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CN111475893B - A method for constructing space fault field model based on product 3D model - Google Patents

A method for constructing space fault field model based on product 3D model Download PDF

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CN111475893B
CN111475893B CN202010258235.6A CN202010258235A CN111475893B CN 111475893 B CN111475893 B CN 111475893B CN 202010258235 A CN202010258235 A CN 202010258235A CN 111475893 B CN111475893 B CN 111475893B
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杨德真
任羿
王自力
冯强
孙博
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Abstract

本发明涉及一种基于产品三维模型的空间故障场模型构建方法。该方法是基于场的概念,将产品组成单元的故障模式分布可视化,并辐射到产品表面形成故障场,进而为产品维修性设计提供依据。它包含三大步骤:(1)计算产品每个单元、每个故障模式的风险指数,即根据故障模式影响及危害性分析结果确定每个组成单元每个故障模式的风险指数;(2)计算产品组成单元的空间故障场强,即结合故障模式风险指数及单元维修难度计算;(3)计算产品表面故障场强,即将不规则单元转化为规则的三维线框体,然后将单元空间故障场强投射到产品表面形成故障场分布模型。基于该模型即可指导产品维修口盖设计。

Figure 202010258235

The invention relates to a method for constructing a space fault field model based on a three-dimensional model of a product. Based on the concept of field, the method visualizes the distribution of failure modes of product components, and radiates to the product surface to form a failure field, thereby providing a basis for product maintainability design. It includes three steps: (1) Calculate the risk index of each unit and each failure mode of the product, that is, determine the risk index of each failure mode of each component unit according to the results of the failure mode impact and criticality analysis; (2) Calculate The spatial fault field strength of the product component unit is calculated by combining the failure mode risk index and the unit maintenance difficulty; (3) calculating the product surface fault field strength, that is, converting the irregular unit into a regular three-dimensional wireframe, and then converting the unit spatial fault field The strong projection to the product surface forms the fault field distribution model. Based on this model, the design of product maintenance cover can be guided.

Figure 202010258235

Description

一种基于产品三维模型的空间故障场模型构建方法A method for constructing space fault field model based on product 3D model

所属技术领域Technical field

本发明一种基于产品三维模型的空间故障场模型构建方法,实现了产品故障信息在产品表面的可视化表达,从而为产品维修口盖设计提供指导。本发明属于可靠性工程技术领域。The present invention is a method for constructing a space fault field model based on a three-dimensional model of a product, which realizes the visual expression of product fault information on the product surface, thereby providing guidance for the design of product maintenance hatches. The invention belongs to the technical field of reliability engineering.

背景技术Background technique

维修口盖设计是设计中的重要方面之一,为了确保产品在使用中得到及时的保养和维修,产品表面需要开设口盖。口盖设计的好坏直接关系到产品的维修性水平。目前的口盖设计虽然简单的考虑了经济性、可达性的问题,但没有一个良好的方法精确指导维修口盖设计。鉴于此,本发明引入“场”的概念,结合产品故障模式影响及危害性分析(FMECA)结果,设计了一种基于产品三维模型的空间故障场模型构建方法。该方法可综合考虑产品每一个故障模式的发生概率、严酷度类别等因素,并将其表现在产品表面,叠加形成故障场强,进而可视化描述故障信息在产品表面的分布情况,以指导设计师合理设计口盖位置和大小。The design of the maintenance cover is one of the important aspects in the design. In order to ensure the timely maintenance and repair of the product during use, the surface of the product needs to have a cover. The quality of the flap design is directly related to the maintainability of the product. Although the current flap design simply considers the issues of economy and accessibility, there is no good method to accurately guide the maintenance flap design. In view of this, the present invention introduces the concept of "field", and designs a method for constructing a spatial fault field model based on a three-dimensional product model in combination with the results of product failure mode influence and criticality analysis (FMECA). This method can comprehensively consider factors such as the occurrence probability and severity category of each failure mode of the product, and express it on the product surface, superimpose it to form the failure field strength, and then visualize the distribution of failure information on the product surface to guide designers. Reasonably design the position and size of the flap.

发明内容SUMMARY OF THE INVENTION

本发明的目的是实现产品故障信息在产品表面的可视化表达,为产品维修口盖设计提供指导。The purpose of the present invention is to realize the visual expression of product failure information on the product surface, and to provide guidance for the design of the product maintenance cover.

本发明提供的一种基于产品三维模型的空间故障场模型构建方法,主要包含以下步骤:A method for constructing a space fault field model based on a three-dimensional model of a product provided by the present invention mainly includes the following steps:

步骤一:计算产品每个单元、每个故障模式的风险指数。Step 1: Calculate the risk index of each unit and each failure mode of the product.

根据产品故障模式影响及危害性分析(FMECA)结果,计算产品每个组成单元、每个故障模式的风险指数,它由故障模式发生概率等级、严酷度等级、检测难度等级确定。According to the results of product failure mode impact and criticality analysis (FMECA), the risk index of each component unit and each failure mode of the product is calculated, which is determined by the failure mode occurrence probability level, severity level, and detection difficulty level.

步骤二:计算产品组成单元的空间故障场强。Step 2: Calculate the space fault field strength of the product component unit.

空间故障场强是综合描述单元故障模式风险指数及维修难度的定量指标。其中,单元空间故障场强又是由单元故障模式的故障场强叠加而来。本步骤包含3个子步骤:Spatial fault field strength is a quantitative index that comprehensively describes the risk index of unit failure mode and maintenance difficulty. Among them, the unit space fault field strength is superimposed by the fault field strength of the unit fault mode. This step contains 3 sub-steps:

步骤1:计算产品组成单元中每个故障模式的故障场强,它由每个故障模式的风险指数确定。Step 1: Calculate the failure field strength of each failure mode in the product constituent unit, which is determined by the risk index of each failure mode.

