CN102927921A - Stumpage breast height diameter measuring method based on optics similar triangle method - Google Patents

Abstract

Translated fromChinese

本发明提供了一种基于光学相似三角形法的立木胸径测量方法，所述方法涉及林木测量，所述方法包括：利用一台相机及激光器构建立木胸径测量系统；利用上述相机对被测立木进行拍照以获取立木图像；对拍摄的立木图像进行预处理后，得到所述激光光斑落在相平面的点(X’，Y’)；根据光学相似三角形原理，将上述激光光斑落在相平面的点(X’，Y’)向三维空间进行坐标转换得到点(X，Y，Z)；对上述所得点(X，Y，Z)进行处理后，向水平方向进行投影得点(X_投，Y_投)；采用最小二乘圆拟合方法对点(X_投，Y_投)进行圆拟合，拟合后所得圆半径参数即为被测立木的胸径大小。本发明所述基于光学相似三角形法的立木胸径测量方法具有算法实现简单，无接触快速测量立木胸径的优点。

The invention provides a method for measuring the diameter of a standing tree based on the optical similar triangle method. The method relates to the measurement of forest trees. The method includes: using a camera and a laser to construct a measuring system for the diameter of a standing tree; using the camera to take pictures of the measured standing tree To obtain the standing tree image; after preprocessing the taken standing tree image, the point (X', Y') where the laser spot falls on the phase plane is obtained; according to the principle of optical similar triangles, the above laser spot falls on the point of the phase plane (X', Y') is converted to the three-dimensional space to obtain the point (X, Y, Z); after the above-mentioned obtained point (X, Y, Z) is processed, it is projected to the horizontal direction to obtain the point (X_projection , Y_projection ); use the least squares circle fitting method to carry out circle fitting on the points (X_projection , Y_projection ), and the circle radius parameter obtained after fitting is the diameter at breast height of the measured standing tree. The method for measuring the diameter at breast height of a standing tree based on the optical similar triangle method of the present invention has the advantages of simple algorithm implementation and non-contact rapid measurement of the diameter at breast height of a standing tree.

Description

Translated fromChinese

基于光学相似三角形法的立木胸径测量方法Measuring method of standing tree diameter at breast height based on optical similar triangle method

技术领域technical field

本发明涉及林木测量，尤其涉及一种基于光学相似三角形法的立木胸径测量方法。The invention relates to forest tree measurement, in particular to a method for measuring the diameter at breast height of a standing tree based on an optical similar triangle method.

背景技术Background technique

森林蓄积量是国家自然资源的重要组成部分，其消长动态是林业经济效益的主要标志，也是制定采伐培育计划的依据。传统的森林林木胸径测量方法如采用胸径测量围尺或轮尺，将长度转换为直径刻度，从而进行林木胸径估测，操作中受环境及人为因素影响大，结果误差极大。为了解决上述问题，发明人提供了一种基于光学相似三角形法的立木胸径测量方法。Forest stock volume is an important part of the country's natural resources, and its growth and decline dynamics are the main symbol of forestry economic benefits and the basis for formulating felling and cultivation plans. The traditional method of measuring forest tree diameter at breast height, such as using a diameter measuring girth or a caliper, converts the length into a diameter scale, so as to estimate the forest tree diameter at breast height. The operation is greatly affected by environmental and human factors, and the result error is huge. In order to solve the above problems, the inventor provides a method for measuring the diameter at breast height of standing trees based on the optical similar triangle method.

发明内容Contents of the invention

鉴于上述现有技术的不足以及实际应用的需要，本发明提供一种基于光学相似三角形法的立木胸径测量方法，实现简单快速、无接触的立木胸径测量。In view of the above-mentioned deficiencies in the prior art and the needs of practical applications, the present invention provides a method for measuring the diameter at breast height of standing trees based on the optical similar triangle method, so as to realize simple, fast and non-contact measurement of the diameter at breast height of standing trees.

