Method for finding way when vehicle passes through bridge floor with explosion pit

Abstract

The invention discloses a way-finding method for a vehicle to pass through a bridge floor with an explosion pit, which is an optimal path planning method based on geometric figures, and belongs to the field of damage assessment. According to the method, an actual problem is converted into a mathematical problem, that is, a plurality of circles are arranged in a rectangular area, the left side and the right side are boundaries, the lower side is an entrance, the upper side is an exit, the area outside the circles is a driving area, and a widest path from the lower side to the upper side is found. The method comprises the following steps: firstly, finding out all maximum channels, including channels between circles and edges and channels between two circles; searching for an entrance and exit channel and an exit channel, and establishing a communication relationship between the channels; finding all paths from the entrance and exit channel to the exit channel; and finding out the widest path.

Description

Technical Field

[0001] The invention belongs to the field of damage effect assessment of warheads striking road surfaces and bridge decks, and relates to an optimal path assessment method based on path search. Background Art

[0002] Damage assessment is mainly used to solve the problem of evaluating and estimating the damage efficiency of ammunition and the damage effect of targets. In particular, in recent years, the strong demand for actual combat exercises and training and the precise customization of damage plans in informationized warfare have put forward new requirements, which will inevitably drive damage assessment technology to become an important, cutting-edge and interdisciplinary development field in the future. The optimal path planning method based on geometric figures proposed in this invention can provide strong support for the evaluation of the blocking effect of ammunition on roads and bridge decks.

[0003] The existing method for calculating the impact effect of ammunition on bridge deck based on grid division is to first divide the undamaged bridge deck into grids, and then use the traversal method to find the path. This method can only find the path, but cannot find the maximum width of the path. In addition, the calculation accuracy is limited by the size of the minimum grid, and the calculation speed is affected by the number of grids.

[0004] The method used in the present invention is a mathematical method for finding channels and paths based on geometric figures. The accuracy is completely unrestricted and the calculation speed is only related to the number of channels, which is obviously much less than the number of grids. Summary of the invention

[0005] The main purpose of the present invention is to provide an optimal path planning method based on geometric figures and path search, which can fully consider the location and size of the explosion pit and improve the rationality, accuracy and practicality of the evaluation results; in addition, by introducing path search, the complex calculation problem of the widest path is solved, thereby significantly improving the calculation efficiency and accuracy.

[0006] The purpose of the present invention is achieved through the following technical solutions.

[0007] The present invention discloses a path finding method for a vehicle passing through a bridge deck with a burst pit, comprising the following steps.

[0008] Step 1: Find all channels.

[0009] A channel is similar to an aisle and is the smallest unit of a path. There are two types of channels: one is the channel between two circles; the other is the channel between a circle and the left and right edges. Since we need to get the maximum width of the path, we only need to solve the channel with the maximum width. For a channel between two circles, the maximum width is the distance between the centers of the two circles minus the radius of the two circles.

[0010] Solve all maximum channels of the first type, that is, solve the maximum channel between any two non-intersecting and non-contained circles. Assume that the line segment connecting the centers of the two circles is L1, and the line segment L1 has two intersections with the two circles, set as P11 and P12. These two intersections are the key points of the maximum channel between the two circles. There is only one straight line through P11 that is perpendicular to the line segment L1, and this line is defined as L11; similarly, there is also only one straight line through P12 that is perpendicular to the line segment L1, and this line is set as L12. The channel between L11 and L12 is the maximum channel between the two circles. Obviously, these two straight lines are tangents to the two circles.

[0011] Solve all the maximum channels of the second type, that is, solve the maximum channel between each circle and the left and right edges. In particular, when the circle intersects with the left edge, the left channel does not exist; when the circle intersects with the right edge, the right channel does not exist. Draw a perpendicular line L2 from the center of the circle to the left and right edges, and the foot of the perpendicular is one of the key points, set as P21; the intersection of the perpendicular line L2 and the circle is another key point, set as P22. There is only one straight line through P21 that is perpendicular to the line segment L2. This straight line is defined as L21. Obviously, this straight line coincides with the left edge or the right edge; similarly, there is only one straight line through P22 that is perpendicular to the line segment L2. This straight line is set as L22. Obviously, this straight line is a tangent to the circle. The channel between L21 and L22 is the maximum channel between this circle and the edge.

[0012] Step 2: Solve the range of all channels, that is, solve the positions of both ends of the channel, that is, solve the range of each side of the channel.

