A method for constructing a transition material model for dam filling on the riverbank based on a web-based platform.

Abstract

This invention provides a web-based method for constructing a transition material model for dam filling, relating to the field of model building. It solves the problem of creating a transition material model for dam filling. The method utilizes a web-based approach to create the transition material model, establishing a 3D terrain foundation model of the dam and employing Boolean algorithms to generate the transition material model. This rapidly generates a transition material model that accurately reflects the actual terrain, facilitating subsequent applications and maximizing the value of BIM models. It also addresses the challenges of creating models for complex and special parts such as dam filling transition materials, playing a crucial role in supporting and guiding construction using BIM models.

Description

Technical Field

[0001] This invention relates to the field of model creation, and in particular to a method for constructing a model of transition material on the bank of a dam filling project based on a web-based platform. Background Technology

[0002] Earth-rock dam construction is characterized by complex material area planning, a wide range of fill material properties, and complex and variable slopes on both banks. Therefore, quality control of fill material in each area during dam construction is crucial for the overall safe operation of the dam. In earth-rock dam construction, the bank transition material, as the transition material at the junction of the large-diameter fill material (such as rockfill) and the slopes on both banks, plays a vital role in the stability of the overall mechanical properties of the dam. Therefore, special treatment is required for the compaction of this area during construction. From the perspective of model construction and application, as the bank transition material is in close contact with the topography on both banks, the topography changes significantly with the continuous changes in the dam's filling elevation. It is difficult to construct a bank transition material model that accurately reflects the actual site conditions using conventional 3D design software. This hinders subsequent model-based statistics on fill material and model application demonstrations. Considering that previous statistics on bank transition material filling volume were mainly estimated based on human field experience, the results often had large deviations and could not effectively guide on-site construction.

[0003] Existing desktop 3D design software suffers from insufficient support for localized and professional applications due to limitations of its native platform, and its operation is complex. Furthermore, given the complex variations in riverbank slopes, it cannot achieve accurate and timely construction of transition material models for riverbanks. On the other hand, due to limitations in its modeling principles, the generated model results occupy a large amount of data space, severely restricting the application scenarios of the models and resulting in low model reuse rates. Summary of the Invention

[0004] This invention proposes a web-based method for constructing a transition material model for dam filling, which solves the problem of creating a transition material model during dam filling and lays an important foundation for the rapid and accurate generation and volume calculation of the transition material filling model.

[0005] This invention provides a method for constructing a transition material model for dam filling on the bank based on a web interface. The method includes the following steps:

[0006] Step 101: Obtain the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain basic model based on the web interface;

[0007] Step 102: Based on the web interface, select at least two different coordinate points Q within the range of the three-dimensional terrain base model on the left bank boundary line L1. L1-n Q L1-(n+1) Extract the l-axis between two adjacent points on L1.1-n Line segment, input the filling width W of the transition material on the left bank, so that the polyline l 1-n Offset by a distance W along the dam axis towards the opposite bank to obtain a polyline l' 1-n First offset path W n Second offset path W` n ; composed of the multi-segment line l 1-n polyline l` 1-n First offset path W n Second offset path W` n Forming the first closed region β n， Where n is greater than or equal to 1.

[0008] Step 103: Based on the web interface, the closed region β n Stretch downward along the Z-axis Z n The distance is used to obtain the second closed region β`. n ; First closed region β n The second closed region β` n Distance Z of the stretching path n Composition of the tensile model V n The stretching model V n Perform a Boolean operation on the left bank geometric surface γ1 of the three-dimensional terrain base model to obtain the left bank transition material model. Repeat the operation to obtain the left bank transition material model {V}. 1-1 `、V 1-2 `、……V1- n Similarly, the transition material model {V} on the right bank is obtained. 2-1 `、V 2-2 `、……V 2-n `}

[0009] Preferably, the stretching model V n Performing Boolean operations with the 3D terrain base model results in the stretched model V. n After performing Boolean operations with the left bank geometry γ1 of the 3D terrain base model, retain the closed volume model V on the side closest to the center of the river channel. n ` ， The closed-loop model V on one side of the river channel center n ` is a transition material model.

[0010] Among them, the stretching model V n After performing Boolean operations with the left bank geometry γ1 of the 3D terrain base model, retain the closed volume model V on the side closest to the center of the river channel. n It involves traversing all triangular faces of the left bank geometry γ1 of the basic 3D terrain model and traversing the stretched model V. n Extract the triangular facets {γ} on the left bank geometric facet γ1 of the three-dimensional terrain base model from all triangular facets.1-1 γ 1-2 γ 1-3 ...γ 1-n}、Stretch model V n The triangular facet on the V 1-1 V 1-2 V 1-3 ...V 1-n}, the triangular facet V of the tension model on the left bank of the dam 1-n Triangular facet γ on the left bank geometric surface γ1 1-n Find the Boolean intersection line l n 1 Similarly, the Boolean intersection line {l} is obtained sequentially. n 2 l n 3 ...l n n}, the Boolean intersection line { l n 1 l n 2 ...l n n Connecting them sequentially from left to right according to the principle of sharing adjacent endpoints, they form the Boolean intersection line L1 on the left bank of the dam. 1 Similarly, the Boolean intersection line L2 on the right bank of the dam is obtained. 1 .

[0011] Preferably, through Boolean intersection line L1 1 The left bank geometry γ1 of the 3D terrain base model and the stretched model V n The triangular facets are cut to form two different types of triangular facets, namely {ξ1, ξ2, ξ3, ... ξ}. n}, triangular patch {ξ`1, ξ`2, ξ`3,…ξ` n}; at the Boolean intersection L1 1 Find the minimum straight line segment SL, and from SL, perform a process on all triangular faces {ξ1, ξ2, ξ3} that it passes through. 2. ...ξ n}, triangular facets {ξ`1, ξ`2, ..., ξ` n} and the uncut triangular face {γ 1-1 γ 1-2 γ 1-3 ...γ 1-n Uncut triangular facets in}, extruded model triangular facets { V 1-1 V 1-2 V 1-3 ...V 1-nBy traversing along the Boolean intersection, m continuous closed surfaces F are formed on the left bank of the dam, where m is greater than or equal to 1.

[0012] Preferably, among m continuous closed surfaces F, surfaces f that can constitute the geometry within the target material area are selected and connected end-to-end to form a closed body V. n `, the closed body V n `Model V1 is the transition material for filling the left bank of the dam; similarly, model V2 is obtained for filling the right bank of the dam.

[0013] Preferably, the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain basic model are obtained through the following steps:

[0014] Step 201: Establish a three-dimensional terrain foundation model and create a three-dimensional reference coordinate system online based on the design data of the dam and its foundation excavation.

[0015] Step 202: After assigning the dam planning elevation value to the three-dimensional reference coordinate system, obtain plane β1, and the left bank geometric surface γ1 and right bank geometric surface γ2 of plane β1 relative to the three-dimensional terrain base model. 1 Perform Boolean operations to obtain the left boundary line L1 and the right bank boundary line L2 of the three-dimensional terrain basic model;

[0016] Preferably, Boolean operations are performed to obtain the left boundary line L1 and right bank boundary line L2 of the 3D terrain base model, which is used to extract the triangular mesh model of the left bank geometric surface γ1 and the right bank geometric surface γ2 of the 3D terrain base model. 1 The triangular mesh model of the left bank geometry γ1 is obtained by traversing all triangular faces on the left bank geometry γ1 and the plane β1, and extracting the triangular faces {γ1} on the left bank geometry γ1. 1-1 γ 1-2 γ 1-3 ...γ 1-n}, the triangular facet {β} on plane β1 1-1 β 1-2 β 1-3 ...β 1-n}, calculate the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n Similarly, we obtain n intersection lines, and then divide the n intersection lines {l1, l2, l3, ... l... n Connect them sequentially from left to right to obtain the left bank boundary line L1; similarly, the algorithm obtains the right bank boundary line L2.

