借鉴螺旋等距点 [英] Draw equidistant points on a spiral

问题描述

我需要一种算法来计算上的螺旋路径点的分布

I need an algorithm to calculate the distribution of points on a spiral path.

该算法的输入参数应该是：

The input parameters of this algorithm should be:

• 宽度循环（从最里面的循环距离）
• 固定点之间的距离
• 在点数画

画螺旋是阿基米德螺线，并获得一定的点数等距从对方。

The spiral to draw is an archimedean spiral and the points obtained must be equidistant from each other.

该算法应该打印出单一的点的直角坐标的序列，例如：

The algorithm should print out the sequence of the Cartesian coordinates of single points, for example:

点1：（0.0） 要点2：（...，...） ........ N点（...，...）

Point 1: (0.0) Point 2: (..., ...) ........ Point N (..., ...)

编程语言并不重要，都有助于大大AP preciated！

The programming language isn't important and all help greatly appreciated!

编辑：

我已经获得和修改这个例子从该网站：

I already get and modify this example from this site:

    //
//
// centerX-- X origin of the spiral.
// centerY-- Y origin of the spiral.
// radius--- Distance from origin to outer arm.
// sides---- Number of points or sides along the spiral's arm.
// coils---- Number of coils or full rotations. (Positive numbers spin clockwise, negative numbers spin counter-clockwise)
// rotation- Overall rotation of the spiral. ('0'=no rotation, '1'=360 degrees, '180/360'=180 degrees)
//
void SetBlockDisposition(float centerX, float centerY, float radius, float sides, float coils, float rotation)
{
    //
    // How far to step away from center for each side.
    var awayStep = radius/sides;
    //
    // How far to rotate around center for each side.
    var aroundStep = coils/sides;// 0 to 1 based.
    //
    // Convert aroundStep to radians.
    var aroundRadians = aroundStep * 2 * Mathf.PI;
    //
    // Convert rotation to radians.
    rotation *= 2 * Mathf.PI;
    //
    // For every side, step around and away from center.
    for(var i=1; i<=sides; i++){

        //
        // How far away from center
        var away = i * awayStep;
        //
        // How far around the center.
        var around = i * aroundRadians + rotation;
        //
        // Convert 'around' and 'away' to X and Y.
        var x = centerX + Mathf.Cos(around) * away;
        var y = centerY + Mathf.Sin(around) * away;
        //
        // Now that you know it, do it.

        DoSome(x,y);
    }
}

但点的配置是错误的，点不相互等距。

But the disposition of point is wrong, the points aren't equidistant from each other.

正确的分布的例子是在左边的图像：

The correct distribution example is is the image on the left:

推荐答案

要第一个近似值 - 这可能是够绘制足够接近块好 - 螺旋是一个圆，由比率和弦/半径。

To a first approximation - which is probably good enough for plotting blocks close enough - the spiral is a circle and increment the angle by the ratio chord / radius.

// value of theta corresponding to end of last coil
final double thetaMax = coils * 2 * Math.PI;

// How far to step away from center for each side.
final double awayStep = radius / thetaMax;

// distance between points to plot
final double chord = 10;

DoSome ( centerX, centerY );

// For every side, step around and away from center.
// start at the angle corresponding to a distance of chord
// away from centre.
for ( double theta = chord / awayStep; theta <= thetaMax; ) {
    //
    // How far away from center
    double away = awayStep * theta;
    //
    // How far around the center.
    double around = theta + rotation;
    //
    // Convert 'around' and 'away' to X and Y.
    double x = centerX + Math.cos ( around ) * away;
    double y = centerY + Math.sin ( around ) * away;
    //
    // Now that you know it, do it.
    DoSome ( x, y );

    // to a first approximation, the points are on a circle
    // so the angle between them is chord/radius
    theta += chord / away;
}

然而，对于一个宽松的螺旋，你将有更准确地解决路径距离的空间过大，其中以<$比较的的差别远的连续点是显著C $ C>和弦：

However, for a looser spiral you will have to solve the path distance more accurately as spaces too wide where the difference between away for successive points is significant compared with chord:

的第二个版本以上使用基于基于使用平均半径为THETA和θ+增量求解增量的步骤：

The second version above uses a step based on solving for delta based on using the average radius for theta and theta+delta:

// take theta2 = theta + delta and use average value of away
// away2 = away + awayStep * delta 
// delta = 2 * chord / ( away + away2 )
// delta = 2 * chord / ( 2*away + awayStep * delta )
// ( 2*away + awayStep * delta ) * delta = 2 * chord 
// awayStep * delta ** 2 + 2*away * delta - 2 * chord = 0
// plug into quadratic formula
// a= awayStep; b = 2*away; c = -2*chord

double delta = ( -2 * away + Math.sqrt ( 4 * away * away + 8 * awayStep * chord ) ) / ( 2 * awayStep );

theta += delta;

有关松散螺旋更好的结果，使用数值迭代求解寻找增量的值，其中计算的距离在一个合适的容差。

For even better results on a loose spiral, use a numeric iterative solution to find the value of delta where the calculated distance is within a suitable tolerance.

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