[0013] First, find the two ends of the line L11, that is, find the closest intersection points among all the intersection points of the rays of P11 in two directions with the four sides of other circles or rectangles. Similarly, the closest intersection points of the two sides of the line L12 can also be obtained.

[0014] There is a special case that needs additional processing, that is, there are other circles between the channels. There are two tangents through this circle that are perpendicular to the straight line L11. Of course, these two tangents are also perpendicular to the straight line L12 and parallel to the line segment L1. The tangent closest to the key point will also limit the range of the channel. If the nearest intersection point of L11 obtained earlier is farther than the intersection point of the tangents, the range is corrected to the intersection point of the tangents. In this way, the final channel range can be obtained.

[0015] Step 3: Channel deduplication.

[0016] Whether it is the first type of channel or the second type of channel, there may be channel duplication. For example, if the centers of four circles can form a rectangle, and there are no other circles inside the rectangle, and the two adjacent circles do not intersect, then there must be two channels that are exactly the same. For another example, if the line connecting the centers of two circles is parallel to the left and right edges, and there are no other circles between the line and the edges, then there must be two channels that are exactly the same.

[0017] Step 4: Mark your entrances and exits.

[0018] If one of the four sides of the channel coincides with the bottom of the rectangle, then the channel is the entrance. If one of the four sides of the channel coincides with the top of the rectangle, then the channel is the exit.

[0019] Obviously, some passages may be both entrances and exits.

[0020] Step 5: Establish the connection relationship between each channel.

[0021] By judging whether each channel is connected to other channels, we can know whether a channel can reach another channel. In particular, if a channel is both an entrance and an exit, there is no need to check the connectivity of the channel with other channels. The connection between two channels is determined by the fact that the quadrilaterals of the two channels have at least 3 intersection points, which means that the two channels are connected.

[0022] Step 6: Remove invalid channels.

[0023] There are two cases of invalid channels: one is an isolated channel, that is, a channel that is not connected to other channels; the other is an entry channel that is only connected to other entry channels, that is, this entry channel can only reach other entry channels, then this channel is also meaningless.

[0024] Step 7: Find all the paths.

[0025] To find all the paths, you must first find all the entryways.

[0026] Traverse each entrance channel and start searching from each entrance channel until the exit channel is found. In this way, find all the paths. The width of the path is the width of the narrowest channel on the path.

[0027] Step 8: Find the widest path.

[0028] Traverse all paths and find the widest path, which is what we want. BRIEF DESCRIPTION OF THE DRAWINGS

[0029] Figure 1 The present invention discloses a flow chart of a path-finding method for a vehicle passing through a bridge deck with a burst pit.

[0030] Figure 2 This is an example diagram of the present invention.

[0031] Figure 3 It is a channel diagram of the path-finding method of the present invention, the blue frame is the entrance channel, the red frame is the exit channel, and the black frame is the ordinary channel.

[0032] Figure 4 This is a diagram of the analysis results of the path-finding method of the present invention, and the filled part is the channel used by the widest path. Implementation

[0033] The present invention is described in detail below with reference to the accompanying drawings.

[0034] like Figure 1 As shown, this example discloses a path-finding method for a vehicle passing through a bridge deck with a burst hole, and the calculation example parameters are shown in the following table.

[0035]

[0036] This example discloses a path-finding method for a vehicle passing through a bridge with a burst pit, and the specific implementation steps are as follows.

[0037] Step 1: Find all channels.

[0038] A channel is the smallest unit of a path. There are two types of channels: one is the channel between two circles; the other is the channel between a circle and the left and right edges. Since we need to get the maximum width of the path, we only need to solve the channel with the maximum width. For a channel between two circles, the maximum width is the distance between the centers of the two circles minus the radius of the two circles.

[0039] Solve all maximum channels of the first type, that is, solve the maximum channel between any two non-intersecting and non-contained circles. Assume that the line segment connecting the centers of the two circles is L1, and the line segment L1 has two intersections with the two circles, set as P11 and P12. These two intersections are the key points of the maximum channel between the two circles. There is only one straight line through P11 that is perpendicular to the line segment L1, and this line is defined as L11; similarly, there is also only one straight line through P12 that is perpendicular to the line segment L1, and this line is set as L12. The channel between L11 and L12 is the maximum channel between the two circles. Obviously, these two straight lines are tangents to the two circles.

