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Worm gear system question from a EE!!

Worm gear system question from a EE!!

Worm gear system question from a EE!!

(OP)
Folks, apologies for this question but I have scanned web pages for several hours trying to find this data and can't.

Can someone help me understand the effective torque transmitted to a worm gear shaft that has a 30 tooth worm gear with 20 degree pitch angle, 7.12 degree lead angle, 18mm (0.625") PD mounted on it and being driven by a 2 lead worm of 8.46mm (0.333") PD? The input torque from the motor/gearbox is 500mNm or 70.8 inoz. The reduction ratio for the worm and gear is 15:1.

Here is the logic that I have at this time regarding the calculation.

Input torque = 70.8 inoz. At PD of 0.333" of worm, effective force at worm/gear interface = 70.8/0.333 = 214.5 oz of force.

214.5 oz of force acting on a 0.625" PD gear = 134in oz (FxD)

but I don't think any of this is correct since the primary purpose of the worm mechanism is speed reduction while providing significant increases in torque. Can anyone point me in the correct direction for this calculation?

Thanks, Tony

RE: Worm gear system question from a EE!!

Try this:

Output shaft torque: ratio x input torque x efficiency

Efficiency in worm gearboxes is approximately 0.6

Then: 15 x 0,5 x 0.6 = 4,5 N·m

RE: Worm gear system question from a EE!!

OK, first of all your torque is determined by the output load, not the input.  Your gearmotor may be designed to give you 70.8 oz-in of torque, but if there's no load you won't get that, even if your motor is capable of a million ounce-inches.

Let's assume that you're designing it to load your motor to its limit.  Your mechanical advantage is 15:1 and worm drives don't have very good efficiency, due to friction, so you'll only get about 55-60% of your total power output.  15:1 x 60% = ~9:1 torque ratio.  So your motor/worm setup can handle an output load of about

70.8 x 9 = 637 oz-in.

The force at your gear teeth, if you're interested, is this torque divided by the pitch *radius* of the gear (not the diameter) and would be

637 / .3125 = 2039 oz = 127 lbs.

Hope this helps.

Don
Kansas City

RE: Worm gear system question from a EE!!

(OP)
Gentlemen, thank you.

Don, yes, I know, I am working this in reverse. The primary issue I have is that the mechanism being driven by this motor/worm gear system is, in my limited understanding of mechanical calculations, very complex and I am unable to calculate the torque required to run it. I have a feeling I know about how much it needs but it's a dynamic problem where the torque required changes.

If I post a picture of the mechanism and a description of what it is doing, do you think you could help me understand the torque requirements of the mechanism?

Best regards,

Tony

RE: Worm gear system question from a EE!!

Possibly.  I'm always an advocate of empirical measurement when the calculations get overly complex.

Do you have a spare motor lying around that you could drive the device with?  You could monitor its current draw and calculate the torque/power requirements of the mechanism from that.

Don
Kansas City

RE: Worm gear system question from a EE!!

(OP)
The mechanism is still in the CAD stage of design and I have not yet started the prototyping of it so a spare motor may not be of value yet.

Let me collect the pictures and post a description - maybe it isn't that difficult to calculate the torque requirements or at least, a magnitude I can start from.

Posts to follow - there may be several so hang on till I have finished.

Thank you, Tony

RE: Worm gear system question from a EE!!

(OP)
first question is how do I post pictures on this forum?

RE: Worm gear system question from a EE!!

(OP)
OK, so, I think the best way to start is with a visual of the mechanism.

Attached you will see two pictures of the mechanism without all of the ancilliary components. The best way to visualize this is that the base of the moving/rotating pylon (grey item) that is captured between a rail framework is somewhat similar to a roller coaster. The base of the pylon is a "shuttle" that traverses from one end of the curved rail to the other using the motor and worm gear we discussed via a toothed drive sprocket and teath machined into the rail. As the shuttle moves along the curved rail, there are guide pins mounted in the shuttle that follow a groove also machined in the rail causing the shuttle to rotate by about 90 degrees total between the left and right sides of the rail.

The distance traversed by the shuttle along the rail system is about 13 inches and the length of the pylon attached to the shuttle is also about 13 inches. At the top of the pylon is a mass of about 18 ounces.

The shuttle must traverse the distance of the rail in about 5 to 6 seconds max.

There are two guide pins, one at the front of the shuttle near its bottom and another one at the back of the shuttle also at the bottom. There are two drive sprockets on either side of the shuttle that are driven by the motor and worm gear. The two pictures show the two end positions of the mechanism. To the left when the plyon is vertical and to the right when the pylon is horizontal.