[0017] Preferably, the triangular facet γ on the left bank geometric surface γ1 is calculated. 1-n With triangular facet β 1-n The intersection line ln Previously, it was also necessary to calculate the triangular facet γ on the left bank geometric surface γ1. 1-n The distance from the vertex to plane β1;

[0018] Triangular facet γ on the left bank geometric surface γ1 1-n triangular facet β on plane β1 1-n The planes they lie in are the γ2 plane and the β1 plane, respectively. The general equation for the β1 plane is expressed as:

[0019] N1·X1+K1=0(1)

[0020] Where N1 is the normal vector of the β1 plane, X1 is any point on β1, and K1 is a constant;

[0021] The plane γ2 can be expressed using the general equation of a plane as: N2·X2+K2=0(2)

[0022] Where N2 is the normal vector of the γ2 plane, X2 is any point on the γ2 plane, and K2 is a constant;

[0023] The equation for the intersection line L of the β1 plane and the γ2 plane is:

[0024] L = D· t+O(3)

[0025] Where D = N1 × N2, D is the direction vector of the intersection line L, t is the parameter of the equation, O is any point of L, N1 is the normal vector of plane β1, and N2 is the normal vector of plane γ2.

[0026] Triangular facet γ 1-n The distance from the vertex to plane β1 is expressed as:

[0027] d Vi1 = (N1·V) i 1 + K1) / | N1|,(4)

[0028] Where i = 0, 1, 2; V i 1 For triangular facet γ 1-n The vertex of β1, and N1 is the normal vector of plane β1;

[0029] The triangular facet γ on the left bank geometric surface γ1 1-n Vertex V i 1 Substituting equation (1) into equation (4) yields the triangular facet γ. 1-n The distance d from the vertex to β1 Vi1 .

[0030] Preferably, the triangular facet γ on the left bank geometric surface γ11-n The distance from the vertex to plane β1 is used to obtain the triangular facet γ. 1-n With triangular facet β 1-n The intersection line l n It is based on the triangular facet γ on the left bank geometric surface γ1. 1-n Vertex V i 1 Distance d to plane β1 Vi1 The calculation results indicate the existence of an intersection line; the triangular facet γ on the left bank geometric surface γ1... 1-n Vertex V i 1 Distance d to plane β1 Vi1 After judging against the preset conditions, the Boolean intersection line l is calculated. n The preset condition is judged as follows:

[0031] When d Vi1 When the result of the operation is not equal to 0 and the sign of the operation is opposite, the triangular facet γ on the left bank geometric surface γ1 is determined. 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1... 1-n It intersects with plane β1;

[0032] When d Vi1 ≠0 (i=0,1,2), and the operation results have the same sign, the triangular facet γ on the left bank geometric surface γ1 1-n Located on one side of plane β1, the triangular facet γ on the left bank geometric surface γ1 is... 1-n It will not intersect with line L;

[0033] When d Vi1 =0 (i=0,1,2), triangular facet γ on the left bank geometric surface γ1 1-n On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It has no intersection with plane β1;

[0034] Based on the triangular facet γ on the left bank geometric surface γ1 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 is then calculated. 1-n The intersecting scalar interval on the intersection line L is projected onto the line through the vertex of the triangle:

[0035] P Vi1 = D · (V i 1 - O), (5)

[0036] i = 0, 1, 2; D is the direction vector of the intersection line L; O is any point on L.

[0037] V i1 Given a point in the β1 plane, its projection onto β1 is given by the principle of similar triangles.

[0038] (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 The equation parameter t1 is derived.

[0039] t1 = P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 (6)

[0040] Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived.

[0041] t2 = P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 (7)

[0042] Substituting parameters t1 and t2 into equation (3), the linear equation L = D·t + O, yields the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n The two endpoints:

[0043] l n-1 = D· t1+O,

[0044] l n-2 = D· t2+O,

[0045] Similarly, we can obtain {l1, l2, l3, ... l} n The intersection line will connect {l1, l2, l3, ... l... n The intersection lines are connected in sequence to obtain the left bank boundary line L1 of plane β1 on the left bank geometric surface γ1 of the three-dimensional terrain basic model;

[0046] Similarly, calculate plane β1 and the right bank geometry γ1 of the three-dimensional terrain base model. 1 The Boolean intersection line is used to obtain the right bank boundary line L2 of the three-dimensional terrain base model.

[0047] Similarly, calculate the triangular facet V of the tension model on the left bank of the dam. 1-n Triangular facet γ on the left bank geometric surface γ1 1-n Find the Boolean intersection line l n 1 In finding the Boolean intersection line l n 1 Before that, it is also necessary to calculate the triangular facet γ on the left bank geometric surface γ1. 1-n Vertex V i 1 Distance d to plane β1 Vi1， After comparing it with preset conditions, the Boolean intersection line l is calculated. n 1 ;

[0048] The preset conditions are judged as follows:

[0049] When d Vi1 When the result of the operation is not equal to 0 and the sign of the operation is opposite, the triangular facet γ on the left bank geometric surface γ1 is determined. 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 1-n It intersects with plane β1;

[0050] When d Vi1 ≠0 (i=0,1,2), and the operation results have the same sign, the triangular facet γ on the left bank geometric surface γ1 1-n Located on one side of plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It will not intersect with line L;

[0051] When d Vi1 =0 (i=0,1,2), the triangular facet γ on the left bank geometric surface γ1 1-n On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It has no intersection with plane β1;

[0052] According to the triangular facet γ on the left bank geometric surface γ1 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 is then calculated. 1-n The intersecting scalar interval on the intersection line L is projected onto the line through the vertex of the triangle:

[0053] P Vi1 = D · (V i1 - O), (5)

[0054] i = 0, 1, 2; D is the direction vector of the intersection line L; O is any point on L.

[0055] V i1 Given a point in the β1 plane, its projection onto β1 is given by the principle of similar triangles.

[0056] (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 The equation parameter t1 is derived.

[0057] t1 = P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 (6)

[0058] Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived.

[0059] t2 = P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 (7)

[0060] Substituting the parameters t1 and t2 into the linear equation L = D·t + O in equation (3), we obtain the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n 1 The two endpoints:

[0061] l n-1 1 = D· t1+O,

[0062] l n-2 1 = D· t2+O,

[0063] Similarly, the intersection line l can be obtained. n 1 Similarly, find the intersection line {l} in turn. n 2 l n 3 ...l n n}

[0064] This invention provides a web-based method for constructing a transition material model for dam filling on the riverbank. Relating to the field of model building, it solves the problem of creating a transition material model for the riverbank during dam filling. Based on a web-based approach, the method establishes a 3D terrain foundation model of the dam and uses Boolean algorithms to obtain the transition material model for the dam filling on the riverbank. This rapidly generates a transition material model that conforms to the actual terrain on site, facilitating subsequent applications and maximizing the application value of BIM models. It also solves the previous difficulties in creating models for special and complex parts such as transition materials for dam filling, playing a crucial role in assisting and guiding construction using BIM models.

[0065] It should be understood that the above general description and the following detailed description are exemplary and explanatory only, and are not intended to limit the invention. Attached Figure Description

[0066] The accompanying drawings, which are incorporated in and form part of this specification, illustrate embodiments consistent with the invention and, together with the description, serve to explain the principles of the invention.

