Will the gear tooth break?

[Matt Adams]

I confess that my 'edit' about the two CIMs working on different teeth had poor logic - they're running on different teeth on the 45 tooth gear, but presumably the 13 tooth gear will have one contact point.

Since Dr. Joe seems to have a bit more confidence, from experience in this working well than I do, I want to take another swing just to see what the numbers show.

First step I think is to be a bit more conservative with the torque these gears should actually see. I'll assume that you'll design the max torque to be right around 40 Amps, which I calced out to be about 14 in-lbs from a pair of CIMs. After a 45:12 ratio, this becomes about 53.25 in-lbs.

Using this number, and the standard AGMA gear equation

Lewis Form Factor (Y) for a 14.5 degree PA gear = 0.223
Radius = 13/20 / 2 inches = 0.325 inches
Force on Tooth (Wt) = 53.25 / .325 = 163.85 lbs.
Face Width (F) = .375
Pitch (P) = 20

Stress = (Wt * P) / (F * Y)

I have this showing 38,266 PSI, or a factor of safety of 1.04 (at a megar 40,000 PSI)

I think you have a few factors to consider:

Positive Effects:

• Run time of a FIRST drive train can typically be measured in hours (relatively short).
• The impact loading on the gears in your drive train isn't too substantial.

Negative Effects:

• You could potentially apply more torque before your 40 A breakers trip if something down the line isn't geared to peak out at 40 Amps.

Another Note:

A roughly equivalent pinion in the FIRST kit drive train has a face with of 0.809 inches. This has been designed to take a single CIM motor's peak torque output with an overall reduction of (50/14 * 50/14 * 50/14) or a totally reduction of 45.5 times.

However, since you have 2 times as many motors, your face width is 2.16 times smaller, and your overall reduction is 3.75, this gear will see (14/12 = 1.16) 1.16 times more torque due to their smaller tooth count, and the kit material (assuming 100ksi yield) is 2.5 stronger than plain steel.

This means the strength of this gear, relative to the one in the drive train, based on application, is (2 * 2.16 * 3.75 * 1.16 * 2.5 = 46.98 ) about (45.5/46.98) 0.96 times as robust. However, there's presumably some factor of safety built in to the kit, so being withing 4% is probably well within their factor safety.

Hence, with average steel, I'm still going to call this border lined, but probably more above the border than below. It sounds like Travis is proposing some stronger material, which is probably the easiest fix.

Good luck,

Matt

[Veselin Kolev]

I had a situation like this two years ago. Lets just say, the calculations all worked out to about a 1.7 safety margin, with a 12 tooth 20 pitch hardened 3/8" face gear. Yet, somehow, the teeth still magically disappeared. Later, we found out that the gears were running misaligned. If properly machined and screwed down, your drivetrain should be fine. However, if I were you I would still make a few spares...

[Joe Johnson]

Originally Posted by Matt Adams

Using this number, and the standard AGMA gear equation

Lewis Form Factor (Y) for a 14.5 degree PA gear = 0.223
Radius = 13/20 / 2 inches = 0.325 inches
Force on Tooth (Wt) = 53.25 / .325 = 163.85 lbs.
Face Width (F) = .375
Pitch (P) = 20



Matt

The Lewis form factors are very very conservative for an intermittent duty or short life application (such as a FIRST robot). I have designed many production actuators for automotive applications (power liftgate actuators, power sliding door actuators, power lock/unlock actuators, power cinching latches, power cinching strikers, power unlatch actuators, power window actuators, etc.) I have never had a case where the Lewis forumulas gave me useful data.

I have had good success approximating a gear tooth as a uniform beam with its cross section equal to the cross section of the base of the tooth and its length equal to the height of the tooth with a load applied at the end of the beam.

