DE102023113599B4 - Method for operating an at least assisted vertical or diagonal parking function for a vehicle - Google Patents

Abstract

Method for operating an at least assisted vertical or inclined parking function (3) for a vehicle (1), which includes a parking maneuver (14) in a vertical or inclined parking space (4) with a forward travel path from a current location (C) of the vehicle (1) to a turning location (F) and an arcuate reverse travel path from the turning location (F) to an end location (g) at which the vehicle (1) is at least partially arranged in the parking space (4), the method comprising:
- Providing (S2) sensor information (12) describing the vertical or inclined parking space (4);
- taking into account the sensor information (12), determining (S3) a first parking maneuver (15) which includes the shortest possible reversing path without collision with an object (6) which laterally borders the parking space (4), and a second parking maneuver (16) which includes the shortest possible forward path,
- determining (S4) whether an orientation angle of the vehicle (1) at the turning location (F) is greater for the first parking maneuver (15) or for the second parking maneuver (16); and
- Operating (S5) the vehicle (1) from its current location (C) to the turning location (F) and from there to the final location (g) at least assisted according to the determined parking maneuver (15, 16) having the larger orientation angle.

Description

The invention relates to a method for operating an at least assisted vertical or diagonal parking function for a vehicle. The at least assisted parking function is configured to perform assisted parking or parking with a higher level of automation, such as semi-automatic or fully automatic parking. The parking function provides a parking maneuver into a vertical or diagonal parking space. The parking maneuver includes a forward travel path from a current location of the vehicle to a turning location and an arcuate reverse travel path from the turning location to an end location at which the vehicle is at least partially arranged in the parking space. The invention further relates to a vehicle, a control unit for a vehicle, and a computer program product for performing such a method.

A vehicle may be equipped with a parking function for at least assisted parking into a perpendicular or diagonal parking space. It is assumed that the vehicle has already approached such a parking space, for example, with a forward or reverse maneuver. A sensor device on the vehicle thereby detects the parking space. The parking function then identifies the parking space as a parking space or parking gap and can offer or automatically activate an assisted, semi-automatic, or even fully automatic parking maneuver into a parking position in the identified parking space. This parking maneuver typically involves a reversing path from outside the parking space into the parking space. Therefore, parking can be referred to as perpendicular or diagonal reverse parking. Typical scenarios for this type of parking are perpendicular or diagonal parking spaces on streets, in garages, and/or on parking areas.

Typically, such a reverse parking maneuver involves reversing toward the parking space and dipping into it until the vehicle almost collides with the parking space's boundaries. Afterward, the vehicle may move forward to align with a target parking position located within the parking space. The vehicle may then continue with consecutive forward and reverse moves until it is parked at a final parking location within the parking space. The number of consecutive moves may depend on the narrowness of the parking space. As the gap narrows or objects bordering the parking space approach, the parking maneuver becomes more difficult, and the number of forward and reverse moves increases.

However, it is advantageous for a vehicle user if the parking function can park the vehicle without many changes in direction, especially with only a single reverse movement toward the parking space. Adapting the parking function to such parking maneuvers is therefore desirable.

The document US 2020 / 0 369 262 A1 discloses a parking assistance method of a parking control device that executes parking control with respect to a free parking space surrounding a host vehicle. The method includes detecting a first frame line and a second frame line of the parking space, setting a parking target depending on the frame lines, and executing the parking control so that a position of the host vehicle is adjusted to the parking target.

The document DE 11 2019 001 125 T5 discloses a method for operating an at least assisted vertical parking function for a vehicle, which includes a parking maneuver in a vertical position with a forward travel path from a current location of the vehicle to a turning location and an arcuate reverse travel path from the turning location to an end location where the vehicle is at least partially positioned in the parking space. The method further includes providing sensor information describing the vertical or inclined parking space.

The document US 2019 / 0 176 888 A1 describes a method for a vehicle for instructing and automatically executing a parking maneuver in a vertical parking space. It discloses the planning of a path sequence to a stopping point of the maneuver, including several straight paths to turning points of the maneuver and several curved paths. The method further includes the acquisition and provision of sensor information for describing the parking space. Detected obstacles are incorporated into the description of the parking space as framework conditions to avoid collisions with the obstacle during path planning, and the shortest parking path is calculated iteratively, taking the detected obstacle into account.

It is the object of the invention to adapt a parking function for parking in a vertical or inclined parking space in such a way that a number of changes in the direction of travel during parking are reduced.

The independent claims solve the problem.

A first aspect of the invention relates to a method for operating an at least assisted vertical or angled parking function for a vehicle. The at least assisted parking function is configured to provide at least assisted parking and thus assisted driving of the vehicle into a parking space. In addition, the parking function can also be configured to provide semi-automatic or, in particular, fully automatic parking. Therefore, there are several automation levels covered by the method, starting with assisted parking.

The parking function provides a parking maneuver into a vertical or angled parking space. The parking maneuver includes a forward travel path from a current location of the vehicle to a turnaround location and an arcuate reverse travel path from the turnaround location to an end location where the vehicle is at least partially positioned within the parking space. The vehicle with the parking function first approaches the parking space until it reaches a current location. A front-facing camera and/or other sensor device of the vehicle, for example, first detects the parking space. The angled parking space may be referred to as a herringbone parking space.

The current location is a position and orientation of the vehicle when the method starts, in particular the position of a center of a rear axle of the vehicle. Consequently, current location information describing the current location may include coordinates and an orientation angle of the vehicle at these coordinates. The current location may be near the parking space, for example, on a street adjacent to the parking space. Besides, it is possible to distinguish between a right-hand parking space and a left-hand parking space. The right-hand parking space is the parking space detected on a right side of the vehicle as it approaches it. The left-hand parking space is the parking space detected on the left side of the vehicle as it approaches it.

One or two objects may delimit the parking space in a parking space width direction. The respective object may be another vehicle parked in a neighboring parking space. Alternatively or additionally, the respective object may be an infrastructure component, in particular a wall of a garage and/or another building. Alternatively or additionally, the respective object may be a floor marking and/or another parking space boundary of the parking space. When approaching the current location, the vehicle first detects the presence of one of the objects that is closest to a travel path of the vehicle when approaching the current location. This object may be referred to as the start object or first detected object. As it approaches the parking space further, the vehicle may detect the other object. This other object may be referred to as the end object or second detected object. The parking space here is an open space between the two objects.

• - Der Vorwärtsfahrpfad soll einen kreisbogenförmigen Teil beinhalten, aber von einer definierten geraden Bewegungslinie starten, die durch den aktuellen Ort festgelegt wird. Die gerade Bewegungslinie entspricht einem Steuerkurs des Fahrzeugs während seiner Suche nach dem Parkplatz.
• - Der kreisbogenförmige Teil des Vorwärtsfahrpfades soll mit einem Objekt, das den Parkplatz seitlich begrenzt, kollisionsfrei sein. Das Objekt ist an einer definierten virtuellen Linie auf der entgegengesetzten Seite angeordnet. Dieses Objekt ist das Objekt auf der entgegengesetzten Seite, das durch die virtuelle Linie auf der entgegengesetzten Seite begrenzt ist.
• - Der Rückwärtsfahrpfad in Richtung des Parkplatzes soll mit einer Parkplatzeinfahrtsecke kollisionsfrei sein, die eine äußerste Ecke eines Objekts sein kann, das den Parkplatz seitlich begrenzt. Dieses Objekt kann ein anderes Objekt als das Objekt sein, das für den Vorwärtsfahrpfad relevant ist. Daher kann dieses Objekt das Endobjekt sein.
• - Der Rückwärtsfahrpfad in Richtung des Parkplatzes soll eine Ziellinie im Parkplatz am Endort erreichen, die vor einem endgültigen Parkort oder Zielort im Parkplatz angeordnet ist. Ein Abstand zwischen dem Endort und dem endgültigen Parkort kann ein minimaler Toleranzabstand sein. Der Abstand zwischen dem Endort und dem Parkort soll daher gleich dem minimalen Toleranzabstand sein. Der minimale Toleranzabstand wird im Folgenden als minimaler Abstandswert bezeichnet. Die Idee besteht hier darin, das Fahrzeug das Parkrückwärtsmanöver mit einer geraden Bewegung abschließen zu lassen, während der Vorderräder des Fahrzeugs sowohl auf die Bewegungsrichtung als auch den Fahrzeugsteuerkurs ausgerichtet werden, so dass die Lenkung fast null ist.

The invention is based on the assumption that the following constraints should be considered when determining the parking maneuver:

• - The forward travel path should include a circular arc, but start from a defined straight line of travel determined by the current location. The straight line of travel corresponds to the vehicle's heading during its search for the parking space.
• - The circular arc of the forward travel path must be collision-free with an object that borders the parking space on the side. The object is located along a defined virtual line on the opposite side. This object is the object on the opposite side, bounded by the virtual line on the opposite side.
• - The reverse path toward the parking space should be collision-free with a parking space entrance corner, which may be the outermost corner of an object that laterally borders the parking space. This object may be different from the object relevant to the forward path. Therefore, this object may be the end object.
• - The reverse path toward the parking space should reach a finish line in the parking space at the end location, which is located before a final parking location or destination in the parking space. A distance between the end location and the final parking location can be a minimum tolerance distance. The distance between the end location and the parking location should therefore be equal to the minimum tolerance distance. The minimum The maximum tolerance distance is referred to below as the minimum distance value. The idea here is to allow the vehicle to complete the parking reversing maneuver in a straight line, while the vehicle's front wheels are aligned with both the direction of travel and the vehicle's heading, so that the steering is almost zero.

It is difficult to consider all constraints simultaneously while determining the parking maneuver and therefore the forward and reverse travel path.

Thus, individual constraints are considered separately. The results are then combined to decide how the vehicle should be operated while parked.

The method includes providing sensor information describing the vertical or sloped parking space. The vehicle's front camera and/or other sensor device can capture the sensor information and transmit it to a control unit of the vehicle. The control unit can be an electronic control unit (ECU). The control unit can receive the sensor information and provide it. The control unit performs the method. Therefore, the method is a computer-implemented method. The sensor information can include sensor data such as camera data, radar data, ultrasound data, and/or lidar data.

The method includes determining a first parking maneuver and a second parking maneuver taking into account the sensor information. Therefore, the method may include applying a determination algorithm to the sensor information. The determination algorithm may include a rule to calculate the respective parking maneuver. Both parking maneuvers follow the parking maneuver described above and therefore include the forward travel path and the reverse travel path. However, the first and second parking maneuvers differ from each other. The first parking maneuver includes the shortest possible reverse travel path without colliding with the object that laterally borders the parking space. This object is preferably the end object. The second parking maneuver includes the shortest possible forward travel path. Therefore, it is preferably circular. The two determined parking maneuvers therefore satisfy some of the restrictions described above. The vehicle's control unit, for example, determines the two parking maneuvers. Each determined parking maneuver may include an individual turning location and an individual end location that is different from the respective location of the other parking maneuver.

Furthermore, the method includes determining whether an orientation angle of the vehicle at the turning location is larger for the first parking maneuver or for the second parking maneuver. This analyzes the respective specific parking maneuver. Either the first parking maneuver or the second parking maneuver has the larger orientation angle compared to the respective other parking maneuver. The orientation angle is, for example, an angle between a central longitudinal direction of the vehicle and the straight line of movement. The vehicle's control unit can perform this analysis.

The method involves operating the vehicle from its current location to the turning location and from there to the final location, at least with assistance, according to the determined parking maneuver having the larger orientation angle. Therefore, it is automatically decided whether parking is performed according to the determined first parking maneuver or the determined second parking maneuver. Both parking maneuvers provide parking with a single reverse path. The provided parking function is thus adapted for parking in the perpendicular or inclined parking space, so that the number of changes in driving direction during parking is reduced compared to other parking functions.

By moving according to the selected one of the two specific parking maneuvers, the at least one assisted vertical or diagonal parking function is operated. In the case of assisted parking, the control unit transmits, for example, commands for manually driving and/or steering and/or braking the vehicle, which are to be displayed by a display device in the vehicle, in order to guide a driver to follow a path according to the first or second parking maneuver with their vehicle. However, the parking function preferably operates the vehicle along the path of the first or second parking maneuver semi-automatically or, in particular, fully automatically.

The procedure can only start after the driver manually activates a reverse parking mode of the parking function and/or depending on user-specific and/or vehicle-specific settings of the parking function. Alternatively or additionally, the procedure can start automatically when the vehicle positions itself in a parking environment with mandatory reverse parking.

One embodiment includes the forward travel path of the respective parking maneuver comprising a straight forward travel section from the current location to an intermediate location and an arcuate travel section from the intermediate location to the turning location. The forward travel path thus comprises two sections, one without steering and the other with steering, i.e., with a steering angle greater than zero. The respective parking maneuver therefore fulfills the constraint that the forward travel path should include an arcuate section, but should start from the defined straight line of movement determined by the current location. This enables reaching a particularly expedient turning location from a variety of possible current locations. The method is therefore versatile.

A further embodiment includes the arcuate reverse travel path and/or the arcuate travel portion of the forward travel path being drivable with a constant steering angle of at least 80 percent, in particular between 90 percent and 95 percent, of a maximum steering angle of a steering system of the vehicle. The steering angle can be any value between more than 80 percent and 100 percent of the maximum steering angle. If the steering system cannot be operated at its maximum steering angle, the vehicle can be operated with the largest possible or permissible steering angle during the arcuate reverse travel path and/or the arcuate travel portion of the forward travel path. Preferably, the vehicle travels with the maximum steering angle and thus with a maximum rotation radius. At least between the turning location and the final location and/or between the intermediate location and the turning location, the vehicle consequently travels with large or even maximum steering and therefore preferably with the smallest possible rotation radius. Therefore, the respective parking maneuver can be performed in parking environments with limited free space for maneuvering, as the vehicle's movement radius is kept as small as possible. The steering angle is kept constant throughout the entire movement between the turning point and the final location and/or between the intermediate location and the turning point. A travel path between these two respective locations therefore follows part of a circular arc.

