Interventional surgery robot force-position control structure and control method thereof

Abstract

The application relates to the technical field of mechanical control, in particular to an interventional surgery robot force-position control structure and a control method thereof, which comprises a linear motor, a clamping wheel, an elastic piece, a rotating shaft, a friction wheel and a rotary motor; the clamping wheel is rotationally connected with the rotating shaft, the clamping wheel and the rotating shaft can freely move along the guide rail direction of the linear motor as a whole, and the two ends of the elastic piece are connected with the linear motor and the rotating shaft respectively; the static friction coefficient of the edge of the clamping wheel is greater than the sliding friction coefficient of the edge of the friction wheel. By adjusting the position of the linear motor and the rotating speed of the rotary motor, the force, the speed and the position of the forward and backward movement of a guide wire can be accurately controlled, the response speed of the guide wire is obviously superior to that of the traditional two-wheel clamping synchronous rotation method, and the problems that the existing technology cannot effectively avoid the slippage and damage of the guide wire coating or cause the damage of the friction wheel and the invasion of the debris into the patient's body are solved.

Description

Technical Field

[0001] This invention relates to the field of mechanical control technology, and in particular to a force-position control structure and control method for an interventional surgical robot. Background Technology

[0002] Interventional surgery requires more precise positional control, particularly in the precise positioning of the guidewire. Traditional wheel-type guidewire clamping methods suffer from slippage if the preload is too low, while excessive preload can damage the guidewire coating or cause the friction wheel to break, releasing debris into the patient's body. Furthermore, if the interventional surgical robot uses a linear pushing device, it becomes too bulky and lacks continuous pushing capability. Other force control methods are affected by sensor installation; sensors cannot be sterilized using traditional methods, and sensors should not be installed on parts that come into direct contact with the human body to avoid harm. Therefore, we propose a force-position control structure and method for interventional surgical robots. Summary of the Invention

[0003] Based on the technical problems existing in the background technology, this invention proposes a force-position control structure and control method for interventional surgical robots. By adjusting the position of the linear motor and the rotation speed of the rotary motor, the force, speed and position of the guidewire's forward and backward movement can be accurately controlled. Moreover, the guidewire response speed is significantly better than the traditional method of synchronous rotation with two clamping wheels. This solves the problem that the existing technology cannot effectively avoid slippage and damage to the guidewire coating, or cause the friction wheel to break and generate debris that enters the patient's body.

[0004] This invention provides the following technical solution: a force and position control structure for an interventional surgical robot, comprising a linear motor, clamping wheels, elastic elements, a rotating shaft, friction wheels, and a rotary motor;

[0005] The friction wheel is coaxial with and rigidly connected to the rotary motor. The gap between the clamping wheel and the friction wheel is the diameter of the guide wire. The clamping wheel is rotatably connected to the rotating shaft, and the clamping wheel and the rotating shaft can move freely along the guide rail direction of the linear motor as a whole. The two ends of the elastic element are respectively connected to the linear motor and the rotating shaft.

[0006] The static friction coefficient of the clamping wheel edge is greater than the sliding friction coefficient of the friction wheel edge. The linear motor pushes the clamping wheel to cooperate with the friction wheel to clamp the guide wire through the elastic element. The rotary motor is used to drive the friction wheel to rotate and push the guide wire.

[0007] Preferably, a speed sensor is provided on the clamping wheel.

[0008] Preferably, the clamping wheel is a grating wheel.

[0009] Preferably, the rotary motor is equipped with an encoder.

[0010] A method for force and position control of an interventional surgical robot includes the following steps:

[0011] S1. Place the guide wire between the grating wheel and the friction wheel;

[0012] S2. By controlling the rotary motor, the friction wheel, which is rigidly connected to the shaft of the rotary motor, rotates at the same speed, but there is no pressure between the guide wire and the friction wheel, so the guide wire does not move;

[0013] S3. By controlling the position of the linear motor, the elastic element is compressed, thereby generating pressure between the friction wheel and the guide wire, and between the guide wire and the clamping wheel. At the moment of clamping, the speed of the guide wire is 0, while there is sliding between the guide wire and the friction wheel, and the friction wheel generates sliding friction force on the guide wire.