[0067] Figure 1 This is a flowchart of a method for constructing a transition material model for dam filling on the bank based on a web interface, provided by an embodiment of the present invention.

[0068] Figure 2 This invention provides a method for constructing a web-based model of transition material for dam filling, specifically for the closed region β. n Schematic diagram;

[0069] Figure 3 This invention provides a method for constructing a web-based model of transition material for dam filling on the riverbank, specifically a stretched model V. n Schematic diagram;

[0070] Figure 4 This invention provides a method for constructing a transition material model for dam filling on the riverbank based on a web-based platform, including a terrain model segmentation and stretching model V. n Schematic diagram;

[0071] Figure 5 This is a schematic diagram of the left and right bank boundary lines of a method for constructing a transition material model for dam filling on a web-based platform, according to an embodiment of the present invention.

[0072] Figure 6 This is a schematic diagram of the geometric relationship between γ2 and β1 in a method for constructing a transition material model for dam filling on the bank based on a web interface, provided by an embodiment of the present invention.

[0073] Figure 7This is a method for constructing a model of transitional material on the bank of a dam filling project based on a web-based interface. (Boolean intersection diagram) Detailed Implementation

[0074] Exemplary embodiments will now be described in detail, examples of which are illustrated in the accompanying drawings. When the following description relates to the drawings, unless otherwise indicated, the same numbers in different drawings denote the same or similar elements. The embodiments described in the following exemplary embodiments do not represent all embodiments consistent with the present invention.

[0075] Example 1

[0076] This invention provides a method for constructing a model of transition material on the bank of a dam filling site based on a web-based interface. The method is as follows: Figure 1 As shown, the method includes the following steps:

[0077] Step 101: Obtain the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain basic model based on the web interface;

[0078] Step 102: Based on the web interface, select at least two different coordinate points Q within the range of the three-dimensional terrain base model on the left bank boundary line L1. L1-n Q L1-(n+1) Extract the l-axis between two adjacent points on L1. 1-n Line segment, input the filling width W of the transition material on the left bank, so that the polyline l on the bank... 1-n Offset by a distance W along the dam axis towards the opposite bank to obtain the polyline l' of the river channel. 1-n First offset path W n Second offset path W` n ; composed of multiple lines l along the shore 1-n Multi-segment line of the river channel 1-n First offset path W n Second offset path W` n Forming the first closed region β n， Where n is greater than or equal to 1.

[0079] 103. Based on the web interface, the first closed region β n Stretch downward along the Z-axis Z n The distance is used to obtain the second closed region β`. n The first closed region β n The second closed region β` n Distance Z of the stretching path n Composition of the tensile model V n The stretching model V nPerform a Boolean operation on the left bank geometric surface γ1 of the three-dimensional terrain base model to obtain the left bank transition material model V1`. Repeat the operation to obtain the left bank transition material model {V 1-1 `、V 1-2 `、……V1- n Similarly, the transition material model {V} on the right bank is obtained. 2-1 `、V 2-2 `、……V 2-n `}

[0080] In one embodiment, the stretching model V n Performing Boolean operations with the 3D terrain base model results in the stretched model V. n After performing Boolean operations with the left bank geometry γ1 of the 3D terrain base model, retain the closed volume model V on the side closest to the center of the river channel. n ` ， The closed-loop model V on one side of the river channel center n ` is a transition material model.

[0081] Among them, the stretching model V n After performing Boolean operations with the left bank geometry γ1 of the 3D terrain base model, retain the closed volume model V on the side closest to the center of the river channel. n ` is the traversal of all triangular faces of the left bank geometric surface γ1 of the three-dimensional terrain base model, and the traversal of the stretched model V n Extract the triangular facets {γ} on the left bank geometric facet γ1 of the three-dimensional terrain base model from all triangular facets. 1-1 γ 1-2 γ 1-3 ...γ 1-n} , stretching model V n The triangular facet on the V 1-1 V 1-2 V 1-3 ...V 1-n}, the triangular facet V of the stretched model 1-n Triangular facet γ on the left bank geometric surface γ1 1-n Find the Boolean intersection line l n 1 Similarly, the Boolean intersection line {l} is obtained sequentially. n 2 l n 3 ...l n n}, the intersection line { l n 1 l n 2 ...l n nConnecting them sequentially from left to right according to the principle of sharing adjacent endpoints, they form the Boolean intersection line L1 on the left bank of the dam. 1 Similarly, the Boolean intersection line L2 on the right bank of the dam is obtained. 1 .

[0082] In one embodiment, through the Boolean intersection line L1 1 The left bank geometry γ1 of the 3D terrain base model and the stretched model V n The triangular facets passed through are cut to form two different triangular facets, namely {ξ1, ξ2, ξ3, ... ξ}. n}, triangular patch {ξ`1, ξ`2, ξ`3,…ξ` n}; at the Boolean intersection L1 1 Find the minimum straight line segment SL, and from SL, perform a process on all triangular faces {ξ1, ξ2, ξ3} that it passes through. 2. ...ξ n}、{ξ`1、ξ`2、……ξ` n} and triangular face {γ 1-1 γ 1-2 γ 1-3 ...γ 1-n Uncut triangular facets in}, extruded model triangular facets { V 1-1 V 1-2 V 1-3 ...V 1-n By traversing along the Boolean intersection, m continuous closed surfaces F are formed on the left bank of the dam, where m is greater than or equal to 1.

[0083] In one embodiment, on the left bank, among the m continuous closed surfaces F, surfaces f that can constitute the geometry within the target material area are connected end to end to form a closed body model V. n The closed-body model V n Model V1` is the transition material for filling the left bank of the dam; similarly, model V2` is obtained for filling the right bank of the dam.

[0084] In one embodiment, the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain base model are obtained via a web interface through the following steps:

[0085] 201. Based on the design data of the dam and dam foundation excavation construction, establish a three-dimensional terrain foundation model and establish a three-dimensional reference coordinate system online;

[0086] 202. After assigning the planned elevation value of the dam to the three-dimensional reference coordinate system, obtain plane β1. Plane β1 corresponds to the left bank geometric surface γ1 and the right bank geometric surface γ1 of the three-dimensional terrain base model. 1 Perform Boolean operations to obtain the left boundary line L1 and the right bank boundary line L2 of the three-dimensional terrain basic model;

[0087] In one embodiment, performing Boolean operations to obtain the left boundary line L1 and right bank boundary line L2 of the 3D terrain base model involves extracting the plane β1 and the left bank geometric surface γ1 triangular mesh model and the right bank geometric surface γ1 of the 3D terrain base model. 1 The triangular mesh model of the left bank geometric surface γ1 is obtained by traversing all triangular faces on the left bank geometric surface γ1 and the plane β1, and extracting n triangular faces {γ1} on the left bank geometric surface γ1. 1-1 γ 1-2 γ 1-3 ...γ 1-n}, n triangular facets {β} on plane β1 1-1 β 1-2 β 1-3 ...β 1-n}, calculate the triangular facet γ on the left bank geometric surface γ1. 1-n triangular facet β on plane β1 1-n The intersection line l n Similarly, we obtain n intersection lines, and then divide the n intersection lines {l1, l2, l3, ... l... n Connect them sequentially from left to right to obtain the left bank boundary line L1; similarly, the algorithm obtains the right bank boundary line L2.