In addition, one embodiment includes the respective parking maneuver involving driving straight backward from the end location to the parking location in the parking space. Compared to the end location, the parking location is closer to an end line that delimits the parking space in a longitudinal direction of the parking space. The parking location is the final parking position in the parking space. Alternatively, it can be referred to as the vehicle's destination location. In the case of a perpendicular parking space, the end line is preferably perpendicular to a finish line. The finish line is a line on which the vehicle is positioned at the end location. The finish line can be parallel to a surface of the end object and/or start object. The finish line connects the end location to the parking space. Alternatively, the finish line can be referred to as the parking line.

In summary, the vehicle can move from a current location to an intermediate location, from there, preferably with the maximum steering angle, forward to the turning location, from there, preferably with the maximum steering angle, backward to the final location, and from there straight backward to the parking location. This means that the parking maneuver can include up to five locations between four individual segments of the parking maneuver. A detailed parking path is therefore provided.

A further embodiment includes that the second parking maneuver includes that the end location is located at a distance from the parking location that is equal to a minimum distance value. The second parking maneuver therefore satisfies the constraint that the reverse travel path should reach the finish line at the end location, which is located in front of the parking location in the parking space, wherein the distance between the end location and the parking location is the minimum tolerance distance and therefore equal to the minimum distance value. The minimum distance value can, for example, be between 20 centimeters and 1 meter. The particularly short forward travel path in combination with the specification of the distance between the end location and the parking location defines the second parking maneuver. Some of the constraints are therefore satisfied by the second parking maneuver, and the other constraints are satisfied by the first parking maneuver. This explains how two parking maneuvers can be provided that are different from each other but optimized to satisfy at least some of the constraints described above. Therefore, two reasonable parking maneuvers are provided, one of which can be selected to operate the vehicle. This results in reliable parking with a reversing path.

According to a preferred embodiment, the method includes determining a further arcuate reversing path that satisfies constraints of the first parking maneuver and the second parking maneuver. It also includes performing a validation procedure. After performing the validation procedure, the vehicle is operated at least assisted according to a third parking maneuver that includes the determined further arcuate reversing path. The validation procedure includes at least one verification step to validate the further arcuate reversing path or a to check or verify another part of the third parking maneuver. The third parking maneuver differs at least in part from the first parking maneuver and the second parking maneuver.

This embodiment is based on the observation that the first and second parking maneuvers can be combined to make them more efficient in terms of the method's runtime. A common reverse circular arc can be calculated, which here is the further arcuate reversing path and which satisfies both constraints of no collision with the end object and reaching the finish line at a distance equal to or, in particular, greater than the minimum distance value. The calculation of the further arcuate reversing path is based on the first and second parking maneuvers. The further arcuate reversing path determined thereby is then checked by the validation procedure to ensure that the parking maneuver only follows the further arcuate reversing path if it is a sufficiently good solution for parking. Consequently, the above-described operation of the vehicle according to the first parking maneuver or the second parking maneuver can be replaced by the third parking maneuver, which includes the further arcuate reversing path. This embodiment therefore provides further optimization of the parking maneuver performed by the parking function, if such optimization is possible.

In addition, one embodiment includes the validation procedure verifying whether the further arcuate reversing path intersects a vehicle passage line that the vehicle follows when traveling straight forward from the current location. Therefore, it is verified whether the vehicle can cross the further arcuate reversing path if it were traveling straight forward from the current location. If so, the third parking maneuver includes a first forward path connecting the intermediate location and the turning location with a circular arc-shaped path. This easily determines the complete third parking maneuver, and the vehicle can be operated accordingly. Therefore, the parking maneuver satisfies many of the aforementioned constraints.

In a further embodiment, it is assumed that the further arcuate reversing path does not intersect the vehicle line of travel. The further arcuate reversing path therefore fails the validation procedure. In this case, the third parking maneuver includes a second forward path connecting the intermediate location and the turnaround location with an S-shaped path. The method thus includes determining or calculating a different forward path for the third parking maneuver if the further arcuate reversing does not intersect the vehicle line of travel. The S-shaped path involves a change in the steering direction to move the vehicle to the turnaround location where the further arcuate reversing portion of the parking maneuver starts. Therefore, the third parking maneuver is slightly adapted to this turnaround location.

According to another embodiment, the validation procedure includes verifying whether the S-shaped path collides with another object that laterally borders the parking space at its start according to the detection sequence other than the aforementioned object. This additional object therefore borders the parking space as the starting object. Here, the additional object is the starting object, and the aforementioned object is the ending object. Only if this is not the case, i.e., only if the S-shaped path does not collide with the additional object, does the third parking maneuver include the second forward travel path. This means that preferably, the S-shaped path is first further verified with regard to the available free space in the vicinity of the parking space, so that there is certainly enough space for the vehicle when following the S-shaped path. Although the S-shaped path may be a possibility to reach the turning point where the further arc-shaped reversing begins, it is not blindly chosen as part of the third parking maneuver unless the drivability of this path has been confirmed. Therefore, the process remains particularly reliable.

In an additional embodiment, it is assumed that the S-shaped path collides with the further object and therefore with the starting object. Then, the third parking maneuver involves a third forward travel path. The third forward travel path is determined taking into account a line on the opposite side that delimits an area in front of the parking space available for maneuvering, so that the vehicle reaches the line on the opposite side when it is at the turning point of the third forward travel path. It then starts the further arc-shaped reverse travel path at this turning point. The area in front of the parking space is the free space in the surroundings of the parking space. Preferably, the line on the opposite side is assumed to be virtually parallel to the vehicle passage line, since it is based on the limited free space confirmed by the vehicle's limited detection system, rather than the complete accurate detection of the objects on the opposite side. Side. Preferably, the line on the opposite side is aligned parallel to the vehicle passage line. The line on the opposite side may be an imaginary line at a specific distance from the parking space. Alternatively or additionally, it may be determined by analyzing the sensor information. In summary, the third parking maneuver is determined by shifting the further arcuate reversing path toward the line on the opposite side. The forward path of the third parking maneuver includes, for example, the straight part to the intermediate location and the arcuate part from there to the turning location. This forward path therefore adapts to the limited free space or the limited area available for maneuvering.

Another embodiment includes the validation procedure verifying whether the area in front of the parking space, and therefore the free space, is large enough for the vehicle to perform the third forward travel path. Only if this is the case will the third parking maneuver involve the third forward travel path. Therefore, it is again checked whether the determined forward travel path for the third parking maneuver remains within the available area or free space in the vicinity of the parking space. Consequently, the vehicle is prevented from colliding with any object or obstacle while performing the parking maneuver.

In one embodiment, it is assumed that the area, and therefore the free space, is not large enough for the vehicle to execute the third forward travel path. The method then includes the third parking maneuver including a fourth forward travel path. The fourth forward travel path connects the intermediate location and the turning location with a symmetrical S-shaped path, wherein the turning location is located on a target line that is tangential to the further curved reversing path. The symmetrical S-shaped path requires equal steering in both directions, i.e., equal steering to the left and right in the direction of travel of the vehicle. In addition, the turning location depends on the curved shape of the further curved reversing path, as it is located at a location that is tangential to the further curved reversing path. This enables a forward travel path that is particularly space-saving and therefore suitable for the relatively small area or free space. Even for such tight parking environments, the third parking maneuver can be determined and then used to operate the parking function.

If the method cannot determine the fourth forward travel path, the parking function may operate the vehicle according to the first or second parking maneuver. Alternatively, the parking function may abort and the vehicle will not be parked.

If the maneuver with the larger orientation angle, which is either the first parking maneuver or the second parking maneuver, collides with the object on the opposite side and/or at least one other object on the opposite side, the S-shaped maneuvers described above, which are either symmetrical or asymmetrical, can be proposed. The respective S-shaped maneuver allows the vehicle with the forward travel path to move closer to the parking space, thus enabling parking along the reverse travel path in a single reverse move.

If the respective S-shaped maneuver collides with the object on the opposite side or is not suitable due to limited space for maneuvering in front of the parking space, a single-stage reverse parking maneuver cannot be provided based on the given current location and the detected parking space structure. A multi-stage reverse parking maneuver can then be calculated and provided.

This means that the symmetrical S-shaped path is calculated to make the maneuver with the end object collision-free. It may not require steering at the maximum steering angle, but rather with tailored steering so that a collision of the vehicle with the end object is avoided. However, the method may fail to provide a solution for the symmetrical S-shaped path if the calculated symmetrical S-shaped path requires a free space for maneuvering that is larger than a predetermined maximum free space. The maximum free space may be a requirement of the parking function and specifies a region or area in front of the parking space that is available for the parking maneuver. Consequently, validating the fourth forward travel path may involve verifying that at least this path lies within the predetermined maximum free space in front of the parking space (not beyond it). If it does not lie within the maximum free space, the calculated symmetrical S-shaped path is rejected, and no solution can be provided for parking in a reverse trajectory.

The vehicle's control unit carries out all the described steps of the procedure, i.e. all the described determinations and/or calculations.

Furthermore, one aspect of the invention relates to a vehicle. The vehicle is configured to perform the method described above. The vehicle is preferably a motor vehicle. The vehicle is, for example, a passenger car, a truck, a bus, a motorcycle, and/or a moped. The motor vehicle includes a control unit for performing the method.

Another aspect of the invention relates to a control unit for a vehicle. The control unit is configured to perform the method described above. The control unit performs the method.

The control unit is a computing unit. The control unit can be understood, in particular, as a data processing device that includes a processing circuit. The control unit can therefore, in particular, process data to perform computing operations. This can also include operations to perform indexed accesses to a data structure, for example, a lookup table (LUT).

In particular, the control unit may include one or more computers, one or more microcontrollers, and/or one or more integrated circuits, for example, one or more application-specific integrated circuits (ASICs), one or more field-programmable gate arrays (FPGAs), and/or one or more systems-on-chip (SoCs). The control unit may also include one or more processors, for example, one or more microprocessors, one or more central processing units (CPUs), one or more graphics processing units (GPUs), and/or one or more signal processors, in particular one or more digital signal processors (DSPs). The control unit may also include a physical or virtual cluster of computers or other of the units.

In various embodiments, the control unit includes one or more hardware and/or software interfaces and/or one or more memory units.

A memory unit can be implemented as volatile data memory, for example dynamic random access memory, DRAM, or static random access memory, SRAM, or as non-volatile data memory, for example read-only memory, ROM, programmable read-only memory, PROM, erasable programmable read-only memory, EPROM, electrically erasable programmable read-only memory, EEPROM, flash memory or flash EEPROM, ferroelectric random access memory, FRAM, magnetoresistive random access memory, MRAM, or phase change random access memory, PCRAM.

Another aspect of the invention relates to a computer program product. In addition, one aspect of the invention relates to a computer program product that includes instructions. When the program is executed by a computer, such as the control unit, the instructions cause the computer to perform the method described above. The computer program product is a computer program.

Furthermore, the invention may include a computer-readable storage medium having instructions which, when executed by a computer such as the control unit, cause the computer to carry out the method described above.

Further features of the invention are evident from the claims, the figures, and the description of the figures. The invention includes combinations of the described embodiments. The embodiments described in connection with the method according to the invention apply individually and in combination with one another, as applicable, to the vehicle according to the invention, the control unit according to the invention, and the computer program product according to the invention.

Respective information within the meaning of the invention may include respective data. This means that the sensor information may include or be sensor data.

• 1 eine schematische Darstellung eines Fahrzeugs, das sich einem Parkplatz nähert;
• 2 eine schematische Darstellung eines Verfahrens zum Betreiben einer wenigstens assistierten Senkrecht- oder Schrägparkfunktion für ein Fahrzeug;
• 3 eine schematische Darstellung eines ersten Parkmanövers in einen Parkplatz;
• 4 eine schematische Darstellung, um die Berechnung des ersten Parkmanövers, wie in 3 gezeigt, zu veranschaulichen;
• 5 eine schematische Darstellung, um die Berechnung des ersten Parkmanövers, wie in 3 gezeigt, weiter zu veranschaulichen;
• 6 eine schematische Darstellung, um die Berechnung des ersten Parkmanövers, wie in 3 gezeigt, noch weiter zu veranschaulichen;
• 7 eine schematische Darstellung eines zweiten Parkmanövers in einen Parkplatz;
• 8 eine schematische Darstellung eines Vergleichs zwischen dem ersten und dem zweiten Parkmanöver;
• 9 eine schematische Darstellung eines dritten Parkmanövers in einen Parkplatz mit einem s-förmigen Pfad;
• 10 eine schematische Darstellung eines dritten Parkmanövers in einen Parkplatz, das in Richtung einer Linie auf der entgegengesetzten Seite verschoben ist; und
• 11 eine schematische Darstellung eines dritten Parkmanövers in einen Parkplatz mit einem symmetrischen s-förmigen Pfad.

The figures show in:

• 1 a schematic representation of a vehicle approaching a parking space;
• 2 a schematic representation of a method for operating at least an assisted vertical or diagonal parking function for a vehicle;
• 3 a schematic representation of a first parking maneuver into a parking space;
• 4 a schematic representation to calculate the first parking maneuver, as in 3 shown, to illustrate;
• 5 a schematic representation to calculate the first parking maneuver, as in 3 shown to further illustrate;
• 6 a schematic representation to calculate the first parking maneuver, as in 3 shown to further illustrate;
• 7 a schematic representation of a second parking maneuver into a parking space;
• 8 a schematic representation of a comparison between the first and second parking maneuvers;
• 9 a schematic representation of a third parking maneuver into a parking space with an S-shaped path;
• 10 a schematic representation of a third parking maneuver into a parking space, shifted towards a line on the opposite side; and
• 11 a schematic representation of a third parking maneuver into a parking space with a symmetrical S-shaped path.

