Random sampling evaluation method for earthwork weight

Abstract

The application relates to a random sampling evaluation method for earthwork weight, belonging to the technical field of weighing. The method comprises the following steps: A, pretreatment; A.1, coding node; A.2, coding triangular plane; A.3, coding survey point; A.4, calculating the density at the eight vertices of a three-linear interpolation cuboid; B, circulating sampling; B.1, generating a random variable point; B.2, calculating the density at the random variable point; B.3, judging whether the random variable point is in an evaluation area; B.4, calculating a three-dimensional integral function at the random variable point; C, estimating the earthwork weight. The application can evaluate the earthwork weight and volume, and the change of the earthwork density in space is considered when the earthwork weight is evaluated.

Description

Technical Field

[0001] This invention belongs to the field of weighing technology, specifically a random sampling evaluation method for the weight of earthwork. Background Technology

[0002] The load capacity of earthwork transportation is affected by the amount of earthwork in the project.

[0003] Methods for calculating earthwork volume in engineering projects include: contour line method, cross-section method, grid method, and digital ground model method. {Ma Zhanlin, Li Jilan, Yu Xiaorong. Comparison of earthwork volume calculation methods [J]. Geological and Mineral Surveying, 2010, 26(1):29-32.}{Li Jianfei, Qin Yanbin, Qi Zhongju. Comparison of multiple earthwork volume calculation methods based on CASS [J]. Surveying and Mapping Engineering, 2012, 21(3):78-80.}{Xue Honggang, Wen Hongxing. Comparison of different earthwork volume calculation methods [J]. Jingwei Tiandi, 2021(4):90-93.} These methods calculate volumetric quantities and do not consider the spatial variation in density of earth and rock, including the following disclosed Chinese patents: 202211597827.6, 202211517272.X, 202211129738.9, 202210439729.3, 202210254264.4, 202210150092.6, 202111527769.5, 202111454739.6, 202111335467.8, 202111299910.0, 202111108805.4, 202110530231.3, 202110246759.8, and 20211019511. 7.X, 202110081178.3, 202110009012.0, 202011441659.2, 202021318903 .1, 201911132676.5, 201910298847.5, 201920375888.5, 201811285383. 6. 201811289966.6, 201830565087.6, 201830218856.5, 201510046808.8 , 201410749287.8, 201410154904.X, 201410007464.5, 201110057508.1.

[0004] From a mathematical perspective, numerical methods for calculating the volume of arbitrary shapes include Newton-Cotes integral, Romberg integral, Gauss integral, Monte Carlo integral, etc. {Wang Yong, Xiong Hua. Multiple Newton-Cotes Integrals and Their Coefficients[J]. Journal of Xihua Normal University (Natural Science Edition), 2010, 31(3):271-274.}{Gou Zhijian, Zheng Fenghua, Yang Xiaochun. Comprehensive Application of Romberg Algorithm and Matlab to Accurately Calculate the Volume of Precision Optical Instruments[J]. Journal of Chengdu University of Information Technology, 2007, 22(3):384-387.}{Xiong Zonghong, Wang Shoucai, Wang Yanhong, et al. A New Proof of the Triple Integral Transformation Formula[J]. Research on Higher Mathematics, 2017, 20(2):43-45.}{Sun Weijun, Qin Hua. Application of Monte Carlo Method in Triple Integrals[J]. Journal of Shandong University of Technology (Natural Science Edition), 2008, 22(1):60-63.}

[0005] To determine the load capacity for earthwork transportation, simply calculating the volume of earthwork is insufficient; its weight must also be calculated. Since the density of earthwork varies spatially in actual engineering projects, its weight cannot be simply calculated by multiplying the volume by a constant density. Therefore, the earthwork volume calculation methods described in the aforementioned literature cannot meet the requirements for evaluating the load capacity of earthwork transportation.

[0006] Content of the invention

[0007] The technical problem to be solved by the present invention is to overcome the shortcomings of existing engineering earthwork volume calculation methods that do not take into account the spatial variation of earthwork density, and to provide a random sampling evaluation method for earthwork weight.

[0008] To address the aforementioned problems, this invention provides a random sampling evaluation method for earthwork weight, comprising the following steps.

[0009] A. Preprocessing

[0010] A.1, Encoding Node

[0011] The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, named Ω, and the coordinates of the nodes in this region are (x, y). i ,y i ,z i ), where i refers to the i-th node, and x, y, z are rectangular coordinates, and x∈[min(x i ),max(x i )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(zi The region is named U. The region is inside U and outside Ω.

[0012] Encode the coordinates of the nodes. Where: N i Number the nodes; n V The total number of nodes in Ω

[0013] A.2, Encoding the Triangular Plane

[0014] Encoding the triangular plane, E j (N i1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j The numbering of the triangular plane, from Looking towards Ω, the three points in the triangle plane are arranged counterclockwise, according to the right-hand rule.

[0015] Check if the coded triangle plane is correct.

[0016] n V -n E +n F =2 (1)

[0017] Where: n E n is the number of edges of Ω; F Let be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded.

[0018] A.3, Coding Survey Points

[0019] Based on the earthwork survey results, the density of at least 8 different survey points was obtained.

[0020] The coordinates of the survey points are encoded.

[0021] N si (x i ,y i ,z i )=1(x1,y1,z1),2(x2,y2,z2),...8(x2,y2,z2),...,n s (x ns ,y ns ,z ns ), where: N siNumber the survey points;

[0022] n s This represents the total number of survey points.

[0023] The density of the survey points is coded, N si (D si )=1(D s1 ),2(D s2 ),...8(D s8 ),...,n s (D sns ),

[0024] Where: D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...,D si ,...,D sns The density at the survey point.

[0025] A.4 Calculate the density at the 8 vertices of the trilinear interpolation cuboid.

[0026] Equations (2), (3), (4), (5), (6), (7), (8), (9), and (10) are used to calculate the density at the eight vertices of the trilinear interpolated cuboid.

[0027]

[0028] Where: x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,zs8 ... represents the z-axis rectangular coordinate of the survey point, x... a For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The lower limit of ,...; y a For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The lower limit of z; a For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The lower limit of ...; x b For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The upper limit of ,...; y b For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The upper limit of ...; z b For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The upper limit of ...

[0029]

[0030]

[0031]

[0032] Where: x si Let x be the rectangular coordinate of the si-th survey point; y si Let z be the y-axis rectangular coordinate of the si-th survey point; si Let ξ be the z-axis rectangular coordinate of the si-th survey point; si η si ζ si These are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid.

[0033]

[0034] Among them: A i1 ~A i8 For the intermediate variable defined by equation (6),

[0035]

[0036] Where: D α =[D 000 D 100 D 010 D 110 D 001 D 101 D 011 D 111 ] T (8)

[0037] A ij =[A i1 A i2 A i3 A i4 A i5 A i6 A i7 A i8 (9)

[0038] D s =[D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...] T (10)

[0039] Where: D 000 D 001 D 010 D 011 D 100 D 101 D 110D 111 The density at the 8 vertices of the trilinear interpolation cuboid; D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ... represents the density at the survey point; D α A is the vector defined by equation (8); ij The matrix defined by equation (9); D s The vector defined by equation (10),

[0040] B. Cyclic Sampling

[0041] The preset total number of sampling times is n.

[0042] If the number of iterations has not reached the preset total number of samples n, steps B.1, B.2, B.3, and B.4 will be executed repeatedly.

[0043] B.1. Generating random variable points

[0044] Take a=min(x i b = max(x) i c = min(y) i ), d = max(y i e = min(z) i g = max(z) i Generate a three-dimensional uniformly distributed random variable X k Y k Z k ,

[0045]

[0046] Where: X k Y k Z k Let X be a three-dimensional uniformly distributed random variable; pdf(X) k ,Y k Z k Let be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution.