本发明是通过以下具体技术方案实现的：The present invention is achieved through the following specific technical solutions:

本发明提供一种基于光学相似三角形法的立木胸径测量方法，所述方法包括以下步骤：The invention provides a method for measuring the diameter of standing trees based on the optical similar triangle method, the method comprising the following steps:

步骤一、对所拍摄的立木图像进行预处理，提取单像素的激光光斑骨架部分后，得到所述激光光斑在相平面的点(X’，Y’)；Step 1. Perform preprocessing on the photographed standing tree image, extract the laser spot skeleton part of a single pixel, and obtain the point (X', Y') of the laser spot on the phase plane;

步骤二、根据光学相似三角形原理，将上述激光光斑落在相平面的点(X’，Y’)向三维空间进行坐标转换得到点(X，Y，Z)；Step 2. According to the principle of optical similar triangles, the point (X', Y') where the above-mentioned laser spot falls on the phase plane is converted to the three-dimensional space to obtain the point (X, Y, Z);

步骤三、对上述所得三维空间点(X，Y，Z)进行处理后，向水平方向进行投影得点(X_投，Y_投)；采用最小二乘圆拟合方法对点(X_投，Y_投)进行圆拟合；Step 3, after processing the above-mentioned three-dimensional space points (X, Y, Z), project to the horizontal direction to get points (X_projection , Y_projection ); use least squares circle fitting method to point (X_projection , Y_projection ) ) for circle fitting;

步骤四、上述拟合后所得圆半径参数即为被测立木的胸径大小；Step 4, the circle radius parameter obtained after the above-mentioned fitting is the diameter at breast height of the measured standing tree;

步骤二中所述根据光学三角原理，将激光光斑落在相平面的坐标点(X’，Y’)向三维空间进行坐标转换得到点(X，Y，Z)包括：According to the principle of optical triangulation described in step 2, the coordinate point (X', Y') where the laser spot falls on the phase plane is converted to the three-dimensional space to obtain the point (X, Y, Z) including:

由以下表达式：by the following expression:

实现立木图像相平面的坐标点(X’，Y’)向三维空间的坐标点(X，Y，Z)的转换；其中，f为实验设备中的焦距，b为激光源到摄像机光轴的距离，θ为激光源射出的光平面与COMS平面所成夹角。Realize the conversion of the coordinate point (X', Y') of the phase plane of the standing tree image to the coordinate point (X, Y, Z) of the three-dimensional space; where, f is the focal length in the experimental equipment, b is the distance from the laser source to the optical axis of the camera Distance, θ is the angle formed between the light plane emitted by the laser source and the COMS plane.

步骤三中所述采用最小二乘圆拟合方法对点(X_投，Y_投)进行拟合包括(如下原理推导式中采用X，Y来表达X_投，Y_投)：In step 3, using the least squares circle fitting method to fit the points (X_projection , Y_projection ) includes (the following principle derivation uses X, Y to express X_projection , Y_projection ):

令：make:

$\mathrm{<span><span>C</span>C</span>}<span><span>=</span>=</span><span><span>(</span>(</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>)</span>)</span>$

$\mathrm{<span><span>D</span>D.</span>}<span><span>=</span>=</span><span><span>(</span>(</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>)</span>)</span>$

$\mathrm{<span><span>E</span>E.</span>}<span><span>=</span>=</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>3</span>3</span>}}<span><span>+</span>+</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}$

$\mathrm{<span><span>G</span>G</span>}<span><span>=</span>=</span><span><span>(</span>(</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>)</span>)</span>$

$\mathrm{<span><span>H</span>h</span>}<span><span>=</span>=</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>3</span>3</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}$

由Depend on

$\mathrm{<span><span>a</span>a</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>HD</span>HD</span>}<span><span>-</span>-</span>\mathrm{<span><span>EG</span>EG</span>}}{\mathrm{<span><span>CG</span>CG</span>}<span><span>-</span>-</span>{\mathrm{<span><span>D</span>D.</span>}}^{\mathrm{<span><span>2</span>2</span>}}}$

$\mathrm{<span><span>b</span>b</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>HC</span>HC</span>}<span><span>-</span>-</span>\mathrm{<span><span>ED</span>ED</span>}}{{\mathrm{<span><span>D</span>D.</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>GC</span>GC</span>}}$

$\mathrm{<span><span>c</span>c</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span><span><span>+</span>+</span>\mathrm{<span><span>a\&Sigma;</span>a\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>b\&Sigma;</span>b\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}}{\mathrm{<span><span>N</span>N</span>}}$

得：have to:

$\mathrm{<span><span>A</span>A</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>a</span>a</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>B</span>B</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>b</span>b</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>R</span>R</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>1</span>1</span>}}{\mathrm{<span><span>2</span>2</span>}}\sqrt{{\mathrm{<span><span>a</span>a</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>b</span>b</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>4</span>4</span>}\mathrm{<span><span>c</span>c</span>}}$

实现最小二乘圆拟合；其中a，b，c为圆一般方程的系数，C，D，E，G，H为中间计算参量，N为所有特征点总数。Realize the least squares circle fitting; where a, b, c are the coefficients of the general equation of the circle, C, D, E, G, H are the intermediate calculation parameters, and N is the total number of all feature points.

