JPS62167902A - Dynamically programmed motor operation valve controller - Google Patents

Abstract

(57) [Summary] This bulletin contains application data before electronic filing, so abstract data is not recorded.

Description

[Detailed description of the invention] Technical field The invention moves objects such as elevator cars.

The present invention relates to a hydraulic valve and a valve controller used in a system such as an elevator having a hydraulic actuator such as a piston.

BACKGROUND OF THE INVENTION Feedback control is used to control hydraulic elevators with accuracy approaching that of indexing elevators, which are more complex than hydraulic elevators and typically more expensive. However, comparable performance has not been achieved using feedback control, and the main problem is fluid dynamics. The viscosity of the fluid changes with ambient temperature and also with the heating that occurs as the elevator car is raised and lowered. These fluctuations can make elevator car movement unpredictable. Various grades of feedback have been used in the past, but such solutions are expensive and reduce efficiency. This is because a sufficient pumping capacity is required.

A feedback technique is shown in US Pat. No. 4,205,592, in which fluid flow through a valve and into an object, such as a hydraulic elevator, is passed through a flow meter that includes a potentiometer. As the flow rate increases, the output voltage associated with the wiper movement of the potentiometer changes to indicate the flow rate. US patent 4
, 381.699 also shows a similar type of valve control device.

U.S. Pat. No. 4,418,794 shows a type of valve used in non-flow sensing systems, which uses a large feedback loop, possibly sensing the elevator car position, and controlling the operation of the valve. Controlling 6 DISCLOSURE OF THE INVENTION Although the invention described herein has been developed from a hydraulic valve control system in an elevator, and is described as such for convenience, it may be useful for other systems requiring similar control. The present invention can be used for

To control the flow between the pump and the hydraulic cylinder of the elevator when an object such as an elevator car is raised, and to control the return flow from the cylinder to the tank when the elevator car is lowered. Therefore, in accordance with the present invention, a stepper motor operates a linear flow control valve. The time-related operation of this valve mirrors the flow into the elevator car and the speed profile or speed change of the car. Valve operation begins by placing the valve in a condition where flow from the pump is completely bypassed from the car. The valve is then gradually closed to reduce the bypass flow. When the pressure being applied to the elevator car exceeds the pressure required to maintain the car, valve movement is programmed to produce the desired elevator speed change.

According to the present invention, the differential pressure created when the output pump pressure exceeds the pressure required to maintain the car in the correct position is absorbed by a check valve in which the pump pressure and the car pressure are applied in opposite directions. Detect from movement. Movement of the check valve into the open position, where the car is specifically about to move, is detected by an electrical switch which produces an electrical control signal that is applied to the main valve controller. This control signal serves as the starting point for the programmed positioning of the main valve which determines the speed change or speed profile of the elevator car as it moves. As the car descends, the valves are initially opened at a rate appropriate for the heavy car and hot fluid. If the actual velocity of the car is less than expected for these conditions then the frequency of the subsequent signal to the stepper motor is increased proportionally and the final velocity is greater because the fluid is colder. It can be adjusted for different flow characteristics that may occur if the car is lighter or lighter.

According to the invention, during descent the valve is reset as a function of the actual speed of the car compared to the desired speed. If the car's speed does not change when the valve is repositioned (this will happen if the car is very light), the valve will be opened progressively until car deceleration is detected. When deceleration is detected, the valve is held in that position.

According to another feature of the invention, perhaps of particular relevance to elevators, each changing portion of the car speed of acceleration pull-in, constant acceleration, acceleration pull-out, deceleration pull-in, constant deceleration and deceleration pull-out comprises:
It is specifically controlled by adjusting the valve window area with the step motor and by providing a constant gain between the volume motor step and window area during the entire elevator operation.

The invention has many features. That is, the present invention provides highly accurate performance because fluid and load characteristics control valve operation. Moreover, the present invention is simple and reliable because it selectively uses feedback to adjust those characteristics. For the most part, valve flow is controlled without feedback.

DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS FIG. 1 shows a hydraulic elevator control system for moving an elevator car 10 between multiple floors. Floors are not shown in the diagram. A car piston (plunger) 11 attached to the car extends from a cylinder 12, and fluid is pumped into the cylinder to raise the car and expelled from the sill to lower the car.

The flow rate is controlled and regulated as described below. The movement of the car is detected by the pickup 13. A pickup associated with the fixed position tape 14 provides a signal (position) on line 15, which signal is transmitted to the pump and valve control system 5! (PVC) Sent to 17.

This position signal represents the car's position and speed. The thus sensed position of the car is used to control the flow of fluid between the cylinder and the pump, which in turn controls the position of the car piston or plunger 11. PVCl 7 controls a hydraulic valve system including pump 21 and fluid tank 5. The pump has check valve 6 (
backflow prevention) to the hydraulic control valve, and this assembly, along with the pump,
7. A pump on/off signal on line 22 turns the pump on (starting) or off (stopping).

The pressurized fluid from the pump passes through the check valve 6 to the first
This port 25 is added to the port 25 of the linear valve 27 (two positions [P
1 and P2 in a straight line leading to a "key-shaped" valve window 26 that is part of the retracting valve). It is completely "opened" at P2 and completely "closed" at Pl.

The position of valve 27 is controlled by a stepper motor 28 which receives a signal (speed) on line 29 from PVCl7. This signal is a series of pulses whose frequency changes the speed of the motor 28 and thus the longitudinal direction (arrow A) of the valve 27.
1) Determine the positioning speed. Each pulse in the velocity signal represents an incremental distance along the distance of movement of valve 27 between points P1 and P2'. The position of the valve is represented by an aggregate count between these point positions. The bulb window 26 consists of a large window 26a and an adjacent narrow window 26b, and is generally "key-shaped". At point P2, the large window 26a is connected to the first inlet port 25.
, and the narrow abutment portion 26b is located next to the second port 31. At this point, valve 27 is "open". The second boat 31 reaches a line 32 leading to the tank 5. At position P1, the small window 26b
are mostly adjacent to port 25, and ports 1 to 31
The passage to is blocked by the solid part of the valve.

In this position, valve 27 remains "closed". In the open position P2, fluid flows from the pump through line 24. This is a "flow up J(FU)," or flow that raises the car. Fluid then flows into the large window 26a and from there returns through the small window 26b to line 32 and then back to the tank. .Thus F
The U flow is bypassed when the pump is started.

However, as valve 27 closes (moves to position ff1P1), pressure in FU4 begins to build up within internal port 1-35, while bypass flow in line 32 flows through port 1-31.
It decreases as the path passing through the window 26b decreases. When the valve 27 moves to the position PL (non-bypass position), the two windows 26a, 26b and the main input robot 25 conform to some extent, so that the path passing through the large window 26a is reduced and the path passing through the small window 26b is reduced. The path increases. However, small window 2
The area of 6b depends more on the longitudinal position of the bulb 27 than in the case of a large window. As a result of this, the area of the small window relative to the outlet port 31 controls the change in flow rate, and the area of the small window controls the area of the small window when the main valve is in the closed position PL (in this position all FU
The flow decreases as it begins to move towards the port (flowing from port 25 to inlet 35). There is no path between ports 1-25 and outlet port 31.

Fluid pressure PS1 in internal port 35 is applied to main check valve (MCV) 40. The small stem 41 of this valve enters a guide 41a. MCV is Poh
35 (pressure Psi) and port 43 (pressure PS2). The pump is turned on and the main valve 27 moves towards position P1, PSL
When exceeds PS2, MCV40 is pushed up, and the FU flow passes through MCV to line 42 (this line is connected to cylinder 1
2). This happens when the bypass flow decreases. The resulting fluid flow displaces the screw 1 hem 11 upwardly, moving the car upwardly.

