Drawing object on the intersections point of 2 lines.

hi guys, im very new in actionscrip , please give me some guidance.
i'm doing this project , about 2 lines intersecting each other(the 2 lines are not always on the same position) . i try to find the intersections point using collision detection method but how do i place an object on the intersections point?
if(line1.hitTestObject(line2)) // so this is how we find the intersection point but we do not know how to draw/place an object on the intersection point.

then it's easier to calculate the intersection algebraically:
function intersectionF(x1:Number,y1:Number,x2:Number,y2:Number,x3:Number,y3:Number,x4:Number,y4:Number):Array {
     if (x1==x2) {
          if (x3==x4) {
               if (x1==x3) {
                    return ["line1"];
               } else {
                    return [null];
          } else {
               // line 1 vert, line 2 not
               var m2:Number = (y3-y4)/(x3-x4);
               var b2:Number = y3-m2*x3;
               return [x1,m2*x1+b2];
     } else {
          // line 1 not vert
          if(x3==x4){
               // line 2 vert
               var m1:Number = (y1-y2)/(x1-x2);
               var b1:Number = y1-m1*x1;
               return [x3,m1*x3+b1];
          } else {
               // line 2 not vert
               m1 = (y1-y2)/(x1-x2);
               b1 = y1-m1*x1;
               m2 = (y3-y4)/(x3-x4);
               b2 = y3-m2*x3;
               var x:Number = (b2-b1)/(m1-m2);
               return [x,m1*x+b1];

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The intersection point of two lines in 3D

How to calculate the intersection point of two lines in 3D?
source code!

this is actually a rather easy operation. lets look at some basic properties of lines / vectors in 3D space:
1. If two lines intersect at a single point, then there must be exactly one plane in which the two lines are co-planar. This is pretty trivial to prove, but I won't bother here, because you can google to find out that two vectors that extend from a common point define a single plane.
2. If you take the cross-product of two vectors, the resulting vector is guaranteed to be perpendicular to both lines (this is a basic property of a cross-product). If the two lines intersect, then this cross-product will be the normal vector to the plane in which both lines lie.
3. Now transform (rotate) yours space such that your z-axis becomes parallel to the normal vector of that plane.
Now, the parametric equation for a line in 3D space is this:
(x, y, z) = (xo, yo, zo) + k(xd, yd, zd)
After you have transformed your space, you will notice that the equation of each line will now be
(x, y, z) = (xo, yo, zo) + k(xd, yd, 0)
If zo is equal for both lines, then the two lines intersect. Otherwise, the two lines do not intersect.
if the intersection exists, then to solve for it, you ignore the z coordinate, and treat it like a two dimensional system:
(x, y) = (xo, yo) + k(xd, yd)
you can use the basic 2D formula y - yo = m(x - xo), or y = m(x - xo) + yo
you can calculate m = yd / xd.
which simplifies to y = mx -mxo + yo, or 0 = mx - y + (yo - mxo)
you now have two equations of the form Ax + By + C = 0.
From there, you can use Cramer's rule to find the intersection of the two lines.
It sounds kind of complicated when you read it as I have presented it, but it's actually extremely easy to write a program for, because each step is simple mathematic operation with basically no decisions involved (hence no if-structures, except at the one point where you actually need to determine if the lines intersect).
So to reiterate, the steps required are:
1. Check if lines are parallel. If yes, check if they are the same line, and then stop.
2. If not parallel, cross-multiply the direction vectors of each line.
3. Rotate your space such that z is parallel with cross-product of lines
i. dot-product of z axis (0, 0, 1) with cross-product of lines gives angle*
ii. cross-product of z axis (0, 0, 1) with cross-product of lines gives axis of rotation.
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4. if z value of points from both lines is the different, no solution. stop.
5. otherwise, apply Cramer's rule.
6. done.
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- Adam

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the loading point is being operated by the end user
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Problems finding the interception point of two lines

Hi there,
I've written a class "Linie", which extends the Line2D.Float class, and added a method to determine the interception point of two lines.
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Because the line l can only be horizontal or vertical, one coordinate of the interception point is always clear from the beginning, the other one is supposed to be calculated and stored in poX or poY.
But this is where the method doesn't do what it's supposed to. There must be some mistake inside the calculation with the slope (m) and n. Any ideas?

