in medium.com article
this article is using formula show below to calculate x and y coordinates and is very similar to this question, yet I can't get the correct solution.
One of my Matlab scripts attempting to solve it:
rot = eul2rotm(deg2rad([102 0 90]));
trans = [500 160 1140]' / 1000; % mm to m
t = [rot trans];
u = 795; % here was typo as pointed out by solstad.
v = 467;
cx = 636;
cy = 548;
fx = 241;
fy = 238;
z = 2100 / 1000 % mm to m
tmp_x = (u - cx) * z / fx;
tmp_y = (v - cy) * z / fy;
% attempt 1
tmp_cords = [tmp_x; tmp_y; z; 1]
linsolve(t', tmp_cords)'
% result is: 1.8913 1.8319 -0.4292
% attempt 2
tmp_cords = [tmp_x; tmp_y; z]
rot * tmp_cords + trans
% result is: 2.2661 1.9518 0.4253
If possible I would like to see the calculation process not any kind of a python code.
Correct answer is under the image.
Correct solution provided by the organisers were 2.030, 1.272, 0.228 m
The task states that the object's euclidean (straight-line) distance is 2.1 m. That doesn't mean its distance along z is 2.1 m. Those two only coincide if there is no x or y component in the object's translation to the camera frame.
The z component of the object's translation will be less than 2.1 meters.
You need to take a ray/vector for the screen space coordinates (normalized) and multiply that by the euclidean distance.
v_x = (u - cx) / fx;
v_y = (v - cy) / fy;
v_z = 1;
v = [v_x; v_y; v_z];
dist = 2.1;
tmp = v / norm(v) * dist;
The rotation may be an issue. Roll happens around X, then pitch happens around Y, and then yaw happens around Z. These operations are applied in that order, i.e. inner to outer.
R_Z * R_Y * R_X * v
My rotation matrix is
[[ 0. 0.20791 0.97815]
[ 1. 0. 0. ]
[ 0. 0.97815 -0.20791]]
That camera, taking the usual (X right, Y down, Z far) frame, would be looking, upside down, out the windshield, and slightly down.
Make sure that eul2rotm() does the right thing (specify axis order as 'XYZ') or that you use something else.
You can use rotvec2mat3d() to build individual rotation matrices from an axis-angle encoding.
Perhaps also review different MATLAB conventions regarding matrix multiplication: https://www.mathworks.com/help/images/migrate-geometric-transformations-to-premultiply-convention.html
I used Python and scipy.spatial.transform.Rotation.from_euler('xyz', [R_roll, R_pitch, R_yaw], degrees=True).as_matrix() to arrive at the sample solution.
Properly, the task should have specified a frame conversion step between vehicle and camera because the differing views are quite confusing, with a car having +X being forward and a camera having +Z being forward...
In addition to Christoph Rackwitz answer, which is correct and should get all the credited, here is a working Matlab script:
rot = eul2rotm(deg2rad([90 0 102]));
trans = [500 160 1140]' / 1000; % mm to m
u = 795;
v = 467;
cx = 636;
cy = 548;
fx = 241;
fy = 238;
v_x = (u - cx) / fx;
v_y = (v - cy) / fy;
v_z = 1;
v = [v_x; v_y; v_z];
dist = 2.1;
tmp = v / norm(v) * dist;
rot * tmp + trans

Patch on a sphere of varying size

Imagine a patch glued to a sphere. How would I manage to make the patch keep its center position and surface area as the sphere is scaled up or down? Normally, only the curvature of the patch should change, as it is « glued » to the sphere. Assume the patch is described as a set of ( latitude, longitude ) coordinates.
One possible solution would consist of converting the geographical coordinates of the patch into gnomonic coordinates (patch viewed perpendicularly directly from above), thereby making a 2D texture, which is then scaled up or down as the sphere changes its size. But I am unsure whether this is the right approach and how close of the desired effect this would be.
I am a newbie so perhaps Unity can do this simply with the right set options when applying a texture. In this case which input map projection should be used for the texture? Or maybe I should use a 3D surface and « nail » it somehow to the sphere.
Thank you!!
EDIT
I’m adding an illustration to show how the patch should be deformed as the sphere is scaled up or down. On a very small sphere, the patch would eventually wrap around. Whereas on a larger sphere, the patch would be almost flat. The deformation of the patch could be thought of as being similar to gluing the same sticker to spheres of different sizes.