When the car 10 is at rest, the pressure in line 42 and the pressure in chamber 44 are the same, at pressure PS2.

When pump 21 is turned off, this pressure pushes MCV 40 down and downflow (FD) in line 42 is blocked, keeping car 10 in position. There is no flow through line 42 and back to tank 5 in this condition. To create this flow, MCV 40 must be lifted, and this is accomplished by operation of main check valve actuator 50.

This actuator includes a rod 50a that contacts the stem 41 when pushed up; a first member 50b that is pressed against the rod; and a second member 50c that moves the first member when pushed up. When the rod 50a is pushed up, the MCV 40 is pushed upward, and at this time the pressure PS
2 fluid is added to inlet line 52 and it only occurs when a down signal is applied to line 53 going to solenoid controlled release valve 55. At this time, fluid pressure in line 52 is applied to the bottom of members (pistons) 50b, 50c. The joint area of these members is the valve 40
It is larger than the area 62 of the upper surface of. The second member 50c moves until it abuts the wall 50d of the chamber 50e. 1st
Member 50b moves together with second member 50c due to flange 50e.

This slight movement (up to wall 50d) breaks MCV 40 open and equalizes pressures PS1 and PS2. The first member then continues to rise, it hits the wall and the MCV
40 fully opened. This allows return flow (FD) to flow from chamber 35 through windows 26a, 26b and into line 3.
It can flow through 2. FD passing through line 24
Flow is blocked by check valve 6. valve 27
This position determines the flow rate of the FD flow and determines the velocity profile of the car as it descends. The valve is moved from the closed position P1 towards the open position P2 by the speed signal. The width and frequency of the speed signal pulses determine the descending speed profile.

A switch 70 is adjacent to MC, V2O, and M
The upward movement of CV40 activates the switch. This action generates a signal (CV) on line 71 going to PVCL7. This Cv double signal indicates that the valve is moving in the upward direction for elevator operation. It indicates that the pressure in chamber 35 slightly exceeds the pressure in chamber 43. Using this signal, the PvC controls the movement of the valve spool by adjusting the r[J and frequency of the pulses that make up the speed signal applied to line 20. Pressure PS1 is pressure PS
The Cv times signal occurs when it exceeds 2, and it occurs before there is actual flow. Therefore, the Cv multiplication signal generation manifests a ``predetermined'' flow.

Stepper motor control valve 27 provides a pressure relief function to boat 35. Stepping motor 28 has an output link 28a and a collar or link 28b is attached to that link. This link and collar fit snugly into the hollow part of the valve 27, but are separated from the basin (windows 26a, 26b) by the valve wall 27a.

Wall 27a faces another wall 27b. (Valve 2
7 is in the form of a hollow cylinder through which the fluid flows) A spring 28C is fitted between the wall and the collar 28b. As the step motor operates, the link moves up and down in steps at a speed of 43 steps. This movement is transmitted via the spring to the wall 27a and to the valve 27, which moves synchronously with the link. When the pressure in the pump output line 21'a is large enough to open the pressure relief valve (r'RV), that pressure is applied to the top of valve 27b, and the entire valve 2
7 is lowered and flow passes from the pump through line 32 to tank 5, eliminating the "overpressure" condition.

To manually lower the car, manual valve 80 is operated to allow fluid to return from the chamber to tank 5.

FIG. 2 shows the car speed and speed signal during "up" elevator operation, elevator response to an up call. The pump is first turned on at time TO, and just before that the linear valve is in the fully open position P2. The pump is started at To and the valve is opened at an initial speed rate SO of steps per second. As used in the text below, II S 71 is the speed signal rate, and II S N II is the individual rate (N is a value from 0 to 4). S4 is a faster rate than SO. The linear valve opens at a constant rate determined by the SMAX frequency. At time T2, C
The v times signal is received and at that time the valve is moved to position fiPO2. The rate at that time is SO
This continues for a predetermined time T. Ray 1~ then becomes a predetermined high rate S1, which is So
lasts for a predetermined time T, as was the case. After this initial time T, Ray i- becomes another high Ray 1-82, which also lasts for a time T. The rate increases at each interval of T, passes through S3, and finally reaches sheet 4 S.
Ends with 4, which is the preset maximum acceleration/deceleration rate for that car. S1, S2 and S3 determine the pull-in characteristics. The position of the valve at any point is To
It is known by counting the steps that have occurred since. The valve position at which constant acceleration/deceleration occurs will be shifted somewhat.