1 shows a possible scenario in which the method according to the invention can be applied. A vehicle 1 can include a control unit 2, such as an electronic control unit (ECU) of the vehicle 1. The control unit 2 provides at least one assisted parking function 3 for vertical or diagonal parking. This parking function 3 can be designed for assisted, semi-automatic and/or fully automatic parking in a parking space 4. The parking space 4 here is a vertical parking space 4. Alternatively, it can be an inclined parking space 4. The parking space 4 is bounded laterally by a first detected object 5 (start object) and a second detected object 6 (end object). In the following, the second detected object 6 is referred to as object 6 and the first detected object 5 as further object 5. Both objects 5, 6 can influence the method.

Here, vehicle 1 approaches parking space 4 from a left side, with its right front corner closest to parking space 4. Therefore, parking space 4 can be referred to as a right gap or right parking space 4. Analogously, a left gap or left parking space 4 can be suitable for the procedure (not sketched here). Parking space 4 is spatially delimited by two imaginary lines, which here are a start line 9 and an end line 10 of parking space 4. End line 10 is a line that delimits parking space 4 in a longitudinal direction of parking space 4. Furthermore, an area 13 or free space available for maneuvering is spatially delimited by a line 8 on the opposite side. Here, several opposing objects 7 are arranged behind line 8 on the opposite side to illustrate that vehicle 1 may collide with such an object 7 if it were to pass line 8 on the opposite side.

The vehicle 1 may have a camera 11 and/or any other sensor device capable of detecting the parking space 4. The camera 11 and/or the sensor device captures an environment of the vehicle 1, which includes, for example, the parking space 4 and the area 13. The other sensor device may be, for example, a radar device, a lidar device, and/or an ultrasonic sensor. The camera 11 or the other front sensor device may capture sensor information 12, which includes, for example, sensor data. The captured sensor information 12 is transmitted to the control unit 2 so that the control unit 2 can provide the sensor information 12 to carry out the method according to the invention, as described below. Alternatively or additionally, the vehicle 1 may receive the sensor information 12 from an external device via vehicle-to-vehicle and/or vehicle-to-infrastructure communication.

In addition, 1 a possible parking maneuver 14 into the parking space 4. The parking maneuver 14 passes several locations. The parking function 3 provides the parking maneuver 14 into the vertical or inclined parking space 4. The parking maneuver 14 always includes a forward travel path from a current location C of the vehicle 1 to a turning location F, in particular via an intermediate location S. Furthermore, the parking maneuver 14 always includes an arcuate reverse travel path from the turning location F to an end location g, at in which the vehicle 1 is at least partially located in the parking space 4. The parking maneuver 14 can involve driving straight backward from the end location g to a parking location G in the parking space 4. Compared to the end location g, the parking location G is closer to the end line 10. Two possible end locations g are outlined here. The parking location G can alternatively be referred to as the destination location.

2 shows steps of a method for operating the at least assisted vertical or inclined parking function 3 for the vehicle 1. In a step S1, the camera 11 of the vehicle 1 and/or the other sensor device can capture the sensor information 12 describing the vertical or inclined parking space 4. In a step S2, the control unit 2 provides the sensor information 12. This means that the sensor information 12 is provided in step S2.

A step S3 includes determining a first parking maneuver 15 and a second parking maneuver 16 taking into account the sensor information 12. Therefore, a parking maneuver determination algorithm 18 can be applied to the sensor information 12, which includes at least one rule or software block to calculate the two parking maneuvers 15, 16. The first parking maneuver 15 includes a narrowest possible reversing path that guides the vehicle 1 so that it tangentially passes the corner point of the object 6 (end object) that laterally borders the parking space 4. The second parking maneuver 16 includes the shortest path of forward and reverse moves to reach a finish line 22 (see reference numeral 22 in 3 ) at the final location g, which is located at a distance from the parking location G that is equal to a minimum distance value 17 (see reference numeral 17 in 7 ) is.

Furthermore, the forward travel path of the respective parking maneuver 15, 16 can include a straight forward travel section from the current location C to the intermediate location S and an arcuate travel section from the intermediate location S to the turning location F. The arcuate reverse travel path and/or the arcuate travel section of the forward travel path can be drivable with a constant steering angle of at least 80 percent, in particular between 90 percent and 95 percent, of a maximum steering angle of a steering system of the vehicle 1. Preferably, the vehicle 1 moves with the maximum steering angle along the arcuate reverse travel path and/or the arcuate travel section of the forward travel path.

In a step S4, the method includes determining whether an orientation angle θ _F of the vehicle 1 at the turning location F is greater for the first parking maneuver 15 or for the second parking maneuver 16. In a step S5, the parking function 3 operates the vehicle 1 from the current location C to the turning location F and from there to the final location g, at least assisted, according to the determined parking maneuver 15, 16 having the larger orientation angle θ _F. Furthermore, step S5 may include operating the vehicle 1 from the final location g to the parking location G.

In the following, the determination of the first parking maneuver 15 is described in detail by 3 until 6 illustrated explained.

$$R_{min}=\frac{Radstand}{\text{tan}\left(Max.Radwinkel\right)}$$

3 represents the calculation of a point of a rotation center of the vehicle 1 for the shortest reverse arc that should guide the vehicle 1 so that it reaches a target line 22 with its orientation θ _G . Consequently, the target line 22 is tangential to the arc-shaped reverse path. This arc-shaped reverse path is represented here as a reverse circular arc 23 of the vehicle 1. Here, a radius of the reverse circular arc 23 is a minimum rotation radius R _min of the vehicle 1, which corresponds to the maximum steering. As described above, the maximum steering can mean a steering angle of 80 percent or higher of the maximum steering angle of the steering system of the vehicle 1. In addition, the reverse circular arc 23 is the shortest possible (due to the minimum rotation radius R _min ). In addition, it is the narrowest possible since it guides the vehicle 1 such that it touches an end object vertex P with coordinates (x _p ,y _p ) of the object 6 within a shortest rotation radius (R _l ) of a point in the vehicle 1. Such a point in the vehicle 1 with the shortest rotation radius R _l is an intersection point of a rear axle of the vehicle 1 with a side of the vehicle 1 that is close to the rotation center P. This side is here a right side of the vehicle 1 for the sketched right gap. For a left gap, it would be a left side of the vehicle 1. Consequently, R _l is related to R _min as follows, taking into account a width w of the vehicle 1 at the rear axle. Consequently, the generated reversing path is optimized as the shortest, drivable and collision-free. $$R_l=R_{min}-\frac{\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">w</figure-callout>}}{2}$$

$$R_{min}=\frac{Radstand}{\text{tan}\left(Max.Radwinkel\right)}$$

The turning location F, which can alternatively be referred to as the final frame (x _F ,y _F ,θ _F ) of the forward path before the reverse path starts, can be calculated as follows.

Given variables defined in global coordinates 20 are: the target line 22, the parking location G (hereinafter also referred to as target position (x _G ,y _G )) and a target orientation (θ _G ) at the parking location G, a vehicle passage line 24 during the search for the parking space 4, the current location C with coordinates (x _C ,y _C ) and orientation (θ _l ), the final object vertex P (gap entry point) with coordinates (x _p ,y _p ) and the minimum rotation radius R _min .

1. 1- Berechnen einer Ziellinienparallele 25 zur Ziellinie 22 in einem Abstand R_min in Richtung des Endobjekts 6.
2. 2- Berechnen des Kreisbogens, dessen Zentrum der Endobjekteckpunkt P ist, mit dem Radius (R_l).
3. 3- Berechnen eines Schnittpunkts näher am Parkort G der zum Ziel parallelen Linie 25 (bei 1 erhalten) mit dem Kreisbogen des Endobjekteckpunkts P (bei 2 erhalten). Dies soll das Rotationszentrum mit den Koordinaten (x_r,y_r) des Rückwärtskreisbogens 23 ergeben.
4. 4- Berechnen eines Berührungspunkts T mit den Koordinaten (x_t,y_t), der ein Punkt einer Berührung des Rückwärtskreisbogens 23 mit einer zur Durchfahrt parallelen Linie 26 ist, die zur gegebenen aktuellen Fahrzeugdurchfahrtslinie 24 parallel ist.
5. 5- Berechnen des Wendeorts F mit Koordinaten (x_F,y_F,θ_F) für den Vorwärtsfahrpfad, an dem der Rückwärtsfahrpfad startet.
6. 6- Berechnen des Zwischenorts S mit Koordinaten (x_S,y_S), an dem der bogenförmige Vorwärtsfahrteil startet.

The procedure may then include the following steps:

1. 1- Calculate a target line parallel 25 to the target line 22 at a distance R _min in the direction of the end object 6.
2. 2- Calculate the arc of a circle whose center is the final object vertex P, with radius (R _l ).
3. 3- Calculate an intersection point closer to the parking location G of the line 25 parallel to the target (obtained at 1) with the circular arc of the final object vertex P (obtained at 2). This should yield the center of rotation with the coordinates (x _r , y _r ) of the reverse circular arc 23.
4. 4- Calculating a point of contact T with the coordinates (x _t ,y _t ), which is a point of contact of the reverse circular arc 23 with a line 26 parallel to the passage, which is parallel to the given current vehicle passage line 24.
5. 5- Calculate the turning point F with coordinates (x _F ,y _F ,θ _F ) for the forward path where the reverse path starts.
6. 6- Calculate the intermediate location S with coordinates (x _S ,y _S ) where the arcuate forward movement starts.

Consequently, the complete path is obtained. The forward path begins with a straight path from coordinates (x _C , y _C ) to (x _S , y _S ), then the curved part begins at coordinates (x _S , y _S ) and ends at coordinates (x _F , y _F ). The reverse path begins at coordinates (x _p , y _p ) and ends at the finish line 22.

$$y-y_0=\text{tan}\left({\theta }_G\right)\left(x-x_0\right)$$

Since (x ₀ ,y ₀ ) is a projection of (x _G ,y _G ) onto the line 25 parallel to the target from the target line 22, which is located at a distance (R _min ) from the target line 22, the following equations can be derived, where (x,y) are represented according to the defined global coordinates 20: $$y-y_G=\text{tan}\left({\theta }_G\right)\left(x-x_G\right)$$

$$y-y_0=\text{tan}\left({\theta }_G\right)\left(x-x_0\right)$$

The projection line is perpendicular to both lines: $$y_G-y_0=\left(\frac{1}{\text{tan}\left({\theta }_G\right)}\right)\left(x_G-x_0\right)$$

$$y_G-y_0=K\left(x_G-x_0\right)$$

$${\left(y_G-y_0\right)}^2+{\left(x_G-x_0\right)}^2=R_{min}{}^2$$

$$\left(K^2+1\right){\left(x_G-x_0\right)}^2=R_{min}{}^2$$

It should apply $$K=\frac{-1}{\text{tan}\left({\theta }_G\right)}$$

$$y_G-y_0=K\left(x_G-x_0\right)$$

$${\left(y_G-y_0\right)}^2+{\left(x_G-x_0\right)}^2=R_{min}{}^2$$

$$\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right){\left(x_G-x_0\right)}^2=R_{min}{}^2$$

$$y_0-y_G=\frac{K\ast R_{min}}{\sqrt{\left(K^2+1\right)}}$$

$$x_0=x_G+\frac{R_{min}}{\sqrt{\left(K^2+1\right)}}$$

$$y_0=y_G+\frac{K\ast R_{min}}{\sqrt{\left(K^2+1\right)}}$$

Since the line 25 parallel to the target is arranged in the direction of the end object 6, then x ₀ > x _G : $$x_0-x_G=\frac{R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right)}}$$

$$y_0-y_G=\frac{K\ast R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right)}}$$

$$x_0=x_G+\frac{R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right)}}$$

$$y_0=y_G+\frac{K\ast R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right)}}$$

$${\left(y_r-r_P\right)}^2+{\left(x_r-x_P\right)}^2=R_l{}^2$$

Since the rotation center (x _r ,y _r ) of the backward circular arc 23 is located on the line 25 parallel to the target and at a distance (R _l ) from the final object vertex P with the coordinates (x _p ,y _p ), then (x _r, y _r ) can be obtained by solving the following two equations as follows. $$y_r-y_0=\text{tan}\left({\theta }_G\right)\left(x_r-x_0\right)$$

$${\left(y_r-r_P\right)}^2+{\left(x_r-x_P\right)}^2={\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2$$

$${\left(y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)-y_P\right)}^2+{\left(x_r-x_P\right)}^2=R_l{}^2$$

$${\left(y_0-y_P\right)}^2+2\left(y_0-y_P\right)tan\left({\theta }_G\right)\left(x_r-x_0\right)+ta{n}^2\left({\theta }_G\right){\left(x_r-x_0\right)}^2+{\left(x_r-x_P\right)}^2=R_l{}^2$$

$$\begin{array}{c}\left(ta{n}^2\left({\theta }_G\right)+1\right)x_r{}^2+2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)x_e+{\left(y_0-y_P\right)}^2-R_l{}^2\\ -2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_P{}^2=0\end{array}$$

$$A{x}_r{}^2+B{x}_r+C=0$$

$$A=\left(ta{n}^2\left({\theta }_G\right)+1\right)$$

$$B=2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)$$

$$C={\left(y_0-y_P\right)}^2-R_l{}^2-2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_P{}^2$$

$$x_r=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

$$x_r=\frac{\left(-B\pm \sqrt{B^2-4AC}\right)/ta{n}^2\left({\theta }_G\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}=\frac{\left(-B/ta{n}^2\left({\theta }_G\right)\pm \sqrt{\left(B^2-4AC\right)/ta{n}^4\left({\theta }_G\right)}\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}$$

$$x_r=\frac{\left(-B/ta{n}^2\left({\theta }_G\right)\pm \sqrt{\left(B^2/ta{n}^4\left({\theta }_G\right)-4AC/ta{n}^4\left({\theta }_G\right)\right)}\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}$$

