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I'm trying to align multiple line objects along a human body circumference depending on the orientation of the triangles from the mesh. I would like to put the lines parallel to the mesh. I correctly assign the position for the lines along the circumference, but I also need to add the rotation of the lines such that to be parallel with the body.
The body is a mesh formed by multiple triangles and every line is "linked" with a triangle.
All I have is:
3 points for the closest triangle from the mesh for every line
The normal of the triangle
The positions for the instantiated lines (2 points, start and end)
I need to calculate the angle for every X, Y, Z axes for the line such that the normal of the triangle is perpendicular with the line mesh. I don't know how to get the desired angle. I really appreciate if someone would like to help me.
input:
FVector TrianglePoints[3];
FVector Triangle_Normal; //Calculated as (B-A)^(C-A), where A,B,C are the points of the triangle
FVector linePosition; //I also have the start line and the endLine position if that helps
ouput:
//FRotator rotation(x,y,z), such that the triangle normal and the line object to be perpendicular.
An overview of the circumference line construction. Now the rotation is calculated using the Start position and End position for each line. When we cross some irregular parts of the mesh we want to rotate the lines correctly. Now the rotation is fixed, depending just on the line start and end position.
If I have understood correctly your goal, here is some related vector geometry:
A,B,C are the vertices of the triangle:
A = [xA, yA, zA],
B = [xB, yB, zB]
C = [xC, yC, zC]
K,L are the endpoints of the line-segment:
K = [xK, yK, zK]
L = [xL, yL, zL]
vectors are interpreted as row-vectors
by . I denote matrix multiplication
by x I denote cross product of 3D vectors
by t() I denote the transpose of a matrix
by | | I denote the norm (magnitude) of a vector
Goal: find the rotation matrix and rotation transformation of segment KL
around its midpoint, so that after rotation KL is parallel to the plane ABC
also, the rotation is the "minimal" angle rotation by witch we need to
rotate KL in order to make it parallel to ABC
AB = B - A
AC = C - A
KL = L - K
n = AB x AC
n = n / |n|
u = KL x n
u = u / |u|
v = n x u
cos = ( KL . t(v) ) / |KL|
sin = ( KL . t(n) ) / |KL|
U = [[ u[0], u[1], u[2] ],
[ v[0], v[1], v[2] ],
[ n[0], n[1], n[2] ],
R = [[1, 0, 0],
[0, cos, sin],
[0, -sin, cos]]
ROT = t(U).R.U
then, one can rotate the segment KL around its midpoint
M = (K + L)/2
Y = M + ROT (X - M)
Here is a python script version
A = np.array([0,0,0])
B = np.array([3,0,0])
C = np.array([2,3,0])
K = np.array([ -1,0,1])
L = np.array([ 2,2,2])
KL = L-K
U = np.empty((3,3), dtype=float)
U[2,:] = np.cross(B-A, C-A)
U[2,:] = U[2,:] / np.linalg.norm(U[2,:])
U[0,:] = np.cross(KL, U[2,:])
U[0,:] = U[0,:] / np.linalg.norm(U[0,:])
U[1,:] = np.cross(U[2,:], U[0,:])
norm_KL = np.linalg.norm(KL)
cos_ = KL.dot(U[1,:]) / norm_KL
sin_ = KL.dot(U[2,:]) / norm_KL
R = np.array([[1, 0, 0],
[0, cos_, sin_],
[0,-sin_, cos_]])
ROT = (U.T).dot(R.dot(U))
M = (K+L) / 2
K_rot = M + ROT.dot( K - M )
L_rot = M + ROT.dot( L - M )
print(L_rot)
print(K_rot)
print(L_rot-K_rot)
print((L_rot-K_rot).dot(U[2,:]))
A more inspired solution was to use a procedural mesh, generated at runtime, that have all the requirements that I need:
Continuously along multiple vertices
Easy to apply a UV map for texture tiling
Can be updated at runtime
Isn't hard to compute/work with it
I have a line (P1, P2), and a point on that line (midpoint). What equation can I used to find the perpendicular line of line (P1, P2), that passes through midpoint. The point labelled with a '?' is unknown. I do not wish to use angles, only the 3 points given (P1, P2, midpoint). The line P1, P2 can be of any orientation/angle.
