Worm gear system question from a EE!!

[Skyman2]

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

 

[zekeman]

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

 

[Skyman2]

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.

 

[desertfox]

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

 

[Skyman2]

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.

 

[desertfox]

Hi Skyman2

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

desertfox