How to determine force required to move a load.

I have a steel framework that holds 240 extruded HDPE 'sticks' that are 4" wide by 2" high by 13 feet long. They are in contact with two steel frame members across the 13 foot dimension. When a single 'stick' is pushed with a force gauge, it measures 2.5 pounds of force to move the 'stick'. If you consider all 240 'sticks', this would mean that to move all of them together it would take approximately 720 pounds of force.

The question! What math do I apply to properly size a gear-motor to move this load(all 240)? The load will move a distance of 24" in 15 seconds, wait and repeat.

I have always sized prime movers in a situation like this in cowboy fashion, but in this instance I need to be a little more in the ballpark, versus the rodeo grounds.

The explanation must be simple...I'm just an old Millwright.

Stuart

atomarc said:

I have a steel framework that holds 240 extruded HDPE 'sticks' that are 4" wide by 2" high by 13 feet long. They are in contact with two steel frame members across the 13 foot dimension. When a single 'stick' is pushed with a force gauge, it measures 2.5 pounds of force to move the 'stick'. If you consider all 240 'sticks', this would mean that to move all of them together it would take approximately 720 pounds of force.

The question! What math do I apply to properly size a gear-motor to move this load(all 240)? The load will move a distance of 24" in 15 seconds, wait and repeat.

I have always sized prime movers in a situation like this in cowboy fashion, but in this instance I need to be a little more in the ballpark, versus the rodeo grounds.

The explanation must be simple...I'm just an old Millwright.

Stuart

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??

You tellin' us there are 240 times 4" (one-third of a foot, each) laid out in one single layer .. on two rails , simple friction, no rollers or carriage .. either eighty or one hundred and sixty feet long?

I mean.. you didn't say it was to be reversed..

...and one presumes that if eighty feet worth are fed IN two-feet worth at a go.... the previous batch have to GO somewhere ... for the next two-feet worth to have a place to advance into?

That's a LOT of space. Annnnd mass.

If I actually had to do it that way? My "high grade" rig would only grab the leading six (two feet worth) at a go.

A cruder rig - a paired set of bog-standard - and repurposable, even rentable - roller conveyor, inclined, would then let ignorant - and "free" gravity advance the rest of the lot up to a powered set of stops.

Or gin-up some sort of magazine feed that fed six at a go off a two-foot-wide stack?

Or bundles of an easier count to handle with simpler and bog-standard MHE? "Gravity" on the list.

And what's the goal, anyway?

Saw, mill, drill, engrave, stamp, coat, measure for QC.. feed some OTHER process, six at a go..or simply bundle for shipment as-had?

I suspect you have a better plan in yah than a PAIR of eighty foot long rails?

I mean being an OLD millwright.., not a naif . .as to the per-SF cost of Kalifornickyah SPACE .... if nothing else..?

I must have missed some info in my post because in a down and dirty test, a little Dewalt 20 volt driver drill lashed up to a Chinese 40:1 boat trailer winch easily pulled the entire 240 'sticks' without a problem.

This drive will consist of a small gearmotor turning a rack and pinion setup which will in turn, move the load. The driver will have a reduction to move the load at 8 FPM, which is 24 inches in 15 seconds. This means a small motor...under 3/4hp. The 'sticks' are slippery HDPE extrusions.

What I am after is some idea of how a person equates torque output of any drive to apply it to moving a load of a known weight.

Stuart

atomarc said:

I must have missed some info in my post because in a down and dirty test, a little Dewalt 20 volt driver drill lashed up to a Chinese 40:1 boat trailer winch easily pulled the entire 240 'sticks' without a problem.

This drive will consist of a small gearmotor turning a rack and pinion setup which will in turn, move the load. The driver will have a reduction to move the load at 8 FPM, which is 24 inches in 15 seconds. This means a small motor...under 3/4hp. The 'sticks' are slippery HDPE extrusions.

What I am after is some idea of how a person equates torque output of any drive to apply it to moving a load of a known weight.

Stuart

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8th-grade science class, Pennsylvania, 1950's. It is the CLASSIC and basic calculation for work @ known mass and a given coefficient of friction.

But you don't NEED it.

You already have a "cowbody" AKA "real world" .. point of reference that is just fine for the ACTUAL load, whether you even know or CARE the mass, contact area, or coefficient of friction.

To wit:

... Dewalt 20 volt driver drill lashed up to a Chinese 40:1 boat trailer winch easily pulled the entire 240 'sticks' without a problem.