步骤2:计算每个产品组成单元的维修难度。对于产品组成单元而言,维修难度可通过可达性和可拆卸性衡量。可达性越好,维修难度越低;可拆卸性越好,维修难度越低。Step 2: Calculate the maintenance difficulty of each product component unit. For product components, maintenance difficulty can be measured by accessibility and disassembly. The better the accessibility, the lower the maintenance difficulty; the better the disassembly, the lower the maintenance difficulty.

步骤3:计算每个产品组成单元的空间故障场强。单元空间故障场强由本步骤子步骤1和子步骤2所确定的故障模式场强及单元维修难度叠加决定,单元故障场强与故障模式场强成正比,与单元维修难度成反比。Step 3: Calculate the spatial fault field strength of each product component unit. The unit space fault field strength is determined by the superposition of the failure mode field strength determined in sub-step 1 and sub-step 2 of this step and the unit maintenance difficulty. The unit fault field strength is proportional to the failure mode field strength and inversely proportional to the unit maintenance difficulty.

步骤三:计算产品表面故障场强。Step 3: Calculate the fault field strength on the product surface.

表面故障场强是由部分产品组成单元空间故障场强在产品表面的故障辐射值叠加而成。The surface fault field strength is formed by the superposition of the fault radiation value of the space fault field strength of some product component units on the product surface.

本步骤包含4个子步骤:This step contains 4 sub-steps:

步骤1:基于ZL 201410778944.1“一种基于3D数字化模型的产品组件故障可视化方法”步骤一的方法将不规则产品转化为规则的三维长方体线框模型。Step 1: Based on the method of step 1 of ZL 201410778944.1 "A Product Component Failure Visualization Method Based on 3D Digital Model", the irregular product is transformed into a regular three-dimensional cuboid wireframe model.

步骤2:运用最小包容原则,将产品组成单元转化为三维圆柱体,进而确定其重心、底面半径、圆柱体高度以及圆柱体的中心轴向,并根据单元空间故障场强确定圆柱体故障场强。Step 2: Using the principle of minimum containment, convert the product component unit into a three-dimensional cylinder, and then determine its center of gravity, bottom surface radius, cylinder height and the central axis of the cylinder, and determine the cylinder fault field strength according to the unit space fault field strength .

(1)以产品重心为原点O,分别沿长方体的长、宽、高三个方向建立三维笛卡尔坐标系D,记作(X,Y,Z);(1) Taking the center of gravity of the product as the origin O, establish a three-dimensional Cartesian coordinate system D along the three directions of the length, width and height of the cuboid, denoted as (X, Y, Z);

(2)以单元重心为原点o,沿坐标系D的X轴、Y轴、Z轴方向建立三维子坐标系d,记作(x,y,z);(2) Taking the center of gravity of the element as the origin o, establish a three-dimensional sub-coordinate system d along the X-axis, Y-axis, and Z-axis directions of the coordinate system D, denoted as (x, y, z);

(3)运用最小包容原则,以单元重心为三维圆柱体重心,分别以x轴、y轴、z轴为中心轴,构建圆柱体,记作Cx、Cy、Cz,对比三个圆柱体的体积,取体积最小的圆柱体作为对应产品单元的三维圆柱体,记作C|k(k=x,y,z),圆柱体半径记作r、高记作HC(3) Using the principle of minimum containment, take the center of gravity of the unit as the center of gravity of the three-dimensional cylinder, and use the x-axis, y-axis, and z-axis as the central axes to construct a cylinder, denoted as Cx, Cy, and Cz, and compare the volumes of the three cylinders , take the cylinder with the smallest volume as the three-dimensional cylinder corresponding to the product unit, denoted as C|k (k=x, y, z), the radius of the cylinder as r, and the height as H C .

注:k=x表示以x轴为中心轴构建的圆柱体体积最小,以此类推。Note: k=x indicates that the volume of the cylinder constructed with the x-axis as the central axis is the smallest, and so on.

(4)将单元空间故障场强赋值给对应圆柱体C|k,即圆柱体C|k的故障场强等于对应产品单元空间故障场强。(4) Assign the unit space fault field strength to the corresponding cylinder C|k, that is, the fault field strength of the cylinder C|k is equal to the unit space fault field strength of the corresponding product.

步骤3:计算圆柱体在产品表面上的投影区域。Step 3: Calculate the projected area of the cylinder on the product surface.

投影方向规定如下:The projection direction is specified as follows:

(1)圆柱体中心横截面S(S是以圆柱体重心为圆心、与中心轴k轴垂直所做的横切面,即圆心为圆柱体中心、半径为r的圆):沿k轴正、负方向直接投射到产品表面,其投影即为横截面投影区域。(1) The cross section S of the center of the cylinder (S is the cross section made with the center of gravity of the cylinder as the center of the circle and perpendicular to the k-axis of the central axis, that is, the circle with the center as the center of the cylinder and the radius of r): positive along the k-axis, The negative direction is projected directly to the product surface, and its projection is the cross-sectional projection area.

(2)圆柱体侧面:以圆柱体中心轴截面(是一个长为HC、宽为2r的长方形),沿横截面S的半径方向投射到产品表面,即360°投射到产品表面。辐射到产品表面时,其投影区域应扣除无法投影区域,如民航客机的窗户、机舱门等。(2) Side of the cylinder: Take the section of the central axis of the cylinder (which is a rectangle with a length of H C and a width of 2r), and project it onto the product surface along the radial direction of the cross-section S, that is, 360° onto the product surface. When radiating to the surface of the product, the projection area should deduct the area that cannot be projected, such as the windows and cabin doors of civil aviation aircraft.

步骤4:计算单元到产品表面的辐射场。Step 4: Calculate the radiation field from the cell to the product surface.

(1)圆柱体中心横截面S的辐射场:首先,计算横截面S的圆心OD沿k轴正、负方向投影面的绝对距离,记作d+、d-;然后,计算横截面S分别在正、负方向的辐射场,与单元空间故障场强成正比、与绝对距离成反比。(1) Radiation field of the cross section S at the center of the cylinder: First, calculate the absolute distance of the projection plane of the center O D of the cross section S along the positive and negative directions of the k-axis, denoted as d + , d - ; then, calculate the cross section S The radiation fields in the positive and negative directions, respectively, are proportional to the unit space fault field strength and inversely proportional to the absolute distance.

