I'd like to get a conversation going around gear manufacturing. As this doesn't apply to any other board, I think it's an advanced topic (if not, I apologize)

Here are some of my assumptions: a 4.11 gear ratio is made by using a 39/9 ring and pinion combo and a 4.10 is made by using a 41/10 combo.

If so, and not particular to any of these ratios, how did the standard of the # of teeth come around? Why were 39/9 or 41/10 # of teeth chosen? Is it a compromise in manufacturing or a strength issue.

Here's my thought, would adding more gears actually help strength and help reduce noise? So instead of 39/9 why not 78/18? or 82/20. The end result will still be a 4.10 or a 4.11 and it would appear you'd have more contact across more teeth?

Got the idea after I hung out at my uncle's machine shop and watched all the machines he has going (end mill business)

 

Very interesting question.

Some thoughts:

Everything is a compromise: strength, economy of manufacture, tooth design, etc. Most RWD gears are HYPOID spiral bevel gears which means 1) the teeth are curved, not straight (spiral), 2) the axes of the gears run at a (90 degree) angle (bevel), and 3) the axes or centerlines of the gears do not intersect (hypoid).

Dropping the pinion axis below the ring gear axis allows a lower floor pan or tunnel. That's the main reason for hypoid gears. Ford Model T's A's up thru '48s (I think) had straight bevel gears with intersecting centerlines. These were cheap to manufacture, but not really strong, but neither were the engines!

Spiral gears have more than one tooth in contact at a time while straight tooth bevel gears have one or at least fewer than a spiral. This spreads the load out and makes the gearset stronger for a given size. Spiral gears are usually quieter running also.

Hypoids have at least as much contact a spirals, but are more difficult to machine, or "generate". Usually a cut is taken on the front of each tooth, and another cut on the rear of each tooth. Both sides of the teeth are in contact, with one side loaded during "drive" and the other side loaded during "coast".

Now if there were twice as many teeth, cutting time would be almost twice as long, so costs would go way up. Additionally the complex cutters would be narrower and have more cutting teeth and would be weaker and more expensive to build and resharpen, and couldn't take as deep a cut, so costs would again climb. How much do you want to pay for your R&P?

Rear gears are not hardened all the way through: they have a "case" or fairly hard shell on the teeth with a core that is medium hard (maybe like a grade 8 bolt or a tad more), and very tough; it'll bend or deform some before it cracks or breaks. With smaller teeth, the .030+ thick case is a higher % of the tooth thickness so core strength decreases, and overall gear is weaker. Some drag gears are made from 9310 steel vs. 8620 for most others. The 9310 has a softer core and maybe a slightly softer case also so it can withstand impact loads better, but it wears out rapidly. These are not really street gears.

The minimum number of teeth on a pinion to maintain reasonable strength is maybe 7. So it you are running a 6.42 at a NASCAR short track, that's a 45:7. Keeping the ring gear tooth count between say 37 and 45 makes the teeth fairly strong, and able to be generated with the same tooling. Economy again. If you/ve ever compared a 7 tooth to a 15 or 17 tooth pinion you can see the apparent strength difference.

BTW, I think you meant 37:9 for 4.11's.

FWIW, you rarely see exact multiples like 3.00 : 1, a 39:13 or 42:14.

Bonus question: Why is that?

My $.02

 

Must have missed the first question....

But that "bonus" one's easy....
To keep the ring and pinion from developing a regular wear pattern. If it were an "even" ratio, the same teeth on the pinion would always mesh with the same teeth on the ring gear..... gearset life would diminish.

Rarely seen... yes, but there are some out there.
Oh yeah, whats the prize?????

Love gears, use to write some nice little gear programs every once in a while..... here's a question for you.

Gear "A" has a mass of 20 kg and its centroidal radius of gyration is 150 mm. The mass of gear "B" is 10 kg and its centroidal radius of gyration is 100 mm. Calculate the angular acceleration of gear "B" when a torque of 12 N·m is applied to the shaft of gear "A". Let's neglect friction.

-Mindgame

 

Thanks for playing Brooke.

Incorrect answer though,
and the units should be rad/sec^2

-Mindgame

 

First question was in another thread.

Yep, the pinion only gets intimate with a few of the ring teeth. Not sure about the gearset life diminishing, but it might develop an interesting noise pattern. If you take it apart and reassemble with different teeth meshing, it can get noisy. It's usually avoidedb, but I have seen 9 in. 3.00s. Some guys need ratios every tenth. 3.00, 3.10, 3.20, etc.

Your question:

Let's see, at 30 kg (66 lb) of ring and pinion gears, since that's the topic of the thread, I'm guessing either a Top Fuel or medium sized truck gears, right? 12 N.m on this gearset is probably less than 1/10% of it's design load. It would probably take more torque than that just to rotate the gears to set them up if friction was considered. How can we neglect friction on a hypoid gearset this size running only at 12 N.m?

Actually I have no clue as to the angular acceleration. Would it not depend on the gear ratio? Guess not, or that would be in the "givens". I don't get the cookie, huh?

 

with your question on strength vs number of teeth, the more teeth you have on the ring or pinion, the thinner you are going to have to make them to make room for the additional teeth. and the thinner the teeth, the weaker it is.

 

I’m fairly certain that you didn’t give enough information to compute the acceleration of gear B. You could find the acceleration of gear A, but without actual tooth counts or relative diameters you can’t calculate how fast B is accelerated…
(well, that wasn't worded right but you get what I'm trying to say, I hope)

I supposed that you could assume that the radius of gyration is proportionate to the actual diameter of the gear, but that would rarely be the case in real life.

On the original question I would bet that the reason is a cost/benefit thing. The fact is that it’s probably cheaper for most manufacturers to step up to the next larger axle (say up the ring gear size ½” or something) then to get marginally better strength by spending much more money in machining costs

 

Angular acceleration is not dependent on the tooth count but I did forget to mention the contact radius I used to make the calcs, lol. Sorry bout that guys.

rA=240mm
rB=180mm

-Mindgame

 

Ah, plane kinetics of rigid bodies, how fun.

Your answer: aB = 25.5 rad/s^2

Direction of rotation would also be imperative to an answer like this.

Take care

 

if you don't mind, could you explain how you got that answer? i'm stuck on a similar question to that

 

You'll never see some of these people on here again, sadly.

 

This is exactly right. I believe the reason you don't have the greater number of teeth is because it's a tradeoff between durability and space limitations. If you are running more teeth to gain the same ratio, you will have to either make room for the same size of teeth (physical dimensions), or you have to make the teeth smaller to fit in the same amount of space, which in turn you have weaker teeth. More teeth would be overall stronger (assuming same size of teeth), but then you have to make up for it somewhere. Secondarily, when you have more teeth and have to make up for it in size, you have more mass, which means you lose some to parasitic losses.

 

If this is the case then how come having more splines (teeth) on items such as input/output shafts and rear axles equate to stronger parts?

 

when they have more splines the shafts are larger diameter

 

And, all the splines are engaged.... not just a few, like the gear teeth on a ring and pinion.