Simple Pulley Calculation Required

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#1 Racer911

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Posted 12 April 2020 - 01:01 AM

Hi guys, I have not had a formal education in physics. But I am designing a machine which involves the use of dancing rollers. i.e. rollers fixed to a swingarm that rotates on an axis at one end. multiple rollers are attached through which a flexible web. eg. paper, plastic film etc will pass through. I want to know how much force would be required to lift the swingarm up. I am attaching a diagram with the weights the swingarm exerts downward at each point through which the web will pass.  I am attaching a file which marks the rotational axis, as well as the weight in Newtons exerted by the swingarm if it is only suspended from a single point. What I would like to calculate or know the answer to is, how much force will be required to lift the swingarm up from the farthest point from the axis, which is exerting a weight of 70 Newtons if the web is run through as shown like a pulley.

Sorry if my explanation is very confusing

#2 GAHD

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Posted 12 April 2020 - 01:47 AM

I'm going to give you a couple tools and see if you can work your way though it. To start with, this is a complex lever(type 3) problem more than a compound pulley problem.

Torque = Force * Radial distance fulcrum * sin (θ)

Your values are 300N(distance a) 175N(distance 120N(distance c) 90N(distance d) 70N(distance e) for Force(resistance)

At "distance e" Force and Distance and sin(of 90 degrees =1) all reach a relative 1 value. In particular the sin value will change every degree as it is raised or lowered from that 90 degree position.

For each Force/distance value (a to e) The Force increases because of the mechanical disadvantage (from e to a).  So you need to calculate and sum the various disadvantage ratios.  To do this you can either assume a unit circle at (distance e) or use real values (not presented).

Once you have the summary mechanical disadvantage information, you can use that combined with sinθ to do the simple pulley calculation.

#3 Racer911

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Posted 12 April 2020 - 06:53 AM

Yes. I absolutely understand the mechanical disadvantage part, albeit practically, as we use this kind of a lever in our industrial machines everyday. So with every rise in the level above 90 degrees, the effort to lift it increases as the web will require more and more tension to lift the lever further because it doesnt stay in a vertical 90 degree rise and fall situation. But, quite frankly, that is something way too theoretical and not really required in the calculation in the concerned practical application as the use of this kind of lever in our machines, though absolutely necessary, is not of a highly critical nature. Although, I appreciate and am impressed you already figured that out by this diagram alone. What I basically want to get at is, If I pass the web through all of the rollers as shown in the diagram, how much force will have to be applied by the web at the exit point where I've pointed the arrow upwards, so that the lever starts rising up. I am only asking this question because right now I am assuming the answer wont be 70N as that is downward force when the lever is suspended from Point E alone, and similarly for all other Points. And in my daily use, we observe that with each roller that we engage the web with, the force or muscle strength required to lift the lever reduces. This is basically used to create a buffer for a servo start stop application because the servo starts and stops at the given length, while the unwind and rewinding of the web reel is a continuous process. So this lever keeps dancing up and down, hence the name dancing roller, to keep the tension maintained and unwinding process continuous while the servo motor starts again to draw the next  length.  In the application for which I want to use this lever, I can safely ignore the mechanical disadvantage factor that is caused once the lever starts rising upwards.

Is there a simple way to get at an approximation??

Again apologies for such long and probably confusing explanation.

#4 GAHD

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Posted 12 April 2020 - 09:45 PM

As a general rule I try to point people in the right direction and let them learn for themselves so it actually "sticks" but I'll give you a 10 minute quickie for the ork way.

If you don't need precision, it's (Force or weight) divided by number of ropes supporting the weight = tension, ignoring the fact that it's a diminishing returns

In this case average a to e (755/5), /10(total passes) = ~15.1N

That's NOT going to be completely accurate but "close enough for the wh*res" and you can just dial in the resistor/voltage controlling the motor doing the lifting over a couple test-runs(or just grab a box of springs and play around the "close enough" if it's a neutral-ish tension thing).

If it was(turns out to be) a critical engineering point you need to do euler-bernouli on the actual lever sections to min/max the weight vs strength (should have been done already if it's from factory) and get a bit more technical with the math(above).

