I am drawing polygon by capturing mouse clicks on canvas and then passing these points to fabric.Polygon. So, in this manner I'm drawing multiple polygons.
What I need know is, I want to get the mouse co-ordinates (pixel points on canvas) for the polygon which is selected now?
I have tried with:
canvas.getActiveObject().get('points');
But this is giving some negative and some positive values.
So, can u please tell me a way to find out the polygon points?
Polygon points are relative to its center so you can get their "absolute" position like so:
var polygon = canvas.getActiveObject();
var polygonCenter = polygon.getCenterPoint();
var translatedPoints = polygon.get('points').map(function(p) {
return {
x: polygonCenter.x + p.x,
y: polygonCenter.y + p.y
};
});
Let's check how this looks:
translatedPoints.forEach(function(p) {
canvas.getContext().strokeRect(p.x-5, p.y-5, 10, 10);
});
I think this will only work if polygon's angle is at 0 (otherwise need to "rotate" points coordinates as well).
It looks like that from version 2.0 they changed the coordinates of the polygon. Before 2.0 points relative to the center of the polygon; after 2.0 they are absolute to the canvas;
Check out my response to the similar questions https://stackoverflow.com/a/53710375/4681279
Related
I am building a Warehouse map in 2D with fabricjs where I am displaying their racking systems as a series of rectangles.
As a first "layer/group", i add all bays as groups containing a rectangle and text, both positioned at the same (x,y). They also have both an angle set to fit their orientation in the space.
As a second "layer/group", i add groups containing a circle and text, representing the bay's number of issues. The (x,y) also fits the bays. This way, all my issues are always on top of the bays and the center of the group fits the rotated corner of the bay.
On the first paint, all is well aligned. Once they're shown on the page, the user can create new issues, so I am trying to position the issue group fitting the original (x,y), but since it can all be panned and zoomed, I am having a hard time positioning it where it should be.
I've been looking at their explanations about transforms, but I can't figure who the boss should be and thinking that having nested groups may also be why I am all mixed up.
By doing:
const gMatrix = matchingBay.group.calcTransformMatrix(false);
const targetToCanvas = fabric.util.transformPoint(matchingBay.aCoords.tl, gMatrix);
I am on the bay, but on the "group" corner, which is not what I am looking for. (x,y) will be one of the corners of the rectangle in the group, that may have been rotated.
By specifying the original (x,y) in this code will get me way off the actual painting zone.
So, my question is, how do I get the transformed (x,y) so I can add my issue group at those coordinates?
[EDIT]
As the suggestion of spring, it made me realize I can use the rotated rectangle's transforms to find its coordinates, so I tried:
const rect = matchingBay.getObjects()[BAY_RECTANGLE_INX];
const gMatrix = rect.calcTransformMatrix();
const targetToCanvas = fabric.util.transformPoint(rect.aCoords.bl, gMatrix);
Bottom left corner is where I wish to add the new Circle. After rotation, the bottom left is now the bottom right. But I am still off. As shown, the red circle should be on the corner. It looks like the rotation has not been applied.
qrDecompose gives me something that seems right:
{angle: -90, scaleX: 1, scaleY: -1, skewX: 0, skewY: 0, translateX: 6099.626027314769, translateY: 4785.016008065199 }
I realized that I was not thinking it the right way. Since I have the rectangle already in hands, I just had to get its own transformation and resolve the corner by my own, the following fixed my issue:
{
[...]
const rect = matchingBay.getObjects()[BAY_RECTANGLE_INX];
const gMatrix = rect.calcTransformMatrix();
const decomposed = fabric.util.qrDecompose(gMatrix);
const trans = fabric.util.transformPoint(new fabric.Point((decomposed.scaleX * -rect.width) / 2, (decomposed.scaleY * rect.height) / 2), gMatrix);
const top = trans.y;
const left = trans.x;
[...]
}
Since the matrix is bringing the center point of the rectangle, I can get the corner by substracting its width and height and then transforming the coordinates to find out where the matrix puts it.
