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How to calculate required motor HP
8

How to calculate required motor HP

How to calculate required motor HP

(OP)
Hi,

I have an application in which I will be driving an 8" diameter roll that will be supported on each side by steel frames with fixed position precision bearings. I will be using a VFD to drive an AC motor with a 60:1 gearbox and an additional 2:1 belt/pulley reduction. Other than having to turn the roll itself at a constant speed, there is no additional load. From past experience, I feel pretty confident that a 1/2 HP motor will be able to do this job.

I'd like to know what calculations I can do based on the information above (and any other information I can request from the motor, gearbox, roll, and bearing manufacturers) that would allow me to determine with a higher level of confidence whether the 1/2 HP motor I have in mind will in fact work for this application.

Any advice will be greatly appreciated. Thanks in advance.

Paul


RE: How to calculate required motor HP

Probably the largest single load will be the friction of the V belts in the pulleys.
If this is between the motor and the gearbox the losses will be much greater than if the belt is between the gearbox and the roller.
Churning the grease in the gearbox may be a load.
Do some Googling to try to find out how much the belts will lose to side friction in the pulleys.
Google again to look for gearbox losses at no load. (Or with the belt load.)
Be aware of any possible binding or side thrust on the bearings.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

(OP)
Hi Bill,

Thank you for your response. Our gearbox will be directly coupled to the motor. The belt will be between the gearbox and the roller. I am pretty ignorant with regard to types of belts as I work in the electrical field more than mechanical, but the belts we used are described as timing belts. I am pretty sure we always use 8 mm pitch in various lengths. A typical part number we use would be "8M-1440-20 HTD TIMING BELT". I have attached an image below...



I will google as you suggested. I am just wondering if you would think this kind of belt would introduce more or less loss that a typical "V-Belt" (if in fact there is a difference between what we use and what you had in mind).

thanks again - I really appreciate your help.

Best regards,
Paul

RE: How to calculate required motor HP

That's a timing belt - has a fairly precise positional relationship between drive and driven pulleys but not desperately efficient. Google 'Poly-Vee' belts for one relatively low-loss belt type.

RE: How to calculate required motor HP

I will assume a motor synchronous speed of 1800 RPM. That is the most common speed of three phase induction motors in North America. (Another assumption, almost.)
That will put your output shaft at 30 RPM.
A little belt friction at that reduction will be almost negligible at the motor.
Not a bad choice at that speed and reduction, but check out Scotty's lead as well.
Timing belts like to run tight.
Consider a strong spring loaded idler near the smaller pulley if you decide on the timing belt.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

(OP)
Hi Bill and ScottyUK,

Thank you for your responses and I apologize I took so long to follow up. Bill, you are correct to assume a motor synchronous speed of 1800 RPM. With my planned 60:1 gearbox, this would make the top speed of my driven roll 30 RPM, as you mentioned. I can do some googling as you suggested to come up with an estimate for belt/pulley losses and for gearbox losses, but if I ignore these losses for now, I'd really like to know how to calculate the HP required to drive my 8 inch diameter roll (assuming lossless gearbox and belt/pulleys).

I assume I will need to know the weight of my 8 inch diameter roll. Is there any other information I will need to do this calculation?

Thanks again,
Paul



RE: How to calculate required motor HP

Assuming lossless belts/pulleys/gearboxes/bearings then you will require 0HP to turn the roller at a constant speed. So, those are pointless assumptions to make.

I suggest you also look at the acceleration and deceleration requirements.

RE: How to calculate required motor HP

You keep saying an "8" roll". Roll of what? Toilet paper? Stainless steel?

8" diameter, is it ten feet long or an inch in length?

Keith Cress
kcress - http://www.flaminsystems.com

RE: How to calculate required motor HP

(OP)
Hi Bill, Keith, and Lionel,

Thank you for your responses. This is an 8 inch diameter, 12 inch wide steel roll of hollow construction as shown in the following drawing. While the hollow 8 inch diameter part is 12 inches wide, the roll does have smaller diameter, solid shafts on each side that allow it to be supported by bearings mounted in front and rear support frames. The longer shaft is used to couple the roll to my drive system.



The article that Bill mentioned was very helpful. I do not have this roll in house yet, but I will be able to weigh it when I do. As my roll does not exactly match the description of a hollow or solid cylinder, I suppose I will have to estimate the radius of gyration, but the formula in the article will allow me to estimate my required acceleration torque.