The geometry of the patch could be any polygonal surface, and as previously mentioned must preserve its center position and surface area when the sphere is scaled up or down.
Assume you have a sphere of radius R1 centered at the origin of the standard coordinate system O e1 e2 e3. Then the sphere is given by all points x = [x[0], x[1], x[2]] in 3D that satisfy the equation x[0]^2 + x[1]^2 + x[2]^2 = R1^2. On this sphere you have a patch and the patch has a center c = [c[0], c[1], c[2]].
First, rotate the patch so that the center c goes to the north pole, then project it onto a plane, using an area preserving map for the sphere of radius R1, then map it back using the analogous area preserving map but for radius R2 sphere and finally rotate back the north pole to the scaled position of the center.
Functions you may need to define:
Function 1: Define spherical coordinates
x = sc(u, v, R):
return
x[0] = R*sin(u)*sin(v)
x[1] = R*sin(u)*cos(v)
x[2] = R*cos(u)
where
0 <= u <= pi and 0 <= v < 2*pi
Function 2: Define inverse spherical coordinates:
[u, v] = inv_sc(x, R):
return
u = arccos( x[2] / R )
if x[1] > 0
v = arccot(x[0] / x[1]) if x[1] > 0
else if x[1] < 0
v = 2*pi - arccot(x[1] / x[0])
else if x[1] = 0 and x[0] > 0
v = 0
else if x[1] = 0 and x[0] < 0
v = pi
where x[0]^2 + x[1]^2 + x[2]^2 = R^2
Function 3: Rotation matrix that rotates the center c to the north pole:
Assume the center c is given in spherical coordinates [uc, vc]. Then apply function 1
c = [c[0], c[2], c[3]] = sc(uc, vc, R1)
Then, find for which index i we have c[i] = min( abs(c[0]), abs(c[1]), abs(c[2])). Say i=2 and take the coordinate vector e2 = [0, 1, 0].
Calculate the cross-product vectors cross(c, e2) and cross(cross(c, e2), c), think of them as row-vectors, and form the 3 by 3 rotation matrix
A3 = c / norm(c)
A2 = cross(c, e2) / norm(cross(c, e2))
A1 = cross(A2, A3)
A = [ A1,
A2,
A3 ]
Functions 4:
[w,z] = area_pres(u,v,R1,R2):
return
w = arccos( 1 - (R1/R2)^2 * (1 - cos(u)) )
z = v
Now if you re-scale the sphere from radius R1 to radius R2 then any point x from the patch on the sphere with radius R1 gets transformed to the point y on the sphere of radius R2 by the following chain of transformations:
If x is given in spherical coordinates `[ux, vx]`, first apply
x = [x[0], x[1], x[2]] = sc(ux, vx, R1)
Then rotate with the matrix A:
x = matrix_times_vector(A, x)
Then apply the chain of transformations:
[u,v] = inv_sc(x, R1)
[w,z] = area_pres(u,v,R1,R2)
y = sc(w,z,R2)
Now y is on the R2 sphere.
Finally,
y = matrix_times_vector(transpose(A), y)
As a result all of these points y fill-in the corresponding transformed patch on the sphere of radius R2 and the patch-area on R2 equals the patch-area of the original patch on sphere R1. Plus the center point c gets just scaled up or down along a ray emanating from the center of the sphere.