From [3], y _r can be obtained as a function in x _r and used in [4] as follows. $$y_r=y_0+\text{tan}\left({\theta }_G\right)\left(x_r-x_0\right)$$

$${\left(y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)-y_P\right)}^2+{\left(x_r-x_P\right)}^2={\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2$$

$${\left(y_0-y_P\right)}^2+2\left(y_0-y_P\right)tan\left({\theta }_G\right)\left(x_r-x_0\right)+ta{n}^2\left({\theta }_G\right){\left(x_r-x_0\right)}^2+{\left(x_r-x_P\right)}^2={\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2$$

$$\begin{array}{c}\left(ta{n}^2\left({\theta }_G\right)+1\right)x_{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">r</figure-callout>}}{}^2+2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)x_e+{\left(y_0-y_P\right)}^2-{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2\\ -2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">P</figure-callout>}}{}^2=0\end{array}$$

$$A{\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_r{}^2+B{x}_r+C=0$$

$$A=\left(ta{n}^2\left({\theta }_G\right)+1\right)$$

$$B=2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)$$

$$C={\left(y_0-y_P\right)}^2-{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">P</figure-callout>}}{}^2$$

$$x_r=\frac{-B\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$x_r=\frac{\left(-B\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}\right)/ta{n}^2\left({\theta }_G\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}=\frac{\left(-B/ta{n}^2\left({\theta }_G\right)\pm \sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC\right)/ta{n}^4\left({\theta }_G\right)}\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}$$

$$x_r=\frac{\left(-B/ta{n}^2\left({\theta }_G\right)\pm \sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2/ta{n}^4\left({\theta }_G\right)-4AC/ta{n}^4\left({\theta }_G\right)\right)}\right)}{\left(2A\right)/ta{n}^2\left({\theta }_G\right)}$$

$$

$$x_r=\frac{\left(2x_0\pm \sqrt{4x_0{}^2-4x_0{}^2}\right)}{2}=x_0$$

To determine a unique solution for x _r , it should be assumed that ${\theta }_G=90°\to tan\left({\theta }_G\right)=\infty \to \frac{1}{tan\left({\theta }_G\right)}=0$

$$

$$x_r=\frac{\left(2x_0\pm \sqrt{4x_0{}^2-4x_0{}^2}\right)}{2}=x_0$$

$$B=-2x_P$$

$$C={\left(y_0-y_P\right)}^2-R_l{}^2+x_P{}^2$$

$$B^2=4x_P{}^2$$

$$4AC=4{\left(y_0-y_P\right)}^2-4R_l{}^2+4x_P{}^2$$

$$B^2-4AC=4\left(R_l{}^2-{\left(y_0-y_P\right)}^2\right)$$

$$x_r=\frac{\left(2x_P\pm \sqrt{4\left(R_l{}^2-{\left(y_0-y_P\right)}^2\right)}\right)}{2}=x_P\pm \sqrt{R_l{}^2-{\left(y_0-y_P\right)}^2}$$

And it is assumed that θ _G = 0° → tan(θ _G ) = 0, then: $$A=1$$

$$B=-2x_P$$

$$C={\left(y_0-y_P\right)}^2-{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2+x_{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">P</figure-callout>}}{}^2$$

$${\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2=4{\mathrm{<figure-callout\; id="2"\; label="x\; P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_P{}^2$$

$$4AC=4{\left(y_0-y_P\right)}^2-4{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2+4{\mathrm{<figure-callout\; id="2"\; label="x\; P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_P{}^2$$

$${\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC=4\left({\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-{\left(y_0-y_P\right)}^2\right)$$

$$x_r=\frac{\left(2x_P\pm \sqrt{4\left({\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-{\left(y_0-y_P\right)}^2\right)}\right)}{2}=x_P\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-{\left(y_0-y_P\right)}^2}$$

Since x _r < x _p for the right gap according to the rotated coordinates 21, which are in 3 shown (which are rotated relative to the global coordinates 20), the unique solution can be specified. For the left gap, the other solution can be chosen, since the center of rotation should lie above the final object vertex P. $$x_r=x_P-\sqrt{{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-{\left(y_0-y_P\right)}^2}$$

$$B=2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)$$

$$C={\left(y_0-y_P\right)}^2-R_l{}^2-2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_P{}^2$$

$$falls\left(rechteL\ddot{u}cke\right)$$

$$x_r=\frac{-B-\sqrt{B^2-4AC}}{2A}$$

$$ansonsten$$

$$x_r=\frac{-B+\sqrt{B^2-4AC}}{2A}$$

$$y_r=y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)$$

Consequently, the center of rotation of the backward arc 23 with the coordinates (x _r ,y _r ) can be calculated as follows: $$A=\left(ta{n}^2\left({\theta }_G\right)+1\right)$$

$$B=2\left(\left(y_0-y_P\right)tan\left({\theta }_G\right)-x_{0}ta{n}^2\left({\theta }_G\right)-x_P\right)$$

$$C={\left(y_0-y_P\right)}^2-{\mathrm{<figure-callout\; id="2"\; label="R\; l"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">R</figure-callout>}}_l{}^2-2\left(y_0-y_P\right)tan\left({\theta }_G\right)x_0+ta{n}^2\left({\theta }_G\right)x_0{}^2+x_{\mathrm{<figure-callout\; id="2"\; label="P"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">P</figure-callout>}}{}^2$$

$$falls\left(rechteL\ddot{u}cke\right)$$

$$x_r=\frac{-B-\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$ansonsten$$

$$x_r=\frac{-B+\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$y_r=y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)$$

The point of contact T with the coordinates (x _t ,y _t ) is the point at which the vehicle 1 can start the optimal smallest backward circular arc 23 with minimum radius in the direction of the target line 22 without collision with the end object vertex P from a line 26 parallel to the passage to the travel direction line (vehicle passage line 24) during the search for the parking space 4.

$${\left(y_t-y_r\right)}^2+{\left(x_t-x_r\right)}^2=R_{min}{}^2$$

The vehicle should not reach the contact point T. Rather, it should start the reverse path earlier at the turning point F with the coordinates (x _F ,y _F ). Since the contact point T is located on the reverse circular arc 23 and is the projection of its rotation center onto the line 26 parallel to the passage, the following two equations can be obtained and solved to obtain the contact point T as follows. $$x_t-x_r=-tan\left({\theta }_l\right)\left(y_t-y_r\right)$$

$${\left(y_t-y_r\right)}^2+{\left(x_t-x_r\right)}^2=R_{min}{}^2$$

$${\left(x_t-x_r\right)}^2=N^2{\left(y_t-y_r\right)}^2$$

$${\left(y_t-y_r\right)}^2\left(N^2+1\right)=R_{min}{}^2$$

$$y_t=y_r+\frac{R_{min}}{\sqrt{\left(N^2+1\right)}}$$

$$x_t=x_r+\frac{N\ast R_{min}}{\sqrt{\left(N^2+1\right)}}$$

It should apply $$N=-tan\left({\theta }_l\right)$$

$${\left(x_t-x_r\right)}^2=N^2{\left(y_t-y_r\right)}^2$$

$${\left(y_t-y_r\right)}^2\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)=R_{min}{}^2$$

$$y_t=y_r+\frac{R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)}}$$

$$x_t=x_r+\frac{N\ast R_{min}}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)}}$$

• Der Projektionsabstand von der berechneten Zielposition T zur gegebenen Fahrzeugdurchfahrtslinie 24 kann berechnet und mit (Δ) dargestellt werden. Da die Bewegung von (x_S,y_S) zu (x_F,y_F) ein Kreisbogen ist und die Bewegung von (x_F,y_F) zu (x_t,y_t) ein umgekehrter Kreisbogen ist, ist dann (x_F,y_F) in der Mitte angeordnet und die folgenden Gleichungen können hergeleitet werden.

$$\frac{\mathrm{<figure-callout\; id="2"\; label="\Delta "\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">\Delta </figure-callout>}}{2}=R_{min}-R_{min}\text{ cos}\left(\Delta \theta \right)$$

$$2R_{min}\left(1-\text{cos}\left(\Delta \theta \right)\right)=\Delta$$

$$\text{cos}\left(\Delta \theta \right)=1-\frac{\Delta }{2R_{min}}$$

After the target position T is calculated, the start and final positions of the forward arc (intermediate location S to the turning location F) can be calculated based on a simple geometry according to 4 be calculated as follows:

• The projection distance from the calculated target position T to the given vehicle passage line 24 can be calculated and represented by (Δ). Since the movement from (x _S ,y _S ) to (x _F ,y _F ) is a circular arc and the movement from (x _F ,y _F ) to (x _t ,y _t ) is an inverse circular arc, then (x _F ,y _F ) is located at the center and the following equations can be derived.

$$\frac{\mathrm{<figure-callout\; id="2"\; label="\Delta "\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">\Delta </figure-callout>}}{2}=R_{min}-R_{min}\text{ cos}\left(\Delta \theta \right)$$

$$2R_{min}\left(1-\text{cos}\left(\Delta \theta \right)\right)=\Delta$$

$$\text{cos}\left(\Delta \theta \right)=1-\frac{\Delta }{2R_{min}}$$

$${\theta }_F={\theta }_l+{\text{cos}}^{-1}\left(1-\frac{\Delta }{2R_{min}}\right)$$

falls θ_l = 0, dann x_t - x_S = 2 R_min sin(Δθ).Since the vehicle passage line 24 has the orientation (θ _l ), coordinates {x _F ,y _F ,θ _F } for the turning point F for the right gap can be calculated as follows: $${\theta }_F={\theta }_l+\Delta \theta$$

$${\theta }_F={\theta }_l+{\text{cos}}^{-1}\left(1-\frac{\Delta }{2R_{min}}\right)$$

if θ _l = 0, then x _t - x _S = 2 R _min sin(Δθ).

falls ${\theta }_l=\mathrm{0,}\text{ dann }{x}_t-x_s=2R_{min}\text{sin}\left(\Delta \theta \right)=\sqrt{4R_{min}\Delta -{\Delta }^2}$

und $x_t-x_F=R_{min}\text{sin}\left(\Delta \theta \right)=\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}$

$$x_F=x_t-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{cos}\left({\theta }_l\right)+\left(\frac{\Delta }{2}\right)\text{sin}\left({\theta }_l\right)$$

$$x_F=x_t-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{sin}\left({\theta }_l\right)-\left(\frac{\Delta }{2}\right)\text{cos}\left({\theta }_l\right)$$

$$x_S=x_F-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{cos}\left({\theta }_l\right)+\left(\frac{\Delta }{2}\right)\text{sin}\left({\theta }_l\right)$$

$$y_S=y_F-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{sin}\left({\theta }_l\right)-\left(\frac{\Delta }{2}\right)\text{cos}\left({\theta }_l\right)$$

Considering cos(Δθ), as in 5 As shown, sin(Δθ) can be calculated as: $$\text{sin}\left(\Delta \theta \right)=\frac{\sqrt{4R_{min}\Delta -{\mathrm{<figure-callout\; id="2"\; label="\Delta "\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">\Delta </figure-callout>}}^2}}{2R_{min}}$$

if ${\theta }_l=\mathrm{0,}\text{ dann }{x}_t-x_s=2R_{min}\text{sin}\left(\Delta \theta \right)=\sqrt{4R_{min}\Delta -{\mathrm{<figure-callout\; id="2"\; label="\Delta "\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">\Delta </figure-callout>}}^2}$

and $x_t-x_F=R_{min}\text{sin}\left(\Delta \theta \right)=\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}$

$$x_F=x_t-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{cos}\left({\theta }_l\right)+\left(\frac{\Delta }{2}\right)\text{sin}\left({\theta }_l\right)$$

$$x_F=x_t-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{sin}\left({\theta }_l\right)-\left(\frac{\Delta }{2}\right)\text{cos}\left({\theta }_l\right)$$

$$x_S=x_F-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{cos}\left({\theta }_l\right)+\left(\frac{\Delta }{2}\right)\text{sin}\left({\theta }_l\right)$$

$$y_S=y_F-\left(\sqrt{R_{min}\Delta -{\left(\frac{\Delta }{2}\right)}^2}\right)\text{sin}\left({\theta }_l\right)-\left(\frac{\Delta }{2}\right)\text{cos}\left({\theta }_l\right)$$

$$y*-y_t=\left(\frac{-1}{tan\left({\theta }_l\right)}\right)\left(x*-x_t\right)$$

The calculation of the projection point (x*,y*) and the perpendicular distance (Δ) from the point of contact T with coordinates (x _t ,y _t ) to the given vehicle passage line 24 at the current location (x _C ,y _C ,θ _l ) can be carried out as follows according to 6 be derived and formulated: $$y^{*}-y_c=tan\left({\theta }_l\right)\left(x*-x_C\right)$$

$$y*-y_t=\left(\frac{-1}{tan\left({\theta }_l\right)}\right)\left(x*-x_t\right)$$

$$x*=\frac{\left(y_t-t_C\right)+\left(tan\left({\theta }_l\right)x_C+\left(\frac{1}{tan\left({\theta }_l\right)}\right)x_t\right)}{\left(tan\left({\theta }_l\right)+\frac{1}{tan\left({\theta }_l\right)}\right)}$$

$$x*=\frac{\left(y_t-y_C\right)tan\left({\theta }_l\right)+ta{n}^2\left({\theta }_l\right)x_C+x_t}{1+ta{n}^2\left({\theta }_l\right)}$$

$$y*=y_C+tan\left({\theta }_l\right)\left(x*-x_C\right)$$

$$\Delta =\sqrt{{\left(x*-x_t\right)}^2+{\left(y*-y_t\right)}^2}$$

By subtracting equation 8 from equation 7: $$y_t-t_C=\left(tan\left({\theta }_l\right)+\frac{1}{tan\left({\theta }_l\right)}\right)x*-\left(tan\left({\theta }_l\right)x_{C}9+\left(\frac{1}{tan\left({\theta }_l\right)}\right)x_t\right)$$

$$x*=\frac{\left(y_t-t_C\right)+\left(tan\left({\theta }_l\right)x_C+\left(\frac{1}{tan\left({\theta }_l\right)}\right)x_t\right)}{\left(tan\left({\theta }_l\right)+\frac{1}{tan\left({\theta }_l\right)}\right)}$$

$$x*=\frac{\left(y_t-y_C\right)tan\left({\theta }_l\right)+ta{n}^2\left({\theta }_l\right)x_C+x_{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">t</figure-callout>}}}{1+ta{n}^2\left({\theta }_l\right)}$$

$$y*=y_C+tan\left({\theta }_l\right)\left(x*-x_C\right)$$

$$\Delta =\sqrt{{\left(x*-x_t\right)}^2+{\left(y*-y_t\right)}^2}$$

In the following, the determination of the second parking maneuver 16 is described in detail by 7 illustrated for the right gap. Therefore, it is explained how the shortest path of circular forward travel (arc-shaped forward travel), followed by circular reversing with maximum steering segments in different directions that are required for the vehicle 1 to reach the finish line 22 at the minimum distance value 17 from the parking location G are to be calculated. In the following, the minimum distance value 17 is referred to as the minimum tolerance distance. Therefore, a new candidate target position, which is the end location g, must be reached at the end of the reversing path in the direction of the finish line 22. This new candidate target has the coordinates (x _g , y _g ) and is located on the finish line 22 at the specific minimum tolerance distance (d) above the parking location G with the coordinates (x _G , y _G ).

