I have two points in my world, P0 and P1 (xyz,..) which basically represent the start of my line and the end of my line. I also have a plane/side with 4 points, ABCD (xyz,..).
What I want to do is send a "ray" (in this case a line) through the plane and check whether or not it intersects with it. I have used this (How to find intersection point of a line in a plane in 3D space using MATLAB) question as a reference and this function mathlabcentral. I tried the example "isp-zax" gave and that does give me the same values as he does, but as soon as I plop in my own values it is clearly wrong. The answers formula;
>> AB = B-A
AB = 0.8660 0.5000 0
>> AD = D-A
AD = 0 0 1
>> n = cross(AB,AD)/sqrt(dot(cross(AB,AD),cross(AB,AD)))
n = 0.5000 -0.8660 0
>> [I,check]=plane_line_intersect(n,A,P0,P1)
I = 1.0961 44.5116 6.6948
check = 3
--
So the first set of line points (P) work perfectly fine which it should (see image) but if I move the end point of the line (P1) outside of the plane it still says it intersects (see 2nd image). What could cause this?
A = [1312.901701, 2360.12542566832, 74.24415425756794];
B = [1312.901701, 2423.58274087539, 88.31230234462595];
C = [1312.901701, 2371.241313396465, 24.103624956470995];
D = [1312.901701, 2434.698628603535, 38.171773043529];
P0 = [1495.468628, 2261.038086, 161.329498];
P1 = [1153.250854, 2479.341797, -14.787056];
--
This is it in javascript;
const ab = subtract(B, A);
const ad = subtract(D, A);
const n = divide(cross(ab, ad), sqrt(dot(cross(ab, ad), cross(ab, ad))));
const intersects = planeLineIntersect(n, A, P0, P1);
function planeLineIntersect(n, v0, p0, p1) {
const u = subtract(p1, p0);
const w = subtract(p0, v0);
const D = dot(n, u);
const N = -dot(n, w);
let check = 0;
if (abs(D) < 0.0000001) {
if (N === 0) {
check = 2;
return check;
}
check = 0;
return check;
}
const sI = divide(N, D);
const I = add(p0, multiply(sI, u)); // not used
if (sI < 0 || sI > 1) {
check = 3;
} else check = 1;
return check;
}
Related
I'd like to programmatically draw a shape like this where there is an underlying spiral and equally spaced objects along it, placed tangent to the spiral as shown in this sketch:
I found an example of how to determine equally spaced points along the spiral here and am now trying to place hemispheres along the spiral. However, I'm not sure how to calculate the angle the shape needs to be rotated.
This is what I have so far (viewable here):
var totalSegments = 235,hw = 320,hh = 240,segments;
var len = 15;
points = [];
function setup(){
createCanvas(640,480);
smooth();
colorMode(HSB,255,100,100);
stroke(0);
noFill();
//println("move cursor vertically");
}
function draw(){
background(0);
translate(hw,hh);
segments = floor(totalSegments);
points = getTheodorus(segments,len);
angles = getAngles(segments, len);
for(var i = 0 ; i < segments ; i++){
let c = color('blue');
fill(c);
noStroke();
// draw shape
if(i % 2){
// console.log(i, ' ', angles[i]);
// try rotating around the object's center
push();
// translate(points[i].x, points[i].y)
rotate(PI/angles[i]);
arc(points[i].x, points[i].y, len*3, len*3, 0, 0 + PI);
pop();
}
// draw spiral
strokeWeight(20);
stroke(0,0,100,(20+i/segments));
if(i > 0) line(points[i].x,points[i].y,points[i-1].x,points[i-1].y);
}
}
function getAngles(segment, len){
let angles = [];
let radius = 0;
let angle = 0;
for(var i =0; i < segments; i++){
radius = sqrt(i+1);
angle += asin(1/radius);
angles[i] = angle;
}
return angles;
}
function getTheodorus(segments,len){
var result = [];
var radius = 0;
var angle = 0;
for(var i = 0 ; i < segments ; i++){
radius = sqrt(i+1);
angle += asin(1/radius);
result[i] = new p5.Vector(cos(angle) * radius*len,sin(angle) * radius*len);
}
return result;
}
Note that your drawing shows Archimedean spiral while link refers to Theodorus one.