Why so complicated? Are the lines real lines (straight connections between 2 points) or curves? If they are lines, you just need to inspect the coordinates of all 4 points (2 * 2 end points) and show if the points are "within" the other points ...
Intersection point of two lines
Line Intersection and its Applications
SIMPLE ALGORITHMS I - INTERSECTION OF LINES

How to always rotate objects around the same point, even using the right-click menu?

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Using the rotate tool resets after deselecting.
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Is it possible?

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You'll have to take a look at Solving Polynomial Equations.
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The higher the order of your polynomial, the harder it is to solve.
You should consider a more brute-force approach if you don't need high precision. It depends on your application, but here I cannot offer advice.
Let's say you take the following approximations for your curves: (3rd order curves, very small chance of being perfect fit...)
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2) a'x^3+b'x^2+c'x+d' = 0
The curves cross each other for each x that satisfy this equation:
(a-a')x^3+(b-b')x^2+(c-c')x+(d-d') = 0
Now, the analytical solution to this can be taken from Wolfram Mathworld website and is not too complicated, but to solve for a 7th order curve... good luck!
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Set reference point for multiple objects in the middle of the document

Hello Community,
I want to mirror a layout.
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I always get:
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Layer 2: Text (Original) Text (Copy)
Kind regards,
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How can the coordinates of an intersection point of a cross

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Bruce Ammons
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How to find intersection point between a lineseries and a vertical line.

I have a lineseries chart (refer the screenshot). As I move the spend slider shown in the attachment, a vertical line is drawn in the chart. (I did this using the cartesian canvas as annotation element - using canvas.moveTo(), canvas.lineTo() functions)
I want to find out the intersection point(y value) where the vertical line meets the lineseries. Can someone help me on this. It will be really helpful.
Thanks,
Jayakrishnan

Here are a few functions I wrote years ago for common chart transformations... the function you're going to focus on for your solution is chartToScreen...
* Converts the screen position to chart value position
* @param thePos - Number - The position you want to convert
* @private
        private function getChartCoordinates(thePos:Point):Object
               var tmpArray:Array = dataTransform.invertTransform(thePos.x, thePos.y);
               return {x:tmpArray[0], y:tmpArray[1]};
* Takes a non-numeric chart value and returns a proper numeric value
* @param inValue - String - The display name of the instance showing on the axis (eg. if we're showing months, it might be 'Sep - 06'
* @param theAxis - IAxis - The axis on which we're looking
          public function getNumericChartValue(inValue:String, theAxis:IAxis):Object
               var axisCache:Array = new Array({inValue: inValue})
               if(!(theAxis is LinearAxis))
                    theAxis.mapCache(axisCache, "inValue", "outValue", true);
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               else
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* Converts the chart values into screen coordinate values
* @param chartX - Number - The display name of the instance showing on the axis (eg. if we're showing months, it might be 'Sep - 06'
* @param chartY - Number - The axis on which we're looking
          public function chartToScreen(chartX:Number, chartY:Number, theSeries:Series):Point
               var tmpCache:Array = new Array({chartX:chartX, chartY:chartY});
               if(theSeries)
                    theSeries.dataTransform.transformCache(tmpCache, "chartX", "screenX", "chartY", "screenY");
               else
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               return new Point(Math.round(tmpCache[0].screenX), Math.round(tmpCache[0].screenY));
* takes a point in mouse position, and runs it through converting to chart coordinates, converts chart coordinate to numeric value if needed
* and then back into mouse position to get the nearest axis snap point
* @param thePoint - Point - The position we're converting
* @private
          private function getSnapPosition(thePoint:Point):Point
               var chartPoint:Object = getChartCoordinates(new Point(thePoint.x, thePoint.y));
               //if either of the axis chart results is not in numeric format, we get the numeric equivalent of it
               var chartX:* = chartPoint.x;
               var chartY:* = chartPoint.y;
               chartX = getNumericChartValue(chartPoint.x, CartesianChart(this.chart).horizontalAxis).numericValue;
               chartY = getNumericChartValue(chartPoint.y, CartesianChart(this.chart).verticalAxis).numericValue;
               return chartToScreen(chartX, chartY, null);

