Calculation of gear positions 2

[GFinCA]

I have a problem that has so far stumped our small project engineering staff. I have a gear train that comprises a pinion (pinion 1) that meshes with two idlers (idlers 1 & 2), that in turn mesh with an internal gear. Another pinion (pinion 2) also meshes with idler 1. Both idlers mesh with the same internal gear. The idlers, pinions, and internal gear have fixed centers. Pinion 1 drives idler 2, and Pinion 2 drives idler 1. Both idlers drive the internal gear. The internal gear is an arc segment. Due to space issues, idler 1 can become disengaged from the internal gear at extreme travel, so Pinion 1 is meshed with Idler 1 to provide timing of idler1 so idler 1 can mesh with the internal gear on return. We would like to split drive torque from the pinions to internal gear through the idlers, so don't want pinion 1 to transmit torque to idler 1. Therefore, our desire is to have greater backlash between pinion 1 and idler 1 than between pinion 1 and idler 2. Backlash between the idlers and internal gear are the same. We are pretty sure this can be done by altering the positions of pinion 1 and idler 2. We can mathematically determine the gear centers with equal backlash, but are scratching our heads about how to do it with unequal backlash. Is there a mathematical way, either closed or iterative, to determine the theoretical positions of the idlers and pinion for this configuration, or can it only be accomplished through trial and error? I have a sketch of the gear configuration that I can email, if desired. Thanks for any help you can give.

G

 

[EnglishMuffin]

Pretty much any gear software will tell you what the center distance is between a single pair of gears , if you know the numbers of teeth, the module or dp, the addendum mod factors (if any), and the backlash. Is that your problem, or have I misunderstood something ? If that's all you need to do, you could for example get the free excel software off this thread: thread406-60088. If its more complicated - enlighten me, and we'll try again.

 

[EnglishMuffin]

I forgot to mention "manufactured pressure angle" and "helix angle" (if any)- but you get the idea.

 

[GFinCA]

EnglishMuffin,

Thanks for the reply. Unfortunately, it's more complicated than that. I've attached a schematic to show the gear configuration. Pinion 1 drives Idler 2, and Pinion 2 drives Idler 1. Pinion 1 is meshed with Idler 1 for timing. We have already determined the gear center locations for equal backlash throughout, but we would like the backlash between Pinion 1 and Idler 1 to be greater than that between Pinion 2 and Idler 1. This requires moving Pinion 1 and Idler 2 slightly, but can you calculate the new positions to avoid binding in the geartrain, or would this be done through trial and error? I hope this clears things up. Thanks for your help.

 

[EnglishMuffin]

Unless I am missing something, its not nore complicated than that! If you know the backlash you want at each mesh, then you simply calculate the center distance between each pair of gears based on that backlash, and then compute the positions for the pinions from the center distances using co-ordinate geometry. I can go into more detail, but first tell me why this won't work. I must be missing something.

 

[EnglishMuffin]

Of course - you can use your CAD system to do the coordinate geometry for you. And you do appear to have a CAD system.

 

[EnglishMuffin]

I'm still racking my brains trying to see what your problem is here. It occurs to me that it is just possible that you might be unaware of some basic properties of involute gears.
For the sake of argument, suppose you make a pair of spur gears so that they are each completely "standard" - in other words, the circular tooth thickness of each gear is exactly half the circular pitch, there are no addendum modifications etc.
Now if you mesh these gears together, the center ditance will be given by (z1+z2)/(2*DP), wher z1 & z2 are the numbers of teeth, and there will be no backlash.
Backlash can be introduced into this gear pair by a number of methods. You can modify the addendum on one of the gears by an amount given approximately by :

blashradial = blashcircum/(2*tan(pressure angle))

where blashcircum = desired circumferential backlash

Or, you can modify the addendum on the other gear by this amount.
Or, you can modify the the addendums on both gears by half this amount (this is the most common method).
Or, finally, you can increase the center distance of the gears by this amount.
It really does not matter which of the above you use - its up to you. The gears will work perfectly in all cases - that's the special property of the involute curve. Its just as easy to make the gears, whichever method you use - it just means you have to specify the overpins dimension (or base tangent) to achieve the correct amount of addendum modification (if any).
The first three methods of backlash adjustment are not normally referred to as addendum modification, but in fact that is what they are.
Now if, for the sake of argument, you use the fourth method, all you have to do is use absolutely standard gears, increase each center distance by the appropriate amount in each case to get the backlash you need, and figure out the gear positions using your CAD system.
The formula above is approximate, and just to illustrate a point. If you use gear software, or slog through the detailed calculations, you can get it exact.
Am I still missing something ?