Archimedean spiral is described by equation in polar coordinates (rho-theta)
r = a + b * Theta
where a is initial angle, b is scale value (describes distance between arms), r is radius.
And angle Theta + Pi/2 describes normal to spiral in point at parameter Theta
If you need an approximation to divide spiral into (almost) equal segments - use Clackson formula (example here)
theta = 2 * Pi * Sqrt(2 * s / b)
for arc length s
I am trying to figure out where a bunch of line-segments clip into a window around them. I saw the Liang–Barsky algorithm, but that seems to assume the segments already clip the edges of the window, which these do not.
Say I have a window from (0,0) to (26,16), and the following segments:
(7,6) - (16,3)
(10,6) - (19,6)
(13,10) - (21,3)
(16,12) - (19,14)
Illustration:
I imagine I need to extend the segments to a certain X or Y point, till they hit the edge of the window, but I don't know how.
How would I find the points where these segments (converted to lines?) clip into the edge of the window? I will be implementing this in C#, but this is pretty language-agnostic.
If you have two line segments P and Q with points
P0 - P1
Q0 - Q1
The line equations are
P = P0 + t(P1 - P0)
Q = Q0 + r(Q1 - Q0)
then to find out where they intersect after extension you need to solve the following equation for t and r
P0 + t(P1 - P0) = Q0 + r(Q1 - Q0)
The following code can do this. ( Extracted from my own code base )
public static (double t, double r )? SolveIntersect(this Segment2D P, Segment2D Q)
{
// a-d are the entries of a 2x2 matrix
var a = P.P1.X - P.P0.X;
var b = -Q.P1.X + Q.P0.X;
var c = P.P1.Y - P.P0.Y;
var d = -Q.P1.Y + Q.P0.Y;
var det = a*d - b*c;
if (Math.Abs( det ) < Utility.ZERO_TOLERANCE)
return null;
var x = Q.P0.X - P.P0.X;
var y = Q.P0.Y - P.P0.Y;
var t = 1/det*(d*x - b*y);
var r = 1/det*(-c*x + a*y);
return (t, r);
}
If null is returned from the function then it means the lines are parallel and cannot intersect. If a result is returned then you can do.
var result = SolveIntersect( P, Q );
if (result != null)
{
var ( t, r) = result.Value;
var p = P.P0 + t * (P.P1 - P.P0);
var q = Q.P0 + t * (Q.P1 - Q.P0);
// p and q are the same point of course
}
The extended line segments will generally intersect more than one box edge but only one of those intersections will be inside the box. You can check this easily.
bool IsInBox(Point corner0, Point corner1, Point test) =>
(test.X > corner0.X && test.X < corner1.X && test.Y > corner0.Y && test.Y < corner1.Y ;
That should give you all you need to extend you lines to the edge of your box.
I managed to figure this out.
I can extend my lines to the edge of the box by first finding the equations of my lines, then solving for the X and Y of each of the sides to get their corresponding point. This requires passing the max and min Y and the max and min X into the following functions, returning 4 values. If the point is outside the bounds of the box, it can be ignored.
My code is in C#, and is making extension methods for EMGU's LineSegment2D. This is a .NET wrapper for OpenCv.
My Code:
public static float GetYIntersection(this LineSegment2D line, float x)
{
Point p1 = line.P1;
Point p2 = line.P2;
float dx = p2.X - p1.X;
if(dx == 0)
{
return float.NaN;
}
float m = (p2.Y - p1.Y) / dx; //Slope
float b = p1.Y - (m * p1.X); //Y-Intercept
return m * x + b;
}
public static float GetXIntersection(this LineSegment2D line, float y)
{
Point p1 = line.P1;
Point p2 = line.P2;
float dx = p2.X - p1.X;
if (dx == 0)
{
return float.NaN;
}
float m = (p2.Y - p1.Y) / dx; //Slope
float b = p1.Y - (m * p1.X); //Y-Intercept
return (y - b) / m;
}
I can then take these points, check if they are in the bounds of the box, discard the ones that are not, remove duplicate points (line goes directly into corner). This will leave me with one x and one y value, which I can then pair to the corresponding min or max Y or X values I passed into the functions to make 2 points. I can then make my new segment with the two points.
Wiki description of Liang-Barsky algorithm is not bad, but code is flaw.