This is because the time for SO lag is determined by the difference between To and T2, which is a function of fluid properties.

At time T4, S4 rate 1~ is interrupted in favor of the earlier rate S3/h. Time T4 is T'
It clearly corresponds to the valve position determined by the number of steps made since O.

At time step T, the rate is decreased through S3 to SO, and finally the rate reaches zero at time T5. This determines accelerated withdrawal. Time T5 and T6
The car is moving at a constant speed VMAX between. The valve is fully open in position P1 and all FU
The flow goes towards the cylinder. There is no bypass flow. Time T6
A deceleration signal is received at It is obtained from a device in the shaft and marks the point at which to start decelerating to the ascending mezzanine. It is obtained from the position signal.

At this point, the valve is gradually moved to the open position (FUi4
to the tank) and reduce car speed at acceptable pull-in, pull-out, and deceleration rates. The travel range between PO and Pl is utilized again in the ascending state. The retraction phase for deceleration, which begins a short time after the speed signal, is started by immediately moving the valve towards the open position at 1/-1-3l, but reversed (opposite polarity). ). This is because the valve has to be opened and forced towards the 2nd position P2. Then, one time T later, the rate is gradually increased for each incrementing sac to the final rate 1 to S4, at which time the car is decelerating at a constant rate of -1 determined by the rate S4. When the valve is in POl, the rate is S4 to SO
decreases to At position P○2, the rate decreases to 0 and the motor stops. However, at position P02, CV
4: Due to the delay until No. 5 was made, the valve was slightly open for the distance iDP. Because of this, Kerr slowly approaches the floor at a low rate. This is because some pump power is being applied to the cylinder.

Upon reaching the outer door zone in the floor, the valve is closed at a high rate S5 and then at a high rate I~86 in the inner door zone. When Kerr is in bed, the pump motor is stopped. The valve is fully open at that point.

Descending from the floor is a different method.

This is because the car's speed is equal to the flow down (FD) speed. And it is completely controlled by the positioning of the linear valve. (In the upward direction, the maximum speed is determined by the pump output.) The descent is shown in FIG. Descent begins by placing the valve in the closed position of Pl.

In that position there is no flow back through the pump due to the check valve. The FD flow path through line 32 is closed at the linear valve. The MC* valve 4o is pushed up in response to the generation of a down signal applied to the solenoid release valve.

This creates a Cv multiplier signal, and in response the valve is moved from Pl to P2 (reversing to open the valve) at an initial rate of -80. The car then starts moving and is given a position signal from the pickup. During two equal intervals 120 ms apart between times To and T1 (during which the speed of the stepper motor is kept at So), the speed of the car, i.e. its rate of descent, is determined from the position signal. measured and compared to the maximum car speed. The SO rate is the worst case rate.6 The rate assumes the fluid is hot and the car is fully loaded. Therefore, the SO is lower than the rate when the car is light or the oil is cold. If the car is lightly loaded and/or the fluid is cold.

That is, if the car's speed is lower than what was predicted, 5
L-34 is increased or decreased in proportion to its overspeed or underspeed. By comparison, two speed error signals (VER
R) and recalculate their rates So'-84' using the average of these two. The motor is gradually advanced in equal time steps of T between Sl' and 84'' between times T1 and T2, which is the final acceleration rate. This rate continues to increase to 54' until time T3. circle.

