Motor Operation

For the mechanical aspect of the chair, two motors will be in use. A linear actuator will be used to control the up and down motion, and a DC motor will be used in conjunction with a timing belt to control left and right swivel.

The linear actuator purchased will be set to use 4.5 Amps, has a speed of about .45 inches/second with a 200 lb. load, and has a force of 1540 lbs. when set at 4.5 amps. The current was set to 4.5 A to possibly increase speed (up to .59 inches/second) if the load if on the lighter side, and also because at the average weight of 220 lbs., the motor will need at least 1 amp to even work. The stroke size of the linear actuator is 10 inches, so this means that that it will take the chair approximately (.45inches/second x 10 inches) 4.5 seconds to reach full extension of retraction. Also this motor was chosen, because it is about 17 inches in length when fully retracted. This 17 inch retracted motor will be small enough to fit under the swivel chair's base as lowest level. The motor should fit nicely into the space between the bottom of the seat and the base of the chair without much necessary mounting. Figure 9 show the graph of the Current vs. Load for this linear actuator model. For our purposes, we are using almost max amps (4.5 A).

 

Figure 10: Current vs. load chart for linear actuator [8]

Our measurements with a dynamometer show that about 50 N of force is required to get the chair assembly in motion with a load of 180 lbs on it. Considering the force of static friction is greater than kinetic friction, once the assembly is set in motion, with a force just sufficient to set the assembly in motion, it is possible to sustain movement with the same amount of force.

The plate that the chair is resting on has a radius of 12 inches, which converts to 0.3048 meters. According to these measurements, approximately 180 N·m of external torque is needed to set the assembly in motion from halt state and keep it turning at a slow and safe rate.

Torque = Linear Force x radius

= 50 N x 12 inches

= 50 N x (0.3048 meters)

Torque = 180 N·m