But heay, you can always add some pointy bits and squiggly bits to get it working, paint it red to go faster, and throw some scented oils to keep the machine spirit happy. ~You'll be fine.

#5 OceanBreeze

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Posted 13 April 2020 - 01:01 AM

If you are not interested in a solution that involves the changing angle, and only want to know how the tension in the tape changes with the number and placement of additional rollers, then I believe the problem reduces to a fairly simple one.

It would have helped if you included the distances from the fulcrum for each of the rollers, but no matter, I will just assume some distances in my solution. You can then apply your actual distances and solve your specific problem accordingly.

I am going to assume that 840 Nm of torque is required to raise the lever. (You use whatever the actual value is needed in your case, based on 70 NL where L is the actual length from fulcrum to point e) That amount of torque is required anywhere along the length of the lever to raise it.

In my example, if the torque is applied at the end where you say 70 N of force is needed, the distance to the fulcrum is 12 meters, obviously. (70 N x 12 m = 840 Nm) I should mention this should be a cross product of vectors, $\vec\Gamma = \vec{r}\times\vec{F}$ but I suspect you don’t want to get into all that.

Taking all the forces you have on your diagram in descending order, they are 300 N, 175 N, 120 N, 90 N and 70 N. The respective distances from fulcrum, in my example, then are 2.8 m, 4.8 m,7 m, 9.33 m, and 12 m. Again, you use your actual lengths, not mine.

The total force, F, that is applied at the end is the tension in the tape, and it will be the same at all of the lifting points. As you say, it must be less than 70 N as the work is distributed among the five lift points.

But the total torque will always be 840 Nm regardless of how that gets distributed, in my example.

So, in this simplified case where we are ignoring vectors, we can just sum over each individual (force x distance) on the lever to get the net torque of 840 Nm

840 Nm = 2.8 F + 4.8 F + 7 F + 9.33 F + 12 F = 36 F

F = 23.3 N

That is a big force advantage from 70 N but the tape will need to be pulled through a longer distance to compensate.

It is always very possible I did not understand your question so let me know if this makes any sense to you. If you give me the actual lengths, I can work the actual problem for you instead of using this example. Or, you can use Gawd’s slip shod method.

#6 Racer911

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Posted 13 April 2020 - 04:01 AM

Ok.

So the point of 70N force is 520mm away from the rotational axis. and each consequent point is at a distance of 100mm.

The only reason I want to get a rough estimate of the weight or force exerted on the tape when all 5 rollers are engaged, is because this particular machine will involve processing a web with perforations at certain intervals and if the force exerted is beyond a certain accepted limit, the web or tape will tear from the perforation causing the machine to stop, create lots of wastage and loss of production. Plus the machine runs at about a 120 metres per minute, its not a lot by any definition from industrial standards, but still repeated breakages are never desirable.

@GAHD, if the exerted force is even roughly near what you've calculated, 15N, I think I am good to move forward because the perforated web easily lifts up 2 to 2.5 kgs of weight in a soft manner if no jerking is present.

@oceanbreeze have written the measurements as you requested. Will be waiting for your calculation too.

Fingers crossed. I am hoping I dont have to make design changes

Edited by Racer911, 13 April 2020 - 04:11 AM.

#7 OceanBreeze

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Posted 13 April 2020 - 04:25 AM

OK We just need to make some changes as follows:

Taking all the forces you have on your diagram in descending order, they are 300 N, 175 N, 120 N, 90 N and 70 N. The respective distances from fulcrum are .12 m, .22 m, .32 m, .42 m, and .52 m. Using your actual lengths

The total force, F, that is applied at the end is the tension in the tape, and it will be the same at all of the lifting points. As you say, it must be less than 70 N as the work is distributed among the five lift points.

But the total torque will always be 36.4 Nm regardless of how that gets distributed ( 70 N x 0.52m )

So, in this simplified case where we are ignoring vectors, we can just sum over each individual (force x distance) on the lever to get the net torque of 36.4 Nm:

36.4 Nm = .12 F + .22 F + .32 F + .42 F + .52 F = 1.6 F

F = 22.75 N

Not much different from my first answer which is what I expected since this is just a matter of proportionality.

What does this machine do, just curious, making copper tape maybe?