本发明的特点及技术效果：Features and technical effects of the present invention:

由上述一种基于光学相似三角形法的立木胸径测量方法可以看出，本发明通过实现相平面坐标点(X’，Y’)向三维空间坐标点(X，Y，Z)的转换，以及采用最小二乘圆拟合方法对点(X_投，Y_投)进行拟合，从而简单快速、无接触的测得立木实际胸径大小；尤其是在对林木实现大量测量的工作中，与传统的测量技术相比，能够在很大程度上减少时间及人力消耗。As can be seen from the above-mentioned standing tree diameter measurement method based on the optical similar triangle method, the present invention realizes the conversion of the phase plane coordinate point (X', Y') to the three-dimensional space coordinate point (X, Y, Z), and adopts The least squares circle fitting method fits the points (X_projection , Y_projection ), so that the actual diameter of standing trees can be measured simply, quickly and without contact; Compared with technology, it can greatly reduce the time and labor consumption.

附图说明Description of drawings

图1为本发明立木胸径测量系统图；Fig. 1 is a standing tree diameter measurement system diagram of the present invention;

图2为本发明具体实施方法的流程图；Fig. 2 is the flow chart of specific implementation method of the present invention;

图3为本发明具体实施方法中的光学相似三角形法原理图；Fig. 3 is the schematic diagram of the optical similar triangle method in the specific implementation method of the present invention;

图4为本发明具体实施方法中的圆拟合原理图；Fig. 4 is a schematic diagram of circle fitting in the specific implementation method of the present invention;

具体实施方式Detailed ways

本发明具体实施方法提供一种基于光学三角形法的立木胸径测量方法，所述方法如图1所示，包括以下步骤：The specific implementation method of the present invention provides a kind of standing tree diameter measuring method based on optical triangulation method, described method as shown in Figure 1, comprises the following steps:

步骤一、对所拍摄的立木图像进行预处理，提取单像素的激光光斑骨架部分后，得到所述激光光斑在相平面的点(X’，Y’)；Step 1. Perform preprocessing on the photographed standing tree image, extract the laser spot skeleton part of a single pixel, and obtain the point (X', Y') of the laser spot on the phase plane;

该步骤中所述立木图像预处理方法包括对图像进行G(绿色)分量及B(蓝色)分量弱化后，对立木图像进行二值化处理；该步骤中所述提取单像素的激光光斑骨架方法包括对图像采用数学形态学的基本算法——膨胀和腐蚀来实现激光光斑的细化，然后通过寻找细化后图像的最大联通区域来去除激光光斑骨架以外的噪声点，从而得到单像素的激光光斑骨架部分。The standing tree image preprocessing method described in this step comprises after carrying out G (green) component and B (blue) component weakening to image, carries out binarization processing to standing tree image; Said in this step extracts the laser spot skeleton of single pixel The method includes using the basic algorithm of mathematical morphology—expansion and erosion to realize the thinning of the laser spot, and then removing the noise points outside the skeleton of the laser spot by finding the largest connected area of the thinned image, so as to obtain a single-pixel image. The skeleton part of the laser spot.

步骤二、根据光学相似三角形原理，将上述激光光斑落在相平面的点(X’，Y’)向三维空间进行坐标转换得到点(X，Y，Z)；Step 2. According to the principle of optical similar triangles, the point (X', Y') where the above-mentioned laser spot falls on the phase plane is converted to the three-dimensional space to obtain the point (X, Y, Z);

该步骤中所述将激光光斑落在相平面的点(X’，Y’)向三维空间转换得到点(X，Y，Z)是通过图3中所示光学三角形原理进行推导计算，得到式1，从而实现相平面的点坐标(X’，Y’)向三维空间转换得到点(X，Y，Z)。In this step, the point (X', Y') where the laser spot falls on the phase plane is converted to the three-dimensional space to obtain the point (X, Y, Z) is derived and calculated by the principle of optical triangle shown in Figure 3, and the formula 1. In this way, the point coordinates (X', Y') of the phase plane can be converted to the three-dimensional space to obtain the point (X, Y, Z).