The rate then decreases from 84'' to 0 by time T5. T3 determines the valve position POI, at which position the car is at a speed of 90% of its maximum speed VMAX. Following this process, the valve , FD flow is about 9 of VMAX
It comes to the final position where it becomes 0%.

The valve is almost completely open, ie at or near position P2. The car is lowered and its speed is monitored via position signals during its descent. The valve is opened or closed by applying a low rate speed signal (correction signal) to keep the speed near VMAX. As the floor is approached, the descending signal is again received at a distance from the floor. At that point, the valve position PO2 is determined by adding or subtracting a correction signal step that moves the valve in either direction to "fine tune" the flow rate to the step made by the motor up to position PO2 (at time T5). It is immediately known from the total number found.

Is the lid completely closed? The final position of valve PIA close to PL is calculated by taking into account delays such as the magnitude of the floor position sensor. By making it somewhat smaller than position P1, the valve will not open prematurely and the car will not stop before reaching the floor.

Calculate the distance between PO3 and PIA and use 10% of that distance increase for the entry and exit stages. The recalculated rate SO"-83" is used to perform the withdrawal stage and the withdrawal stage. These are increased proportionately to reduce the rate S within the band that determines the 10% entry and exit portions.
In the 6 position PIA with O4''' the valve is not fully closed and the car slowly approaches the floor a short distance. By closing the MV and removing the descending signal from it, By closing the Cv valve, the car will stop on that floor.

A computer-based system for performing this type of valve operation is shown in FIG.

That is, the pvc includes a processor 17a, which includes a CPU 17a1, a CPU clock 17a2, and a CPU
RAM 17a3, and input/output terminal 17a
4, receives a signal from the CPU via this terminal,
Then it is sent to the CPU. CPU via input/output ports
receives car calls, hall calls, location signals, and Cv double signals. Through the input/output water-1, the CPU provides a falling signal through the buffer triper 17d. Similarly,
It provides a speed signal via buffer 17c and a pump on/off signal via buffer 17b. Ray 1-81, S2, S at the start of the elevator run
3, and in order to calculate S4 by 31, the CPU is connected to an EPROM 17c containing the stored parameters regarding the valve movement. These calculations of ray 1- are E r'
It is simply done from the basic speed profile stored in ROM. The mathematical procedures or algorithms that carry out the calculations are well known and easy to implement for those skilled in the computer processing arts, and as such, the details of the calculation process will not be described.

These rates are first calculated at the start of the run and then "read out" in order to carry out the special sequence that characterizes the invention. The valve positions as the valve 27 opens and closes are stored in the EPROM (in the form of the number of motor 28 steps associated with each position). Connect a backup position sensor to the valve to determine its open and closed position (so that valve movement has no appreciable effect on fluid flow rate)
r Indicates the position of the dead zone. The flow chart shown in FIGS. 4A and 4B describes the process used to program the CPU 11 to achieve the desired elevator control described above.

