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Give curved line a deeper arc in turf.js and mapbox

There's no simple curved-line tool in turf.js, nor is there an easy way to do it in mapbox (so far as I can see), so I've created a workaround based on this answer in this thread.
However, the curve it creates isn't very smooth or satisfying or has an inconsistent hump based on angle/length.
Ideally, I'd like an arc that is always in a nice, rounded form.
and drawing a line between them. I then offset the midpoint by distance / 5 and apply a bearing. I then connect up the three points with a turf.bezierSpline.
const start = [parseFloat(originAirport.longitude), parseFloat(originAirport.latitude)];
const end = [
parseFloat(destinationAirport.longitude),
parseFloat(destinationAirport.latitude),
];
const distance = turf.distance(start, end, { units: 'miles' });
const midpoint = turf.midpoint(start, end);
const destination = turf.destination(midpoint, distance / 5, 20, { units: 'miles' });
// curvedLine gets rendered to the page
const curvedLine = turf.bezierSpline(
turf.lineString([start, destination.geometry.coordinates, end]),
);
Desired curvature:

Well, that question was created a very long time ago, but I recently encounter this problem.
If anybody is still wondering - this code is good is general, but you've missed one detail. We can't use hardcoded bearing value 20 in turf.destination method, because it's incorrect for most cases. We need our moved midpoint to be right at the middle of our geometry, so we have to find the right angle.
const bearing = turf.bearing(start, end);
Then - if we want our arc to be on the left side of our line, we need to add 90 degrees to our calculated bearing. If on the right side - substract 90 degrees, so:
const leftSideArc = bearing + 90 > 180 ? -180 + (bearing + 90 - 180) : bearing + 90;
NOTE!!! Bearing is value between -180 and 180 degrees. Our value have to be calculated properly in case it'll exceed this range.
And then we can pass our bearing to destination method:
const destination = turf.destination(midpoint, distance / 5, leftSideArc, { units: 'miles' });
Now we have a perfect arc.

Unevenly space in D3 axis using rangePoints

I'm using D3 (v3) to build a plot with two axis. One of them it isn't working correctly when I add two or more objects. My piece of code is:
//Set upperLevelyScale
var previousValues = 0;
var upperLevelyScale = d3.scale.ordinal()
.domain(Object.keys(valuesY_numbers).sort())
.rangePoints((function(){
var values = Object.values(valuesY_numbers).map(function(x){
previousValues += (x * itemSize);
return previousValues});
values.unshift(0);
return values;
})());
when domain is ["MyValue1"] and rangePoints is [0, 170], the tick of the axis shows perfectly at the middle of the axis. But if domain is ["MyValue1", "MyValue2"] and rangePoints is [0,170,320] the ticks are not really covering their part but other.
What I'm trying to set up is just: MyValue1 is from 0 to 170, MyValue2 is from 170 to 320, etc. I have tried with '.range' also but still not working.
Thanks in advance.

SVG find rotation angle of a path

I've a problem with my SVG map.
I use jVectorMap to create a custom map and I need to write the name of every field in the center of the field.
The example is: JSFiddle Example (zoom in the right side to see the text)
I can find the center of every field with this function:
jvm.Map.prototype.getRegionCentroid = function(region){
if(typeof region == "string")
region = this.regions[region.toUpperCase()];
var bbox = region.element.shape.getBBox(),
xcoord = (bbox.x + bbox.width/2),
ycoord = (bbox.y + bbox.height/2);
return [xcoord, ycoord];
};
but my problem is that I want to rotate the text for align it with the top line of the relative field.
I've tried with getCTM() function but it give me always the same values for every field.
How can I find the right rotation angle of every field?
Thank you to all!