式1I

Formula 1I

其中，f为实验设备中的焦距，b为激光源到摄像机光轴的距离，θ为激光源射出的光平面与COMS平面所成夹角，(X’，Y’)为相平面的点M’坐标，(X，Y，Z)为三维空间点M坐标；Among them, f is the focal length in the experimental equipment, b is the distance from the laser source to the optical axis of the camera, θ is the angle between the light plane emitted by the laser source and the COMS plane, (X', Y') is the point M of the phase plane 'coordinates, (X, Y, Z) is the M coordinate of the three-dimensional space point;

步骤三、对上述所得三维空间点(X，Y，Z)进行处理后，向水平方向进行投影得点(X_投，Y_投)；采用最小二乘圆拟合方法对点(X_投，Y_投)进行圆拟合；Step 3, after processing the above-mentioned three-dimensional space points (X, Y, Z), project to the horizontal direction to get points (X_projection , Y_projection ); use least squares circle fitting method to point (X_projection , Y_projection ) ) for circle fitting;

该步骤中所述三维空间的点(X，Y，Z)向水平方向进行投影得点(X_投，Y_投)包括将三维空间点所在坐标平面向水平面投影，所述水平面与三维空间点所在坐标平面之间的夹角为90°-θ；所述投影所得点(X_投，Y_投)与三维空间点中的(X，Y)之间存在如关系：The points (X, Y, Z) in the three-dimensional space described in this step are projected to the horizontal direction to get points (X_projection , Y_projection ) comprising projecting the coordinate plane where the three-dimensional space point is located to the horizontal plane, and the horizontal plane and the coordinates where the three-dimensional space point is located The angle between the planes is 90°-θ; there is such a relationship between the projected point (X_projection , Y_projection ) and (X, Y) in the three-dimensional space point:

X_投＝XX_vote = X

Y_投＝Y·cos(90°-θ)Y_cast = Y cos(90°-θ)

该步骤中所述采用最小二乘圆拟合方法对点(X_投，Y_投)进行圆拟合包括：最小二乘圆拟合方法是通过最小误差的平方和找到一组数据的最佳函数匹配，可通过下述方法实现最小二乘圆拟合方法(如下原理推导式中采用X，Y来表达X_投，Y_投)；Said in this step adopts the least squares circle fitting method to carry out circle fitting to the point (X_throws , Y_throws ) and comprises: the least squares circle fitting method is to find the optimal function of a group of data by the square sum of minimum error Matching, the least squares circle fitting method can be realized by the following method (in the following principle derivation formula, X and Y are used to express X_cast and Y_cast );

圆曲线方程式为：The equation of a circular curve is:

R²＝(X-A)²+(Y-B)²R² =(XA)² +(YB)²

展开式得：Expanded to get:

R²＝X²-2AX+A²+Y²-2BY+B²R² ＝X² -2AX+A² +Y² -2BY+B²

令make

a＝-2Aa=-2A

b＝-2Bb=-2B

c＝A²+B²-R²c＝A² +B² -R²

故可得圆曲线一般方程式：Therefore, the general equation of the circular curve can be obtained:

X²+Y²+aX+bY+c＝0X² +Y² +aX+bY+c＝0

只要求出参数a，b，c即可求得圆心半径的参数：Only the parameters a, b, and c are required to obtain the parameters of the radius of the center of the circle:

$\mathrm{<span><span>A</span>A</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>a</span>a</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>B</span>B</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>b</span>b</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>R</span>R</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>1</span>1</span>}}{\mathrm{<span><span>2</span>2</span>}}\sqrt{{\mathrm{<span><span>a</span>a</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>b</span>b</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>4</span>4</span>}\mathrm{<span><span>c</span>c</span>}}$

点集(X_i，Y_i)i∈(1，2，3...N)中点到圆心距离为d_i：The distance from the point to the center of the point set (X_i , Y_i )i∈(1, 2, 3...N) is d_i :

d_i＝(X_i-A)²+(Y_i-B)²d_i =(X_i -A)² +(Y_i -B)²

点(X_i，Y_i)到圆边缘的距离的平方与半径平方的差为：The difference between the square of the distance from the point (X_i , Y_i ) to the edge of the circle and the square of the radius is:

${\mathrm{<span><span>\&delta;</span>\&delta;</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>=</span>=</span>{\mathrm{<span><span>d</span>d</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>{\mathrm{<span><span>R</span>R</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>=</span>=</span>{<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>-</span>-</span>\mathrm{<span><span>A</span>A</span>}<span><span>)</span>)</span>}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{<span><span>(</span>(</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>-</span>-</span>\mathrm{<span><span>B</span>B</span>}<span><span>)</span>)</span>}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>{\mathrm{<span><span>R</span>R</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>=</span>=</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>\mathrm{<span><span>a</span>a</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>b</span>b</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>c</span>c</span>}$