[0088] Preferably, the triangular facet γ on the left bank geometric surface γ1 is calculated. 1-n triangular facet β on plane β1 1-n The intersection line l n Previously, it was also necessary to calculate the triangular facet γ on the left bank geometric surface γ1. 1-n The distance from the vertex to plane β1;

[0089] Triangular facet γ on the left bank geometric surface γ1 1-n triangular facet β on plane β1 1-n The planes they lie in are the γ2 plane and the β1 plane, respectively. The general equation of the β1 plane is expressed as: N1·X1+K1=0(1)

[0090] Where N1 is the normal vector of the β1 plane, X1 is any point on β1, and K1 is a constant;

[0091] The plane γ2 can be expressed using the general equation of a plane as: N2·X2+K2=0(2)

[0092] Where N2 is the normal vector of the γ2 plane, X2 is any point on the γ2 plane, and K2 is a constant;

[0093] The equation for the intersection line L of the β1 plane and the γ2 plane is:

[0094] L = D· t+O(3)

[0095] Where D = N1 × N2, D is the direction vector of the intersection line L, t is the parameter of the equation, O is any point of L, N1 is the normal vector of plane β1, and N2 is the normal vector of plane γ2.

[0096] Triangular facet γ 1-n The distance from vertex to β1 is expressed as:

[0097] d Vi1 = (N1·V) i 1 + K1) / | N1|,(4)

[0098] Where i = 0, 1, 2; V i 1 For triangular facet γ 1-n The vertex of β1, and N1 is the normal vector of plane β1;

[0099] The triangular facet γ on the left bank geometric surface γ1 1-n Vertex V i 1 Substituting equation (1) into equation (4) yields the triangular facet γ. 1-n The distance d from the vertex to β1 Vi1 .

[0100] In one embodiment, the triangular facet γ on the left bank geometric surface γ1 1-n The distance from the vertex to plane β1 is used to obtain the triangular facet γ on the left bank geometry γ1. 1-n With triangular facet β 1-n The intersection line l n It is based on the triangular facet γ on the left bank geometric surface γ1. 1-n Vertex V i 1 Distance d to plane β1 Vi1 The calculation results indicate the existence of an intersection line; the triangular facet γ on the left bank geometric surface γ1... 1-n Vertex V i 1 Distance d to plane β1 Vi1 After judging against the preset conditions, the intersection line l is calculated. n The preset condition is judged as follows:

[0101] When d Vi1 When the result of the operation is not equal to 0 and the sign of the operation is opposite, the triangular facet γ on the left bank geometric surface γ1 is determined. 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1... 1-n It intersects with plane β1;

[0102] When d Vi1 ≠0 (i=0,1,2), and the operation results have the same sign, the triangular facet γ on the left bank geometric surface γ1 1-n Located on one side of plane β1, the triangular facet γ on the left bank geometric surface γ1 is... 1-n It will not intersect with line L;

[0103] When d Vi1 =0 (i=0,1,2), triangular facet γ on the left bank geometric surface γ1 1-n On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It has no intersection with plane β1;

[0104] Based on the triangular facet γ on the left bank geometric surface γ1 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 is then calculated. 1-n The intersecting scalar interval on the intersection line L is projected onto the line through the vertex of the triangle:

[0105] P Vi1 = D · (V i 1 - O), (5)

[0106] i = 0, 1, 2; D is the direction vector of the intersection line L; O is any point on L.

[0107] V i 1 Given a point in the β1 plane, its projection onto β1 is given by the principle of similar triangles.

[0108] (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 The equation parameter t1 is derived.

[0109] t1 = P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 (6)

[0110] Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived.

[0111] t2 = P V21 + (P V11- P V21 ) ·d v21 / ( d v21 - d v11 (7)

[0112] Substituting parameters t1 and t2 into equation (3), the linear equation L = D·t + O, yields the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n The two endpoints:

[0113] l n-1 = D· t1+O,

[0114] l n-2 = D· t2+O,

[0115] Similarly, we can obtain {l1, l2, l3, ... l} n The intersection line will connect {l1, l2, l3, ... l... n The intersection lines are connected in sequence to obtain the left bank boundary line L1 of plane β1 on the left bank geometric surface γ1 of the three-dimensional terrain basic model;

[0116] Similarly, calculate plane β1 and the right bank geometry γ1 of the three-dimensional terrain base model. 1 The Boolean intersection line is used to obtain the right bank boundary line L2 of the three-dimensional terrain basic model.

[0117] Similarly, calculate the triangular facet V of the tension model on the left bank of the dam. 1-n Triangular facet γ on the left bank geometric surface γ1 1-n Find the Boolean intersection line l n 1 The intersection line {l} is obtained sequentially. n 2 l n 3 ...l n n}, the intersection line { l n 1 l n 2 ...l n n}

[0118] In finding the intersection line l n 1 Before that, it is also necessary to calculate the triangular facet γ on the left bank geometric surface γ1. 1-n Vertex V i 1 Distance d to plane β1 Vi1， After comparing it with preset conditions, the intersection line l is calculated.n 1 ;

[0119] The preset conditions are judged as follows:

[0120] When d Vi1 When the result of the operation is not equal to 0 and the sign of the operation is opposite, the triangular facet γ on the left bank geometric surface γ1 is determined. 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 1-n It intersects with plane β1;

[0121] When d Vi1 ≠0 (i=0, 1, 2), and the operation results have the same sign, the triangular facet γ on the left bank geometric surface γ1 1-n Located on one side of plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It will not intersect with line L;

[0122] When d Vi1 =0 (i=0,1,2), the triangular facet γ on the left bank geometric surface γ1 1-n On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It has no intersection with plane β1;

[0123] According to the triangular facet γ on the left bank geometric surface γ1 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 is then calculated. 1-n The intersecting scalar interval on the intersection line L is projected onto the line through the vertex of the triangle:

[0124] P Vi1 = D · (V i1 - O), (5)

[0125] i = 0, 1, 2; D is the direction vector of the intersection line L; O is any point on L.

[0126] V i 1 Given a point in the β1 plane, its projection onto β1 is given by the principle of similar triangles:

[0127] (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 The equation parameter t1 is derived as follows:

[0128] t1 = P V01 + (P V11 - P V01 ) ·dv01 / ( d v01 - d v11 (6)

[0129] Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived as follows:

[0130] t2 = P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 (7)

[0131] Substituting the parameters t1 and t2 into the linear equation L = D·t + O in equation (3), we obtain the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n 1 The two endpoints:

[0132] l n-1 1 = D· t1+O,

[0133] l n-2 1 = D· t2+O,

[0134] Similarly, the intersection line l can be obtained. n 1 Similarly, find the intersection line {l} in turn. n 2 l n 3 ...l n n}

[0135] This invention provides a web-based method for constructing a transition material model for dam filling, relating to the field of model building. It solves the problem of creating a transition material model for the dam filling process. Based on a web-based approach, the method establishes a 3D terrain foundation model of the dam and uses Boolean algorithms to obtain the transition material model, quickly generating a transition material model that conforms to the actual site terrain. This facilitates subsequent applications and maximizes the application value of BIM models. It also solves the previous difficulties in creating models for special and complex parts such as dam filling transition materials, playing a crucial role in assisting and guiding construction using BIM models.

[0136] Based on the above Figure 1 The corresponding embodiment describes a method for constructing a transition material model for dam filling on the bank based on a web interface. The following is an embodiment of the present invention, which can be used to execute the method embodiment of the present invention.