Please suggest how to achieve the flying arcs in the force layout of d3.JS.
As in the image
Do something like this
http://fiddle.jshell.net/cyril123/ov1bjrq9/4/
The idea is to make the curve path between two nodes.
The mechanism to make curve(or the d attribute of a path)
Lives here inside the tick function:
Feel free to change it as per your choice.
link.attr("d",
function(d){
var s=d.source;
var t = d.target;
return lineFunction([
{x:s.x, y: s.y},//start point of curve
{x:(s.x+t.x)/2, y: (s.y+t.y)/2+30}, //middle point of curve
{x:s.x, y: t.y} ////end point of curve
])
}
);
I have a polygon element in my SVG file. It actually is a clipping path for a floor plan of an apartment. Currently the clipping path doesn't capture the walls of the floor plan and I need it to do so.
Is there a way to 'expand' the polygon by some value? I've tried scaling it from center with transform=translate() scale(), but the effect is different from 'expanding'.
To illustrate the issue, here is a picture:
The original polygon is filled with gray and I have the coordinates for the blue points. I want the polygon to become like dashed black one and get the coordinates of the green points.
As you have already pointed out, scaling is not the answer - unless you are happy with a very rough-fitting clip.
A simpler solution (than trying to manipulate the polygon coordinates) is to change the clip-path to a mask and draw the polygon with a stroke thick enough to reveal your walls.
Your polygon is not a convex polygon, therefore its 'center' for scaling is best chosen from the center of its bounding box. You can compute the scale based on how many units you want the larger polygon to exceed the original polygon. myPolygon is a clone of your base polygon. I edited this to fix a few typos. Sorry:( ... In the future I'll use jsFiddle.
var myScale=1.025
var myPolygon=basePolygon.cloneNode(true)
//---append to you root SVG---
mySVG.appendChild(myPolygon)
var bb=myPolygon.getBBox()
var bbx=bb.x
var bby=bb.y
var bbw=bb.width
var bbh=bb.height
var cx=bbx+.5*bbw
var cy=bby+.5*bbh
myPolygon,setAttribute("transform","translate("+(cx)+" "+(cy)+")scale("+myScale+")translate("+(-cx)+" "+(-cy)+")")
Then to return the screen points of the scaled polygon(myPoly). (the mySVG is the root svg)
//---changes all transformed points to screen points---
function screenPolygon(myPoly,mySVG)
{
var sCTM = myPoly.getCTM()
var pointsList = myPoly.points;
var n = pointsList.numberOfItems;
for(var m=0;m<n;m++)
{
var mySVGPoint = mySVG.createSVGPoint();
mySVGPoint.x = pointsList.getItem(m).x
mySVGPoint.y = pointsList.getItem(m).y
mySVGPointTrans = mySVGPoint.matrixTransform(sCTM)
pointsList.getItem(m).x=mySVGPointTrans.x
pointsList.getItem(m).y=mySVGPointTrans.y
}
//---force removal of transform--
myPoly.setAttribute("transform","")
myPoly.removeAttribute("transform")
}
For a project we are trying to make a circle into a line (and back again) while it is rotating along a linear path, much like a tire rotates and translates when rolling on a road, or a curled fore finger is extended and recurled into the palm.
In this Fiddle, I have a static SVG (the top circle) that rotates along the linear black path (which is above the circle, to mimic a finger extending) that is defined in the HTML.
I also use d3 to generate a "circle" that is made up of connected points (and can unfurl if you click on/in the circle thanks to #ChrisJamesC here ), and is translated and rotated
in the function moveAlongLine when you click on the purple Line:
function moveAlongLine() {
circle.data([lineData])
.attr("transform", "translate(78.5,0) rotate(-90, 257.08 70) ")
.duration(1000)
circle.on("click", transitionToCircle)
}
The first problem is that the .duration(1000) is not recognized and throws a Uncaught TypeError: Object [object Array] has no method 'duration' in the console, so there is a difference between the static definition of dur in SVG and dynamically setting it in JS/D3, but this is minor.