The general idea behind this appriach is that, basically, the area element of the R1 sphere is R1^2*sin(u) du dv and we can look for a transformation of the latitude-longitude coordinates [u,v] of the R1 sphere into latitude-longitude coordinates [w,z] of the R2 sphere where we have the functions w = w(u,v) and z = z(u,v) such that
R2^2*sin(w) dw dz = R1^2*sin(u) du dv
When you expand the derivatives of [w,z] with respect to [u,v], you get
dw = dw/du(u,v) du + dw/dv(u,v) dv
dz = dz/du(u,v) du + dz/dv(u,v) dv
Plug them in the first formula, and you get
R2^2*sin(w) dw dz = R2^2*sin(w) * ( dw/du(u,v) du + dw/dv(u,v) dv ) wedge ( dz/du(u,v) du + dz/dv(u,v) dv )
= R1^2*sin(u) du dv
which simplifies to the equation
R2^2*sin(w) * ( dw/du(u,v) dz/dv(u,v) - dw/dv(u,v) dz/du(u,v) ) du dv = R^2*sin(u) du dv
So the general differential equation that guarantees the area preserving property of the transformation between the spherical patch on R1 and its image on R2 is
R2^2*sin(w) * ( dw/du(u,v) dz/dv(u,v) - dw/dv(u,v) dz/du(u,v) ) = R^2*sin(u)
Now, recall that the center of the patch has been rotated to the north pole of the R1 sphere, so you can think the center of the patch is the north pole. If you want a nice transformation of the patch so that it is somewhat homogeneous and isotropic from the patch's center, i.e. when standing at the center c of the patch (c = north pole) you see the patch deformed so that longitudes (great circles passing through c) are preserved (i.e. all points from a longitude get mapped to points of the same longitude), you get the restriction that the longitude coordinate v of point [u, v] gets transformed to a new point [w, z] which should be on the same longitude, i.e. z = v. Therefore such longitude preserving transformation should look like this:
w = w(u,v)
z = v
Consequently, the area-preserving equation simplifies to the following partial differential equation
R2^2*sin(w) * dw/du(u,v) = R1^2*sin(u)
because dz/dv = 1 and dz/du = 0.
To solve it, first fix the variable v, and you get the ordinary differential equation
R2^2*sin(w) * dw = R1^2*sin(u) du
whose solution is
R2^2*(1 - cos(w)) = R1^2*(1 - cos(u)) + const
Therefore, when you let v vary, the general solution for the partial differential equation
R2^2*sin(w) * dw/du(u,v) = R^2*sin(u)
in implicit form (equation that links the variables w, u, v) should look like
R2^2*(1 - cos(w)) = R1^2*(1 - cos(u)) + f(v)
for any function f(v)
However, let us not forget that the north pole stays fixed during this transformation, i.e. we have the restriction that w= 0 whenever u = 0. Plug this condition into the equation above and you get the restriction for the function f(v)
R2^2*(1 - cos(0)) = R1^2*(1 - cos(0)) + f(v)
R2^2*(1 - 1) = R1^2*(1 - 1) + f(v)
0 = f(v)
for every longitude v
Therefore, as soon as you impose longitudes to be transformed to the same longitudes and the north pole to be preserved, the only option you are left with is the equation
R2^2*(1 - cos(w)) = R1^2*(1 - cos(u))
which means that when you solve for w you get
w = arccos( 1 - (R1/R2)^2 * (1 - cos(u)) )
and thus, the corresponding area preserving transformation between the patch on sphere R1 and the patch on sphere R2 with the same area, fixed center and a uniform deformation at the center so that longitudes are transformed to the same longitudes, is
w = arccos( 1 - (R1/R2)^2 * (1 - cos(u)) )
z = v
Here I implemented some of these functions in Python and ran a simple simulation:
import numpy as np
import math
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
def trig(uv):
return np.cos(uv), np.sin(uv)
def sc_trig(cos_uv, sin_uv, R):
n, dim = cos_uv.shape
x = np.empty((n,3), dtype=float)
x[:,0] = sin_uv[:,0]*cos_uv[:,1] #cos_u*sin_v
x[:,1] = sin_uv[:,0]*sin_uv[:,1] #cos_u*cos_v
x[:,2] = cos_uv[:,0] #sin_u
return R*x
def sc(uv,R):
cos_uv, sin_uv = trig(uv)
return sc_trig(cos_uv, sin_uv, R)
def inv_sc_trig(x):
n, dim = x.shape
cos_uv = np.empty((n,2), dtype=float)
sin_uv = np.empty((n,2), dtype=float)
Rad = np.sqrt(x[:,0]**2 + x[:,1]**2 + x[:,2]**2)