[0137] Example 2

[0138] This invention provides a method for constructing a model of transitional material on the bank of a dam filling site based on a web-based interface, such as... Figure 2 As shown:

[0139] Step 1: Obtain the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain basic model based on the web interface;

[0140] Step 2, Boundary division is given, such as Figure 2 As shown. At least two distinct coordinate points Q within the range of the aforementioned three-dimensional terrain model are selected on the left bank boundary line L1 of the dam. L1-n Q L1-(n+1) Where n≥1. Extract a polyline l from the bank between two adjacent points on the left bank boundary line L1 of the dam. 1-n Input the filling width W of the transition material on the bank, so that the multi-segment line l 1-n Offset by a distance W along the dam axis towards the opposite bank to obtain the polyline l' of the river channel. 1-n And the first offset path Wn, the second offset path W`n. From the polyline l on the shore. 1-n Multi-segment line of the river channel 1-n First offset path W n Second offset path W` n Forming a closed region β n .

[0141] Step 3, Extrusion Modeling, such as Figure 3 As shown, the closed region β n (A`B`C`D`) Stretching downwards along the Z-axis Z n Calculate the distance to obtain the closed region ABCD;

[0142] The stretching model Vn consists of closed regions ABCD, closed regions (A`B`C`D`), and the stretching path. n Perform Boolean operations with the 3D terrain base model to obtain the shore transition material model.

[0143] In one embodiment, the Boolean operation process is illustrated using the left bank as an example. Figure 4 As shown

[0144] Let the geometric surface of the three-dimensional terrain base model on the left bank be γ;

[0145] Traverse all triangular faces of the geometric surface γ1 of the 3D terrain foundation model on the left bank of the dam, traverse all triangular faces of the extruded model Vn, and extract the triangular face {γ} on the 3D terrain foundation model γ1. 1-1 γ 1-2 γ 1-3 、…γ 1-n}, the triangular facets {V} on V1 1-1 V 1-2 V 1-3 ...V 1-n}, using γ 1-1 triangular facets, V 1-1 Find the intersection line l1 of the triangular facets. 1 Similarly, we can obtain {l2} 1 l3 1 ...l n 1}, will { l1 1 l2 1 l3 1 ...l n 1 Connect them sequentially from left to right to form a continuous polyline PL 1 ,Right now Figure 4 The lines in the text are ae, eh, hd, da;

[0146] Connect lines ae, eh, hd, and da according to the principle of sharing adjacent endpoints to form the Boolean intersection line L1 on the left bank of the dam. 1 .

[0147] Through the L1 intersection on the left bank of the dam 1 Cut the triangular facets that γ1 and V1 pass through to form the cut triangular facets {ξ1, ξ2, ξ3, ..., ξn};

[0148] Boul intersection line L1 is formed on the left bank of the dam. 1 Find the smallest straight line segment SL, and starting from SL, process all the cutting triangular facets {ξ1, ξ2, ξ3}. 2. ...ξ n}、{ξ`1、ξ`2、……ξ` n} and the uncut triangular face {γ 1-1 γ 1-2 γ 1-3 、…γ 1-n},{V 1-1 V 1-2 V 1-3 ...V 1-n} Perform recursive traversal to form multiple continuous closed surfaces F, such as Figure 4The closed surfaces shown are eklh, aehd, aijd, aie, ijhe, djh, aefb, ehgf, dhgc, adcb, bfgc.

[0149] Among the m continuous closed surfaces F formed on the left bank, select the surface f that can constitute the geometry within the target material area, where m ≥ 1. Figure 4 The closed surfaces aehd, aefb, ehgf, dhgc, adcb, bfgc shown in the diagram connect multiple f's end to end to form a closed body model V1. Closed body model V1 is the model V of the transition material on the bank of the dam filling. m `.

[0150] In one embodiment, the left bank boundary line L1 and right bank boundary line L2 of the 3D terrain basic model obtained from the web interface in step 1 are used; for example... Figure 5 As shown, the steps are as follows:

[0151] Step 201: Preparation of planning reference elements; Using CATIA in conjunction with construction drawings, establish a terrain model of the dam foundation after excavation is completed, and create an online XYZ three-dimensional reference coordinate system consistent with the construction drawings.

[0152] Step 202: Select the planned elevation; Input the planned starting elevation value Z1 of the dam's fill, and use the Z1 value to select a point Q1 with coordinates (0, 0, Z1) on the Z-axis of the three-dimensional coordinate system from Step 1. Draw a plane β1 perpendicular to the Z-axis through point Q1. Let the left bank geometry of the terrain model be γ1 and the right bank geometry be γ2. 1 The left and right bank geometric surfaces γ1 and γ1' of the terrain model are obtained through plane β1 and the topography model. 1 Perform Boolean operations to obtain the left and right bank boundary lines L1 and L2 of the terrain model.

[0153] In one embodiment, the Boolean operation process, taking the operation of the left bank boundary line L1 as an example, extracts the triangular mesh models of plane β1 and the left bank geometric surface γ1 of the terrain model, respectively. For example... Figure 6 As shown, traverse all triangular faces on γ1 and β1 respectively, and extract the triangular face {γ1} on γ1. 1-1 γ 1-2 γ 1-3 、…γ 1-n}, the triangular facet {β} on β1 1-1 β 1-2 β 1-3 ...β 1-n}, using γ 1-1 β 1-1 Find the intersection line l1, and similarly find {l2, l3 ...l n}, and { l1, l2, l3, ... l n Connect them sequentially from left to right to form a Boolean intersection line L1.

[0154] Among them, such as Figure 7 As shown, γ 1-1 β 1-1 The algorithm for finding the intersection line l1 is as follows:

[0155] Let triangle γ 1-1 β 1-1 The vertices are as follows:

[0156] V0 1 (x0) 1 y0 1 z0 1 V1 1 (x1) 1 y1 1 z1 1 V2 1 (x2) 1 y2 1 z2 1 )

[0157] V0 2 (x0) 2 y0 2 z0 2 V1 2 (x1) 2 y1 2 z1 2 V2 2 (x2) 2 y2 2 z2 2 ),

[0158] triangle γ 1-1 β 1-1 The planes in question are γ2 and β1, respectively; the resulting geometric relationships are as follows: Figure 7 As shown.

[0159] β1 can be expressed using the general form of a plane equation as follows:

[0160] N1·X1+K1=0(1)

[0161] N1 is the normal vector of the β1 plane:

[0162] N1= (V1 2 - V0 2 ) × (V2 2 - V0 2 )

[0163] =(x12 - x0 2 y1 2 - y0 2 z1 2 - z0 2 ) ×( x2 2 - x0 2 y2 2 - y0 2 z2 2 - z0 2 )

[0164] ={(y1 2 - y0 2 (z2) 2 - z0 2 (-(y2)) 2 - y0 2 (z1) 2 - z0 2 ) i+{(x1) 2 - x0 2 (z2) 2 - z0 2 (-(x2)) 2 -x0 2 (z1) 2 - z0 2 )} j+{(x1) 2 -x0 2 (y2) 2 -y0 2 (-(x2)) 2 -x0 2 (y1) 2 -y0 2 k

[0165] =[{(y1 2 - y0 2 (z2) 2 - z0 2 (-(y2)) 2 - y0 2 (z1) 2 - z0 2 )} , {(x1 2 - x0 2 (z2) 2 - z0 2 (-(x2)) 2 -x0 2 (z1) 2 - z0 2 )}, {(x1 2 -x0 2(y2) 2 -y0 2 )-(x2 2 -x0 2 (y1) 2 -y0 2 )}]

[0166] X1 is any point on β1.

