Gear and gearbox design - circular pitch

[Noobita]

Hi everyone! Just joined the fórum after a lot of hours searching the web and Reading some books...I have one question, didn't found na answer to it...but let's give it a try ;) How can you determine the optimal circular pitch and in consequence the optimal diametral pitch of a gear (increasing its strength)?

I need to design a gearbox (just want to design it for my electric motor) that will reduce the input speed of 11000 to 55 rpm, so my gearbox ratio will be 200:1. The input torque is about 20mNm. I'm thinking about a 5/6 stage gearbox (don't know if this is the appropriate designation, the gearbox will have 5/6 gears, counting the pinion in motor as a gear and the shaft as a gear).

To know the required number of teeth, as the gearbox will have space constraints (30x40mm) how can i know determine the circular pitch of each gear?

And do you have also some recommendations? I already saw here in the forum that the gears have a prime number of teeths and that contacting gears shouldn't have the same number of teeths (but gears that are in contacting can have the same number of teeths?) .

Grateful

 

[OldEngr63]

Seems to me that finding circular pitch is getting a bit ahead of yourself. You really should be looking for tooth numbers to get your required gear ratio before you try to design the individual gears.

It is too limiting to say that all gears must have tooth numbers that are prime numbers. Rather, for any pair that are in mesh, the tooth numbers should be relatively prime. This means simply that they have no common factors. Thus, for example, a 27 tooth gear (prime factors are 3,3,3) in mesh with a 28 tooth gear (prime factors are 2,2,7) is quite acceptable; they are relatively prime, even though neither 27 nor 28 are prime numbers.

This looks like a pretty big project. Do you intend to carry it through to hardware, or is this only a paper design?

 

[Baluncore]

An economic gearbox design for a high reduction ratio will have several stages, each stage will have a reduction ratio of between about 3 and 4. You can therefore expect to have n stages where, 200 = 3.5^n, n = 4.229, say 4 stages. With 4 stages each gear pair would have a reduction ratio near √( √(200) ) = 3.76

Remember that power is RPM * torque. Each stage of your gearbox must handle the same power. As the RPM falls through the stages, the torque will rise and so need bigger section or longer gear teeth to handle the torque. Gear tooth section is proportional to force. Select the tooth section needed, put 20 of those teeth around a circle and you have some idea of the DP required.

If the number of teeth falls below about 19 you will start to get undercutting and so weaker teeth. That means your final bull gear will have between 57 and about 73 teeth. Your high input RPM will probably require a different lubricant to the final output stage.

 

[Noobita]

Seems to me that finding circular pitch is getting a bit ahead of yourself. You really should be looking for tooth numbers to get your required gear ratio before you try to design the individual gears.

It is too limiting to say that all gears must have tooth numbers that are prime numbers. Rather, for any pair that are in mesh, the tooth numbers should be relatively prime. This means simply that they have no common factors. Thus, for example, a 27 tooth gear (prime factors are 3,3,3) in mesh with a 28 tooth gear (prime factors are 2,2,7) is quite acceptable; they are relatively prime, even though neither 27 nor 28 are prime numbers.

This looks like a pretty big project. Do you intend to carry it through to hardware, or is this only a paper design?

I'm trying to carry it to hardware. You said that i should look for the teeth number. I have done a preliminar test, and got these numbers (gears are double spur gears): pinion - 13, gear 1 - 29/13, gear 2 - 39/13, gear 3 - 37/13, gear 4 - 37/13, shaft - 48, but as i said, i have space constrains, so i needed to know the circular pitch to see if the gears will fit in the space...so how can i find the optimal circular pitch?

An economic gearbox design for a high reduction ratio will have several stages, each stage will have a reduction ratio of between about 3 and 4. You can therefore expect to have n stages where, 200 = 3.5^n, n = 4.229, say 4 stages. With 4 stages each gear pair would have a reduction ratio near √( √(200) ) = 3.76

Remember that power is RPM * torque. Each stage of your gearbox must handle the same power. As the RPM falls through the stages, the torque will rise and so need bigger section or longer gear teeth to handle the torque. Gear tooth section is proportional to force. Select the tooth section needed, put 20 of those teeth around a circle and you have some idea of the DP required.

If the number of teeth falls below about 19 you will start to get undercutting and so weaker teeth. That means your final bull gear will have between 57 and about 73 teeth. Your high input RPM will probably require a different lubricant to the final output stage.

Thanks for your answer...from what i understood, a stage means a gear (as i thought). You said that the gearbox should have 4 stages, if it has more, there isn't any problema, right? Just more gears, more space ocupied and the cost will be bigger, right? And how can i calculate the tooth section?

Grateful

 

[OldEngr63]

With an involute tooth form, a 13 tooth pinion is almost certain to undercut.

