I need to move a platform on wheels containing a 20lb object back and forth on a flat plane using a stepper motor and a shaft. It needs to move a maximum of 100 rpm, 4 cm. What would the wheel radius be and what would the torque required to move such a thing be? Also any recommendations on what to buy would be good
I think that this depends entirely on the rolling resistance.
It needs to move a maximum of 100 rpm, 4 cm
I have no idea what RPM means in relation to a distance.
Whoops. I mean the motor needs to constantly move at 100 rpm and move an object 4 cm everytime it rotates. And what would the typical rolling resistance be for say a plastic wheel?
As well as the rolling resistance the force needed to accelerate and decelerate the load will be important. Much more so than for a steady movement in one direction. It is possible to reduce friction to a low level but the acceleration forces depend entirely on the mass being moved.
I suspect you need to do some research on calculating accleration when you know the acceleration the force can be calculated from the formula force = mass * acceleration (Force in Newtons, mass in kgs and acceleration in m/s/s). To calculate the acceleration use one of the formulas v=u + at, v2 = u2 + 2as or s = ut + 0.5at2. U is the initial velocity, v is the final velocity, a is the acceleration and s is the distance
The only practical way to know the rolling resistance is to measure it - and allow a good margin of error. And be sure to measure it in the fully loaded condition.
A diagram of the machine you want to make would be a big help.
...R
Somark:
I need to move a platform on wheels containing a 20lb object back and forth on a flat plane using a stepper motor and a shaft. It needs to move a maximum of 100 rpm, 4 cm. What would the wheel radius be and what would the torque required to move such a thing be? Also any recommendations on what to buy would be good
Not sure what you mean by "using a stepper motor and a shaft" - draw your mechanical setup for
us if you want to avoid confusion.
"It needs to move a maximum of 100 rpm, 4 cm"
rpm is a measure of (angular) velocity. Without a known wheel radius that doesn't make sense.
Can you describe the setup better?
I'm afraid the picture conveys nothing to me.
...R
Robin2:
I'm afraid the picture conveys nothing to me....R
So as the motor turns, it in turn spins this wheel, attached to a rod. And as the wheel rotates the rod moves back and forth along with the base
Somark:
So as the motor turns, it in turn spins this wheel, attached to a rod. And as the wheel rotates the rod moves back and forth along with the base
The word you want to use is 'crank' instead of wheel. As pictured, the connecting rod and the flat plain will also be moving up and down as the crank is turned. Is this what you want?
Paul
Paul_KD7HB:
The word you want to use is 'crank' instead of wheel. As pictured, the connecting rod and the flat plain will also be moving up and down as the crank is turned. Is this what you want?Paul
In real life they won't be rigidly connected. It'll be attached like an axle so the flat plane remains on the ground yet can still move back and forth
Somark:
I need to move a platform on wheels containing a 20lb object back and forth on a flat plane using a stepper motor and a shaft. It needs to move a maximum of 100 rpm, 4 cm. What would the wheel radius be and what would the torque required to move such a thing be? Also any recommendations on what to buy would be good
Which wheel are you referring to? Those under your platform or the "crank" attached to your motor? If crank is the right answer, then the length of the crank arm, obviously, must be 1/2 the distance the the platform must move. This requires the platform to be exactly on the same plane as the shaft of the motor.
You also need to factor in the resistance of the bearings on the ends of your connecting rod.
Do you have a shop where you can make all the necessary components?
Paul
At least put the motor vertically perpendicular to the platform plane if this is the contraption you're going for.
Looking at the picture some more I wonder if the bar under the platform is intended to be a roller that is rotated by the motor?
Although it is hard to reconcile that idea with the apparent huge gear-up between the motor shaft and the roller. It looks like the roller would rotate several times for a single rotation of the motor.
...R
radius 0.02m, angular velocity ~10rad/s (100rpm), so platform position:
s = 0.02 sin(10t)
v = 0.2 cos(10t)
a = -2 sin(10t)
ie the amplitude of the acceleration is 2m/s/s, with a payload of about 8kg.
So max linear force needed = 2 x 8 = 16N
So max torque ~= 16 x 0.02 = 0.32Nm
[ actually its a bit less than this as the crank geometry gives mechanical advantage for
part of the stroke ]
The penny has finally dropped. I had no notion from that drawing coupled with the description in the Original Post that the idea was for a push-pull mechanism with the plate moving from right to left. I was thinking of it moving from front to back on rollers.
...R
MarkT:
So max linear force needed = 2 x 8 = 16N
I am wary of challenging @MarkT but is there a * 9.81 missing?
I was correct to be wary. I realized I was confusing kilogram mass and kilogram weight.
...R
Paul_KD7HB:
Which wheel are you referring to? Those under your platform or the "crank" attached to your motor? If crank is the right answer, then the length of the crank arm, obviously, must be 1/2 the distance the the platform must move. This requires the platform to be exactly on the same plane as the shaft of the motor.You also need to factor in the resistance of the bearings on the ends of your connecting rod.
Do you have a shop where you can make all the necessary components?
Paul
Yes, I have access to a machine shop where I can do everything
MarkT:
radius 0.02m, angular velocity ~10rad/s (100rpm), so platform position:s = 0.02 sin(10t)
v = 0.2 cos(10t)
a = -2 sin(10t)ie the amplitude of the acceleration is 2m/s/s, with a payload of about 8kg.
So max linear force needed = 2 x 8 = 16N
So max torque ~= 16 x 0.02 = 0.32Nm
[ actually its a bit less than this as the crank geometry gives mechanical advantage for
part of the stroke ]
Thank you so much for this.
INTP:
At least put the motor vertically perpendicular to the platform plane if this is the contraption you're going for.
I don't think I'm correctly interpreting what you mean. Could you perhaps draw a picture?
Somark:
Yes, I have access to a machine shop where I can do everythingThank you so much for this. I don't think I'm correctly interpreting what you mean. Could you perhaps draw a picture?
That means instead of the motor pulling the platform up and down, it pulls from side to side as it rotates. Probably easier to control the platform movement.
Paul
Robin2:
I am wary of challenging @MarkT but is there a * 9.81 missing?I was correct to be wary. I realized I was confusing kilogram mass and kilogram weight.
...R
Indeed. SI units use kg only for mass, and newtons only for force, so never any confusion 
When gravity does come into a calculation I usually make the assumption that the surface
of the earth is the intended environment, but its still an assumption...