r_xy = np.sqrt(x[:,0]**2 + x[:,1]**2)
cos_uv[:,0] = x[:,2]/Rad #cos_u = x[:,2]/R
sin_uv[:,0] = r_xy/Rad #sin_v = x[:,1]/R
cos_uv[:,1] = x[:,0]/r_xy
sin_uv[:,1] = x[:,1]/r_xy
return cos_uv, sin_uv
def center_x(x,R):
n, dim = x.shape
c = np.sum(x, axis=0)/n
return R*c/math.sqrt(c.dot(c))
def center_uv(uv,R):
x = sc(uv,R)
return center_x(x,R)
def center_trig(cos_uv, sin_uv, R):
x = sc_trig(cos_uv, sin_uv, R)
return center_x(x,R)
def rot_mtrx(c):
i = np.where(c == min(c))[0][0]
e_i = np.zeros(3)
e_i[i] = 1
A = np.empty((3,3), dtype=float)
A[2,:] = c/math.sqrt(c.dot(c))
A[1,:] = np.cross(A[2,:], e_i)
A[1,:] = A[1,:]/math.sqrt(A[1,:].dot(A[1,:]))
A[0,:] = np.cross(A[1,:], A[2,:])
return A.T # ready to apply to a n x 2 matrix of points from the right
def area_pres(cos_uv, sin_uv, R1, R2):
cos_wz = np.empty(cos_uv.shape, dtype=float)
sin_wz = np.empty(sin_uv.shape, dtype=float)
cos_wz[:,0] = 1 - (R1/R2)**2 * (1 - cos_uv[:,0])
cos_wz[:,1] = cos_uv[:,1]
sin_wz[:,0] = np.sqrt(1 - cos_wz[:,0]**2)
sin_wz[:,1] = sin_uv[:,1]
return cos_wz, sin_wz
def sym_patch_0(n,m):
u = math.pi/2 + np.linspace(-math.pi/3, math.pi/3, num=n)
v = math.pi/2 + np.linspace(-math.pi/3, math.pi/3, num=m)
uv = np.empty((n, m, 2), dtype=float)
uv[:,:,0] = u[:, np.newaxis]
uv[:,:,1] = v[np.newaxis,:]
uv = np.reshape(uv, (n*m, 2), order='F')
return uv, u, v
uv, u, v = sym_patch_0(18,18)
r1 = 1
r2 = 2/3
r3 = 2
limits = max(r1,r2,r3)
p = math.pi
x = sc(uv,r1)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x[:,0], x[:,1], x[:,2])
ax.set_xlim(-limits, limits)
ax.set_ylim(-limits, limits)
ax.set_zlim(-limits, limits)
ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')
plt.show()
B = rot_mtrx(center_x(x,r1))
x = x.dot(B)
cs, sn = inv_sc_trig(x)
cs1, sn1 = area_pres(cs, sn, r1, r2)
y = sc_trig(cs1, sn1, r2)
y = y.dot(B.T)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(y[:,0], y[:,1], y[:,2])
ax.set_xlim(-limits, limits)
ax.set_ylim(-limits, limits)
ax.set_zlim(-limits, limits)
ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')
plt.show()
cs1, sn1 = area_pres(cs, sn, r1, r3)
y = sc_trig(cs1, sn1, r3)
y = y.dot(B.T)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(y[:,0], y[:,1], y[:,2])
ax.set_xlim(-limits, limits)
ax.set_ylim(-limits, limits)
ax.set_zlim(-limits, limits)
ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')
plt.show()
One can see three figures of how a patch gets deformed when the radius of the sphere changes from radius 2/3, through radius 1 and finally to radius 2. The patch's area doesn't change and the transformation of the patch is homogeneous in all direction with no excessive deformation.
You could e.g. do something like
public class Example : MonoBehaviour
{
public Transform sphere;
public float latitude;
public float longitude;
private void Update()
{
transform.position = sphere.position
+ Quaternion.AngleAxis(longitude, -Vector3.up)
* Quaternion.AngleAxis(latitude, -Vector3.right)
* sphere.forward * sphere.lossyScale.x / 2f;
transform.LookAt(sphere);
transform.Rotate(90,0,0);
}
}
The pin would not be a child of the sphere. It would result in a pin (in red) like:
Alternatively as said you could make the pin a child of the sphere in a structure like
Sphere
|--PinAnchor
|--Pin
So in order to change the Pin position you would rotate the PinAnchor. The Pin itself would update its own scale so it has always a certain target scale e.g. like
public class Example : MonoBehaviour
{
public float targetScale;
private void Update()
{
var scale = transform.parent.lossyScale;
var invertScale = new Vector3(1 / scale.x, 1 / scale.y, 1 / scale.z);
if (float.IsNaN(invertScale.x)) invertScale.x = 0;
if (float.IsNaN(invertScale.y)) invertScale.y = 0;
if (float.IsNaN(invertScale.z)) invertScale.z = 0;
transform.localScale = invertScale * targetScale;
}
}
I am going to add another answer, because it is possible you may decide that different properties are important for your patch transformation, more specifically having minimal (in some sense) distortion, and the area preservation of the patch is not as important.