#8 Racer911

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Posted 13 April 2020 - 04:46 AM

What does this machine do, just curious, making copper tape maybe?

No No. Much Simpler. Its a actually a plastic bag on roll making machine. so theres basically a reel of plastic at the unwind side, the servo motor draws lengths as per the operator setting and seal and perforate at that point, and then those bags are rewound into smaller rolls, say 10 bags per roll or 20 bags per roll for consumer use. This is where the above dancing roller swingarm comes into effect. as the servo motor starts and stops to draw the appropriate length and seal the bag, the rewind process is a continuous one and it is neither feasible nor efficient to stop the rewind process everytime the servo motor stops. so this swingarm goes up and down, effectively creating a buffer of extra web while the web draw and sealing cycle continuously repeats itself. just search youtube for garbage bag or trash bag on roll making machine, you'll know what I'm talking about.

Thanks very much for the help.

WATCH THIS.

https://www.youtube....h?v=Sn50vV_ilvw

Edited by Racer911, 13 April 2020 - 04:49 AM.

#9 Racer911

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Posted 13 April 2020 - 10:57 AM

Guys,  off topic. Wanted some information on PID Control Loop for Motor Speed Control. Can PID Control Loop help in averaging out or say smoothen the motor speed output when the input speed reference is continously oscillating. Again, the oscillation being caused by the part mentioned in the earlier query. the Dancing roller which is connected with a potentiometer that gives a 0 to 10VDC signal to a VFD. Normally, cheaper machine manufacturers just dont care about fine tuning. Motor speed keeps on oscillating with the oscillating input reference. Some level of averaging in actual speed is achieved by increasing acceleration and deacceleration time parameters on the VFD or inverter which basically creates a scenario where the output is oscillating just as much but the motor acceleration times and de acceleration times are increased to a higher value, the variation in speed is kind of averaged to an extent because by the time the motor can achieve the high point reference speed, the dancer comes down again causing the motor to begin to de accelerate and vice versa. But again, this is a very very inefficient solution as when a machine needs to be emergency stopped or stopped in general, the motor will not come to a complete stop in the required time as the de acceleration time is quite high.

I  hope you guys dont mind such elaborate explanations. Its just that you guys caught on so fast to the earlier problem, I felt I can throw all my queries at you and have them solved to my satisfaction.

Edited by Racer911, 13 April 2020 - 10:59 AM.

#10 GAHD

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Posted 14 April 2020 - 02:32 AM

...sounds to me you'd be better served with a triggering mechanism on the Estop that applies a mechanical brake to the motor output if deceleration is a problem. That or a way to disengage the motor from the mechanism quickly like a clutchpack.

The analog to VFD is effectively already doing the job a digital PID would be doing. Do you think real time analog response is causing you some output problems?

#11 OceanBreeze

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Posted 14 April 2020 - 03:12 AM

I will answer the last part of your question first, as that is the simplest part. To get the machine to come to a complete stop quickly, as in the case of an emergency stop, you use a braking resistor. There should be terminals on the front panel of the inverter where you can connect one. These are high wattage, low ohmage resistors; a typical value may be 13.6 ohms, 9600 watts but you will need to refer to the technical manual for your particular inverter.

As for the PID loop, you really need to refer to your technical manual and not accept any suggestions from an anonymous source on the Internet!

I am a marine enginner on a research vessel and it sometimes seems my life is dictated to by various types and sizes of inverters, but whenever I feel the temptation to tweak any of the parameters, I stick my hands in my pockets until the urge passes.

Seriously, don’t mess with the PID parameters, or any other inverter parameters, unless you know exactly what you are doing.

Having said that, I will only speak here in generalities, Don’t do anything without consulting your technical manual or even better have a technical representative from the inverter manufacturer on site with you or on the phone with you.

First, is the PID loop enabled or disabled on your inverter? If it is disabled you might consider enabling it as it is there precisely to solve the sort of problem you seem to be having, that is, to stabilize an oscillating system.

Proportional Gain (P):  the gain of the feedback loop. If the gain is large, the response will be strong and immediate (If the gain is too large, vibration may occur).