(I do not require fast acceleration - I'd say if I can get the roll up to 20 rpm in 5 seconds, that would be great. After the initial acceleration, normal operation would be running at a constant speed for an extended time and then stopping).

If I use this method to calculate/estimate my required acceleration torque, the next thing I would like to do is learn how to relate this to required motor horsepower.

According to the engineering information presented in a Martin Sprocket and Gear catalog, the formula for HP is :

HP = Torque (in lb-ft) * RPM / 5252

... If I use the formula in the article mentioned by Bill to calculate my required acceleration torque, and plug it in to the HP formula above, I suppose this might give me a clue regarding my required motor HP if the motor was going to drive the Roll directly, but I am not sure how this is complicated by the fact that I will be using a 60:1 gearbox.

If any of you guys can give me a clue what information I need about my 60:1 gearbox in order to determine how this will affect my required motor HP, I would greatly appreciate it.

Thanks again for all your help.

Best regards,
Paul

RE: How to calculate required motor HP

Acklands-Grainger offers a gearmotor with a 30:1 reduction driven by a 1/15 HP motor.
You are belt driving and the belt will have more friction than the direct drive.
Seat of the pants and a WAG I suggest a 1/4 HP motor and move on.
1/4 HP motors are often used to belt drive fans and your greatest loss will be belt friction.
Your won't have much inertia at 30 RPM.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

The power to accelerate and and rotate the roll will be close to zero. What is the purpose of the roll that would require work? Why not make it an idler roll with no motor?

RE: How to calculate required motor HP

pffft!

I concur with Bill and Comp. That will take nothing to turn.

I'd make every effort to directly drive it with a 1/4hp motor and run it with a VFD. Probably $50 VFD will let you set it at between 10 and 1725RPM.

If this thing is actually going to do something like move a conveyor then it and it's bearings at 20RPM is still so little as to be completely negligible. The only thing you need to compute is the tension and hence work that roller will have to impart on the driven belt of whatever it's tasked with moving.

In that case then yeah a drive reduction would be in order. Do it with only a belt and a VFD.

Keith Cress
kcress - http://www.flaminsystems.com

RE: How to calculate required motor HP

(OP)
Hi Bill, Comp, and Keith,

To give you a better idea of my entire process, the 8 inch diameter roll I have described in previous posts is the one that I have labeled "Coat Roll" on the following illustration. The entire machine/process will take a spool of flat material (typically a 0.003" thick poly film) and will unwind it at constant speed. As it is being unwound, it will pass underneath a coating head that will apply a thin (adhesive based) coating to the film before the film is passed through a dryer. After the Dryer, the coated film passes through a pair of "Pull/Laminating" Rolls where a backing film is applied on top of the coating. After the Pull/Laminating Rolls, the three layer finished product (two films with a coating in between) is wound up at at a Rewind.



The Pull Rolls determine the speed at which the film is conveyed through the machine. The Pull Roll on the left is chrome-plated steel, mounted in fixed position bearings, and directly driven. The Pull Roll on the right is silicone covered and mounted in pneumatically actuated swingarms. The Silicone Covered roll is not directly driven - it is driven only by making surface contact with the chrome-plated Pull Roll (or the three layer finished product). By setting the gap (Nip) between the Pull Rolls properly, the Pull Rolls provide the necessary traction to pull the films at a constant speed (the surface speed at which we drive the Chrome-Plated Pull Roll).

In this scenario, the Coat Roll is chrome-plated. We have tried making this an idler in the past, but found that the film did not provide enough traction to make it turn reliably. Having the film slide over the Coat Roll without it turning generated static electricity, which lead to some issues. This is why we typically drive the Coat Roll now at the same surface speed as the Pull Rolls (and the speed of the film), just not to create drag.

Based on what you have said, it seems as though the HP required of my Coat Roll drive will be very small. We have been building this type of machine for many years, and we have always used gearboxes rather than directly driving our rolls. I think our original reasoning was that (due to required drying time), we run these machines very slowly. Back in the days of DC motors, we used 20:1 as a rule-of-thumb speed range as far as what you can hope to get out of a DC motor without worrying about cogging (although in actual experience, I have had reasonably good results up to 50:1).

I assume the HP required of my Chrome-Plated Pull Roll will be higher than that required for my Coat Roll because the Pull Roll has to create the force that pulls the films through the machine (against the resistance created by the Primary Unwind and the Backing Film Unwind, for which we use electro-magnetic brakes to create our desired level of tension.