(2)圆柱体侧面的辐射场:首先,计算圆柱体重心沿横切面S半径方向到产品表面的距离,此时可以设定一个角度ε(ε可整除360,且ε需足够大以保证同一表面不会被同一侧面重复投射),横切面S的半径沿横切面每转动ε°,则计算一次圆柱体重心沿横切面S半径方向到产品表面的距离,由此可得到一个距离数组

Figure BDA0002428236370000031
然后,计算圆柱体故障场强在产品表面的辐射场,它与圆柱体故障场强成正比,与di成反比。(2) Radiation field on the side of the cylinder: First, calculate the distance from the center of gravity of the cylinder to the surface of the product along the radial direction of the cross-section S. At this time, an angle ε can be set (ε is divisible by 360, and ε must be large enough to ensure the same The surface will not be repeatedly projected by the same side), every time the radius of the cross-section S rotates along the cross-section ε°, the distance from the center of gravity of the cylinder along the radial direction of the cross-section S to the product surface is calculated, and a distance array can be obtained.
Figure BDA0002428236370000031
Then, the radiation field of the cylinder fault field strength on the product surface is calculated, which is proportional to the cylinder fault field strength and inversely proportional to di.

步骤5:计算产品表面故障场强。Step 5: Calculate the fault field strength on the product surface.

由于产品组成单元众多,各个组成单元的空间故障场强投射到产品表面形成辐射场,还需进一步叠加计算产品表面故障场强,进而为产品维修口盖设计提供依据。Due to the large number of product components, the spatial fault field strength of each component unit is projected onto the product surface to form a radiation field. It is necessary to further superimpose the product surface fault field strength to provide a basis for the design of product maintenance covers.

(1)取产品表面上任意一个点f(f=1,2,…,n),计算所有投射到该点的辐射值之和

Figure BDA0002428236370000032
(1) Take any point f (f=1,2,...,n) on the surface of the product, and calculate the sum of all the radiation values projected to this point
Figure BDA0002428236370000032

(2)归一化处理,确定产品表面点f(f=1,2,…,n)的故障场强

Figure BDA0002428236370000033
(2) Normalization processing to determine the fault field strength of the product surface point f (f=1,2,...,n)
Figure BDA0002428236370000033

(3)将故障场强

Figure BDA0002428236370000034
与故障场强颜色条进行对比,取产品表面上点f(f=1,2,…,n)的故障场强对应的颜色值,对产品对应位置进行着色,从而得到产品表面故障场强分布。(3) Set the fault field strength
Figure BDA0002428236370000034
Compare with the color bar of the fault field strength, take the color value corresponding to the fault field strength of the point f (f=1,2,...,n) on the product surface, and color the corresponding position of the product, so as to obtain the fault field strength distribution on the product surface .

基于上述步骤一到步骤三,即可为产品着色,进而得到产品表面故障场强分布。基于该分布,设计师可进一步进行维修口盖设计。Based on the above steps 1 to 3, the product can be colored, and then the fault field intensity distribution on the surface of the product can be obtained. Based on this distribution, the designer can further design the service hatch.

附图说明Description of drawings

图1方法流程图Fig. 1 Method flow chart

图2产品三维长方体线框模型示例(a)产品(b)长方体线框Figure 2 Example of a 3D cuboid wireframe model of a product (a) Product (b) Cuboid wireframe

图3基于产品三维长方体线框模型的笛卡尔坐标系示例Fig. 3 Example of Cartesian coordinate system based on product 3D cuboid wireframe model

图4单元三维子坐标系示例Figure 4 Example of a 3D sub-coordinate system of a unit

图5最小包容圆柱体C|k示例Fig. 5 Example of minimum accommodating cylinder C|k

图6圆柱体横截面S在x轴正、负方向投影区域示例Figure 6 Example of the projection area of the cross section S of the cylinder in the positive and negative directions of the x-axis

图7圆柱体侧面在产品表面的投影区域示例(a)ε°投影区域(b)360°投影区域Figure 7 Example of the projection area of the cylinder side on the product surface (a) ε° projection area (b) 360° projection area

图8故障场强颜色条Figure 8 Fault field strength color bar

图9产品表面故障场分布图Fig. 9 Distribution of fault field on product surface

具体实施方式Detailed ways

本发明方法流程如图1所示。为使本发明的特征及优点得到更清楚的了解,以下结合附图及案例进行详细说明。本发明所选择的案例为一个包含3个组成单元的产品P。具体实施步骤是:The process flow of the method of the present invention is shown in FIG. 1 . In order to make the features and advantages of the present invention more clearly understood, the following detailed description is given in conjunction with the accompanying drawings and cases. The selected case of the present invention is a product P containing 3 constituent units. The specific implementation steps are:

步骤一:计算产品每个单元、每个故障模式的风险指数。Step 1: Calculate the risk index of each unit and each failure mode of the product.

根据产品故障模式影响及危害性分析(FMECA)结果,计算产品每个组成单元、每个故障模式的风险指数,它由故障模式发生概率等级、严酷度等级、检测难度等级确定。According to the results of product failure mode impact and criticality analysis (FMECA), the risk index of each component unit and each failure mode of the product is calculated, which is determined by the failure mode occurrence probability level, severity level, and detection difficulty level.

产品FMECA结果可通过表1表示,从中可以获取每个单元、每个故障模式的发生概率等级、严酷度等级、检测难度等级,进而计算风险指数RPN。RPN的计算公式如下:The product FMECA results can be expressed in Table 1, from which the probability level, severity level, and detection difficulty level of each unit and each failure mode can be obtained, and then the risk index RPN can be calculated. The formula for calculating RPN is as follows:

表1产品FMECA结果表示意Table 1. FMECA results of products

Figure BDA0002428236370000041
Figure BDA0002428236370000041

注:在GJB/1391中,有定义发生概率等级、严酷度等级、检测难度等级的取值范围,均为{1,2,3,…,10}。Note: In GJB/1391, the value ranges of occurrence probability level, severity level, and detection difficulty level are defined, all of which are {1,2,3,…,10}.