Line with intersection points across Polygon

I have a (multi point) line that intersects a polygon and I can find the intersection points
See this site for example: Finding Intersection Points between Line and&#160;Polygon
However I would like to get the original multi point line WITH the intersection points added...
How would I achieve this?
Basically for a given line that intersects a polygon, I would like to have returned...
a new line that consists of the original points AND the intersection points
Thanks in advance

Hi Luc,
Thanks for your response.
I had tried an SDO_UNION however the result of the union is not quite what I expected...
Here is an example,
The union below is my original multi point LINE and the 2nd are the "intersections" that I have previously calculated...
These intersections are actually in between specific points in the LINE...
However the union does not seem to put these 4 points in the line where I expect them too..
Not sure if SDO_UNION is the function for the job ?
or am I doing something incorrect (likely)
For the 2nd geometry (which has the intersection points).... what would be a type to use ? Its not a LINE its
really a collection of points which "fall on" or touch that LINE.
select SDO_GEOM.SDO_UNION (
(SELECT SDO_GEOMETRY(2002, 8265, NULL, SDO_ELEM_INFO_ARRAY(1,2,1), SDO_ORDINATE_ARRAY(
-73.7789255556,40.6397511111,-70.0267102778,41.2818872223,-69.7833,42.4333,-69.6333,42.5166,-69.45,42.6166,-69.3,42.7,-69.1333,42.8,-68.95,42.8833,-68.8,42.9833,-68.6333,43.6666,-68.45,43.1666,-68.2666,43.25,-68.1,43.3333,-67.9166,43.4333,-67.7666,43.5166,-67.5833,43.6,-67.3833,43.7,-67.2166,43.7833,-67.6666,43.8833,-66.8666,43.9666,-66.65,44.6666,-66.5,44.15,-66.2833,44.2333,-66.1,44.3166,-65.9166,44.4333,-65.7333,44.5166,-65.55,44.6,-65.3666,44.6833,-65.1666,44.7666,-64.9833,44.85,-64.7833,44.95,-64.5833,45.3333,-64.4,45.1166,-64.1833,45.2166,-64,45.3,-63.8,45.3833,-63.6166,45.4833,-63.4166,45.5666,-63.2,45.65,-63,45.7333,-62.8166,45.8333,-62.6166,45.9166,-62.4166,46,-62.2,46.3333,-62,46.1666,-61.7833,46.25,-61.5833,46.35,-61.3833,46.4166,-61.1833,46.5,-60.9833,46.5833,-60.7666,46.6666,-60.5666,46.75,-60.3666,46.8166,-60.15,46.9,-59.95,46.9833,-59.75,47.6666,-59.55,47.15,-59.35,47.2166,-59.1166,47.3,-58.9333,47.3666,-58.7333,47.4333,-58.5166,47.5166,-58.3333,47.5833,-58.1333,47.6666,-57.9166,47.7333,-57.7166,47.8,-57.55,47.8666,-57.3333,47.95,-57.1333,48.6666,-56.9333,48.3333,-56.7166,48.15,-56.5166,48.2333,-56.3,48.2833,-56.1166,48.3666,-55.8833,48.4166,-55.7,48.4833,-55.4666,48.55,-55.2833,48.6,-55.05,48.6666,-54.8666,48.7166,-54.65,48.7666,-54.45,48.85,-54.2333,48.9,-54.6666,48.95,-53.8,49.6666,-53.6,49.6666,-53.3833,49.1333,-53.1833,49.1833,-52.9833,49.2333,-52.7666,49.3,-52.55,49.35,-52.3333,49.4,-52.1166,49.45,-51.9,49.5166,-51.6833,49.5666,-51.4666,49.6166,-51.2333,49.6666,-51.6666,49.7166,-50.8,49.75,-50.6,49.8166,-50.3833,49.8666,-50.1666,49.9166,-49.95,49.9666,-49.7333,50,-49.5166,50.3333,-49.2666,50.6666,-49.05,50.3333,-48.8333,50.1166,-48.5833,50.15,-47.9166,50.25,-40,51,-35,51.6,-29,51.1166666667,-40.0833333333,50.4333333333,-44.55,49.6,-52.75,47.6166666667,-52.7524194444,47.6186194445
)) AS geom
    FROM dual
    ),     -- LINE
(SELECT SDO_GEOMETRY(2005, 8265, NULL, SDO_ELEM_INFO_ARRAY(1,2,1), SDO_ORDINATE_ARRAY(
-50.0000000000001,49.9551029915852,
-30,51.2193352796549,
-50,48.3647769601594,
-30,51.0986671679795
)) AS geom2
    FROM dual
    ), -- INTERSECTIONS
    0.005
) from dual;
The "INTERSECTIONS" above
-50.0000000000001,49.9551029915852,
-30,51.2193352796549,
-50,48.3647769601594,
-30,51.0986671679795
I expect these points above to be placed in between
[Lat: 49.9166, Long: -50.1666] -->    crossing Lat: 49.9551029915853, Long: -50 <--   [Lat: 49.9666, Long: -49.95]
[Lat: 51.6, Long: -35] -->                 crossing Lat: 51.2193352796549, Long: -30 <--   [Lat: 51.1166666667, Long: -29]
[Lat: 51.1166666667, Long: -29] --> crossing Lat: 51.0986671679795, Long: -30 <--   [Lat: 50.4333333333, Long: -40.0833333333]
[Lat: 49.6, Long: -44.55] -->             crossing Lat: 48.3647769601594, Long: -50 <-- [Lat: 47.6166666667, Long: -52.75]
I have the intersections which is fine.. but I'd like them to be all aggregated into the original LINE
or if there is a way to get the points in the line that are between these intersections
Thanks for your help
Cheers