[0014] The static friction coefficient of the clamping wheel edge is greater than the sliding friction coefficient of the friction wheel edge. When the guide wire moves under frictional force, it can always drive the clamping wheel to move at the same speed.

[0015] S4. When the clamping wheel and the friction wheel are not moving at the same speed, the current force applied to the guide wire is calculated by calculating the position of the linear motor, the pressure of the elastic element, and the speed difference of the friction wheel.

[0016] When the clamping wheel and the friction wheel move at the same speed, the guide wire is in a state of uniform displacement.

[0017] S5. By adjusting the position of the linear motor and the speed of the rotary motor, the force, speed, and position of the guide wire's forward and backward movement are controlled.

[0018] Preferably, the position control method for the guidewire during startup is as follows:

[0019] The rotational speed of the rotary electric motor, v6, gradually increases from 0 to v6. m ;

[0020] A linear motor starts moving to the right at a speed of v1(t). When it reaches time t1, the magnitude of the elastic force F compressing the elastic element is... elastic for:

[0021]

[0022] Since the linear motor is connected to the elastic element, the elastic element is compressed. The elastic element applies a spring force to the bearing, the rotating shaft, and the grating wheel. The magnitude of the sliding friction force between the friction wheel and the guide wire is:

[0023]

[0024] The speed of the clamping wheel, i.e., the guide wire, is:

[0025]

[0026] The clamping wheel, i.e., the current position of the guidewire, is:

[0027]

[0028] In the above formula, v6 represents the rotational speed of the rotating motor, m g v1 represents the guide wire mass, v2 represents the clamping wheel speed, and is equal to the guide wire speed, F elastic The value represents the elastic force of the elastic element, k represents the elastic coefficient of the elastic element, μ5 represents the friction coefficient between the guide wire and the friction wheel, and v1 represents the moving speed of the linear motor.

[0029] Preferably, the position control method for the guidewire during operation is as follows:

[0030] During guide wire acceleration or deceleration, in order to maintain the differential speed of sliding, the rotary motor also needs to accelerate to ensure that the speed difference meets the requirements.

[0031] v6-v2=v m ;

[0032] Therefore, the rotary motor also needs to increase its speed v6 to [a certain value] within the same operating time.

[0033]

[0034] To keep the magnitude of the guide wire's acceleration constant, if the guide wire needs to increase its acceleration, the linear motor is controlled to move forward; otherwise, the linear motor is controlled to move backward.

[0035] When a transition to uniform motion is required, the rotary motor reduces the speed from v6 to v2 with an acceleration a6. The transition time Δt is:

[0036]

[0037] Preferably, the guidewire thrust control method is as follows:

[0038] When the guidewire encounters resistance as it advances, without damaging the blood vessel, the guidewire loses one degree of freedom in its direction of movement but gains a degree of freedom in its thrust, thus changing the magnitude of the force F at the tip of the guidewire. end for:

[0039]

[0040] Where f5 represents the frictional force between the guide wire and the friction wheel.

[0041] This invention provides a force-position control structure and method for an interventional surgical robot. It controls the pressure between the clamping wheel, guide wire, and friction wheel using a linear motor and elastic components. During uniform motion, the linear motor gradually retracts, reducing spring compression until a critical position is reached where the speed difference between the two wheels causes a problem. The linear motor typically pauses slightly ahead of the critical position during uniform motion to enhance system robustness. Although there may be a slight overshoot for a very short time, the system can accurately stabilize at the expected speed. The advantage of this speed adjustment method is that the response speed is significantly faster than the traditional synchronous rotation method with two clamping wheels. It also avoids the slippage problem inherent in traditional guide wire feeding and prevents excessive preload from damaging the guide wire coating or causing friction wheel breakage, resulting in debris entering the patient's body. Attached Figure Description

[0042] Figure 1 This is a schematic diagram of the structure of the present invention.

[0043] In the diagram: 1. Linear motor; 2. Grating wheel; 3. Spring; 4. Shaft; 5. Friction wheel; 6. Rotary motor. Detailed Implementation

[0044] The technical solutions of the embodiments of the present invention will be clearly and completely described below with reference to the accompanying drawings. Obviously, the described embodiments are only some embodiments of the present invention, and not all embodiments. Based on the embodiments of the present invention, all other embodiments obtained by those skilled in the art without creative effort are within the scope of protection of the present invention.

