EP1880970B1 - Method for optimising the weight of a counterweight of a lift facility and lift facility with such a counterweight - Google Patents

Description

The present invention relates to a method for optimizing the weight of a counterweight of an elevator system according to the preamble of claim 1.

An elevator system generally comprises a cab for carrying persons and / or loads which is raised, lowered or held at a height by means of a propellant, for example a traction cable. For this purpose, a drive means applies a corresponding tensile force to a propellant.

For this purpose, the drive means comprises a motor whose output torque or lifting force is converted into a tensile force on the propellant. Due to the design, this engine can apply a certain maximum lifting force during continuous or time operation. For example, the heat dissipation limits the continuous power of electric motors in continuous operation. In the time mode, during which the motor can apply a higher lifting force for a short time, the maximum current consumption limits the maximum lifting force.

The static holding force for holding the cabin at a height may also be applied by the engine or, advantageously, by a brake which may be integrated with the engine or separately apply a holding force to the propellant. Since brakes can apply high braking (holding) moments with simple means, the static holding force generated by the brake is generally greater than the (permanent) lifting force that can be applied by the motor.

To reduce the drive means to be provided holding or lifting force, it is for example from the US 5,984,052 It is known to couple a counterweight to the cab via a suspension means so that it lifts as the cab lowers and lowers as the cab rises. The suspension can with be the same or separate from the propellant and be firmly connected to the cabin and / or the drive. For the sake of simplicity, the term suspension means is also understood to mean blowing agent.

Usually, the weight of this counterweight is chosen so that it substantially corresponds to the sum of the empty weight and half of the maximum permissible payload of the cabin. This minimizes the maximum tractive force that the drive means must lift, hold or lower the cab. At half payload, the elevator system is balanced, i. the drive means does not have to apply a holding force and overcome only frictional forces when lifting or lowering. The maximum tractive effort then occurs with the cab empty (with the counterweight pulling down) and the cab full (with the cab pulling down). The drive means is chosen so that on the one hand apply this maximum tensile force as a static holding force and on the other hand also compensate for the occurring at a nominal speed profile inertial forces of the cabin including payload and the counterweight in the continuous or time-lifting operation.

Deviating from this suggests the US 5,984,052 to select the counterweight so that it corresponds to the sum of the curb weight and a statistical average of the payload distribution, which is assumed in the embodiment with 30% of the maximum payload. Such an elevator system is balanced on a statistical average, ie it requires only little holding or lifting forces during a large proportion of daily operations. However, if the cabin carries more than 40% of the maximum payload in the exemplary embodiment, the tractive force to be applied by the drive means increases with respect to the previously described 50% balanced elevator system and exceeds 80% of the maximum payload with the maximum applied pulling force of the 50% balanced elevator system.

In this area, the same drive means can no longer compensate for the same inertial forces. Accordingly, the US 5,984,052 before, off to change the nominal speed profile for a given payload value and only to work with lower accelerations.

The disadvantage of the requires US 5,984,052 proposed balancing the elaborate empirical determination of the payload mean value. If the payload distribution deviates from the distribution used in the design of the weight of the counterweight in actual operation, the elevator system will operate suboptimal. Even with a large standard deviation from the mean value, ie when payloads that frequently deviate significantly from the mean value, the efficiency of this elevator system deteriorates.

The conventional 50% balance requires relatively large counterweights. These are unfavorable in the production, assembly and maintenance. In particular, large counterweights disadvantageously require additional space in the elevator shaft. Balancing with a statistical payload average considerably reduces the transport capacity during full load operation, since the nominal speed is reduced precisely in this operating state.

The object of the present invention is therefore to provide an elevator system which avoids the above-mentioned disadvantages. In particular, it is an object of the present invention to provide a method and an elevator system which is more favorable in terms of manufacturing, assembly, maintenance and / or the required installation space in the elevator shaft.

To solve this problem, a method according to the preamble of claim 1 is further developed by its characterizing features. It is a method specified, with a counterweight can be optimized accordingly.