Assume you want to create a transformation from a patch (an open subset of the sphere with relatively well-behaved boundary, e.g. piecewise smooth or even piecewise geodesic boundary) on a sphere of radius R1 to a corresponding patch on a sphere of radius R2. However, you want the transformation to not distort the original patch on R1 wen mapping it to R2. Assume the patch on R1 has a distinguished point c, called the center. This could be its geometric center, i.e. its center of mass (barycenter), or a point selected in another way.
For this discussion, let us assume the center c is at the north pole of the sphere R1. If it is not, we can simply rotate it to the north pole (see my previous post for one way to rotate the center), so that the standard spherical coordinates [u, v] (latitude and longitude) naturally apply, i.e.
for sphere R1:
x[0] = R1*sin(u)*cos(v)
x[1] = R1*sin(u)*sin(v)
x[2] = R1*cos(u)
for sphere R2:
y[0] = R2*sin(w)*cos(z)
y[1] = R2*sin(w)*sin(z)
y[2] = R2*cos(w)
with point c being with coordinates [0,0] (or any [0,v] for that matter, as these coordinates have a singularity at the pole). Ideally, if you construct an isometric transformation between the two patches (isometry is a transformation that preserves distances, angles and consequently area), then you are done. The two spheres, however, have different radii R1 and R2 and so they have different intrinsic curvature, so there can be no isometry between the patches. Nevertheless, let us see what an isometry would have done: An isometry is a transformation that transforms the metric tensor (the line element, the way we measure distance on the sphere) of the first sphere to the metric tensor of the second, i.e.
Metric tensor of R1:
R1^2 * ( du^2 + (sin(u))^2 dv^2 )
Metric tensor of R2:
R2^2 * ( dw^2 + (sin(w))^2 dz^2 )
An isometry: [u,v] --> [w,z] so that
R1^2 * ( du^2 + (sin(u))^2 dv^2 ) = R2^2 * ( dw^2 + (sin(w))^2 dz^2 )
What an isometry would do, fist it would send spherical geodesics (great circles) to spherical geodesics, so in particular longitudinal circles of R1 should be mapped to longitudinal circles of R2, because we want the north pole of R1 to be mapped to the north pole of R2. Also, an isometry would preserve angles, so in particular, it would preserve angles between longitudinal circles. Since the angle between the zero longitudinal circle and the longitudinal circle of longitude v is equal to v (up to a translation by a constant if a global rotation of the sphere around the north pole is added, but we don't want that), then v should be preserved by an isometry (i.e. the isometry should preserve the bearing at the north pole). That implies that the desired isometric map between the patches should have the form
Map between patch on R1 and patch on R2,
which maps the north pole of R1 to the north pole of R2:
w = w(u, v)
z = v
Furthermore, since the sphere looks the same at any point and in any direction (it is homogeneous and isotropic everywhere), in particular this is true for the north pole and therefore an isometry should transform identically in all direction when looking from the north pole (the term is "isometric transformations should commute with the with the group of isometric automorphisms of the surfaces") which yields that w = w(u, v) should not depend on the variable v:
Map between patch on R1 and patch on R2,
which maps the north pole of R1 to the north pole of R2:
w = w(u)
z = v
The final steps towards finding an isometric transformation between the patches on R1 and R2 is to make sure that the metric tensors before and after the transformation are equal, i.e.:
R2^2 * ( dw^2 + (sin(w))^2 dz^2 ) = R1^2 * ( du^2 + (sin(u))^2 dv^2 )
dw = (dw/du(u)) du and dz = dv
R2^2 * ( (dw/du(u))^2 du^2 + (sin( w(u) ))^2 dv^2 ) = R1^2 * ( du^2 + (sin(u))^2 dv^2 )
set K = R1/R2
( dw/du(u) )^2 du^2 + (sin( w(u) ))^2 dv^2 = K^2 du^2 + K^2*(sin(u))^2 dv^2
For the latter equation to hold, we need the function w = w(u) to satisfy the following two restrictions
dw/du(u) = K
sin(w(u)) = K * sin(u)
