Finding x, y coordinates of a point on an Archimedean spiral
I am trying to plot data on a timeline using an Archimedean spiral as an axis using D3.js.
So I need a Javascript function where I pass it
- D distance for each step
- S few steps
- X distance between each spiral arm
The function will traverse the spiral arc at a distance s * d and give me x and y cartesian coordinates (point S in the diagram, where s = 10). The first point in the center of the spiral is at 0.0.
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2 answers
Thank you for your help. I tried to plot your example, but it is a little strange when I plotted 5 consecutive points (see screenshot below).
I managed to find the answer from the link below. It sounds like you were very close.
Algorithm to solve points of uniformly distributed / even breaks of a spiral?
My final implementation is based on the link above.
function archimedeanSpiral(svg,data,circleMax,padding,steps) {
var d = circleMax+padding;
var arcAxis = [];
var angle = 0;
for(var i=0;i<steps;i++){
var radius = Math.sqrt(i+1);
angle += Math.asin(1/radius);//sin(angle) = opposite/hypothenuse => used asin to get angle
var x = Math.cos(angle)*(radius*d);
var y = Math.sin(angle)*(radius*d);
arcAxis.push({"x":x,"y":y})
}
var lineFunction = d3.svg.line()
.x(function(d) { return d.x; })
.y(function(d) { return d.y; })
.interpolate("cardinal");
svg.append("path")
.attr("d", lineFunction(arcAxis))
.attr("stroke", "gray")
.attr("stroke-width", 5)
.attr("fill", "none");
var circles = svg.selectAll("circle")
.data(arcAxis)
.enter()
.append("circle")
.attr("cx", function (d) { return d.x; })
.attr("cy", function (d) { return d.y; })
.attr("r", 10);
return(arcAxis);
}
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Wouldn't hurt to try: (sorry my javascript newbie)
function spiralPoint(dist, sep, step) {
this.x = 0;
this.y = 0;
var r = dist;
var b = sep / (2 * Math.PI);
var phi = r / b;
for(n = 0; n < step-1; ++n) {
phi += dist / r;
r = b * phi;
}
this.x = r * Math.cos(phi);
this.y = r * Math.sin(phi);
this.print = function() {
console.log(this.x + ', ' + this.y);
};
}
new spiralPoint(1,1,10).print();
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