A method according to the invention comprises, in a manner known per se, a car with an empty weight MK which moves a maximum payload MLmax can. Attached to the cab is a support means on which a drive means can apply a pulling force such that the cab rises, lowers or is held at a height. In this case, the drive means can apply a maximum tensile force MFmax as a static holding force MFmaxA, as a dynamic permanent lifting force MFmaxUD and / or as a time-lifting force MfmaxUZ.

As a rule, the dynamic lifting force, which must compensate for inertial and frictional forces in addition to weight, is greater than the static holding force. In this case, the time-lifting force that can generate the drive means for a short time is generally greater than the duration of the lifting force which the drive means can apply over a longer period of time. Conversely, in particular, provided that the drive means advantageously comprises a brake, which may be integrated into or formed separately from a motor, the maximum static holding force MFmaxA which can be generated by the drive means may also exceed the dynamic lifting force MFmaxU. In particular, safety brakes in elevator systems can exceed the nominal powers of the drive motors in order to be able to decelerate and hold the cars safely in the event of a failure of the engines. In order to compensate for the inertial forces that occur during such emergency braking, which can exceed the dynamic loads in normal operation, safely, the brakes can be correspondingly strong dimensions.

An elevator system further comprises a counterweight coupled to the cab via a suspension means so as to rise as the cab lowers and to lower as the cab rises.

According to the invention, it is now proposed that the weight MG of the counterweight essentially corresponds to the sum of the empty weight MK and the difference between the maximum tractive force MFmax of the drive means and the maximum payload MLmax of the car, in equation form $$\mathrm{MG}\approx \mathrm{MK}+\left(\mathrm{MLmax}-\mathrm{MFmax}\right)$$

The weight of the counterweight does not necessarily correspond exactly to the sum of the empty weight and the difference between the maximum traction force and the maximum payload. In particular, the counterweight, as will be explained below, be chosen somewhat larger, to take into account inertial and frictional forces and additional weights of the suspension means, so that applies $$\mathrm{MG}\ge \mathrm{MK}+\left(\mathrm{MLmax}-\mathrm{MFmax}\right)$$

The drive means can apply a maximum of a tensile force MFmax due to the design. This is always at least greater than half the maximum payload MLmax, otherwise the drive means could not hold or raise and lower either the full or empty cab. $$\mathrm{MFmax}>0.5\times \mathrm{MLmax}$$

Now, according to the invention, the weight of the counterweight is selected such that the drive means with this maximum traction force can hold the cab with coupled counterweight straight or lift or lower it with the nominal velocity profile. In this case, the safety factors required for elevator systems can be taken into account, for example, by setting a quotient of the design-related maximum tensile force of the drive means and a corresponding factor as the maximum tensile force MFmax in equation (1) or (2). A typical value range of this security range is 1.1 to 2.0. This makes it possible to take into account the usual effects of acceleration and inertia, friction losses, suspension element displacements or overload reserves. This safety factor is usually set for certain elevator categories. Preferably, this safety factor is about 1.3. This value is proven in passenger elevators with up to 10 storeys.
Of course, this safety factor can already be included in the specification of the maximum tensile force MFmax of the drive means. In this case This safety factor does not need to be taken into account when optimizing the counterweight.

In contrast to the previous interpretation of the weight of the counterweight where either the required maximum traction of the drive means is minimized (50% balance) or the required traction of the drive means is minimized by statistical means, it is proposed according to the invention, provided by a drive means available To fully exploit traction while optimizing or minimizing the weight of the counterweight.

This makes it advantageously possible to select the drive means from a series with predetermined graded tensile forces. In a first step, the drive means with the smallest maximum traction force is selected, which is sufficient to raise, lower or hold the cabin at a 50% balance. Because with a 50% balance, the required maximum tensile force is minimal, so that a drive means in any case must be able to afford these depending on the balance of the smallest possible maximum traction.

In graded series, the maximum tensile force of the individual types will usually not match exactly with the determined thus, the empty and payload weight of the cabin, friction coefficients, weights of the suspension, safety factors and the like, the lowest maximum tensile force for a specific application. Accordingly, in the first step that drive means selected from the series whose maximum tensile force exceeds this minimum required maximum tensile force.