[0047] B.2 Calculate the density at points of random variables

[0048] Let x = X k y = Y k z = Z kCalculate equations (12) to (21) to obtain the earthwork volume at the random variable point (X). k ,Y k Z k Density at )

[0049]

[0050]

[0051]

[0052] D 00 =D 000 (1-ξ)+D 100 ξ (15)

[0053] D 01 =D 001 (1-ξ)+D 101 ξ (16)

[0054] D 10 =D 010 (1-ξ)+D 110 ξ (17)

[0055] D 11 =D 011 (1-ξ)+D 111 ξ (18)

[0056] D0 = D 00 (1-η)+D 10 η (19)

[0057] D1 = D 01 (1-η)+D 11 η (20)

[0058] D=D0(1-ζ)+D1ζ (21)

[0059] Where: ξ, η, and ζ are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid, respectively; D 00 D 01 D 10 D 11 D0 and D1 are intermediate variables defined by equations (15) to (20); D is the density at (x,y,z).

[0060] B.3 Determine whether the random variable point is within the evaluation region.

[0061] Calculate the random variable point (X) k ,Y k Z k ) and node N i (xi ,y i ,z i The distance between them

[0062]

[0063] Where: d i For random variable point (X) k ,Y k Z k ) and the i-th node N i (x i ,y i ,z i The distance between them

[0064] Find the nearest node N nearest (x nearest ,y nearest ,z nearest ),

[0065] d nearest =min{d i} (twenty three)

[0066] Where: d nearest For random variable point (X) k ,Y k Z k The distance between N and the nearest neighbor node; nearest The nearest node number; (x nearest ,y nearest ,z nearest ) represents the coordinates of the nearest node.

[0067] Find the node containing node N nearest (x nearest ,y nearest ,z nearest The triangular plane E) near.j (N nearest N i2 N i3 ) or E near.j (N i1 N nearest N i3 ) or E near.j (N i1 N i2 N nearest ), whose set is {E near.j |E near.j (N i1 N i2 N i3 ),N nearest ∈N i1 ∪Ni2 ∪N i3}, determine the point (X) respectively k ,Y k Z k Whether a point (X) is on the Ω side of each element in the set depends on the point (X). k ,Y k Z k If the random variable point (X) is located on one side of Ω of all elements in the set, then determine the random variable point (X). k ,Y k Z k If the value is within Ω, then determine the random variable point (X). k ,Y k Z k )exist Inside,

[0068] B.4 Calculate the three-dimensional integrand at points of the random variable.

[0069]

[0070] Where: f(X) k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at ()

[0071] End of loop,

[0072] C. Estimate the weight of earthwork.

[0073] Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral.

[0074]

[0075] Where: n is the total number of samplings; k refers to the kth sampling; It is an estimate of the three-dimensional Monte Carlo integral over the three-dimensional exact integral, i.e., the weight of the earthwork being evaluated.

[0076] The judgment of random variable point (X) k ,Y k Z k ) in the Ω or triangle plane The specific method on one side is as follows:

[0077] Given three nodes N in the plane of a triangle. L (x L ,y L ,z L ), N M (x M,y M ,z M ), N N (x N ,y N ,z N Viewed from Ω to Ω, these three nodes are arranged counterclockwise, and the triangular plane extends from Ω to... The components of the normal vector in the direction of are,

[0078]

[0079] in: They are respectively Components in the x, y, and z directions; For the triangular plane from Ω to The normal vector in the direction; (x) L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (x M ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N Cartesian coordinates,

[0080] When equation (27) is 1, the sampling point is at When equation (27) is -1, the sampling point is within Ω.

[0081]

[0082] Where: sign[·] is the sign function.

[0083] The aforementioned random sampling method for evaluating the weight of earthwork, which evaluates the volume of earthwork, includes the following steps.

[0084] Set D = 1, and execute steps A, A.1, A.2, B, B.1, B.3, B.4, and C. Numerically, it is equal to the volume of earth and stone.

[0085] The principle of this invention:

[0086] I. Principles of Encoding Earthwork Weight Evaluation Zones

[0087] The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, and the interior of this region is named Ω. The coordinates of the nodes in this region are (x... i ,yi ,z i ), where i refers to the i-th node. x, y, z are rectangular coordinates. Let x∈[min(x i ),max(x i )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(z i The region is named U. The region is inside U and outside Ω.

[0088] Encode the coordinates of the nodes. Where: N i Number the nodes; n V Ω represents the total number of nodes.

[0089] Encoding the triangular plane, E j (N i1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j This refers to the numbering of the triangular planes. For example, 1(1,2,3), 2(2,4,3), ..., 4(2,1,4). From... Looking towards Ω, the three points in the triangle plane are arranged counterclockwise, which is the right-hand rule.

[0090] To ensure the correct execution of the steps, it is necessary to check whether the coded triangle plane is correct. According to Euler's theorem for polyhedra, we have...

[0091] n V -n E +n F =2 (1)

[0092] Where: n E n is the number of edges of Ω; F Let be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded.

[0093] II. Principle for calculating the density of earthwork at random variable points

[0094] When evaluating the weight of earthwork, f(X) k ,Y k Z k ) represents earthwork in X kY k Z k The density at each point is usually not known in advance. Based on the earthwork survey results, the density of at least 8 different survey points is obtained, and the coordinates of the survey points are coded, N. si (x i ,y i ,z i )=1(x1,y1,z1),2(x2,y2,z2),...8(x2,y2,z2),...,n s (x ns ,y ns ,z ns The density of the survey points is coded, N si (D si )=1(D s1 ),2(D s2 ),...8(D s8 ),...,n s (D sns ), where: N si For the survey point number; n s D represents the total number of survey points. s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...,D si ,...,D sns This represents the density at the survey point. s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8... represents the z-axis rectangular coordinates of the survey points. Based on the density of the survey points, the density of earth and stone at any point is calculated by interpolation. The trilinear interpolation formulas are equations (12) to (21), which represent the mapping from the density at the vertex of the cuboid to the density at (x,y,z), where D(vertex of the cuboid) → D(x,y,z). Figure 1 As shown.

[0095]

[0096]

[0097]

[0098] D 00 =D 000 (1-ξ)+D 100 ξ (15)

[0099] D 01 =D 001 (1-ξ)+D 101 ξ (16)

[0100] D 10 =D 010 (1-ξ)+D 110 ξ (17)

[0101] D 11 =D 011 (1-ξ)+D 111 ξ (18)

[0102] D0 = D 00 (1-η)+D 10 η (19)

[0103] D1 = D 01 (1-η)+D 11 η (20)

[0104] D=D0(1-ζ)+D1ζ (21)

[0105] Where: x a For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The lower limit of ,...; y a For y s1 ,y s2 ,y s3 ,y s4 ,ys5 ,y s6 ,y s7 ,y s8 The lower limit of z; a For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The lower limit of ...; x b For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The upper limit of ,...; y b For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The upper limit of ...; z b For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The upper limit of ...; ξ, η, ζ are the relative coordinates of points x, y, z in the trilinear interpolation cuboid; D 000 D 001 D 010 D 011 D 100 D 101 D 110 D 111 The density at the 8 vertices of the trilinear interpolation cuboid; D 00 D 01 D 10 D 11 D0 and D1 are intermediate variables defined by equations (15) to (20); D is the density at (x,y,z).

[0106] However, the density at the eight vertices of the trilinear interpolation cuboid cannot be known in advance. Therefore, based on the density at the survey points, the density at the eight vertices of the trilinear interpolation cuboid is obtained through the inverse mapping D(vertices of the cuboid) → D(x,y,z). The density at the survey points and the coordinates D... si (x si ,y si ,z si Substituting into equations (12) to (21), we get

[0107]

[0108]

[0109]

[0110] D 00 =D 000 (1-ξ si )+D 100 ξ si (28)

[0111] D 01 =D 001 (1-ξ)+D 101 ξ (29)

[0112] D 10 =D 010 (1-ξ si )+D 110 ξ si (30)

[0113] D 11 =D 011 (1-ξ si )+D 111 ξ si (31)

[0114] D0 = D 00 (1-η si )+D 10 η si (32)

[0115] D1 = D 01 (1-η si )+D 11 η si (33)

[0116] D si =(1-ζ si )D0+ζ si D1 (34)

[0117] Where: x siLet x be the rectangular coordinate of the si-th survey point; y si Let z be the y-axis rectangular coordinate of the si-th survey point; si Let ξ be the z-axis rectangular coordinate of the si-th survey point; si η si ζ si These are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid.

