- #1
TJS
Hey guys, I'm working on a product that rotates cups around a shaft, I am trying to calculate the torque required to do so. I have given it my best shot below, does this seem right to you?
The Main Shaft Motor will need to rotate holder one from the filling cups position to the pressing lids position within a 2 second interval (see attached for an illustrtaion). The angle of rotation between these two points is 29deg. As the motor will be required to rotate the shaft from a static equilibrium, an acceleration is required, the acceleration will be at it's max at the center point of this rotation just before it begins to decelerate. Therefore, selecting the motor to cope with the worst-case scenario of ∝ = maximum I will use the angular velocity at the point of half rotation.
The angle at the half rotation point is :
θ = 14.26 deg
The time required to get there will be half the time of a full rotation, therefore:
∆t=1-0
To find ω, θ needs to be in radians, therefore converting degrees to radians.
θ rad=14.26*0.01745329= 0.2489rad
Finding the angular velocity ω at the center point.
ω =( 0.2489)/(1 )
Using ω to calculate ∝ = maximum:
∝ = ( ∆ω )/(∆t )= ( 0.2489-0)/(1-0 )= 0.2489 m⁄s^2
Stepper motors can be selected based on Torque and Power requirements. Using angular acceleration and velocity the power and torque requirements can be calculated using the formulas below;
τ = I × ∝ and P = τ × ω
From the solid-works CAD model, the value for moment of inertia about the axis of the motor shaft is taken to be 0.197 Kgm^2 and therefore;
Multiplying the moment of inertia by a safety factor of five to ensure reliability, the moment of inertia I will take is:
0.9850 Kgm^2
τ = I × ∝ = 0.9850×0.2489=0.2452Nm
P = τ × ω= 0.2452 ×0.2489= 0.0610 W
Based on these calculations I have selected the NEMA Stepper Motor model number 42BYGHW60. This stepper motor can produce a torque between 0.16-0.56Nm.
The Main Shaft Motor will need to rotate holder one from the filling cups position to the pressing lids position within a 2 second interval (see attached for an illustrtaion). The angle of rotation between these two points is 29deg. As the motor will be required to rotate the shaft from a static equilibrium, an acceleration is required, the acceleration will be at it's max at the center point of this rotation just before it begins to decelerate. Therefore, selecting the motor to cope with the worst-case scenario of ∝ = maximum I will use the angular velocity at the point of half rotation.
The angle at the half rotation point is :
θ = 14.26 deg
The time required to get there will be half the time of a full rotation, therefore:
∆t=1-0
To find ω, θ needs to be in radians, therefore converting degrees to radians.
θ rad=14.26*0.01745329= 0.2489rad
Finding the angular velocity ω at the center point.
ω =( 0.2489)/(1 )
Using ω to calculate ∝ = maximum:
∝ = ( ∆ω )/(∆t )= ( 0.2489-0)/(1-0 )= 0.2489 m⁄s^2
Stepper motors can be selected based on Torque and Power requirements. Using angular acceleration and velocity the power and torque requirements can be calculated using the formulas below;
τ = I × ∝ and P = τ × ω
From the solid-works CAD model, the value for moment of inertia about the axis of the motor shaft is taken to be 0.197 Kgm^2 and therefore;
Multiplying the moment of inertia by a safety factor of five to ensure reliability, the moment of inertia I will take is:
0.9850 Kgm^2
τ = I × ∝ = 0.9850×0.2489=0.2452Nm
P = τ × ω= 0.2452 ×0.2489= 0.0610 W
Based on these calculations I have selected the NEMA Stepper Motor model number 42BYGHW60. This stepper motor can produce a torque between 0.16-0.56Nm.