The other is should the transform attributes be abstracted from one another like in the static circle? in the static circle, the translate is one animation, and the rotation is another, they just have the same star and duration, so they animate together. How would you apply both in d3?
The challenge that I can not get, is how to let it unroll upwards(and also re-roll back), with the static point being the top center of the circle also being the same as the leftmost point on the line.
these seem better:
I should try to get the unfurl animation to occur while also rotating? This seems like it would need to be stepwise/sequential based...
Or Consider an octogon (defined as a path), and if it were to rotate 7 of the sides, then 6, then 5.... Do this for a rather large number of points on a polyhedron? (the circle only needs to be around 50 or so pixels, so 100 points would be more than enough) This is the middle example in the fiddle. Maybe doing this programmatically?
Or This makes me think of a different way: (in the case of the octogon), I could have 8 line paths (with no Z, just an additional closing point), and transition between them? Like this fiddle
Or anything todo with keyframes? I have made an animation in Synfig, but am unsure ho get it to SVG. The synfig file is at http://specialorange.org/filedrop/unroll.sifz if you can convert to SVG, but the xsl file here doesn't correctly convert it for me using xsltproc.
this seems really complicated but potential:
Define a path (likely a bézier curve with the same number of reference points) that the points follow, and have the reference points dynamically translate as well... see this for an concept example
this seems complicated and clunky:
Make a real circle roll by, with a growing mask in front of it, all while a line grows in length
A couple of notes:
The number of points in the d3 circle can be adjusted in the JS, it is currently set low so that you can see a bit of a point in the rendering to verify the rotation has occurred (much like the gradient is in the top circle).
this is to help students learn what is conserved between a number line and a circle, specifically to help learn fractions. For concept application, take a look at compthink.cs.vt.edu:3000 to see our prototype, and this will help with switching representations, to help you get a better idea...
I ended up using the same function that generates the circle as in the question, and did a bit of thinking, and it seemed like I wanted an animation that looked like a finger unrolling like this fiddle. This lead me to the math and idea needed to make it happen in this fiddle.
The answer is an array of arrays, with each nested array being a line in the different state, and then animate by interpolating between the points.
var circleStates = [];
for (i=0; i<totalPoints; i++){
//circle portion
var circleState = $.map(Array(numberOfPoints), function (d, j) {
var x = marginleft + radius + lineDivision*i + radius * Math.sin(2 * j * Math.PI / (numberOfPoints - 1));
var y = margintop + radius - radius * Math.cos(2 * j * Math.PI / (numberOfPoints - 1));
return { x: x, y: y};
})
circleState.splice(numberOfPoints-i);
//line portion
var lineState = $.map(Array(numberOfPoints), function (d, j) {
var x = marginleft + radius + lineDivision*j;
var y = margintop;
return { x: x, y: y};
})
lineState.splice(i);
//together
var individualState = lineState.concat(circleState);
circleStates.push(individualState);
}
and the animation(s)
function all() {
for(i=0; i<numberOfPoints; i++){
circle.data([circleStates[i]])
.transition()
.delay(dur*i)
.duration(dur)
.ease("linear")
.attr('d', pathFunction)
}
}
function reverse() {
for(i=0; i<numberOfPoints; i++){
circle.data([circleStates[numberOfPoints-1-i]])
.transition()
.delay(dur*i)
.duration(dur)
.ease("linear")
.attr('d', pathFunction)
}
}
(Note: This should be in comments but not enough spacing)
Circle Animation
Try the radial wipe from SO. Need to tweak it so angle starts at 180 and ends back at same place (line#4-6,19) and move along the X-axis (line#11) on each interation. Change the <path... attribute to suit your taste.
Line Animation Grow a line from single point to the length (perimeter) of the circle.
Sync both animation so that it appears good on all browsers (major headache!).
we are programming a 2D game in XNA. Now we have polygons which define our level elements. They are triangulated such that we can easily render them. Now I would like to write a shader which renders the polygons as outlined textures. So in the middle of the polygon one would see the texture and on the border it should somehow glow.