What do you mean by gear 1 - 29/13? Are saying that this is a 29 tooth gear in mesh with your 13 tooth pinion? This is awkward notation!

Are you working in SI or in USCustomary units? In SI, you need to select a trial module and make the calculations to find the size of your gears. In USC units, you need to select a trial diametral pitch and make the calculations to find the gear sizes. If you don't like the result, select another value for module (or diametral pitch) and try again. Design is an iterative process, and rarely to we achieve an optimum. Most folks will settle for a combination that works in the sense that meets all the requirements. Do not neglect the tooth strength calcs in your work. It is very helpful to write a program for all the calcs so that you can iterate with ease.

 

[Noobita]

With an involute tooth form, a 13 tooth pinion is almost certain to undercut.

What do you mean by gear 1 - 29/13? Are saying that this is a 29 tooth gear in mesh with your 13 tooth pinion? This is awkward notation!

Are you working in SI or in USCustomary units? In SI, you need to select a trial module and make the calculations to find the size of your gears. In USC units, you need to select a trial diametral pitch and make the calculations to find the gear sizes. If you don't like the result, select another value for module (or diametral pitch) and try again. Design is an iterative process, and rarely to we achieve an optimum. Most folks will settle for a combination that works in the sense that meets all the requirements. Do not neglect the tooth strength calcs in your work. It is very helpful to write a program for all the calcs so that you can iterate with ease.

I don't know a lot about gears so, with the notation 29/13 i wanted to explain that the double gear as 29 tooths and the the pinion on the gear has 13 tooths, yes. There is a right way? I was working on SI (so all the units are in m).

And what do you mean it will undercut? You mean that i'll have to decrease the circular pitch in the pinion gear and increase in the mesh gear (+0.5)?

Do you know any software with which i can do that?

 

[Baluncore]

With SI units you are now better to refer to Metric Module = MM = 25.4 / D.P.
https://en.wikipedia.org/wiki/Gear#Standard_pitches_and_the_module_system

For “stage” read “gear pair”. 4 stages needs 4 * 2 = 8 gearwheels.
4 stages has 4 pairs of gears. (13/29)*(13/39)*(13/37)*(13/48) = 0.0142189 = 1 / 70.328
The “( / )” represents a meshed gear pair.They have the same DP or MM.
Each asterisk, “)*(“ represents a “shaft” with two fixed gears. They will have different DPs or MM.
(19/72)*(19/72)*(19/71)*(19/71) = 1 / 200.52

https://en.wikipedia.org/wiki/List_of_gear_nomenclature
https://en.wikipedia.org/wiki/Gear

 

[OldEngr63]

If the pinion is undercut, it means that, at some point in the engagement, the gear contacts the pinion inside the base circle of the pinion. Of necessity, this means that involute action is not in effect while that happens. It literally undercuts the root of the tooth if this happens in the forming process, weakening the tooth form.

As for software, get the necessary standard and an editor to write the calcs you need in Fortran, BASIC, C, MatLab, Scilab, Maple, Mathematica, etc, etc. -- In short, write your own. Why be eternally dependent on the work of others? How will you know if they are correct?

 

[OldEngr63]

PS: the plural of tooth is teeth, not tooths.

 

[Baluncore]

If the pinion is undercut, it means that, at some point in the engagement, the gear contacts the pinion inside the base circle of the pinion.

Where you have root-tip interference, relief grinding the tips should reduce the need to cut inside the base circle. “Undercut” refers to a tooth profile that is narrower near the root than at some point at a greater radius. Tooth strength is analysed based on the tooth being modeled as a cantilevered beam. There is no advantage to be gained by giving either a beam or a gear tooth a “wasp waist”.

 

[OldEngr63]

Relief grinding the tooth tips will destroy proper involute action just as well as undercutting. You can't have it both ways.

And yes, undercutting does reduce the tooth thickness near the root, as the name rather clearly implies. That is why I mentioned that this weakens the tooth.

The whole point is not to use tooth numbers so small that undercutting cannot be avoided. This is bad practice, and the fact that you can get away with it in some circumstance is hardly a recommendation in favor of it.

The AGMA states very clearly in their standards the minimum number of teeth on a pinion of a specified module (or diametral pitch) to avoid undercutting. Using any number smaller is simply bad practice, the work of a rank amateur.

 

[Baluncore]

Relief grinding the tooth tips will destroy proper involute action just as well as undercutting. You can't have it both ways.

Relief grinding is essential for quiet or heavily loaded gears as it allows the gear teeth to come into contact gently without an impact against the meshing teeth surfaces. During handover from one tooth to the next, there is a period while both teeth are carrying load. Tip relief grinding smooths that transfer of load.

