如何在画布上绘制一个简单的五角大楼

问题描述

我想创建一个五边形，我成功创建了五边形.

I want to create a Pentagon, I succeeded in creating such pentagon.

但是我的五边形是不正确的，因为它在表面上不正确.

But my pentagon is not right, because it doesn't stand right on the surface.

我该如何解决?我需要一个简洁的答案，而不仅仅是快速修复.

How can I fix it? I need an elegant answer not just a quick fix.

我还想知道另一件事:

如何仅使用坐标绘制五边形，我是指五边形的5个坐标?

How can I draw a Pentagon using coordinates only, I mean the 5 coordinates of the pentagon?

我想基于五个已知坐标(v1，v2..v5)绘制一个五边形而且没有任何循环，可以在五个点之间画出某种路径.

I want to draw a pentagon based on five known cordinates(v1,v2..v5)and without any loop,to draw some kind of path between five points.

$(function(){
var canvas=document.getElementById("canvas");
var cxt=canvas.getContext("2d");
    // hexagon
    var numberOfSides = 5,
        size = 100,
        Xcenter = 150,
        Ycenter = 150;

    cxt.beginPath();
    cxt.moveTo (Xcenter +  size * Math.cos(0), Ycenter +  size *  Math.sin(0));          

    for (var i = 1; i <= numberOfSides;i += 1) {
      cxt.lineTo (Xcenter + size * Math.cos(i * 2 * Math.PI / numberOfSides), Ycenter + size * Math.sin(i * 2 * Math.PI / numberOfSides));
    }

    cxt.strokeStyle = "#000000";
    cxt.lineWidth = 1;
    cxt.stroke();
  });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<canvas id="canvas" width=650 height=500></canvas>

推荐答案

有趣的问题，您可以移动sin/cos值，使底线水平对齐:

Interesting question, you can shift the sin/cos values so the bottom line aligns horizontally:

$(function(){
   var canvas=document.getElementById("canvas");
   var cxt=canvas.getContext("2d");
    // hexagon
    var numberOfSides = 5,
        size = 100,
        Xcenter = 150,
        Ycenter = 150,
        step  = 2 * Math.PI / numberOfSides,//Precalculate step value
        shift = (Math.PI / 180.0) * -18;//Quick fix ;)

    cxt.beginPath();
    //cxt.moveTo (Xcenter +  size * Math.cos(0), Ycenter +  size *  Math.sin(0));          

    for (var i = 0; i <= numberOfSides;i++) {
    	var curStep = i * step + shift;
        cxt.lineTo (Xcenter + size * Math.cos(curStep), Ycenter + size * Math.sin(curStep));
    }

    cxt.strokeStyle = "#000000";
    cxt.lineWidth = 1;
    cxt.stroke();
  });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<canvas id="canvas" width=650 height=500></canvas>

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