# Engineering anomalies



## Dranreb (Aug 30, 2013)

*Engineering anomalies...need enlightenment please explain.*

Pondering on how two gear teeth became broken on this shaft, it seemed to me that there was a strange miss-match between the two pins involved.

On one end the quill feed handle is held on by a huge pin, but the gear is held on by a relatively tiny taper pin which is obviously up to the job as it hasn't failed under obvious abuse. 

So my question is what is the point of the huge pin the other end, is this just to look butch and strong to impress buyers, or is it a necessity?




All a bit strange to me...suggestions welcome..

More anomalies to follow 

Bernard


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## RandyM (Aug 30, 2013)

Are you sure the gear and the sleeve are the same part? My guess is they are seperate pieces and just look like one.


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## Dranreb (Aug 30, 2013)

Sorry randy, didn't describe that well did I...the shaft is all one piece with a separate gear against a shoulder on that shaft, hence the taper pin to prevent it rotating.


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## Walt (Aug 30, 2013)

*Re: Engineering anomalies...need enlightenment please explain.*



Dranreb said:


> Pondering on how two gear teeth became broken on this shaft, it seemed to me that there was a strange miss-match between the two pins involved.
> 
> On one end the quill feed handle is held on by a huge pin, but the gear is held on by a relatively tiny taper pin which is obviously up to the job as it hasn't failed under obvious abuse.
> 
> ...



My uninformed guess: the gear-shaft-handwheel assembly was designed by committee. Or the one you have is an update of an earlier design. Perhaps the original had the gear+shaft as a single part, joined to a handwheel via a rather large pin. Then some bright intern came along and said, "Hey, we can save $0.15 by machining the gear as a sleeve and this little pin will me more than strong enough to keep it from spinning on the shaft."

Walt


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## OldMachinist (Aug 30, 2013)

My guess is that the gear was originally made on the shaft and was damaged. Someone machined off the original gear and pinned on a new one. The large pin may also not be original.


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## FanMan (Aug 30, 2013)

The small pin holding the gear is actually taking less load if I understand correctly, because the shear faces are on a smaller diameter... though not enough to explain the large difference.

I suspect the gear pin is sized to be strong enough, while the larger pin on the handwheel is also intended to take abuse, odd direction loading when people push and pull on it, and for it to survive repeated removal and reinstallation.

Or the manufacturer just had a bunch of handwheels with big holes in stock...


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## Dranreb (Aug 30, 2013)

Walt said:


> My uninformed guess: the gear-shaft-handwheel assembly was designed by committee. Or the one you have is an update of an earlier design. Perhaps the original had the gear+shaft as a single part, joined to a handwheel via a rather large pin. Then some bright intern came along and said, "Hey, we can save $0.15 by machining the gear as a sleeve and this little pin will me more than strong enough to keep it from spinning on the shaft."
> 
> Walt





Great theory Walt, and one that had passed though my brain also, but I've had a look at a few parts lists and it seems the gear was always a  separate part, but sold already pinned to the shaft, although some  ambiguity exists....I did find an early hand drawn parts list with the taper pin clearly marked but couldn't copy it. We are talking Atlas here, who were not well known  for wasting ink paper or metal!:LOL:




OldMachinist said:


> My guess is that the gear was originally made  on the shaft and was damaged. Someone machined off the original gear  and pinned on a new one. The large pin may also not be original.



Makes sense OM considering this machine has had a hard life, but the separate gear and large pin are definitely original spec.














FanMan said:


> The small pin holding the gear is actually taking less load if I understand correctly, because the shear faces are on a smaller diameter... though not enough to explain the large difference.
> 
> I suspect the gear pin is sized to be strong enough, while the larger pin on the handwheel is also intended to take abuse, odd direction loading when people push and pull on it, and for it to survive repeated removal and reinstallation.
> 
> Or the manufacturer just had a bunch of handwheels with big holes in stock...



That's most likely as near enough as makes no odds FM, maybe someone has the calcs for these things handy?

I had thought maybe the small pin was originally a shear pin, which would  save the gear, but then there would be spares listed and they're not...

Thanks for the input chaps, not sure why this has become an obsession, this forum has done this to me, is there an antidote? :biggrin:

Bernard


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## pdentrem (Aug 30, 2013)

Engineering people over engineer gizmos all the time. The manufacturing dept is told to make the gizmo for as cheap as possible. One way is not re-inventing the wheel. One way to do so is to use what is at hand and save money for the bean counters.
Likely used a handle from another machine or old stock patterns.
Pierre


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## epanzella (Aug 30, 2013)

The loads on the two pins are very different. They are both resisting the torque applied to a lever but the levers are different lengths. The lever length at the pinion end is from the shaft center to the center of pressure on the pinion teeth. Looks to be about 3/4" or less. The lever length on the handle end is from the center of the shaft to the center of the ball ends. Looks to be eight or ten inches. So the load difference on the two pins is directly proportional to the length difference of the two levers. I'm guessing the ratio is about ten to one.


