Create equilateral triangle in the middle of canva

问题:

I want to draw an equilateral triangle in the middle of canvas. I tried this:

ctx.moveTo(canvas.width/2, canvas.height/2-50);
ctx.lineTo(canvas.width/2-50, canvas.height/2+50);
ctx.lineTo(canvas.width/2+50, canvas.height/2+50);
ctx.fill();

But the triangle looks a bit too tall.

How can I draw an equilateral triangle in the middle of canvas?

Someone told me you have to find the ratio of the height of an equilateral triangle to the side of an equilateral triangle.

h:s

What are the two numbers?

回答1:

You have to do it with the height of the triangle

var h = side * (Math.sqrt(3)/2);

or

var h = side * Math.cos(Math.PI/6);

So the ratio h:s is equal to:

sqrt( 3 ) / 2 : 1 = cos( π / 6 ) : 1 ≈ 0.866025

See : http://jsfiddle.net/rWSKh/2/

回答2:

The equation for the three corner points is

x = r*cos(angle) + x_center
y = r*sin(angle) + y_center

where for angle = 0, (1./3)*(2*pi), and (2./3)*(2*pi); and where r is the radius of the circle in which the triangle is inscribed.

回答3:

A simple version where X and Y are the points you want to top of the triangle to be:

var height = 100 * (Math.sqrt(3)/2);
context.beginPath();
context.moveTo(X, Y);
context.lineTo(X+50, Y+height);
context.lineTo(X-50, Y+height);
context.lineTo(X, Y);
context.fill();
context.closePath();

This makes an equilateral triange with all sides = 100. Replace 100 with how long you want your side lengths to be.

After you find the midpoint of the canvas, if you want that to be your triangle's midpoint as well you can set X = midpoint's X and Y = midpoint's Y - 50 (for a 100 length triangle).

回答4:

The side lengths will not be equal given those coordinates.

The horizontal line constructed on the bottom has a length of 100, but the other sides are actually the hypotenuse of a 50x100 triangle ( approx. 112).

回答5:

I can get you started with drawing an equilateral triangle but I don't have the time to get it centered.

jsFiddle

var ax=0;
var ay=0;
var bx=0;
var by=150;

var dx=bx-ax
var dy=by-ay;
var dangle = Math.atan2(dy, dx) - Math.PI / 3;
var sideDist = Math.sqrt(dx * dx + dy * dy);

var cx = Math.cos(dangle) * sideDist + ax;
var cy =  Math.sin(dangle) * sideDist + ay;

var canvas = document.getElementById('equ');
var ctx = canvas.getContext('2d');

ctx.beginPath();  
ctx.moveTo(ax,ay);  
ctx.lineTo(bx,by);  
ctx.lineTo(cx,cy);  

ctx.fill(); 

回答6:

my code for drawing triangle also depending on direction (for lines). code is for Raphael lib.

drawTriangle(x2 - x1, y2 - y1, x2, y2);
function drawTriangle(dx, dy, midX, midY) {
        var diff = 0;
        var cos = 0.866; 
        var sin = 0.500; 

        var length = Math.sqrt(dx * dx + dy * dy) * 0.8; 
        dx = 8 * (dx / length); 
        dy = 8 * (dy / length); 
        var pX1 = (midX + diff) - (dx * cos + dy * -sin); 
        var pY1 = midY - (dx * sin + dy * cos); 
        var pX2 = (midX + diff) - (dx * cos + dy * sin); 
        var pY2 = midY - (dx * -sin + dy * cos); 

        return [
                    "M", midX + diff, midY,
                    "L", pX1, pY1,
                    "L", pX2, pY2,
                    "L", midX + diff, midY
                ].join(","); 
    }