我開始使用 d3。還有一個(gè)問題，如何沿著填充路徑正確設(shè)置三角形動(dòng)畫？我為儀表圖創(chuàng)建了一個(gè)基礎(chǔ)并對其進(jìn)行了動(dòng)畫處理。var chart = d3.select("#speedometer");const arc = d3    .arc()    .outerRadius(120)    .innerRadius(90)    .startAngle(-Math.PI / 2);chart    .append("path")    .datum({        endAngle: Math.PI / 2    })    .attr("transform", "translate(160, 180)")    .attr("class", "background")    .style("fill", "#495270")    .attr("d", arc);const triangle = chart.append('g').attr('transform', 'translate(30, 180) rotate(90)').style('width', '100%').append("path").attr("d", "M3.937,0,7.873,14H0Z");const newAngle = (70 / 100) * Math.PI - Math.PI / 2;const foreground = chart    .append("path")    .datum({ endAngle: -Math.PI / 2 })    .style("fill", "rgb(50, 188, 228)")    .attr("transform", "translate(160, 180)")    .attr("d", arc);foreground    .transition()    .duration(3000)    .attrTween("d", function (d) {        const interpolate = d3.interpolate(d.endAngle, newAngle);        return function (t) {            d.endAngle = interpolate(t);            return arc(d);        };    });triangle    .transition()    .duration(3000)function pathTween(path) {    const length = path.node().getTotalLength(); // Get the length of the path    const r = d3.interpolate(0, length); // Set up interpolation from 0 to the path length    return function (t) {        const point = path.node().getPointAtLength(r(t)); // Get the next point along the path        d3            .select(this) // Select the circle            .attr("transform", `translate(${point.x}, ${point.y})`);    };}.main-wrapper{  max-width: 80%;  margin: 20px auto;}.element{  display: flex;  flex-flow: column nowrap;  margin-bottom: 20px;    border: 1px solid rgba(0,0,0,.4);    padding: 20px;    border-radius: 6px;}.title{  margin-bottom: 4px;  font-weight: 500;}.description{  margin-bottom: 10px;  color: rgba(0,0,0,.4);}#speedometer {    width: 300px;    height: 300px;}canvas{    width: 100%;    height: 100%;}問題是，如何將三角形鏈接到動(dòng)畫弧的邊緣，如圖所示？我知道我應(yīng)該使用tween('path'....)