However, we have only one function w(u) and two equations which are satisfied only when K = 1 (i.e. R1 = R2) which is not the case. This is where the isometric conditions break and that is why there is no isometric transformation between a patch on sphere R1 and a patch on R2 when R1 != R2. One thing we can try to do is to find a transformation that in some reasonable sense minimizes the discrepancy between the metric tensors (i.e. we would like to minimize somehow the degree of non-isometricity of the transformation [w = w(u), z = v] ). To that end, we can define a Lagrangian discrepancy function (yes, exactly like in physics) and try to minimize it:
Lagrangian:
L(u, w, dw/du) = ( dw/du - K )^2 + ( sin(w) - K*sin(u) )^2
minimize the action:
S[w] = integral_0^u2 L(u, w(u), dw/du(u))du
or more explicitly, find the function `w(u)` that makes
the sum (integral) of all discrepancies:
S[w] = integral_0^u2 ( ( dw/du(u) - K )^2 + ( sin(w(u)) - K*sin(u) )^2 )du
minimal
In order to find the function w(u) that minimizes the discrepancy integral S[w] above, one needs to derive the Euler-Lagrange equations associated to the Lagrangian L(u, w, dw,du) and to solve them. The Euler-Lagrange equation in this case is one and it is second derivative one:
d^2w/du^2 = sin(w)*cos(w) - K*sin(u)*cos(w)
w(0) = 0
dw/du(0) = K
or using alternative notation:
w''(u) = sin(w(u))*cos(w(u)) - K*sin(u)*cos(w(u))
w(0) = 0
w'(0) = K
The reason for the condition w'(0) = K comes from imposing the isometric identity
( dw/du(u) )^2 du^2 + (sin( w(u) ))^2 dv^2 = K^2 du^2 + K^2*(sin(u))^2 dv^2
When u = 0, we already know w(0) = 0 because we want the north pole to be mapped to the north pole and so the latter identity simplifies to
( dw/du(0) )^2 du^2 + (sin(0))^2 dv^2 = K^2 du^2 + K^2*(sin(0))^2 dv^2
( dw/du(0) )^2 du^2 = K^2 du^2
( dw/du(0) )^2 = K^2
which holds when
dw/du(0) = u'(0) = K
Now, to obtain a north -pole respecting transformation between circular patches on two spheres of radii R1 and R2 respectively, that has as little distortion as possible (with respect to the error Lagrnagian), we have to solve the non-linear initial value problem
d^2w/du^2 = sin(w)*cos(w) - K*sin(u)*cos(w)
w(0) = 0
dw/du(0) = K
or written as a system of two first-derivative differential equations (Hamiltonain form):
dw/du = p
dp/du = sin(w)*cos(w) - K*sin(u)*cos(w)
w(0) = 0
p(0) = K
I seriously doubt that this is an exactly solvable (integrable) system of ordinary differential equations, but a numerical integration with a reasonably small integration step can give an excellent discrete solution, which combined with a good interpolation scheme, like cubic splines, can give you a very accurate solution.
Now, if you do not care too much about exactly equal areas between the patches, but reasonably close areas and would actually prefer to have a smallest possible (in some sence) geometric deformation, you can simply use this model and stop here. However, if you really insist on the equal area between the two patches, you can continue further, by splitting your original patch (call it D1) on sphere R1 into a subpatch C1 inside D1 with the same center as D1, such that the difference D1 \ C1 is a narrow frame surrounding C1. Let the image of C1 under the map w = w(u), z = v, defined above, be denoted by C2. Then to find a transformation (a map) from the patch D1 onto a patch D2 on the sphere R2, which has the same area as D1 and includes C2, you can piece together one map from two submaps:
w = w(u)
z = v
for [u,v] from C1 ---> [w,z] from C2
w = w_ext(u, v)
z = v
for [u,v] from D1 \ C1 ---> [w,z] from D2 \ C2
The question is how to find the extension transfromation w_ext(u). For the area of D2 to be equal to the area of D1, you need to choose w_ext(u) so that
integra_(D1 \ C1) sin(w_ext(u)) dw_ext/du(u) du dv = (R1/R2)^2 Area(D1) - Area(C2) ( = the areas on the right are constants )
Now, pick a suitable function (you can start with a cosntant if you want) f(u), say a polynomial with adjustable coefficients, so that
integra_(D1 \ C1) f(u) du dv = (R1/R2)^2 Area(D1) - Area(C2)
e.g.