RPN=ESR*OPR*DDR (1)RPN=ESR*OPR*DDR (1)

其中ESR表示严酷度等级、OPR表示发生概率等级、DDR表示检测难度等级。Among them, ESR represents the severity level, OPR represents the occurrence probability level, and DDR represents the detection difficulty level.

【示例】案例产品P的FMECA结果如表2所示。[Example] The FMECA results of case product P are shown in Table 2.

表2产品P的FMECA结果示意Table 2 shows the FMECA results of product P

Figure BDA0002428236370000042
Figure BDA0002428236370000042

步骤二:计算产品组成单元的空间故障场强。Step 2: Calculate the space fault field strength of the product component unit.

空间故障场强是综合描述单元故障模式风险指数及维修难度的定量指标。其中,单元空间故障场强又是由单元故障模式的故障场强叠加而来。本步骤包含3个子步骤:Spatial fault field strength is a quantitative index that comprehensively describes the risk index of unit failure mode and maintenance difficulty. Among them, the unit space fault field strength is superimposed by the fault field strength of the unit fault mode. This step contains 3 sub-steps:

步骤1:计算产品组成单元i中每个故障模式的故障场强。它由每个故障模式的风险指数确定。风险指数越高,故障后果越严重,故障场强越高。故障模式故障场强计算公式如下:Step 1: Calculate the failure field strength for each failure mode in product component unit i. It is determined by the risk index for each failure mode. The higher the risk index, the more serious the failure consequences and the higher the failure field strength. The formula for calculating the fault field strength in the fault mode is as follows:

Figure BDA0002428236370000051
Figure BDA0002428236370000051

其中,

Figure BDA0002428236370000052
表示为单元i故障模式FMij的故障场强,RPNij为故障模式的风险指数。in,
Figure BDA0002428236370000052
Denoted as the fault field strength of unit i failure mode FM ij , RPN ij is the risk index of failure mode.

【示例】案例产品P的故障模式故障场强计算结果如表3所示。[Example] The calculation results of the failure mode failure field strength of the case product P are shown in Table 3.

表3产品P的故障模式故障场强Table 3 Failure Mode Failure Field Strength of Product P

单元名称unit name 故障模式failure mode 严酷度等级severity level 发生概率等级Occurrence probability level 检测难度等级Difficulty level of detection 风险指数risk index 故障场强fault field strength 单元1Unit 1 FM11FM11 55 44 99 180180 0.180.18 单元1Unit 1 FM12FM12 66 44 22 4848 0.0480.048 单元1Unit 1 FM13FM13 66 44 44 9696 0.0960.096 单元1Unit 1 FM14FM14 88 11 11 88 0.0080.008 单元2Unit 2 FM21FM21 22 88 11 1616 0.0160.016 单元2Unit 2 FM22FM22 66 66 66 216216 0.2160.216 单元2Unit 2 FM23FM23 55 44 55 100100 0.10.1 单元3Unit 3 FM31FM31 22 88 99 144144 0.1440.144 单元3Unit 3 FM32FM32 11 77 66 4242 0.0420.042 单元3Unit 3 FM33FM33 88 66 88 384384 0.3840.384 单元3Unit 3 FM34FM34 44 66 22 4848 0.0480.048 单元3Unit 3 FM35FM35 88 88 88 512512 0.5120.512

步骤2:计算每个产品组成单元i的维修难度。对于产品组成单元而言,维修难度可以通过可达性和可拆卸性衡量。可达性越好,维修难度越低;可拆卸性越好,维修难度越低。Step 2: Calculate the maintenance difficulty of each product component unit i. For product components, maintenance difficulty can be measured by accessibility and disassembly. The better the accessibility, the lower the maintenance difficulty; the better the disassembly, the lower the maintenance difficulty.

单元i的维修难度MDi具体定义如下:The maintenance difficulty MD i of unit i is specifically defined as follows:

Figure BDA0002428236370000053
Figure BDA0002428236370000053

其中,REDi为单元i的可达难度等级,RMDi为单元i的可拆卸难度等级。具体评分标准如表4和表5所示。Among them, RED i is the attainable difficulty level of unit i, and RMD i is the detachable difficulty level of unit i. The specific scoring criteria are shown in Tables 4 and 5.

表4单元可达难度等级定义Table 4 Definition of the attainable difficulty level of the unit

Figure BDA0002428236370000054
Figure BDA0002428236370000054

表5单元可拆卸难度等级定义Table 5 Definition of Unit Detachable Difficulty Level

Figure BDA0002428236370000055
Figure BDA0002428236370000055

Figure BDA0002428236370000061
Figure BDA0002428236370000061

【示例】案例产品P的故障模式故障场强计算结果如表6所示。[Example] The calculation results of the failure mode failure field strength of the case product P are shown in Table 6.

表6产品P的维修难度Table 6 Maintenance difficulty of product P

单元名称unit name 可达难度等级Achievable difficulty level 可拆卸难度等级Detachable difficulty level 维修难度Maintenance difficulty 单元1Unit 1 极难(10)extremely difficult (10) 较难(8)more difficult (8) 0.800.80 单元2Unit 2 一般(6)General (6) 较难(8)more difficult (8) 0.480.48 单元3Unit 3 较易(4)easier (4) 一般(6)General (6) 0.240.24

步骤3:计算每个产品组成单元i的空间故障场强。单元空间故障场强由本步骤子步骤1和子步骤2所确定的故障模式场强及单元维修难度叠加决定,单元故障场强与故障模式场强成正比,与单元维修难度成反比。其计算公式如下:Step 3: Calculate the spatial fault field strength of each product component unit i. The unit space fault field strength is determined by the superposition of the failure mode field strength determined in sub-step 1 and sub-step 2 of this step and the unit maintenance difficulty. The unit fault field strength is proportional to the failure mode field strength and inversely proportional to the unit maintenance difficulty. Its calculation formula is as follows:

Figure BDA0002428236370000062
Figure BDA0002428236370000062

其中,

Figure BDA0002428236370000063
表示单元i的故障场强,Ji表示单元i的故障模式总数。in,
Figure BDA0002428236370000063
represents the fault field strength of unit i, and J i represents the total number of failure modes of unit i.