 

[diamondjim]

English Muffin
Notice that the the two idlers
must be sufficiently apart to not
interfere with each other.
It is not as simple as you are implying.
No question that the center distance are
as simple for the idlers. As to pinion
number one it is a geometry problem yes
in that it is dependent on how far apart
the idlers will be.
As to pinion number two, it can be in many
places around idler one and still function.

GFinCA,
I am curious how you posted the illustrated
gears to the forum. Very impressive. Will
you share how this was done?
the Muffin is right in that you can draw
concentric circles based on the pitch diameters
and move these around to achieve this in acad
except you would increase the center distance
by a small amount only for the two which you
want to have the extra backlash. The tangent
formula is close enough for this small change.

 

[GFinCA]

EnglishMuffin, diamondjim, thanks for your great posts. You've obviously put some thought into my question, and I really appreciate it. I think diamondjim understands more what my issue is. The two idlers have specific places they can be, and still mesh with both the internal gear and pinion 1. The position of pinion 2 is not critical. We've determined the postions of the idlers and pinion 1, base on all the pitch circles being tangent. If we want more backlash between pinion 1 and idler 1, is it really just a matter of increasing the center distance between the two gears (for small distances)? Does Idler 2 need to move also? It would be nice if it was just a matter of moving pinion 1 away from Idler 1. Here's another question. Since our goal is to not transmit torque from pinion 1 to Idler 1, how much more backlash is needed between pinion 1 and idler 1? In other words, if the "standard" backlash is .005 inch, what should the backlash be between pinion 1 and idler 1? I know ideally you could say .006 inch, but what is a practical number to be certain?

diamondjim, as to the posting of the picture, not all that impressive, I followed instructions on the website on how to do this, which are at the bottom of the page when you preview your post before posting it. Basically, its a matter of typing (img "image web address&quot in your post (with brackets instead of parentheses), where the "image web address" is the location of a website where the image sits. If you don't have a website to host the image, you'll need to set up an account to do so. I just registered with a free photo album hosting website like ofoto.com or fotki.com. The "image web address" then is the web address of your image. My solid modeling software lets me generate a jpeg of the image on the screen, which is what I posted.

Thanks, again, for all your help.

G

 

[EnglishMuffin]

Well - of course the OD's have to miss, but that is relatively easy to achieve - you can always reduce the blank diameter a tiny amount without affecting the meshing properties very much. I have even done that on regular gears on occasion to balance various properties and/or miss things. Ans yes, GFinCA, it is just as simple as increasing the center distance ! The only thing that happens when you do that with a pair of involute gears is that the working pressure angle changes very slightly. When you alter the backlash by thinning the teeth it doesn't change. But we are talking miniscule amounts here. Good luck !

 

[EnglishMuffin]

Regarding how much to make the other backlash amount - it depends on how you tolerance yourt everpins dimension when you specify the gears. If you go back to the tangent formula, you will see that for a 20 deg pressure angle gear, a .005 circumferential backlash is about equivalent to a .007 radial backlash. So if you put a +/- .001 overpins dimension on each gear, you should be OK with .005" backlash at one mesh and .010" at the other.
Another thing you might want to consider doing, just for the prototype, is to make the centerdistances adjustable. You could accomplish that using eccentric shafts, for example, such as you see on roller followers. but that might introduce more complexity than it was worth. If its a one shot deal, I would definitely consider doing that, but I dont know what the other implications of that might be for the rest of your design. Good luck !

 

[EnglishMuffin]

And one final point - (unless something else comes up) - you will need a tight positional tolerance on your shaft locations. +/- .001" (or the GDT equivalent) is not unusual in precision gear arrangements, - but it all depends on who is doing your manufacturing.

 

[Pekelder]

Here's something else:
I made a design where two pinions meshed with one rack (straight), and in which the rack could become disengaged from ither pinion. The two pinions where coupled with a timingbelt. The meshing was fairly easy to get right, BUT, mostly through the elasticity of the timingbelt, the engaging of the rack in the pinion proved to be rather difficult, given the high speeds, high loads and the inertia to considder. In your case, please do a calculation (or, a detailed CAD-drawing, that worked for us) where you work out the worst-case, the one pinion driving the load, the other being driven at the moment the internal gear is about to engage. It seams unlikely, but it happened in our system that 99 out of 100 it worked perfectly, and 1 time with a big CLUNK one tooth was reduced with a tiny scrap by the hardened tooth of the pinion.