$$y_g=y_G+d\ast \text{sin}\left({\theta }_G\right)$$

Considering the parking location G with (x _G ,y _G ), the candidate target position (x _g ,y _g ) (final location g) can be calculated as follows: $$x_g=x_G+d\ast \text{cos}\left({\theta }_G\right)$$

$$y_g=y_G+d\ast \text{sin}\left({\theta }_G\right)$$

$$y_{r2}=y_g-R_{min}\ast \text{cos}\left({\theta }_G\right)$$

Then, from the obtained final position g, the center of rotation (x _{r 2} ,y _{r 2} ) of a second circle of the reverse arc 23 can be calculated, since its projection onto the target line 22 is the final location g with (x _g ,y _g ). It is also located at a distance (R _min ) from the target line 22: $${\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_r=x_g+R_{min}\ast \text{sin}\left({\theta }_G\right)$$

$${\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_r=y_g-R_{min}\ast \text{cos}\left({\theta }_G\right)$$

$$y_m=y_C+R_{min}\ast \text{cos}\left({\theta }_l\right)$$

Similarly, (x _m ,y _m ), which is the projection of the current location C with (x _C ,y _C ) onto the line 26 parallel to the vehicle passage, can be calculated, since the line 26 parallel to the vehicle passage is located here at a distance (R _min ) from the vehicle passage line 24 in the direction of the line 8 on the opposite side: $$x_m=x_C-R_{min}\ast \text{sin}\left({\theta }_l\right)$$

$$y_m=y_C+R_{min}\ast \text{cos}\left({\theta }_l\right)$$

$$y_{r_1}-y_m=tan\left({\theta }_l\right)\left(x_{r_1}-x_m\right)$$

The next step is to find the center of rotation (x _{r 1} ,y _{r 1} ) of the first forward circular path located on the line 26 parallel to the vehicle passage, indicated by the point (x _m ,y _m ) and the orientation (θ _l ), and at a distance (2R _min ) from the center of rotation (x _{r 2} ,y _{r 2} ) of the second circular backward path. Consequently, (x _{r 1} ,y _{r 1} ) can be calculated by solving the following two equations: $${\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_1}\right)}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_1}\right)}^2={\left(2R_{min}\right)}^2$$

$${\mathrm{<figure-callout\; id="1"\; label="y\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m=tan\left({\theta }_l\right)\left({\mathrm{<figure-callout\; id="1"\; label="x\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_m\right)$$

$$y_{r_1}=y_m+K\left(x_{r_1}-x_m\right)$$

It should apply $$K=tan\left({\theta }_l\right)$$

$${\mathrm{<figure-callout\; id="1"\; label="y\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}=y_m+K\left(x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}-x_m\right)$$

$$\begin{array}{c}{x}_{r_2}{}^2-\left(2x_{r_2}\right)x_{r_1}+x_{r_1}{}^2+{\left(y_{r_2}-y_m\right)}^2-2K\left(y_{r_2}-y_m\right)x_{r_1}+2K\left(y_{r_2}-y_m\right)x_m+\left(K^2\right)x_{r_1}{}^2\\ -\left(2K^2{x}_m\right)x_{r_1}+K^2{x}_m{}^2-4R_{min}{}^2=0\end{array}$$

$$\begin{array}{c}{x}_{r_2}{}^2-\left(2x_{r_2}\right)x_{r_1}+x_{r_1}{}^2+{\left(y_{r_2}-y_m\right)}^2-2K\left(y_{r_2}-y_m\right)x_{r_1}+2K\left(y_{r_2}-y_m\right)x_m+\left(K^2\right)x_{r_1}{}^2\\ -\left(2K^2{x}_m\right)x_{r_1}+K^2{x}_m{}^2-4R_{min}{}^2=0\end{array}$$

$$\begin{array}{l}\left(K^2+1\right)x_{r_1}{}^2-\left(2K\left(y_{r_2}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)x_{r_1}\hfill \\ +\left(x_{r_2}{}^2+{\left(y_{r_2}-y_m\right)}^2+2K\left(y_{r_2}-y_m\right)x_m+K^2{x}_m{}^2-4R_{min}{}^2\right)=0\hfill \end{array}$$

$$A=K^2+1$$

$$B=-\left(2K\left(y_{r_2}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)$$

$$C=\left(x_{r_2}{}^2+{\left(y_{r_2}-y_m\right)}^2+2K\left(y_{r_2}-y_m\right)x_m+K^2{x}_m{}^2-4R_{min}{}^2\right)$$

$$x_{r_1}=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

Inserting equation [10] into equation [9]: $${\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_1}\right)}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m-K\left(x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}-x_m\right)\right)}^2={\left(2R_{min}\right)}^2$$

$$\begin{array}{c}{x}_{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">r</figure-callout>}}_2}{}^2-\left(2x_{r_2}\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2-2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_m+\left(K^2\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}{}^2\\ -\left(2K^2{x}_m\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+K^2{x}_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}{}^2-4R_{min}{}^2=0\end{array}$$

$$\begin{array}{c}{x}_{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">r</figure-callout>}}_2}{}^2-\left(2x_{r_2}\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2-2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_m+\left(K^2\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}{}^2\\ -\left(2K^2{x}_m\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+K^2{x}_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}{}^2-4R_{min}{}^2=0\end{array}$$

$$\begin{array}{l}\left({\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}{}^2-\left(2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}\hfill \\ +\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2+2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_m+K^2{x}_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}{}^2-4R_{min}{}^2\right)=0\hfill \end{array}$$

$$A={\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1$$

$$B=-\left(2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)$$

$$C=\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2+2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_m+K^2{x}_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}{}^2-4R_{min}{}^2\right)$$

$$x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}=\frac{-B\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$A=1$$

$$B=-2x_{r_2}$$

$$C=x_{r_2}{}^2+{\left(y_{r_2}-y_m\right)}^2-16R_{min}{}^2$$

$$B^2-4AC=4x_{r_2}{}^2-4x_{r_2}{}^2-4{\left(y_{r_2}-y_m\right)}^2+16R_{min}{}^2=16R_{min}{}^2-4{\left(y_{r_2}-y_m\right)}^2$$

$$x_{r_1}=\frac{-B\pm \sqrt{B^2-4AC}}{2A}=x_{r_2}\pm \sqrt{{\left(2R_{min}\right)}^2-{\left(y_{r_2}-y_m\right)}^2}$$

To find the unique solution for x _{r 1} To determine, we can assume: $$K=0\leftrightarrow tan\left({\theta }_l\right)=0\leftrightarrow {\theta }_l=0$$

$$A=1$$

$$B=-2{\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}$$

$$C={\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2-16R_{min}{}^2$$

$${\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC=4{\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2-4{\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2-4{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2+16R_{min}{}^2=16R_{min}{}^2-4{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2$$

$$x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}=\frac{-B\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}={\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}\pm \sqrt{{\left(2R_{min}\right)}^2-{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2}$$

$$A=K^2+1$$

$$B=-\left(2K\left(y_{r_2}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)$$

$$C=\left(x_{r_2}{}^2+{\left(y_{r_2}-y_m\right)}^2+2K\left(y_{r_2}-y_m\right)x_m+K^2{x}_m{}^2-4R_{min}{}^2\right)$$

$$x_{r_1}=\frac{-B-\sqrt{B^2-4AC}}{2A}$$

$$y_{r_1}=y_m+K\left(x_{r_1}-x_m\right)$$

Since x _{r 1} < x _{r 2} holds for both right and left gaps, then the unique solution for x _{r 1} be defined: $${\mathrm{<figure-callout\; id="1"\; label="x\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}=\frac{-B-\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}={\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-\sqrt{{\left(2R_{min}\right)}^2-{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2}$$

$$A={\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">K</figure-callout>}}^2+1$$

$$B=-\left(2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)+\left(2K^2{x}_m\right)+\left(2x_{r_2}\right)\right)$$

$$C=\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}{}^2+{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)}^2+2K\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_m\right)x_m+K^2{x}_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}{}^2-4R_{min}{}^2\right)$$

$$x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}=\frac{-B-\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$${\mathrm{<figure-callout\; id="1"\; label="y\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}=y_m+K\left(x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}-x_m\right)$$

$$x_F=x_{r_1}+R_{min}\ast \text{sin}\left({\theta }_F\right)$$

$$y_F=y_{r_1}-R_{min}\ast \text{cos}\left({\theta }_F\right)$$

$$x_S=x_{r_1}+R_{min}\ast \text{sin}\left({\theta }_l\right)$$

$$y_S=y_{r_1}-R_{min}\ast \text{cos}\left({\theta }_l\right)$$

Since vehicle 1 is tangential to both forward and reverse arcs at the turning point F, the heading of the vehicle (θ _F ) is perpendicular to the line connecting the two centers of the arcs. Consequently, it can be calculated as follows. Then, the positions of both the intermediate point S with (x _S ,y _S ) as well as the turning point F with (x _F ,y _F ) of the forward path can be calculated, since both are located on the first forward arc and the points of contact with the orientation lines are (θ _l ) and (θ _F ) respectively: $${\theta }_F={\text{tan}}^{-1}\left(\frac{-\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_1}\right)}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_1}}\right)={\text{tan}}^{-1}\left(\frac{{\mathrm{<figure-callout\; id="1"\; label="x\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_2}}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_1}}\right)$$

$$x_F=x_{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}+R_{min}\ast \text{sin}\left({\theta }_F\right)$$

$$y_F={\mathrm{<figure-callout\; id="1"\; label="y\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-R_{min}\ast \text{cos}\left({\theta }_F\right)$$

$$x_S={\mathrm{<figure-callout\; id="1"\; label="x\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}+R_{min}\ast \text{sin}\left({\theta }_l\right)$$

$$y_S={\mathrm{<figure-callout\; id="1"\; label="y\; r"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-R_{min}\ast \text{cos}\left({\theta }_l\right)$$

So far, the first parking maneuver 15 and the second parking maneuver 16 can be determined based on the above equations. The circular forward and reverse paths are preferably set with maximum steering (minimal radius and maximum curvature) to ensure that the circular paths are the shortest possible, as this is very advantageous in tight parking situations and could have additional benefits in less tight parking situations to reduce the traveled distance and the associated parking space 4.

8 shows how the orientation change over the circular forward step affects the depth of the reached position on the finish line 22. As the length of the circular forward path increases, the orientation change increases and vehicle 1 gets closer to the line 8 on the opposite side. Meanwhile, the length of the circular backward step to the finish line 22 becomes shorter and vehicle 1 reaches the finish line 22 earlier. 8 The orientation angles and lengths on the target line 22 are sketched as examples for the first parking maneuver 15 and the second parking maneuver 16. As the orientation change in the forward travel path decreases (as shown in the orientation for the first parking maneuver 15 compared to the orientation for the second parking maneuver 16), the position reached on the target line 22 becomes lower (as shown in the positions for the respective parking maneuver 15, 16).

On the basis of this 8 From the rule shown, it can be deduced that the orientation at the turning location F, which is the orientation angle (θ _F ) achieved in the first parking maneuver 15, is the minimum orientation that can be achieved in the forward travel step to cause the vehicle 1 to reach the target line 22 in the reverse travel step without colliding with the end object vertex P. Moreover, the orientation of the turning location F reached in the second parking maneuver 16 is the minimum orientation that can be achieved in the forward travel step to cause the vehicle 1 to reach the target line 22 in the reverse travel step at the maximum acceptable depth, which lies at the minimum tolerance distance d from the parking location G.

Consequently, the selection criteria for the appropriate forward step between the one with minimal distance and minimal orientation change to avoid collision with a gap entry corner and the one with minimal distance and minimal orientation change to achieve the minimum tolerance distance d to the parking location G consist of simply selecting the one with a larger orientation change (the maximum of them) to ensure the other. The idea of the maximum selection is to guarantee that the generated forward step is the minimum possible forward path toward the line 8 on the opposite side, which guarantees that the following reverse step reaches the parking position G without collision. Consequently, the final generated forward step, which here is the second parking maneuver 16, is the optimized shortest path leading to safe parking in one step.

The turning location F with {x _F ,y _F ,θ _F } of such a generated forward travel path can be checked for collision with the line 8 on the opposite side. If it collides with the line 8 on the opposite side or exceeds the acceptable limit for a virtual line 8 on the opposite side, then it should be deduced that no possible solution for a one-step reverse parking by starting with the forward circular motion towards the line 8 on the opposite side exists. In such a case, the vehicle 1 shall start parking with the conventional reverse step towards the parking space 4 without any need to start with the forward circular path.