[0045] like Figure 1 As shown, the present invention provides a technical solution: a force position control structure for an interventional surgical robot, including a linear motor 1, a clamping wheel, an elastic element, a bearing and a rotating shaft 4, a friction wheel 5, and a rotary motor 6.

[0046] The clamping wheel is a grating wheel 2, which facilitates the sensor to detect the rotational speed of the grating wheel 2. The elastic element is a spring 3, or other elements that play an elastic role can be used. The static friction coefficient of the edge of the grating wheel 2 is greater than the sliding friction coefficient of the edge of the friction wheel 5.

[0047] Friction wheel 5 is coaxial with and rigidly connected to the rotating shaft. The distance between grating wheel 2 and friction wheel 5 is the diameter of the guide wire. Grating wheel 2, bearing, and rotating shaft 4 can move freely back and forth along the guide rail of linear motor 1. One end of spring 3 is connected to linear motor 1, and the other end is fixed to rotating shaft 4. Rotating shaft 4 is also connected to grating wheel 2 through bearing. Spring 3 is in a fully relaxed state.

[0048] As the linear motor 1 moves forward, the spring 3 is compressed because it is connected to the linear motor 1. The spring 3 applies a spring force to the bearing, the rotating shaft 4, and the grating wheel 2. The bearing, the rotating shaft 4, and the grating wheel 2 gain acceleration to the right until they contact the right guide wire, at which point they apply a pressure to the guide wire equal to the spring force generated by the compression of the spring 3.

[0049] This structure controls the rotational speed of the rotary motor 6 and the position of the linear motor 1, thereby adjusting the force, speed, and position of the guide wire held between the grating wheel 2 and the friction wheel 5 to move back and forth.

[0050] A method for force and position control of an interventional surgical robot includes the following steps:

[0051] The guide wire is placed between the grating wheel 2 and the friction wheel 5, without the need to adjust the linear motor 1 to compress the spring 3.

[0052] By controlling the rotary motor 6, the friction wheel 5, which is rigidly connected to the shaft 4 of the rotary motor 6, rotates at the same speed. Since the spring 3 is still in a relaxed state, there is no pressure between the guide wire and the friction wheel 5, so the guide wire does not move.

[0053] By controlling the position of the linear motor 1, the spring 3 is compressed, thereby generating pressure between the friction wheel 5 and the guide wire, and between the guide wire and the grating wheel 2. Note that the guide wire speed is 0 at the moment of clamping, while the friction wheel 5 already has an initial speed in step 2, so there is still slippage between the guide wire and the friction wheel 5. Consequently, the friction wheel 5 generates sliding friction on the guide wire.

[0054] Because the static friction coefficient of the edge of the grating wheel 2 is greater than the sliding friction coefficient of the edge of the friction wheel 5, when the guide wire moves due to friction, it can always drive the grating wheel 2 to move at the same speed. The rotational speed of the grating wheel 2 can be obtained using a sensor installed near it.

[0055] When the grating wheel 2 and the friction wheel 5 are not moving at the same speed, the current force applied to the guide wire can be determined by calculating the position of the linear motor 1, using the pressure of the spring 3 and the speed difference of the friction wheel 5.

[0056] When the grating wheel 2 and the friction wheel 5 move at the same speed, the guide wire is in a state of uniform displacement.

[0057] To accurately control the rotational speed of the rotary motor 6 and the position of the linear motor 1, and thus adjust the force, speed, and position of the guide wire held between the grating wheel 2 and the friction wheel 5, it is necessary to calculate the correspondence between the rotational speed of the rotary motor 6, the position of the linear motor 1, and the magnitude of the force, speed, and position of the guide wire's forward and backward movement.

[0058] In the calculation, the following assumptions are made: Assumption 1: Since the size and mass of the grating wheel 2 are relatively small, its rotational inertia can be approximately ignored. Assumption 2: Spring 3 satisfies Hooke's Law. Assumption 3: The viscous term is ignored in the frictional force generated when the friction wheel 5 slides relative to the guide wire. Assumption 4: The assembly clearance of the equipment is ignored. Assumption 5: The friction of the bearing rotation is ignored. Assumption 6: The steel core of the guide wire is assumed to be incompressible.