Looks like squeamish ossifrage has beaten me to this one, and what they've said would be exactly my approach too...
Solution
Essentially find the longest line segment in each region's path and then orient your text to align with that line segment whilst trying to ensure that the text doesn't end up upside-down(!)
Example
Here's a sample jsfiddle
In the $(document).ready() function of the fiddle I'm adding labels to all the regions but you will note that some of the regions have centroids that aren't within the area or non-straight edges that cause problems - Modifying your map slightly might be the easiest fix.
Explanation
Here are the 3 functions I've written to demonstrate the principles:
addOrientatedLabel(regionName) - adds a label to the named region of the map.
getAngleInDegreesFromRegion(regionName) - gets the angle of the longest edge of the region
getLengthSquared(startPt,endPt) - gets length squared of line seg (more efficient than getting length).
addOrientatedLabel() places the label at the centroid using a translate transform and rotates the text to the same angle as the longest line segment in the region. In SVG transforms are resolved right to left so:
transform="translate(x,y) rotate(45)"
is interpreted as rotate first, then translate. This ordering is important!
It also uses text-anchor="middle" and dominant-baseline="middle" as explained by squeamish ossifrage. Failing to do this will cause the text to be misaligned within its region.
getAngleInDegreesFromRegion() is where all the work is done. It gets the SVG path of the region with a selector, then loops through every point in the path. Whenever a point is found that is part of a line segment (rather than a Move-To or other instruction) it calculates the squared length of the line segment. If the squared length of the line segment is the longest so far it stores its details. I use squared length because that saves performing a square root operation (its only used for comparison purposes, so squared length is fine).
Note that I initialise the longestLine data to a horizontal one so that if the region has no line segments at all you'll at least get horizontal text.
Once we have the longest line, I calculate its angle relative to the x axis with Math.atan2, and convert it from radians to degrees for SVG with (angle / Math.PI) * 180. The final trick is to identify if the angle will rotate the text upside down, and if so, to rotate another 180 degrees.
Note
I've not used SVG before so my SVG code might not be optimal, but it's tested and it works on all regions that consist mostly of straight line segments - You will need to add error checking for a production application of course!
Code
function addOrientatedLabel(regionName) {
var angleInDegrees = getAngleInDegreesFromRegion(regionName);
var map = $('#world-map').vectorMap('get', 'mapObject');
var coords = map.getRegionCentroid(regionName);
var svg = document.getElementsByTagName('g')[0]; //Get svg element
var newText = document.createElementNS("http://www.w3.org/2000/svg","text");
newText.setAttribute("font-size","4");
newText.setAttribute("text-anchor","middle");
newText.setAttribute("dominant-baseline","middle");
newText.setAttribute('font-family', 'MyriadPro-It');
newText.setAttribute('transform', 'translate(' + coords[0] + ',' + coords[1] + ') rotate(' + angleInDegrees + ')');
var textNode = document.createTextNode(regionName);
newText.appendChild(textNode);
svg.appendChild(newText);
}
Here's my method to find the longest line segment in a given map region path:
function getAngleInDegreesFromRegion(regionName) {
var svgPath = document.getElementById(regionName);
/* longest edge will default to a horizontal line */
/* (in case the shape is degenerate): */
var longestLine = { startPt: {x:0, y:0}, endPt: {x:100,y:0}, lengthSquared : 0 };
/* loop through all the points looking for the longest line segment: */
for (var i = 0 ; i < svgPath.pathSegList.numberOfItems-1; i++) {
var pt0 = svgPath.pathSegList.getItem(i);
var pt1 = svgPath.pathSegList.getItem(i+1);
if (pt1.pathSegType == SVGPathSeg.PATHSEG_LINETO_ABS) {
var lengthSquared = getLengthSquared(pt0, pt1);
if( lengthSquared > longestLine.lengthSquared ) {
longestLine = { startPt:pt0, endPt:pt1, lengthSquared:lengthSquared};
}
}/* end if dealing with line segment */
}/* end loop through all pts in svg path */
/* determine angle of longest line segement relative to x axis */
var dY = longestLine.startPt.y - longestLine.endPt.y;
var dX = longestLine.startPt.x - longestLine.endPt.x;
var angleInDegrees = ( Math.atan2(dY,dX) / Math.PI * 180.0);
/* if text would be upside down, rotate through 180 degrees: */
if( (angleInDegrees > 90 && angleInDegrees < 270) || (angleInDegrees < -90 && angleInDegrees > -270)) {
angleInDegrees += 180;
angleInDegrees %= 360;
}
return angleInDegrees;
}
Note that my getAngleInDegreesFromRegion() method will only consider the longest straight line in a path if it is created with the PATHSEG_LINETO_ABS SVG command... You'll need more functionality to handle regions which don't consist of straight lines. You could approximate by treating curves as straight lines with:
if (pt1.pathSegType != SVGPathSeg.PATHSEG_MOVETO_ABS )
But there will be some corner cases, so modifying your map data might be the easiest approach.
And finally, here's the obligatory squared distance method for completeness:
function getLengthSquared(startPt, endPt ) {
return ((startPt.x - endPt.x) * (startPt.x - endPt.x)) + ((startPt.y - endPt.y) * (startPt.y - endPt.y));
}
Hope that is clear enough to help get you started.