[0167] K1 = -N1· V0 2 (K1 is a constant)

[0168] =- [{y1 2 - y0 2 (z2) 2 - z0 2 )-(y2 2 -y0 2 (z1) 2 - z0 2 )} , {(x1 )} , {(x1 ) 2 - x0 2 (z2) 2 -z0 2 )-(x2 2 -x0 2 (z1) 2 - z0 2 )}, {(x1 )}, {(x1 )} 2 -x0 2 (y2) 2 -y0 2 )-(x2 2 -x0 2 (y1) 2 -y0 2 )}]·[x0 2 y0 2 z0 2 ]

[0169] =-{(y1 2 - y0 2 (z2) 2 - z0 2 )-(y2 2 - y0 2 (z1) 2 - z0 2 )} x0 2 -{(x1 2 - x0 2 (z2) 2 - z0 2 )-(x2 2 -x0 2 (z1) 2 - z02 )} y0 2 -{(x1 2 -x0 2 (y2) 2 -y0 2 )-(x2 2 -x0 2 (y1) 2 -y0 2 )} z0 2

[0170] γ2 can be expressed in general form as a plane equation as follows:

[0171] N2·X2+K2=0(2)

[0172] N2 is the normal vector of the γ2 plane:

[0173] N2= (V1 1 - V0 1 ) × (V2 1 - V0 1 )

[0174] =( x1 1 - x0 1 , y1 1 - y0 1 , z1 1 - z0 1) ×( x2 1 - x0 1 y2 1 - y0 1 z2 1 - z0 1 )

[0175] ={(y1 1 -y0 1 (z2) 1 -z0 1 )-(y2 1 -y0 1 (z1) 1 -z0 1 )} i+{(x1 )} 1 -x0 1 (z2) 1 -z0 1 )-(x2 1 -x0 1 (z1) 1 -z0 1 )} j+{(x1 )} 1 -x0 1 (y2)1 -y0 1 )-(x2 1 -x0 1 (y1) 1 -y0 1 )} k

[0176] = [{y1 1 -y0 1 (z2) 1 -z0 1 )-(y2 1 -y0 1 (z1) 1 -z0 1 )}, {(x1 )}, {(x1 )} 1 -x0 1 (z2) 1 -z0 1 )-(x2 1 -x0 1 (z1) 1 -z0 1 )}, {(x1 )}, {(x1 )} 1 -x0 1 (y2) 1 -y0 1 )-(x2 1 -x0 1 (y1) 1 -y0 1 )}]

[0177] X2 is any point on γ2.

[0178] K2 = -N2· V0 1 (K2 is a constant)

[0179] =- [{y1 1 -y0 1 (z2) 1 -z0 1 )-(y2 1 -y0 1 (z1) 1 -z0 1 )} , {(x1 )} , {(x1 ) 1 -x0 1 (z2) 1 -z01)-(x2) 1 -x0 1 (z1) 1 -z0 1 )}, {(x1 )}, {(x1 )} 1 -x0 1 (y2) 1 -y0 1 )-(x21 -x0 1 (y1) 1 -y0 1 )}]·[x0 1 y0 1 z0 1 ]

[0180] =-{(y1 2 - y0 2 (z2) 2 - z0 2 )-(y2 2 - y0 2 (z1) 2 - z0 2 )} x0 1 -{(x1 2 - x0 2 (z2) 2 - z0 2 )-(x2 2 -x0 2 (z1) 2 - z0 2 )} y0 1 -{(x1 2 -x0 2 (y2) 2 -y0 2 )-(x2 2 -x0 2 (y1) 2 -y0 2 )} z0 1

[0181] The equation of the line containing the intersection of β1 and γ2 can be derived from the equations of the β1 plane and the γ2 plane. The parameterized expression of the intersection line L is as follows:

[0182] L = D· t + O(3)

[0183] Where D = N1 × N2, D is the direction vector of the intersection line L, N2 is the normal vector of plane γ2, N1 is the normal vector of plane β1, O is any point on L; t is the parameter of the equation.

[0184] Then γ 1-1 The distance from the vertex to β1 is:

[0185] d Vi1 = (N2·V) i1 + K1) / | N2|, i=0,1,2; (4)

[0186] judge:

[0187] ① When d Vi1 If γ ≠ 0 (i = 0, 1, 2) and the results of the operations have the same sign, then γ 1-1 If they are located on one side of β1, they will not intersect.

[0188] ② When d Vi1 =0 (i=0,1,2), then the triangular facet γ on the left bank geometric surface γ1 1-1 On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n There is no intersection with plane β1.

[0189] ③ When d Vi1 When the result of the operation is ≠0 (i=0,1,2) and the sign of the result is opposite, the triangular facet γ on the left bank geometric surface γ1 is... 1-n It intersects with plane β1 and determines the triangular facet γ on the left bank geometric surface γ1. 1-n Intersects with line L;

[0190] Excluding the first two cases mentioned above, the triangular facet γ on the left bank geometric surface γ1 1-1 It intersects with β1 and determines the triangular facet γ on the left bank geometric surface γ1. 1-1 It intersects with line L.

[0191] In one embodiment, for this situation, it would be necessary to calculate the triangular facet γ on the left bank geometry γ1. 1-1 In the intersecting scalar interval on the intersection line L, we can assume V0 1 V2 1 On the same side, V1 1 On the other side of β1 (other cases ① and ② have been excluded).

[0192] The equation for the projection of a triangle vertex onto a line is:

[0193] P Vi1 = D ·(V i1 - O), (5)

[0194] i = 0, 1, 2; D is the direction vector of the intersection line L; O is any point on the intersection line L, and the geometric relationship is as follows: Figure 4 As shown;

[0195] Use K i1 V represents i1 Based on the principle of similar triangles, we find a similar triangle △V by projecting it onto β1. 01 BK 01 and △V 11 BK 11 Therefore, (t1 - PV) 01 ) / ( t1- PV 11 )=dv 01 / dv 11 It is derived that

[0196] t1 = P V01 + (P V11 - P V01 ) ·dv 01 / ( d v01 - d v11 (6)

[0197] For the same reason (t2-P) V21) / ( t2- P V11 )=dv 21 / dv 11 It is derived that

[0198] t2 = P V21 + (P V11 - P V21 ) ·dv 21 / ( d v21 - d v11) (7)

[0199] Substituting the parameters t1 and t2 obtained above into equation (4), the linear equation L = D·t+O, we can obtain γ. 1-1 With β 1-1 Intersection l n The two endpoints:

[0200] l n-1 = D· t1+O,

[0201] l n-2 = D· t2+O.

[0202] like Figure 7 As shown, similarly, {l2, l3, ... l} can be obtained. n The intersection lines {l1, l2, l3, ..., ln} are connected sequentially to obtain the left bank boundary line L1 of plane β1 on the terrain model γ1.

[0203] Similarly, calculate plane β1 and the right bank geometry γ1 of the 3D terrain model. 1 The Boolean intersection line, i.e., the right bank boundary line L2 of the terrain model, is obtained.

[0204] Similarly, calculate the triangular facet V of the tension model on the left bank of the dam. 1-n Triangular facet γ on the left bank geometric surface γ1 1-n Find the Boolean intersection line l n 1 The intersection line {l} is obtained sequentially. n 2 l n 3...l n n}, the intersection line { l n 1 l n 2 ...l n n}

[0205] In finding the intersection line l n 1 Before that, it is also necessary to calculate the triangular facet γ on the left bank geometric surface γ1. 1-n Vertex V i 1 Distance d to plane β1 Vi1， After comparing it with preset conditions, the intersection line l is calculated. n 1 ;

[0206] The preset conditions are judged as follows:

[0207] When d Vi1 When the result of the operation is not equal to 0 and the sign of the operation is opposite, the triangular facet γ on the left bank geometric surface γ1 is determined. 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 1-n It intersects with plane β1;

[0208] When d Vi1 ≠0 (i=0, 1, 2), and the operation results have the same sign, the triangular facet γ on the left bank geometric surface γ1 1-n Located on one side of plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It will not intersect with line L;

[0209] When d Vi1 =0 (i=0,1,2), the triangular facet γ on the left bank geometric surface γ1 1-n On plane β1, the triangular facet γ on the left bank geometric surface γ1 1-n It has no intersection with plane β1;

[0210] According to the triangular facet γ on the left bank geometric surface γ1 1-n Intersecting with the line of intersection L, the triangular facet γ on the left bank geometric surface γ1 is then calculated. 1-n The intersecting scalar interval on the intersection line L is projected onto the line through the vertex of the triangle:

[0211] P Vi1 = D · (V i1 - O), (5)

[0212] i = 0, 1, 2; D is the direction vector of line L; O is any point on L.

