Trying to find the torque required to turn the crankshaft in my mechanism

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Discussion Overview

The discussion revolves around calculating the torque required to operate a crankshaft mechanism designed to lift a 150 lb object vertically. Participants explore various factors affecting torque, including the mechanism's design, weight distribution, and operational speed, while seeking to determine if the selected motor and gearbox can handle the required torque.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to determine the torque needed for a mechanism that lifts a 150 lb load vertically 7 inches using a crankshaft.
  • Another participant questions whether the term "crankshaft" is used correctly, suggesting it may refer to a linear screw actuator instead.
  • It is proposed that the maximum torque occurs when the cam is horizontal, approximating torque as the product of the crank length (3.5 inches) and the load (150 lbs).
  • Concerns are raised about the speed requirement of 200 RPM and whether a higher ratio gearbox could be used to reduce the lifting speed.
  • Suggestions include increasing counterweights or incorporating a spring to assist with lifting the load.
  • One participant estimates that rapid direction changes could result in a force of approximately 450 lbs, leading to a calculated torque of 1575 in-lb.
  • Another participant independently calculates a torque of 525 in-lb but acknowledges the need to consider acceleration effects.
  • Further calculations suggest that accounting for the angle between the crank and connecting rod reduces the effective lifting force, leading to a revised torque requirement of 2100 in-lb.
  • Participants discuss the implications of friction, safety factors, and the design considerations for the crankshaft mechanism.

Areas of Agreement / Disagreement

Participants express varying estimates for the required torque, with some calculations converging around 525 in-lb while others suggest higher values when considering additional factors like acceleration and friction. No consensus is reached on the exact torque requirement, and multiple competing views remain regarding the calculations and design considerations.

Contextual Notes

Participants note the complexity of the mechanism's operation, including the effects of inertia, friction, and the angle of force application, which complicate the torque calculations. There are also unresolved assumptions regarding the specific design and operational parameters of the mechanism.

TURNIGY
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Hello!, First post and I am excited.

I am an artist and want to see if the motor I have selected for this mechanism will work for a project I am working on. I am trying to see if the gearbox/motor I have sourced for this mechanism is powerful enough to drive it.

The weight of the object I wish to move is 150 lbs and it moves in a completely linear motion up and down. The crankshaft is 3.5 inches from the rotation point to the bottom of the push rod and the push rod is 13.17 inches long. Attached is the power output chart for the motor and a diagram I made for my crankshaft design.

I am trying to find the torque required in in-lbs but not sure how to start.
 

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I need it to move 7 inches vertically. Crankshaft, here is a photo of what I am working on.
cranksahft 2.png
unnamed (4).jpg
 
TURNIGY said:
I am trying to find the torque required in in-lbs but not sure how to start.
I think the maximum torque will be when the cam is horizontal (half-way through the lifting motion, but it's complicated a bit by how the shaft will move. Still, it's a good approximation to multiply the 3.5" length by the 150 pound load, because that shows that the current motor can't supply that torque.

How fast do you need to lift this load? Can you lift it 10x slower (via a higher ratio gearbox)?
 
berkeman said:
How fast do you need to lift this load? Can you lift it 10x slower (via a higher ratio gearbox)?

Unfortunately I cannot, I need it to cycle at least 200 RPM.

I can choose a different gearbox/motor setup if I know the torque or the calculation required to get the torque with this setup.
 
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The crank should not need to lift the entire mass.
Can you increase the crank counterweights ?
Can you include a spring somewhere that supports half the weight ?
 
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Welcome, TURNIGY :cool:

What guides the 150-lb object linearly up and down?
What is the approximate weight of the con-rod?
Is there considerable sliding friction to consider?
Any requirements for starting rotation up's?

The main torque requirement comes in this case from the inertial resistance of the 150-lb object plus con-rod to accelerate and decelerate 400 times per minute or 6.7 times per second.
 
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Just a back-of-the-envelope estimate here. If I'm all wet, I'm sure others here will make the needed corrections. :wink:

Please consider:
Changing direction that rapidly is around 2g, g being the acceleration of gravity. That shows the force pushing the weight up will be in the neighborhood of 450lbs... along with a 150lb force needed to pull it down fast enough.

Converting to inch-lbs as you asked is:
3.5 x 450 = 1575in-lb

Also, since the weight has to be pulled down with 150lbs, your motor/crank better be securely fastened down.

With a cyclic load on the connecting rod bearings, crankshaft bearings, and the sideways force on the weight, design by a mechanical engineer is strongly recommended.
 
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A 4 bar mechanism guides the object up and down, I can add more weight to the end of the counterweights of the crankshaft. Cool, I did some sleuthing on my own and am also getting a ~similar force, solving I get 525 in-lb but yea, forgot about the acceleration in terms of direction, opps.
unnamed (6).jpg
 
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OK, we match on the 525in-lb. That's what I got before the acceleration. :oldbiggrin:

But I did forget the angle between crank and connecting rod. About 1/4 of the force generated by the crank will try to move the connecting rod to the left in the drawing, leaving 75% as lifting force.

1575 / 0.75 = 2100in-lb

Plus friction, plus safety factor, plus x4 (at least) since you will be reversing the motor every quarter turn.

Can you put the crank on the unsupported end of the shaft? That will allow continuous motor rotation and reduce the crank throw by half, which also reduces the torque by half and increases bearing lifetime.

Cheers,
Tom
 
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