The reason I started out only by describing my Coat Roll in my question is because this is the simplest of my drive systems. I figured that if I could learn how to calculate my required motor HP (assuming a 60:1 gearbox), I could get a feel for the basics before digging in to the more complicated scenarios of the Pull Roll drive and then the Rewind Drive.

My feeling is that over the years, not knowing how to do these calculations has lead to us greatly over-sizing our drive systems. I have already ordered the 1/2 HP motor and 60:1 drive box for the Coat Roll drive. I am sure this will work based on experience, but based on your feedback, I now realize I most likely could have gotten away with something smaller.

Assuming that I will be able to estimate my required acceleration torque for the Coat Roll (after I know its weight), and assuming the fact that I will have a 60:1 gearbox, if any of you can tell me what other information I need to calculate my required motor HP (even if it is ridiculously low), I think this would be a good start towards my getting a better understanding of this stuff.

I greatly appreciate your help.

Best regards,
Paul




RE: How to calculate required motor HP

Thanks for the detailed description of your process.
I am a little concerned about the high reduction ratio.
I can see the motor and reduction gear setting the speed of the roll rather than the actual film speed.
We had a somewhat similar problem many years ago with a veneer lathe.
The lathe was used to slice a thin sheet of veneer from a turning log to be used in the manufacture of plywood.
The knife was backed up by a small diameter roll that set the sheet thickness.
The forces were much greater than in your application but the shared problem was that the roll needed help turning at varying speeds.
The solution was to drive the roll with a wound rotor motor with a lot of rotor resistance.
With more than normal resistance in the rotor circuit the motor developed a small amount of torque over a large speed range.

I don't think that your application warrants a wound rotor motor, but the point is a torque based input rather than a speed based input may be a solution.
The thought comes to mind that a weak magnetic clutch may be a solution that will allow the roll to exactly match the speed of the film.
The reduction gear would run just a little fast and the weak clutch would slip enough to allow the roll speed to match the film speed exactly.
Any other comments or suggestions friends?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

Oh my... That is really something.
1/2hp thru 60:1 could probably lift a Buick.

I think Bill makes a very good point about the cart driving the donkey. The method you describe -gear box and motor- is going guarantee that idler pulley will never (ever) run at the synchronous speed of the poly film. You need a "loose drive" that simply supplies the energy needed to make up the bearing losses which is almost nothing at all and not calculable in any useful manner and will change with time and temperature and how someone nearby holds their mouth.

You want a very very slippy drive that would in essence start the roll spinning if nothing was touching it and eventually get it up to about 10% faster than you'd ever run the poly. If you were to grab it with your hand it would stop then start back up when you released it. Nothing like a 60:1 gear box. If you insist on the gear box method you will need a servo drive system and some serious sensors and sack of money that makes your 1/2hp/gearbox look cheap.

The motor should be very very small. We need to think about this for a bit.

Keith Cress
kcress - http://www.flaminsystems.com

RE: How to calculate required motor HP

I spent over 20 years designing and operating roll coaters. While they seem very simple there are many complicated interactions between speed synchronization, material stretch, surface friction, rubber compression, temperature, fluid viscosity, gauge variation across the width, and a host of other factors. An engineering degree only begins to allow you to understand these things. However, a 0.5 hp motor is probably a very good choice for your application. Anything smaller is almost a toy in the industrial world. My applications usually used D.C. servomotors and controls. A VFD on an A.C. induction motor can be very crude when you are trying to accomplish electronic gearing. You have to maintain synchronization under all operating conditions, such as starting and stopping.

RE: How to calculate required motor HP

Paul - Well explained post. This looks to me like the good old cassette recorder (may it RIP) where the tape has to travel at constant linear speed and the spools have to spin fast/slow depending on the diameter of tape build up. What's the mechanism used in a tape recorder?

Muthu
www.edison.co.in

RE: How to calculate required motor HP

The tape recorder uses a pinch roll to pinch the tape against the drive capstain.
You may be on to something.
A resilient pinch roll about 90 degrees ahead of the Coating Head may develop enough friction to allow the film to drive the roll.
BTW What HP do you use on the pull roll?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

Dancers. You need dancers on the spool off and dancers on the spool on. Maybe multiple dancers. They are your friends. Spring loaded dancers will make the whole system much less complex and allow a fair bit more slop in the inner control loop.