[0118] Substituting equations (32) and (33) into equation (34) yields

[0119] D si =(1-ζ si )[D 00 (1-η si )+D 10 η si ]+ζ si [D 01 (1-η si )+D 11 η si (35)

[0120] Substituting equations (28) to (6) into equation (29) and rearranging, we get...

[0121]

[0122] Perform variable substitution

[0123]

[0124] Among them: A i1 ~A i8 Let be the intermediate variable defined by equation (6). Then equation (36) is replaced by .

[0125] D si =A i1 D 000 +A i2 D 100 +A i3 D 010 +A i4 D 110 +A i5 D 001 +A i6 D 101 +A i7 D 011 +A i8 D 111 (37)

[0126] Written in matrix form as

[0127] D s =A ijD α (38)

[0128] Where: D α =[D 000 D 100 D 010 D 110 D 001 D 101 D 011 D 111 ] T (8)

[0129] A ij =[A i1 A i2 A i3 A i4 A i5 A i6 A i7 A i8 (9)

[0130] D s =[D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...] T (10)

[0131] Where: D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ... represents the density at the survey point; D α A is the vector defined by equation (8); ij The matrix defined by equation (9); D s The vector defined by equation (10).

[0132] According to the least squares method, multiplying both sides of equation (38) by A ij The transpose is obtained

[0133]

[0134] Multiply both sides of equation (39) by The inverse matrix is ​​obtained

[0135]

[0136] Equation (7) maps D(survey point) to D(vertices of the cuboid). Take...

[0137]

[0138] Where: x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 ... represents the z-axis rectangular coordinates of the survey point.

[0139] Equations (2), (3), (4), (5), (6), (7), (8), (9), and (10) are used to obtain the density at the eight vertices of the trilinear interpolation cuboid. Then, x = X is taken. k y = Y k z = Z k Calculate equations (12) to (21) to obtain the earthwork volume at the random variable point (X). k ,Y k Z k The density at ().

[0140] III. Principles for estimating three-dimensional exact integrals

[0141] Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral.

[0142]

[0143] Where: n is the total number of samplings; k refers to the kth sampling; X is an estimator of the three-dimensional Monte Carlo integral over the three-dimensional exact integral; k Y k Z k f(X) is a three-dimensional random variable;k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at ); pdf(X k ,Y k Z k ) is the probability density function of a three-dimensional random variable.

[0144] Prove that the three-dimensional Monte Carlo integral is an unbiased estimate of the three-dimensional exact integral:

[0145]

[0146] Where E(·) is the expected value. Q.E.D.

[0147] The more sampling points there are, the closer the 3D Monte Carlo integral is to the 3D exact integral:

[0148]

[0149] Where: σ 2 Or D(·) represents the variance. The sign of D(·) is not italicized, distinguishing it from the density D at (x,y,z).

[0150] Clearly, the larger n is, the smaller the variance, and the closer the three-dimensional Monte Carlo integral is to the three-dimensional exact integral. Q.E.D.

[0151] When X k Y k Z k When X is a three-dimensional uniformly distributed random variable, pdf(X) k ,Y k Z k Let be the probability density function of a three-dimensional uniformly distributed random variable, i.e.

[0152]

[0153] Where: pdf(X) k ,Y k Z k Let be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution. Substituting equation (11) into equation (40) yields...

[0154]

[0155] Take a=min(x i b = max(x) i c = min(y) i), d = max(y i e = min(z) i g = max(z) i Generate a three-dimensional uniformly distributed random variable X k Y k Z k Substituting into equation (25), the three-dimensional exact integral can be estimated. This refers to the weight of the earth and stone work being evaluated.

[0156] IV. Methods for determining whether a random variable point is within the earthwork weight evaluation area

[0157] Since the required quantity is the weight of earth and stone within Ω, in the calculation formula (25), take...

[0158]

[0159] At this point, it is necessary to determine the random variable point (X). k ,Y k Z k Is it within Ω? Set the distance point (X) to be within Ω. k ,Y k Z k The nearest triangle plane faces point (X). k ,Y k Z k When the normal vectors of all points simultaneously into Ω, determine the random variable point (X). k ,Y k Z k The distance is within Ω. This is equivalent to the distance from point (X) k ,Y k Z k The nearest triangle plane faces point (X). k ,Y k Z k The unit normal vector of the triangle is perpendicular to the plane of the triangle. The direction of the unit normal vector is opposite.

[0160] Proof: As Figure 2 As shown. Assume point (X) k ,Y k Z k ) is inside Ω, and point (X) k ,Y k Z k The distance between (X) and the triangular plane F is the smallest among all triangular planes. If F is oriented towards (X) k ,Y k Z k The unit normal vector of ) and the orientation of F unit normal vector If the directions are the same, then there must exist a triangular plane between (X and Y). k ,Y k Z k Between ) and F, such that point (X) k ,Y k Z k ) remains within Ω. This is consistent with "point (X) k ,Y k Z k The assumption that the distance between point (X) and triangle plane F is the smallest among all triangle planes contradicts this premise. Therefore, when point (X) k ,Y k Z k ) is inside Ω, and point (X) k ,Y k Z k The distance between (X) and the triangular plane F is the smallest among all triangular planes, and F is oriented towards (X). k ,Y k Z k The unit normal vector of ) and the orientation of F The direction of the unit normal vector is opposite. Q.E.D.

[0161] V. Derivation of the unit normal vector of the triangular plane facing outside the earthwork weight evaluation area

[0162] like Figure 3 As shown. Given that the three nodes on the triangular plane are N... L (x L ,y L ,z L ), N M (x M ,y M ,z M ), N N (x N ,y N ,z N ),from Looking towards Ω, these three nodes are arranged counterclockwise. Therefore, the plane of the triangle extends from Ω towards... The normal vector in the direction is,

[0163]

[0164] in: For the triangular plane from Ω to The normal vector in the direction; Let x, y, and z be the unit vectors in the x, y, and z directions, respectively; (x L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (xM ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N The rectangular coordinates.

[0165] The triangular plane from Ω to The components of the normal vector in the direction are:

[0166]

[0167] in: They are respectively Components in the x, y, and z directions.

[0168]

[0169] in: for The model.

[0170]

[0171] in: Oriented towards the triangular plane The unit normal vector.

[0172] VI. Derivation of the unit normal vector passing through the sampling points outside the triangular plane

[0173] Represent the triangular plane using the point-normal form of the plane equation.

[0174]

[0175] Passing point (X) k ,Y k Z k The parametric equation of the line perpendicular to the plane of the triangle is:

[0176]

[0177] Where t is the parameter of the straight line equation. Substituting equation (47) into equation (46) and solving, we get...

[0178]

[0179] Substituting equation (48) into equation (47) yields,

[0180]

[0181]

[0182]

[0183] Where: x0, y0, z0 are points (X... k ,Y k Z k The perpendicular projection of ) onto the triangular plane. Assume it passes through a point (X) outside the triangular plane. k ,Y k Z k The normal vector of ) is n sam Each component is

[0184]

[0185]

[0186]

[0187] Where: n sam.x n sam.y n sam.z n sam Components in the x, y, and z directions.

[0188]

[0189] Where: d sam For n sam The model.

[0190]

[0191] Where: e sam The triangle plane faces point (X) k ,Y k Z k The unit normal vector of ); sign[·] is the sign function. Therefore, when equation (27) is 1, e sam and The directions are the same, that is, the sampling points are in Inside; when equation (27) is -1, e sam and The direction is opposite, meaning the sampling point is within Ω.

[0192]

[0193] VII. Principles for Evaluating Earthwork Volume

[0194] When the density D is taken to be 1 within Ω Numerically, it is equal to the volume of earth and stone.

[0195] The beneficial effects of this invention are as follows:

[0196] 1. Steps A through C enable the evaluation of earthwork weight. Steps A.3, A.4, and B.2 consider the spatial variation in earthwork density when evaluating the weight. When the earthwork density is non-uniform, the method of this invention can determine the weight, whereas conventional analytical methods cannot. This overcomes the technical deficiency of conventional analytical methods in failing to consider the spatial variation in earthwork density.

[0197] 2. Through steps A, A.1, A.2, B, B.1, B.3, B.4, and C, the volume of earthwork can be evaluated. The total volume of earthwork obtained using the method of this invention does not differ by more than 0.7%, meeting the requirements of engineering applications. Attached Figure Description

[0198] Figure 1 This is a schematic diagram of a cuboid for trilinear interpolation.