Thanks in advance.
Let define vector
D = P2 - P1 (dx=x2-x1, dy = y2-y1)
and middle point
mx = (x2+x1)/2
my = (y2+y1)/2
Perpendicular to D vector
PD = (-dy, dx)
Unit (normalized) perpendicular vector
U = (-dy / L, dx / L)
where
L = Sqrt (dx * dx + dy * dy)
And coordinates of point lying at distance F from the middle are
x = mx + U.x * F
y = my + U.y * F
or (for point at another side)
x = mx - U.x * F
y = my - U.y * F
Coordinates of P1: (x1,y1)
Coordinates of P2: (x2,y2)
Coordinates of midpoint: ( (x1+x2)/2 , (y1+y2)/2)
Slope of the P1P2 line: (y1-y2)/(x1-x2)
Slope of any perpendicular line to P1P2: (x2-x1)/(y1-y2)
Equation of the red line: y - (y1+y2)/2 = ((x2-x1)/(y1-y2))*(x - (x1+x2)/2)
If you have the actual values of coordinates of P1 y P2, then just make a substitution.
I try to rotate and translate an equilateral triangle in 3D until his vertices reach some coordinates.
The vertices coordinates F,G,H and F',G',H' are known :
I was able to find the new centroid c' coordinates like this :
c'.x = ( F'.x + G'.x + H'.x ) / 3
c'.y = ( F'.y + G'.y + H'.y ) / 3
c'.z = ( F'.z + G'.z + H'.z ) / 3
So no problem to translate the triangle. But I can't find a way to calculate the rotations needed to put F'G'H' triangle in the right position...
I have to know by how much the triangle F'G'H' has to be rotated in degrees, around each axis (x,y,z), knowing that the rotations of the initial triangle are 0°.
By rotation for each axis, I'm talking about this:
Any ideas?
trick is to find the normal vectors of the triangles using cross product b4 and after rotations
v1 = (F.x - G.x, F.y - G.y, F.z - G.z)
v2 = (F.x - H.x, F.y - H.y, F.z - H.z)
n = cross_prod(v1, v2) # see http://en.wikipedia.org/wiki/Cross_product
n = n / norm(n) # normalize to unit vector
v'1 = (F'.x - G'.x, F'.y - G'.y, F'.z - G'.z)
v'2 = (F'.x - H'.x, F'.y - H'.y, F'.z - H'.z)
n' = cross_prod(v'1, v'2)
n' = n' / norm(n')
rot = arc_cos(n.x * n'.x + n.y * n'.y + n.z * n'.z)
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I have a triangle and I know the coordinates of two vertices: A=(x1,y1),B=(x2,y2)
All the angles: ABC=90∘,CAB=30∘ and BCA=60∘ and all the edge lengths. How can I find the coordinates of the third vertex C=(x3,y3)?
I know there are two solutions (I want both).
You know p1 and p2. You know the internal angles.
Make a ray from p1 trough p2, and rotate it CW or CCW 30° around p1.
Make a line trough p1 and p2, and rotate it 90° around p2.
Calculate the intersections.
You get the points:
x3 = x2 + s*(y1 - y2)
y3 = y2 + s*(x2 - x1)
and
x3 = x2 + s*(y2 - y1)
y3 = y2 + s*(x1 - x2)
where s = 1/sqrt(3) ≈ 0.577350269
In a 30-60-90 right triangle, smallest leg (the smallest side adjacent the 90 degree angle) has length of 1/2 of the hypotenuse (the side opposite to 90 degree angle), so since you have the side lengths, you can determine which leg is the line segment AB.
From that you deduce where do the angles go.
Then to compute the coordinate you just need to pick the point on the circle of the radius with the correct radius length at the correct angle.
Two solutions come from measuring the angle clock-wise or counter-clockwise, and result in symmetrical triangles, with the edge AB being the line of symmetry.
Since you already have given the angles, compute the length of AB via quadratic formula
L(AB) = Sqrt[(x1-x2)^2 + (y1-y2)^2].