[0059] The symbols in the formula have the following meanings: v6 represents the rotational speed of rotary motor 6, p1 represents the position of linear motor 1, and m g Indicates guidewire quality, v m v2 represents the maximum speed difference between friction wheel 5 and clamping wheel, v2 represents the speed of clamping wheel, which is equal to the guide wire speed, t represents the current running time of the system, and F represents the maximum speed difference between friction wheel 5 and clamping wheel. elastic The elastic force of the elastic element is represented by k, the elastic coefficient of the elastic element is represented by μ5, the friction coefficient between the guide wire and friction wheel 5 is represented by μ2, the static friction coefficient between the guide wire and clamping wheel is represented by v1, the moving speed of the linear motor 1 is represented by f5, the friction force between the guide wire and friction wheel 5 is represented by Δt, and the transition time of the guide wire from acceleration to uniform motion is represented by F. end This indicates the magnitude of the force applied to the end of the guidewire.

[0060] Position control during guide wire startup: The rotational speed v6 of the rotary motor 6 gradually increases from 0 to v m ;

[0061] Linear motor 1 starts moving to the right at a speed of v1(t). When it reaches time t1, the magnitude of the elastic force F compressing the elastic element is... elastic for:

[0062]

[0063] Since the linear motor 1 is connected to the elastic element, the elastic element is compressed. The elastic element applies a spring force to the bearing, the rotating shaft 4, and the grating wheel 2. The magnitude of the sliding friction force between the friction wheel 5 and the guide wire is:

[0064]

[0065] The speed of the grating wheel 2 (guide wire) is:

[0066]

[0067] The current position of grating wheel 2 (guide wire) is:

[0068]

[0069] The position control method during guidewire operation is as follows:

[0070] During the acceleration (deceleration) of the guide wire, in order to maintain the differential speed of the sliding motion, the rotary motor 6 also needs to accelerate to ensure that the speed difference meets the requirements.

[0071] v6-v2=v m ;

[0072] Therefore, the rotary motor 6 also needs to increase its speed v6 to [a certain value] within the same operating time.

[0073]

[0074] To keep the magnitude of the guide wire's acceleration constant, if the guide wire needs to increase its acceleration, linear motor 1 is controlled to move forward; otherwise, linear motor 1 is controlled to move backward.

[0075] Suppose that at a certain time t0, the system needs to transition to uniform motion, and the rotating motor 6 reduces the velocity from v6 to v2 with acceleration a6. Then the transition time Δt is:

[0076]

[0077] The guidewire thrust control method is as follows:

[0078] When the guidewire encounters resistance as it advances, without damaging the blood vessel, it loses one degree of freedom in its direction of movement, but gains a degree of freedom in its thrust. Since the guidewire core is assumed to be incompressible, the frictional force f5 exerted by the friction wheel 5 on the guidewire can be approximated as the force exerted by the end of the guidewire core on the spring 3 connected to it. Therefore, changing the magnitude of the force F at the end of the guidewire... end for:

[0079]

[0080] Example 1:

[0081] The parameters are set as follows:

[0082] Diameter of friction wheel 5 and grating wheel 2: 2 × 10 -2 m

[0083] The coefficient of sliding friction between friction wheel 5 and the guide wire: μ5 = 0.4

[0084] The coefficient of static friction between the grating wheel 2 and the guide wire: μ2 = 0.7

[0085] The spring constant of spring 3 is: k = 12.5 N / m

[0086] The speed difference between friction wheel 5 and grating wheel 2 is set as: v m =2πrad / s

[0087] Guide wire mass: m g =100g

[0088] Step 1: At the beginning, ensure that spring 3 is in an uncompressed state, and gradually increase the speed of the rotary motor 6.