【示例】案例产品P的故障模式故障场强计算结果如表7所示。[Example] The calculation results of the failure mode failure field strength of the case product P are shown in Table 7.

表7产品P各单元故障场强Table 7. The fault field strength of each unit of product P

单元名称unit name 故障模式故障场强之和The sum of the fault field strength of the fault mode 维修难度Maintenance difficulty 故障场强fault field strength 单元1Unit 1 0.3320.332 0.800.80 0.41500.4150 单元2Unit 2 0.3320.332 0.480.48 0.69170.6917 单元3Unit 3 1.131.13 0.240.24 4.70834.7083

步骤三:计算产品表面故障场强。表面故障场强是由部分产品组成单元空间故障场强在产品表面的故障辐射值叠加而成。本步骤包含4个子步骤:Step 3: Calculate the fault field strength on the product surface. The surface fault field strength is formed by the superposition of the fault radiation value of the space fault field strength of some product component units on the product surface. This step contains 4 sub-steps:

步骤1:基于ZL 201410778944.1“一种基于3D数字化模型的产品组件故障可视化方法”步骤一的方法将不规则产品转化为规则的三维长方体线框模型。Step 1: Based on the method of step 1 of ZL 201410778944.1 "A Product Component Failure Visualization Method Based on 3D Digital Model", the irregular product is transformed into a regular three-dimensional cuboid wireframe model.

示例图形见图2。An example graph is shown in Figure 2.

步骤2:运用最小包容原则,将产品组成单元i转化为三维圆柱体,进而确定其重心、底面半径、圆柱体高度以及圆柱体的中心轴向,并根据单元空间故障场强确定圆柱体故障场强。Step 2: Using the principle of minimum containment, convert the product component unit i into a three-dimensional cylinder, and then determine its center of gravity, bottom surface radius, cylinder height, and the central axis of the cylinder, and determine the cylinder fault field according to the unit space fault field strength. powerful.

(1)以产品重心为原点O,分别沿长方体的长、宽、高三个方向建立三维笛卡尔坐标系D,记作(X,Y,Z)。(1) Taking the center of gravity of the product as the origin O, establish a three-dimensional Cartesian coordinate system D along the three directions of the length, width and height of the cuboid, denoted as (X, Y, Z).

示例图形见图3。An example graph is shown in Figure 3.

(2)以单元i的重心为原点o,沿坐标系D的X轴、Y轴、Z轴方向建立三维子坐标系d,记作(x,y,z)。(2) Take the center of gravity of the unit i as the origin o, and establish a three-dimensional sub-coordinate system d along the X-axis, Y-axis, and Z-axis directions of the coordinate system D, denoted as (x, y, z).

示例图形见图4。An example graph is shown in Figure 4.

(3)运用最小包容原则,以单元i的重心为三维圆柱体重心,分别以x轴、y轴、z轴为中心轴,构建圆柱体,记作Cx、Cy、Cz,对比三个圆柱体的体积,取体积最小的圆柱体作为对应产品单元的三维圆柱体,记作C|k(k=x,y,z),圆柱体半径记作r、高记作HC(3) Using the principle of minimum containment, take the center of gravity of the unit i as the center of gravity of the three-dimensional cylinder, and take the x-axis, y-axis, and z-axis as the central axes, respectively, to construct a cylinder, denoted as Cx, Cy, and Cz, and compare the three cylinders Take the cylinder with the smallest volume as the three-dimensional cylinder corresponding to the product unit, denoted as C|k (k=x, y, z), the radius of the cylinder as r, and the height as H C .

注:k=x表示以x轴为中心轴构建的圆柱体体积最小,以此类推。Note: k=x indicates that the volume of the cylinder constructed with the x-axis as the central axis is the smallest, and so on.

最小包容圆柱体C|k示例如图5所示,其中k=x。An example of a minimum containing cylinder C|k is shown in Figure 5, where k=x.

(4)将单元空间故障场强赋值给单元i对应的圆柱体C|k,即圆柱体C|k的故障场强等于单元i的空间故障场强。(4) Assign the unit space fault field strength to the cylinder C|k corresponding to the unit i, that is, the fault field strength of the cylinder C|k is equal to the space fault field strength of the unit i.

步骤3:计算单元i的圆柱体在产品表面上的投影区域。投影方向规定如下:Step 3: Calculate the projected area of the cylinder of unit i on the product surface. The projection direction is specified as follows:

(1)圆柱体中心横截面S(S是以圆柱体重心为圆心、与中心轴k轴垂直所做的横切面,即圆心为圆柱体中心、半径为r的圆):沿k轴正、负方向直接投射到表面,其投影即为横截面投影区域。(1) The cross section S of the center of the cylinder (S is the cross section made with the center of gravity of the cylinder as the center of the circle and perpendicular to the k-axis of the central axis, that is, the circle with the center as the center of the cylinder and the radius of r): positive along the k-axis, The negative direction is projected directly to the surface, and its projection is the cross-sectional projection area.

示例图形见图6。An example graph is shown in Figure 6.