The thus selected drive means thus provides more maximum traction available than would be required for the specific application. This excess is used according to the invention to optimize the weight of the counterweight as much as possible, that is to minimize. Because a counterweight, which is not balanced with 50%, requires in the In the limiting case of an empty or maximally loaded cab, a higher tractive force for lifting, lowering or holding the cab. However, this higher tractive force can be achieved by the oversized drive means selected from the series.

On the other hand, it's not like that US 5,984,052 necessary to change the nominal speed profile at higher payloads, since according to the invention the weight of the counterweight is minimized only to the extent that the car can drive over its full payload distribution with the desired nominal speed profile. Because according to the invention, the weight of the counterweight is reduced only so far that the drive means can raise or lower the cabin in all operating conditions with the desired speed profiles. This increases the transport capacity during full load operation.

Thus, the choice of counterweight weight according to the invention represents an optimum compromise between 50% balancing with, in the extreme case, minimal traction, and balancing on the statistical payload average, where the traction is minimal on a statistical average. It allows in particular to select the drive means from a series with predetermined graded tensile forces and thus makes it possible to resort to cost-effective series drive means, these nevertheless optimally exploit and minimize costs of the elevator system.

A minimal counterweight brings a number of advantages: on the one hand, material costs are already saved during production. On the other hand, the handling of a smaller counterweight in the production, transport to the place of use, installation in the elevator shaft, maintenance and dismantling is much easier. Finally, a smaller counterweight advantageously requires less space in the elevator shaft (or a separate shaft). In a limiting case, the weight of the counterweight could even be made so light that the counterweight is equal to the weight of the empty car. As Stawinoga in the journal Liftreport from Sept./Okt. 1996 could show in this Case of further measures to protect against uncontrolled upward movements are waived.

The suspension means may comprise one or more cables and / or one or more belts. As a rule, the carrying and blowing agents are identical, i.e. Rope (s) and / or straps attached to the cab and the counterweight and diverted by means of loose and / or fixed rollers and / or one or more traction sheaves.

One or more cables and / or belts of the suspension element are preferably coated with an elastomer, in particular polyurethane. This increases in particular the traction or traction of the suspension element. As is known, in the case of a deflection via a traction sheave, the counterweight according to the Euler-Eytelwein equation must be at least e ^{μα of} the cabin ^weight with the coefficient of friction μ between the traction sheave and the suspension angle α. An increase in the coefficient of friction by the advantageous coating thus allows a reduction in the weight of the counterweight.

The drive means preferably comprises a motor, in particular a frequency-controlled electric motor, and may have at least one traction sheave for converting an output torque of the motor into a tensile force on the suspension element. Integrated into the engine or separated from this, a brake can be provided which can apply a static holding torque on the at least one traction sheave. As brakes all known friction and / or positive brakes come into consideration.

Preferably, as the maximum traction force MFmax of the drive means, the smaller value of the static holding force MFmaxA at which the drive means holds the cabin at a height, the dynamic endurance lift force MFmaxUD at which the drive means can lift the cabin for a longer period of time, and / or the dynamic time-lifting force MFmaxUZ with which the drive means can lift the cabin briefly, scheduled. As described in the introduction, especially with safety brakes, the static holding force MFmaxA can exceed the dynamic lifting force MFmaxU. Conversely, for example, with pure engine brakes, the static endurance force can fall below the dynamic (time) lifting force. In order to ensure both a safe lifting and lowering, ie a sufficient dynamic lifting force of the drive means, as well as a safe holding of the car at a height, ie a sufficient static lifting force of the drive means, it is proposed that the smallest of these values in the design of the weight of the Counterweight.