A. Sykes. Gear Hobbing and Shaving, 1956, writes: " It is the usual practice not to make the hob straight sided throughout the whole depth of tooth corresponding to the involute profile, but to introduce rounding near the root of the hob to give tip relief. This allows for deflection of the teeth of a gear under working load without causing the tips to strike hard at the point of entry into contact.
Whilst tip relief is essential for heavily loaded spur gears it can be of lesser amount in the case of helical gears, where end relief is more important since each tooth enters the contact zone at the leading end. Such relief can conveniently be carried out by shaving, preferably on the pinion only ".

Undercutting does not destroy proper involute action, it extends contact while sacrificing tooth strength.

I was pointing out that your interpretation of "undercutting" as contact inside the base circle was misplaced. You wrote "If the pinion is undercut, it means that, at some point in the engagement, the gear contacts the pinion inside the base circle of the pinion". That is incorrect. It is obvious that there is no need for the gear tooth tips to contact inside the base circle of the pinion as shortening the excessive length of the teeth is so much easier.

 

[Noobita]

With SI units you are now better to refer to Metric Module = MM = 25.4 / D.P.
https://en.wikipedia.org/wiki/Gear#Standard_pitches_and_the_module_system

For “stage” read “gear pair”. 4 stages needs 4 * 2 = 8 gearwheels.
4 stages has 4 pairs of gears. (13/29)*(13/39)*(13/37)*(13/48) = 0.0142189 = 1 / 70.328
The “( / )” represents a meshed gear pair.They have the same DP or MM.
Each asterisk, “)*(“ represents a “shaft” with two fixed gears. They will have different DPs or MM.
(19/72)*(19/72)*(19/71)*(19/71) = 1 / 200.52

https://en.wikipedia.org/wiki/List_of_gear_nomenclature
https://en.wikipedia.org/wiki/Gear

My calcs were with a pinion (connected to the motor) and a shaft. So, the gearbox would have 6 gears (2 spur gears and 4 double spur gears).

PS: the plural of tooth is teeth, not tooths.

Sorry about that, english is not my first language, and some times with the hurry, some error slips away...

So, you don't recommend to use 13 teeth for the gear? And my only way of finding the circular pitch is my experimente and error?

Gratefuk you all

 

[Baluncore]

How are you going to make your gearwheels ?
If you will buy them, then where from ? The catalogue and guidelines will help the design.

What is the RPM and torque of your motor ?
What gear ratio, or what RPM output range do you need ?
Please attach a drawing of your design.

 

[Noobita]

How are you going to make your gearwheels ?
If you will buy them, then where from ? The catalogue and guidelines will help the design.

What is the RPM and torque of your motor ?
What gear ratio, or what RPM output range do you need ?
Please attach a drawing of your design.

I've posted that info in the first post...i have access to a lathe and i was thinking about doing them by myself or in a initial phase, print them in 3d.

But here is the info:

"I need to design a gearbox (just want to design it for my electric motor) that will reduce the input speed of 11000 to 55 rpm, so my gearbox ratio will be 200:1. The input torque is about 24mNm. I'm thinking about a 5/6 stage gearbox (don't know if this is the appropriate designation, the gearbox will have 5/6 gears, counting the pinion in motor as a gear and the shaft as a gear)."

 

[Baluncore]

It is very difficult to cut a spur gear in a lathe. You actually need a gear tooth cutter of the correct profile in a milling machine, with an indexing head to hold the gearwheel being cut and set the tooth angular spacing around the circumference.

A 3D printer will not have the accuracy necessary to print gear wheels to operate over 600 RPM. It will be necessary to finish cut the final product, so you might as well start with a blank disc.

I've posted that info in the first post

The way you are using terms makes it difficult to work out what you mean. For example:

I'm thinking about a 5/6 stage gearbox (don't know if this is the appropriate designation, the gearbox will have 5/6 gears, counting the pinion in motor as a gear and the shaft as a gear)

The information provided here suggests that “two spur gear wheels” with different D.P. joined together and able to rotate together represent “one gear”, even though the two joined gears do not mesh and are members of different stages. I was hoping that you had done some reading and so had changed your use of gear terminology to be more consistent. That is why I ask for a drawing, sketch or diagram of your conceptual gear-train.

A spur gear is a single gear wheel with teeth that are cut with the profile in the plane of the wheel.
Each stage of a gear train comprises two gear wheels. The ratio of their tooth counts gives the ratio of that stage.
Each stage of a spur gear train requires an input axis and an output axis. An axis can be a shaft or a free axle.

Cutting two single gear wheels and joining them together on one axis is much easier than cutting the two gear profiles with different diameters on a single blank. To cut double gear wheels requires special equipment because of the wall at one end of the teeth on the smaller gear.

When building a special gear-train, gears are selected from available stock. Mass production of gear-trains makes it possible to reduce the cost by manufacturing pairs of gear wheels from different stages together.