Calculating optimal weight efficiency ratios

[asid61]

Here's an interesting problem I thought of today:

Given a gearbox with 2 or more stages, in general the first stage of the gearbox tends to have the best bull gear teeth-to-reduction ratio, because the motor pinion is usually much smaller than the pinions for the other stages.
However, if you make the bull gear (downstream mating gear) for the first gear larger by 1 tooth, you vary the pitch diameter, not the area.
The weight of a gear is almost directly proportional to the area, so at what point does it become more weight-efficient to use a second stage in a gearbox, disregarding the weight of shafts, bearings, and plate? What is the optimal ratio between the number of teeth in the bull gears?

The variables in play are the first stage pinion teeth or pitch diameter, and the second stage pinion teeth or pitch diameter. I have a suspicion the optimal tooth relationship between first and second stage bull gears is equal to the ratio between the squares of the pinion pitch diameters, but I'm not sure where to start to work this out. Does anybody have any ideas?

[GeeTwo]

While weight of the bull gear would eventually be an issue, the decision to go from single-stage to double (or triple) stage is more commonly driven by geometric considerations (the SIZE of the gearbox, esp. relative to the wheels). Pinions for most of the FRC motors typically have 10-15 teeth. Smaller numbers of (spur gear) teeth would result in poor pressure angles. Gears with more than about 50 or 60 teeth (again, in FRC-typical sizes) are often larger than the wheels they drive, leading to untenable gearboxes. That is, at somewhere around 3.5:1, a two-stage gearbox is preferable to drive wheels of 4" or larger. At somewhere around 14:1, a three-stage gearbox is preferable to drive 4" or larger wheels.

With planetary gearboxes, these ratios are somewhat different, of course. Planetary gearboxes can reduce up to 10:1 (theoretically a bit higher, but no more than 14.9:1) per stage.

[asid61]

Originally Posted by GeeTwo

While weight of the bull gear would eventually be an issue, the decision to go from single-stage to double (or triple) stage is more commonly driven by geometric considerations (the SIZE of the gearbox, esp. relative to the wheels). Pinions for most of the FRC motors typically have 10-15 teeth. Smaller numbers of (spur gear) teeth would result in poor pressure angles. Gears with more than about 50 or 60 teeth (again, in FRC-typical sizes) are often larger than the wheels they drive, leading to untenable gearboxes. That is, at somewhere around 3.5:1, a two-stage gearbox is preferable to drive wheels of 4" or larger. At somewhere around 14:1, a three-stage gearbox is preferable to drive 4" or larger wheels.

With planetary gearboxes, these ratios are somewhat different, of course. Planetary gearboxes can reduce up to 10:1 (theoretically a bit higher, but no more than 14.9:1) per stage.

This is a purely academic exercise, not really one that's practical.
I totally agree with you though. I'm working a geared flipped-cim gearbox right now, and my main constraint is the combined area of the gears and the mounting on the 2x1, not the weight.
I usually run 2 stages for 14:1, something like a 12:50 x 14:50, but it varies based on the application.

Why do you say a planetary can only go up to 14.9:1? The formula should just be R/S + 1, and as R (teeth on ring) gets larger it better approximates a gear rack, so S (teeth on sun) can get smaller while avoiding interference.

[sanddrag]

In my 15 years in FIRST, most of which consisted of reading Chief Delphi and designing gearboxes, and my 5 years spent studying mechanical engineering, I have never once heard the term "bull gear." Is this some kind of new slang or do I just live under a rock? Would someone care to fill me in?

[asid61]

Originally Posted by sanddrag

In my 15 years in FIRST, most of which consisted of reading Chief Delphi and designing gearboxes, and my 5 years spent studying mechanical engineering, I have never once heard the term "bull gear." Is this some kind of new slang or do I just live under a rock? Would someone care to fill me in?

I heard Joey from 192 use it once. I have never been sure what to call the downstream mating gear in a gearbox, so bull gear fits nicely.

[GeeTwo]

Originally Posted by sanddrag

In my 15 years in FIRST, most of which consisted of reading Chief Delphi and designing gearboxes, and my 5 years spent studying mechanical engineering, I have never once heard the term "bull gear." Is this some kind of new slang or do I just live under a rock? Would someone care to fill me in?

The bull gear is the gear which handles the greatest torque in a drive train. I am most familiar with the term in relation to submarine drive trains, but I have seen it in other contexts.

Originally Posted by asid61

Why do you say a planetary can only go up to 14.9:1? The formula should just be R/S + 1, and as R (teeth on ring) gets larger it better approximates a gear rack, so S (teeth on sun) can get smaller while avoiding interference.

See this thread. Essentially, near or beyond this limit, the sun gear is too small to engage the planetary gears. (I assumed that there are at least three; if you only have two, you can go higher).

[chadr03]

If weight is your only concern larger gears can be machined thinner below the teeth and spoked to be much lighter. A gear rim will typically have significant stiffness and rating if there is at least two tooth depths of solid material below the root but beyond that most of the material is not very highly stressed. So you can get quite high ratios without significant addition of weight when compared to a solid gear.