令Q(α，b，c)为δ_i的平方和：Let Q(α,b,c) be the sum of the squares of_δi :

$\mathrm{<span><span>Q</span>Q</span>}<span><span>(</span>(</span>\mathrm{<span><span>a</span>a</span>}<span><span>,</span>,</span>\mathrm{<span><span>b</span>b</span>}<span><span>,</span>,</span>\mathrm{<span><span>c</span>c</span>}<span><span>)</span>)</span><span><span>=</span>=</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>\&delta;</span>\&delta;</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>=</span>=</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{<span><span>[</span>[</span><span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>\mathrm{<span><span>a</span>a</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>b</span>b</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>c</span>c</span>}<span><span>)</span>)</span><span><span>]</span>]</span>}^{\mathrm{<span><span>2</span>2</span>}}$

求解参数a，b，c使得Q(a，b，c)的值最小即可；Solve the parameters a, b, c so that the value of Q(a, b, c) is the smallest;

令make

$\mathrm{<span><span>C</span>C</span>}<span><span>=</span>=</span><span><span>(</span>(</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>)</span>)</span>$

D＝(N∑X_iY_i-∑X_i∑Y_i)D＝(N∑X_i Y_i -∑X_i ∑Y_i )

$\mathrm{<span><span>E</span>E.</span>}<span><span>=</span>=</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>3</span>3</span>}}<span><span>+</span>+</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}$

$\mathrm{<span><span>G</span>G</span>}<span><span>=</span>=</span><span><span>(</span>(</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>)</span>)</span>$

$\mathrm{<span><span>H</span>h</span>}<span><span>=</span>=</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>N\&Sigma;</span>N\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>3</span>3</span>}}<span><span>-</span>-</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span>\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}$

可以解得：can be solved:

$\mathrm{<span><span>a</span>a</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>HD</span>HD</span>}<span><span>-</span>-</span>\mathrm{<span><span>EG</span>EG</span>}}{\mathrm{<span><span>CG</span>CG</span>}<span><span>-</span>-</span>{\mathrm{<span><span>D</span>D.</span>}}^{\mathrm{<span><span>2</span>2</span>}}}$

$\mathrm{<span><span>b</span>b</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>HC</span>HC</span>}<span><span>-</span>-</span>\mathrm{<span><span>ED</span>ED</span>}}{{\mathrm{<span><span>D</span>D.</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>GC</span>GC</span>}}$

$\mathrm{<span><span>c</span>c</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>\&Sigma;</span>\&Sigma;</span>}<span><span>(</span>(</span>{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>)</span>)</span><span><span>+</span>+</span>\mathrm{<span><span>a\&Sigma;</span>a\&Sigma;</span>}{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>i</span>i</span>}}<span><span>+</span>+</span>\mathrm{<span><span>b\&Sigma;</span>b\&Sigma;</span>}{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>i</span>i</span>}}}{\mathrm{<span><span>N</span>N</span>}}$

得：have to:

$\mathrm{<span><span>A</span>A</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>a</span>a</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>B</span>B</span>}<span><span>=</span>=</span><span><span>-</span>-</span>\frac{\mathrm{<span><span>b</span>b</span>}}{\mathrm{<span><span>2</span>2</span>}}$

$\mathrm{<span><span>R</span>R</span>}<span><span>=</span>=</span>\frac{\mathrm{<span><span>1</span>1</span>}}{\mathrm{<span><span>2</span>2</span>}}\sqrt{{\mathrm{<span><span>a</span>a</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>b</span>b</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>-</span>-</span>\mathrm{<span><span>4</span>4</span>}\mathrm{<span><span>c</span>c</span>}}$

步骤四、上述拟合后所得圆半径参数即为被测立木的胸径大小；Step 4, the circle radius parameter obtained after the above-mentioned fitting is the diameter at breast height of the measured standing tree;

通过上述步骤的实现，可以完成立木胸径的快速测量；本发明所提供的实施方案可以简单快速、无接触的完成立木胸径测量；本发明所述的实施方案具体实例形式灵活，故任何熟悉本领域的技术人员，在本发明揭示的技术范围内，可以轻易想到的变换或替换，都应涵盖在本发明之内。因此，本发明的保护范围应以权利要求的保护范围为准。Through the realization of the above steps, the rapid measurement of the standing tree diameter at breast height can be completed; the embodiment provided by the present invention can be simple, fast and non-contact to complete the standing tree diameter at breast height measurement; the embodiment of the present invention is flexible in form, so anyone familiar with the art Within the technical scope disclosed in the present invention, any changes or substitutions that can be easily conceived by those skilled in the art shall be covered by the present invention. Therefore, the protection scope of the present invention should be based on the protection scope of the claims.

