在螺旋线上画等距点

我需要一种算法来计算螺旋路径上的点的分布。

该算法的输入参数应为：

• 环路宽度（距最内环路的距离）
• 点之间的距离固定
• 绘制点数

要绘制的螺旋是阿基米德螺线并且获得的积分必须是等距离来自彼此。

该算法应该打印出单点的笛卡尔坐标序列，例如：

点 1：(0.0) 第 2 点：（...，...） ………… N 点 (..., ...)

编程语言并不重要，非常感谢所有帮助！

EDIT:

我已经从这个网站获取并修改了这个示例：

    //
//
// 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，点之间的距离不等。

正确的分布示例是左图：

对于第一个近似值 - 这对于绘制足够近的块来说可能足够好 - 螺旋是一个圆，并按比例增加角度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;
}

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:

上面的第二个版本使用基于使用 theta 和 theta+delta 的平均半径求解 delta 的步骤：

// 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;

为了在松散螺旋上获得更好的结果，请使用数值迭代解决方案来查找 delta 值，其中计算的距离在合适的公差范围内。