[0213] V i 1 Given a point in the β1 plane, its projection onto β1 is given by the principle of similar triangles.

[0214] (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 The equation parameter t1 is derived.

[0215] t1 = P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 (6)

[0216] Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived.

[0217] t2 = P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 (7)

[0218] Substituting the parameters t1 and t2 into the linear equation L = D·t + O in equation (3), we obtain the triangular facet γ on the left bank geometric surface γ1. 1-n With triangular facet β 1-n The intersection line l n 1 The two endpoints:

[0219] l n-1 1 = D· t1+O,

[0220] l n-2 1 = D· t2+O,

[0221] Similarly, the intersection line l can be obtained. n 1 Similarly, find the intersection line {l} in turn. n 2 l n 3 ...ln n}

[0222] This invention provides a web-based method for constructing a transition material model for dam filling, relating to the field of model building. It solves the problem of creating a transition material model for the dam filling process. Based on a web-based approach, the method establishes a 3D terrain foundation model of the dam and uses Boolean algorithms to obtain the transition material model, quickly generating a transition material model that conforms to the actual site terrain. This facilitates subsequent applications and maximizes the application value of BIM models. It also solves the previous difficulties in creating models for special and complex parts such as dam filling transition materials, playing a crucial role in assisting and guiding construction using BIM models.

[0223] The components, structures, and processes not described in detail in this embodiment are well-known components, common structures, or common methods in this industry, and will not be described in detail here.

Claims

1. A method for constructing a web-based dam fill bank transition material model, characterized in that, The method comprises the following steps: Step 101. Obtain the left bank boundary line L1 and the right bank boundary line L2 of the three-dimensional terrain base model based on a web terminal; Step 102. Based on the web, select at least two different coordinate points Q within the range of the three-dimensional terrain base model on the left bank boundary line L1 L1-n 、 L1-(n+1) , intercept l 1-n between the adjacent two points on L1 1-n , input the filling width W of the left bank transition material, so that the multi-segment line l 1-n of the bank is offset by W distance in the direction of the dam axis to the opposite bank direction, and the multi-segment line l` n of the river channel is obtained n ; the first offset path W` 1-n ; the multi-segment line l 1-n of the bank, the multi-segment line l` n of the river channel, the first offset path W` n , and the second offset path W` n， form a first closed area β n is greater than or equal to 1; Step 103. Based on the web side, the first closed area β n Stretch Z downward along the Z axis n The distance between the first closed area β n The second closed area β n , the second closed area β n The distance Z n The stretching model V n , the stretching model V n Boolean operation with the left bank geometry γ1 of the three-dimensional terrain base model, obtain the left bank edge transition material model V1`, repeat the operation in turn, obtain the left bank edge transition material model {V 1-1 `、V 1-2 `、……V1- n `}, and obtain the right bank edge transition material model {V 2-1 `、V 2-2 `、……V 2-n `}.

2. The method for constructing a web-based dam fill bank transition material model according to claim 1, wherein, The stretch model V n Boolean operation with the left bank geometric face γ1 of the three-dimensional terrain base model is the stretch model V n After the Boolean operation with the three-dimensional terrain base model geometric face γ1, the closed body model V n `on the side close to the river center is the stretch model V n `is the transition material model.

3. The method for constructing a web-based dam fill bank transition material model according to claim 2, wherein, The stretching model V n After the Boolean operation with the left bank geometry face γ1 of the three-dimensional terrain base model, the closed body model V n ` is retained near the center of the river channel n All triangular facets of the left bank geometry face γ1 of the three-dimensional terrain base model are traversed, all triangular facets of the stretching model V 1-1 are traversed, and the triangular facets {γ 1-2 , γ 1-3 , ……γ 1-n} on the left bank geometry face γ1 of the three-dimensional terrain base model and the triangular facets {V n , V 1-1 , V 1-2 , ……V 1-3} on the stretching model V 1-n are extracted, respectively 1-n The triangular facets V 1-n of the left bank stretching model and the triangular facets γ n on the left bank geometry face γ1 are subjected to a Boolean intersection to obtain a line l 1 n 2 , ……l n 3 , ……l n n} are obtained in sequence, and the Boolean intersection lines {l n 1 , l n 2 , ……l n n} are connected in sequence from left to right according to the principle of sharing adjacent endpoints to form the left bank Boolean intersection line L1 1 ; the right bank Boolean intersection line L2 1 is obtained in the same way.

4. The method for constructing a web-based dam fill bank transition material model according to claim 3, wherein, By the Boolean intersection line L1 1 The left bank geometry face γ1 and the stretch model V n The triangular patches that pass through the upper are cut, respectively forming two different triangular patches after cutting, which are triangular patches {ξ1, ξ2, ξ3, … ξ n}, triangular patches {ξ`1, ξ`2, ξ`3, … ξ` n}; In the Boolean intersection line L1 1 Get the smallest straight line segment SL from SL, respectively to all the triangular patches {ξ1, ξ 2. ……ξ n}, triangular patches {ξ`1, ξ`2, … ξ` n} and triangular patches {γ 1-1 , γ 1-2 , γ 1-3 , … γ 1-n} that are not cut, stretch model triangular face {V 1-1 , V 1-2 , V 1-3 , … V 1-n} along the Boolean intersection line traversal, forming m continuous closed faces F on the left bank of the dam, wherein m is greater than or equal to 1.

5. The method for constructing a web-based dam fill bank transition material model according to claim 4, wherein, Selecting the faces f that can constitute the geometry of the target material zone in the m consecutive closed faces F to form a closed volume model V n `The closed volume model V n `is the left bank transition material model V1` for the dam filling; similarly, the right bank transition material model V2` for the dam filling is obtained.

6. The method for constructing a web-based dam fill bank transition material model according to claim 3, wherein, The left bank boundary line L1 and the right bank boundary line L2 of the three-dimensional terrain base model based on the web terminal are obtained through the following steps: Step 201. Establish a three-dimensional terrain base model and an online three-dimensional reference coordinate system according to dam and dam foundation excavation construction design data information; Step 202. Obtain a plane β1 after assigning the dam planning elevation value to the three-dimensional reference coordinate system, the plane β1 is on the left bank geometric surface γ1 and the right bank geometric surface γ1 of the three-dimensional terrain base model 1 Respectively, the Boolean operation is obtained to obtain the left boundary line L1 and the right bank boundary line L2 of the three-dimensional terrain base model.