[0040] Solve all the maximum channels of the second type, that is, solve the maximum channel between each circle and the left and right edges. In particular, when the circle intersects with the left edge, the left channel does not exist; when the circle intersects with the right edge, the right channel does not exist. Draw a perpendicular line L2 from the center of the circle to the left and right edges, and the foot of the perpendicular is one of the key points, set as P21; the intersection of the perpendicular line L2 and the circle is another key point, set as P22. There is only one straight line through P21 that is perpendicular to the line segment L2. This straight line is defined as L21. Obviously, this straight line coincides with the left edge or the right edge; similarly, there is only one straight line through P22 that is perpendicular to the line segment L2. This straight line is set as L22. Obviously, this straight line is a tangent to the circle. The channel between L21 and L22 is the maximum channel between this circle and the edge.

[0041] Step 2: Solve the range of all channels, that is, solve the positions of both ends of the channel, that is, solve the range of each side of the channel.

[0042] First, find the two ends of the line L11, that is, find the closest intersection points among all the intersection points of the rays of P11 in two directions with the four sides of other circles or rectangles. Similarly, the closest intersection points of the two sides of the line L12 can also be obtained.

[0043] There is a special case that needs additional processing, that is, there are other circles between the channels. There are two tangents through this circle that are perpendicular to the straight line L11. Of course, these two tangents are also perpendicular to the straight line L12 and parallel to the line segment L1. The tangent closest to the key point will also limit the range of the channel. If the nearest intersection point of L11 obtained earlier is farther than the intersection point of the tangents, the range is corrected to the intersection point of the tangents. In this way, the final channel range can be obtained.

[0044] Step 3: Channel deduplication.

[0045] Whether it is the first type of channel or the second type of channel, there may be channel duplication. For example, if the centers of four circles can form a rectangle, and there are no other circles inside the rectangle, and the two adjacent circles do not intersect, then there must be two channels that are exactly the same. For another example, if the line connecting the centers of two circles is parallel to the left and right edges, and there are no other circles between the line and the edges, then there must be two channels that are exactly the same.

[0046] Step 4: Mark your entrances and exits.

[0047] If one of the four sides of the channel coincides with the bottom of the rectangle, then the channel is the entrance. If one of the four sides of the channel coincides with the top of the rectangle, then the channel is the exit.

[0048] Obviously, some passages may be both entrances and exits.

[0049] like Figure 3As shown, the blue quadrilateral represents the entrance channel, the red quadrilateral represents the exit channel, and the black quadrilateral represents the ordinary channel.

[0050] Step 5: Establish the connection relationship between each channel.

[0051] By judging whether each channel is connected to other channels, we can know whether a channel can reach another channel. In particular, if a channel is both an entrance and an exit, there is no need to check the connectivity of the channel with other channels. The connection between two channels is determined by the fact that the quadrilaterals of the two channels have at least 3 intersection points, which means that the two channels are connected.

[0052] Step 6: Remove invalid channels.

[0053] There are two cases of invalid channels: one is an isolated channel, that is, a channel that is not connected to other channels; the other is an entry channel that is only connected to other entry channels, that is, this entry channel can only reach other entry channels, then this channel is also meaningless.

[0054] Step 7: Find all the paths.

[0055] To find all the paths, you must first find all the entryways.

[0056] Traverse each entrance channel and start searching from each entrance channel until the exit channel is found. In this way, find all the paths. The width of the path is the width of the narrowest channel on the path.

[0057] Step 8: Find the widest path.

[0058] Traverse all paths and find the widest path, which is what we want.

[0059] like Figure 4 As shown, the filled portion is the channel used by the widest path.

