翼型轮廓几何图 [英] airfoil profile geometry plotting

问题描述

我在计算坐标方面有一点问题。给出两个列表中的机翼轮廓，并具有以下示例坐标：

I have a little issue with calculating coordinates. Given are airfoil profiles in two lists with the following exemplary coordinates:

示例：

x_Coordinates = [1,   0.9,   0.7,   0.5,   0.3,   0.1, 0.0, ...] 
y_Coordinates = [0, -0.02, -0.06, -0.08, -0.10, -0.05, 0.0, ...]

图1：

关于配置文件的唯一已知信息是上面和下面的列表事实：

The only known things about the profile are the lists above and the following facts:

• 第一个坐标始终是后沿，在上面的示例中为（x = 1，y = 0）


• 坐标始终在底部/底端到前缘，在上面的示例中为（0,0），然后从那里回到后缘


• 配置文件未标准化，并且可以旋转形式存在

现在我要确定

1. 前沿和


2. 外倾线。

直到现在，我一直都使用最小的x-coodinate作为前沿。但是，这在示例性轮廓之后的
中不起作用，因为最小的x坐标位于轮廓的上表面。

Until now, I have always used the smallest x-coodinate as the leading edge. However, this would not work in the following exemplary profile, since the smallest x-coordinate is located on the upper surface of the profile.

图2：

diagram 2:

有任何人都有一个主意，我如何轻松地计算/确定这些数据？

Does anybody have an idea, how I could easily calculate/determine this data?

编辑

一个完整的样本数组数据

one full sample array data

(1.0, 0.95, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.25, 0.2, 0.15, 0.1, 0.075, 0.05, 0.025, 0.0125, 0.005, 0.0, 0.005, 0.0125, 0.025, 0.05, 0.075, 0.1, 0.15, 0.2, 0.25, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.95, 1.0)
(0.00095, 0.00605, 0.01086, 0.01967, 0.02748, 0.03423, 0.03971, 0.04352, 0.04501, 0.04456, 0.04303, 0.04009, 0.03512, 0.0315, 0.02666, 0.01961, 0.0142, 0.0089, 0.0, -0.0089, -0.0142, -0.01961, -0.02666, -0.0315, -0.03512, -0.04009, -0.04303, -0.04456, -0.04501, -0.04352, -0.03971, -0.03423, -0.02748, -0.01967, -0.01086, -0.00605, -0.00095)

推荐答案

好几年了，我用翅膀做一些事情。

Well it was quite a few years I do something with wings.

我没有任何倾斜的翅膀数据，因为在您的图像上我发现的最接近的东西是：

I do not have any skewed wings data as on your image the closest thing I found was this:

1. 前缘对不平凡的翅膀不正确

1. leading edge not correct for nontrivial wings

仅找到 dx 符号的点c $ c>正在翻转并计算

just find point where the sign of dx is flipping and compute

dx(i)=x(i)-x(i-1)

然后标记 dx 为正或为负的区域找到它们之间的中间位置（通常是该区域的 dx == 0 ）。将边缘点标记为 ix1

then mark zones where dx is positive or negative and find the middle between them (usually dx==0 for that zone). Mark the edge point as ix1

弧线

要获得精确的几何图形，您需要从两侧投射法线的交点，以便：

for precise geometry you will need intersections of normals casted from each side so:

• 开始轮廓点 i


• 在机翼内投射法线


• 在对侧搜索。


• 找到点，它是法线与相反的法线相交并将两个法线划分为相同的距离

• start on outline point i
• cast normal inside wing
• search opposite side.
• find point, It's normal intersect the opposite normal and divide both normals to the same distance

这是可行的但是复杂性很差

This is doable but with insane complexity

近似外倾线

不太精确的方法但要快得多：

less precise way but much much faster so:

1. 从轮廓点 i
开始

2. 在另一侧找到最接近的点


3. 计算它们之间的中点并将其存储为不正确的 axis0 点。对所有 i =（0-ix1）（红线）


4. 的点都执行相同的操作，但从对面存储开始 axis1 （深红色）


5. 完成后，只需找到 axis0，axis1



1. start on outline point i
2. find closest point to it on the opposite side
3. compute midpoint between them and store it as inaccurate axis0 points. Do this for all points i=(0-ix1) (Red line)
4. do the same but start from opposite side store as axis1 (dark red)
5. when done then just find the average between axis0,axis1

这可以用相同的方式完成，结果是蓝轴折线

This can be done in the same way result is blue axis polyline

C ++来源：

    List<double> pnt;   // outline 2D pnts = {x0,y0,x1,y1,x2,y2,...}
    List<double> axis;  // axis line 2D pnts = {x0,y0,x1,y1,x2,y2,...}
    int ix0,ix1;        // edge points

void compute()
    {
    int i,i0,i1;
    double d,dd;
    double *p0,*p1,*p2;
    double x0,x1,y0,y1;
    List<double> axis0,axis1;

