Suppose that I have the following graph. I know A,B,C,D,E location(x,y). My problem is how to find which line is the nearest with point E. And also I want to draw a new line (blue color) between point and the nearest line. How can I found the (x,y) so that I can draw the line?
You need to make projection of point E onto line AB, determine whether that projection lies in the range of AB segment and find length of perpendicular to projection point. Same for CD segment.
To find projection point, get direction vector of AB and AE
AB = (B - A)
AE = (E - A)
and find parameter using scalar product of
t = (AB * AE) / (AB * AB)
if t lies out of rnage 0..1, then the closest point of segment is one of the ends. Otherwise find propjection point and length of projection
P = A + AB * t
Len = length(E - P)
Related
Here, line segment ab is cast upward on arbitrary vector n where I do somethings to find the black point on the line segment cd. My question is, how do I find the point on ab that intersects with the inverted n vector coming down from the new point?
Looks like it will have the same x-coordinate as the black point (call this x). The slope of ab is m = (by - ay) / (bx - ax), so the y coordinate is mx + ay.
If the projection is parallel, by the Thales theorem the ratios are preserved.
|ae| / |ab| = |cf| / |cd| = r
which is known.
The searched point is, vectorially
e = a + r.ab = a + |cf|/|cd|.ab
Is there an algorithm for detecting if a polygon has a line of mirror symmetry, for example, for a polygon on the image below:
Calculate image moments for vertex set of polygon.
For example - sum all x[i] and y[i] for vertices and divide by number of vertices to get centroid coordinates (cx=m10/m00 and cy=m01/m00).
Do similar for central moments. For example:
mu11 = sum((x[i]-cx)*(y[i]-cy))
Build line with angle Theta (in example section) through centroid.
If polygon has mirror symmetry, every vertex will have another vertex as mirrored image - it is enough to find any pair, then check another pairs in order)
For reference - how to get mirror point against line
L = A + AB * ScalarProduct(AB, AP) / ScalarProduct(AB, AB)
Mirror point
P' = P + 2*(L-P) = 2*L-P
I have drawn a 2D representation of the problem, but I will eventually have to solve this in 3 dimensions.
A line is drawn from an origin point to infinity, in a direction given by pitch and yaw. There is an axis-aligned box "in front of" the point.
I want to get the coordinates of the point on box that is closest to the line, or, if it intersects, closest to the origin point.
I.e., if the line were 'turned' towards the box, which point of the box would intersect with the line first?
Make parametric represenation of the ray with base point P0, direction vector D and parameter t
P = P0 + t * D
Get t for intersections of the ray with rectangle edges like this (similar in 3d):
Rect.Right = X0 + t * D.X
Find what intersection occurs first (smaller t), check coordinates of intersection. If inside edge - point found. If not, analyze intersection parameters with edgr continuations to determine what corner (perhaps edge in 3d) is the closest
Note that in 2d case you need to check only two possible edges - depending on ray direction. For example - left and bottom for your right picture. When you see that intersections are out of edges - check what of two corners is closer. The same for 3d - but intersection is possible for three faces and closest for more edges or corners.
Let the position of the point be (x0,y0,z0) and the box have corners (x1,y1,z1) and (x2,y2,z2) with x1 < x2, y1 < y2, z1 < z2. In terms of yaw ψ and pitch θ a unit vector along the line will be give by
(u,v,w) = (cos ψ sin θ, sin ψ, cos ψ cos θ)
The line is (x0,y0,z0) + t (u,v,w)
Finding intersection with one of the plane containing a face of the box is trivial. Say to find the intersection with the plane x=x1, just requires solving
x0 + u t = x1 so t = (x1-x0)/u. Once found its easy to check if the intersection is contained in the face.
The tricky situation happens if the line does not intersect the faces. Here we have a pair of skew lines, and wish to find the closest pair of points one on each line.
Consider the closest point to the edge from (x1,y1,z1) to (x2,y1,z1).
We want to find the parameter s,t such that that the points
(x0,y0,z0)+s(u,v,w)
(x1,y1,z1)+t(1,0,0)
are the closest. The segment joining these points must be perpendicular to both lines. A vector along that line is the cross product
N = (u,v,w) X (1,0,0) = (0,w,-v)
Now consider the plane through (x1,y1,z1) spanned by (1,0,0) and N, this has normal
N2 = (1,0,0) X N
= (1,0,0) X (0,w,-v)
= (0,v,w)
and the plane is defined by
P . N2 = (x1,y1,z1) . N2
Take a point on our ray
( (x0,y0,z0)+s(u,v,w) ) . N2 = (x1,y1,z1) . N2
(x0,y0,z0) . N2 + s (u,v,w) . N2 = (x1,y1,z1) . N2
s (u,v,w) . N2 = ((x1,y1,z1)-(x0,y0,z0)) . N2
s (v^2+w^2) = (y1-y0) v + (z1-z0) w
so
s = [ (y1-y0) v + (z1-z0) w ] / (v^2+w^2)
We can repeat the above for each edge on the box, find the closest points and select the smallest.
