[从头学数学] 第171节 点、直线、平面之间的位置关系

剧情提要：
[机器小伟]在[工程师阿伟]的陪同下进入了结丹初期的修炼，
这次要修炼的目标是[点、直线、平面之间的位置关系]。

正剧开始：

星历2016年04月10日 13:48:09, 银河系厄尔斯星球中华帝国江南行省。
[工程师阿伟]正在和[机器小伟]一起研究[点、直线、平面之间的位置关系]。

来画个平面玩玩吧。

<span style="font-size:18px;">//平面

	if (1) {
		var r = 20;
		config.setSector(1,1,1,1);
		config.graphPaper2D(0, 0, r);
		config.axis3D(0, 0, 0, 180);
		
		var transform = new Transform();
		
		var a = 4*r, b = 3*r, h = 2*r;
		var edges = 4;
		var array = shape.rect(a, b);
		
		var size = array.length;
		
		var topFace = [], bottomFace = [];
		for (var i = 0; i < size; i++) {
			topFace.push(shape.point3D(array[i][0], 0, array[i][1]));
		}
		
		topFace = transform.flipY(topFace);
		
		
		shape.strokeDraw([].concat(topFace), 'pink',5);
		shape.pointDraw([].concat(topFace), 'red',5, 1);

	
	
	}</span>

再来画两个平面相交：

<span style="font-size:18px;">	if (1) {
		var r = 20;
		config.setSector(1,1,1,1);
		config.graphPaper2D(0, 0, r);
		config.axis3D(0, 0, 0, 180);
		
		var transform = new Transform();
		
		var a = 4*r, b = 3*r, h = 2*r;
		var edges = 4;
		var array = shape.rect(a, b);
		
		var size = array.length;
		
		var topFace = [], bottomFace = [];
		var thita = 2*Math.PI/3;
		var x = y = 0;
		for (var i = 0; i < size; i++) {
			x = -array[i][0]*Math.cos(thita)*0.6;
			y = array[i][0]*Math.sin(thita)*0.6;
			
			topFace.push(shape.point3D(array[i][0], 0, array[i][1]));
			bottomFace.push(shape.point3D(x, y, array[i][1]));
		}
		
		topFace = transform.flipY(topFace);
		bottomFace = transform.flipY(bottomFace);
		
		
		shape.strokeDraw([].concat(topFace), 'pink',5);
		shape.pointDraw([].concat(topFace), 'red',5, 1);
		shape.strokeDraw([].concat(bottomFace), 'orange',5);
		shape.pointDraw([].concat(bottomFace), 'orange',5, 1, ['A\'', 'B\'', 'C\'', 'D\'']);

	
	
	}
	
</span>

由于这些公理都是在一定的假设前提下证明的，所以小伟没办法多加深入，

但小伟想知道的是：一个投影点的坐标，它可能是哪些空间点产生的，

比如在斜二测画法的情况下？

于是做了一个研究：

<span style="font-size:18px;">def point3D(x, y, z):
    z = z/2;
    x = -(x-z*0.707);
    y = y+z*0.707;
    return [x, y];

def tmp():
    x0 = y0 = z0 = 5;
    b = point3D(x0, y0, z0);
    epsilon = 0.01;

    rMin = 0;
    rMax = 10;

    delta = 0.1;

    x = y = z = 0;
    while x < rMax:
        x += delta;
        y = rMin;

        while y < rMax:
            y += delta;
            z = rMin;

            while z < rMax:
                z += delta;                
                a = point3D(x, y, z);
                if abs(a[0] - b[0])< epsilon and abs(a[1] - b[1])<epsilon:
                    print('PROJ[{0:.2f}, {1:.2f}, {2:.2f}] = PROJ[{3}, {4}, {5}] = {6:.2f}, {7:.2f}'\
                          .format(x, y, z, x0, y0, z0, a[0], a[1]));</span>

这是结果：

<span style="font-size:18px;">>>> 
PROJ[3.30, 6.70, 0.20] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[3.40, 6.60, 0.50] = PROJ[5, 5, 5] = -3.22, 6.78
PROJ[3.70, 6.30, 1.30] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[3.80, 6.20, 1.60] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[3.90, 6.10, 1.90] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[4.30, 5.70, 3.00] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[4.40, 5.60, 3.30] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[4.50, 5.50, 3.60] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[4.90, 5.10, 4.70] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[5.00, 5.00, 5.00] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[5.10, 4.90, 5.30] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[5.50, 4.50, 6.40] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[5.60, 4.40, 6.70] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[5.70, 4.30, 7.00] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[6.10, 3.90, 8.10] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[6.20, 3.80, 8.40] = PROJ[5, 5, 5] = -3.23, 6.77
PROJ[6.30, 3.70, 8.70] = PROJ[5, 5, 5] = -3.22, 6.78
PROJ[6.60, 3.40, 9.50] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[6.70, 3.30, 9.80] = PROJ[5, 5, 5] = -3.24, 6.76
PROJ[6.80, 3.20, 10.10] = PROJ[5, 5, 5] = -3.23, 6.77</span>

尽管精度很低，范围也很小，但还是看得出来，在任一种投影运算下，

都可能会有很多点被投影到同一个图像位置。

其实这次小伟想来算一下点阵围成图形的面积。

<span style="font-size:18px;">	if (1) {
		var r = 20;
		config.setSector(1,1,1,1);
		config.graphPaper2D(0, 0, r);
		config.axis2D(0, 0, 180);
		
		var transform = new Transform();
		
		var a = 4, b = 3, h = 2;
		var edges = 4;
		var array = shape.trapezoid(0, 0, 0, a, b, Math.PI/3);
		
		var size = array.length;
		
		//document.write(areaCalc(array));
		//由阵列分割出的三角形
		var tri = [];
		var xCenter = yCenter = 0;
		var S = 0;
		var scale = 2*r;
		var measure = 0;
		
		for (var i = 1; i < size-1; i++) {
			tri = [];
			tri.push(array[0]);
			tri.push(array[i]);
			tri.push(array[i+1]);
			
			xCenter = (tri[0][0]+tri[1][0]+tri[2][0])/3*scale;
			yCenter = -(tri[0][1]+tri[1][1]+tri[2][1])/3*scale;
			
			S = areaCalc(tri);
			
			if (S > 0) {
				shape.strokeDraw(tri, 'red', scale);
				
				var text = 'S='+S.toFixed(2);
				measure = plot.measureText(text);
				plot.fillText(text, xCenter-measure/2, yCenter, measure);
			}
		}
			
		shape.angleDraw(array, 'blue', scale);
	
	
	}
	
	
}



function areaCalc(array) {
	//array中包含组成三角形的三个点的坐标
	
	array = shape.angularSort(array);
	
	var S = 0;
	
	if (array.length >=3) {
		var x1 = array[0][0], y1 = array[0][1],
			x2 = array[1][0], y2 = array[1][1],
			x3 = array[2][0], y3 = array[2][1];
			
		S = 0.5*(x1*y2+x2*y3+x3*y1-x1*y3-x2*y1-x3*y2);
	}
	
	return S;
}</span>

效果如下：

有了这个工具，以后应该会省不少事的。

本节到此结束，欲知后事如何，请看下回分解。