[0089] Step 2: Linear motor 1 begins to compress spring 3. At this time, spring 3 drives grating wheel 2 to contact and squeeze the guide wire. Since the speed cannot change abruptly, there is sliding friction between the guide wire and friction wheel 5, and static friction between the guide wire and grating wheel 2. Because the speed at which linear motor 1 compresses spring 3 is relatively slow, the acceleration and deceleration time periods are extremely small. In actual calculations, the acceleration process can be ignored. It can be approximated as uniform motion. Assuming linear motor 1 is moving at a speed of 0.02 m / s, Hooke's Law can be used to deduce the force F exerted by grating wheel 2 contacting and squeezing the guide wire at a running time τ. g2 And the force F of the guide wire squeezing friction wheel 5 g5 They are respectively:

[0090]

[0091] At this time, the friction wheel 5 uses sliding friction to drive the guide wire forward, and the guide wire drives the grating wheel 2 to accelerate forward in sync.

[0092] After the grating wheel 2 starts rotating in Step 3, in order to ensure that the guide wire maintains uniform acceleration until the speed reaches the expected value, a speed difference must be maintained between the friction wheel 5 and the grating wheel 2.

[0093] v m =v6-v2=2πrad / s;

[0094] Therefore, the rotary motor 6 needs to maintain its speed based on the feedback speed from the grating wheel 2.

[0095] v6 = v2 + 2π;

[0096] Since sliding friction is constant and static friction is maintained between the guide wire and the grating wheel 2, and assuming that rotational inertia is neglected according to assumption 1, the rotational acceleration a2 of the grating wheel 2 at time τ satisfies the following condition.

[0097]

[0098] The speed of grating wheel 2 is

[0099] v2=τ 2 ;

[0100] Therefore, the rotary motor 6 also needs to accelerate simultaneously with an acceleration a5 = τ. Therefore, the speed of the rotary motor 6 at this time is...

[0101] v6=2π+τ 2 ;

[0102] Note that when time τ satisfies τ2 When the speed reaches 2π, the grating wheel 2 reaches the expected speed. The rotary motor 6 can quickly reduce the speed from 4π rad / s to 2π rad / s. At this time, the system operates at a speed of 5 × 10⁻⁶ rad / s. -2 The linear motor 1 moves at a constant speed of m / s. During this constant speed motion, the linear motor 1 can gradually retreat, reducing the compression of the spring 3 until it reaches the critical position where a speed difference exists between the two wheels. The linear motor 1 typically stops slightly ahead of the critical position during constant speed operation to enhance the robustness of the system. Although there may be a slight overshoot for a very short time, it can accurately stabilize at the expected speed. The advantage of this speed adjustment method is that the response speed is significantly better than the traditional synchronous rotation method with two clamping wheels. It also avoids the slippage problem inherent in traditional guide wire feeding.

[0103] Example 2:

[0104] Parameters are set according to Example 1:

[0105] The current position of linear motor 1 is at the critical position and moving forward: Δx = 1 × 10 -2 m.

[0106] First, during interventional surgery, when the guidewire encounters an obstacle, the guidewire tip undergoes elastic deformation, resulting in a force opposite to friction. This force is similar in nature to the force of spring 3 and obeys Hooke's law. Let the elastic constant k of this force be... term =10N / m.

[0107] When the guide wire moves forward at a constant speed, the static friction force f between the guide wire and the friction wheel 5 can be calculated by setting parameters. s5 satisfy

[0108] f s5 <k×Δx=0.125N;

[0109] At this point, after the tip touches the obstacle, it moves forward 1.25 × 10⁻⁶ times. -2 After m, the friction wheel 52 and the grating wheel 25 experience a speed difference due to resistance, meaning that at this time...

[0110] v2 ≠ v5.

[0111] At this point, the system can determine whether the guidewire is obstructed based on this condition, and thus stop the guidewire from advancing in time to prevent unexpected situations such as puncture during surgery.

[0112] Example 3:

[0113] A force of 2N is continuously applied forward. Assuming the guidewire stops due to an obstruction, if the physician needs to continue applying force to the blood vessel using the guidewire, the specific steps are as follows:

[0114] Step 1: Confirm that spring 3 is in an uncompressed state, and gradually increase the speed of rotary motor 6 to...