That being said a limitation in large reduction gearing is deflection across the pinion under load. The larger the ratio the the pinon selected will be as small as possible causing it to be relatively flexible. When the pinion deflects all of the loading will move toward one end of the tooth effectively reducing the face width.

[asid61]

Originally Posted by GeeTwo

The bull gear is the gear which handles the greatest torque in a drive train. I am most familiar with the term in relation to submarine drive trains, but I have seen it in other contexts.


See this thread. Essentially, near or beyond this limit, the sun gear is too small to engage the planetary gears. (I assumed that there are at least three; if you only have two, you can go higher).

"You must spread some Reputation around before giving it to GeeTwo again"
I never considered that particular issue before, very neat! Seems obvious in retrospect...
Of course one could solve it by placing the gears in front or behind each other, but that seems useless when you could just stack two stages.

Originally Posted by chadr03

If weight is your only concern larger gears can be machined thinner below the teeth and spoked to be much lighter. A gear rim will typically have significant stiffness and rating if there is at least two tooth depths of solid material below the root but beyond that most of the material is not very highly stressed. So you can get quite high ratios without significant addition of weight when compared to a solid gear.

That being said the main limitation in very large reduction single stage gearing is deflection across the pinion under load. The larger the ratio the greater the gear forces and the larger the gear forces the more deflection along the shaft. When the pinion deflects all of the loading will move toward one end of the tooth effectively reducing the face width. If there isn't significant loading then large reductions can be made with a single stage. If there is large loading deflection starts to become a real issue when you start getting much higher than the 4:1 ratios that are common in single reduction boxes.

If I lighten the gears, then the problem becomes a lot more complex due to the need to calculate the volume of leftover material with respect to the radius of the gear needed to preserve strength. I'm really looking for a mathematical solution here, not a realistic one.

Correct me if I'm off, but shouldn't the pinion tooth load only be proportional to the mechanical power the motor is outputting and the pitch radius of the pinion, and be independent of the gear ratio?

[FrankJ]

Originally Posted by asid61

Correct me if I'm off, but shouldn't the pinion tooth load only be proportional to the mechanical power the motor is outputting and the pitch radius of the pinion, and be independent of the gear ratio?

Right concept, but wrong terms. Actual it is related to torque. A brushed DC motor at stall is transmitting maximum gear load but no power. The actual force transmitted is the same on both the big gear and small gear. The small gear wears faster because there are fewer of them. There is a practical limit on how small of a gear you can make where the geometry falls apart.

[chadr03]

Originally Posted by asid61

Correct me if I'm off, but shouldn't the pinion tooth load only be proportional to the mechanical power the motor is outputting and the pitch radius of the pinion, and be independent of the gear ratio?

I need to quit posting after its time for be in bed and my brain shuts off. You are correct the gear loading isn't a function of the ratio. The real issue with deflection is typically due to having to use a small pinion and large gears. The small pinion is quite flexible as was stated earlier there are practical limitations on how small a pinion can be. I edited my previous post to correct the errors.

[FrankJ]

Originally Posted by sanddrag

In my 15 years in FIRST, most of which consisted of reading Chief Delphi and designing gearboxes, and my 5 years spent studying mechanical engineering, I have never once heard the term "bull gear." Is this some kind of new slang or do I just live under a rock? Would someone care to fill me in?

Bull gear is an old term. It is the larger of the two gears in a gear set. The smaller gear is called the pinion. I suspect the term has its roots in the conversion from animal to machine power. IE the high torque output shaft would have a large slow turning "bull gear" on it.

[hrench]

Originally Posted by sanddrag

I have never once heard the term "bull gear." Is this some kind of new slang or do I just live under a rock?

This was me too!

After a quick Google search, I feel pretty stupid that I didn't know the word.

This seems mostly like a great question, but the constraint:

Originally Posted by asid61

disregarding the weight of shafts, bearings, and plate?

seems to ruin this question. When the bull gear is really large, you don't need another gear pass, which if we recall, each gear pass consumes 1-2% (or more) of your power, decreasing your efficiency. So I can make any ratio I want--even 1000:1 with only one gear pass, but the size of the transmission would be silly. So you can't disregard that stuff.

Also, each gear pass consumes dollars. Fewer passes are less money.

I remember the year we used andymark's "cimplebox"http://www.andymark.com/CIMplebox-p/am-0734.htm which only has one gear pass in it. I was hoping that the gain in efficiency would help to compensate for a really bad gear ratio (4.67 to 1 max). We ended up with an amazing top speed but it was a terrible robot to position for frisbees.

Also don't forget that most of us that are designing transmissions are still using off-the-shelf gears from robotics vendors, but if we were custom designing gears, we could make them look like the beauties in the Chairman's trophy,
http://www.chiefdelphi.com/media/img...3509e9d6_l.jpg
with spans and webs and lots of weight reduction.
But for the complexity of obtaining your own gears, the weight reduction just isn't worth it.

Also, a 'real' gear designer will vary the modulus of elasticity (materials) of the gears in the train so point contact doesn't increase wear. Never see robots do this. We use steel gears.

