在圆石中将圆分成n个相等的一半的点的坐标 [英] Coordinates of points dividing circle into n equal halves in Pebble

问题描述

我正在尝试使用直线将圆分成n个相等的部分.想法是获取圆圆周上端点的坐标，然后绘制直线.

I am trying to divide the circle into n equal parts using straight lines. Idea is to get the coordinates of the end points on the circumference of the circle and then to draw the line.

要找到我的配伍酸盐，我正在使用以下代码:

to find coodrinates I am using following code:

static void getPoints(int x0,int y0,int r)
{

   double angle = 0;
   double angleToRadian = 0;

   for(int i = 0 ; i < noOfDividingPoints  ;i++)
   {   
      angle = i * (360/noOfDividingPoints);

      angleToRadian = ( angle * 3.141 ) / 180;

      APP_LOG(APP_LOG_LEVEL_DEBUG, "LOG: Angle: %lf, AngleToRadian: %lf", angle, angleToRadian );

      x[i] = (int) (x0 + r * cos_lookup(angleToRadian));
      y[i] = (int) (y0 + r * sin_lookup(angleToRadian));
      APP_LOG(APP_LOG_LEVEL_DEBUG, "LOG: i: %d,  x[i]: %d, y[i]: %d", i, (int)x[i], (int)y[i] );
   }
}

使用此getPoints方法，我要填充数组x []和y []，然后遍历这些数组以在x [i]和y [i]之间绘制线

Using this getPoints method I am filling the array x[] and y[] and then iterate over these arrays to draw lines between x[i] and y[i]

但是，从上面的代码计算出的值很奇怪，下面是打印x []和y []内容的日志

However, the values calculated from the above code comes out to be weird, below are the logs printing the contents of x[] and y[]

[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f

[14:18:57] WatchFace.c:45> LOG: i: 0,  x[i]: 4587522, y[i]: 84
[14:18:57] WatchFace.c:36> LOG: Angle: 90
[14:18:57] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 1,  x[i]: 4587522, y[i]: 84
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 2,  x[i]: 4587452, y[i]: 504
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 3,  x[i]: 4587452, y[i]: 924
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 4,  x[i]: 4587452, y[i]: 1344
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 5,  x[i]: 4587452, y[i]: 1344
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 6,  x[i]: 4587452, y[i]: 1834
[14:18:58] WatchFace.c:36> LOG: Angle: 90
[14:18:58] WatchFace.c:41> LOG: Angle: f, AngleToRadian: f
[14:18:58] WatchFace.c:45> LOG: i: 7,  x[i]: 4587452, y[i]: 2254

请指出我在这里想念的东西.

Please point out what I am missing here.

谢谢.

推荐答案

lookup- Pebble中的正弦和余弦函数不采用弧度或度数表示角度；他们使用归一化为TRIG_MAX_ANGLE的整数角，它对应于一个完整的圆或360°.

The lookup-functions for sine and cosine in Pebble don't take the angle in radians or degrees; they use an integer angle normalized to TRIG_MAX_ANGLE, which corresponds to a full circle or 360°.

这同样适用于返回值，它是一个带符号的int，必须使用TRIG_MAX_RATIO进行标准化才能获得有效的余弦或正弦.

The same applies to the returned value, which is a signed int that must be normalized with TRIG_MAX_RATIO to get a valid cosine or sine.

因此，您应该可以执行类似的操作.

So you should be able to do something like this.

for (int i = 0; i < n ; i++) {
    int32_t angle = i * TRIG_MAX_ANGLE / n;

    x[i] = x0 + (r * cos_lookup(angle)) / TRIG_MAX_RATIO;
    y[i] = y0 + (r * sin_lookup(angle)) / TRIG_MAX_RATIO;
}

请注意，所有计算都是使用整数完成的.请注意先相乘然后相除，以免截断结果.例如

Note that all calculations are done with integers. Take care to multiply first and then divide, so that you don't truncate results. For example,

cos_lookup(angle) / TRIG_MAX_RATIO

将始终产生零，除非角度等于0°或180°.

will always yield zero, except when the angle corresponds to 0° or 180°.

您可能还想看看上面链接的页面上的时针示例.

You also might want to have a look at the clock hand example on the page linked above.

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