Find the Right Motor for a Vertical Load Torque

[SALMAN22]

If I'm to choose a motor to suit a vertical load (torque)-see the attached picture for more illustration-. What is the criteria should i follow?

After i have done the math for the weight of the object to be rotated, the torque to rotate the load is to be horizontal, and i only have the weight -which is vertical-.

 

[jackwhirl]

What sort of shaft is supporting gear 2?

 

[SALMAN22]

What sort of shaft is supporting gear 2?

I don't need to directly choose a motor, i need the formulas to calculate it.

The issue is that I'm confused between the vertical and horizontal load- i.e. in this case the load is vertical (weight) and the torque to rotate the gear is horizontal. How to combine these two forces in one relation.

And I'm already using this relation to calculate the torque, which is Torque = mL

m = mass
L = distance from the center of gravity to the center of rotation.

is this formula enough to calculate the torque or there are further considerations.
Thank you inn advance.

 

[SALMAN22]

What sort of shaft is supporting gear 2?

Cylinder

 

[jackwhirl]

There are more. If the motor and its shaft don't have to support gear 2, then the vertical load is only important for its contribution to the system's moment of inertia.
How quickly do you want to rotate gear 2, and how long do you want it to take to reach its maximum rotation speed? These are required inputs for calculating the torque needed, which tells you how big a motor you need.

 

[Dr.D]

No, the torque to rotate #2 gear us vertical, parallel to the axis of rotation.

 

[SALMAN22]

No, the torque to rotate #2 gear us vertical, parallel to the axis of rotation.

Can you please give more explanation, using the current example. I mean where is the horizontal force contribution here ?

 

[SALMAN22]

There are more. If the motor and its shaft don't have to support gear 2, then the vertical load is only important for its contribution to the system's moment of inertia.
How quickly do you want to rotate gear 2, and how long do you want it to take to reach its maximum rotation speed? These are required inputs for calculating the torque needed, which tells you how big a motor you need.

The motor is to rotate in constant speed- no need for acceleration-.

 

[Dr.D]

Torque is a vector cross product, T = r x F. A force tending to turn the gear must be in the plane of the gear as is r also. The cross product is perpendicular to each factor.

 

[SALMAN22]

Torque is a vector cross product, T = r x F. A force tending to turn the gear must be in the plane of the gear as is r also. The cross product is perpendicular to each factor.

It is the right-hand rule. I remember now. Thank you

 

[Dr.D]

Yes, the cross product follows the right-hand rule.

 

[hutchphd]

The motor is to rotate in constant speed- no need for acceleration-.

Then the only requirements on the motor are to overcome various "frictional" forces...by which I mean air resistance, bearing drag and motor resistances. The weight does not directly matter ( it will likely increase bearing drag ). These are very hard to know without measuring.
In fact you will need slightly more torque than "steady state" value to get up to speed...

 

[SALMAN22]

The weight does not directly matter ( it will likely increase bearing drag ). These are very hard to know without measuring.

Can you please explain why the weight does not affect the whole operation of rotation.

Are you saying it is a matter of bearing drag? How to calculate it in this case? and what is its significance. Thank you.

 

[Baluncore]

The weight of the shaft and rotating load will be carried on a thrust bearing.
The vertical rotating shaft will need to be held in position so that it remains vertical with the gears meshed. Depending on the rotating weight, a pair of tapered roller bearings might do both jobs. They would probably be part of the gear box assembly.

The motor torque needed to rotate the shaft and load is a separate issue. It must initially accelerate the mass and then maintain the rotation against windage and friction.

Forces between the gears will be countered through the gearbox and bearings.

 

[Dr.D]

Looks like a free body diagram is needed here.

 

[SALMAN22]

Looks like a free body diagram is needed here.

Here is the application to be implemented. It is a solar tracker, and i need to choose the proper motors for it, and still confused how the vertical load will affect motor's torque- I'm talking about the vertical-shaft motor.

And as our friend @Baluncore here said,

The weight of the shaft and rotating load will be carried on a thrust bearing.

Is it better done with a gearbox assembly or can be manually crafted?

 

[Baluncore]

Is it better done with a gearbox assembly or can be manually crafted?

It depends on what you mean by "manually crafted".