RE: How to calculate required motor HP

Since the main problem seems to be static why not coat the roller with rubber. Now there will be no sliding of the film across the roller. But now you say "there's more static"!

There are a huge number of static killers out there. You mount an ionizing bar where ever you need the static zapped and it's 100% eliminated.

http://www.exair.com/index.php/products/static-eliminators/gen4-ib/p8003.html

Ultra-High-Speed Static Ionizers


Check the two movies on static issues and note that the ionization bar kicks butt in the second video. Note the plethora of bars available.

https://www.ccsteven.com/cc/product-category/static-bars/

Keith Cress
kcress - http://www.flaminsystems.com

RE: How to calculate required motor HP

(OP)
Hi Bill, Keith, CompositePro, and Muthu,

Thank you for your continued support. The suggestions of using a clutch or other "loose" or "slippy" drive mechanism makes total sense to me if given my 60:1 gearbox reduction and my roll diameter, I was running the motor at a speed that (without slipping) resulted in the roll surface speed being faster than my web speed. This is not the case (and I apologize for not adequately explaining my application). As I believe Keith and Compositepro alluded to, if I insist on using the gearbox, I would have to implement a drive system with some type of closed-loop, master/follower, electronic gearing type speed controls. I believe this is exactly what I am doing. I will try to explain using my original illustration, which I have pasted again below...



...I actually have two separate drive systems with closed-loop speed controls, one for the Pull/Laminating Rolls and one for the Coat Roll. As it is the Pull/Laminating Rolls that determine web speed, I configure the Pull/Laminating Roll Drive System as the "Master" Drive. The Pull/Laminating Roll Drive System consists of a DC motor fitted with a 60 PPR ring-kit along with a DC regenerative drive board and a closed-loop speed controller. The speed controller allows operators to enter a speed set point in feet/minute. Internally, the speed controller does some scaling using roll diameter, gear reduction, and ring-kit PPR to relate ft/min to motor RPM, but the end result is that it controls motor RPM so that the actual roll surface speed will equal the entered set-point. As long as the gap between the two Pull/Laminating Rolls is set properly so that there is no web slippage, this ensures that my web speed will be the same as the entered speed set point.

Just like the Pull/Laminating Roll Drive System, the Coat Roll Drive System consists of a DC motor fitted with a 60 PPR ring-kit along with a DC regenerative drive board and a closed-loop speed controller. The only difference is that I configure the Coat Roll Drive System as a "Follower" Drive. In this configuration, the Coat Roll speed controller allows the operators to enter a speed set point as a ratio of the Pull/Laminating Roll speed. Typically this is always 1:1 because while the Pull/Laminating Rolls actually determine web speed, we just want to match the surface speed of the Coat Roll to the web speed so as not to create drag. (Sometimes we might adjust this to 0.999:1 or 1.001:1 just to compensate for any stretching or shrinking that might happen in the dryer). Internally, the Coat Roll speed controller monitors the feedback not only from the Coat Roll motor, but also the master feedback from the Pull/Laminating Roll motor, and (doing the necessary scaling), controls Coat Roll motor RPM to achieve the desired speed match.

Anyway, this control scheme works fine. I have absolutely no problem getting the speeds to match using this type of drive system with gearboxes. I have gotten this to work many times with different gearbox ratios, sometimes with an additional belt/pulley reduction in addition to the gearbox reduction. I choose my gearbox and belt-pulley reduction based on the particular customer's required speed range.

The reason I originally posted this question is because I strongly suspect that I routinely over-size the motors I use because I do not know how to calculate required motor horsepower based on what I am asking the motors to do.

Just to complete the description of my application, Muthu touched on the fact that this application sounds similar to a cassette recorder where the tape has to travel at a constant speed, but the Unwind and Rewind spools have to vary in speed as diameter changes. I think this is a good analogy. While the Pull/Laminating Roll Drive (assisted slightly by the Coat Roll Drive) handles my speed control function, I implement tension control (rather than speed control) for my Unwind and Rewind Shafts as described below.

For unwind tension, I mount an electro-magnetic brake on the Unwind Shaft and I have a tension sensing roll (not shown in my illustration) in the Unwind Web path. I use a closed-loop tension controller to monitor the feedback from the tension sensing roll and to regulate the brake voltage (allowing it to slip) as necessary to keep tension at set point while web speed remains constant.