[0199] Figure 2 There exists a triangular plane between (X) k ,Y k Z k A diagram showing the relationship between F and F.

[0200] Figure 3 Oriented towards the triangular plane A schematic diagram of the unit normal vector.

[0201] Figure 4 The earthwork evaluation areas are for Examples 1 and 2.

[0202] Figure 5 This is the evaluation area for the earthwork weight in Example 3. Detailed Implementation

[0203] The technical solution of the present invention will be further described in detail below with reference to the accompanying drawings and embodiments.

[0204] Example 1

[0205] This embodiment is an example of a random sampling evaluation method for earthwork weight according to the present invention, including the following steps: A. Pre-processing

[0206] A.1, Encoding Node

[0207] The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, named Ω, and the coordinates of the nodes in this region are (x, y). i ,y i ,z i ), where i refers to the i-th node, and x, y, z are rectangular coordinates, and x∈[min(x i ),max(xi )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(z i The region is named U. The region is inside U and outside Ω.

[0208] Encode the coordinates of the nodes. Where: N i Number the nodes; n V The total number of nodes in Ω

[0209] 1(0,0,0),2(1,0,0),3(1,1,0),4(0,1,0),5(1,1,1),6(0,1,1), as shown Figure 4 As shown, the unit of length is meters (m).

[0210] A.2, Encoding the Triangular Plane

[0211] Encoding the triangular plane, E j (N i1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j The numbering of the triangular plane, from Looking towards Ω, the three points in the triangle plane are arranged counterclockwise, according to the right-hand rule: 1(1,2,5), 2(1,5,6), 3(2,3,5), 4(3,4,5), 5(4,6,5), 6(1,6,4), 7(1,3,2), 8(1,4,3).

[0212] Check if the coded triangle plane is correct.

[0213] n V -n E +n F =2 (1)

[0214] Where: n E n is the number of edges of Ω; F Let n be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded. V =6,n E =12,n F =8,n V -nE +n F =2

[0215] A.3, Coding Survey Points

[0216] Based on the earthwork survey results, the density of at least 8 different survey points was obtained.

[0217] The coordinates of the survey points are encoded.

[0218] N si (x i ,y i ,z i )=1(x1,y1,z1),2(x2,y2,z2),...8(x2,y2,z2),...,n s (x ns ,y ns ,z ns ), where: N si For the survey point number; n s This represents the total number of survey points, and the coordinates of the survey points are as follows:

[0219] 1(0,0,0),2(1,0,0),3(1,1,0),4(0,1,0),5(0,0,1),6(1,0,1),7(1,1,1),8(0,1,1)

[0220] The density of the survey points is coded, N si (D si )=1(D s1 ),2(D s2 ),...8(D s8 ),...,n s (D sns ),

[0221] Where: D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...,D si ,...,D sns The density at the survey points is given by the following values: 1(20), 2(20), 3(20), 4(20), 5(20), 6(20), 7(20), 8(20). The unit of density is kN / m³. 3 .

[0222] A.4 Calculate the density at the 8 vertices of the trilinear interpolation cuboid.

[0223] Equations (2), (3), (4), (5), (6), (7), (8), (9), and (10) are used to calculate the density at the eight vertices of the trilinear interpolation cuboid.

[0224]

[0225] Where: x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 ... represents the z-axis rectangular coordinate of the survey point, x... a For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The lower limit of ,...; y a For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The lower limit of z; a For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The lower limit of ...; x b For xs1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The upper limit of ,...; y b For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The upper limit of ...; z b For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The upper limit of ..., x a =0,x b =1,y a =0,y b =1,z a =0,z b =1

[0226]

[0227]

[0228]

[0229] Where: x si Let x be the rectangular coordinate of the si-th survey point; y si Let z be the y-axis rectangular coordinate of the si-th survey point; si Let ξ be the z-axis rectangular coordinate of the si-th survey point; si η si ζ si These are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid.

[0230] (ξ) of the survey point si η si ζ si The values ​​are: 1(0,0,0), 2(1,0,0), 3(1,1,0), 4(0,1,0), 5(0,0,1), 6(1,0,1), 7(1,1,1), 8(0,1,1)

[0231]

[0232] Among them: A i1 ~A i8 Formula (6) fixed intermediate change, calculation point A i1 ~A i8 Separation:

[0233] 1(A i1 = 1.0, A i2 = 0.0, A i3 = 0.0, A i4 = 0.0, A i5 = 0.0, A i6 = 0.0, A i7 = 0.0, A i8 = 0.0)

[0234] 2(A i1 = 0.0, A i2 = 1.0, A i3 = 0.0, A i4 = 0.0, A i5 = 0.0, A i6 = 0.0, A i7 = 0.0, A i8 = 0.0)

[0235] 3(A i1 = 0.0, A i2 = 0.0, A i3 = 0.0, A i4 = 1.0, A i5 = 0.0, A i6 = 0.0, A i7 = 0.0, A i8 = 0.0)

[0236] 4(A i1 = 0.0, A i2 = 0.0, A i3 = 1.0, A i4 = 0.0, A i5 = 0.0, A i6 = 0.0, A i7 = 0.0, A i8 = 0.0)

[0237] 5(A i1 = 0.0, A i2 = 0.0, A i3 = 0.0, A i4 = 0.0, A i5 = 1.0, A i6 = 0.0, A i7=0.0, A i8 =0.0)

[0238] 6(A i1 =0.0, A i2 =0.0, A i3 =0.0, A i4 =0.0, A i5 =0.0, A i6 =1.0, A i7 =0.0, A i8 =0.0)

[0239] 7(A i1 =0.0, A i2 =0.0, A i3 =0.0, A i4 =0.0, A i5 =0.0, A i6 =0.0, A i7 =0.0, A i8 =1.0)

[0240] 8(A i1 =0.0, A i2 =0.0, A i3 =0.0, A i4 =0.0, A i5 =0.0, A i6 =0.0, A i7 =1.0, A i8 =0.0)

[0241]

[0242] Wherein: D α =[D 000 , D 100 , D 010 , D 110 , D 001 , D 101 , D 011 , D 111 T (8)

[0243] A ij =[A i1 , A i2 , A i3 , A i4 , A i5 , A i6 , A i7 , A i8 (9)

[0244] D s =[D​s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...] T (10)

[0245] Where: D 000 D 001 D 010 D 011 D 100 D 101 D 110 D 111 The density at the 8 vertices of the trilinear interpolation cuboid; D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ... represents the density at the survey point; D α A is the vector defined by equation (8); ij The matrix defined by equation (9); D s The vector defined by equation (10),

[0246] D s =[20 20 20 20 20 20 20] T D α =[20 20 20 20 20 2020 20] T

[0247] B. Cyclic Sampling

[0248] The preset total number of samplings is n, and the preset n = 10000

[0249] If the number of iterations has not reached the preset total number of samples n, steps B.1, B.2, B.3, and B.4 will be executed repeatedly.

[0250] B.1. Generating random variable points

[0251] Take a=min(x i b = max(x) i c = min(y) i ), d = max(y i e = min(z) i g = max(z) iGenerate a three-dimensional uniformly distributed random variable X k Y k Z k ,

[0252]

[0253] Where: X k Y k Z k Let X be a three-dimensional uniformly distributed random variable; pdf(X) k ,Y k Z k Let f(x) be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution, where a = 0, b = 1, c = 0, d = 1, e = 0, and g = 1.

[0254] B.2 Calculate the density at points of random variables

[0255] Let x = X k y = Y k z = Z k Calculate equations (12) to (21) to obtain the earthwork volume at the random variable point (X). k ,Y k Z k Density at )

[0256]

[0257]

[0258]

[0259] D 00 =D 000 (1-ξ)+D 100 ξ (15)

[0260] D 01 =D 001 (1-ξ)+D 101 ξ (16)

[0261] D 10 =D 010 (1-ξ)+D 110 ξ (17)

[0262] D 11 =D 011 (1-ξ)+D 111 ξ (18)

[0263] D0 = D 00 (1-η)+D10 η (19)

[0264] D1 = D 01 (1-η)+D 11 η (20)

[0265] D=D0(1-ζ)+D1ζ (21)

[0266] Where: ξ, η, and ζ are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid, respectively; D 00 D 01 D 10 D 11 D0 and D1 are intermediate variables defined by equations (15) to (20); D is the density at (x,y,z), with (X k ,Y k Z k )=(0.937,0.117,0.305) as an example, ξ=0.937, η=0.117, ζ=0.305, D 00 =20,D 01 =20,D 10 =20,D 11 =20, D0=20, D1=20, D=20

[0267] B.3 Determine whether the random variable point is within the evaluation region.