Claims (8)

1. the standing tree tree breast diameter survey method based on optics similar triangles method is characterized in that, described measuring method comprises the steps:

Step 1: captured standing tree image is carried out pre-service, extract the laser facula skeleton part of single pixel after, obtain described laser facula at the point of phase plane (X ', Y ');

Step 2: according to the optics similar triangle theory, the point (X ', Y ') that above-mentioned laser facula is dropped on phase plane carries out coordinate conversion to three dimensions and obtains point (X, Y, Z);

Step 3: after above-mentioned gained three dimensions point (X, Y, Z) processed, carry out projection invocation point (X to horizontal direction_Throw, Y_Throw); Adopt the least square circle approximating method to point (X_Throw, Y_Throw) justify match;

Step 4: gained radius of circle parameter is the diameter of a cross-section of a tree trunk 1.3 meters above the ground size of tested standing tree after the above-mentioned match;

2. described a kind of standing tree tree breast diameter survey method based on optics similar triangles method according to claim 1, it is characterized in that, described in the step 2 according to the optics similar triangle theory, with laser facula drop on phase plane point (X ', Y ') carries out coordinate conversion to three dimensions and obtain point (X, Y, Z) comprising:

By following formula:

Realization is by the conversion of the coordinate points of the phase plane of standing tree image (X ', Y ') to three-dimensional coordinate points (X, Y, Z); Wherein, f is the focal length in the experimental facilities, b be lasing light emitter to the distance of camera optical axis, θ is optical plane and the COMS plane angle that lasing light emitter A penetrates.

3. described a kind of standing tree tree breast diameter survey method based on optics similar triangles method according to claim 2, described with laser facula drop on phase plane point (X ', Y ') carries out coordinate conversion to three dimensions and obtain point (X, Y, the foundation of Coordinate Transformation Models Z) comprises: the foundation of photo coordinate system X ' O ' Y ' is parallel with XOY plane in the three-dimensional system of coordinate, and three-dimensional system of coordinate initial point O passes the camera optical axis position.

4. the foundation of Coordinate Transformation Models according to claim 3 comprises: the position of lasing light emitter A is positioned at the below of camera optical axis.

5. the foundation of Coordinate Transformation Models according to claim 3 comprises: it is just parallel with surface level that lasing light emitter is launched formed laser plane, and optical plane and COMS plane angle θ angle that above-mentioned lasing light emitter A penetrates can be surveyed.

6. described a kind of standing tree tree breast diameter survey method based on optics similar triangles method according to claim 1, it is characterized in that, described in the step 3 to gained three dimensions point (X, Y, Z) process, comprise to the judgement of abnormity point in the three dimensions point (X, Y, Z) and to satisfying the point that coordinate conversion and circle match require and screening.

7. described a kind of standing tree tree breast diameter survey method based on optics similar triangles method according to claim 1 is characterized in that, the three dimensions point (X, Y, Z) in the step 3 after the above-mentioned processing carries out projection invocation point (X to horizontal direction_Throw, Y_Throw), comprising three dimensions is put the place coordinate plane to the surface level projection, the angle between described surface level and the three dimensions point place coordinate plane is 90 °-θ; The described projection invocation point (X of institute_Throw, Y_Throw) with three dimensions point in (X, Y) between exist as concern:

X_Throw=X

Y_Throw=Ycos (90 °-θ).

8. described a kind of standing tree tree breast diameter survey method based on optics similar triangles method according to claim 1 is characterized in that, adopts the least square circle approximating method to point (X_Throw, Y_Throw) justify match and may further comprise the steps that (adopt X in the following principle derivation formula, Y expresses X_Throw, Y_Throw):

Order:

D＝(N∑X_iY_i-∑X_i∑Y_i)

By

:

Realize the Least Square Circle match; A wherein, b, c is the coefficient of circle general equation, C, D, E, G, H are the intermediate computations parameter, N be that all unique points are total.

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