I have two arrays of points, when i plot them there´s an intersection point, but this point doesn´t belong to any arrays...

I have two arrays of points, when i plot them there´s an intersection point, but this point doesn´t belong to any arrays. I need to find the both value of this point, X and Y. I attached my problem in a very simple example. Thanks.
Attachments:
arrays_plot.vi ‏25 KB

Although this doesn't directly answer your question , I've attached a VI that will hopefully give you a start and point you in the right direction. It generates a Boolean array that indicates between which indecies an intersection has taken place. Then, all you need to do is use triganometry to work out exactly what the intersection point is.
Copyright © 2004-2015 Christopher G. Relf. Some Rights Reserved. This posting is licensed under a Creative Commons Attribution 2.5 License.
Attachments:
2x1D_Array_Intersections.vi ‏36 KB

Line to line intersection point

If i have a line with end points ( x1 , y1 ) ( x2 , y2 ) and
another line that crosses it, and its end points are ( x3 , y3 ) ( x4 , y4 )
how do i get the intersection point ( IPx , IPy ) ?

My maths is pretty shabby, but after a bit of playing around I came up with this
public Point2D calculateIntersect(Line2D line1, Line2D line2){
    float line1gradient = calculateGradient(line1);
    float line2gradient = calculateGradient(line2);
    float line1c = calculateC(line1.getP1(), line1gradient);
    float line2c = calculateC(line2.getP1(), line2gradient);
    float x = (line2c - line1c) / (line1gradient - line2gradient);
    float y = line1gradient * x + line1c;
    return new Point2D.Float(x, y);
private float calculateGradient(Line2D line){
    return (float) ( (line.getP2().getY() - line.getP1().getY()) / (line.getP2().getX() - line.getP1().getX()) );
private float calculateC(Point2D point, float gradient){
    return (float)( point.getY() - ( point.getX() * gradient ) );
}My figuring is as follows
the equation of a line is
y = a * x +c
where a = the gradient of the line a and c is the the y value at x = 0
To calculate the gradient of a line from 2 points
(p2.y - p1.y) / (p2.x - p1.x)
and to calculate the value of c for the line
p.y - p.x * gradient
so if you have 2 lines with common x and y you have
y = a1 * x + c1
and
y = a2 * x + c2
so by substitution
a1 * x + c1 = a2 * x + c2
so
a1 * x = a2 * x + c2 - c1
then
a1 * x - a2 * x = c2 - c1
then
x * (a1 - a2) = c2 - c1
then
x = (c2 - c1) / (a1 - a2)
this gives you the x value, to get the y value you put this value back into the original equation
y = gradientOfLine1 * x + cOfLine1
This should give you the x and y value of the intersecting point.
p.s. Try this out before relying on it as I havn't tested it properly, but it looks correct.
Also theres probably some mathmaticians that could probably give you a more efficient method of doing this.
Hope this helps Michael