Now, let x = L(AC) = 2*L(BC) so since it is the right triangle,
L(AC)^2 = L(BC)^2 + L(AB)^2,
x^2 = (0.5x)^2 + L(AB)^2, so L(AB) = x*Sqrt[3]/2,
and since you already computed L(AB) you now have x.
The angle of the original AB is a = arctan([y2-y1]/[x2-x1]).
Now you can measure 30 degrees up or down (use a+30 or a-30 as desired)
and mark the point C on the circle (centered at A) of radius x (which we computed above) at the angle a +/- 30.
Then, C has coordinates
x3 = x1 + x*cos(a+30)
y3 = y1 + x*sin(a+30)
or you can use (a-30) to get the symmetrical triangle.
Here is the code to return points of full polygon if two points and number of sides are provided as input.
This is written for Android(Java) and the logic can be re-used for other languages
private static final float angleBetweenPoints(PointF a, PointF b) {
float deltaY = b.y - a.y;
float deltaX = b.x - a.x;
return (float) (Math.atan2(deltaY, deltaX));
}
private static PointF pullPointReferenceToLineWithAngle(PointF a, PointF b,
float angle) {
float angleBetween = angleBetweenPoints(b, a);
float distance = (float) Math.hypot(b.x - a.x, b.y - a.y);
float x = (float) (b.x + (distance * Math.cos((angleBetween + angle))));
float y = (float) (b.y + (distance * Math.sin((angleBetween + angle))));
return new PointF(x, y);
}
private static List<PointF> pullPolygonPointsFromBasePoints(PointF a,
PointF b, int noOfSides) {
List<PointF> points = new ArrayList<>();
points.add(a);
points.add(b);
if (noOfSides < 3) {
return points;
}
float angleBetweenTwoSides = (float) ((((noOfSides - 2) * 180) / noOfSides)
* Math.PI / 180);
for (int i = 3; i <= noOfSides; i++) {
PointF nextPoint = pullPointReferenceToLineWithAngle(
points.get(i - 3), points.get(i - 2), angleBetweenTwoSides);
points.add(nextPoint);
}
return points;
}
Usage is onDraw method:
PointF a = new PointF(100, 600);
PointF b = new PointF(300, 500);
int noOfSides = 3;
List<PointF> polygonPoints = pullPolygonPointsFromBasePoints(a, b,
noOfSides);
drawPolyPoints(canvas, noOfSides, polygonPoints);
This is a right angled triangle. The angle ABC is 90 degrees, so calculate the vector joining A to B and call this AA and normalise it:
AA = (x2-x1,y2-y1) / |(x2-x1,y2-y1)|
A unit vector perpendicular to AA is given by
BB = (-(y2-y1),x2-x1) / |(x2-x1,y2-y1)|
Because AC is perpendicular to AB all you can obtain your first point P1 as
P1 = (x2,y2) + K * BB
where K is the scalar value equal to the length of side AC (which you say you already know in the question). Your second solution point P2 is then simply given by going in the negative BB direction
P2 = (x2,y2) - K * BB
I want to calculate a point on a given line that is perpendicular from a given point.
I have a line segment AB and have a point C outside line segment. I want to calculate a point D on AB such that CD is perpendicular to AB.
I have to find point D.
It quite similar to this, but I want to consider to Z coordinate also as it does not show up correctly in 3D space.
Proof:
Point D is on a line CD perpendicular to AB, and of course D belongs to AB.
Write down the Dot product of the two vectors CD.AB = 0, and express the fact D belongs to AB as D=A+t(B-A).
We end up with 3 equations:
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)
(Dx-Cx)(Bx-Ax)+(Dy-Cy)(By-Ay)=0
Subtitute the first two equations in the third one gives:
(Ax+t(Bx-Ax)-Cx)(Bx-Ax)+(Ay+t(By-Ay)-Cy)(By-Ay)=0
Distributing to solve for t gives:
(Ax-Cx)(Bx-Ax)+t(Bx-Ax)(Bx-Ax)+(Ay-Cy)(By-Ay)+t(By-Ay)(By-Ay)=0
which gives:
t= -[(Ax-Cx)(Bx-Ax)+(Ay-Cy)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
getting rid of the negative signs:
t=[(Cx-Ax)(Bx-Ax)+(Cy-Ay)(By-Ay)]/[(Bx-Ax)^2+(By-Ay)^2]
Once you have t, you can figure out the coordinates for D from the first two equations.