Similarly, for rewind tension, I mount an electro-magnetic clutch on the Rewind Shaft and I have a tension sensing roll (not shown in my illustration) in the Rewind Web path. I drive the input to the clutch slightly faster than what would be necessary to keep up with web speed at minimum spool diameter). Then I use a closed loop tension controller to monitor the feedback from the tension sensing roll and to regulate the clutch voltage (allowing it to slip) as necessary to keep tension at setpoint while web speed remains constant.

Again, this all has worked fine for me for quite some time. I would just like to become better able to estimate my motor HP requirements to (hopefully) avoid over-sizing as I am sure I have done in the past.

As I mentioned, for the project that sparked my question, I have already ordered the 1/2 HP motor and 60:1 gearbox for the Coat Roll, and it seems like I could have gotten away with something smaller. The project that requires this Coat Roll only includes an add-on coating station for an existing machine (built by others). The information I have provided about the Pull/Laminating Rolls and Rewind are based on what I consider typical for the machines we build. In this case, my Pull Roll motor (with a 60:1) gearbox would be directly driving the (chrome-plated) Pull/Laminating roll on the left, and indirectly driving (by surface contact) the (silicone covered) Pull/Laminating Roll on the right. It would also be pulling the two films (one from the Primary Unwind and one from the Backing Film Unwind). In some cases, I use the same motor to also drive the Rewind Shaft (with electro-magnetic clutch). Based on past experience, I would be totally confident that a 1-1/2 HP motor would be more than capable of this task, and I would be reasonably confident that 1 HP would also be fine. While I expect I might be able to go even smaller, not knowing how to do the necessary calculations would make me hesitant to try.

It seems to me that there must be a way that I can calculate a rough estimate of my required motor HP in an application like this. I have a feeling that using the weight of my rolls, my tension ranges, my diameter ranges, and possibly other factors, I should be able to calculate my torque requirement for the driven roll. What is unclear to me is considering that I will be using a gearbox, how do I relate my torque requirement for the driven roll to a HP requirement for my motor.

I am going to do some googling/reading on gearbox sizing (with a focus on learning how application torque requirements relate to gearbox specs and motor specs). If anyone here can point towards some useful material or provide any additional advice, I'd greatly appreciate it.

You all have been amazingly helpful.

Best regards,
Paul




RE: How to calculate required motor HP

Paul

The torque is proportional to the speed reduction in your gear box. Half the speed, twice the torque; one third the speed, three times the torque and so on.

You have meticulously designed and successfully time tested a special purpose machine for a specific application. Why chance it by chasing the 'power saving' squirrel down that rabbit hole? At such fractional HP's, the power saving would be minuscule since it saves only the no-load loss (aka iron loss) and since the load loss is fixed more by the load than by the motor.

As a matter of fact, an oversized motor in your case provides the cushion for those occasional overloads without burning out the motor. Also, an oversized motor runs cooler thereby extending its useful life considerably.

To coin a cliché, let sleeping dogs lie. :)

Muthu
www.edison.co.in

RE: How to calculate required motor HP

The idea with the torque drive is to supply enough torque to almost start the roller turning. Then the film may be able to turn the roll.
More torque than that and you may over-drive the roll and cause the film to droop.
On the other hand, if the film is able to turn the roll, you may be able to fit a pinch roll and let the film turn the roll without a motor.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

(OP)
Thank you Bill and Muthu for your follow-ups. The idea of the torque drive is very interesting, and I think this and/or a pinch roll is worth considering in situations where the budget is not there for the motor, gearbox, and drive/control system capable of a master/follower electronic gearing type implementation like I am accustomed to doing.

Muthu - your point about letting a sleeping dog lie is well taken. I wouldn't change a proven design that has worked in the past just to save a little money on motor cost or energy savings.

The reason I am still interested in learning how to do these calculations is because I am rarely fortunate enough to get to repeat a proven design over and over again. All of the equipment we design/build performs basically the same function (roll-to-roll coating), but it is all custom and the specific requirements for every project is different.