In the design of the weight of the counterweight, the weight of the counterweight and / or the unladen weight of the cabin and the maximum payload of the cabin is reduced from the laws known for pulleys according to the number of loose rollers around which the support means is deflected. Thus, in equation (1) or (2), the weight of the counterweight MG or the empty weight MK and the maximum payload MLmax can be divided by a suspension factor of 2, for example, if the suspension element is one or more loose on the cabin and counterweight sides Roll (simply) is deflected. In a multiple suspension (i.e., 4-fold, 5-fold, etc.), the divisor changes accordingly for the design of the weights. In a direct suspension, without loose roles, this divisor is eliminated, or he is equal to 1.

In a manner known per se, for equation (1) or (2) the empty weight of the car and / or the maximum tractive force of the drive means and / or the maximum payload of the car can be increased by the safety factor to take account of the inertial forces occurring during operation. Likewise, in equation (1) or (2) friction and / or the weight of the support and / or suspension means are taken into account.

The present invention proposes a method for designing the weight of the counterweight of an elevator system with which this weight can be optimized for a drive means with a predetermined maximum tensile force.

Likewise, the present invention relates to an elevator system having a counterweight designed according to this method.

Fig. 1

schematisch den Aufbau eines Aufzugsystems nach einer Ausführung der vorliegenden Erfindung.

Fig. 2

schematisch den Aufbau eines weiteren Aufzugsystems nach einer Ausführung der vorliegenden Erfindung.

Die Figuren verwenden gleiche Bezugszeichen für vergleichbare BauteileOther objects, features and advantages will become apparent from the dependent claims and the following embodiments. This shows:

Fig. 1

schematically the structure of an elevator system according to an embodiment of the present invention.

Fig. 2

schematically the structure of another elevator system according to an embodiment of the present invention.

The figures use the same reference numerals for comparable components

An elevator system according to an embodiment of the present invention comprises, as in Fig. 1 schematically illustrated a cabin 1 with an in Fig. 1 indicated empty weight MK,
which can raise, lower or keep a payload ML at a certain height. The payload ML must not exceed MLmax.

On the cabin 1, a suspension element 2 is attached via a loose roller 20, which is indicated here as a single rope. This is fixed at one end in a shaft area, is then deflected over the loose roller 20, wraps around in the sequence of a traction sheave 30, is deflected at its other end on a counterweight-side loose roll 20.1 and again firmly connected to the shaft.

A drive means 3 comprises a motor and a brake (not shown in each case) which can apply a lifting or holding torque to the traction sheave 30. This torque is frictionally converted into a tensile force in the traction sheave 30 looping around the rope 2, so that the car 1 is raised due to the lifting or holding torque, lowers or held at a level.

By way of design, the drive means 3 can, by means of its brake, apply a maximum static holding force MFmaxA and, by means of its motor, apply a maximum dynamic continuous lifting force and a maximum dynamic lifting time force MFmaxUD or MFmaxUZ. The static holding force that can apply the brake in the embodiment, depending on the type of drive means is greater or less than the dynamic time-lifting force that can apply the engine for a short time. This is due to the limited heat dissipation in turn greater than the dynamic continuous lifting force that the engine can afford over a longer period

As from the schematic representation of Fig. 1 it can be seen, the counterweight 4 is on the support means 2, which is identical to the propellant in the embodiment, so coupled to the car 1 that it rises when the car 1 lowers, and lowers when the car 1 rises. As a result of this balancing, the tensile force which the drive means 3 applies to the suspension element 2 is reduced or has to be transmitted in a known manner.

In the exemplary embodiment, the in Fig. 1 sketched elevator system as follows: first, the empty weight MK of the car 1 and the maximum allowable payload MLmax of the elevator system are determined. In the embodiment weigh empty car 1 1600kg, the maximum payload is estimated to be 2000 kg.

Due to the loose rollers 20, 20.1, these weights are halved in the following calculations, since the drive means due to the pulley only half the tensile force must apply (MK = 800kg, MLmax = 1000kg).