    // find leading edge point
    ix0=0; ix1=0;
    for (p0=pnt.dat,p1=p0+2,p2=p1+2,i=2;i<pnt.num;i+=2,p0=p1,p1=p2,p2+=2)
     if ((p1[0]-p0[0])*(p2[0]-p1[0])<=0.0) { ix1=i; break; }
    // find axis0: midpoint of i0=(0-ix1) i1=find closest from (ix1,pnt.num)
    for (i0=2,i1=pnt.num-2;i0<ix1-2;i0+=2)
        {
        x0=pnt[i0+0];
        y0=pnt[i0+1];
        for (d=-1.0,i=i1;i>ix1+2;i-=2)
            {
            x1=pnt[i+0];
            y1=pnt[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||(dd<=d)) { i1=i; d=dd; }
            }
        if (d>=0.0)
            {
            x1=pnt[i1+0];
            y1=pnt[i1+1];
            axis0.add(0.5*(x0+x1));
            axis0.add(0.5*(y0+y1));
            }
        }
    // find axis1: midpoint of i0=(ix1,pnt.num) i1=find closest from (0,ix1)
    for (i1=2,i0=pnt.num-2;i0>ix1+2;i0-=2)
        {
        x0=pnt[i0+0];
        y0=pnt[i0+1];
        for (d=-1.0,i=i1;i<ix1-2;i+=2)
            {
            x1=pnt[i+0];
            y1=pnt[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||(dd<=d)) { i1=i; d=dd; }
            }
        if (d>=0.0)
            {
            x1=pnt[i1+0];
            y1=pnt[i1+1];
            axis1.add(0.5*(x0+x1));
            axis1.add(0.5*(y0+y1));
            }
        }
    // find axis: midpoint of i0=<0-axis0.num) i1=find closest from <0-axis1.num)
    axis.add(pnt[ix0+0]);
    axis.add(pnt[ix0+1]);
    for (i0=0,i1=0;i0<axis0.num;i0+=2)
        {
        x0=axis0[i0+0];
        y0=axis0[i0+1];
        for (d=-1.0,i=i1;i<axis1.num;i+=2)
            {
            x1=axis1[i+0];
            y1=axis1[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||(dd<=d)) { i1=i; d=dd; }
            }
        if (d>=0.0)
            {
            x1=axis1[i1+0];
            y1=axis1[i1+1];
            axis.add(0.5*(x0+x1));
            axis.add(0.5*(y0+y1));
            }
        }
    axis.add(pnt[ix1+0]);
    axis.add(pnt[ix1+1]);
    }

• List< double> ; xxx; 只是我的动态列表模板，与 double xxx []; c


• xxx.add（5）;在列表末尾添加5


• xxx [7] 访问数组元素


• xxx.num 是数组的实际使用大小


• xxx.reset（）清除数组并设置xxx.num = 0


• List<double> xxx; is just mine dynamic list template the same as double xxx[];
• xxx.add(5); adds 5 to end of the list
• xxx[7] access array element
• xxx.num is the actual used size of the array
• xxx.reset() clears the array and set xxx.num=0

[edit1]正确的前沿点

对此进行了疯狂的思考，以找到运行中的边缘点，并对代码进行了一些调整，结果对我来说已经足够好了：），所以首先来解释一下：

Have an insane thought about this to find the edge point on the run plus some code tweaking and the outcome is good enough for me :) so first some explaining:

轴的算法保持不变，但不使用 ix1 绑定，仅使用尚未使用的点...也仅计算有效的最近点（在相反）（如果没有找到）（顶部图像盒）。从这一点开始，找到离最后一个轴点最远的点，这就是最前沿。

algorithm for axis stays the same but instead of ix1 bound use only points that was not yet used ... Also count only valid closest points (on the opposite side) if none found stop (top image case). From this point find the most far point from last axis point this is the leading edge point.