f(u) = L (constant) such that
integra_(D1 \ C1) L du dv = (R1/R2)^2 Area(D1) - Area(C2)
i.e.
L = ( (R1/R2)^2 Area(D1) - Area(C2) ) / integra_(D1 \ C1) du dv
Then solve the differential eqution
sin(w) dw/du = f(u)
e.g.
sin(w) dw/du = L
w(u) = arccos(L*u + a)
But in this case it is imortant to glue this solution with the previous one, so the initial condition of w_ext(u) matters, possibly depending on the direction v, i.e
w_ext(u, v) = arccos(L*u + a(v))
So there exists a somewhat more laborious approach, but it has a lot of details and is more comlicated.

How to find two points on a line perpendicular to a given line

From this image:
I have a line P0P1 with both points known. How can I find two points, say Q0 and Q1 on a line perpendicular to P0P1?
Supposing you are working in 2D. To compute Q0Q1, a perpendicular vector to P0P1, you will need a "temporary" perpendicular vector in the 3D space (either this vector "enters your screen" or "leaves" it). Thanks to this temporary vector, you will be able to get Q0Q1.
Vector3 P0 = ... ;
Vector3 P1 = ... ;
Vector3 P0P1 = P1 - P0 ;
Vector3 cross = Vector3.Cross( P0.normalized, P0P1.normalized ); // Temporary vector
Vector3 Q0Q1 = Vector3.Cross( cross.normalized, P0P1.normalized );
Vector3 Q0 = (P0 + P1) * 0.5f + cross ;
Vector3 Q1 = (P0 + P1) * 0.5f - cross ;
Check the manual if you want to understand how the Vector3.Cross operation works.

Point in plane closest to point

I'm trying to find a point in the plane closest to a given point.
I have the equation of the plane, the point and the distance between them.
How do I find the point in the plane which is closest to the given point?
I have a tetrahedron with sides:
bcx=0, acy=0, abz=0, x/a+y/b+z/c=1 (a,b,c are not to mix with Ax+By+Cz=D in the planes equation, they are to be entered when running the script).
function [d n]=tetradist(x,y,z,a,b,c)
if z>0 && y>0 && x>0 && z<c && y<b && x<a && x/a+y/b+z/c<1
d1=abs(a*b*z)/sqrt((a*b)^2);
d2=abs(b*c*x)/sqrt((b*c)^2);
d3=abs(a*c*y)/sqrt((a*c)^2);
d4=abs(b*c*x+a*c*y+a*b*z-a*b*c)/sqrt((b*c)^2 + (a*c)^2 + (a*b)^2);
A = [d1 d2 d3 d4];
B = sort(A,'ascend');
d = B(1);
point=[x y z];
if d==d1
normalv=[0 0 a*b]';
elseif d==d2
normalv=[b*c 0 0]';
elseif d==d3
normalv=[0 a*c 0]';
else
normalv=[b*c a*c a*b]';
end
end
So now I have the shortest distance, my point as a vector and the normal vector of the closest plane. Now how do I find the point in said plane which is closest to my point 'point'?
Thanks in advance!
If the equation of your plane is Ax + By + Cz = D and the location of the point is (P, Q, R) then the location in the plane that is closest to the point is
(P,Q,R) + λ * (A,B,C)
where
λ = (D - P*A - B*Q - C*R) / (A^2 + B^2 + C^2)
The following Matlab code calculates this point
function x = closestpoint(n, d, p)
# n is the vector [A,B,C] that defines the plane
# d is the distance of the plane from the origin
# p is the point [P,Q,R]
v = (d - sum(p.*n)) / sum(n.*n);
x = p + v * n;

How to find the bisector of an angle in MATLAB

I have a question connected to this code:
t = -20:0.1:20;
plot3(zeros(size(t)),t,-t.^2);
grid on
hold on
i = 1;
h = plot3([0 0],[0 t(i)],[0 -t(i)^2],'r');
h1 = plot3([-1 0],[0 0],[-400 -200],'g');
for(i=2:length(t))
set(h,'xdata',[-1 0],'ydata',[0 t(i)],'zdata',[-400 -t(i)^2]);
pause(0.01);
end
In this code, I plot two intersecting lines. H1, and H2. H1 is fixed, H2 moves as a function of time. H2 happens to trace a parabola in this example, but its movement could be arbitrary.