(2)圆柱体侧面:以圆柱体中心轴截面(是一个长为HC、宽为2r的长方形),沿横截面S的半径方向投射到产品表面,即360°投射到产品表面。辐射到产品表面时,其投影区域应扣除无法投影区域,如民航客机的窗户、机舱门等。(2) Side of the cylinder: Take the section of the central axis of the cylinder (which is a rectangle with a length of H C and a width of 2r), and project it onto the product surface along the radial direction of the cross-section S, that is, 360° onto the product surface. When radiating to the surface of the product, the projection area should deduct the area that cannot be projected, such as the windows and cabin doors of civil aviation aircraft.

示例图形见图7。An example graph is shown in Figure 7.

步骤4:计算单元i到产品表面的辐射场。Step 4: Calculate the radiation field from unit i to the product surface.

(1)圆柱体中心横截面S的辐射场:首先,计算横截面S的圆心OD沿k轴正、负方向投影面的绝对距离,记作

Figure BDA0002428236370000071
然后,计算横截面S分别在正、负方向的辐射场,与单元空间故障场强成正比、与绝对距离成反比。(1) Radiation field of the cross section S of the center of the cylinder: First, calculate the absolute distance of the projection plane of the center O D of the cross section S along the positive and negative directions of the k-axis, denoted as
Figure BDA0002428236370000071
Then, the radiation fields of the cross section S in the positive and negative directions are calculated, which are proportional to the unit space fault field strength and inversely proportional to the absolute distance.

·正方向辐射场

Figure BDA0002428236370000072
的计算公式如下:· Positive radiation field
Figure BDA0002428236370000072
The calculation formula is as follows:

Figure BDA0002428236370000081
Figure BDA0002428236370000081

其中,

Figure BDA0002428236370000082
表示单元i的故障场强,
Figure BDA0002428236370000083
表示单元i的横截面S的圆心OD沿k轴正方向投影面的绝对距离。in,
Figure BDA0002428236370000082
represents the fault field strength of unit i,
Figure BDA0002428236370000083
It represents the absolute distance of the projection plane of the center OD of the cross-section S of the unit i along the positive direction of the k-axis.

·负方向辐射场

Figure BDA0002428236370000084
的计算公式如下:·Negative radiation field
Figure BDA0002428236370000084
The calculation formula is as follows:

Figure BDA0002428236370000085
Figure BDA0002428236370000085

其中,

Figure BDA0002428236370000086
表示单元i的横截面S的圆心OD沿k轴负方向投影面的绝对距离。in,
Figure BDA0002428236370000086
It represents the absolute distance of the projection plane of the center OD of the cross-section S of the unit i along the negative direction of the k-axis.

(2)圆柱体侧面的辐射场:首先,计算单元i的圆柱体重心沿横切面S半径方向到产品表面的距离,此时可以设定一个角度εii可整除360,且εi需足够大以保证同一表面不会被同一侧面重复投射),横切面S的半径沿横切面每转动ε°,则计算一次圆柱体重心沿横切面S半径方向到产品表面的距离,由此可得到一个距离数组

Figure BDA0002428236370000087
然后,计算圆柱体侧面故障场强在产品表面的辐射场,它与圆柱体故障场强成正比,与die成反比。(2) Radiation field on the side of the cylinder: First, calculate the distance from the center of gravity of the cylinder of unit i to the product surface along the radial direction of the cross-section S. At this time, an angle ε i can be set (ε i is divisible by 360, and ε i It needs to be large enough to ensure that the same surface will not be repeatedly projected by the same side), every time the radius of the cross-section S rotates along the cross-section ε°, the distance from the center of gravity of the cylinder to the product surface along the radial direction of the cross-section S is calculated. get a distance array
Figure BDA0002428236370000087
Then, calculate the radiation field of the cylinder side fault field strength on the product surface, which is proportional to the cylinder fault field strength and inversely proportional to die.

i×e)°方向的辐射场计算公式如下:The formula for calculating the radiation field in the (ε i ×e)° direction is as follows:

Figure BDA0002428236370000088
Figure BDA0002428236370000088

其中,

Figure BDA0002428236370000089
表示单元i的圆柱体侧面在(εi×e)°方向的辐射场;die表示单元i的中心轴截面的中心沿(εi×e)°方向到产品表面的距离。in,
Figure BDA0002428236370000089
Represents the radiation field of the cylindrical side of the unit i in the (ε i ×e)° direction; d ie represents the distance from the center of the central axis section of the unit i to the product surface along the (ε i ×e)° direction.

步骤5:计算产品表面故障场强。Step 5: Calculate the fault field strength on the product surface.

由于产品组成单元众多,各个组成单元的空间故障场强投射到产品表面形成辐射场,还需进一步叠加计算表面故障场强,进而为产品维修口盖设计提供依据。Due to the large number of product components, the spatial fault field strength of each component unit is projected to the product surface to form a radiation field. It is necessary to further superimpose and calculate the surface fault field strength to provide a basis for the design of product maintenance covers.

(1)取产品表面上任意一点f(f=1,2,…,n),计算所有投射到该点的辐射值之和

Figure BDA00024282363700000810
公式如下:(1) Take any point f (f=1,2,...,n) on the surface of the product, and calculate the sum of all the radiation values projected to this point
Figure BDA00024282363700000810
The formula is as follows:

Figure BDA00024282363700000811
Figure BDA00024282363700000811

其中,If表示在产品表面f点有辐射场的单元集合;μi=1表示单元i的横截面S沿k轴正方向在f点有辐射场,沿k轴负方向在f点无辐射场;μi=0表示单元i的横截面S沿k轴正方向在f点无辐射场,沿k轴负方向在f点有辐射场Among them, If represents the unit set with radiation field at point f on the product surface; μ i =1 means that the cross section S of unit i has radiation field at point f along the positive direction of the k-axis, and no radiation at point f along the negative direction of the k-axis Field; μ i = 0 means that the cross section S of unit i has no radiation field at point f along the positive direction of the k-axis, and there is a radiation field at point f along the negative direction of the k-axis

(2)进一步进行归一化处理,确定产品表面上点f(f=1,2,…,n)的故障场强

Figure BDA0002428236370000093
处理公式如下:(2) Further normalization is performed to determine the fault field strength of the point f (f=1,2,...,n) on the product surface
Figure BDA0002428236370000093
The processing formula is as follows:

Figure BDA0002428236370000091
Figure BDA0002428236370000091

(3)将故障场强

Figure BDA0002428236370000092
与故障场强颜色条(见图8)进行对比,取产品表面点f(f=1,2,…,n)的故障场强对应的颜色值,并对产品表面上的投影区域进行着色。(3) Set the fault field strength
Figure BDA0002428236370000092
Compare with the color bar of fault field strength (see Figure 8), take the color value corresponding to the fault field strength of the product surface point f (f=1,2,...,n), and color the projected area on the product surface.