If no end object vertex P is present, then generally the second parking maneuver 16 is selected for operating the vehicle 1 according to the available free space between the vehicle 1 and the virtual line 8 on the opposite side.

9 until 11 show optimized selection criteria for single-stage perpendicular or diagonal reverse parking. The method may thus include determining a further arcuate reverse path 30 that satisfies constraints of the first parking maneuver 15 and the second parking maneuver 16. The method may also include performing a validation procedure, wherein, after performing the validation procedure, the vehicle 1 is operated at least assisted according to a third parking maneuver 31 that includes the determined further arcuate reverse path 30.

The validation procedure may include checking whether the further arcuate reversing path 30 intersects the vehicle passage line 24 that the vehicle follows when traveling straight forward from the current location (C). If this is the case, the third parking maneuver 31 includes a first forward path connecting the intermediate location S and the turning location F with a circular arc-shaped path. However, if the further arcuate reversing path 30 does not intersect the vehicle passage line 24, the third parking maneuver 31 may include a second forward path 32 connecting the intermediate location S and the turning location F with an S-shaped path. This S-shaped path is in 9 shown.

The validation procedure may include checking whether the S-shaped path collides with the additional object 5 that laterally borders the parking space 4 as the starting object. Only if this is not the case does the third parking maneuver 31 include the second forward travel path 32. It should be required that the parking space 4 be detected early enough so that there is sufficient space for the S-shaped maneuver to be collision-free.

If the S-shaped path collides with the further object 5, the third parking maneuver 31 may include a third forward travel path 33, which is determined taking into account the line 8 on the opposite side, which delimits the area 13 in front of the parking space 4 that is available for maneuvering. Consequently, the vehicle 1 reaches the line 8 on the opposite side when it is located at the turning location F of the third forward travel path 33 and starts the further curved reverse travel path 30 at this turning location F. The line 8 on the opposite side is preferably aligned parallel to the vehicle passage line 24. This shifted turning location F according to the third forward travel path 33 is in 10 shown.

The validation procedure may include checking whether the area 13 in front of the parking space 4 is large enough for the vehicle 1 to execute the third forward travel path 33. Only if this is the case can the third parking maneuver 31 include the third forward travel path 33. If the area is not large enough, the third parking maneuver 31 may include a fourth forward travel path 34 connecting the intermediate location S and the turning location F with a symmetrical S-shaped path, wherein the turning location F is arranged on a finish line 35 that is tangential to the further curved reversing path 30. The third parking maneuver 31 with such a fourth forward travel path 34 is in 11 shown.

In other words, the methods for determining the first parking maneuver 15 and the second parking maneuver 16 can be combined in a more efficient manner in terms of running time. This allows a common reverse circular arc, which is the further arc-shaped reverse path 30, to be generated, which satisfies both the constraints of no collision with the end object 6 and reaching the target line 22 at a distance from the parking location G that is greater than or equal to the minimum tolerance distance (d). The calculation of the further arc-shaped reverse path 30, which satisfies the above-defined constraints, can be detailed as follows according to the following common conditions. The obtained further arc-shaped reverse path 30 is then compared with the vehicle passage line 24. If the vehicle passage line 24 intersects the further arc-shaped reverse path 30, then they should be connected with a forward circular arc. Otherwise, they should be connected with the S-shaped path. Alternatively, the further curved reversing path 30 shall be shifted away from the parking space 4 and therefore move towards the line 8 on the opposite side, so that the vehicle 1 shall reach the destination line 22 earlier and park at a more limited driving distance.

The s-shaped forward path solution (as in 9 shown) is an optimized one for parking at a very limited driving distance. However, it must be checked whether it is collision-free with the gap structure (basically the end object 6 and a possible start object 5). It is determined based on an s-shaped starting point so that it is reached by the vehicle as early as possible. The starting point can be the current location C or the intermediate location S. In particular, these two locations can coincide here.

When the S-shaped path collides with the gap structure, the forward circular arc is calculated based on the virtual line 8 on the opposite side. This is done by shifting the further arc-shaped reversing path 30 toward the line 8 on the opposite side, so that at the turning location F, the vehicle 1 reaches a location at the end of the forward arc where it is just outside of a collision with the virtual line 8 on the opposite side, as shown in 10 Finally, if there is not enough free space between the vehicle 1 and the virtual line 8 on the opposite side, the only possible remaining solution to have a single-stage vertical reversing parking is to plan a symmetrical S-shaped long path from the vehicle passage line 24 to the finish line 35. This finish line 35 is tangential to the obtained further curved reversing path 30, as shown in 11 shown.

Given variables are: the finish line 22 and the parking location with coordinates {x _G ,y _G ,θ _G }, the final object vertex P with (x _p ,y _p ) and the radii R _min and R _l , as described above.

1. 1. Berechnen des Rotationszentrums (x_r2,y_r2) des weiteren bogenförmigen Rückwärtsfahrpfades 30 auf der Basis des Schnittpunkts des Kreisbogens von (x_p,y_p) mit dem Abstand R_l und der zum Ziel parallelen Linie 25 im Abstand R_min in Richtung des Endobjekts 6.
2. 2. Berechnen eines Projektionspunkts (x_rp ,y_rp ) des Rotationszentrumspunkts (x_r2 ,y_r2 ) auf der Ziellinie 22.
3. 3. Berechnen des Endorts g mit (x_g,y_g) , der auf der Ziellinie 22 im minimalen Toleranzabstand d zum Parkort G mit (x_G,y_G) angeordnet ist.
4. 4. Prüfen, ob der Projektionspunkt (x_rp ,y_rp ) näher ist als der Endort g mit (x_g,y_g) in Richtung der Lückeneinfahrt, das heißt darüber für die rechte Lücke oder darunter für die linke Lücke. Falls (x_rp,y_rp ) näher zur Lückeneinfahrt ist, dann ist keine Änderung des berechneten Rotationszentrumspunkts (x_r2 ,y_r2 ) erforderlich. Ansonsten wird der Rotationszentrumspunkt (x_r2 ,y_r2 ) verschoben, so dass seine Projektion auf die Ziellinie 22 der Endort g mit (x_g,y_g) ist.
5. 5. Vergleichen der Fahrzeugdurchfahrtslinie 24 mit dem erhaltenen weiteren bogenförmigen Rückwärtsfahrpfad 30, um zu prüfen, ob sie sich schneiden oder nicht, durch Berechnen des Projektionspunkts (x_n,y_n) des Rotationszentrums auf der Fahrzeugdurchfahrtslinie 24 und Vergleichen seines Abstandes zum Rotationszentrum mit dem Radius (R_min) des weiteren bogenförmigen Rückwärtsfahrpfades 30. Wenn sie sich schneiden, dann wird der Vorwärtskreisbogen, der die Fahrzeugdurchfahrtslinie 24 mit dem weiteren bogenförmigen Rückwärtsfahrpfad 30 verbindet, auf der Basis der Sequenz von Berechnungen berechnet, die für das erste Parkmanöver 15 detailliert dargestellt sind (der Berührungspunkt T mit (x_t,y_t), dann sowohl Zwischenort S mit (x_S,y_S) als auch der Wendeort mit (x_F,y_F) auf dem Vorwärtskreisbogen können bestimmt werden). Ansonsten wird die Fahrzeugdurchfahrtslinie 24 mit dem erhaltenen weiteren bogenförmigen Rückwärtsfahrpfad 30 durch eine asymmetrische s-Form verbunden, die den verfügbaren Platz innerhalb des Parkplatzes 4 ausnutzen soll, um das Fahrzeug 1 in einem Schritt und mit weniger gefahrenem Abstand parken zu lassen.

Required calculations are:

1. 1. Calculating the center of rotation (x _r2 ,y _r2 ) of the further arcuate backward travel path 30 on the basis of the intersection point of the circular arc of (x _p ,y _p ) with the distance R _l and the line 25 parallel to the target at the distance R _min in the direction of the end object 6.
2. 2. Calculate a projection point (x _{r p} ,y _{r p} ) of the rotation center point (x _{r 2} ,y _{r 2} ) on the finish line 22.
3. 3. Calculate the final location g with (x _g ,y _g ) , which is located on the finish line 22 at the minimum tolerance distance d to the parking location G with (x _G ,y _G ).
4. 4. Check whether the projection point (x _{r p} ,y _{r p} ) is closer than the final location g with (x _g ,y _g ) in the direction of the gap entrance, that is, above it for the right gap or below it for the left gap. If (x _{r p ,} y _{r p} ) is closer to the gap entrance, then no change in the calculated rotation center point (x _{r 2} ,y _{r 2} ) is required. Otherwise, the rotation center point (x _{r 2} ,y _{r 2} ) so that its projection onto the target line 22 is the final location g with (x _g ,y _g ).
5. 5. Comparing the vehicle passage line 24 with the obtained further arcuate reversing path 30 to check whether they intersect or not by calculating the projection point (x _n , y _n ) of the rotation center on the vehicle passage line 24 and comparing its distance to the rotation center with the radius (R _min ) of the further arcuate reversing path 30. If they intersect, then the forward circular arc connecting the vehicle passage line 24 with the further arcuate reversing path 30 is calculated based on the sequence of calculations detailed for the first parking maneuver 15 (the point of contact T with (x _t , y _t ), then both intermediate location S with (x _S , y _S ) and turning location with (x _F , y _F ) on the forward circular arc can be determined). Otherwise, the vehicle passage line 24 is connected to the remaining further curved reversing path 30 by an asymmetric S-shape, which is intended to utilize the available space within the parking space 4 in order to allow the vehicle 1 to park in one step and with less traveled distance.

$$y_n=y_c+tan\left({\theta }_l\right)\left(x_n-x_C\right)$$

Similar to equations 7 and 8 and their corresponding 6 the projection point (x _n ,y _n ) of the rotation center (x _{r 2} ,y _{r 2} ) on the vehicle passage line 24, which can start at the coordinates {x _C ,y _C ,θ _l }, can be calculated as follows: $$x_n=\frac{\left({\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_C\right)tan\left({\theta }_l\right)+ta{n}^2\left({\theta }_l\right)x_c+{\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}}{1+ta{n}^2\left({\theta }_l\right)}$$

$$y_n=y_c+tan\left({\theta }_l\right)\left(x_n-x_C\right)$$

1. 1- In Anbetracht von (x_S,y_S) und (R_min) wird das Rotationszentrum (x_r1a ,y_r1a ) des ersten Kreisbogens (a) berechnet.
2. 2- Das Rotationszentrum (x_r1b ,y_r1b ) des zweiten Kreisbogens (b) wird berechnet, da es im Abstand (2R_min) vom Rotationszentrum des Rückwärtskreisbogens (x_r2 ,y_r2 ) und im Abstand $\sqrt{{\left(2R_{min}\right)}^2+\mathrm{<figure-callout\; id="2"\; label="c\; c"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">c</figure-callout>}{c}^{}{}^2}$
   
   vom Rotationszentrum (x_r1a ,y_r1a ) des ersten Kreisbogens (a) angeordnet ist. Somit ist es der Schnittpunkt der zwei Kreisbögen a, b von den Mittelpunkten (x_r2 ,y_r2 ) und (x_r1a ,y_r1a ) mit Radien (2R_min) beziehungsweise $\sqrt{{\left(2R_{min}\right)}^2+\mathrm{<figure-callout\; id="2"\; label="c\; c"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">c</figure-callout>}{c}^{}{}^2}.$
   
3. 3- Sobald die Rotationszentren der zwei Kreisbögen a, b des s-Form-Manövers (x_r1a ,y_r1a ) und (x_r1b ,y_r1b ) berechnet sind, können die Punkte (x_a,y_a) und (x_b,y_b) berechnet werden. Die Linie, die (x_r1a ,y_r1a ) mit (x_a,y_a) verbindet, ist zur Linie, die (x_r1b ,y_r1b ) mit (x_b,y_b) verbindet, parallel, wobei beide denselben Winkel (α + β) aufweisen. Werte der Winkel α und β können berechnet werden, dann können die Punkte (x_a,y_a), (x_b,y_b) und (x_F,y_F) und die Orientierung (θ_F) am Wendeort F wie folgt berechnet werden: $$\beta ={\text{tan}}^{-1}\left(\frac{y_{r_{1b}}-y_{r_{1a}}}{x_{r_{1b}}-x_{r_{1a}}}\right)$$
   
   $$\alpha ={\text{tan}}^{-1}\left(\frac{cc}{2R_{min}}\right)$$
   
   $$x_a=x_{r_{1a}}+R_{min}\ast \text{cos}\left(\alpha +\beta \right)$$
   
   $$y_a=y_{r_{1a}}+R_{min}\ast \text{sin}\left(\alpha +\beta \right)$$
   
   $$x_b=x_{r_{1b}}-R_{min}\ast \text{cos}\left(\alpha +\beta \right)$$
   
   $$y_b=y_{r_{1b}}-R_{min}\ast \text{sin}\left(\alpha +\beta \right)$$
   
   $${\theta }_F={\text{tan}}^{-1}\left(\frac{-\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_{{}_{1b}}}\right)}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_{1b}}}\right)={\text{tan}}^{-1}\left(\frac{x_{r_{1b}}-x_{r_2}}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_{1b}}}\right)$$
   
   $$falls\left(RechteL\ddot{u}cke\right)$$
   
   $$x_F=x_{r_{1b}}+R_{min}\ast sin\left({\theta }_F\right)$$
   
   $$y_F=y_{r_{1b}}-R_{min}\ast cos\left({\theta }_F\right)$$
   
   $$ansonsten$$
   
   $$x_F=x_{r_{1b}}-R_{min}\ast \text{sin}\left({\theta }_F\right)$$
   
   $$y_F=y_{r_{1b}}+R_{min}\ast \text{cos}\left({\theta }_F\right)$$
   
   $$Ende$$
   
   $$y_r=y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)$$
   

9 shows the S-shaped maneuvering situation, where the vehicle passage line 24 lies above the obtained reverse circular arc 23, which here is the further curved reversing path 30. The S-shaped The forward path or the s-shaped forward maneuver consists of a circular arc (a), a straight segment (c), and a circular arc (b). All circular arcs a, b, c are of the minimum radius (R _min ). The straight segment c has a defined length (cc) to connect the two circular arcs a, b in reverse directions, so that the control unit has enough distance to adjust the steering. The starting position of the forward path here is the intermediate location S with (x _S , y _S ), which is set as early as possible once the gap is fully detected and offered by the system, so that there is enough space for the forward s-shape. The center of rotation of the first circular arc a in the forward s-shape maneuver is (x _{r 1a} ,y _{r 1a} ), while the center of rotation of the second arc b in the forward s-shape maneuver (x _{r 1b} ,y _{r 1b} ). The first arc (a) starts at (x _S ,y _S ) and ends at (x _a ,y _a ), while the second arc (b) starts at (x _b ,y _b ) and ends at (x _F ,y _F ). The calculation of the positions forming the s-shape maneuver {(x _a ,y _a ), (x _b ,y _b ), (x _F ,y _F )} can be performed as follows, given (x _S ,y _S ), cc, and the radius of all the arcs (R _min ).