The valve control process begins by entering a call, either an up call or a down call. Step S10 determines whether it is a descending call or an ascending call. If it is a down call, the test in step SIO is yes, and the procedure begins in step S90, which will be described below. If it is an up call,
It is assumed that test 1 of the down call is negative, and the procedure proceeds to step S12, in which the valve 27 is moved toward position P2, where it is fully open. In step S14, the pump is turned on,
The fluid then returns to the tank through the valve. The first stepper motor rate SMAX is read in step 16 by indicating N=O, and in step 18 the computer clock is set to TO. In step 2° the step motor speed signal is commanded at rate S to N=O, and in step S27 a test 1- is made to determine if the Cv signal is occurring and if not, the speed The signal is SMA
It's X. If the answer to this test in S22 is yes, it indicates that the C■ times signal is generated, and the process advances to step S24, where N=1.
Select N by using the formula +X (or initially selected as zero, so N=1). Step S
In 2G, the computer is asked to determine the speed rate for S with N equal to 1 (SL used earlier in this description). In the next step 328 the time counter is started at T1 and in step S30 81 is added to the speed signal. In step S32 a measurement is taken to determine the duration of the speed signal (which should be T). Continue to generate the speed signal until the time T. When the duration T ends, step: I
In II, go to Test 1 and decide at what stage the retraction speed becomes 3-Progera I. There is one stage beyond the SO stage and, as already mentioned, Sll determines the constant acceleration part. If N is in step S36, t;/
If not, X is increased by 1 and the process returns to step 32G so that S2 becomes the speed signal rate. When N equals 4, it becomes S4
has been used for a duration T. S4 is generated, as indicated by step 336, and tested in step 338 to determine if time T3 has arrived. That is the time when the withdrawal stage should start. Until T3 occurs, the speed rate remains at S (N), where N is 4.
be equivalent to. If the answer to the test in step 338 is yes, the process proceeds to step 4o, which in turn creates the reverse of the programmed sequence of speed signals from SO to 84. Step S 4. In O, N is determined to be equal to X-1, and X is initially given a value of 4. In step S42, the speed signal is given the value of N for S determined by the step S/100 equation with N=3. The speed (i) is maintained until a positive answer is given to the test at S/1.4 that the duration is equal to T.

In step 346, a test 1- is made to determine whether N is equal to zero, which is the last rate in the withdrawal phase. If the answer is no, X is decremented by one in step 348 and the process then returns to step S42 where the speed signal is given a new value, which in this case would be S2. If the answer to step 46 is yes, it indicates that the withdrawal phase is complete and the process continues in step 5.
0, this step asks if the deceleration flag is obtained. The deceleration flag is an accumulated signal indicating that the deceleration position has been reached. At this point, the elevator car is moving at maximum speed in the upward direction and is approaching the point of deceleration. Thus, step 40 yields a negative answer. An additional test is made at step 42 to determine if downhill travel has begun. This is a climb run, and therefore the answer is no, and the process continues to step 44, at which time the stepper motor is turned off. Therefore, at this point the valve position is fixed and the valve is effectively in position P1 due to the number of increments that have occurred in the retraction acceleration stage and the withdrawal stage. At step 46, a test 1- is performed to determine whether the deceleration position has been reached. If the answer is no, the motor must remain off.

If the answer is yes, the process proceeds to step 58 which performs an initialization whereby the speed signal is reversed for the purpose of moving the valve in the opposite direction in response to the speed rate signal.
minus S). This is required as already explained. This is because at this stage the valve has to be moved from the closed position to the open position in order to slow down the car and bring it level with the landing. Step S6
0 establishes an initial value for N. As before, N is determined by 1+X, where X is initially equal to 1. Using this calculated parameter for N, the procedure returns to step 26. At step 56, a deceleration flag is stored in response to the deceleration signal. Thus, step 4
Upon completion of step 6 (which occurs during the deceleration withdrawal phase), a yes answer occurs at step 50. The process then proceeds from step 50 to step 62, where the motor is turned off. At this point the car approaches the floor and a determination is made as to whether the car has reached the outer zone, which occurs in step 64.

If the answer is yes, the process proceeds to step 66, in which the speed signal is previously set to ? 5 multiplied value - 85
, which is a preselected high inversion rate. This reversal rate -85 is Test 1- of step 68.
Test 1~ determines whether the car has reached the inside zone and gives a yes answer. after that.

In step 60, the velocity is increased to a higher reversal rate 1--
56, which occurs in step 70.

When this floor level is reached, the test in step 72 returns a yes, which causes the pump to be turned off and the motor to be turned off in step 74.

The climb run is then complete and the process ends.

If step 10 returns a yes (which indicates that the car is falling), the process proceeds from step 10 to step 90.