Pcircular = 2 Pi/P = .31"
Circular Tooth Thickness = Pcircular/2 = .16
Tooth Height = 2.25/P = .11

This can be approximated as a beam
L = .11 in
H = Circ Tooth Thickness = .16 in (this is conservative but a first order approx)
T = .375 in
F = 164 lbsf
I = (1/12) T * H^3 = (1/12) * (.375) * (.16)^3 = .000128 in^4
c = H/2 = .08


Going to the efunda.com site I linked to above, I get the
Stress = 11,000 psi



Unless your gear is plastic, I don't think you have to worry about 11,000psi stress levels.



Joe J.

[Matt Adams]

Originally Posted by Joe Johnson

The Lewis form factors are very very conservative for an intermittent duty or short life application (such as a FIRST robot). I have designed many production actuators for automotive applications (power liftgate actuators, power sliding door actuators, power lock/unlock actuators, power cinching latches, power cinching strikers, power unlatch actuators, power window actuators, etc.) I have never had a case where the Lewis forumulas gave me useful data.
Joe J.

Hmph. Sometimes I lothe what academia has (not) taught me.

Thank you, Joe!

Matt

[Matt Adams]

Originally Posted by Joe Johnson

Pcircular = 2 Pi/P = .31"
Circular Tooth Thickness = Pcircular/2 = .16
Tooth Height = 2.25/P = .11

This can be approximated as a beam
L = .11 in
H = Circ Tooth Thickness = .16 in (this is conservative but a first order approx)
T = .375 in
F = 164 lbsf
I = (1/12) T * H^3 = (1/12) * (.375) * (.16)^3 = .000128 in^4
c = H/2 = .08

Joe J.

I confess I double checked your circular tooth thickness. Perhaps we're just using different terminology, but if you mean the thickness of the gear tooth on the circular pitch, I think the formulas are different.

I have:
pCircular * Pdiametral = pi

and
Circular tooth thickness:
t = pi / (2 * P) or 1.5708 / P

So I get a tooth height (thickness) of (1.57 / 20) = 0.07854 in. or exactly half of what you found.

Calcs:
I = (1/12) T * H^3 = (1/12) * (.375) * (.07854)^3 = 0.0000151 in^4
c = H/2 = .0392

Hence, the stresses are substantialy higher.

I know that there's lot of terminology floating around, are we just using a differnent lingo?

Matt

[Joe Johnson]

Originally Posted by Matt Adams

I confess I double checked your circular tooth thickness. Perhaps we're just using different terminology, but if you mean the thickness of the gear tooth on the circular pitch, I think the formulas are different.

I have:
pCircular * Pdiametral = pi

and
Circular tooth thickness:
t = pi / (2 * P) or 1.5708 / P

So I get a tooth height (thickness) of (1.57 / 20) = 0.07854 in. or exactly half of what you found.

Hence:
I = (1/12) T * H^3 = (1/12) * (.375) * (.07854)^3 = 0.0000151 in^4
c = H/2 = .0392

and well... to put it lightly, the stresses are substantial and it doesn't work out so well.

I know that there's lot of terminology floating around, are we just using a differnent lingo?

Matt

Matt,

Good, thanks for checking the calcs.

Nope, no new lingo, I just put an extra factor of 2 in my formula.

The correct formula is Pcirc = pi / Pdiametral not "2 pi" as I said in my note.

Recalculating we get the stress as:
45,000psi -- which is to be expected because the stress goes up like the square of the beam thickness so we would expect 4X the stress for halving the thickness.

45Kpsi is highish but not too bad. While it is higher than a typical Low Carbon steel, it is sort of middle of the road for a good steel.

If all you can use in low carbon steel, I think it would be fine if you could case harden the pinion (any reasonable depth of case hardening will effectively harden the high stressed areas of the tooth -- the edged of a beam take the highest stress -- if you can get the upper 1/3 and lower 1/3 hardened the fact that the middle is soft will not matter very much because the stress level in the middle 1/3 is so much less than the outer 1/3's).

Also, note that there is conciderable safety factor even in this calculation since the geartooth is not a beam but closer to a parabola and the force is not all on one tooth AND at the tip. So... ...you've got some margin even in this calculation, though I would not depend on it too much. My experience is that if the material can take the stress you calculate in this manner you will be fine and not have too much safety margin in the design.

Joe J.