[0268] Calculate the random variable point (X) k ,Y k Z k ) and node N i (x i ,y i ,z i The distance between them

[0269]

[0270] Where: d i For random variable point (X) k ,Y k Z k ) and the i-th node N i (x i ,y i ,z i The distance between them

[0271] d i The values ​​are: 0.9923, 0.3326, 0.9363, 1.3231, 1.1254, and 1.4631, respectively.

[0272] Find the nearest node N nearest(x nearest ,y nearest ,z nearest ),

[0273] d nearest =min{d i} (twenty three)

[0274] Where: d nearest For random variable point (X) k ,Y k Z k The distance between N and the nearest neighbor node; nearest The nearest node number; (x nearest ,y nearest ,z nearest ) represents the coordinates of the nearest node, d nearest =0.3326, N nearest =2

[0275] Find the node containing node N nearest (x nearest ,y nearest ,z nearest The triangular plane E) near.j (N nearest N i2 N i3 ) or E near.j (N i1 N nearest N i3 ) or E near.j (N i1 N i2 N nearest ), whose set is {E near.j |E near.j (N i1 N i2 N i3 ),N nearest ∈N i1 ∪N i2 ∪N i3}, determine the point (X) respectively k ,Y k Z k Whether a point (X) is on the Ω side of each element in the set depends on the point (X). k ,Y k Z k If the random variable point (X) is located on one side of Ω of all elements in the set, then determine the random variable point (X). k ,Y k Z k If the value is within Ω, then determine the random variable point (X). k ,Y k Z k)exist Inside, there is node N. nearest (x nearest ,y nearest ,z nearest The triangle planes are numbered 1, 3, and 7. Point (X) k ,Y k Z k In the plane of triangle 1 One side, on the Ω side of the plane of triangle 3, and on the Ω side of the plane of triangle 7. Determine the random variable point (X). k ,Y k Z k It is not within Ω.

[0276] B.4 Calculate the three-dimensional integrand at points of the random variable.

[0277]

[0278] Where: f(X) k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at point (X) is f(X). k ,Y k Z k ) = 0

[0279] End of loop,

[0280] C. Estimate the weight of earthwork.

[0281] Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral.

[0282]

[0283] Where: n is the total number of samplings; k refers to the kth sampling; It is an estimate of the three-dimensional Monte Carlo integral over the three-dimensional exact integral, i.e., the weight of the earthwork being evaluated. The unit of weight is kN.

[0284] The judgment of random variable point (X) k ,Y k Z k ) in the Ω or triangle plane The specific method on one side is as follows:

[0285] Given three nodes N in the plane of a triangle. L (x L ,y L ,zL ), N M (x M ,y M ,z M ), N N (x N ,y N ,z N ),from Looking towards Ω, these three nodes are arranged counterclockwise, and the triangular plane extends from Ω towards... The components of the normal vector in the direction of are,

[0286]

[0287] in: They are respectively Components in the x, y, and z directions; For the triangular plane from Ω to The normal vector in the direction; (x) L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (x M ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N Cartesian coordinates,

[0288] With (X) k ,Y k Z k Taking (x) = (0.937, 0.117, 0.305) and the plane of triangle 1 as examples. L ,y L ,z L )=(0,0,0),(x M ,y M ,z M )=(1,0,0),(x N ,y N ,z N ) = (1,1,1).

[0289] When equation (27) is 1, the sampling point is at When equation (27) is -1, the sampling point is within Ω.

[0290]

[0291] Where: sign[·] is the sign function. Sampling points at Inside.

[0292] The maximum load capacity of a certain trolley is 0.6 tons. The total weight of the earth and rock excavated is 10.0692 kN, approximately equal to 1.00692 tons. It would take two trips to complete the transportation of all the earth and rock using this trolley.

[0293] Assuming the earth and rock have uniform density, if we use the traditional analytical method, we can first calculate the total volume of the earth and rock: (0.5 × 1 × 1) × 1 = 0.5 m³. 3 Its density is 20 kN / m³ 3 Therefore, the total weight of the earthwork is 10 kN. Thus, under the condition of uniform earthwork density, the difference between the total weight of the earthwork obtained using the method of this invention and the previous analytical method is no more than 0.692%, meeting the requirements of engineering applications.

[0294] Example 2

[0295] An example of a random sampling evaluation method for earthwork weight, evaluating earthwork volume, includes the following steps: taking D=1, executing steps A, A.1, A.2, B, B.1, B.3, B.4, and C. Numerically, it is equal to the volume of earth and stone.

[0296] A. Preprocessing

[0297] A.1, Encoding Node

[0298] The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, named Ω, and the coordinates of the nodes in this region are (x, y). i ,y i ,z i ), where i refers to the i-th node, and x, y, z are rectangular coordinates, and x∈[min(x i ),max(x i )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(z i The region is named U. The region is inside U and outside Ω.

[0299] Encode the coordinates of the nodes. Where: N i Number the nodes; n V The total number of nodes in Ω

[0300] 1(0,0,0),2(1,0,0),3(1,1,0),4(0,1,0),5(1,1,1),6(0,1,1), as shown Figure 4 As shown, the unit of length is meters (m).

[0301] A.2, Encoding the Triangular Plane

[0302] Encoding the triangular plane, E j (N i1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j The numbering of the triangular plane, from Looking towards Ω, the three points in the triangle plane are arranged counterclockwise, according to the right-hand rule: 1(1,2,5), 2(1,5,6), 3(2,3,5), 4(3,4,5), 5(4,6,5), 6(1,6,4), 7(1,3,2), 8(1,4,3).

[0303] Check if the coded triangle plane is correct.

[0304] n V -n E +n F =2 (1)

[0305] Where: n E n is the number of edges of Ω; F Let n be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded. V =6,n E =12,n F =8,n V -n E +n F =2

[0306] B. Cyclic Sampling

[0307] The preset total number of samplings is n, and the preset n = 10000

[0308] If the number of iterations has not reached the preset total number of samples n, steps B.1, B.2, B.3, and B.4 will be executed repeatedly.

[0309] B.1. Generating random variable points

[0310] Take a=min(x ib = max(x) i c = min(y) i ), d = max(y i e = min(z) i g = max(z) i Generate a three-dimensional uniformly distributed random variable X k Y k Z k ,

[0311]

[0312] Where: X k Y k Z k Let X be a three-dimensional uniformly distributed random variable; pdf(X) k ,Y k Z k Let f(x) be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution, where a = 0, b = 1, c = 0, d = 1, e = 0, and g = 1.

[0313] B.3 Determine whether the random variable point is within the evaluation region.

[0314] Calculate the random variable point (X) k ,Y k Z k ) and node N i (x i ,y i ,z i The distance between them

[0315]

[0316] Where: d i For random variable point (X) k ,Y k Z k ) and the i-th node N i (x i ,y i ,z i The distance between (X) k ,Y k Z k Taking (0.937, 0.117, 0.305) as an example, d i The values ​​are: 0.9923, 0.3326, 0.9363, 1.3231, 1.1254, and 1.4631, respectively.