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## FanMan (Aug 31, 2013)

epanzella said:


> The loads on the two pins are very different. They are both resisting the torque applied to a lever but the levers are different lengths. The lever length at the pinion end is from the shaft center to the center of pressure on the pinion teeth. Looks to be about 3/4" or less. The lever length on the handle end is from the center of the shaft to the center of the ball ends. Looks to be eight or ten inches. So the load difference on the two pins is directly proportional to the length difference of the two levers. I'm guessing the ratio is about ten to one.



Actually, no.  The loads are only slightly different, according to the shaft diameter, since you can't put any more torque on the shaft via the handwheel than the pinion can react (plus the torque from the return spring, which is minor by comparison).


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## epanzella (Aug 31, 2013)

FanMan said:


> Actually, no.  The loads are only slightly different, according to the shaft diameter, since you can't put any more torque on the shaft via the handwheel than the pinion can react (plus the torque from the return spring, which is minor by comparison).


Essentially you have a shaft with a 3/4" lever at one end and about a 9 inch lever at the other end. The short end enjoys a 10 to 1 mechanical advantage.


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## Codered741 (Aug 31, 2013)

epanzella said:


> Essentially you have a shaft with a 3/4" lever at one end and about a 9 inch lever at the other end. The short end enjoys a 10 to 1 mechanical advantage.



While the gear and handle act as a second class lever, which will give you a mechanical advantage (from your example actually a 12:1 advantage), the forces on the shaft are the same.  However, the shaft is acting on the pins at different radii, so the forces on the pins are slightly different.  

Because the shaft is acting as the fulcrum in the lever, it encounters torque developed by the handwheel/return spring.  This torque is expressed in Ft/Lbs, metric N/m, and is calculated by multiplying the force applied perpendicular to the handle, to the distance at which it is applied.  Eg. you have a 1' long handle attached to a shaft, and apply 10 lbs of force to the handle, you will create 10 ft/lbs of torque on the shaft that it is connected to.  (1ft * 10lbs = 10ft/lbs)

If you were to exert force closer to the shaft, say at 8in from the center, you would develop less torque.  8in * 10lbs = 80 IN/lbs / 12 = 6 2/3 Ft/Lbs.  This torque is then transmitted undiminished, assuming the shaft is sturdy enough to handle the loads, for the entire length of the shaft.  

If you want to find out the force that would be acting on the gear, you simply run the calculation in reverse.  Continuing with our example, 10Ft/lbs, we will say that the gear OD is 3/4" (you would actually calculate this with the pitch diameter).  Therefore, 10Ft/Lbs * 12 = 120IN/Lbs / 0.75IN = 90Lbs.  

This is the number that you would use as the force when calculating the necessary cross-section for shear strength.  The shear strength of mild steel is classified by ASTM as a minimum of 36,000 psi, and say our pin is 0.125" dia. The cross-sectional area of this pin is 0.012in^2.  This number will be doubled, as the pin passes through the shaft twice, allowing twice the area to resist the force.  Therefore, 
0.024 in^2 * 36,000psi = 864Lbs.  

This is the double shear strength of the pin.  However, we have one more step to take in our calculations, as the pin is reacting to torque, not a direct force.  To find the maximum allowable force on the pin, we need to calculate what torque would be developed on the shaft, in order to develop 864lbs at the shear plane.  

First we need the diameter of the shear plane, say 0.375in (3/8" bore on a 3/4" gear seems reasonable).  So we just do our calculation again, using 864lbs at 0.1875in, calculating the force at 1ft.  

.1875in * 864lbs = 162 in/Lbs / 12 = 13.5lbs.  

This is how much force you will need to apply to the handle to break the pin.  

As you can see, when the shear diameter decreases, the forces acting on the part increase.  I am not saying that these numbers are correct, as I do not have all of the dimensions of the actual shaft, or the actual material types used.  If the pin was an alloy of steel, say 4130, even in its normalized and annealed state, its shear strength will increase to 70,000psi, making the applicable force 1680lbs, force at the handle being about 26lbs.  

As for the reason for the mis-matched pins, I would guess it was probably for ease of assembly, as the gear was made in a machine shop, with craftsmen who dealt with those pins all the time, and the handle was put on by an assembly line worker, or even the consumer, to make it easier for assembly.  

-Cody


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## Dranreb (Sep 1, 2013)

Those are some interesting thoughts on this subject folks, thanks all for taking the time to reply!

 Cody, your detailed explanation is just what I was after, much appreciated, although I had to Google 'second class lever' 
	

		
			
		

		
	




I'm quite surprised at just how strong the taper pin is, it's nice to know that I can hang on the handle like gorilla a when a drill bit starts to screech and turn blue without it failing on me...:biggrin:

Bernard


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