Goal
I want to determine if a test point is within a defined quadrilateral. I'm probably going to implement the solution in Matlab so I only need pseudo-code.
Inputs
Corners of quadrilateral : (x1,y1) (x2,y2) (x3,y3) (x4,y4)
Test point : (xt, yt)
Output
1 - If within quadrilateral
0 - Otherwise
Update
It was pointed out that identifying the vertices of the quadrilateral is not enough to uniquely identify it. You can assume that the order of the points determines the sides of the quadrilateral (point 1 connects 2, 2 connects to 3, 3 connects to 4, 4 connects to 1)
You can test the Point with this condition. Also you can treat quadrilateral as 2 triangles to calculate its area.
Use inpolygon. Usage would be inpolygon(xt,yt,[x1 x2 x3 x4],[y1 y2 y3 y4])
Since it's a simple quadrilateral you can test for a point in triangle for each end and a point in rectangle for the middle.
EDIT Here is some pseudo code for point in triangle:
function SameSide(p1,p2, a,b)
cp1 = CrossProduct(b-a, p1-a)
cp2 = CrossProduct(b-a, p2-a)
if DotProduct(cp1, cp2) >= 0 then return true
else return false
function PointInTriangle(p, a,b,c)
if SameSide(p,a, b,c) and SameSide(p,b, a,c)
and SameSide(p,c, a,b) then return true
else return false
Or using Barycentric technique:
A, B, and C are the triangle end points, P is the point under test
// Compute vectors
v0 = C - A
v1 = B - A
v2 = P - A
// Compute dot products
dot00 = dot(v0, v0)
dot01 = dot(v0, v1)
dot02 = dot(v0, v2)
dot11 = dot(v1, v1)
dot12 = dot(v1, v2)
// Compute barycentric coordinates
invDenom = 1 / (dot00 * dot11 - dot01 * dot01)
u = (dot11 * dot02 - dot01 * dot12) * invDenom
v = (dot00 * dot12 - dot01 * dot02) * invDenom
// Check if point is in triangle
return (u > 0) && (v > 0) && (u + v < 1)
If the aim is to code your own test, then pick any classic point in polygon test to implement. Otherwise do what Jacob suggests.
assuming you the given coordinates are arranged s.t.
(x1,y1) = rightmost coordinate
(x2,y2) = uppermost coordinate
(x3,y3) = leftmost coordinate
(x4,y4) = botoom-most coordinate
You can do the following:
1. calculate the 4 lines of the quadrilateral (we'll call these quad lines)
2. calculate 4 lines, from the (xt, yt) to every other coordinate (we'll call these new lines)
3. if any new line intersects any of the quad lines, then the coordinate is outside of the quadrilateral, otherwise it is inside.
Assume A,B,C,D are the vertices of the quadrilateral and P is the point.
If P is inside the quadrilateral then all dot products dot(BP,BA), dot(BP,BC), dot(AP,AB), dot(AP,AD), dot(DP,DC), dot(DP,DA), dot(CP,CB) and dot(CP,CD) will be positive.
If P is outside the quadrilateral at least one of these products will be negative.
The solution I used to solve this problem was to get the angle of P (in the diagrams the OP posted) for each of the 4 triangles it makes with each side of the quadrilateral. Add the angles together. If they equal (or nearly equal, depending on the error tolerance of the code) 360, the point is inside the quadrilateral. If the sum is less than 360, the point is outside. However, this might only work with convex quadrilaterals.
I am having two Vectors (X,Y,Z), one above Y=0 and one below Y=0.
I want to find the Vector (X,Y,Z) where the line between the two original vectors intersects with the Y=0 level.
How do I do that?
Example Point A:
X = -43.54235
Y = 95.2679138
Z = -98.2120361
Example Point B:
X = -43.54235
Y = 97.23531
Z = -96.24464
These points read from two UnProjections from a users click and I'm trying to target the unprojection to Y=0.
(I found 3D line plane intersection, with simple plane but didn't understand the accepted answer as it's for 2D)
I suspect that by two vectors, you really mean two points, and want to intersect the line connecting those two points with the plane defined by Y=0.
If that's the case, then you could use the definition of a line between two points:
<A + (D - A)*u, B + (E - B)*u, C + (F - C)*u>
Where <A,B,C> is one of your points and <D,E,F> is the other point. u is an undefined scalar that is used to calculate the points along this line.
Since you're intersecting this line with the plane Y=0, you simply need to find the point on the line where the "Y" segment is 0.
Specifically, solve for u in B + (E - B)*u = 0, and then feed that back into the original line equation to find the X and Z components.
The equation for the line is
(x–x1)/(x2–x1) = (y–y1)/(y2–y1) = (z–z1)/(z2–z1)
So making y=0 yields your coordinates for the intersection.
x = -y1 * (x2-x1)/(y2-y1) + x1
and
z = -y1 * (z2-z1) /(y2-y1) + z1