Our "most standard" machine has an eight inch diameter (16 inch wide) Coat Roll and 8 inch diameter (16 inch wide) Pull/Laminating Rolls, with two 3 inch diameter rewind Air Shafts capable of winding spools up to 20 inch diameter. As we have built this a few times, I know that a 1 HP motor (w/ 60:1 gearbox) for the Coat Roll, a 1-1/2 HP motor (w/ 60:1 gearbox) for the Pull/Laminating Rolls, and a 1-1/2 HP motor to drive both rewinds (using electro-magnetic clutches) works fine. Even though I feel like my motors might have more horsepower than I truly need, I would not change this if I am fortunate enough to have an opportunity to build this machine again.

Quite often, however, we take on projects that are quite different than our standard machine. Currently, we are beginning our design phase for a very small machine which would have just a single drive. My Pull Rolls (no laminating function on this one) will be 3 inch diameter and 12 inches wide. My single Rewind will only have to handle spools from 6 - 10" diameter. Where as with the "standard" machine I mentioned above I used three separate motors/drives, for this small machine, I hope to use the same motor/drive for the Rewind that I will use for the Pull Rolls. On this machine, there will not be a driven Coat Roll. This will be a 2020 version of a machine that was built by others in 1990 (just two inches wider). The 1990 machine drives uses a 1/8 HP motor, so I am leaning towards 1/4 HP just to be on the safe side considering the extra 2 inches of width. While I feel comfortable in making this decision, I still wish I knew how to back it up with some calculations.

Another project I am working on at the moment is to add a dedicated motor/drive to handle two Rewind Shafts on an existing machine. These rewind shafts currently share the motor/drive that handles the Pull/Laminating Rolls, but a new process requirement will make it necessary for the Rewinds to have an independent drive. In this case, space on the existing machine is very limited. For this reason, it will be to my advantage to use as (physically) small a motor/gearbox that I can comfortably get away with.

For reasons like this, I really want to make an effort to get an understanding of how to (mathematically rather than gut feeling) estimate my required HP in various circumstances. I will always try to err on the side of caution (over-sizing), but I feel making this effort will be well worth my while, even if for no reason other than trying to get a better understanding of the systems/equipment I work with.

You guys have all been so helpful. If you don't mind, I may try to get started on some calculations for the project on which I will be adding a rewind drive and post them here to see if you think I am on the right track.

I really appreciate your help.

Thanks again,
Paul

RE: How to calculate required motor HP

I guess none of us who have replied have answered your your basic question of how you calculate required horsepower. This is because the answer is so basic to engineers that we assumed you were interested in more subtle details. Horsepower is a unit of power. Power is the rate at which energy is consumed. Energy is force times distance, ft-lb (ft. traveled x pounds force pushing). Do not get confused by the fact that torque has apparently the same units of lb-ft (pounds force twisting x ft. of lever arm).

For your application hp is torque (required to turn your roll) x angular speed (rpm). Get a basic physics book and the first chapter on mechanics will explain this. You can also find tutorial on the internet like those at Khan Academy.

RE: How to calculate required motor HP

To further Comp's statement we're not really answering your fundamental question because it's not a mathematically straight forward calculation. Answering the question of, "how much horsepower does it take to turn this roller" has many possibilities of answers based on more specific details.

1) Are you seeking steady state hp?
The bearings are all that matters in your case.

2) The hp to accelerate the roller up to what speed?
When you accelerate the roller you are storing energy in it that once speed is achieved no longer demands further energy input. But your app likely requires no real acceleration.

3) How fast to that speed? This increases the torque demanded.
The faster you want to get the roller to speed the more torque demanded. More torque can be achieved most simply by a bigger motor. However it can also be achieved with a gear reduction.

4) Drive train efficiency is never 100%. One set of gears is at least a 3 percent loss. Belts have losses too. So this muddies the water further.

5) Then you have the load contribution of the material across the roller. Is the roller contributing to driving the material? If it is, that number for the drag is critical to the answer. No math is going to provide that number for you! It has to be empirically measured.

Your particular situation is not amenable to "calculation" because the variability of the needed hp/torque calculation is completely swamped by the motion system itself as compared to the actual work needed to be done.

As an example, a sand conveyor is relatively easy to calculate. You can pile W amount of weight on a foot of conveyor belt. The conveyor belt is d long. The belt weight can also be calculated though it could be a wash due to the returning belt paying back with gravity. You then have the angle of the belt which provides the height the sand is being hoisted. These few facts directly gives the fundamental hp required to do the job.

The designers then find the drag of the rollers (probably from a book, rule of thumb, or a belt provider) which they add into the hp requirement. When done they multiply in a safety factor of perhaps 1.3 to cover any unexpected drag.