7. The method for constructing a web-based dam fill bank transition material model according to claim 6, wherein, The Boolean operation to obtain the left boundary line L1 and right bank boundary line L2 of the three-dimensional terrain base model involves extracting the plane β1 and the left bank geometric surface γ1 triangular mesh model and the right bank geometric surface γ1 of the three-dimensional terrain base model. 1 The triangular mesh model of the left bank geometric surface γ1 is obtained by traversing all triangular faces on the left bank geometric surface γ1 and the plane β1, and extracting the triangular faces {γ1} on the left bank geometric surface γ1. 1-1 γ 1-2 γ 1-3 ...γ 1-n }, the triangular facet {β} on plane β1 1-1 β 1-2 β 1-3 ...β 1-n }, calculate the triangular facet γ on the left bank geometric surface γ1. 1-n triangular facet β on plane β1 1-n The intersection line l n Similarly, the intersection lines {l1, l2, l3, ... l} are obtained. n }, the intersection lines { l1, l2, l3, ... l n Connect them sequentially from left to right to obtain the left bank boundary line L1; similarly, the algorithm obtains the right bank boundary line L2.

8. The method for constructing a web-based dam fill bank transition material model according to claim 7, wherein, the triangles γ on the left bank geometry face γ1 1-n intersecting the triangles β on the plane β1 1-n n the distance of the vertices of the triangles γ on the left bank geometry face γ1 1-n to the plane β1​ The triangular facet γ on the left bank geometry facet γ1 1-n The triangular facet β on the plane β1 1-n The planes where the triangular facets are located are respectively the γ2 plane and the plane β1, and the general equation of the plane β1 is: N1· X1+ K1= 0(1) Wherein, N1 is the normal vector of the β1 plane, X1 is an arbitrary point on the β1 plane, and K1 is a constant; The plane γ2 is expressed by the general equation of the plane as: N2· X2+ K2= 0(2) Wherein, N2 is the normal vector of the γ2 plane, X2 is an arbitrary point on the γ2 plane, and K2 is a constant; The intersection line L equation of the β1 plane and the γ2 plane is: L = D· t+O(3) Wherein, D = N1× N2, D is the direction vector of the intersection line L, t is the parameter of the equation, O is an arbitrary point of L, N1 is the normal vector of the plane β1, and N2 is the normal vector of the plane γ2; The distance of the vertices of the triangle patch γ 1-n to the plane β1is expressed by the formula: d Vi1 = (N1· V i 1 + K1) / | N1|, (4) where i = 0, 1, 2; V i 1 is a vertex of the triangle patch γ 1-n is a normal vector of the plane β1; The triangular facet γ on the left bank geometric surface γ1 1-n Vertex V i 1 Substituting equation (1) into equation (4) yields the triangular facet γ. 1-n The distance d from the vertex to β1 Vi1 .

9. The method of claim 8, wherein the method further comprises: the triangular patches γ on the left bank geometric face γ1 1-n the distance of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 1-n the intersection line l of the triangular patch β 1-n and the triangular patch γ n is determined according to the distance d of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 1-n i 1 the distance d of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 Vi1 is determined according to the distance d of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 1-n i 1 the distance d of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 Vi1 is determined according to the distance d of the vertex V of the triangular patch γ on the left bank geometric face γ1 to the plane β1 n after the preset condition is determined; the preset condition is determined as follows:​​ When d Vi1 ≠ 0, and the sign of the result of the operation is opposite, and it is determined that the triangular patch γ 1-n on the left bank geometry face γ1 intersects the intersection line L, then the triangular patch γ 1-n on the left bank geometry face γ1 has an intersection line with the plane β1. When d Vi1 ≠ 0 (i = 0, 1, 2), and the operation result signs are same, the triangular facets γ 1-n on the left bank geometric face γ1are located on one side of the plane β1, then the triangular facets γ 1-n on the left bank geometric face γ1will not intersect with the intersection line L; When d Vi1 = 0 (i = 0, 1, 2), the triangular facets γ 1-n On the plane β1, the triangular facets γ 1-n have no intersection with the plane β1. According to the triangle patch γ on the left bank geometry face γ1 1-n With the intersection line L, and then calculate the triangle patch γ on the left bank geometry face γ1 1-n In the intersection scalar interval on the intersection line L, through the projection of the triangle vertex to the straight line: P Vi1 = D · (V i1 - O), (5) i=0,1,2; D is the direction vector of the intersection line L; O is an arbitrary point on L, V i 1 For a point of the plane β1, the projection on β1, according to the principle of similar triangles (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 derived equation parameters t1 t1= P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 ), (6) t2= PV 21 ) / ( t2- P V11 )=d v21 / d v11 t2= PV t2= P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 ), (7) Substituting the parameters t1, t2 into the straight line equation L = D-t + O of equation (3) gives the triangle patch γ on the left bank geometry γ1 1-n and the intersection line I of the triangle patch β 1-n and the triangle patch γ n has the two end points: l n-1 = D· t1+O, l n-2 = D· t2+O, Similarly, the intersection lines { l1, l2, l3,... l n} are obtained, and the intersection lines { l1, l2, l3,... l n} are sequentially connected to obtain the left bank boundary line L1 of the plane β1 on the left bank geometric surface γ1 of the three-dimensional terrain base model. The right bank geometric face γ1 of the three-dimensional terrain base model is calculated in the same way as the plane β1 1 The right bank boundary line L2 of the three-dimensional terrain base model is obtained by the Boolean intersection of the plane β1 and the right bank geometric face γ1 of the three-dimensional terrain base model.

10. The method of claim 8, wherein the method further comprises: The triangular facet V of the left bank stretch model 1-n The triangular facet γ on the left bank geometric face γ1 1-n The intersection line l n 1 The intersection line l n 1 The triangular facet γ on the left bank geometric face γ1 1-n The vertex V of the triangular facet γ i 1 The distance d to the plane β1 Vi1， The Boolean intersection line l is calculated after the preset condition is judged n 1 ; The preset condition is judged as: When d Vi1 ≠ 0, and the sign of the result of the operation is opposite, and the triangle patch γ 1-n intersecting the intersection line L, then the triangle patch γ 1-n on the left bank geometry face γ1 intersects the plane β1. When d Vi1 ≠0 (i=0, 1, 2), and the operation result signs are same, the triangular facets γ 1-n on the left bank geometric face γ1are located on one side of the plane β1, then the triangular facets γ 1-n on the left bank geometric face γ1will not intersect with the intersection line L; When d Vi1 = 0 (i = 0, 1, 2), the triangular facets γ 1-n On the plane β1, the triangular facets γ 1-n have no intersection with the plane β1. According to the determination of the triangular facet γ on the left bank geometry facet γ1 1-n With the intersection line L, and then calculate the triangular facet γ on the left bank geometry facet γ1 1-n In the intersection scalar interval on the intersection line L, through the projection of the triangular vertex to the straight line: P Vi1 = D · (V i1 - O), (5) i=0,1,2; D is the direction vector of the intersection line L; O is an arbitrary point on L, V i 1 For a point of the plane β1, the projection on β1, according to the principle of similar triangles (t1- P V01 ) / ( t1- P V11 )=d v01 / d v1 derived equation parameters t1 t1= P V01 + (P V11 - P V01 ) ·d v01 / ( d v01 - d v11 ), (6) Same reason (t2- PV 21 ) / ( t2- P V11 )=d v21 / d v11 The equation parameter t2 is derived. t2= P V21 + (P V11 - P V21 ) ·d v21 / ( d v21 - d v11 ), (7) Substituting the parameters t1, t2 into the straight line equation L = D - t + O of equation (3) gives the triangle patch γ on the left bank geometry γ1 1-n of the intersection line I of the triangle patch β 1-n n 1 the two end points of the intersection line I of the triangle patch β​ l n-1 1 = D·t1+O, l n-2 1 = D·t2+O, Similarly, the Boolean intersection line l is obtained n 1 Similarly, the Boolean intersection line { l n 2 , l n 3 , … l n n}.

Images