Reputation: 6657
I'm using a d3 attrTween to translate a circle over a path smoothly, similar to this example and as shown in the picture below:
The circle's transition is defined here:
function transition() {
circle.transition()
.duration(2051)
.ease("linear")
.attrTween("transform", translateAlong(path.node()))
}
And the attribute tween is shown here:
function translateAlong(path) {
var l = path.getTotalLength();
return function (d, i, a) {
return function (t) {
var p = path.getPointAtLength(t * l);
return "translate(" + p.x + "," + p.y + ")";
};
};
}
This works well thanks to the SVG method getPointAtLength
, which allows us to retrieve coordinates at different lengths of the path. However, I need a different kind of behavior and I've been unable to come up with a solution so far.
I need the circle to animate along the path, but at a steady horizontal speed. Meaning that the circle ought to take as much time to navigate this slice:
As it does with this slice:
Because both slices encompass the same width. On a low level, what I need is to be able to translate any X coordinate with its corresponding Y coordinate along the path. I've looked at all the SVG path methods and I haven't found anything particularly useful here. I'm hoping there's some way in D3 to feed an X coordinate to a d3 line and retrieve its corresponding Y coordinate.
Here's a JSFiddle working as described above. I'd really appreciate any help I can get on this. Thanks!
Upvotes: 1
Views: 938
Reputation: 6657
I ended up creating a lookup array for all my points along the line using getPointAtLength
:
var lookup = [];
var granularity = 1000;
var l = path.node().getTotalLength();
for(var i = 1; i <= granularity; i++) {
var p = path.node().getPointAtLength(l * (i/granularity))
lookup.push({
x: p.x,
y: p.y
})
}
Once I had all those points in my lookup table, I used a bisector in my translate tween:
var xBisect = d3.bisector(function(d) { return d.x; }).left;
function translateAlong(path) {
var l = path.getTotalLength();
return function (d, i, a) {
return function (t) {
var index = xBisect(lookup, l * t);
var p = lookup[index];
return "translate(" + p.x + "," + p.y + ")";
};
};
}
And it works as expected! Yahoo!
Upvotes: 2