In your very different case you're turning the roller so slowly that acceleration, inertia, and bearing drag are all essentially nothing or are so fleeting as to be completely inconsequential. By far the largest consumer of hp in your case is the gearbox/belts aspect followed by the fan on the motor. The material across the roller is not even a load.

So, asking how to calculate the hp requirements in your application is.. incalculable. Instead it would normally be done one of three ways.

1) You'd assemble the roller and load up the machine and pull the web with a spring scale to get the work needed to be done. Then you'd double that and work back to the motor size.

Or

2) You'd work from experience having a feel for the size needed.

or

3) You'd go with the smallest size normally found in the industry.

Note that none of these embraces "calculations".

Hence, you see why we aren't showing you a way do it. We ourselves wouldn't be bothering.

Now, if you aren't price sensitive and want an excellent source for everything related to motion control talk to Mike at http://kilroywashere.com/. He is a great guy with vast motion control knowledge. He helped me thru a huge switch from active pneumatics to servo drives. He helped me -in detail- with how to use their Kollmorgen servo systems. Those systems have outstanding user interfaces with amazing built-in metrics and setup wizards that help with any conceivable use of a servo drive system including crazy variable electronic geared applications. Contact Mike with your application info and you'll get a concise solution quickly. Warn Mike that I sent you.

Keith Cress
kcress - http://www.flaminsystems.com

RE: How to calculate required motor HP

A Watt-meter may be a good investment. Watts is a much better indication of motor load than Amps.
Measure the Watts draw on your successful machines.
This will help you to develop some judgement as too approximately how much power you motors are using.
In the end, judgement and experience may outweigh calculations.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

Paul

The manufacturers of commonly used driven equipments like fans, pumps and compressors do lot of theoretical calculations, model studies and actual load tests before they publish their performance curves.

In the absence of such a professional set up to do these complex studies, measuring the power from your already supplied equipments and present ones in your factory is a quick guide for your fixing the power of your motor. As Bill said, a good power quality analyzer would give you plenty of info about your equipment power needs.

As I said before, at such a fractional HP range, neither the cost nor the size is going to be the deal breaker in your overall system cost and space. Over-sizing the motors is quite normal in industries since they are the critical parts and are most likely to break down early and often.

Muthu
www.edison.co.in

RE: How to calculate required motor HP

(OP)
Thank you Muthu, Bill, and Keith for your continued support. It is greatly appreciated.

I now understand the complexity of what I have been hoping to "calculate". A watt-meter or power quality analyzer sounds like a great idea.

For my current project in which I am adding a motor/gearbox to drive two rewind Air Shafts, I am leaning towards erring on the side of caution (over-sizing) and going with a 1 HP motor and a 30:1 gearbox. The rating data (based on 1750 RPM input) listed for the (Dodge/Baldor Tiger-2 Size 20) gearbox I have in mind is as follows:

30:1 gearbox:
Mechanical input HP = 0.96
Thermal input HP = 1.49
Output torque (lb.in) = 802
Mechanical output HP = 0.74
Output overhung load (lbs) = 1560

I am curious about the meaning of the "Mechanical input HP" specification. This Tigear-2 Size 20 gearbox is available in different reductions starting at 5:1 and going all the way up to 60:1 (whereas I am leaning towards 30:1).

Anyway, closest to my preferred 30:1 option, there is a 25:1 option and a 40:1 option. The rating data for these is listed as follows:

25:1 gearbox:
Mechanical input HP = 1.11
Thermal input HP = 1.59
Output torque (lb.in) = 788
Mechanical output HP = 0.88
Output overhung load (lbs) = 1560

40:1 gearbox:
Mechanical input HP = 0.76
Thermal input HP = 1.29
Output torque (lb.in) = 801
Mechanical output HP = 0.56
Output overhung load (lbs) = 1560

Does this "Mechanical input HP" spec mean that I should avoid using a motor with more HP than the "Mechanical input HP" spec? (i.e. make sure that motor HP <= Mechanical Input HP)

Or does it mean that I should use a motor with at least as much HP the "Mechanical input HP"? (i.e. make sure that motor HP >= Mechanical Input HP)

Thanks again and best regards,
Paul

RE: How to calculate required motor HP

You should be able to apply 1/2 HP or 323 W by hand. You'll get tired pretty fast but you should be able to get it moving at least. Riding a bicycle at 20 mph is about 180 watts. It will be hard but if you can't move it by hand I would be worried that 1/2 hp is not enough. I don't know if that helps you in design, though.