基于上述步骤一到步骤三,即可为产品表面着色,进而得到产品表面故障场强分布图,如图9所示。基于该分布,设计师可进一步进行维修口盖设计。如图9所示,其左右中心区域A1、A2故障场强相较于周围区域高很多,因此需在A1、A2分别设计一个维修口盖。Based on the above steps 1 to 3, the surface of the product can be colored, and then the fault field intensity distribution map on the surface of the product can be obtained, as shown in Figure 9. Based on this distribution, the designer can further design the service hatch. As shown in Figure 9, the fault field strength of the left and right central areas A1 and A2 is much higher than that of the surrounding areas, so it is necessary to design a maintenance cover in A1 and A2 respectively.

Claims (1)

1.一种基于产品三维模型的空间故障场模型构建方法,其特征在于它包含以下步骤:1. a method for constructing a space fault field model based on a three-dimensional model of a product, is characterized in that it comprises the following steps: 步骤一:计算产品每个单元、每个故障模式的风险指数;Step 1: Calculate the risk index of each unit and each failure mode of the product; 根据产品故障模式影响及危害性分析FMECA结果,计算产品每个组成单元、每个故障模式的风险指数,它由故障模式发生概率等级、严酷度等级、检测难度等级确定;According to the FMECA results of product failure mode impact and hazard analysis, calculate the risk index of each component unit and each failure mode of the product, which is determined by the failure mode occurrence probability level, severity level, and detection difficulty level; 步骤二:计算产品组成单元的空间故障场强;Step 2: Calculate the space fault field strength of the product component unit; 空间故障场强是综合描述单元故障模式风险指数及维修难度的定量指标,其中,单元空间故障场强又是由单元故障模式的故障场强叠加而来,本步骤包含3个子步骤:The spatial fault field strength is a quantitative index that comprehensively describes the unit failure mode risk index and maintenance difficulty. Among them, the unit spatial fault field strength is superimposed by the fault field strength of the unit failure mode. This step includes 3 sub-steps: 步骤1:计算产品组成单元中每个故障模式的故障场强,它由每个故障模式的风险指数确定;Step 1: Calculate the failure field strength of each failure mode in the product component unit, which is determined by the risk index of each failure mode; 步骤2:计算每个产品组成单元的维修难度,对于产品组成单元而言,维修难度可通过可达性和可拆卸性衡量:可达性越好,维修难度越低;可拆卸性越好,维修难度越低;Step 2: Calculate the maintenance difficulty of each product component unit. For a product component unit, the maintenance difficulty can be measured by accessibility and disassembly: the better the accessibility, the lower the maintenance difficulty; the better the disassembly, the lower the maintenance difficulty. The lower the maintenance difficulty; 步骤3:计算每个产品组成单元的空间故障场强,单元空间故障场强由本步骤子步骤1和子步骤2所确定的故障模式场强及单元维修难度叠加决定,单元故障场强与故障模式场强成正比,与单元维修难度成反比;Step 3: Calculate the space fault field strength of each product component unit. The unit space fault field strength is determined by the superposition of the failure mode field strength and unit maintenance difficulty determined in sub-step 1 and sub-step 2 of this step. It is directly proportional to the strength and inversely proportional to the maintenance difficulty of the unit; 步骤三:计算产品表面故障场强;Step 3: Calculate the fault field strength on the surface of the product; 表面故障场强是由部分产品组成单元空间故障场强在产品表面的故障辐射值叠加而成,本步骤包含5个子步骤:The surface fault field strength is formed by the superposition of the fault radiation values of the spatial fault field strength of some product component units on the product surface. This step includes 5 sub-steps: 步骤1:运用最小包容原则,将不规则的产品三维物理模型转化为规则的三维长方体线框模型,其长、宽、高分别用L、W、H表示,记作(L,W,H);Step 1: Using the principle of minimum containment, convert the irregular 3D physical model of the product into a regular 3D cuboid wireframe model, whose length, width and height are represented by L, W, and H respectively, denoted as (L, W, H) ; 步骤2:继续运用最小包容原则,将产品组成单元转化为三维圆柱体,进而确定其重心、底面半径、圆柱体高度以及圆柱体的中心轴向,并根据单元空间故障场强确定圆柱体故障场强;Step 2: Continue to apply the principle of minimum containment, convert the product component unit into a three-dimensional cylinder, and then determine its center of gravity, bottom surface radius, cylinder height, and the central axis of the cylinder, and determine the cylinder fault field according to the unit space fault field strength powerful; (1)以产品重心为原点O,分别沿长方体(L,W,H)的长、宽、高三个方向建立三维笛卡尔坐标系D,记作(X,Y,Z);(1) Taking the center of gravity of the product as the origin O, establish a three-dimensional Cartesian coordinate system D along the three directions of the length, width and height of the cuboid (L, W, H), denoted as (X, Y, Z); (2)以单元重心为原点o,沿坐标系D的X轴、Y轴、Z轴方向建立三维子坐标系d,记作(x,y,z);(2) Taking the center of gravity of the element as the origin o, establish a three-dimensional sub-coordinate system d along the X-axis, Y-axis, and Z-axis directions of the coordinate system D, denoted as (x, y, z); (3)运用最小包容原则,以单元重心为三维圆柱体重心,分别以x轴、y轴、z轴为中心轴,构建圆柱体,记作Cx、Cy、Cz,对比三个圆柱体的体积,取体积最小的圆柱体作为对应产品单元的三维圆柱体,记作C|k(k=x,y,z),圆柱体半径记作r、高记作HC(3) Using the principle of minimum containment, take the center of gravity of the unit as the center of gravity of the three-dimensional cylinder, and use the x-axis, y-axis, and z-axis as the central axes to construct a cylinder, denoted as Cx, Cy, and Cz, and compare the volumes of the three