这种方法的输出结果非常准确（ axis0， axis1 靠在一起）

This approach has much much accurate output (axis0,axis1 are closer together)

现在使用C ++代码：

void compute()
    {
    int i,i0,i1,ii,n=4;
    double d,dd;
    double x0,x1,y0,y1;
    List<double> axis0,axis1;
    ix0=0; ix1=0;

    // find axis0: midpoint of i0=(0-ix1) i1=find closest from (ix1,pnt.num)
    for (i0=0,i1=pnt.num-2;i0+n<i1;i0+=2)
        {
        x0=pnt[i0+0];
        y0=pnt[i0+1];
        i=i1+n; if (i>pnt.num-2) i=pnt.num-2; ii=i1;
        for (d=-1.0;i>i0+n;i-=2)
            {
            x1=pnt[i+0];
            y1=pnt[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||((dd<=d)&&(dd>1e-10))) { i1=i; d=dd; }
            if ((d>=0.0)&&(dd>d)) break;
            }
        if (d>=0.0)
            {
            if (i1-i0<=n+2) { i1=ii; break; } // stop if non valid closest point found
            x1=pnt[i1+0];
            y1=pnt[i1+1];
            axis0.add(0.5*(x0+x1));
            axis0.add(0.5*(y0+y1));
            }
        }
    // find leading edge point (the farest point from last found axis point)
    x0=axis0[axis0.num-2];
    y0=axis0[axis0.num-1];
    for (d=0.0,i=i0;i<=i1;i+=2)
        {
        x1=pnt[i+0];
        y1=pnt[i+1];
        dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
        if (dd>d) { ix1=i; d=dd; }
        }
    axis0.add(pnt[ix1+0]);
    axis0.add(pnt[ix1+1]);

    // find axis1: midpoint of i0=(ix1,pnt.num) i1=find closest from (0,ix1)
    for (i1=0,i0=pnt.num-2;i0+n>i1;i0-=2)
        {
        x0=pnt[i0+0];
        y0=pnt[i0+1];
        i=i1-n; if (i<0) i=0; ii=i1;
        for (d=-1.0;i<i0-n;i+=2)
            {
            x1=pnt[i+0];
            y1=pnt[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||((dd<=d)&&(dd>1e-10))) { i1=i; d=dd; }
            if ((d>=0.0)&&(dd>d)) break;
            }
        if (d>=0.0)
            {
            if (i0-i1<=n+2) { i1=ii; break; } // stop if non valid closest point found
            x1=pnt[i1+0];
            y1=pnt[i1+1];
            axis1.add(0.5*(x0+x1));
            axis1.add(0.5*(y0+y1));
            }
        }
    // find leading edge point (the farest point from last found axis point)
    x0=axis1[axis1.num-2];
    y0=axis1[axis1.num-1];
    for (d=0.0,i=i1;i<=i0;i+=2)
        {
        x1=pnt[i+0];
        y1=pnt[i+1];
        dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
        if (dd>d) { ix1=i; d=dd; }
        }
    axis1.add(pnt[ix1+0]);
    axis1.add(pnt[ix1+1]);

    // find axis: midpoint of i0=<0-axis0.num) i1=find closest from <0-axis1.num)
    for (i0=0,i1=0;i0<axis0.num;i0+=2)
        {
        x0=axis0[i0+0];
        y0=axis0[i0+1];
        for (d=-1.0,i=i1;i<axis1.num;i+=2)
            {
            x1=axis1[i+0];
            y1=axis1[i+1];
            dd=((x1-x0)*(x1-x0))+((y1-y0)*(y1-y0));
            if ((d<0.0)||(dd<=d)) { i1=i; d=dd; }
            }
        if (d>=0.0)
            {
            x1=axis1[i1+0];
            y1=axis1[i1+1];
            axis.add(0.5*(x0+x1));
            axis.add(0.5*(y0+y1));
            }
        }
    }

恒定 n = 4 只是为了安全地重叠搜索最接近的点，它应该是 pnt.num 的一部分。有时，最接近的点在最后找到的最接近的点之前，这取决于喷漆室侧面的曲率。 n 太大会导致速度变慢，如果 n> pnt.num / 4 也会使输出无效。

constant n=4 is just for safety overlapped search for closest points it should be a fraction of pnt.num. Sometimes the closest point is before the last found closest point this depends on the curvature of booth sides. Too big n will cause slowdowns and if n>pnt.num/4 it could also invalidate output.

如果太小，则对于较小的曲率半径会降低精度，这种方法取决于足够的点覆盖率。如果机翼采样点计数太低，可能会导致不准确。源代码几乎是您选择要记住的 ix1 （通过第一次搜索或第二次搜索）它们是相邻点的3倍

If too small then for smaller radius of curvature will lower the accuracy this approach is dependent on sufficient point coverage. If the wing is sampled with too low point count it can lead to inaccuracy. The source code is 3 times almost the same thing you can chose which ix1 to remember (from first or second search) they are neighboring points