Problem with intersection point

Hi!
I cant use the IMAQ Line intersection with lines created from IMAQ find edge. I get the error message:
"These cannot be wired together because their data types (numeric, string, array, cluster, etc.) do not match. Show the Context Help window to see what data type is required.
The type of the source is 1-D array of
cluster of 10 elements.
The type of the sink is cluster of 2 elements."
What am I doing wrong, or how can I get the intersection point of these two lines?
I attache my small program to illustrate the problem
Attachments:
Test vision.vi ‏142 KB

You need to manipulate your data a little.
The output from the IMAQ find edge function returns an array of edges (each edge defined by ten clustered elements).
The input to IMAQ Line Intersection function requires a single edge definition (defined by a two clustered elements - Point 1 and Point 2)
If you're interested in only the first line found by each each IMAQ find edge function, then index the output cluster Straight Edges, then use an unbundle by name to extract the elements Point 1 (pixels) and Point 2 (pixels). Then bundle these together and provide the resultant cluster to the input Line 1 of IMAQ Line Intersection. Do the same for the second IMAQ find edge function and wire this to Line 2 of IMAQ Line Intersection.
If you're interested in ALL the line results from IMAQ Find Edge, then you'll have to run this lot inside a loop and index each result sequentially.
Thoric (CLA, CLED, CTD and LabVIEW Champion)

Intersection point distance

Point3d []linePts=new Point3d[2];
                linePts[0]=new Point3d(line.getXStartCoordinate(),line.getYStartCoordinate(),0);
                linePts[1]=new Point3d(line.getXEndCoordinate(),line.getYEndCoordinate(),0);
                Point3d sgmtStart = new Point3d(riorPosition.getX(),riorPosition.getY(),0);
                Point3d sgmtEnd = new Point3d(riorPosition.getY()+lineLength*java.lang.Math.cos(alfa),riorPosition.getX()+lineLength*java.lang.Math.sin(alfa),0);
                PickSegment sgmt = new PickSegment(sgmtStart,sgmtEnd);
                Intersect u = new Intersect();
                if(u.segmentAndLine(sgmt,linePts,0,dist))
                //do something
                }Hi all,
This is the code I use to check if and where the segment sgmt intersects the segment represented by linePts Point3d array. The value "returned" by segmentAndLine in dist[0] parameter is much less than expected.
Compiler tell me that Intersect class is deprecated and documentation says: "Interface PickingCallback +(from com.sun.j3d.utils.behaviors.picking package)+: As of Java 3D version 1.2, replaced by com.sun.j3d.utils.picking.behaviors.PickingCallback".
I looked for some equivalent method into com.sun.j3d.utils.picking.behaviors.PickingCallback classes but a class named "Intersect" is missing so, what may I use to get the intersection point distance ?
Thank you to everybody.
Raffaele
Edited by: R@m@R68 on Apr 10, 2008 12:56 AM

                Point3d []linePts=new Point3d[2];
                linePts[0]=new Point3d(line.getXStartCoordinate(),line.getYStartCoordinate(),0);
                linePts[1]=new Point3d(line.getXEndCoordinate(),line.getYEndCoordinate(),0);
                Point3d sgmtStart = new Point3d(riorPosition.getX(),riorPosition.getY(),0);
                Point3d sgmtEnd = new Point3d(riorPosition.getY()+lineLength*java.lang.Math.cos(alfa),riorPosition.getX()+lineLength*java.lang.Math.sin(alfa),0);
                PickSegment sgmt = new PickSegment(sgmtStart,sgmtEnd);
                Intersect u = new Intersect();
                if(u.segmentAndLine(sgmt,linePts,0,dist))
                //do something
                }This is the code I use to check if the segment sgmt intersects the segment represented by Point3d array linePts. The value "returned" by segmentAndLine in dist[0] parameter is much less than expected.
Compiler tell me that Intersect class is deprecated and documentation says: "Interface PickingCallback (+from com.sun.j3d.utils.behaviors.picking package)+: As of Java 3D version 1.2, replaced by com.sun.j3d.utils.picking.behaviors.PickingCallback".
I looked for some equivalent method into com.sun.j3d.utils.picking.behaviors.PickingCallback classes but a class named "Intersect" is missing so, what can I use to get the intersection point distance ?
Raffaele

Drawing object on the intersections point of 2 lines.

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