cylinders , take the cylinder with the smallest volume as the three-dimensional cylinder corresponding to the product unit, denoted as C| k (k=x, y, z), the radius of the cylinder as r, and the height as H C ; 其中,k=x表示以x轴为中心轴构建的圆柱体体积最小,以此类推;Among them, k=x indicates that the volume of the cylinder constructed with the x-axis as the central axis is the smallest, and so on; (4)将单元空间故障场强赋值给对应圆柱体C|k,即圆柱体C|k的故障场强等于对应产品单元空间故障场强;(4) Assign the unit space fault field strength to the corresponding cylinder C| k , that is, the fault field strength of the cylinder C| k is equal to the unit space fault field strength of the corresponding product; 步骤3:计算圆柱体在产品表面上的投影区域;Step 3: Calculate the projection area of the cylinder on the product surface; 投影方向规定如下:The projection direction is specified as follows: (1)圆柱体中心横截面S:S是以圆柱体重心为圆心、与中心轴k轴垂直所做的横切面,即圆心为圆柱体中心、半径为r的圆,沿k轴正、负方向直接投射到产品表面,其投影即为横截面投影区域;(1) The cross section of the center of the cylinder S: S is the cross section made by the center of gravity of the cylinder and perpendicular to the k-axis of the central axis, that is, the circle with the center as the center of the cylinder and the radius of r, positive and negative along the k-axis The direction is directly projected onto the product surface, and its projection is the cross-sectional projection area; (2)圆柱体侧面:以一个长为HC、宽为2r的长方形圆柱体中心轴截面,沿横截面S的半径方向投射到产品表面,即360°投射到产品表面,辐射到产品表面时,其投影区域应扣除无法投影区域;(2) Cylinder side: a rectangular cylinder with a length of H C and a width of 2r is projected onto the product surface along the radial direction of the cross-section S, that is, 360° projected onto the product surface, and when it radiates to the product surface , its projection area should deduct the unprojectable area; 步骤4:计算单元到产品表面的辐射场;Step 4: Calculate the radiation field from the unit to the product surface; (1)圆柱体中心横截面S的辐射场:首先,计算横截面S的圆心OD沿k轴正、负方向投影面的绝对距离,记作d+、d-;然后,计算横截面S分别在正、负方向的辐射场,与单元空间故障场强成正比、与绝对距离成反比;(1) Radiation field of the cross section S at the center of the cylinder: First, calculate the absolute distance of the projection plane of the center O D of the cross section S along the positive and negative directions of the k-axis, denoted as d + , d - ; then, calculate the cross section S The radiation fields in the positive and negative directions, respectively, are proportional to the unit space fault field strength and inversely proportional to the absolute distance; (2)圆柱体侧面的辐射场:首先,计算圆柱体重心沿横切面S半径方向到产品表面的距离,此时可以设定一个角度ε,其中ε可整除360,且ε需足够大以保证同一表面不会被同一侧面重复投射,横切面S的半径沿横切面每转动ε°,则计算一次圆柱体重心沿横切面S半径方向到产品表面的距离,由此可得到一个距离数组
Figure FDA0002944491560000021
然后,计算圆柱体故障场强在产品表面的辐射场,它与圆柱体故障场强成正比,与di成反比;
(2) Radiation field on the side of the cylinder: First, calculate the distance from the center of gravity of the cylinder to the surface of the product along the radial direction of the cross-section S. At this time, an angle ε can be set, where ε is divisible by 360, and ε must be large enough to ensure The same surface will not be repeatedly projected by the same side. For every ε° of the radius of the cross-section S along the cross-section, the distance from the center of gravity of the cylinder to the product surface along the radial direction of the cross-section S is calculated, and a distance array can be obtained.
Figure FDA0002944491560000021
Then, calculate the radiation field of the cylinder fault field strength on the surface of the product, which is proportional to the cylinder fault field strength and inversely proportional to di;
步骤5:计算产品表面故障场强;Step 5: Calculate the fault field strength on the product surface; 由于产品组成单元众多,各个组成单元的空间故障场强投射到产品表面形成辐射场,还需进一步叠加计算产品表面故障场强,进而为产品维修口盖设计提供依据,其过程如下:Due to the large number of product components, the spatial fault field strength of each component unit is projected onto the product surface to form a radiation field. It is necessary to further superimpose the product surface fault field strength to provide a basis for the design of the product maintenance cover. The process is as follows: (1)取产品表面上任意一个点f,其中f=1,2,…,n,计算所有投射到该点的辐射值之和
Figure FDA0002944491560000022
(1) Take any point f on the surface of the product, where f=1,2,...,n, and calculate the sum of all the radiation values projected to this point
Figure FDA0002944491560000022
(2)归一化处理,确定产品表面点f的故障场强
Figure FDA0002944491560000023
其中f=1,2,…,n;
(2) Normalization processing to determine the fault field strength of the product surface point f
Figure FDA0002944491560000023
where f=1,2,...,n;
(3)将故障场强
Figure FDA0002944491560000024
与故障场强颜色条进行对比,取产品表面上点f的故障场强对应的颜色值,其中f=1,2,…,n,对产品对应位置进行着色,从而得到产品表面故障场强分布。
(3) Set the fault field strength
Figure FDA0002944491560000024
Compare with the color bar of fault field strength, take the color value corresponding to the fault field strength of point f on the product surface, where f=1,2,...,n, and color the corresponding position of the product, so as to obtain the fault field strength distribution on the product surface .
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