Querying getCTM() won't help. All that gives you is a transformation matrix for the shape's coordinate system (which, as you discovered, is the same for every shape). To get a shape's vertex coordinates, you'll have to examine the contents of region.element.shape.pathSegList.
This can get messy. Although a lot of the shapes are drawn using simple "move-to" and "line-to" commands with absolute coordinates, some use relative coordinates and other types of line. I noticed at least one cubic curve. It might be worth looking for an SVG vertex manipulation library to make life easier.
But in general terms, what you need to do is fetch the list of coordinates for each shape (converting relative coordinates to absolute where necessary), and find the segment with the longest length. Be aware that this may be the segment between the two end points of the line. You can easily find the orientation of this segment from Math.atan2(y_end-y_start,x_end-x_start).
When rotating text, make life easy for yourself by using a <g> element with a transform=translate() attribute to move the coordinate origin to where the text needs to be. Then the text won't shoot off into the distance when you add a transform=rotate() attribute to it. Also, use text-anchor="middle" and dominant-baseline="middle" to centre the text where you want it.
Your code should end up looking something like this:
var svg = document.getElementsByTagName('g')[0]; //Get svg element
var shape_angle = get_orientation_of_longest_segment(svg.pathSegList); //Write this function
var newGroup = document.createElementNS("http://www.w3.org/2000/svg","g");
var newText = document.createElementNS("http://www.w3.org/2000/svg","text");
newGroup.setAttribute("transform", "translate("+coords[0]+","+coords[1]+")");
newText.setAttribute("font-size","4");
newText.setAttribute("text-anchor","middle");
newText.setAttribute("dominant-baseline","middle");
newText.setAttribute("transform","rotate("+shape_angle+")");
newText.setAttribute('font-family', 'MyriadPro-It');
var textNode = document.createTextNode("C1902");
newText.appendChild(textNode);
newGroup.appendChild(newText);
svg.appendChild(newGroup);