[0317] Find the nearest node N nearest (xnearest ,y nearest ,z nearest ),

[0318] d nearest =min{d i} (twenty three)

[0319] Where: d nearest For random variable point (X) k ,Y k Z k The distance between N and the nearest neighbor node; nearest The nearest node number; (x nearest ,y nearest ,z nearest ) represents the coordinates of the nearest node, d nearest =0.3326, N nearest =2

[0320] Find the node containing node N nearest (x nearest ,y nearest ,z nearest The triangular plane E) near.j (N nearest N i2 N i3 ) or E near.j (N i1 N nearest N i3 ) or E near.j (N i1 N i2 N nearest ), whose set is {E near.j |E near.j (N i1 N i2 N i3 ),N nearest ∈N i1 ∪N i2 ∪N i3}, determine the point (X) respectively k ,Y k Z k Whether a point (X) is on the Ω side of each element in the set depends on the point (X). k ,Y k Z k If the random variable point (X) is located on one side of Ω of all elements in the set, then determine the random variable point (X). k ,Y k Z k If the value is within Ω, then determine the random variable point (X). k ,Y k Z k )exist Inside,

[0321] Includes node N nearest (x nearest ,y nearest ,z nearest The triangle planes are numbered 1, 3, and 7. Point (X) k ,Y k Z k In the plane of triangle 1 One side, on the Ω side of the plane of triangle 3, and on the Ω side of the plane of triangle 7. Determine the random variable point (X). k ,Y k Z k It is not within Ω.

[0322] B.4 Calculate the three-dimensional integrand at points of the random variable.

[0323]

[0324] Where: f(X) k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at point (X) is f(X). k ,Y k Z k ) = 0

[0325] End of loop,

[0326] C. Estimate the weight of earthwork.

[0327] Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral.

[0328]

[0329] Where: n is the total number of samplings; k refers to the kth sampling; It is an estimate of the three-dimensional Monte Carlo integral over the three-dimensional exact integral, i.e., the weight of the earthwork being evaluated. The unit of volume is m. 3 .

[0330] The judgment of random variable point (X) k ,Y k Z k ) in the Ω or triangle plane The specific method on one side is as follows:

[0331] Given three nodes N in the plane of a triangle. L(x L ,y L ,z L ), N M (x M ,y M ,z M ), N N (x N ,y N ,z N ),from Looking towards Ω, these three nodes are arranged counterclockwise, and the triangular plane extends from Ω towards... The components of the normal vector in the direction of are,

[0332]

[0333] in: They are respectively Components in the x, y, and z directions; For the triangular plane from Ω to The normal vector in the direction; (x) L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (x M ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N Cartesian coordinates,

[0334] With (X) k ,Y k Z k Taking (x) = (0.937, 0.117, 0.305) and the plane of triangle 1 as examples. L ,y L ,z L )=(0,0,0),(x M ,y M ,z M )=(1,0,0),(x N ,y N ,z N ) = (1,1,1).

[0335] When equation (27) is 1, the sampling point is at When equation (27) is -1, the sampling point is within Ω.

[0336]

[0337] Where: sign[·] is the sign function. Sampling points at Inside.

[0338] The total volume of earthwork obtained using the method of this invention is 0.5035 m³. 3 .

[0339] If we use the traditional analytical method, the total volume of earthwork is calculated as: (0.5 × 1 × 1) × 1 = 0.5 m³ 3 Therefore, the difference between the total volume of earthwork obtained by the method of this invention and the previous analytical method is no more than 0.7%, which meets the requirements of engineering applications.

[0340] Example 3

[0341] This embodiment is an example of a random sampling evaluation method for earthwork weight according to the present invention, including the following steps: A. Pre-processing

[0342] A.1, Encoding Node

[0343] The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, named Ω, and the coordinates of the nodes in this region are (x, y). i ,y i ,z i ), where i refers to the i-th node, and x, y, z are rectangular coordinates, and x∈[min(x i ),max(x i )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(z i The region is named U. The region is inside U and outside Ω.

[0344] Encode the coordinates of the nodes. Where: N i Number the nodes; n V The total number of nodes in Ω

[0345] 1(1,1,0),2(3,1,0),3(2,3,0),4(2,2,-2), as shown Figure 5 As shown, the unit of length is meters (m).

[0346] A.2, Encoding the Triangular Plane

[0347] Encoding the triangular plane, E j (Ni1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j The numbering of the triangular plane, from Looking towards Ω, the three points in the triangle plane are arranged counterclockwise, according to the right-hand rule: 1(1,2,3), 2(1,4,2), 3(2,4,3), 4(1,3,4).

[0348] Check if the coded triangle plane is correct.

[0349] n V -n E +n F =2 (1)

[0350] Where: n E n is the number of edges of Ω; F Let n be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded. V =4,n E =6,n F =4,n V -n E +n F =2

[0351] A.3, Coding Survey Points

[0352] Based on the earthwork survey results, the density of at least 8 different survey points was obtained.

[0353] The coordinates of the survey points are encoded.

[0354] N si (x i ,y i ,z i )=1(x1,y1,z1),2(x2,y2,z2),...8(x2,y2,z2),...,n s (x ns ,y ns ,z ns ), where: N si For the survey point number; n sThe coordinates of the survey points are: 1(1.5,1.5,0), 2(3,1,0), 3(2.5,2.5,0), 4(1,3,0), 5(1.5,1.5,-2), 6(3,1,-2.5), 7(2.5,2.5,-3), 8(1,3,-3.5), 9(2,2,-4).

[0355] The density of the survey points is coded, N si (D si )=1(D s1 ),2(D s2 ),...8(D s8 ),...,n s (D sns ),

[0356] Where: D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...,D si ,...,D sns The density at the survey points is given by the following values: 1(15), 2(16), 3(15.5), 4(15.4), 5(18), 6(19), 7(19.5), 8(20), 9(20.4). The unit of density is kN / m³. 3 .

[0357] A.4 Calculate the density at the 8 vertices of the trilinear interpolation cuboid.

[0358] Equations (2), (3), (4), (5), (6), (7), (8), (9), and (10) are used to calculate the density at the eight vertices of the trilinear interpolation cuboid.

[0359]

[0360] Where: x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,ys6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 ... represents the z-axis rectangular coordinate of the survey point, x... a For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The lower limit of ,...; y a For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The lower limit of z; a For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The lower limit of ...; x b For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The upper limit of ,...; y b For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The upper limit of ...; z b For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,zs6 ,z s7 ,z s8 The upper limit of ..., x a =1,x b =3,y a =1,y b =3,z a =-4,z b =0

[0361]

[0362]

[0363]

[0364] Where: x si Let x be the rectangular coordinate of the si-th survey point; y si Let z be the y-axis rectangular coordinate of the si-th survey point; si Let ξ be the z-axis rectangular coordinate of the si-th survey point; si η si ζ si These are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid.

[0365] (ξ) of the survey point si η si ζ si The values ​​are: 1(0.25,0.25,1.0), 2(1.0,0.0,1.0), 3(0.75,0.75,1.0), 4(0.0,1.0,1.0), 5(0.25,0.25,0.5), 6(1.0,0.0,0.375), 7(0.75,0.75,0.25), 8(0.0,1.0,0.125), 9(0.5,0.5,0.0).

[0366]

[0367] Among them: A i1 ~A i8 As an intermediate variable defined by equation (6), the A of the survey point i1 ~A i8 They are respectively:

[0368] 1(A i1 =0.0,A i2 =0.0,A i3 =0.0,A i4 =0.0,A i5 =0.5625,A i6 =0.1875,A i7=0.1875,A i8 =0.0625)

[0369] 2(A i1 =0.0,A i2 =0.0,A i3 =0.0,A i4 =0.0,A i5 =0.0,A i6 =1.0,A i7 =0.0,A i8 =0.0)

[0370] 3(A i1 =0.0,A i2 =0.0,A i3 =0.0,A i4 =0.0,A i5 =0.0625,A i6 =0.1875,A i7 =0.1875,A i8 =0.5625)

[0371] 4(A i1 =0.0,A i2 =0.0,A i3 =0.0,A i4 =0.0,A i5 =0.0,A i6 =0.0,A i7 =1.0,A i8 =0.0)

[0372] 5(A i1 =0.28125,A i2 =0.09375,A i3 =0.09375,A i4 =0.03125,A i5 =0.28125,A i6 =0.09375,A i7 =0.09375,A i8 =0.03125)

[0373] 6(A i1 =0.0,A i2 =0.625,A i3 =0.0,A i4 =0.0,A i5 =0.0,A i6 =0.375,A i7 =0.0,A i8 =0.0)

[0374] 7(A i1 =0.046875, A i2 =0.140625, A i3 =0.140625, A i4 =0.421875, A i5 =0.015625, A i6 =0.046875, A i7 =0.046875, A i8 =0.140625)