Step 90 sets a descending travel flag indicating that the car is descending in response to a floor call for descending or an in-car call for descending. Thereafter, in step 93, the Cv valve is opened by the CPU which provides a lowering number 48. In step 94, the CPU reads its accumulated value S for N equal to O, and the time is set to To in step 96. As before, its speed is S
(N) rate (N equal to zero) or a previously determined SO rate. At this point, the car begins to gather speed and the valve opens at Leh SO. In step 100, test 1~ is performed to determine whether 120 milliseconds have elapsed since time TO. 12
After 0 milliseconds, the difference between the desired elevator speed and the speed represented by the position signal is accumulated. That is VEL
Known as ERRI. If the time elapsed since TO is 2/10 milliseconds as measured in step 104, another velocity error signal, VEL, ERR2, is provided in step 106. In step 108, the average of ■ELERRI and VELERR2 is obtained,
It is stored as a percentage number. Step 110 is an initialization process to provide N, which is used, as previously described, to determine which of the rate signals to use. This happens at the beginning, and X starts out as zero. Then, in step 112, the speed rate signal S (N) is read and since X is zero, this is Sl. When the motor speed is commanded, adjust Sl according to the percentage of the error signal to a larger or smaller value. If the car is moving faster than expected, decrease Sl.

If the car is moving slower than expected, S
Increase l. The result of this modification is S' (N), and in step 116 the velocity signal is S' (
N), which in this case is 81 plus overspeed or underspeed. A test I- is made in step 118 to determine the duration of the velocity signal. When this duration is T, a test is made in step 120 to determine whether N is 4. This is because there are four steps beyond SO in the lead-in fuse.

Since N is equal to one in this example, X is increased by one step in step 122 and the process is repeated until N equals four. At that point the speed signal is S'4, which is the adjusted maximum acceleration rate. Step 124 identifies the procedure for holding S'4.

A test is made in step 126 to determine if the car speed V has reached 90% of its accumulated VMAX. This VMAX is the maximum descending speed of the car. If the answer to this test is yes, the procedure proceeds to step 40, which includes an extraction phase during acceleration. This procedure has already been described, except that the number used for the draw is now SIN. Once the withdrawal phase is complete and the car is found not to be in the level position at step 7G, the valve position is stored as VPA at step 77.

This shows the position of the valve just after the withdrawal phase. In step 578 the error between the accumulated speed and the reference speed is obtained and stored as plus or minus SC, and the SC signal is the speed controller d
is commanded to move the valve between positions P1 and P2 in small increments to determine the difference between its speed and the reference speed within the error limits of the closed-loop system being used during this mode of operation. I keep it inside. In fact, step 8o yields a yes, indicating that the deceleration position has been reached. Step 84
At , the deceleration flag is set.

In step 86, the signal S' (N) is corrected C○
Can be multiplied by RR. This correction increases or decreases the rail l- of the step to move the valve to position PIA (see FIG. 3) so that it spends about 10% of its time in the inlet and outlet stages. Upon completion of step 86, the velocity values S'' (N) are inverted in step 58 (they are made negative).

because its value must move in the opposite direction,
Then, from step 58, the withdrawal phase continues as before, but with a new value -5'' (N).

) In fact, the test in step 76 shows that the car is in the level zone near the floor, and the test in step 8
A yes answer at 8 causes a fixation or termination of the down signal, at which point the car stops. Then the process is finished and the car is on that floor.

The present invention has been described as being applied to an elevator. However, it is clear that the present invention can be used in other hydraulic control one-to-roll systems that require similar speed and positioning accuracy.

Furthermore, although a preferred embodiment of the present invention has been described, those skilled in the art will be able to modify part or all of this embodiment within the scope of the spirit of the present invention.