As possible drive means 3, the following four types of a drive series are available: Type maximum holding force MFmaxA maximum continuous lifting force MFmaxUD maximum time lifting force MFmaxUZ Type I 1250kg 1250kg 1500kg Type II 1250kg 1000kg 1200kg Type III 500kg 750kg 800kg Type IV 500kg 450kg 600kg

As can be seen from the second spade, types I and II or III and IV respectively have the same mechanical brake but different drive motors. As can be seen from the fourth column, the lifting forces which a drive means 3 can apply for a short time exceed those which it provides in continuous operation.

First, in this example, all of the above values are reduced by a factor of 1.3 to account for a safety factor equal to 1.3 (as explained above) in the design. This factor takes into account, for example, frictional influences, inertial forces, special requirements, etc. Subsequently, for each drive means 3, the smallest maximum tensile force from the holding, the duration and the time-lifting force is determined (underlined in the above table). This is compared with the half payload MLmax / 2 = 500kg according to equation (3), since the drive means 3 would have to apply this half payload even with a 50% balance. $$\begin{array}{cc}\mathrm{MFmax}& >0.5\times \mathrm{MLmax}\hfill \\ & >500\mathrm{kg}\hfill \end{array}$$

While type III with MFmaxA / 1.3 (= safety factor) = 384kg is still insufficient, the drive type II with MFmaxUD / 1.3 = 769kg is the one with the smallest sufficient tensile force satisfying the condition according to equation (3) and is selected.

$$\mathrm{MG}=2\times 1031\mathrm{kg}=2062\mathrm{kg}\mathrm{.}$$

However, since this selected drive means 3 can lift a load of 769kg even while in continuous operation, while only 500kg would be required with a 50% balance, the weight MG of the counterweight 4 can be correspondingly reduced according to Equation (1) - taking into account the above Safety factor 1.3 - on, due to the counterweight-side loose roller 20.1, the weight of the counterweight is again doubled. $$\begin{array}{cc}\mathrm{MG}/2& =\mathrm{MK}+\left(\mathrm{MLmax}-\mathrm{MFmax}/1.3\right)\hfill \\ & =800\mathrm{kg}+\left(1000\mathrm{kg}-769\mathrm{kg}\right)\hfill \\ & =1031\mathrm{kg}\hfill \end{array}$$

$$\mathrm{MG}=2\times 1031\mathrm{kg}=2062\mathrm{kg}\mathrm{,}$$

It makes sense, the counterweight 4 is preferred, chosen slightly larger according to a weight grading, in the present case, for example, to 2075kg.

Thus, compared to a conventional 50% balance, where the weight of the counterweight would be 2x (MK + MLmax / 2) = 2600kg, the counterweight 4 will be minimized, unlike a 30% balancing as seen in the embodiment of FIGS US 5,984,052 It is known that for all payloads, even with maximum payload, the same nominal speed profile can be used. Thus, the tensile force of the drive means 3 is optimally utilized while minimizing the counterweight 4, or optimized.

At the in Fig.2 example shown is only the car 1 is attached via a loose roller 20. The support means 2 is fixed at one end in the shaft area, is then deflected over the loose roller 20, wraps around in the sequence of a traction sheave 30 and is connected at its other end fixed to the counterweight 4. In this example, the cab side empty weight MK as well as the allowable payload MLmax is halved because of the cab side loose roller 20. However, the mass or weight of the counterweight 4 does not have to be doubled again, as no counterweight loose roll is used. The calculation of the weight of the counterweight 4 thus takes place as explained above, wherein only, due to the missing roller 20.1, the weight of the counterweight 4 need not be doubled. $$\begin{array}{cc}\mathrm{MG}& =\mathrm{MK}+\left(\mathrm{MLmax}-\mathrm{MFmax}/1.3\right)\hfill \\ & =800\mathrm{kg}+\left(1000\mathrm{kg}-769\mathrm{kg}\right)\hfill \\ & =1031\mathrm{kg}\hfill \end{array}$$

Because of the weight grading, the counterweight 4 would preferably have been chosen to be somewhat larger, in the present case for example 1050 kg.
This example serves to illustrate the influence of the loose roller 20, 20.1, wherein it should be noted that, of course, the routes of counterweight 4 and cab 1 result differently, which must be taken into account in the design of the shaft.