RE: Worm gear system question from a EE!!

(OP)
This thid picture shows a line drawing representation of the mechanism at varying positions along the rail. Note the black circle represents the drive sprockets that are mounted to the base of the pylon or shuttle and motor drives the shaft that these sprockets are on.

So, in summary, the motor drives the shaft that drives the sprockets. The sprockets engage the teeth that are machined into the rails and the forward and rear pins (noted as small circles at the two tips of the green triangle) engage into a slot machined in the rail system either side of the shuttle.

RE: Worm gear system question from a EE!!

(OP)
So, the mechanism is essentially a sliding, rotating lever arm and the maximum torque required, I believe is when the shuttle is to the far right and has to be moved to the far left. When the shuttle is to the far right, the full 18 mass on the end of the pylon is acting on the pins. If we ignore the friction components of the pins which I am currently working to overcome, and assiming the 12.5 inches is actually 13, I need to figure out what torque is required on the drive sprocket to move the arm from its horizontal position to the vertical. I believe the greatest torque requirement is in poistion six noted in the post before last.

Any ideas how I can calculate this to give me a start point?

RE: Worm gear system question from a EE!!

Hi Skyman2

Firstly I concur with eromlignod in respect of efficiency
and torque, I calculated 65% efficiency based on a velocity
of 0.055m/s which would give a torque output of 4.875Nm.
He is also correct in that you have to determine the load output requirements and work backwards.
The speed of operation is fairly slow so you can analyse
this arm movement by a series of positions from start to finish statically, pretty much as your diagrams you have posted without worrying about dynamic effects such as inertia etc. If it were me I would find the centre of gravity of the pylon and for various positions I would calculate the torque required to keep the pylon in the position at each stage of its travel by balancing the static forces. From this you should see where the maximum torque is required and that can be used in sizing the drive.
In addition as you rightly point out you need to consider friction, so each position of the arm you have drawn you need to work out the load on each guide pin and multiply by the appropriate coefficient of friction to get the force required to slide or roll the guide pins along the track.
At this point though I have a concern that your design might not work because when I look at the guide pins I feel that the guide pins will against each other at some point along the track and just stop dead, in order to check this you need to divide the pylon movement into very small angular movements and study the reactions and angles that they make relative to the track and with each other.
I am not sure whether your software your using can do simulations of mechanism movement and alert you to any binding problems etc.
Finally I would like to know how the worm gear drives the pylon as far as I can see your design requires the motor drive and worm gearing to be attached to drive sprockets which means that all the drive moves with the pylon is that correct or am I missing something?

best
regards

desertfox

RE: Worm gear system question from a EE!!

Hi again

At this point though I have a concern that your design might not work because when I look at the guide pins I feel that the guide pins will WORK against each other at some point along the track and just stop dead,

I missed the word WORK at line 15

regards

desertfox

RE: Worm gear system question from a EE!!

(OP)
Desertfox, yes, the motor is mounted in the plyon base as is the worm gear and drive sprockets. The motor movs with the pylon. The teeth for the sprocket are machined along the curved rail and are stationary - just like a curved rack and pinion where the rack is on the rail.

The software I have does not allow me to simulate unfortunately so I am unable to do this.

Also, I am concerned about your comment of the pins working against each other and creating a jam. Where exactly would you see this occuring along the travel of the rail?

As I mentioned in my earlier post, I am working to reduce the frictional content of the pins in the guide slots and am looking at various strategies that include roller bearings on the pins between the pins and the guide rail surfaces (with clearance built in), thrust washers on the pins working in unison with the roller bearings to control lateral movement and to eliminate frictional forces of the outer case of the roller bearing contacting the shuttle internal structure/frame. I believe that I have the frictional component of the pins down to about a constant of 0.01 or thereabouts. I am still fighting a lateral movement control issue that I think will be present and am trying to identify solutions for this also.

Let me know where you think this jamming might occur.

Lastly, calculating the static forces is not something I have ever done and I do not know quite where to start in this analysis. how do I go about this?

Thanks in advance

RE: Worm gear system question from a EE!!

Hi Skyman2

You will only get the friction coefficient down to 0.01 by using rolling friction any sliding friction would have a lot higher coefficient, but even if you use roller bearings you need to ensure bearings will rotate and maintain rolling
contact between guides and rail throughout the cycle.

Not easy to say where else it might lock up but look at position 6 in your last but one picture, imagine reversing
the worm to take the pylon back to the vertical position, the bottom guide pin looks like it will be driven hard against the bottom of the guide rail almost as though it wants to burst through the guide rail I see that as one area
there may well be others.

