Placing a six-axis robot on a linear track greatly expands its reach envelope, allowing it to tend more than one machine or transfer materials across a much greater distance. For example, instead of three robots each tending a machine at separate locations, a single robot on a linear track—also known as a seventh axis—could potentially tend all three, saving money and increasing efficiency. Or, a single material handling robot on a rail could replace multiple robots with intermediate handoff locations.

The overall cycle time of a robotic cell is defined as the time it takes to produce a single unit. There are many factors that affect the cycle time of a cell.

To begin with, the layout of the workcell will have a major effect on overall cycle time. The placement of the robot and other equipment, such as machine tools, conveyors and pallet racks, should be done with the sequence of operations in mind.

For a workcell utilizing a linear axis, first try to minimize the distance that the robot must move along the track between one step and the next. For a workcell with multiple operations, place each operation as close to the previous operation as possible.

Reducing or removing robot motions that are unnecessary or excessive is the next step. Examine intermediate points in the robot’s motion path to determine if movements and rotations can be reduced or eliminated.

Some robot controllers allow for easy optimization of robot motions. For example, changing the path accuracy settings to coarse instead of fine for motions that do not require precision will increase the robot’s speed.

Where allowable, increase the acceleration, deceleration and velocity settings to their maximum levels, except where this might be detrimental to the payload (such as for fragile, long or large parts). This includes both robot movements and the linear axis movements. (Ensure that the acceleration and velocity settings of the linear axis are within design parameters.)

Finally, try to have the robot perform operations in parallel instead of in sequence. For example, the robot could open an end-effector during an approach motion instead of waiting until that motion has been completed.

##
**Moving From Point A to Point B**

For a robot on a track, minimizing the distance between operations is the first step to reduce overall cycle time. Once this has been done, a number of factors will determine how quickly the robot is able to move through its sequence of operations.

The carriage motion profile describes the speed and acceleration of the robot carriage over time. To move from point A to point B, the carriage accelerates to a certain velocity, maintains that travel velocity for some amount of time, and then decelerates to a stop. On a speed-time graph, this looks like a trapezoid, so this is known as a “trapezoidal” motion profile. If the travel distance is sufficiently long—greater than the “critical distance”—the carriage will reach its maximum velocity (or rated speed).

This maximum velocity is a function of several factors, including the rated speed of the servomotor, the rated speed of the gearbox, and the pinion size. For longer tracks, increasing the maximum velocity of the seventh axis can significantly decrease the cycle time of your cell.

##
**Determining Maximum Velocity**

For a linear axis driven by a rack and pinion system, the maximum velocity (V_{rack}) is determined simply by the effective pitch radius of the pinion gear (r_{pinion}) and the maximum angular velocity (w_{pinion}) at which it is able to turn.

V_{rack} = w_{pinion} x r_{pinion}

On Güdel’s linear tracks, the pinion pitch radii are 25.47 millimeters for Models TMF-1 and TMF-2, 33.96 millimeters for Model TMF-3, and 42.44 millimeters for Model TMF-4.

Each gearbox has a maximum allowable input speed. For the Güdel AE060 gearbox (used for TMF-1 and TMF-2) this is 6,000 rpm. For the AE090 (TMF-3) and AE120 (TMF-4) gearboxes, it is 4,500 rpm. Also, each gearbox is available in ratios from 2-to-1 to 24-to-1. Dividing the maximum input speed (w_{input}) by the gear ratio (i) will provide the maximum output speed (w_{output}) at which the pinion gear can turn.

w_{output} = w_{input}/i

To increase the maximum velocity of a given system, engineers should choose a smaller gearbox ratio. (However, this will have an inverse effect on maximum acceleration.)

Thus, the maximum possible output velocity (using a 2-to-1 gearbox ratio and the maximum input velocity) is 8 meters per second for the TMF-1, TMF-2, and TMF-3, and 10 meters per second for TMF-4. However, it is not advisable to operate a gearbox at maximum input speed. Also, finding a servomotor that can provide the required torque at such high speeds may be problematic, if not impossible.

Therefore, it is safe to assume that the actual maximum velocity will be no more than about 60 percent to 70 percent of those theoretical values. Finally, keep in mind that choosing a small gearbox ratio (such as 2-to-1) will have a negative effect on the maximum possible acceleration.

##
**Determining Maximum Acceleration**

Determining the maximum possible acceleration of the carriage requires quite a bit more information. According to Newton’s famous equation, acceleration equals force divided by mass. The forces acting upon the carriage include both the force between the pinion and the rack, as well as various frictional forces, such as the rolling resistance of the rollers and friction from the guideway scrapers and lubrication system. The mass of the system includes the carriage itself, the robot, as well as everything else that is connected to it, such as cables and connectors. Also, because the carriage involves rotational elements (rollers, motor shaft, couplings, gearbox elements, pinion), rotational momentum and rotational friction also must be factored into the equations of acceleration.

For the sake of simplicity, let us focus on the three main variables affecting maximum carriage acceleration: robot mass, servomotor torque and gearbox ratio.