Glitches with Triangulation + Linear Interpolation + 3D projection

Intro
Hey!
Some weeks ago, I did a small demo for a JS challenge. This demo was displaying a landscape based on a procedurally-generated heightmap. To display it as a 3D surface, I was evaluating the interpolated height of random points (Monte-Carlo rendering) then projecting them.
At that time, I was already aware of some glitches in my method, but I was waiting for the challenge to be over to seek some help. I'm counting on you. :)
Problem
So the main error I get can be seen in the following screenshot:
Screenshot - Interpolation Error? http://code.aldream.net/img/interpolation-error.jpg
As you can see in the center, some points seem like floating above the peninsula, forming like a less-dense relief. It is especially obvious with the sea behind, because of the color difference, even though the problem seems global.
Current method
Surface interpolation
To evaluate the height of each point of the surface, I'm using triangulation + linear interpolation with barycentric coordinates, ie:
I find in which square ABCD my point (x, y) is, with A = (X,Y), B = (X+1, Y), C = (X, Y+1) and D = (X+1, Y+1), X and Y being the truncated value of x, y. (each point is mapped to my heightmap)
I estimate in which triangle - ABD or ACD - my point is, using the condition: isInABD = dx > dy with dx, dy the decimal part of x, y.
I evaluate the height of my point using linear interpolation:
if in ABD, height = h(B) + [h(A) - h(B)] * (1-dx) + [h(D) - h(B)] * dy
if in ACD, height = h(C) + [h(A) - h(C)] * (1-dy) + [h(D) - h(C)] * dx, with h(X) height from the map.
Displaying
To display the point, I just convert (x, y, height) into the world coordinates, project the vertex (using simple perspective projection with yaw and pitch angles). I use a zBuffer I keep updated to check if I draw or not the obtained pixel.
Attempts
My impression is that for some points, I get a wrong interpolated height. I thus tried to search for some errors or some non-covered boundaries cases, in my implementation of the triangulation + linear interpolation. But if there are, I can't spot them.
I use the projection in other demos, so I don't think the problem comes from here. As for the zBuffering, I can't see how it could be related...
I'm running out of luck here... Any hints are most welcome!
Thank for your attention, and have a nice day!
Annexe
JsFiddle - Demo
Here is a jsFiddle http://jsfiddle.net/PWqDL/ of the whole slightly simplified demo, for those who want to tweak around...
JsFiddle - Small test for the interpolation
As I was writing down this question, I got an idea to have a better look at the results of my interpolation. I implemented a simple test in which I use a 2x2 matrix containing some hue values, and I interpolate the intermediate colors before displaying them in the canvas.
Here is the jsFiddle: http://jsfiddle.net/y2K7n/
Alas, the results seem to match the expected behavior for the kind of "triangular" interpolation I'm doing, so I'm definitly running out of ideas.
Code sample
And here is the simplified most-probably-faulty part of my JS code describing my rendering method (but the language doesn't matter much here I think), given a square heightmap "displayHeightMap" of size (dim x dim) for a landscape of size (SIZE x SIZE):
for (k = 0; k < nbMonteCarloPointsByFrame; k++) {
// Random float indices:
var i = Math.random() * (dim-1),
j = Math.random() * (dim-1),
// Integer part (troncated):
iTronc = i|0,
jTronc = j|0,
indTronc = iTronc*dim + jTronc,
// Decimal part:
iDec = i%1,
jDec = j%1,
// Now we want to intrapolate the value of the float point from the surrounding points of our map. So we want to find in which triangle is our point to evaluate the weighted average of the 3 corresponding points.
// We already know that our point is in the square defined by the map points (iTronc, jTronc), (iTronc+1, jTronc), (iTronc, jTronc+1), (iTronc+1, jTronc+1).
// If we split this square into two rectangle using the diagonale [(iTronc, jTronc), (iTronc+1, jTronc+1)], we can deduce in which triangle is our point with the following condition:
whichTriangle = iDec < jDec, // ie "are we above or under the line j = jTronc + distanceBetweenLandscapePoints - (i-iTronc)"
indThirdPointOfTriangle = indTronc +dim*whichTriangle +1-whichTriangle, // Top-right point of the square or bottm left, depending on which triangle we are in.
// Intrapolating the point's height:
deltaHeight1 = (displayHeightMap[indTronc] - displayHeightMap[indThirdPointOfTriangle]),
deltaHeight2 = (displayHeightMap[indTronc+dim+1] - displayHeightMap[indThirdPointOfTriangle]),
height = displayHeightMap[indThirdPointOfTriangle] + deltaHeight1 * (1-(whichTriangle? jDec:iDec)) + deltaHeight2 * (!whichTriangle? jDec:iDec),
posX = i*distanceBetweenLandscapePoints - SIZE/2,
posY = j*distanceBetweenLandscapePoints - SIZE/2,
posZ = height - WATER_LVL;
// 3D Projection:
var temp1 = cosYaw*(posY - camPosY) - sinYaw*(posX - camPosX),
temp2 = posZ - camPosZ,
dX = (sinYaw*(posY - camPosY) + cosYaw*(posX - camPosX)),
dY = sinPitch*temp2 + cosPitch*temp1,
dZ = cosPitch*temp2 - sinPitch*temp1,
pixelY = dY / dZ * minDim + canvasHeight,
pixelX = dX / dZ * minDim + canvasWidth,
canvasInd = pixelY * canvasWidth*2 + pixelX;
if (!zBuffer[canvasInd] || (dZ < zBuffer[canvasInd])) { // We check if what we want to draw will be visible or behind another element. If it will be visible (for now), we draw it and update the zBuffer:
zBuffer[canvasInd] = dZ;
// Color:
a.fillStyle = a.strokeStyle = EvaluateColor(displayHeightMap, indTronc); // Personal tweaking.
a.fillRect(pixelX, pixelY, 1, 1);
}
}