1. 1- Considering (x _S ,y _S ) and (R _min ), the center of rotation (x _{r 1a} ,y _{r 1a} ) of the first circular arc (a) is calculated.
2. 2- The center of rotation (x _{r 1b} ,y _{r 1b} ) of the second circular arc (b) is calculated since it is at a distance (2R _min ) from the center of rotation of the backward circular arc (x _{r 2} ,y _{r 2} ) and at a distance $\sqrt{{\left(2R_{min}\right)}^2+\mathrm{<figure-callout\; id="2"\; label="c\; c"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">c</figure-callout>}{c}^{}{}^2}$
   
   from the center of rotation (x _{r 1a} ,y _{r 1a} ) of the first circular arc (a). Thus, it is the intersection point of the two circular arcs a, b from the centers (x _{r 2} ,y _{r 2} ) and (x _{r 1a} ,y _{r 1a} ) with radii (2R _min ) respectively $\sqrt{{\left(2R_{min}\right)}^2+\mathrm{<figure-callout\; id="2"\; label="c\; c"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">c</figure-callout>}{c}^{}{}^2}.$
   
3. 3- Once the centers of rotation of the two circular arcs a, b of the s-shape maneuver (x _{r 1a} ,y _{r 1a} ) and (x _{r 1b} ,y _{r 1b} ) are calculated, the points (x _a ,y _a ) and (x _b ,y _b ) can be calculated. The line that (x _{r 1a} ,y _{r 1a} ) with (x _a ,y _a ) is the line connecting (x _{r 1b} ,y _{r 1b} ) with (x _b ,y _b ), both having the same angle (α + β). Values of the angles α and β can be calculated, then the points (x _a ,y _a ), (x _b ,y _b ) and (x _F ,y _F ) and the orientation (θ _F ) at the turning point F can be calculated as follows: $$\beta ={\text{tan}}^{-1}\left(\frac{y_{r_{1b}}-y_{r_{1a}}}{x_{r_{1b}}-x_{r_{1a}}}\right)$$
   
   $$\alpha ={\text{tan}}^{-1}\left(\frac{cc}{2R_{min}}\right)$$
   
   $$x_a=x_{r_{1a}}+R_{min}\ast \text{cos}\left(\alpha +\beta \right)$$
   
   $$y_a=y_{r_{1a}}+R_{min}\ast \text{sin}\left(\alpha +\beta \right)$$
   
   $$x_b=x_{r_{1b}}-R_{min}\ast \text{cos}\left(\alpha +\beta \right)$$
   
   $$y_b=y_{r_{1b}}-R_{min}\ast \text{sin}\left(\alpha +\beta \right)$$
   
   $${\theta }_F={\text{tan}}^{-1}\left(\frac{-\left({\mathrm{<figure-callout\; id="2"\; label="x\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">x</figure-callout>}}_{r_{}}-x_{r_{{}_{1b}}}\right)}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_{1b}}}\right)={\text{tan}}^{-1}\left(\frac{x_{r_{1b}}-x_{r_2}}{{\mathrm{<figure-callout\; id="2"\; label="y\; r"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_{r_{}}-y_{r_{1b}}}\right)$$
   
   $$falls\left(RechteL\ddot{u}cke\right)$$
   
   $$x_F=x_{r_{1b}}+R_{min}\ast sin\left({\theta }_F\right)$$
   
   $$y_F=y_{r_{1b}}-R_{min}\ast cos\left({\theta }_F\right)$$
   
   $$ansonsten$$
   
   $$x_F=x_{r_{1b}}-R_{min}\ast \text{sin}\left({\theta }_F\right)$$
   
   $$y_F=y_{r_{1b}}+R_{min}\ast \text{cos}\left({\theta }_F\right)$$
   
   $$Ende$$
   
   $$y_r=y_0+tan\left({\theta }_G\right)\left(x_r-x_0\right)$$
   

$${\left(x-x_2\right)}^2+{\left(y-y_2\right)}^2=L_2{}^2$$

To formulate a function for calculating the intersection points of the two circular arcs a, b, we should assume a general case of two centers of rotation (x ₁ ,y ₁ ) and (x ₂ ,y ₂ ) with corresponding radii L ₁ and L ₂ , then the general formula for the intersection points can be obtained as follows: $${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2={\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{}^2$$

$${\left(x-x_2\right)}^2+{\left(y-y_2\right)}^2={\mathrm{<figure-callout\; id="2"\; label="L"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">L</figure-callout>}}_2{}^2$$

$$x=\left(\frac{A^2-B^2+x_2{}^2-x_1{}^2+y_2{}^2-y_1{}^2}{2\left(x_2-x_1\right)}\right)-\left(\frac{y_2-y_1}{x_2-x_1}\right)y$$

$$x=k+my$$

$$x^2=k^2+2kmy+m^2{y}^2$$

$$x^2-2x_{1}x+x_1{}^2+y^2-2y_{1}y+y_1{}^2=L_1{}^2$$

$$k^2+2kmy+m^2{y}^2-2x_1\left(k+my\right)+x_1{}^2+y^2-2y_{1}y+y_1{}^2=L_1{}^2$$

$$\left(m^2+1\right)y^2+2\left(km-x_{1}m-y_1\right)y-2x_{1}k+x_1{}^2+y_1{}^2+k^2-L_1{}^2=0$$

$$A=\left(m^2+1\right)$$

$$B=2\left(km-x_{1}m-y_1\right)$$

$$C=-2x_{1}k+x_1{}^2+y_1{}^2+k^2-L_1{}^2$$

$$y_{UP}=\frac{-B+\sqrt{B^2-4AC}}{2A}$$

$$y_{DOWN}=\frac{-B-\sqrt{B^2-4AC}}{2A}$$

By subtracting [11] and [12]: $$2\left(x_2-x_1\right)x+x_1{}^2-x_2{}^2+2\left({\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_2-y_1\right)y+{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2-{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_2{}^2=A^2-{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2$$

$$x=\left(\frac{A^2-{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2+x_2{}^2-x_1{}^2+{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_2{}^2-{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2}{2\left(x_2-x_1\right)}\right)-\left(\frac{{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}_2-y_1}{x_2-x_1}\right)y$$

$$x=k+my$$

$$x^2={\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">k</figure-callout>}}^2+2kmy+m^2{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}^2$$

$$x^2-2x_{1}x+x_1{}^2+{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}^2-2y_{1}y+{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2={\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{}^2$$

$${\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">k</figure-callout>}}^2+2kmy+m^2{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}^2-2x_1\left(k+my\right)+x_1{}^2+{\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}^2-2y_{1}y+{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2={\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{}^2$$

$$\left({\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}^2+1\right){\mathrm{<figure-callout\; id="2"\; label="y"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">y</figure-callout>}}^2+2\left(km-x_{1}m-y_1\right)y-2x_{1}k+x_1{}^2+{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2+{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">k</figure-callout>}}^2-{\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{}^2=0$$

$$A=\left({\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">m</figure-callout>}}^2+1\right)$$

$$B=2\left(km-x_{1}m-y_1\right)$$

$$C=-2x_{1}k+x_1{}^2+{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">y</figure-callout>}}_1{}^2+{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">k</figure-callout>}}^2-{\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{}^2$$

$$y_{UP}=\frac{-B+\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$y_{DOWN}=\frac{-B-\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

• Gegebene Variablen sind:
  • L: Fahrzeuglänge
  • w: Fahrzeugbreite
  • rr: Abstand von der Hinterachse zum hinteren Stoßfänger
  • R_min: minimaler Rotationsradius am Zentrum der Hinterachse
  • R_Fmin : minimaler Rotationsradius am vorderen linken Eckpunkt $$R_{F_{min}}=\sqrt{{\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2}$$
    
    $$\alpha ={\text{tan}}^{-1}\left(\frac{L-rr}{R_{min}-\frac{w}{2}}\right)$$
    
    $$\mathrm{\Delta }\theta ={\theta }_F-{\theta }_l$$
    

On the other hand, as in 10 As shown, the forward circular arc (here, the third forward path 33) can be calculated based on the virtual line 8 on the opposite side when the s-shaped path collides and there is sufficient distance between the vehicle passage line 24 and the virtual line 8 on the opposite side. It should be noted that the given virtual line 8 on the opposite side is assumed here to be parallel to the vehicle passage line 24. The forward circular arc is performed with the minimum radius until the front left corner 36 reaches the virtual line 8 on the opposite side. The vehicle orientation change along this forward circular arc is (Δθ), which is calculated based on the geometry of the given right gap defined in 10 shown can be calculated as follows:

• Given variables are:
  • L: Vehicle length
  • w: vehicle width
  • rr: Distance from the rear axle to the rear bumper
  • R _min : minimum rotation radius at the center of the rear axle
  • R _{F min} : minimum rotation radius at the front left corner point $$R_{F_{min}}=\sqrt{{\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2}$$
    
    $$\alpha ={\text{tan}}^{-1}\left(\frac{L-rr}{R_{min}-\frac{w}{2}}\right)$$
    
    $$\mathrm{\Delta }\theta ={\theta }_F-{\theta }_l$$
    

$$cos\left(\alpha \right)-cos\left(\alpha \right)cos\left(\mathrm{\Delta }\theta \right)+sin\left(\alpha \right)sin\left(\mathrm{\Delta }\theta \right)=\frac{\mathrm{\Delta }{y}_{OPP-Init}-\frac{w}{2}}{R_{F_{min}}}$$

$$cos\left(\alpha \right)=\frac{R_{min}-\frac{w}{2}}{R_{F_{min}}}$$

$$sin\left(\alpha \right)=\frac{L-rr}{R_{F_{min}}}$$

$$\left(R_{min}-\frac{w}{2}\right)-\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)=\mathrm{\Delta }{y}_{OPP-Init}\frac{w}{2}$$

$$\left(R_{min}-\frac{w}{2}\right)-\mathrm{\Delta }{y}_{OPP-Init}+\frac{w}{2}+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)=\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)$$

$$\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)=\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)$$

$$\begin{array}{l}{\left(R_{min}-\frac{w}{2}\right)}^{2}co{s}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ ={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2+{\left(L-rr\right)}^{2}si{n}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ +2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)\hfill \end{array}$$

$$\begin{array}{l}{\left(R_{min}-\frac{w}{2}\right)}^2\left(1-si{n}^2\left(\mathrm{\Delta }\theta \right)\right)\hfill \\ ={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2+{\left(L-rr\right)}^{2}si{n}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ +2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)\hfill \end{array}$$

$$\begin{array}{c}\left({\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2\right)si{n}^2\left(\mathrm{\Delta }\theta \right)+\left[2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)\right]sin\left(\mathrm{\Delta }\theta \right)\\ +{\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2-{\left(R_{min}-\frac{w}{2}\right)}^2\end{array}$$

$$Asi{n}^2\left(\mathrm{\Delta }\theta \right)+Bsin\left(\mathrm{\Delta }\theta \right)+C=0$$

$$A=\left({\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2\right)$$

$$B=\left[2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)\right]$$

$$C={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2-{\left(R_{min}-\frac{w}{2}\right)}^2$$

$$sin\left(\mathrm{\Delta }\theta \right)=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