[0375] 8(A i1 =0.0, A i2 =0.0, A i3 =0.875, A i4 =0.0, A i5 =0.0, A i6 =0.0, A i7 =0.125, A i8 =0.0)

[0376] 9(A i1 =0.25, A i2 =0.25, A i3 =0.25, A i4 =0.25, A i5 =0.0, A i6 =0.0, A i7 =0.0, A i8 =0.0)

[0377]

[0378] Wherein: D α =[D 000 , D 100 , D 010 , D 110 , D 001 , D 101 , D 011 , D 111 T (8)

[0379] A ij =[A i1 , A i2 , A i3 , A i4 , A i5 , A i6 , A i7 , A i8 (9)

[0380] D s ​=[D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...] T (10)

[0381] Where: D 000 D 001 D 010 D 011 D 100 D 101 D 110 D 111 The density at the 8 vertices of the trilinear interpolation cuboid; D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ... represents the density at the survey point; D α A is the vector defined by equation (8); ij The matrix defined by equation (9); D s The vector defined by equation (10),

[0382]

[0383] D s =[15 16 15.5 15.4 18 19 19.5 20 20.4] T

[0384] D α =[20.4542 20.7347 20.6275 20.6768 14.6298 16.0134 15.4031 15.5107] T

[0385] B. Cyclic Sampling

[0386] The preset total number of samplings is n, and the preset n = 10000

[0387] If the number of iterations has not reached the preset total number of samples n, steps B.1, B.2, B.3, and B.4 will be executed repeatedly.

[0388] B.1. Generating random variable points

[0389] Take a=min(x ib = max(x) i c = min(y) i ), d = max(y i e = min(z) i g = max(z) i Generate a three-dimensional uniformly distributed random variable X k Y k Z k ,

[0390]

[0391] Where: X k Y k Z k Let X be a three-dimensional uniformly distributed random variable; pdf(X) k ,Y k Z k Let f(x) be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution, where a = 1, b = 3, c = 1, d = 3, e = -2, and g = 0.

[0392] B.2 Calculate the density at points of random variables

[0393] Let x = X k y = Y k z = Z k Calculate equations (12) to (21) to obtain the earthwork volume at the random variable point (X). k ,Y k Z k Density at )

[0394]

[0395]

[0396]

[0397] D 00 =D 000 (1-ξ)+D 100 ξ (15)

[0398] D 01 =D 001 (1-ξ)+D 101 ξ (16)

[0399] D 10 =D 010 (1-ξ)+D 110 ξ (17)

[0400] D 11 =D 011 (1-ξ)+D 111 ξ (18)

[0401] D0 = D 00 (1-η)+D 10 η (19)

[0402] D1 = D 01 (1-η)+D 11 η (20)

[0403] D=D0(1-ζ)+D1ζ (21)

[0404] Where: ξ, η, and ζ are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid, respectively; D 00 D 01 D 10 D 11 D0 and D1 are intermediate variables defined by equations (15) to (20); D is the density at (x,y,z).

[0405] With (X) k ,Y k Z k )=(2,2,-1.5) as an example, ξ=0.5, η=0.5, ζ=0.625, D 00 =20.5945,D 01 =15.3216,D 10 =20.6521,D 11 =15.4569,D0=20.6233,D1=15.3893,D=17.3520

[0406] B.3 Determine whether the random variable point is within the evaluation region.

[0407] Calculate the random variable point (X) k ,Y k Z k ) and node N i (x i ,y i ,z i The distance between them

[0408]

[0409] Where: d i For random variable point (X) k ,Y k Z k ) and the i-th node N i (x i ,y i,z i The distance between them

[0410] d i The values ​​are: 2.0615, 2.0615, 1.8027, and 0.5 respectively.

[0411] Find the nearest node N nearest (x nearest ,y nearest ,z nearest ),

[0412] d nearest =min{d i} (twenty three)

[0413] Where: d nearest For random variable point (X) k ,Y k Z k The distance between N and the nearest neighbor node; nearest The nearest node number; (x nearest ,y nearest ,z nearest ) represents the coordinates of the nearest node, d nearest =0.5, N nearest =4

[0414] Find the node containing node N nearest (x nearest ,y nearest ,z nearest The triangular plane E) near.j (N nearest N i2 N i3 ) or E near.j (N i1 N nearest N i3 ) or E near.j (N i1 N i2 N nearest ), whose set is {E near.j |E near.j (N i1 N i2 N i3 ),N nearest ∈N i1 ∪N i2 ∪N i3}, determine the point (X) respectively k ,Y k Z k Whether a point (X) is on the Ω side of each element in the set depends on the point (X). k ,Y k Zk If the random variable point (X) is located on one side of Ω of all elements in the set, then determine the random variable point (X). k ,Y k Z k If the value is within Ω, then determine the random variable point (X). k ,Y k Z k )exist Inside, there is node N. nearest (x nearest ,y nearest ,z nearest The triangle planes are numbered 2, 3, and 4. Point (X) k ,Y k Z k The point (X) lies on one side of the plane of triangle 2, on one side of the plane of triangle 3, and on one side of the plane of triangle 4. Determine the random variable point (X). k ,Y k Z k It is not within Ω.

[0415] B.4 Calculate the three-dimensional integrand at points of the random variable.

[0416]

[0417] Where: f(X) k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at point (X) is f(X). k ,Y k Z k ) = 17.3520

[0418] End of loop,

[0419] C. Estimate the weight of earthwork.

[0420] Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral.

[0421]

[0422] Where: n is the total number of samplings; k refers to the kth sampling; It is an estimate of the three-dimensional Monte Carlo integral over the three-dimensional exact integral, i.e., the weight of the earthwork being evaluated. The unit of weight is kN.

[0423] The judgment of random variable point (X) k ,Y k Zk ) in the Ω or triangle plane The specific method on one side is as follows:

[0424] Given three nodes N in the plane of a triangle. L (x L ,y L ,z L ), N M (x M ,y M ,z M ), N N (x N ,y N ,z N ),from Looking towards Ω, these three nodes are arranged counterclockwise, and the triangular plane extends from Ω towards... The components of the normal vector in the direction of are,

[0425]

[0426] in: They are respectively Components in the x, y, and z directions; For the triangular plane from Ω to The normal vector in the direction; (x) L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (x M ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N Cartesian coordinates,

[0427] With (X) k ,Y k Z k Taking the plane of triangle 2 (x) = (2, 2, -1.5) as an example. L ,y L ,z L )=(1,1,0),(x M ,y M ,z M )=(2,2,-2),(x N ,y N ,z N ) = (3,1,0).

[0428] When equation (27) is 1, the sampling point is at When equation (27) is -1, the sampling point is within Ω.

[0429]

[0430] Where: sign[·] is the sign function. The sampling point is within Ω.

[0431] The maximum load capacity of a certain trailer is 1 ton. The total weight of the earth and rock excavated is 21.3169 kN, approximately equal to 2.13169 tons. It would take 3 trips to complete the transportation of all the earth and rock using this trailer.

[0432] When the density of the earth and stone is uneven, if the traditional analytical method is used, the total volume of the earth and stone is first calculated to be 8m³. 3 However, the densities at each survey point are different, specifically: 1(15), 2(16), 3(15.5), 4(15.4), 5(18), 6(19), 7(19.5), 8(20), and 9(20.4), in kN / m³. 3 Since density is not constant, previous analytical methods cannot be used for further calculations. Therefore, previous analytical methods cannot account for the spatial variation of earthwork density.

[0433] In summary, the method of this invention can take into account the spatial variation of earthwork density, thus meeting the needs of engineering applications.