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My Milwaukee cordless goods have their inch/ounce specs published. I'm sure DeWalt does as well. Rack & pinion, geartrains in general, lead-screws, synchronous belt drives, cable winders, etc, typically waste ten percent or less. Sync belting a LOT less and could be a sound alternative to rack & pinion. Easier to make "endless" or continuous as well-as - or instead-off reversible, if nothing else.

So JF use a NICER gearbox.... and a motor with more reserve ... so it lasts long enough.. to justify a higher price? And won't bust your chops on service work?

Off-the shelf winch - a 12 or 24 VDC one would be easy to externally control as to speed of advance - and some Gilmer belting instead of wire rope, mayhap?

Who cares if it has four times - or TEN times - the power it needs in reserve?

Electrical motation goods in general only DRAW the power the load actually imposes.

Unless you are a Smothers Brother's fan?

In which case, one of "the rules" was that if it ain't broke BREAK it!

California thing, is that?

Why are you dragging across rails?

Your sketch resembles many assorted factory things where product is moved along.

Many ways of configuring a moving surface, from simple roller chain, they make it with mounting tabs to attach brackets or plate to many more elaborate things.

A roller chain with proper support would have very little friction and the weight would be supported by the frame and support system using bearing supported wheels or ???

Now your power only needs to overcome the friction of the conveyer.

Sent from my SAMSUNG-SM-G930A using Tapatalk

Thank you...the design of the conveyor is set, and while unconventional, is applicable to its intended use, so there will be no re-designing there. My question is what is the relationship between output torque of a 'prime mover' and the sliding friction load. I know the value of the load to be moved..using that figure, how do I spec a gearmotor output torque to acceptably move that load.

Stuart

Stick /slip, friction coefficient is gonna screw with any calculations you might consider (But do consider a rigorous math treatment based on materials friction coefficient and contact patch loading. It always gives insight!)

A bit of "jiggle" in the structure, such as being attached to an operating machine, will change (decrease) the effort required to shift the load appreciably.

If moving by worm screw, the screw builders (aka motion control) will provide easy to understand worksheets on calculations. Inertia, friction and all that to-do!

Smooth steel and HDPE is self lubricating,and roofing tile conveyors use HDPE liners along the long beams and plain steel slats running along the poly strip.....any additional lube must be dry moly type or grit will cut into the poly...Big cranes also use polymer wear pads for the boom sections to extend under load .....the blocks last for ages ,but a bit of grease is needed to prevent rust of the steel boom s.

In the absence of gravity and other outside forces, the only consideration you have in moving any mass is acceleration and time. You have a certain mass and a given force will accelerate that mass at an easily calculated rate. This also neglects any time or reverse force needed to stop that mass when it reaches the desired position.

But, with gravity you will have friction. In a real world situation this friction will very likely be a lot greater than the force needed to accelerate the mass. So it will be the controlling factor. Unfortunately, the frictional force will depend on the coefficient of friction between the mass and whatever it is resting on. This coefficient of friction would be a lot different with different materials. Two rough surfaces will have a large coefficient and two smooth surfaces will have a lower one. Any lubrication between them will have a large effect: this is why we add oil to bearings. If there are cylindrical or spherical rollers between them then the coefficient will be VERY SMALL. This is why we have ball and roller bearings.

Just saying what the two materials are (steel and HDPE) is not enough to accurately determine that coefficient of friction. For one thing, the surface roughness or smoothness will be a factor. I know that it is generally said that friction does not depend on the area in contact, but only on the amount of normal (perpendicular to the surface) force involved. I do not think that is strictly true but it can probably be used if the HDPE sticks are stacked to a reasonable height.

So some actual measurements would be needed to get an accurate figure. It sounds like you have already have a head start with that. That is your best course.

Since you are going to have to experiment anyway, so a purely experimental method that includes the time needed to move the HDPE would probably be the easiest. But you could actually measure the coefficient of friction involved and then do the math needed to accelerate the mass to a velocity that would move them in the required time. That sounds like too much work to me.

In any case I would add a generous safety factor, perhaps 2X. That way real world things like wear and tear or "Oh, I haven't oiled the motor in 5 years, was I supposed to." would have less of an effect.

atomarc said:

What I am after is some idea of how a person equates torque output of any drive to apply it to moving a load of a known weight.


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I'm no engineer, but if you're merely converting rotational force to linear force, simply account for the radius of your driving interface (gear pitch circle or maximum lay diameter on a winch drum) and leave a few percent on top for acceleration, friction losses and stick-slip. Adequate motor service factor will accommodate stick-slip but steady drag from friction will have to be figured as continuous loading.