On the basis of the circular arc connecting the front left corner 36 with the rear left corner 37, which in 10 shown, the following equations can be derived: $$R_{F_{min}}cos\left(\alpha \right)-R_{F_{min}}cos\left(\alpha +\mathrm{\Delta }\theta \right)=\mathrm{\Delta }{y}_{OPP-Init}-\frac{\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">w</figure-callout>}}{2}$$

$$cos\left(\alpha \right)-cos\left(\alpha \right)cos\left(\mathrm{\Delta }\theta \right)+sin\left(\alpha \right)sin\left(\mathrm{\Delta }\theta \right)=\frac{\mathrm{\Delta }{y}_{OPP-Init}-\frac{w}{2}}{R_{F_{min}}}$$

$$cos\left(\alpha \right)=\frac{R_{min}-\frac{w}{2}}{R_{F_{min}}}$$

$$sin\left(\alpha \right)=\frac{L-rr}{R_{F_{min}}}$$

$$\left(R_{min}-\frac{w}{2}\right)-\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)=\mathrm{\Delta }{y}_{OPP-Ini\mathrm{<figure-callout\; id="2"\; label="t\; w"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">t</figure-callout>}}\frac{w}{}\frac{}{2}$$

$$\left(R_{min}-\frac{w}{2}\right)-\mathrm{\Delta }{y}_{OPP-Init}+\frac{\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">w</figure-callout>}}{2}+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)=\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)$$

$$\left(R_{min}-\frac{w}{2}\right)cos\left(\mathrm{\Delta }\theta \right)=\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)+\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)$$

$$\begin{array}{l}{\left(R_{min}-\frac{w}{2}\right)}^{2}co{s}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ ={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2+{\left(L-rr\right)}^{2}si{n}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ +2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)\hfill \end{array}$$

$$\begin{array}{l}{\left(R_{min}-\frac{w}{2}\right)}^2\left(1-si{n}^2\left(\mathrm{\Delta }\theta \right)\right)\hfill \\ ={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2+{\left(L-rr\right)}^{2}si{n}^2\left(\mathrm{\Delta }\theta \right)\hfill \\ +2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)sin\left(\mathrm{\Delta }\theta \right)\hfill \end{array}$$

$$\begin{array}{c}\left({\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2\right)si{n}^2\left(\mathrm{\Delta }\theta \right)+\left[2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)\right]sin\left(\mathrm{\Delta }\theta \right)\\ +{\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2-{\left(R_{min}-\frac{w}{2}\right)}^2\end{array}$$

$$Asi{n}^2\left(\mathrm{\Delta }\theta \right)+Bsin\left(\mathrm{\Delta }\theta \right)+C=0$$

$$A=\left({\left(R_{min}-\frac{w}{2}\right)}^2+{\left(L-rr\right)}^2\right)$$

$$B=\left[2\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)\right]$$

$$C={\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2-{\left(R_{min}-\frac{w}{2}\right)}^2$$

$$sin\left(\mathrm{\Delta }\theta \right)=\frac{-B\pm \sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$\mathrm{\Delta }\theta ={\text{sin}}^{-1}\left(\frac{-\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)+\left(R_{min}-\frac{w}{2}\right)\sqrt{R_{F_{min}}{}^2-{\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2}}{R_{F_{min}}{}^2}\right)$$

Since B > 0 and Δθ < 180°, then sin(Δθ) > 0 $$sin\left(\mathrm{\Delta }\theta \right)=\frac{-B+\sqrt{{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">B</figure-callout>}}^2-4AC}}{2A}$$

$$\mathrm{\Delta }\theta ={\text{sin}}^{-1}\left(\frac{-\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)\left(L-rr\right)+\left(R_{min}-\frac{w}{2}\right)\sqrt{R_{F_{min}}{}^2-{\left(R_{min}-\mathrm{\Delta }{y}_{OPP-Init}\right)}^2}}{R_{F_{min}}{}^2}\right)$$

Based on the calculated (Δθ), the distance (K) that is 10 shown, can be calculated as follows: the distance (K) represents the location of the vehicle center of the rear axle such that the vehicle 1 is just in contact with the virtual line 8 on the opposite side. Such a distance is the closest distance that the vehicle 1 should reach in the direction of the virtual line 8 on the opposite side after the circular forward pull (here the third forward travel path 33). The location of the vehicle 1 at this closest location to the line 8 on the opposite side is in 10 shown along a line 38. $$K=\left(L-rr\right)\text{sin}\left(\mathrm{\Delta }\theta \right)+\left(\frac{w}{2}\right)\text{cos}\left(\mathrm{\Delta }\theta \right)$$

Since the nearest location of the vehicle 1 (at the center of the rear axle) is determined to the virtual line 8 on the opposite side, the corresponding center of rotation for the circle with minimum radius can also be obtained, which corresponds to a line 39 in 10 is shown. Line 39 is parallel to line 38 and is located at a distance (R _min cos(Δθ)) in the direction of the parking lot.

The center of rotation of the new shifted backward arc (x _{r 2} ',y _{r 2} ') can be calculated by obtaining the intersection point of the line 39, which is arranged at a distance (R _min cos(Δθ) + K) from the virtual line 8 on the opposite side, with the line 25 parallel to the target, which is arranged at a distance (R _min ) from the target line 22.

$$x_{OPP}\text{'}=x_{OPP}-\frac{N\ast M}{\sqrt{\left(N^2+1\right)}}$$

$$falls\left(RechteL\ddot{u}cke\right)$$

$$y_{OPP}\text{'}=y_{Opp}-\frac{M}{\sqrt{\left(N^2+1\right)}}$$

$$ansonten$$

$$y_{Opp}\text{'}=y_{Opp}+\frac{M}{\sqrt{\left(N^2+1\right)}}$$

$$Ende$$

Since the virtual line 8 on the opposite side is assumed to be parallel to the vehicle passage line 24, it can be assumed to be defined with the frame {x _Opp ,y _Opp ,θ _l }. Then, the line 39, which is parallel to the virtual line 8 on the opposite side at a distance (M = R _min cos(Δθ) + K), can be defined with the frame {x _Opp ',y _Opp ',θ _l }, which can be calculated as follows: $$N=-tan\left({\theta }_l\right)$$

$$x_{OPP}\text{'}=x_{OPP}-\frac{N\ast M}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)}}$$

$$falls\left(RechteL\ddot{u}cke\right)$$

$$y_{OPP}\text{'}=y_{Opp}-\frac{M}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)}}$$

$$ansonten$$

$$y_{Opp}\text{'}=y_{Opp}+\frac{M}{\sqrt{\left({\mathrm{<figure-callout\; id="2"\; label="N"\; filenames="DE102023113599B4\_0176.png,DE102023113599B4\_0177.png"\; state="\{\{state\}\}">N</figure-callout>}}^2+1\right)}}$$

$$Ende$$

Since the line 25 parallel to the target is already defined with the frame {x ₀ ,y ₀ ,θ _G }, the center of rotation of the new backward arc (x _{r 2} ',y _{r 2} ') can be calculated as the intersection point of the two line frames {x _Opp ',y _Opp ',θ _l } and {x ₀ ,y ₀ ,θ _G }.

As in 11 Finally, as shown, if there is insufficient space between the vehicle and the virtual line 8 on the opposite side, the remaining final solution that can be considered to have a one-step reversing path for perpendicular parking is to establish a symmetrical long S-shaped path. This symmetrical long S-shaped path connects the vehicle starting position (current location C) on the vehicle passage line 24 with a turning location F on the parallel finish line 35, which is tangent to the reversing circular arc 23, which here is the further arc-shaped reversing path 30.

$$y_F=y_c+\mathrm{\Delta }x\text{ sin}\left({\theta }_l\right)-\mathrm{\Delta }{y}_1\text{ cos}\left({\theta }_l\right)$$

$$x_a=x_c+R\text{ sin}\left(\theta \right)\text{cos}\left({\theta }_l\right)+R\left(1-\text{cos}\left(\theta \right)\right)\text{ sin}\left({\theta }_l\right)$$

$$y_a=y_c+R\text{ sin}\left(\theta \right)\text{sin}\left({\theta }_l\right)+R\left(1-\text{cos}\left(\theta \right)\right)\text{ cos}\left({\theta }_l\right)$$

$${\theta }_a={\theta }_l-\theta$$

$$x_b=x_a+cc\text{ cos}\left({\theta }_a\right)$$

$$y_b=y_a+cc\text{ sin}\left({\theta }_a\right)$$

$${\theta }_b={\theta }_a$$

The turning location F of this symmetric s-shaped path (x _p ,y _p ) can be calculated from the given current location C with (x _C ,y _C ) by calculating the rotation radius (R) and the orientation change (θ) for each circular segment of the s-shaped path. The s-shaped path consists of two circular arcs (a and b) that are inverted to each other and connected by a straight segment of length (cc) to guarantee steering prediction by the control unit. The rotation radius (R) is calculated such that the second circular arc b of the s-shaped path is tangent to a (virtual or real) road surface line (start line 9) located at a vertical distance (Δy ₂ ) from the intended finish line 35. Based on (R), (cc) and the given vertical distance of the s-shaped path (Δy ₁ ) between the vehicle passage line 24 and the target line 35, the longitudinal distance of the s-shaped path (Δx) and the orientation change (θ) for each circle segment can then be calculated: $$x_F=x_C+\mathrm{\Delta }x\text{ cos}\left({\theta }_l\right)+\mathrm{<figure-callout\; id="1"\; label="\Delta \; y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">\Delta </figure-callout>}{y}_{}{}_1\text{ sin}\left({\theta }_l\right)$$

$$y_F=y_c+\mathrm{\Delta }x\text{ sin}\left({\theta }_l\right)-\mathrm{<figure-callout\; id="1"\; label="\Delta \; y"\; filenames="DE102023113599B4\_0178.png,DE102023113599B4\_0179.png"\; state="\{\{state\}\}">\Delta </figure-callout>}{y}_{}{}_1\text{ cos}\left({\theta }_l\right)$$

$$x_a=x_c+R\text{ sin}\left(\theta \right)\text{cos}\left({\theta }_l\right)+R\left(1-\text{cos}\left(\theta \right)\right)\text{ sin}\left({\theta }_l\right)$$

$$y_a=y_c+R\text{ sin}\left(\theta \right)\text{sin}\left({\theta }_l\right)+R\left(1-\text{cos}\left(\theta \right)\right)\text{ cos}\left({\theta }_l\right)$$

$${\theta }_a={\theta }_l-\theta$$

$$x_b=x_a+cc\text{ cos}\left({\theta }_a\right)$$

$$y_b=y_a+cc\text{ sin}\left({\theta }_a\right)$$

$${\theta }_b={\theta }_a$$

Note that θ _a and θ _b are the orientations of vehicle 1 at locations a and b, respectively. The orientation of the starting line 9, which passes through the final object vertex P with (x _p ,y _p ), is θ _p .

In summary, starting a vertical reverse parking maneuver with a forward turn to park in a single step was described.

Claims (15)

A method for operating an at least assisted vertical or inclined parking function (3) for a vehicle (1), which includes a parking maneuver (14) into a vertical or inclined parking space (4) with a forward travel path from a current location (C) of the vehicle (1) to a turning location (F) and an arcuate reverse travel path from the turning location (F) to an end location (g) at which the vehicle (1) is at least partially located in the parking space (4), the method including: - providing (S2) sensor information (12) describing the vertical or inclined parking space (4); - taking into account the sensor information (12), determining (S3) a first parking maneuver (15) that includes the shortest possible reverse travel path without collision with an object (6) that laterally borders the parking space (4), and a second parking maneuver (16) that includes the shortest possible forward travel path, - determining (S4) whether an orientation angle of the vehicle (1) at the turning location (F) is greater for the first parking maneuver (15) or for the second parking maneuver (16); and - operating (S5) the vehicle (1) from its current location (C) to the turning location (F) and from there to the final location (g) at least assisted according to the determined parking maneuver (15, 16) that has the greater orientation angle. Procedure according to Claim 1 , wherein the forward travel path of the respective parking maneuver (15, 16) includes a straight forward travel section from the current location (C) to an intermediate location (S) and an arcuate travel section from the intermediate location (S) to the turning location (F). Procedure according to Claim 1 or 2 , wherein the curved reverse travel path and/or the curved travel part of the forward travel path can be driven with a constant steering angle of at least 80 percent, in particular between 90 percent and 95 percent, of a maximum steering angle of a steering system of the vehicle (1). Method according to one of the preceding claims, wherein the respective parking maneuver (15, 16) includes driving straight backward from the end location (g) to a parking location (G) in the parking lot (4), wherein, compared to the end location (g), the parking location (G) is closer to an end line (10) which delimits the parking lot (4) in a longitudinal direction of the parking lot (4). Procedure according to Claim 4 , wherein the second parking maneuver (16) includes that the end location (g) is arranged at a distance from the parking location (G) which is equal to a minimum distance value (17). Method according to one of the preceding claims, wherein the method includes determining a further arcuate reversing path (30) which satisfies constraints of the first parking maneuver (15) and the second parking maneuver (16), and performing a validation procedure, wherein after performing the validation procedure the vehicle (1) is operated at least assisted according to a third parking maneuver (31) which includes the determined further arcuate reversing path (30). Procedure according to Claim 6 in his references to Claim 2 , wherein the validation procedure includes checking whether the further arcuate reversing path (30) intersects a vehicle passage line (24) followed by the vehicle (1) when driving straight forward from the current location (C), and if so, the third parking maneuver (31) includes a first forward travel path connecting the intermediate location (S) and the turning location (F) with a circular arc-shaped path. Procedure according to Claim 7 , wherein, if the further curved reversing path (30) does not intersect the vehicle passage line (24), the third parking maneuver (31) includes a second forward travel path (32) connecting the intermediate location (S) and the turning location (F) with an S-shaped path. Procedure according to Claim 8 , wherein the validation procedure includes checking whether the S-shaped path collides with another object (5) that laterally delimits the parking space (4) as a starting object, wherein, only if this is not the case, the third parking maneuver (31) includes the second forward travel path (32). Procedure according to Claim 9 , wherein, when the S-shaped path collides with the further object (5), the third parking maneuver (31) includes a third forward travel path (33) which, taking into account a line (8) on the opposite side is determined, which delimits an area (13) in front of the parking space (4) which is available for maneuvering, so that the vehicle (1) reaches the line (8) on the opposite side when it is at the turning point (F) of the third forward travel path (33) and starts the further arcuate reverse travel path (30) at this turning point (F). Procedure according to Claim 10 , wherein the validation procedure includes checking whether the area (13) in front of the parking space (4) is large enough for the vehicle (1) to perform the third forward travel path (33), and only if this is the case does the third parking maneuver (31) include the third forward travel path (33). Procedure according to Claim 11 , wherein, if the area (13) is not large enough, the third parking maneuver (31) includes a fourth forward travel path (34) connecting the intermediate location (S) and the turning location (F) with a symmetrical S-shaped path, the turning location (F) being arranged on a finish line (35) which is tangential to the further arcuate reverse travel path (30). Vehicle (1) which is adapted to carry out a method according to one of the preceding claims. Control unit (2) for a vehicle (1), wherein the control unit (2) is designed to carry out a method according to one of the Claims 1 until 12 to carry out. A computer program product containing instructions which, when executed by a computer, cause the computer to perform a method according to any one of the Claims 1 until 12 executes.

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