How can I calculate and draw the bisector of the angle between these two intersecting lines for every position of the line H2? I would like to see in the plot the bisector and the line H2 moving at the same time.
Solving this problem for one position of H2 is sufficient, since it will be the same procedure for all orientations of H2 relative to H1.
I am not a geometry genius, there is likely an easier way to do this. As of now, no one has responded though, so this will be something.
You have three points in three space:
Let A be the common vertice of the two line segments.
Let B and C be two known points on the two line segments.
Choose an arbitrary distance r where
r <= distance from A to B
r <= distance from A to C
Measure from A along line segment AB a distance of r. This is point RB
Measure from A along line segment AC a distance or r. This is point RC
Find the mid point of line segment connecting RB and RC. This is point M
Line segment AM is the angular bisector of angle CAB.
Each of these steps should be relatively easy to accomplish.
Here is basically MatlabDoug's method with some improvement on the determination of the point he calls M.
t = -20:0.1:20;
plot3(zeros(size(t)),t,-t.^2);
grid on
hold on
v1 = [1 0 200];
v1 = v1/norm(v1);
i = 1;
h = plot3([-1 0],[0 t(i)],[-400 -t(i)^2],'r');
h1 = plot3([-1 0],[0 0],[-400 -200],'g');
l = norm([1 t(i) -t(i)^2+400]);
p = l*v1 + [-1 0 -400];
v2 = (p + [0 t(i) -t(i)^2])/2 - [-1 0 -400];
p2 = [-1 0 -400] + v2/v2(1);
h2 = plot3([-1 p2(1)],[0 p2(2)],[-400 p2(3)],'m');
pause(0.1)
for(i=2:length(t))
l = norm([1 t(i) -t(i)^2+400]);
p = l*v1 + [-1 0 -400];
v2 = (p + [0 t(i) -t(i)^2])/2 - [-1 0 -400];
p2 = [-1 0 -400] + v2/v2(1);
set(h,'xdata',[-1 0],'ydata',[0 t(i)],'zdata',[-400 -t(i)^2]);
set(h2,'xdata',[-1 p2(1)],'ydata',[0 p2(2)],'zdata',[-400 p2(3)]);
pause;
end
I just use the following:
Find the normalized vectors AB, and AC, where A is the common point of the segments.
V = (AB + AC) * 0.5 // produces the direction vector that bisects AB and AC.
Normalize V, then do A + V * length to get the line segment of the desired length that starts at the common point.
(Note that this method does not work on 3 points along a line to produce a perpendicular bisector, it will yield a vector with no length in that case)
I have added a C# implementation (in the XZ plane using Unity 3D Vector3 struct) that handles Perpendicular and Reflex bisectors in case someone that knows MATLAB would translate it.
public Vector3 GetBisector(Vector3 center, Vector3 first, Vector3 second)
{
Vector3 firstDir = (first - center).normalized;
Vector3 secondDir = (second - center).normalized;
Vector3 result = ((firstDir + secondDir) * 0.5f).normalized;
if (IsGreaterThan180(-firstDir, secondDir))
{
// make into a reflex vector
(result.x, result.z) = (-result.x, -result.z);
}
if (result.sqrMagnitude < 0.99f)
{
// we have a colinear set of lines.
// return the perpendicular bisector.
result = Vector3.Cross(Vector3.up, -firstDir).normalized;
}
return result;
}
bool IsGreaterThan180(Vector3 dir, Vector3 dir2)
{
// < 0.0 for clockwise ordering
return (dir2.x * dir.z - dir2.z * dir.x) < 0.0f;
}
Also note that the returned bisector is a vector of unit length. Using "center + bisector * length" could be used to place it into worldspace.