Regards,

Pekelder

 

[gearguru]

GFinCA,
here are my 2 pennies:
E.Muff is right in the calculation of the backlash. This is perfectly valid for 2 meshing gears.
Now imagine that you increase the center distance (let's call it "a&quot and locate the new Pinion1 center somewhere else. This new center must be located on the circle with the radius of the center distance pinion1-idler2. This circle has a center point in the center of idler2, because you do not change that mesh's backlash. But this also means, that your Pinion1 twisted little bit. (I assume that the Idler2 and internal gear do not move).
If the pinion1 twists, then the clearance between Pinion1 and Idler1 is not dispersed equally on both sides of the tooth.
Therefore - if you want the clearances to be equal on both sides of the Pinion1 teeth you have to move also the idler1 and you still need to keep the distance "a".
I am sure you can find the appropriate location; if you were able to calculate the "standard" centers (not a simple task, I made my math to find the locations) you can master this last step also.
Anyway - what is the No of teeth on the internal gear? Just curious, I want to test my math for locating the centers...
Have a nice weekend, thanks for nice brain teaser!
gearguru

 

[EnglishMuffin]

Pekelder

It sounds as though you did a deflection/dynamic analysis, not just a kinemetic one, since you mention an elasticity problem. So in that case, a CAD drawing alone would not appear to be sufficient, unless combined with assumptions and/or measurements regarding the deflections.

 

[EnglishMuffin]

Well, I'm sorry GearGuru, but I fail to see what all the fuss is about! It appears to be the case that DFinCA is happy about everything except the backlash between idler1 and pinion1 - although I could be wrong about that since there have been so many posts. So all he has to do is compute a new position for pinion1 such that the center distance between idler1 and pinion1 is greater than idler2 and pinion1 by an amount of his choice depending on the approximate equation above, (or a precise equation if he feels like it). It is simply a matter of finding the intersection of two arcs, or some similar construction. If he doesn't want to move pinion1, he can do a similar thing by moving one of the other gears. I think this all came about simply because he did not realize that he could increase his center distances at will. Once you realize that,its just a geometry problem.

 

[EnglishMuffin]

And one other comment - If I were designing this, making the gears from scratch, I'd be inclined to use absolutely standard gears throughout with no backlash thinning, and figure all the locations out starting from that. I'd be less likely to make a mistake that way. If you are using ready made purchased gears, of course, its another matter.

 

[gearguru]

I believe he asked, if also the idler has to move. I am sure it has to, if he wants the clearance on both sides of the tooth.
It's clear that the starting point in the calculations is the standard gear. Nothing new. From that location the new centerpoint has to be found. If you move pinion only, you also twist it. That's my "fuss".
gearguru

 

[EnglishMuffin]

Well, I guess I'll shut up for a bit! I expect after everyones enlightening comments and your calculations, FGinCA will be just fine. But the only thing that teased my brain was trying to figure out why there was a problem in the first place!

Cheers

 

[EnglishMuffin]

I know I said I would shut up for a bit - and I did – for a bit!
But if you absolutely HAVE to have the backlash evenly distributed on both sides of the teeth at every mesh - tho' I'm not quite convinced you really have to - here's how you could do it - EXACTLY. (Although I admit it is maybe a little bit more complicated than coordinate geometry):

t1 = no. of teeth in idler1 or idler2
t2 = no of teeth in pinion1
L = existing center distance between idler1 and idler2
r1= pitch radius of idler 1 or idler 2
r2 = pitch radius of pinion1
theta = existing angle in radians between the line joining centers of idler1 and idler2 , and the line joining centers of idler2 and pinion1 - (you know what the value of this is - I don't)
delta = desired increase in radial backlash between pinion1 and idler1 (to get approximate circumferential backlash increase use the tangent equation given earlier).

deltheta = the angle in radians through which you must swing pinion 1 around idler 2 in order to get the new location – you can compute the new coordinates from this angle.

Then deltheta is given by the solution of the following transcendental equation :

[L+r1*deltheta*(1+t2/t1)]^2-2*[L+r1*deltheta*(1+t2/t1)]*(r1+r2)*cos(theta+deltheta)-2*(r1+r2)*delta-delta^2 = 0

You can solve this with TK solver or something similar if you’ve got it, (I wish I did - used to), or you can write a Basic program to solve it with a Newton-Raphson iterative procedure.

In addition to repositioning pinion1, you must spread the centers between idlers 1 and 2 by an amount :

deltaL = r1*(1+t2/t1)*deltheta

I’ve left the equations in program style so you can cut and paste them. I have no idea if they work - there might be a sign or two wrong, or perhaps something more serious.

I’ll be interested to see what GearGuru comes up with - it can probably be simplified a good deal. At least he should be able to confirm or refute the second equation.