For most linear track systems, the most significant contribution to mass will be the robot. Therefore, choosing the smallest robot that can perform the required tasks is essential for maximizing the acceleration of the system. This also applies to any robot riser or payload attached to the carriage. The lighter the overall system, the more quickly it can accelerate with a given applied force.

A wide range of servomotors are compatible with Güdel gearboxes. Each motor has its own speed and torque characteristics, usually illustrated by a speed vs. torque graph. For electric motors, maximum torque will decrease as the speed of the output shaft increases.

Servomotors typically have a “peak” or intermittent torque capacity, as well as a rated torque capacity, for continuous operation, for a given speed. The motor can operate for an indefinite period within the “continuous” zone without overheating. Therefore the motor’s RMS speed and torque must lie within the continuous region.

To increase the acceleration of the carriage system, choose a servomotor with a high intermittent torque rating. (Keep in mind that the maximum servomotor torque must not exceed the maximum torque rating of the gearbox coupling, or slipping will occur.)

Because the gearbox reduces the rotational velocity going from the motor to the pinion, it has the effect of increasing the output torque (t_{output}) by a corresponding amount.

t_{output} = t_{input} x i

Therefore, choosing a higher gearbox ratio allows for a higher output torque for a given motor speed. However, increasing the gearbox ratio has the effect of reducing the maximum velocity! There is a trade-off between maximum velocity and maximum acceleration for a given servomotor.

##
**Balancing Speed and Acceleration**

What is the ideal tradeoff between speed and acceleration for a particular application? At travel distances greater than the critical distance (d_{cr}), the robot carriage will peak at its maximum velocity. The critical distance is a function of the maximum velocity (V_{max}) and acceleration (a) of the system.

d_{cr} = (V_{max})^{2}/a

For travel distances less than this critical distance, where the robot carriage never achieves its maximum velocity before decelerating, the travel time (t) is simply a function of distance (d) and the rate of acceleration (a).

t = 2 x √(d/a)

For short distances, increasing the maximum velocity has no effect on the travel time! Therefore, when travel distances are short enough such that the robot does not reach its maximum speed, the strategy to reduce travel time is to increase acceleration by choosing a servomotor with higher torque or by choosing a higher gearbox ratio. (Keep in mind that maximum acceleration can also be improved by lightening the payload on the carriage, so look for ways to reduce mass wherever possible.)

For distances above the critical distance, the travel time (t) is a function of distance (d), maximum velocity (V_{max}) and acceleration (a).

t = (d/V_{max}) + (V_{max} /a)

For travel distances above the critical distance, the first component of this equation will be the larger factor in the travel time. Therefore, once the critical distance has been exceeded, the first strategy to reduce cycle time is to increase maximum velocity by choosing a servomotor with a greater maximum rated speed or a lower gearbox ratio. Once an attempt has been made to increase the maximum velocity, further cycle time improvements may be achieved by increasing acceleration.

##
**Workflow Scheduling**

For robotic cells in which more than one operation is performed on each part, several choices can affect cycle time. First, engineers must decide whether to implement a forward or reverse cycle. Secondly, implementing a dual gripper instead of a single gripper can greatly reduce cycle time. Lastly, the number of parallel machines simultaneously performing each operation must be determined for optimum efficiency.

In a forward cycle, the cycle begins with all machines empty. The robot picks up a part from the input buffer, carries it to the first machine, waits for the first machine to finish, then carries the part to the next machine, and so on until all processes are completed, and the part is dropped off at the output buffer.

In a reverse cycle, the cycle begins with all machines except the first one full. (The first machine must be empty to receive an incoming part.) The robot picks up a raw part from the input buffer and carries it to the first machine. Then the robot travels to the last machine, picks up a finished part, and transfers it to the output buffer. Then the robot goes to the second-to-the-last machine, picks up a finished part, and transfers it to the last machine. This process continues until the robot returns to the beginning.

The reverse cycle allows for more machines to be occupied by parts simultaneously, which increases efficiency. However, it increases the overall move distance required by the robot per part.

So, for single-gripper robots, when the processing time for each machine is always less than the time required for the robot to travel from that machine to the next, the forward cycle is preferable. In all other situations, the reverse cycle is more efficient.

Implementing a dual-gripper system can greatly reduce overall cycle time for most systems in which machine processing time is greater than the robot travel time. A dual-gripper system allows for all machines to be occupied at the beginning of each cycle, and it reduces the overall travel distance per part. The cycle begins with all machines occupied and both grippers empty. The robot picks up a part from the input device with the first gripper and then travels to the first machine.

The tooling then rotates to pick up the completed part with the empty gripper. Next, the tooling rotates again and drops off the new part from the first gripper.

The robot then travels to machine No. 2. The tooling rotates to pick up the finished part with the empty gripper, and rotates again to drops off the part from the first machine.

This sequence continues until the robot drops off a finished part to the output device. It then returns to the beginning and starts the sequence over.

In situations where one operation takes longer than another, the overall efficiency of the cell can be increased by running more than one of the slower operation in parallel. For example, if operation No. 1 takes 10 seconds per part, but operation No. 2 takes 30 seconds per part, there will always be wasted time as the robot waits for operation No. 2 to complete, limiting the overall capacity of the system. However, this wasted time can be reduced by adding additional machines performing the slower operation.