[0115] Step 2: Linear motor 1 begins to compress spring 3. At this time, spring 3 drives grating wheel 2 to contact and squeeze the guide wire. Since the speed cannot change abruptly, there is sliding friction between the guide wire and friction wheel 5. Because the guide wire is obstructed and its position remains unchanged, the sliding friction is transmitted to the end of the guide wire. Therefore, force control can be approximately regarded as the control of the magnitude of sliding friction between the guide wire and friction wheel 5.

[0116] Step 3: To achieve a frictional force of 2N, linear motor 1 needs to move forward to compress spring 3, increasing friction. The movement distance d1 is...

[0117]

[0118] When the linear motor 1 advances a distance d1, the end of the guide wire is subjected to a force of 2N.

[0119] The above description is only a preferred embodiment of the present invention, but the scope of protection of the present invention is not limited thereto. Any equivalent substitutions or modifications made by those skilled in the art within the scope of the technology disclosed in the present invention, based on the technical solution and inventive concept of the present invention, should be covered within the scope of protection of the present invention.

Claims

1. A force-position control method for an interventional surgical robot, characterized in that, Includes the following steps: S1. Place the guide wire between the clamping wheel and the friction wheel; S2. By controlling the rotary motor, the friction wheel, which is rigidly connected to the shaft of the rotary motor, rotates at the same speed, but there is no pressure between the guide wire and the friction wheel, so the guide wire does not move; S3. By controlling the position of the linear motor, the elastic element is compressed, thereby generating pressure between the friction wheel and the guide wire, and between the guide wire and the clamping wheel. At the moment of clamping, the speed of the guide wire is 0, while there is sliding between the guide wire and the friction wheel, and the friction wheel generates sliding friction force on the guide wire. The static friction coefficient of the clamping wheel edge is greater than the sliding friction coefficient of the friction wheel edge. When the guide wire moves under frictional force, it can always drive the clamping wheel to move at the same speed. S4. When the clamping wheel and the friction wheel are not moving at the same speed, the current force applied to the guide wire is calculated by calculating the position of the linear motor, the pressure of the elastic element, and the speed difference of the friction wheel. When the clamping wheel and the friction wheel move at the same speed, the guide wire is in a state of uniform displacement. S5. By adjusting the position of the linear motor and the speed of the rotary motor, the force, speed, and position of the guide wire's forward and backward movement are controlled. The position control method for guidewire activation is as follows: Rotary motor speed Gradually improve from 0 to ; The linear motor starts moving to the right at a speed Movement, running to the time The magnitude of the elastic force compressing the elastic element. for: ； Because the linear motor is connected to the elastic element, the elastic element is compressed. The elastic element applies a spring force to the bearing, the rotating shaft, and the clamping wheel. The magnitude of the sliding friction force between the friction wheel and the guide wire is: ； The speed of the clamping wheel, i.e., the guide wire, is: ； The clamping wheel, i.e., the current position of the guidewire, is: ； In the above formula, Indicates the rotational speed of the rotating motor. Indicates guidewire quality. This indicates the speed of the clamping wheel, which is equal to the speed of the guide wire. Indicates the magnitude of the elastic force of the elastic element. This represents the elastic modulus of an elastic element. This represents the coefficient of friction between the guide wire and the friction wheel. This indicates the moving speed of the linear motor.

2. The force and position control method for an interventional surgical robot according to claim 1, characterized in that: The position control method for the guidewire during operation is as follows: During guide wire acceleration or deceleration, in order to maintain the differential speed of sliding, the rotary motor also needs to accelerate to ensure that the speed difference meets the requirements. ； Therefore, rotating electric motors also need to increase their speed within the same operating time. Upgraded to ； To keep the magnitude of the guide wire's acceleration constant, if the guide wire needs to increase its acceleration, the linear motor is controlled to move forward; otherwise, the linear motor is controlled to move backward. When a transition to uniform motion is needed, the rotary motor accelerates... speed from Reduce to The transition time for: 。 3. The force and position control method for an interventional surgical robot according to claim 2, characterized in that: The guidewire thrust control method is as follows: When the guidewire encounters resistance as it advances, without damaging the blood vessel, the guidewire loses one degree of freedom in its direction of movement but gains a degree of freedom in its thrust, thus changing the magnitude of the force on the guidewire tip. for: ； in, This represents the frictional force between the guide wire and the friction wheel.

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