Got it. And it was as stupid a mistake as expected: I was reinitializing my zBuffer each frame...
Usually it's what you should do, but in my case, each frame (ie call of my Painting() function) adds details to the same frame (ie drawed static scene from a constant given point of view).
If I reset my zBuffer at each call of Painting(), I lose the depth information of the points drawn during the previous calls. The corresponding pixels are thus considered as blank, and will be re-painted for any projected points, without any regard for their depth.
Note: Without reinitiliazation, the zBuffer gets quite big. Another fix I should have done earlier was thus to convert the pixel's positions of the projected point (and thus the indices of the zBuffer) into integer values:
pixelY = dY / dZ * minDim + canvasHeight +.5|0,
pixelX = dX / dZ * minDim + canvasWidth +.5|0,
canvasInd = pixelY * canvasWidth*2 + pixelX;
if (dZ > 0 && (!zBuffer[canvasInd] || (dZ < zBuffer[canvasInd]))) {
// We draw the point and update the zBuffer.
}
Fun fact
If the glitches appeared more obvious for relief with the sea behind, it wasn't only for the color difference, but because the hilly parts of the landscape need much more points to be rendered than flat areas (like the sea), given their stretched surface.
My simplistic Monte-Carlo sampling of points doesn't take this characteristic into account, which means that at each call of Painting(), the sea gains statistically more density than the lands.
Because of the reinitialization of the zBuffer each frame, the sea was thus "winning the fight" in the picture's areas where mountains should have covered it (explaining the "ghostly mountains" effect there).
Corrected JsFiddle
Corrected version for those interested: http://jsfiddle.net/W997s/1/

RaphaelJs Pie Chart Animation on Hover

i am customizing the pie chart given on the raphael site below link
http://raphaeljs.com/pie.html
this chart shows animation when hover a slice, this animation simply expand the slice a little
what i want is to separate the slice from the chart
i played with the transform property of following lines of code but could not make it as i want.
p.mouseover(function () {
var xis= p.getBBox(true);
p.stop().animate({transform: "s1.1 1.1 "+ cx + " " + cy }, ms, "linear");
txt.stop().animate({opacity: 1}, ms, "linear");
}).mouseout(function () {
p.stop().animate({transform: ""}, ms, "linear");
txt.stop().animate({opacity: 0}, ms);
});
Changing the line 3's cx and cy actually fixed the animation for every slice in the same manner, that is, on hover every slice will change the position constantly.
http://imageshack.us/photo/my-images/690/sliced1.png
anyone please help me out to solve this problem

If I understand your question correctly, you want the slice to completely disconnect from the pie chart when somebody hovers over it.
To do this, you want to translate the segment, which allows you to move an SVG object in a given direction, toward x, y co-ordinates. I'm no SVG pro, so I'd suggest taking a look into the full functionality of this yourself; regardless, to do these types of operations with Raphael, you can use the Element.transform, or can provide transform values in an animate call.
The second of these is what is happening in the example you provided, except a scale transformation is being used, as indicated by the leading s in transform: "s1.1 1.1.. A scale will make the object bigger.
Here, you want to use a translation which moves the object but doesn't make it bigger - it uses a t.
Here is a slightly edited block of code that will do this:
p.mouseover(function () {
var distance = 20;
var xShiftTo = distance*Math.cos(-popangle * rad);
var yShiftTo = distance*Math.sin(-popangle * rad);
p.stop().animate({transform: "t" + xShiftTo + " " + yShiftTo}, ms, "bounce");
txt.stop().animate({opacity: 1}, ms, "bounce");
}).mouseout(function () {
p.stop().animate({transform: ""}, ms, "bounce");
txt.stop().animate({opacity: 0}, ms);
});
In the example, distance refers to how far the slice should move away (so feel free to adjust this), xShiftTo and yShiftTo calculate the x, y values that the object should shift by, relative to where they currently are. Note that this is a little complicated - you need to figure out the x, y values at a given angle away from the pie centre. The positioning of the text also does something like this, so I just took the required maths from there. Also, I just left the bounce animation, but you can change that to linear or whatever you want. Hope that helps.

You should probably be using .hover which is built into Raphael. See documentation here: http://raphaeljs.com/reference.html#Element.hover
Working off of Oli's example I was able to figure out most of the basic animation principles. Not being a math guru there were a lot of gaps to fill in for the example. Here is a fully functioning version (tested). Enjoy!
pie.hover(function () {
// Stop all existing animation
this.sector.stop();
// Collect/Create variables
var rad = Math.PI / 180;
var distance = 50; // Amount in pixels to push graph segment out
var popangle = this.sector.mangle; // Pull angle right out of segment object
var ms = 300; // Time in milliseconds
// Setup shift variables to move pie segments
var xShiftTo = distance*Math.cos(-popangle * rad);
var yShiftTo = distance*Math.sin(-popangle * rad);
this.sector.animate({transform: "t" + xShiftTo + " " + yShiftTo}, ms, "linear");
}, function () {
// Passing an empty transform property animates the segment back to its default location
this.sector.animate({ transform: '' }, ms, "linear");
});