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Power = Force X Velocity = Torque X Angular Velocity
Velocity = Distance / Time Measurement of speed
Angular Velocity = Rotation / Time Measurement of rotational speed
Torque = Force X Distance Measurement of rotational force
Power (electric) = Voltage X Current Electric power of motor (power source)
EXAMPLE PROBLEM
Lift a 5 lbs big ball 7ft in the air on top of the goal within 6 seconds, and you decide to use a 3.5 ft long 20 lbs arm by rotating it 180 degrees. How much power is needed?
STEP 1: Calculate Power Requirement
Power needed:
Power = Torque X Angular Velocity
Therere two torque in this device: torque from the ball, and torque from the arm itself
Torque of ball = 5 lbs (how heavy) X 3.5 ft (how far from center of rotation)
Torque of arm = 20 lbs (how heavy) X 1.75 ft (how far is the arms center of mass from center of rotation)
Total torque = torque of ball + torque of arm = (5 lbs X 3.5 ft + 20 lbs X 1.75 ft)
= 630 in-lbs = 69 N-m
**This is the worst case when arm is parallel with the ground It take less torque to rotate arm when it is point toward the ground or upward**
Angular Velocity = 180 degrees / 6 sec. = 5 RPM = 0.5 Rad/sec
Power = 69 N-m X 0.5 Rad/sec = 34.5 Watt
STEP 2: Choosing the right motors
This can be done with the Fisher Price Motor w/gearbox (140W), Sliding Door Motor (69W), and Globe Motor w/ gearbox (50W).
Try using Globe motor w/ gearbox.
STEP 3: Calculating working torque and applying gear ratio
Maximum Torque required for this task = (5 lbs X 3.5 ft + 20 lbs X 1.75 ft)
= 630 in-lb
Best to design a gear ratio such that the load reflected back to motor is around 20%~50%
stall torque when motors are most happy. Globe motor w/ gearbox has a stall torque: 170 in-lbs
20%~50% S.T. = 34 in-lb~85 in-lb
Max. torque / working torque = 630 in-lb / 34in-lb ~ 630 in-lb / 85 in-lb
Gear ratio requires to increase torque up to Max. torque= 18.5:1 ~ 12.6:1
We will run motor at 45% of stall torque with a little room before maximum motor power:
Working Torque = 170 in-lb * .45 = 77 in-lb
Gear ratio = 630 in-lb / 77 in-lb = 8.2:1
STEP 4: Choosing the gears/sprockets
Many combinations between gears or sprockets can be used to obtain a certain ratio.
Two of many options: (Basically, the less gears, the more efficient)
One stage of 10:1 gear ratio with 0.9 efficiency, or
Two stages of 3.3:1 with 0.9 efficiency per stage.
Effective gear ratio = 3.3 * .9 * 3.3 * .9 = 8.82:1
STEP 5: Effective Power
Calculate effective power to double check if motor really has enough power for task, since no gear/sprocket system has 100% efficiency. 100% efficiency only when direct drive from motor.
Globe motor with gearbox @ 45W ~ 50W
Effective power = motor power * total component efficiency We will just consider the
gears efficiency for this purpose
.
Effective power = 45W ~ 50W * (.9 * .9)
= 36.5 W ~ 40.5 W
It take 34.5 W to complete the task, and the way we set up this motor, we will be getting at least 36.5 W out of the globe motor w/ gear box. So we know this motor is good enough for this component.
(50W only true at near peak power range with perfect gearing. Usually power loses cause by heat and friction.)
This worksheet was taken from WRRFs A Part of Motor mounts/Drive train/Motors selection workshop Lecture Notes by Mark Dunn, Charlie Steele & Ken Leung
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