My first idea was to walk along the polygon and draw a quad on each line segment with a specific texture. This works but looks strange for small corners where the textures are forced to overlap.
My second approach was to mark all border vertices with some kind of normal pointing out of the polygon. Passing this to the shader would interpolate the normals across edges of the triangulation and I could use the interpolated "normal" as a value for shading. I could not test it yet but would that work? A special property of the triangulation is that all vertices are on the border so there are no vertices inside the polygon.
Do you guys have a better idea for what I want to achieve?
Here A picture of what it looks right now with the quad solution:
You could render your object twice. A bigger stretched version behind the first one. Not that ideal since a complex object cannot be streched uniformly to create a border.
If you have access to your screen buffer you can render your glow components into a rendertarget and align a full-screen quad to your viewport and add a fullscreen 2D silhouette filter to it.
This way you gain perfect control over the edge by defining its radius, colour, blur. With additional output values such as the RGB values from the object render pass you can even have different advanced glows.
I think rendermonkey had some examples in their shader editor. Its definetly a good starting point to work with and try out things.
Propaply you want calclulate new border vertex list (easy fill example with triangle strip with originals). If you use constant border width and convex polygon its just:
B_new = B - (BtoA.normalised() + BtoC.normalised()).normalised() * width;
If not then it can go more complicated, there is my old but pretty universal solution:
//Helper function. To working right, need that v1 is before v2 in vetex list and vertexes are going to (anti???) cloclwise!
float vectorAngle(Vector2 v1, Vector2 v2){
float alfa;
if (!v1.isNormalised())
v1.normalise();
if (!v2.isNormalised())
v2.normalise();
alfa = v1.dotProduct(v2);
float help = v1.x;
v1.x = v1.y;
v1.y = -help;
float angle = Math::ACos(alfa);
if (v1.dotProduct(v2) < 0){
angle = -angle;
}
return angle;
}
//Normally dont use directly this!
Vector2 calculateBorderPoint(Vector2 vec1, Vector2 vec2, float width1, float width2){
vec1.normalise();
vec2.normalise();
float cos = vec1.dotProduct(vec2); //Calculates actually cosini of two (normalised) vectors (remember math lessons)
float csc = 1.0f / Math::sqrt(1.0f-cos*cos); //Calculates cosecant of angle, This return NaN if angle is 180!!!
//And rest of the magic
Vector2 difrence = (vec1 * csc * width2) + (vec2 * csc * width1);
//If you use just convex polygons (all angles < 180, = 180 not allowed in this case) just return value, and if not you need some more magic.
//Both of next things need ordered vertex lists!
//Output vector is always to in side of angle, so if this angle is.
if (Math::vectorAngle(vec1, vec2) > 180.0f) //Note that this kind of function can know is your function can know that angle is over 180 ONLY if you use ordered vertexes (all vertexes goes always (anti???) cloclwise!)
difrence = -difrence;
//Ok and if angle was 180...
//Note that this can fix your situation ONLY if you use ordered vertexes (all vertexes goes always (anti???) cloclwise!)
if (difrence.isNaN()){
float width = (width1 + width2) / 2.0; //If angle is 180 and border widths are difrent, you cannot get perfect answer ;)
difrence = vec1 * width;
//Just turn vector -90 degrees
float swapHelp = difrence.y
difrence.y = -difrence.x;
difrence.x = swapHelp;
}
//If you don't want output to be inside of old polygon but outside, just: "return -difrence;"
return difrence;
}
//Use this =)
Vector2 calculateBorderPoint(Vector2 A, Vector2 B, Vector2 C, float widthA, float widthB){
return B + calculateBorderPoint(A-B, C-B, widthA, widthB);
}
Your second approach can be possible...
mark the outer vertex (in border) with 1 and the inner vertex (inside) with 0.
in the pixel shader you can choose to highlight, those that its value is greater than 0.9f or 0.8f.
it should work.