Claims

1. A path planning method based on geometric figures, characterized in that: The steps include: Step 1: Find all channels: A channel is similar to an aisle and is the smallest unit of a path. There are two types of channels: one is the channel between two circles; the other is the channel between a circle and the left and right edges. Since we need to get the maximum width of the path, we only need to solve the channel with the maximum width. For the channel between two circles, the maximum width is the distance between the centers of the two circles minus the radii of the two circles. Solve all maximum channels of the first type, that is, solve the maximum channel between any two non-intersecting and non-contained circles. Assume that the line segment connecting the centers of the two circles is L1, and this line segment L1 has two intersection points with the two circles, set as P 11 and P 12 , these two intersection points are the key points of the maximum channel between the two circles. 11 There is only one straight line perpendicular to line segment L1, which is defined as L 11 ; Similarly, over P 12 There is also only one straight line perpendicular to line segment L1, which is L 12 .L 11 With L 12 The channel between is the maximum channel between the two circles. Obviously, these two lines are tangents to the two circles. Solve all the maximum channels of the second type, that is, solve the maximum channel between each circle and the left and right edges. In particular, when the circle intersects with the left edge, the left channel does not exist; when the circle intersects with the right edge, the right channel does not exist. Draw a perpendicular line L2 from the center of the circle to the left and right edges, and the foot of the perpendicular is one of the key points, set as P 21 ; The intersection of the vertical line L2 and the circle is another key point, set as P 22 . 21 There is only one straight line perpendicular to line segment L2, which is defined as L 21 Obviously, the line coincides with the left edge or the right edge; similarly, through P 22 There is also only one straight line perpendicular to line segment L2, which is L 22 , it is obvious that this line is a tangent to the circle. 21 With L 22 The channel between is the largest channel between this circle and the edge. Step 2: Solve the range of all channels, that is, solve the positions of the two ends of the channel, that is, solve the range of each side of the channel: first solve the straight line L 11 The two ends of P 11 The closest intersection of the rays in both directions with all the intersections of the four sides of other circles or rectangles. Similarly, the straight line L can also be obtained. 12 The closest intersection of two sides. There is a special case that requires additional processing, that is, there are other circles between the channels. There are two tangent lines through this circle and the straight line L 11 Perpendicular, of course, these two tangent lines are also directly L 12 The tangent line closest to the key point will also limit the range of the channel. 11 If the nearest intersection point is farther than the tangent intersection point, the range is corrected to the tangent intersection point. In this way, the final channel range can be obtained. Step 3: Channel deduplication: Whether it is the first type of channel or the second type of channel, there may be channel duplication. For example, if the centers of the four circles can just form a rectangle, and there are no other circles inside the rectangle, and the two adjacent circles do not intersect, there must be two channels that are exactly the same. For another example, if the line connecting the centers of the two circles is parallel to the left and right edges, and there are no other circles between the line and the edges, there must be two channels that are exactly the same. Step 4: Mark the entrance and exit: If one of the four sides of the channel coincides with the bottom edge of the rectangle, then the channel is the entrance. If one of the four sides of the channel coincides with the top edge of the rectangle, then the channel is the exit. Obviously, some channels may be both entrances and exits. Step 5: Establish the connection relationship between each channel: Determine whether each channel is connected to other channels, so as to know whether one channel can reach another channel. In particular, if a channel is both an entrance and an exit, there is no need to check the connectivity of the channel with other channels. The connection between two channels is determined by the fact that the quadrilaterals of the two channels have at least three intersections, which means that the two channels are connected. Step 6: Remove invalid channels: There are two types of invalid channels: one is an isolated channel, that is, a channel that is not connected to other channels; The other is an entryway that is only connected to other entryways, that is, this entryway can only reach other entryways, so this channel is also meaningless. Step 7: Find all paths: To find all paths, you must first find all entryways. Traverse each entryway and start searching from each entryway until you find the exit channel. In this way, find all paths. The width of a path is the width of the narrowest channel on the path. Step 8: Find the widest path: Traverse all paths and find the widest path, which is what you want.

2. A method for finding a path for a vehicle passing through a bridge with a blast pit as claimed in claim 1, characterized in that: The method comprises step 7: finding all paths. All paths through the bridge deck are obtained according to step 7, which can be used as a basis for finding the widest path. The widest path is found among these paths, providing a basis for combating the restriction of traffic on the bridge deck.

3. A method for finding a path for a vehicle passing through a bridge with a blast pit as claimed in claim 1 or 2, characterized in that: Find the widest passage between blast holes and the passage between blast holes and the bridge deck.

4. A method for finding a path for a vehicle passing through a bridge with a blast pit as claimed in claim 1 or 2, characterized in that: Establish the connection relationship between each channel, that is, whether one channel can reach another channel.

5. A method for finding a path for a vehicle passing through a bridge with a blast pit as claimed in claim 1 or 2, characterized in that: Based on the channels and connections, find all the paths that can pass through the bridge deck.

6. A method for finding a path for a vehicle passing through a bridge with a blast pit as claimed in claim 1 or 2, characterized in that: Based on all the paths found that can pass through the bridge deck, find the widest path.

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