Different approaches in the application of the formulas are possible, so a number of loose rollers 20.20.1 may be taken into account in the weights of car 1 and / or counterweight 4 or their influence may be taken into account in the holding force table. Likewise, safety factors may be taken into account directly in determining the holding forces, or they may be taken into account in determining the actual weight of the counterweight 4.

Claims (9)

1. Method of optimising the weight of the counterweight (4) of a lift system, the lift system comprising a cage (1) which has an empty weight (MK) and which can move a permissible rated load (MLmax); a counterweight (4) which has a weight (MG) and which is so coupled with the cage (1) by way of a support means (2) that it rises when the cage (1) lowers and lowers when the cage (1) rises; and a drive means (3) which can apply a maximum traction force (MFmax) to the support means (2), characterised in that

   - in a first step the drive means (3) is selected from a product line (I - IV) with stepped, predetermined maximum traction force (MFmax), wherein the maximum traction force (MFmax) is at least greater than half the rated load (MLmax): $$\mathrm{MFmax}\mathrm{>}\mathrm{0.5}\mathrm{x\; MLmax}\mathrm{;}$$

   

   - and in a further step the weight (MG) of the counterweight (4) is so optimised that the weight (MG) of the counterweight (4) is substantially equal to the empty weight (MK) and the difference between the rated load (MLmax) of the cage (1) and the maximum traction force (MFmax) of the selected drive means (3): $$\mathrm{MG}\approx \mathrm{MK}+\left(\mathrm{MLmax}-\mathrm{MFmax}\right)\mathrm{.}$$

   
2. Method of optimising the weight of the counterweight (4) of a lift system according to claim 1, wherein the smaller value of a static holding force (MFmaxA) by which the drive means (3) holds the cage (1) at a height, a dynamic time-extended lifting force (MFmaxUD) by which the drive means (3) can lift the cage (1) over a longer period of time and/or a dynamic time-limited lifting force (MFmaxUZ) by which the drive means (3) can lift the cage (1) over a short time is or are defined as maximum traction force (MFmax) of the drive means (3).
3. Method of optimising the weight of the counterweight (4) of a lift system according to claim 1 or 2, wherein the weight of the counterweight (4) and/or the empty weight of the cage (1) and the permissible rated load of the cage (1) is or are reduced in correspondence with the number of floating rollers (20, 20.1) around which the support means (2) is deflected or the maximum traction force of the drive means (3) is increased in correspondence with the number of floating rollers (20, 20.1) around which the support means (2) is deflected.
4. Method of optimising the weight of the counterweight (4) of a lift system according to any one of claims 1 to 3, wherein the empty weight of the cage (1) and the permissible rated load of the cage (1) and/or the weight of the counterweight (4) is or are increased by a safety factor for consideration of the frictional and inertial forces occurring in operation or the maximum traction force of the drive means (3) is reduced by a safety factor for consideration of the frictional and inertial forces occurring in operation.
5. Method of optimising the weight of the counterweight (4) of a lift system according to claim 4, wherein the safety factor is 1.1 to 2.0, preferably 1.3.
6. Method of optimising the weight of the counterweight (4) of a lift system according to any one of claims 1 to 5, wherein the support means (2) comprises one or more cables and/or one or more belts and wherein one or more cables and/or belts is or are coated with an elastomer, particularly polyurethane.
7. Method of optimising the weight of the counterweight (4) of a lift system according to any one of claims 1 to 6, wherein a motor, particularly a frequency-regulated electric motor, and at least one drive pulley (30) for converting a drive output torque of the motor into a traction force on the support means (2) are used as drive means (3).
8. Method of optimising the weight of the counterweight (4) of a lift system according to any one of claims 1 to 7, wherein a brake which is integrated in the motor or separate therefrom and which can apply a static holding moment to the at least one drive pulley is arranged in the drive means (3).
9. Method of optimising the weight of the counterweight (4) of a lift system according to claim 7 or 8, wherein the motor and/or the brake is or are selected from a product line with stepped, predetermined holding or lifting moments.

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