[Brief explanation of drawings]

FIG. 1 is a functional block diagram of an elevator controller (-roll system) including a hydraulic valve according to the invention, showing the valve in a longitudinal section. FIG. 2 shows two waveforms with a common time axis, one of which Figure 3 shows the speed of the car between two floors for an elevator up call, and the other shows the step motor drive signal applied to the valve step motor to create the car's speed profile. Flowchart 1- of the processor routine used to control the stepper motor to achieve the desired car speed profile with and downward travel. In the figure: A: Hydraulic valve 5: Tank 6: Check valve. 10: Car 11: Car piston 12 Niji cylinder 13: Pick up 14: Hoist way tape 17: Pump and valve control 22 Two lines 24 Two lines 25: First port 26: Valve window 27: Wall 28: Motor 28a: Output link 28b: Ring 28C Ni spring 31: Outlet port 35: Chamber 40: Main check valve 41a Ni guide 43: Port 50a: Ronod 52: Inlet line 53 Ni line

Claims (7)

[Claims] (1) Object 10; A hydraulic actuator 12 having a piston 11 that extends and retracts the object to raise and lower it; A sensor 13 that detects the position of the object and provides a position signal indicating the speed and position of the object; Hydraulic fluid tank 5; Hydraulic fluid pump 21; Adjusts the flow rate of fluid between the pump and the actuator to raise the object, and adjusts the flow rate of fluid between the actuator and the tank to lower the object. a hydraulic valve A for regulating the hydraulic valve A; processing means 17 for controlling the operation of the hydraulic valve and the pump in response to a position signal; said valve moving in a first direction from the pump to the actuator; and simultaneously decrease the flow bypassed from the pump to the tank to control the rate of rise of the object when the pump is on, and move in the opposite second direction to remove the object from the actuator when the pump is off. A single flow control valve 26 reduces the flow to the tank to control the rate of descent of the object, and the pressure of the pump output applied to the car exceeds the pressure required to hold the object in place. means for providing a control signal indicating that the velocity signal is of one polarity and moving the valve in a first direction when the velocity signal is of one polarity; an electric actuator 28 responsive to the velocity signal by moving the valve in a second direction opposite the first direction when polarized; said processing means responsive to the velocity signal of a first magnitude after activation of the pump; Hydraulic system characterized in that it comprises means (17a1, 17a4) for providing a series of velocity signals of different magnitudes for determining the velocity profile of the object over time in response to the control signal. (2) the electric actuator comprises a step motor, the speed signal having a first frequency, each starting in response to a control signal and continuing for a predetermined time interval until a predetermined maximum frequency; providing a speed signal of said first polarity according to a first sequence having a first series of higher frequencies, maintaining the speed signal at its maximum frequency for a predetermined number of steps, and then having a frequency equal to that of the first sequence; Then, the speed signals of successive frequencies are reduced in a second sequence decreasing from the maximum to the first frequency (the duration of the speed signal of each frequency in this second sequence is the aforementioned predetermined time interval). 2. A hydraulic system as claimed in claim 1, wherein said processing means comprises means for providing. (3) The means for providing a control signal includes a check valve (MCV) connected to the actuator and a switch 70 that activates the check valve when the check valve opens to allow flow to the actuator; 4. A hydraulic system as claimed in claim 3 providing a check valve signal. (4) the processing means comprises means for providing a series of high frequency speed signals of said second polarity after stopping the motor following deceleration of the object when the pump is on; 3. A hydraulic system according to claim 2, wherein the velocity signal is provided at the highest of these frequencies until the object is at a predetermined position. (5) providing a speed signal at a first rate of change during a constant sampling interval, then at various rates each lasting the same time interval, these rates being the product of the predetermined rate and the adjustment signal); the processing means provides a velocity signal in a sequence that determines the velocity of the object as it is lowered by giving 5. Hydraulic system according to any one of claims 1 to 4, comprising: said adjustment signal representing a ratio between the speed of the object and a reference speed. (6) A claim in which the object is an elevator curve, and the processing means controls the stopping and starting of the car on the floor of the building and the speed of the car in response to the car call and the floor call. Hydraulic system according to clause 5. (7) Claims in which the object is an elevator car, and the processing means controls the stopping and starting of the car on the floor of the building and the speed of the car in response to the car's call and the floor's call. Hydraulic system according to items 1 to 4.