**Joseph Kaplan****Vincent Spitaleri****Machine Components****Plainview, N.Y.**

Machines that raise loads against gravity are used throughout industry. Typical examples include aircraft cargo lifting systems, warehouse lifts for parts storage, and robotic lifts for part transfers. Designing such a lift should always include the possibility of overhauling loads. This problem crops up by interrupting power to the lift mechanism which lets gravity accelerate loads downward and drive transmissions and prime movers backward.

Concerns with overhauling loads include the safety of personnel working on or near a lift, equipment damage due to possible shock loading, and loss of accuracy in positioning systems.

For an analytical look at overhauling loads, consider a common ball leadscrew arrangement. A typical system consists of an electric motor, ball screw, a weight driven by the ball-screw nut and, possibly, a coupling between motor and ball screw. Also, a bidirectional drag brake prevents overhauling loads. The salient feature of this arrangement is that the brake insignificantly loads the motor when raising the weight. Conversely, when the weight moves downward the brake provides a drag torque that exceeds the overhauling torque, to decelerate and stop the load.

Torque for the lift operation is expressed by

*T _{l} = FL(100)/2πδ*

and power requirements as

*P _{l} = 0.01183T_{l} N.*

Speed of the screw comes from

*N = 60V/L*

where *V* = maximum operating speed of the weight, ips, and *L* = screw lead, in./rev.

The overhauling torque *T _{o}* produced when lifting is

*T _{o} = FLδ/2π(100).*

Note that a drag brake with torque equal to *T _{o}* produces a static balance that holds the weight and ball screw in equilibrium. However, if the weight is traveling downward when motor power is cut, the precarious static balance between the weight and dragbrake torque will be upset by the kinetic energy stored in the weight, motor, and transmission, permitting the weight to work itself down before reaching equilibrium. To avoid this undesirable condition actual brake torque should be somewhat higher than the gravity balancing torque,

*T*.

_{o}To determine the torque required to stop an overhauling load in a specified distance take into account the kinetic energy of each drive-system component, as well as the change in potential energy of the weight. If power to the motor is cut, dragbrake torque needed to stop the weight when the load moves downward at full speed is

The weight’s kinetic energy *Ew = FV ^{2}/2g*. Kinetic energy of motor, ball screw, brake, and coupling at full speed

*E*= 0.00548

_{s}*N*. In the above equation

^{2}I_{s}*FC*= change in potential energy of the weight, where

_{d}*C*= overtravel of the weight after power is cut.

_{d}A similar equation spells out dynamic characteristics of the system with the weight traveling upward at full speed,

The overtravel of the weight upwards after power is cut, *C _{u}*, is found from

As an example, a ball screw with a 0.20-in. lead and 90% efficiency moves a 50-lb weight up and down over a 10-in. distance. Speed of the weight in either direction is 4 ips. Travel time is 2.5 sec up and 2.5 sec down with a dwell of 10 sec after each move. The equivalent moment of inertia Is of the motor, ball screw, brake, and coupling is 650 310^{–6} lb-in.-sec^{2}.

Torque required for lifting is

*T _{l} = 50(0.20)(100)/2π(90)*

= 1.77 lb-in.

Maximum ball screw speed is *N* = 60(4)/0.20 = 1,200 rpm. The motor power required for lifting is

*P _{l}* = 0.01183(1.77)(1,200)

= 25 W.

Note that motor selection should allow for an additional 25 to 50% power capacity. The overhauling torque is

*T _{o} = 50(0.20)(90)/2π(100) = 1.43 lb-in.*

A drag brake with a holding torque of 1.43 lb-in. will keep the weight in static equilibrium after it comes to rest. However, designers generally need to know the braking torque required to stop the weight in a specific location, especially in positioning systems or when it is necessary to minimize the average power dissipated in the brake.

An array chart provides a simple means for comparing brake torque versus weight overtravel. Calculating brake torque *T _{d}* in the example (neglecting transmission efficiency) for overtravel distances of 0.25, 0.5, 1, 2, and 3 in., where

*E*= 1.04 and

_{w}*E*= 5.13, results in

_{s}Average power dissipated when the brake operates dynamically is

where *T _{u}* = drag torque of the brake in the upward direction. A typical value is 0.1 of

*T*, the downward torque. The number of slip revolutions per minute,

_{d}*q*is found from

*q = 60m/T.*

Slip revolutions per cycle comes from* m = 2a/L* where a = weight travel distance. In the example, *a* = 10 in. and *m* = 2(10)/0.2 = 100 slip rev/cycle.

Duration of one cycle is from *T = 2t + t _{1} + t_{2}* where

*T*= cycle time, sec,

*t*= time for the weight to travel up or down,

*t*= dwell time in the up position, and

_{1}*t*= dwell time in the down position. For the example,

_{2}I = 2(2.5) + 10 + 10 = 25 sec

and

*q* = 60(100)/25 = 240 slip rev/min. Substituting the above values,

*P _{a}* = 0.01183(240)((2.4 + 0.24)/2)

= 3.7 W.

Average power consumption for other overtravel distances can be tabulated as shown.

Overtravel distance of the weight correlates directly with the magnitude of the brake torque. A practical range for torque is *T _{d}* = 1.15 to 1.50 of the overhauling torque,

*T*. If positioning accuracy of the weight is the primary consideration, then

_{o}*T*may approach 1.5 To. On the other hand, in applications requiring substantial power dissipation, such as when the weight moves continuously, reducing drag torque to 1.15

_{d}*T*may be advisable.

_{o}This analysis is also valid for other mechanical configurations used in lift mechanisms. Here’s a brief review of the characteristics of three other leadscrew lifting systems.

In designs with a self-locking leadscrew, typically an Acme thread, a drag brake is not required to hold the weight in place when the drive motor loses power. Instead, the selflocking feature is obtained at the expense of transmission efficiency that must be less than 50%. Assuming an efficiency of 40% for the example, *T _{l}* = 3.98 lb-in. and motor power

*P*required = 56.5 W.

_{l} Another common setup uses a ball-screw arrangement with a drag brake that has equal torque in both rotation directions. The brake’s drag torque must exceed the overhauling torque, *T _{o}* = 1.43 lb-in. for system stability. Allowing for a 25% increase, the torque becomes 1.79 lb-in. Power demand from the motor increases to 50 W, 25 W to raise the weight and 25 W to overcome the drag torque of the brake when lifting.

Some lift mechanisms use electromagnetic fail-safe brakes. Energizing the electromagnetic brake lets the motor freely lift the weight without power penalty to the motor. When the motor stops, the brake actuates friction pads with built-in springs or magnets to provide the stopping and holding torque. Power demand from the motor for lifting the weight is 25 W. However, the need for associated circuitry and a power supply for energizing the electromagnetic brake adds to power requirements and complexity of the lift system.