Claims

1. A random sampling evaluation method for earthwork weight, characterized in that: Includes the following steps, A. Preprocessing A.1, Encoding Node The earthwork weight evaluation area is simplified to a region enclosed by at least four triangular planes, named Ω, and the coordinates of the nodes in this region are (x, y). i ,y i ,z i ), where i refers to the i-th node, and x, y, z are rectangular coordinates, and x∈[min(x i ),max(x i )]、y∈[min(y i ),max(y i )]、z∈[min(z i ),max(z i The region is named U. The region is inside U and outside Ω. Encode the coordinates of the nodes. Where: N i Number the nodes; n V The total number of nodes in Ω A.2, Encoding the Triangular Plane Encoding the triangular plane, E j (N i1 N i2 N i3 ), where: j refers to the j-th triangular plane; N i1 N i2 N i3 Represents the node number, N i1 ≠N i2 ≠N i3 E j The numbering of the triangular plane, from Looking towards Ω, the three points within the triangle plane are arranged counter-clockwise, according to the right-hand rule. Check if the coded triangle plane is correct. n V -n E +n F =2 (1) Where: n E n is the number of edges of Ω; F Let be the number of faces of the triangle plane of Ω. If equation (1) is satisfied, then the triangle plane is correctly encoded. A.3, Coding Survey Points Based on the earthwork survey results, the density of survey points at least 8 different locations was obtained, and the coordinates of the survey points were coded. N si (x i ,y i ,z i )=1(x1,y1,z1),2(x2,y2,z2),...8(x2,y2,z2),...,n s (x ns ,y ns ,z ns ), Where: N si For the survey point number; n s This represents the total number of survey points. The density of the survey points is coded, N si (D si )=1(D s1 ),2(D s2 ),...8(D s8 ),...,n s (D sns ), Where: D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ,...,D si ,...,D sns The density at the survey point. A.4 Calculate the density at the 8 vertices of the trilinear interpolation cuboid. Equations (2), (3), (4), (5), (6), (7), (8), (9), and (10) are used to calculate the density at the eight vertices of the trilinear interpolation cuboid. Where: x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 ... represent the x-axis rectangular coordinates of the survey points; y... s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 ... represents the y-axis rectangular coordinates of the survey point; z... s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 ... represents the z-axis rectangular coordinate of the survey point, x... a For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The lower limit of ,...; y a For y s1 ,y s2 ,y s3 ,y s4 ,y s5 ,y s6 ,y s7 ,y s8 The lower limit of z; a For z s1 ,z s2 ,z s3 ,z s4 ,z s5 ,z s6 ,z s7 ,z s8 The lower limit of ...; x b For x s1 ,x s2 ,x s3 ,x s4 ,x s5 ,x s6 ,x s7 ,x s8 The upper limit of ,...; y b For y s1 , y s2 , y s3 , y s4 , y s5 , y s6 , y s7 , y s8 , an upper limit value of ...; z b is z s1 , z s2 , z s3 , z s4 , z s5 , z s6 , z s7 , z s8 , an upper limit value of ..., Where: x si Let x be the rectangular coordinate of the si-th survey point; y si Let z be the y-axis rectangular coordinate of the si-th survey point; si Let ξ be the z-axis rectangular coordinate of the si-th survey point; si η si ζ si These are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid. Among them: A i1 ~A i8 For the intermediate variable defined by equation (6), Wherein: D α = [D 000 , D 100 , D 010 , D 110 , D 001 , D 101 , D 011 , D 111 T (8) A ij =[A i1 ,A i2 ,A i3 ,A i4 ,A i5 ,A i6 ,A i7 ,A i8 ] (9) D s =[D s1 ,D s2 ,D s3 ,D s4 ,D s5 ,D s6 ,D s7 ,D s8 ,...] T (10) Where: D 000 D 001 D 010 D 011 D 100 D 101 D 110 D 111 The density at the 8 vertices of the trilinear interpolation cuboid; D s1 D s2 D s3 D s4 D s5 D s6 D s7 D s8 ... represents the density at the survey point; D α A is the vector defined by equation (8); ij The matrix defined by equation (9); D s The vector defined by equation (10), B. Cyclic Sampling The preset total number of sampling times is n. If the number of iterations has not reached the preset total number of samples n, steps B.1, B.2, B.3, and B.4 will be executed repeatedly. B.

1. Generating random variable points Take a=min(x i b = max(x) i c = min(y) i ), d = max(y i e = min(z) i g = max(z) i Generate a three-dimensional uniformly distributed random variable X k Y k Z k , Where: X k Y k Z k Let X be a three-dimensional uniformly distributed random variable; pdf(X) k ,Y k Z k Let be the probability density function of a three-dimensional uniformly distributed random variable; a, c, and e are the lower bound parameters of the three-dimensional uniform distribution; b, d, and g are the upper bound parameters of the three-dimensional uniform distribution. B.2 Calculate the density at points of random variables Let x = X k y = Y k z = Z k Calculate equations (12) to (21) to obtain the earthwork volume at the random variable point (X). k ,Y k Z k Density at ) D 00 =D 000 (1-ξ)+D 100 ξ (15) D 01 =D 001 (1-ξ)+D 101 ξ (16) D 10 =D 010 (1-ξ)+D 110 ξ (17) D 11 =D 011 (1-ξ)+D 111 ξ (18) D0=D 00 (1-h)+D 10 the (19) D1=D 01 (1-h)+D 11 the (20) D=D0(1-ζ)+D1ζ (21) Where: ξ, η, and ζ are the relative coordinates of points x, y, and z within the trilinear interpolation cuboid, respectively; D 00 D 01 D 10 D 11 D0 and D1 are intermediate variables defined by equations (15) to (20); D is the density at (x,y,z). B.3 Determine whether the random variable point is within the evaluation region. Calculate the random variable point (X) k ,Y k Z k ) and node N i (x i ,y i ,z i The distance between them Where: d i For random variable point (X) k ,Y k Z k ) and the i-th node N i (x i ,y i ,z i The distance between them Find the nearest node N nearest (x nearest ,y nearest ,z nearest ), the nearest =min{d i } (23) Where: d nearest For random variable point (X) k ,Y k Z k The distance between N and the nearest neighbor node; nearest The nearest node number; (x nearest ,y nearest ,z nearest ) represents the coordinates of the nearest node. Find the node containing node N nearest (x nearest ,y nearest ,z nearest The triangular plane E) near.j (N nearest N i2 N i3 ) or E near.j (N i1 N nearest N i3 ) or E near.j (N i1 N i2 N nearest ), whose set is {E near.j |E near.j (N i1 N i2 N i3 ),N nearest ∈N i1 ∪N i2 ∪N i3 }, determine the point (X) respectively k ,Y k Z k Whether a point (X) is on the Ω side of each element in the set depends on the point (X). k ,Y k Z k If the random variable point (X) is located on one side of Ω of all elements in the set, then determine the random variable point (X). k ,Y k Z k If the value is within Ω, then determine the random variable point (X). k ,Y k Z k )exist Inside, B.4 Calculate the three-dimensional integrand at points of the random variable. Where: f(X) k ,Y k Z k ) is a random variable point (X) k ,Y k Z k The three-dimensional integrand at () End of loop, C. Estimate the weight of earthwork. Three-dimensional Monte Carlo integration is used to estimate the three-dimensional exact integral. Where: n is the total number of samplings; k refers to the kth sampling; It is an estimate of the three-dimensional Monte Carlo integral over the three-dimensional exact integral, i.e., the weight of the earthwork being evaluated.

2. The random sampling evaluation method for earthwork weight according to claim 1, characterized in that: Determine the point (X) of the random variable k ,Y k Z k ) in the Ω or triangle plane The specific method on one side is as follows: Given three nodes N in the plane of a triangle. L (x L ,y L ,z L ), N M (x M ,y M ,z M ), N N (x N ,y N ,z N ),from Looking towards Ω, these three nodes are arranged counterclockwise, and the triangular plane extends from Ω towards... The components of the normal vector in the direction of are, in: They are respectively Components in the x, y, and z directions; For the triangular plane from Ω to The normal vector in the direction; (x) L ,y L ,z L ) is a node N on the triangular plane. L Cartesian coordinates; (x M ,y M ,z M ) is a node N on the triangular plane. M Cartesian coordinates; (x N ,y N ,z N ) is a node N on the triangular plane. N Cartesian coordinates, When equation (27) is 1, the sampling point is at When equation (27) is -1, the sampling point is within Ω. Where: sign[·] is the sign function.

3. The random sampling evaluation method for earthwork weight according to claim 1, characterized in that: Evaluating the volume of earthwork includes the following steps. Set D = 1, and execute steps A, A.1, A.2, B, B.1, B.3, B.4, and C. Numerically, it is equal to the volume of earth and stone.

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