As regards working out static forces I can only suggest that you look in some mechanics books for free body diagrams.
Have to get ready for work now I will try and post later.

regards

desertfox

RE: Worm gear system question from a EE!!

Hi Skyman2

I have another question if the whole of the motor and worm gear travese with the pylon how is it supported during operation ie starting at the pylon vertical position how is
the motor, worm etc prevented from falling down the slope?
I am assuming also that the pylon can rotate independently
of the worm gear shaft and you use the guide pins to control its position during movement am I correct in this
assumption?


Regards

Desertfox

RE: Worm gear system question from a EE!!

If I'm understanding the mechanism correctly, the impossibility of back-driving a worm gear would pretty well prevent the thing from falling.

RE: Worm gear system question from a EE!!

(OP)
Handleman, you got it. The worm gear is mounted in the base of the pylon and creates a self-locking mechanism at each end of the rail. Still to be tested to see how well that works, but that is the plan. If this does not work, a locking mechanism will be introduced into the base of the pylon to lock it into position at the extremes.

There are four states of the mechanism:
1). Fully down and locked via the worm
2). Fully up and locked via the worm
3). Traversing from down to up
4). Traversing from up to down.

The rail system is mounted into a structure that holds everything in place.

I think my plan is going to have to be to buy a drive motor with maximum sustainable, countinuous operating torque of 70 in oz and test the mechanism via monitoring the current drawn by the motor and then adjust from there.

RE: Worm gear system question from a EE!!

I think that if you keep the rear pin at a low friction level ( the front pin is not a problem since it is directly under the sprocket), your potential for lockup is zero; even as you have it, with a friction coefficient of say .1, it still don't think it will lock assuming your sketch is reasonably accurate
In order to show this (not considering the weight of the motor assy) you find the normal forces on the pins,approximately 12.5/distpins*1 lb , 4 lb. Now putting the pylon in equilibrium, the forces on it are:
1) normal rail  forces of the pins, 4 lb
2) tangential (to the rail) pin friction forces, say mhu*4=.4lb assuming a static coefficient of .2
3) tangential force at sprocket bearings (reaction force from sprocket on rail)to be found)
4) weight of motor assembly
5) 1 lb weight at end of pylon
At each position, you take moments about the bearing centerline and note the force vector ( rear pin) normal to rail + static friction perpendiclar to that normal. If that force is in a position to rotate the mechanism in the proper direction , the system won't lock up. If the static friction is high enough there is a potential for lockup.
The maximum torque on the motor is in the extreme horizontal position and is found from the energy eq
Fdy+friction forces*ds= Tm*d@ and
Tm=Fdy/d@+friction*ds/d@
dy/d@  = about 12.5
I estimated d@/ds=1
F = 1 lb
Tm = 1*12.5+.4*1 +.4*1 =13.3 lb in approximately
If you include the weight of the assembly then friction will add modestly to this.

where:
@ is the incremental angular motion
ds is the incremental motion along the rail corresponding to d@.
Tm is maximum motor torque required
friction forces-mhu(dynamic)*Fn for each pin pins

RE: Worm gear system question from a EE!!

(OP)
zekeman, many, many thanks. That's the data I was looking for. I will look at numerous positions across the traverse but as I suspected, the worst case position is when the arm is horizontal and you have provided what I need to make the calculations.

The drawings are very accurate in scale positioning of pin, sprocket and weight location so your estimate really helps me make furhter decisions relating to impact of increasing the weigh at the end of the pylon and its effect on the mechanism as a whole.

Again, very much appreciate your effort and time.

Thank you.

RE: Worm gear system question from a EE!!

Hi Skyman2

Can you post a pick with some dimensions between the guides and also to the drive gear.
also a pic or sketch giving the guide pin dia's and the slots there inserted into.
I ask because I tried to superimpose on one of your diagrams
the reactions of the guide pins with a torque on the drive of 4.5Nm.
Where does the 1lb mass come from is it just the weight of the arm or is the arm carrying a component of that weight.

regards

desertfox

RE: Worm gear system question from a EE!!

(OP)
See attached.

The 1lb mass is on the top of the pylon. The pylon itself does have weight but only about 6 oz or so and evenly distributed over its length.

The 1lb mass acts somewhere behind the vertical pylon line but not very far - say 20mm or so. Hopefully you can see the details and there are enough of them for the calculation.

I am going to be traveling next week and most of the week after so will not have access to this PC to post further data but can on my return if needed. I will however, be able to view your responses during my travels.

Thanks desertfox - very much appreciated.

RE: Worm gear system question from a EE!!

Hi Skyman2

Thanks for the post I'll see what I can do

desertfox

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