Dx=Ax+t(Bx-Ax)
Dy=Ay+t(By-Ay)
function getSpPoint(A,B,C){
var x1=A.x, y1=A.y, x2=B.x, y2=B.y, x3=C.x, y3=C.y;
var px = x2-x1, py = y2-y1, dAB = px*px + py*py;
var u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
var x = x1 + u * px, y = y1 + u * py;
return {x:x, y:y}; //this is D
}
There is a simple closed form solution for this (requiring no loops or approximations) using the vector dot product.
Imagine your points as vectors where point A is at the origin (0,0) and all other points are referenced from it (you can easily transform your points to this reference frame by subtracting point A from every point).
In this reference frame point D is simply the vector projection of point C on the vector B which is expressed as:
// Per wikipedia this is more efficient than the standard (A . Bhat) * Bhat
Vector projection = Vector.DotProduct(A, B) / Vector.DotProduct(B, B) * B
The result vector can be transformed back to the original coordinate system by adding point A to it.
A point on line AB can be parametrized by:
M(x)=A+x*(B-A), for x real.
You want D=M(x) such that DC and AB are orthogonal:
dot(B-A,C-M(x))=0.
That is: dot(B-A,C-A-x*(B-A))=0, or dot(B-A,C-A)=x*dot(B-A,B-A), giving:
x=dot(B-A,C-A)/dot(B-A,B-A) which is defined unless A=B.
What you are trying to do is called vector projection
Here i have converted answered code from "cuixiping" to matlab code.
function Pr=getSpPoint(Line,Point)
% getSpPoint(): find Perpendicular on a line segment from a given point
x1=Line(1,1);
y1=Line(1,2);
x2=Line(2,1);
y2=Line(2,1);
x3=Point(1,1);
y3=Point(1,2);
px = x2-x1;
py = y2-y1;
dAB = px*px + py*py;
u = ((x3 - x1) * px + (y3 - y1) * py) / dAB;
x = x1 + u * px;
y = y1 + u * py;
Pr=[x,y];
end
I didn't see this answer offered, but Ron Warholic had a great suggestion with the Vector Projection. ACD is merely a right triangle.
Create the vector AC i.e (Cx - Ax, Cy - Ay)
Create the Vector AB i.e (Bx - Ax, By - Ay)
Dot product of AC and AB is equal to the cosine of the angle between the vectors. i.e cos(theta) = ACx*ABx + ACy*ABy.
Length of a vector is sqrt(x*x + y*y)
Length of AD = cos(theta)*length(AC)
Normalize AB i.e (ABx/length(AB), ABy/length(AB))
D = A + NAB*length(AD)
For anyone who might need this in C# I'll save you some time:
double Ax = ;
double Ay = ;
double Az = ;
double Bx = ;
double By = ;
double Bz = ;
double Cx = ;
double Cy = ;
double Cz = ;
double t = ((Cx - Ax) * (Bx - Ax) + (Cy - Ay) * (By - Ay)) / (Math.Pow(Bx - Ax, 2) + Math.Pow(By - Ay, 2));
double Dx = Ax + t*(Bx - Ax);
double Dy = Ay + t*(By - Ay);
Here is another python implementation without using a for loop. It works for any number of points and any number of line segments. Given p_array as a set of points, and x_array , y_array as continues line segments or a polyline.