Note: this algorithm intended to throw out lines without intersection as soon as possible. If most of lines intersect the rectangle, then approach from your answer might be rather effective, otherwise L-B algorithm wins.
This page describes approach in details and contains concise effective code:
// Liang-Barsky function by Daniel White # http://www.skytopia.com/project/articles/compsci/clipping.html
// This function inputs 8 numbers, and outputs 4 new numbers (plus a boolean value to say whether the clipped line is drawn at all).
//
bool LiangBarsky (double edgeLeft, double edgeRight, double edgeBottom, double edgeTop, // Define the x/y clipping values for the border.
double x0src, double y0src, double x1src, double y1src, // Define the start and end points of the line.
double &x0clip, double &y0clip, double &x1clip, double &y1clip) // The output values, so declare these outside.
{
double t0 = 0.0; double t1 = 1.0;
double xdelta = x1src-x0src;
double ydelta = y1src-y0src;
double p,q,r;
for(int edge=0; edge<4; edge++) { // Traverse through left, right, bottom, top edges.
if (edge==0) { p = -xdelta; q = -(edgeLeft-x0src); }
if (edge==1) { p = xdelta; q = (edgeRight-x0src); }
if (edge==2) { p = -ydelta; q = -(edgeBottom-y0src);}
if (edge==3) { p = ydelta; q = (edgeTop-y0src); }
if(p==0 && q<0) return false; // Don't draw line at all. (parallel line outside)
r = q/p;
if(p<0) {
if(r>t1) return false; // Don't draw line at all.
else if(r>t0) t0=r; // Line is clipped!
} else if(p>0) {
if(r<t0) return false; // Don't draw line at all.
else if(r<t1) t1=r; // Line is clipped!
}
}
x0clip = x0src + t0*xdelta;
y0clip = y0src + t0*ydelta;
x1clip = x0src + t1*xdelta;
y1clip = y0src + t1*ydelta;
return true; // (clipped) line is drawn
}
I'm working on an openCL kernel that loads up some points, decides which is the highest, and returns it. All good there, but I want to add a calculation before the highest evaluation. This compares the point to a pair of lines. I have it written and working to a degree, as follows:
size_t i = group_id * group_stride + local_id;
while (i < n){
//load up a pair of points using the index to locate them within a massive dataSet
int ia = LOAD_GLOBAL_I1(input, i);
float4 a = LOAD_GLOBAL_F4(dataSet, ia);
int ib = LOAD_GLOBAL_I1(input, i + group_size);
float4 b = LOAD_GLOBAL_F4(dataSet, ib);
//pre-assess the points relative to lines
if(pass == 0){
float px = a.x;
float py = a.y;
int checkAnswer;
//want to write this section as a function
float x1 = tri_input[0].x; float y1 = tri_input[0].y;
float x2 = tri_input[2].x; float y2 = tri_input[2].y;
float check = sign((x1-x2) * (py-y1) - (y2-y1) * (px-x1));
if(check != tri_input[3].x){ //point is outside line 1
checkAnswer = 1;
}
else{
x1 = tri_input[2].x; y1 = tri_input[2].y;
x2 = tri_input[1].x; y2 = tri_input[1].y;
check = sign((x1-x2)*(py-y1) - (y2-y1)*(px-x1));
if(check != tri_input[3].y){ //point is outside line 2
checkAnswer = 2;
}
else{
checkAnswer = 0; //point is within both lines
}}}
//later use the checkAnswer result to change the following
//find the highest of the pair
float4 result;
if(a.z>b.z) result = a;
else result = b;
//load up the previous highest result locally
float4 s = LOAD_LOCAL_F4(shared, local_id);
//if the previous highest beat this, stick, else twist
if(s.z>result.z){ STORE_LOCAL_F4(shared, local_id, s);}
else{ STORE_LOCAL_F4(shared, local_id, result);}
i += local_stride;
}
What I would like to do is call the line check twice as a function, i.e the code becomes:
size_t i = group_id * group_stride + local_id;
while (i < n){
//load up a pair of points using the index to locate them within a massive dataSet
int ia = LOAD_GLOBAL_I1(input, i);
float4 a = LOAD_GLOBAL_F4(dataSet, ia);
int ib = LOAD_GLOBAL_I1(input, i + group_size);
float4 b = LOAD_GLOBAL_F4(dataSet, ib);
//pre-assess the points relative to lines
if(pass == 0){
float px = a.x;
float py = a.y;
int checkA = pointCheck( px, py, tri_input);
px = b.x;
py = b.y;
int checkB = pointCheck( px, py, tri_input);
}
//later use the checkAnswer result to change the following
//find the highest of the pair
float4 result;
if(a.z>b.z) result = a;
else result = b;
//load up the previous highest result locally