I.E. to apply 1000 lbs of linear force to a load in an ideal world, you could use a gear motor with an 83.33 ft-lb torque rating and a rack & pinion system with a 1" radius pinion gear. (1000 lbs / 12" per foot = 83.33 ft-lbs of torque required with a 1/12th foot (1") radius gear. An e.g. 4" radius gear would require 83.33 * 4 = 333.33 ft-lbs of driving torque to do the same job, but would do it much faster provided the driving motor and gearbox are up to the task.

For reference, you can probably figure a modern, brand-name 18 volt drill at somewhere around roughly 1/3 peak horsepower. That's a short-time rating obviously and with much more pull-up torque than an induction motor, but it gives you a frame of reference.

It's 550 pound-feet per second. I believe the hp would be about 0.175. That seems about correct if the drill was able to move the load.

atomarc said:

.... a little Dewalt 20 volt driver drill lashed up to a Chinese 40:1 boat trailer winch easily pulled the entire 240 'sticks' without a problem.
..... The driver will have a reduction to move the load at 8 FPM, which is 24 inches in 15 seconds. This means a small motor...under 3/4hp.
Stuart

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All in favor of real world test or hacks.
Did the drill/trailer winch do the speed of 96 inches per minute? 3/8 or 1/2 inch drill?
Is this stop start or continuous? Getting it moving takes lots more power than moving and you should have seen that in your one piece force test.
Yon need to move a friction load. Heck I can move a 8000 lb CMM head with a 9 volt battery and tiny motor on air bearings.
If good gearing to load and speed I get about 3/8 HP needed and can for sure be done with much less.
Down in here brush motors are often used. If high stop start machine one needs to add a big factor. Great with running, not so happy starting.
My basic design parameter I try is twenty years, 8760 hours per year without a failure or replacement.
Bob

You have all the info you need in the replies. A summary may be helpful.

Force required is total weight x friction coefficient. You know this = 720#.

1 hp is lifting 550# 1 foot in one second or applying a 550#ft torque for 1 second.

Moving 720# 2 ft in on second is 720/550= 1.31 hp. In 15 seconds 1.31/15 = 0.09hp

What is not mentioned in the replies is a consideration of inertia. I’m guessing your initial force measurement was to keep the one stick moving. To start it moving it will take an instantaneous force of 2x 2.5#= 5#. So throw in a factor of 2 and hp requirement is 0.18 hp to start and 0.09hp to move the 24”.

Where is some one that is motion guru,?
I know the problem is so badly defined but..
Some things here make me ..... lets go with wonder.
Bob

Well Mr. Bob,

I included just about all the information I had, sans the kitchen sink, so I'm not sure what you mean by 'badly defined'?

To the rest who came forward with intelligent, succinct answers, thank you..I now have enough information to proceed.

Stuart

If you are building quantity 1, which must work first time, every time, size your winch as if you are lifting it straight up. Forget about friction.
The difference between lifting and real world dragging (friction) will be "undocumented safety margin".

Rickyb said:

You have all the info you need in the replies. A summary may be helpful.

Force required is total weight x friction coefficient. You know this = 720#.

1 hp is lifting 550# 1 foot in one second or applying a 550#ft torque for 1 second.

Moving 720# 2 ft in on second is 720/550= 1.31 hp. In 15 seconds 1.31/15 = 0.09hp

What is not mentioned in the replies is a consideration of inertia. I’m guessing your initial force measurement was to keep the one stick moving. To start it moving it will take an instantaneous force of 2x 2.5#= 5#. So throw in a factor of 2 and hp requirement is 0.18 hp to start and 0.09hp to move the 24”.

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"2x 2.5#= 5#": are you saying it will take 5 lbs to start it moving? how do you come up with that?

dian said:

"2x 2.5#= 5#": are you saying it will take 5 lbs to start it moving? how do you come up with that?

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Sorry for the confusion. The 5# is to get one stick moving to break inertia if it takes 2.5# to keep it moving.

I am ignorant of the math, but my force gauge has a min/max hold feature, and the highest the gadget read was 2 1/2 lbf.

So to add a final chapter to this silly saga, I found a 'tarp motor' used on commercial dump trucks, etc that is 12v DC, 40 RPM out and has a whopping 160 ft/lb of torque. I'm going to give it a go as I think it's going to be more than enough.

Stuart