RE: How to calculate required motor HP

(OP)
Thank you DM61850 for trying to put HP requirements in easy to understand terms and Muthu for the thermal rating articles.

Both of these articles make it sound like it is more typical for a gearbox to have a thermal input rating that is lower than its mechanical input rating. In my case, the Dodge Tigear-2 size 20, 30:1 reducer I am considering lists "Thermal Input HP" (1.49) as being higher than the "Mechanical Input HP" (0.96).

We are considering using this reducer with a 1 HP motor. It seems to me I am ok with the Thermal Input rating, but I am concerned about what would be the potential drawback (if any) of using a motor HP that is slightly higher than the Mechanical Input HP rating of the reducer. I called Dodge technical support, and the person I spoke to informed me that if we apply more than 0.96 HP to the reducer input, we will over-load the reducer.

The datasheet on the DC motor we have in mind lists HP = 1750 and torque = 36 lb-in (3 lb-ft). This is exactly what I would expect based on the formula HP = torque x RPM / 5252. What I am not sure of is for a "typical" DC motor, whether I would expect the same torque (3 lb-ft) at lower speeds or if I would expect greater torque at lower speeds. (I guess the question could be rephrased as whether I would expect the motor to produce constant torque at all speeds or constant HP at all speeds). I reached out to the motor manufacturer (Leeson) to get an answer to this.

It seems to me that if the motor produces no more torque than 3 lb-ft at lower speeds, then my HP at the reducer input would be less than 1 HP at lower speeds. As my application will never require motor speeds greater than 1000-1100 rpm, it seems like I might be safe to assume that I would not be in jeopardy of over-loading the reducer.

If anyone here can tell me if my logic sounds reasonable or if I am off-track, I would greatly appreciate it.

Thanks and best regards,
Paul




RE: How to calculate required motor HP

The load demands torque and HP.
The motor produces whatever the load demands, within reason.
The motor torque rating is the limit of safe operation without overheating.
Many motors are capable of about 200% of rated torque.
At that loading, they will overheat and burn out if the protection does not take them off-line.
The gear-box must be rated to handle the load.
If he motor and gearbox are fully loaded it may be wise to select a gearbox from the next column.
If the assembly will be running lightly loaded the gearbox may be roughly matched to the motor HP.
Consider a gear-head motor.
It will be quicker to mount and the gearbox will be properly sized by the manufacturer.
It looks better and there is one less guard to to build.
Have you acquired a Watt=meter yet?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: How to calculate required motor HP

Take Bill's advice. Go for geared motor for your HP and final speed. Cost and space saver. No need to worry about thermal input HP and all that crap. :)

Muthu
www.edison.co.in

RE: How to calculate required motor HP

(OP)
Hi Bill and Muthu,

Thank your for your responses. When you say gearhead-motor or geared motor, does this simply mean a motor with a built-in gearbox? (As opposed to buying a motor and a gearbox separately?)

Bill - You say the motor produces whatever torque the load demands. Considering that the torque rating of my DC motor is 36 lb-in, if it turns out that my load (the rolls I am turning, coupled to the motor using a gearbox with belts and pulleys, with whatever load is added by the film I am pulling through the machine) demands less than 36 lb-in at whatever speed I am running (considering that I am using a closed loop speed controller to regulate the DC voltage to my motor to ensure a constant motor speed that is less than 1750 RPM), that the motor will produce only the torque I need (something less than 36 lb-in) to run at this speed?

Intuitively, this seems fortunate because if the motor produced more torque than required to run at this speed, I imagine that there would be some undesired effect (possibly unwanted acceleration or unnecessary strain on my drive system mechanical components?).

I have not purchased watt-meter yet, but I want to. I have googled watt-meters and it seems as though there is a bewildering variety of choices. Is there a particular brand/type that you would recommend?

Thanks again for all your help.

Best regards,
Paul




RE: How to calculate required motor HP

"Wattmeter"
If this will be used only to measure motor loads occasionally, you may use a single phase, 240 Volt Watt-Meter for $15 or $25.
For three phase measure two phases and multiply by 1.73.
OR
You may wish to spend more for a good quality power analyzer that may be used for more than checking motor loads.

Bill
--------------------
"Why not the best?"
Jimmy Carter

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