float4 s = LOAD_LOCAL_F4(shared, local_id);
//if the previous highest beat this, stick, else twist
if(s.z>result.z){ STORE_LOCAL_F4(shared, local_id, s);}
else{ STORE_LOCAL_F4(shared, local_id, result);}
i += local_stride;
}
In this instance the function is:
int pointCheck( float *px, float *py, float2 *testLines){
float x1 = testLines[0].x; float y1 = testLines[0].y;
float x2 = testLines[2].x; float y2 = testLines[2].y;
float check = sign((x1-x2) * (py-y1) - (y2-y1) * (px-x1));
if(check != testLines[3].x){ //point is outside line 1
return 1;
}
else{
x1 = testLines[2].x; y1 = testLines[2].y;
x2 = testLines[1].x; y2 = testLines[1].y;
check = sign((x1-x2)*(py-y1) - (y2-y1)*(px-x1));
if(check != testLines[3].y){ //point is outside line 2
return 2;
}
else{
return 0; //point is within both lines
}}}
Whilst the longhand version runs fine and returns a normal 'highest point' result, the function version returns an erroneous result (not detecting the highest point I have hidden in the data set). It produces a wrong result even though the function as yet has no overall effect.
What am I doing wrong?
S
[Update]:
This revised function works as far as the commented out line, then hangs on something:
int pointCheck(float4 *P, float2 *testLines){
float2 *l0 = &testLines[0];
float2 *l1 = &testLines[1];
float2 *l2 = &testLines[2];
float2 *l3 = &testLines[3];
float x1 = l0->x; float y1 = l0->y;
float x2 = l2->x; float y2 = l2->y;
float pX = P->x; float pY = P->y;
float c1 = l3->x; float c2 = l3->y;
//float check = sign((x1-x2) * (pY-y1) - (y2-y1) * (pX-x1)); //seems to be a problem with sign
// if(check != c1){ //point is outside line 1
// return 1;
// }
// else{
// x1 = l2->x; y1 = l2->y;
// x2 = l1->x; y2 = l1->y;
// check = sign((x1-x2) * (pY-y1) - (y2-y1) * (pX-x1));
// if(check != c2){ //point is outside line 2
// return 2;
// }
// else{
// return 0; //point is within both lines
// }}
}
One immediate issue is how you pass the parameters to the called function:
int checkA = pointCheck( px, py, tri_input);
whereas the function itself expects pointers for px and py. You should instead call the function as:
int checkA = pointCheck(&px, &py, tri_input);
It is surprising that OpenCL does not give build errors for this kernel.
In my experience, some OpenCL runtimes do not like multiple return statements in a single function. Try to save the return value into a local variable and use a single return statement at the end of the function. This is because OpenCL does not support real function calls, but rather inlines all functions directly into the kernel. A best practice is therefore to mark all non __kernel functions as inline, and treat them as such (i.e. make it easier for the compiler to inline your function by not using multiple return statements).
Except using look-up tables, is there another way to optimize the parameterization algorithm of a cubic Bézier curve like this? (5000 steps for a good parameterization is simply too much for a slower PC, as I need to call this function many times in 1 second):
function parameterizeCurve(path, partArc, initialT)
{
// curve length is already known and globally defined
// brute force
var STEPS = 5000; // > precision
var t = 1 / STEPS;
var aX=0;
var aY=0;
var bX=path[0], bY=path[1];
var dX=0, dY=0;
var dS = 0;
var sumArc = 0;
var arrT = new Array(Math.round(partArc));
var z = 1;
arrT[0] = -1;
var oldpartArc = partArc;
partArc = partArc - initialT;
var j = 0;
for (var i=0; i<STEPS; j = j + t) {
aX = bezierPoint(j, path[0], path[2], path[4], path[6]);
aY = bezierPoint(j, path[1], path[3], path[5], path[7]);
dX = aX - bX;
dY = aY - bY;
// deltaS. Pitagora
dS = Math.sqrt((dX * dX) + (dY * dY));
sumArc = sumArc + dS;
if (sumArc >= partArc) {
arrT[z] = j; // save current t
z++;
sumArc = 0;
partArc = oldpartArc;
}
bX = aX;
bY = aY;
i++;
}
return arrT;
}
function bezierPoint(t, o1, c1, c2, e1) {
var C1 = (e1 - (3.0 * c2) + (3.0 * c1) - o1);
var C2 = ((3.0 * c2) - (6.0 * c1) + (3.0 * o1));
var C3 = ((3.0 * c1) - (3.0 * o1));
var C4 = (o1);
return ((C1*t*t*t) + (C2*t*t) + (C3*t) + C4)
}
If I've guessed correctly, you're trying to come up with a cubic Bezier curve parameterization that moves at a constant speed along the curve.