This uses the equation Y = mX + n and considering that the m factor for a perpendicular line segment is -1/m.
import numpy as np
def ortoSegmentPoint(self, p_array, x_array, y_array):
"""
:param p_array: np.array([[ 718898.941 9677612.901 ], [ 718888.8227 9677718.305 ], [ 719033.0528 9677770.692 ]])
:param y_array: np.array([9677656.39934991 9677720.27550726 9677754.79])
:param x_array: np.array([718895.88881594 718938.61392781 718961.46])
:return: [POINT, LINE] indexes where point is orthogonal to line segment
"""
# PENDIENTE "m" de la recta, y = mx + n
m_array = np.divide(y_array[1:] - y_array[:-1], x_array[1:] - x_array[:-1])
# PENDIENTE INVERTIDA, 1/m
inv_m_array = np.divide(1, m_array)
# VALOR "n", y = mx + n
n_array = y_array[:-1] - x_array[:-1] * m_array
# VALOR "n_orto" PARA LA RECTA PERPENDICULAR
n_orto_array = np.array(p_array[:, 1]).reshape(len(p_array), 1) + inv_m_array * np.array(p_array[:, 0]).reshape(len(p_array), 1)
# PUNTOS DONDE SE INTERSECTAN DE FORMA PERPENDICULAR
x_intersec_array = np.divide(n_orto_array - n_array, m_array + inv_m_array)
y_intersec_array = m_array * x_intersec_array + n_array
# LISTAR COORDENADAS EN PARES
x_coord = np.array([x_array[:-1], x_array[1:]]).T
y_coord = np.array([y_array[:-1], y_array[1:]]).T
# FILAS: NUMERO DE PUNTOS, COLUMNAS: NUMERO DE TRAMOS
maskX = np.where(np.logical_and(x_intersec_array < np.max(x_coord, axis=1), x_intersec_array > np.min(x_coord, axis=1)), True, False)
maskY = np.where(np.logical_and(y_intersec_array < np.max(y_coord, axis=1), y_intersec_array > np.min(y_coord, axis=1)), True, False)
mask = maskY * maskX
return np.argwhere(mask == True)
As Ron Warholic and Nicolas Repiquet answered, this can be solved using vector projection. For completeness I'll add a python/numpy implementation of this here in case it saves anyone else some time:
import numpy as np
# Define some test data that you can solve for directly.
first_point = np.array([4, 4])
second_point = np.array([8, 4])
target_point = np.array([6, 6])
# Expected answer
expected_point = np.array([6, 4])
# Create vector for first point on line to perpendicular point.
point_vector = target_point - first_point
# Create vector for first point and second point on line.
line_vector = second_point - first_point
# Create the projection vector that will define the position of the resultant point with respect to the first point.
projection_vector = (np.dot(point_vector, line_vector) / np.dot(line_vector, line_vector)) * line_vector
# Alternative method proposed in another answer if for whatever reason you prefer to use this.
_projection_vector = (np.dot(point_vector, line_vector) / np.linalg.norm(line_vector)**2) * line_vector
# Add the projection vector to the first point
projected_point = first_point + projection_vector
# Test
(projected_point == expected_point).all()
Since you're not stating which language you're using, I'll give you a generic answer:
Just have a loop passing through all the points in your AB segment, "draw a segment" to C from them, get the distance from C to D and from A to D, and apply pithagoras theorem. If AD^2 + CD^2 = AC^2, then you've found your point.
Also, you can optimize your code by starting the loop by the shortest side (considering AD and BD sides), since you'll find that point earlier.
Here is a python implementation based on Corey Ogburn's answer from this thread.
It projects the point q onto the line segment defined by p1 and p2 resulting in the point r.
It will return null if r falls outside of line segment:
def is_point_on_line(p1, p2, q):
if (p1[0] == p2[0]) and (p1[1] == p2[1]):
p1[0] -= 0.00001
U = ((q[0] - p1[0]) * (p2[0] - p1[0])) + ((q[1] - p1[1]) * (p2[1] - p1[1]))
Udenom = math.pow(p2[0] - p1[0], 2) + math.pow(p2[1] - p1[1], 2)
U /= Udenom
r = [0, 0]
r[0] = p1[0] + (U * (p2[0] - p1[0]))
r[1] = p1[1] + (U * (p2[1] - p1[1]))
minx = min(p1[0], p2[0])
maxx = max(p1[0], p2[0])
miny = min(p1[1], p2[1])
maxy = max(p1[1], p2[1])
is_valid = (minx <= r[0] <= maxx) and (miny <= r[1] <= maxy)
if is_valid:
return r
else:
return None