So though I'd like to see some kind of comparison, but with no changing material weight and no lightening holes in the gears, it would be hard to be fair in the comparison. I think multiple passes would win every time because the bull gears would be so heavy.

A funny aside...once a supervisor asked me to put some 'lightening' holes in a part and I drew a blank. I responded "how big of hole does lightning make?"

[asid61]

Originally Posted by hrench

This was me too!

After a quick Google search, I feel pretty stupid that I didn't know the word.

This seems mostly like a great question, but the constraint:



seems to ruin this question. When the bull gear is really large, you don't need another gear pass, which if we recall, each gear pass consumes 1-2% (or more) of your power, decreasing your efficiency. So I can make any ratio I want--even 1000:1 with only one gear pass, but the size of the transmission would be silly. So you can't disregard that stuff.

Also, each gear pass consumes dollars. Fewer passes are less money.

I remember the year we used andymark's "cimplebox"http://www.andymark.com/CIMplebox-p/am-0734.htm which only has one gear pass in it. I was hoping that the gain in efficiency would help to compensate for a really bad gear ratio (4.67 to 1 max). We ended up with an amazing top speed but it was a terrible robot to position for frisbees.

Also don't forget that most of us that are designing transmissions are still using off-the-shelf gears from robotics vendors, but if we were custom designing gears, we could make them look like the beauties in the Chairman's trophy,
http://www.chiefdelphi.com/media/img...3509e9d6_l.jpg
with spans and webs and lots of weight reduction.
But for the complexity of obtaining your own gears, the weight reduction just isn't worth it.

Also, a 'real' gear designer will vary the modulus of elasticity (materials) of the gears in the train so point contact doesn't increase wear. Never see robots do this. We use steel gears.

So though I'd like to see some kind of comparison, but with no changing material weight and no lightening holes in the gears, it would be hard to be fair in the comparison. I think multiple passes would win every time because the bull gears would be so heavy.

A funny aside...once a supervisor asked me to put some 'lightening' holes in a part and I drew a blank. I responded "how big of hole does lightning make?"

Originally Posted by FrankJ

Right concept, but wrong terms. Actual it is related to torque. A brushed DC motor at stall is transmitting maximum gear load but no power. The actual force transmitted is the same on both the big gear and small gear. The small gear wears faster because there are fewer of them. There is a practical limit on how small of a gear you can make where the geometry falls apart.

Wouldn't the rotation speed of the gear also come into effect (hence why I said mech. power)?

Also, I must reiterate- this is a mathematical question, not a practical one. Not really sure how I can make that more clear. I have designed gearboxes before (look through my submitted images) and weight is rarely my constraint. If you don't like the constraints, then I can't help you.

[hrench]

Originally Posted by asid61

Wouldn't the rotation speed of the gear also come into effect (hence why I said mech. power)?

I think you're being fooled because the rotation speed is a backwards-way to correlate torque. I agree with FrankJ, motor torque is the only concern and it doesn't matter the speed. In my gear design handbook, even when figuring the max horsepower a gear train can do, it takes it straight back to Force on the gear teeth, which is all from torque.

Originally Posted by asid61

this is a mathematical question, not a practical one. Not really sure how I can make that more clear

Yeah I understand, but if you're optimizing the weight of the gears, I think you'll still be concerned about the weight of the plates and shafts. Unless you're mostly concerned about the angular momentum of rotating mass, which can be important sometimes.
Yes, you can write excel sheets to compare different areas of gear-sets and it might help us all learn a few things, but you'll have to make assumptions about lightening and material (or probably assume no change on both).
Another interesting side effect of one-pass (or fewer passes) is with a very large bull gear you'll decrease the Force on each tooth appreciably, such that you'll need less gear face width, making the gear lighter in another way. But the interior of the bull gear will still need to be as strong for the torque--you'll only save weight near the circumference.
Multiple passes won't benefit from this, they'll still need wide face-width gears for higher torque.

So you'll need equations for area, but also thickness (which can be varied from face to center) because weight is from volume.

Originally Posted by asid61

I have designed gearboxes before (look through my submitted images) and weight is rarely my constraint. If you don't like the constraints, then I can't help you.

I've looked at your gear boxes and they're amazing and creative. I particularly like the cycloidal and drill press swerve. But it does look like you pay attention to weight.

[FrankJ]

Originally Posted by asid61

Wouldn't the rotation speed of the gear also come into effect (hence why I said mech. power)?

I think we are really quibbling on terms. As you said knowing speed and power will start you on the way. Ultimately we need to get to contact force on the gear tooth which give will you the stress and strain in the material. Way more complicated than my ability to explain in a short post. For FRC (Brushed DC motor) worst case scenario would be at near stall where you are transmitting maximum torque but relatively little power. By comparison at max power the gear set is relatively lightly load.

Other threads on CD are talking about the WCP gearboxes stripping their third stages. I have not seen this personally so I am just speculating. But I think with big wheels, six cim drives, and near stall conditions, we are seeing the limits gear sets we are using.