测试配置文件：

1.000000 0.000000
0.990000 0.006719
0.980000 0.013307
0.970000 0.019757
0.960000 0.026064
0.950000 0.032223
0.940000 0.038228
0.930000 0.044075
0.920000 0.049759
0.910000 0.055276
0.900000 0.060623
0.890000 0.065795
0.880000 0.070790
0.870000 0.075604
0.860000 0.080234
0.850000 0.084678
0.840000 0.088935
0.830000 0.093001
0.820000 0.096876
0.810000 0.100558
0.800000 0.104046
0.790000 0.107339
0.780000 0.110438
0.770000 0.113342
0.760000 0.116051
0.750000 0.118566
0.740000 0.120887
0.730000 0.123016
0.720000 0.124954
0.710000 0.126702
0.700000 0.128262
0.690000 0.129637
0.680000 0.130829
0.670000 0.131839
0.660000 0.132672
0.650000 0.133331
0.640000 0.133818
0.630000 0.134137
0.620000 0.134292
0.610000 0.134287
0.600000 0.134127
0.590000 0.133815
0.580000 0.133356
0.570000 0.132755
0.560000 0.132016
0.550000 0.131146
0.540000 0.130148
0.530000 0.129030
0.520000 0.127795
0.510000 0.126450
0.500000 0.125000
0.490000 0.123452
0.480000 0.121811
0.470000 0.120083
0.460000 0.118275
0.450000 0.116392
0.440000 0.114441
0.430000 0.112429
0.420000 0.110361
0.410000 0.108244
0.400000 0.106085
0.390000 0.103889
0.380000 0.101663
0.370000 0.099414
0.360000 0.097148
0.350000 0.094870
0.340000 0.092589
0.330000 0.090309
0.320000 0.088037
0.310000 0.085779
0.300000 0.083541
0.290000 0.081329
0.280000 0.079149
0.270000 0.077006
0.260000 0.074906
0.250000 0.072855
0.240000 0.070858
0.230000 0.068920
0.220000 0.067047
0.210000 0.065242
0.113262 0.047023
0.110002 0.042718
0.106385 0.038580
0.102428 0.034615
0.098146 0.030832
0.093556 0.027239
0.088673 0.023844
0.083516 0.020652
0.078101 0.017670
0.072448 0.014904
0.066574 0.012361
0.060499 0.010044
0.054241 0.007958
0.047820 0.006108
0.041256 0.004497
0.034569 0.003129
0.027779 0.002005
0.020907 0.001129
0.013972 0.000502
0.006997 0.000126
0.000000 0.000000
0.000000 0.000000
-0.003997 0.000126
-0.007972 0.000502
-0.011907 0.001129
-0.015779 0.002005
-0.019569 0.003129
-0.023256 0.004497
-0.026820 0.006108
-0.030241 0.007958
-0.033499 0.010044
-0.036574 0.012361
-0.039448 0.014904
-0.042101 0.017670
-0.044516 0.020652
-0.046673 0.023844
-0.048556 0.027239
-0.050146 0.030832
-0.051428 0.034615
-0.052385 0.038580
-0.053002 0.042718
-0.053262 0.047023
-0.053153 0.051484
-0.052659 0.056093
-0.051768 0.060841
-0.050467 0.065717
-0.048744 0.070711
-0.046588 0.075813
-0.043988 0.081012
-0.040935 0.086297
-0.037420 0.091658
-0.033435 0.097082
-0.028972 0.102558
-0.024025 0.108074
-0.018589 0.113618
-0.012657 0.119178
-0.006228 0.124741
0.000704 0.130295
0.008139 0.135828
0.016079 0.141326
0.024525 0.146777
0.033475 0.152169
0.042930 0.157488
0.052885 0.162722
0.063339 0.167858
0.074287 0.172883
0.085723 0.177784
0.097643 0.182549
0.110038 0.187166
0.122902 0.191621
0.136226 0.195903
0.150000 0.200000
0.164214 0.203899
0.178856 0.207590
0.193914 0.211059
0.209376 0.214297
0.225227 0.217291
0.241453 0.220032
0.258039 0.222509
0.274968 0.224711
0.292223 0.226629
0.309787 0.228254
0.327641 0.229575
0.345766 0.230585
0.364142 0.231274
0.382749 0.231636
0.401566 0.231662
0.420570 0.231345
0.439740 0.230679
0.459054 0.229657
0.478486 0.228274
0.498015 0.226525
0.517615 0.224404
0.537262 0.221908
0.556930 0.219032
0.576595 0.215775
0.596231 0.212132
0.615811 0.208102
0.635310 0.203684
0.654700 0.198876
0.673956 0.193679
0.693050 0.188091
0.711955 0.182115
0.730644 0.175751
0.749091 0.169002
0.767268 0.161869
0.785149 0.154357
0.802706 0.146468
0.819913 0.138207
0.836742 0.129580
0.853169 0.120591
0.869166 0.111246
0.884707 0.101553
0.899768 0.091518
0.914322 0.081149
0.928345 0.070455
0.941813 0.059445
0.954701 0.048128
0.966987 0.036514
0.978646 0.024614
0.989658 0.012439
1.000000 0.000000

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