Integral Time (I):  to eliminate the deviation by setting response time. Be careful not to set (I) too small, since a rapid response may cause oscillation in the PID loop.

Differential Time (D): to restrain vibration by having a damping effect.

As you can see, all of these parameters interact with the others and it is a fine balancing act to get them just right. If you get it wrong, you will be entering the world of destructive testing, so don’t touch them unless you know what you are doing.

#12 OceanBreeze

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Posted 14 April 2020 - 08:05 AM

The above post ^^^^ Is all the more reason for me to stress this: As for the PID loop, you really need to refer to your technical manual and not accept any suggestions from an anonymous source on the Internet!

#13 Racer911

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Posted 15 April 2020 - 07:18 AM

...sounds to me you'd be better served with a triggering mechanism on the Estop that applies a mechanical brake to the motor output if deceleration is a problem. That or a way to disengage the motor from the mechanism quickly like a clutchpack.

The analog to VFD is effectively already doing the job a digital PID would be doing. Do you think real time analog response is causing you some output problems?

The only problem with real time analogue response is that there is always a certain degree of continuous acceleration deceleration scenario. While this is absolutely comfortable with reels that are not extremely heavy or large in diameter, in this particular area the problem arises that unless an expensive eletronically controlled breaking or tension creating mechanism is not applied, the reel maintains its angular momentum even as the motor slightly decelerates causing the web between the reel and the rubber rolls pulling the web to loosens effectively causing them to first loose their traction, as in the line of the web starts shifting, not an acceptable scenario, second, when the motor accelerates again, it will always create a jerk, kind of a sudden tug on the web once the motor speed gets higher than the reel rotation speed that it has maintained due to its inertia.

So with lighter reels, a slightly lower line speed, an extremely low cost and time proven solution is to just hang a friction belt over the reel so that keeps a certain amount of breaking in check. and its a wonderful system because its hanging around the circumference of the reel, so as the reel diameter reduces, so does the effect of breaking caused by the belt, effectively creating a  self adjusting mechanism of reduction in breaking as the reel gets lighter and lighter. This will even work wonderfully in my application, it already is on one of my machines from China, the only problem is when the machine needs to Estop.

So basically, if the PID can smoothen out the rising and falling graph of input signal, and basically output an averaged speed reference that would match the total distance of reel unwound, that would be the perfect solution.

#14 GAHD

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Posted 15 April 2020 - 09:50 PM

If i'm reading you correctly and the mechanisms are close to what my mental picture of them is (pictures being worth thousands of words)...Clutchpack transmission would be my go-to.
Firstly it lets you near-instantly stop drive force in Estop situations, though adding mechanical brakes to any high-mass things is always a good precaution as well.
Second, if you dial-in the compression of a clutchpack properly you can allow a bit of over-torque slippage to smooth-out sudden acceleration events. Kinda like a LSD.

ED: sounds like you know what sort of programming you need for a PID or other controller. Might want to consider a "reel sizing" sensor on the PID interface to account for that rotating mass you're describing.

Edited by GAHD, 15 April 2020 - 09:52 PM.

#15 Flummoxed

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Posted 17 April 2020 - 09:29 AM

A report by Rockwell automation, which is mildly related https://literature.r...wp014_-en-p.pdf

A motion controller can control multiple axis with extreme accuracy. If you are using flying shears or rotating knifes with a motion controller why would you need to stop start and use a dancing arm to maintain tension.

PID is probably not a good idea in this application, PID is more suitable for process control.

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#16 OceanBreeze

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Posted 17 April 2020 - 11:09 AM

Nice link! The bag making machine is very interesting.

#17 Racer911

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Posted 19 April 2020 - 12:25 AM

A report by Rockwell automation, which is mildly related https://literature.r...wp014_-en-p.pdf

A motion controller can control multiple axis with extreme accuracy. If you are using flying shears or rotating knifes with a motion controller why would you need to stop start and use a dancing arm to maintain tension.

PID is probably not a good idea in this application, PID is more suitable for process control.

Hi. thanks for this. This article is actually not mildly but fully related to my application. Such bag making machines is exactly what we manufacture. Am just trying to upgrade control techniques to make the machine logic control more self sufficient and reduce operator skill and intuition required to run these.