Your diagram shows a gearwheel halfway along the horizontal shaft where maximum deflection will occur. That needs to be redesigned with the gear and motor mounted at the end of the shaft where there is a solid mount available for the motor and the driven gear.

When two gears are meshed, with a tooth force, there is a reaction due to the 20° tooth face angle that forces the two gear wheels apart. Meshed gears push shafts apart. Sprockets with a roller chain, or stepped belt, pull the shafts together.
Maybe you should be using roller chain rather than meshed gear wheels.

This also has environmental implications. Sunlight damages plastic and paint. Wind blows dust into gaps, damaging bearings and lubrication. Rain, salt and moisture cause rusting.

 

[jrmichler]

Solar trackers run at very low speed, so you can ignore acceleration forces. Wind loads on the panels are orders of magnitude greater than friction loads, assuming halfway reasonable bearings. Therefore, you can design for just the maximum wind load and ignore other loads. The design wind load is the worst case wind from the worst case direction with the panels in the worst orientation. Wind exerts a torque on the panels because the center of pressure is not the center of the panel when the wind is not perpendicular to the panel.

You need to find or specify the maximum wind velocity and the most vertical angle that the panel could be oriented. Then calculate the torque about the vertical axis.

 

[berkeman]

Maybe you should be using roller chain rather than meshed gear wheels.

I agree; that was my thought as well.

This also has environmental implications. Sunlight damages plastic and paint. Wind blows dust into gaps, damaging bearings and lubrication. Rain, salt and moisture cause rusting.

It would be good to enclose the gear/chain mechanisms to keep them cleaner, and plan on a regular periodic maintenance (PM) schedule like once a year to clean and lube them.

 

[SALMAN22]

@Baluncore
I think this is a great analysis

Your diagram shows a gearwheel halfway along the horizontal shaft where maximum deflection will occur

This can be solved by supporting the horizontal shaft by equally-distant vertical pillars- using journal bearings, of course.

Sprockets with a roller chain, or stepped belt, pull the shafts together

Should this be done also with the horizontal shaft, because i think this will be a bit complex in our case. And if it is possible to make an edit on this design, what would you suggest considering the horizontal shaft?

And thank you for being so responsive and giving really helpful solutions.

 

[Baluncore]

Where do I start.
Re: The drawing in post #16. The top part of the assembly, the tilting frame.

The tilted frame appears to be made from tubular material, so will be rigid = torque tube.
There is no advantage in having the long horizontal round shaft.
There is no advantage in having isolated supports on the long flexible shaft.
If there were three bearings, then you could use 3 short pins or bolts. That would make assembly and maintenance much easier since you do not have to thread the long pin through all the eyes.

The tilting frame has a wide middle bay. Add a tubular bay divider in the middle. Space the bays equally.
Narrow the U fork so the ends are inside, not outside, the tilting frame.
The U does not need to be so wide. Bring the ends into be inside the bay separators.
Put a support bearing in the middle above the post. That will bear part of the weight of the tilting frame on the middle divider. Put simple pin bearings at the ends.

Mount the tilt drive at one of the three pin bearings, sheltered below the tilting frame.

 

[Dr.D]

It seems that Baluncore may be taking the sketch in #16 too literally.

For example, critiquing the use of a single long shaff rather than three pins (short shafts) maybe simply reading too much into the figure. What the figure does make clear, and this is signiificant, is that the three bearings are colinear. If drawn as three pins, this might not be evident.

The width of the middle bay is also criticized. That is not really germane to the discussion of the drive mechanism, but is simply a matter of structural stiffness.

There are lots more things of this sort that one might criticize, but the OP has not really asked for that additional guidance, as I see it.

 

[Baluncore]

The width of the middle bay is also criticized. That is not really germane to the discussion of the drive mechanism, but is simply a matter of structural stiffness.

The bull-gear, or chain sector size, and the location of the drive mechanism are important. The drive mechanism connects, constrains and bridges between the two sides of an axis.

The azimuth bearing could be at the base as shown in #16, part way up, or at the top of a fixed post. The maximum wind moment will be at the base of the post. I would lift the AZ bearing and actuator to the top of the post, close to the similar elevation mechanism. Not only are wind induced moments on the azimuth bearing reduced, but the mechanism is then also clear of the ground, children and animals. It can bring all the mechanisms into one housing.

Once the placement of the azimuth bearing has been decided, the FBD can be drawn.