So, why do you need 5000 steps? The minimum one can move along a curve is one pixel. A Bezier stays within the convex hull of its four control points, so the length of the curve will be less than that of the polyline P0 -> P1 -> P2 -> P3. So find that length in pixels, and use it (instead of 5000).
Let me know if that speeds things up enough.
I've got the following grid of numbers centered around 0 and increasing in spiral. I need an algorithm which would receive number in spiral and return x; y - numbers of moves how to get to that number from 0. For example for number 9 it would return -2; -1. For 4 it would be 1; 1.
25|26|... etc.
24| 9|10|11|12
23| 8| 1| 2|13
22| 7| 0| 3|14
21| 6| 5| 4|15
20|19|18|17|16
This spiral can be slightly changed if it would help the algorithm to be better.
Use whatever language you like. I would really appreciate mathematical explanation.
Thank you.
First we need to determine which cycle (distance from center) and sector (north, east, south or west) we are in. Then we can determine the exact position of the number.
The first numbers in each cycle is as follows: 1, 9, 25
This is a quadratic sequence: first(n) = (2n-1)^2 = 4n^2 - 4n + 1
The inverse of this is the cycle-number: cycle(i) = floor((sqrt(i) + 1) / 2)
The length of a cycle is: length(n) = first(n+1) - first(n) = 8n
The sector will then be:
sector(i) = floor(4 * (i - first(cycle(i))) / length(cycle(i)))
Finally, to get the position, we need to extrapolate from the position of the first number in the cycle and sector.
To put it all together:
def first(cycle):
x = 2 * cycle - 1
return x * x
def cycle(index):
return (isqrt(index) + 1)//2
def length(cycle):
return 8 * cycle
def sector(index):
c = cycle(index)
offset = index - first(c)
n = length(c)
return 4 * offset / n
def position(index):
c = cycle(index)
s = sector(index)
offset = index - first(c) - s * length(c) // 4
if s == 0: #north
return -c, -c + offset + 1
if s == 1: #east
return -c + offset + 1, c
if s == 2: #south
return c, c - offset - 1
# else, west
return c - offset - 1, -c
def isqrt(x):
"""Calculates the integer square root of a number"""
if x < 0:
raise ValueError('square root not defined for negative numbers')
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = 2**(a+b)
while True:
y = (x + n//x)//2
if y >= x:
return x
x = y
Example:
>>> position(9)
(-2, -1)
>>> position(4)
(1, 1)
>>> position(123456)
(-176, 80)
Do you mean something like this? I did not implement any algorithm and the code can be written better but it works - that's always a start :) Just change the threshold value for whatever you wish and you'll get the result.
static int threshold=14, x=0, y=0;
public static void main(String[] args) {
int yChange=1, xChange=1, count=0;
while( !end(count) ){
for (int i = 0; i < yChange; i++) {
if( end(count) )return;
count++;
y--;
}
yChange++;
for (int i = 0; i < xChange; i++) {
if( end(count) )return;
count++;
x++;
}
xChange++;
for (int i = 0; i < yChange; i++) {
if( end(count) )return;
count++;
y++;
}
yChange++;
for (int i = 0; i < xChange; i++) {
if( end(count) )return;
count++;
x--;
}
xChange++;
}
}
public static boolean end(int count){
if(count<threshold){
return false;
}else{
System.out.println("count: "+count+", x: "+x+", y: "+y);
return true;
}
}