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.

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

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.

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.

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

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 ?

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.

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") 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

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 !

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 !

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.

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

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") 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

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.

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.

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.

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

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

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.

G,
I am quite curious why you are using the
system that you have shown.  What is the
advantage?  Are you using pinion number
2 to try to take out the backlash in the
system by having is run in reverse?
English Muffin,
You seem to be caught in detail, I think
G is asking for principles only.  I do not
think gearguru is attacking you but is
focused on the problem.  We all learn on
this forum.  We all have opinions based
on our experience or interpretations of
that experience which may or may not be
correct.  We all are in a learning mode.
I do admire your tenacity.  You would
probably make or are a good researcher.

diamondjim
Funny - I hadn't thought about what this thing was actually supposed to do. I'm as curious as you are.
And attacking me? Gearguru? Why, only the other day he gave me a star! You mustn't take my remarks too seriously -I have a rather warped sense of humor.
As far as learning goes - you got that right, I don't think a day goes by that I don't learn something, either on this site or somewhere else.
I think my last post is simply my best shot at answering the original question as I now understand it. I thought the problem was that I didn't get into enough detail ! It probably can't be answered to GFinCA's satisfaction without getting into the details!
Its not tenacity, by the way, - I type fast and I just don't have anything better to do right now!

diamondjim :
Oh - by the way, on your "researcher" remark, I think "might have made" would be more like it. At my age, its rather bitter-sweet to be told that you "might make" anything! I sound young though, don't I? (I wouldn't want it any other way!).

DiamondJim, E.Muffin:
It is probably my non-Oxford English, what creates problems. Sorry, in another language I could express it better, but you would not be able to read it... Anyway, here is my point again:
Let's keep the internal gear and both idlers stationary and in their theoretical tight-mesh position.
Now if we swing the pinion1 around the idler2, the pinion1 will also rotate around it's own center, because it remains in tight mesh with the pinion2. But idler1 can not move with pinion1 - it is held in position by the internal gear. Therefore if we want the clearance between pinion1 and idler1, especially on both sides of the teeth, WE ALSO HAVE TO MOVE IDLER1 (or idler 2 in the opposite direction).
This way (grounding the gears as mentioned above) it is the easiest to visualize what happens in this gear train when relocating the pinion.
I'll look at E.M.'s calculations, will post my finding later.
Have a nice weekend, gentlemen!
gearguru
   

GearGuru :
I didn't post my derivation of that equation (too much DETAIL!)- its just a solution of the scalene triangle formed by the lines joining the centers of idlers 1 and 2 and pinion 1, taking into account the fact that the gears have to roll, as you say.  Pinion 1 rolls like a planetary, and idler 1 just rolls in a linear direction. The roll angles are coupled by the tooth ratios. The pitch radii are operating radii, in other words the solution starts by assuming that GFinCA's existing equal backlash solution is good and takes it from there. It also assumes that the ring gear is actually a straight rack - which looks like a reasonable assumption for such a small increase in spread between idler 1 and 2. I'm sure that this solution, even if correct, could be simplified with further approximations - that delta^2 at the end could probably be dispensed with for example. Don't see any way of avoiding the trancendental nature of the equation (whatever it turns out to be), and an iterative solution - when you've got pure angles and trigonometric ratios all in one equation it would seem inevitable.

GearGuru - one other thing - I considered idler 2 fixed and allowed idler 1 to move - I just found it easier to visualize that way. I think you are visualizing idler1 fixed and letting idler 2 move. But the solution should come out the same either way.

E. M.
The equations are OK. I was ready to criticize that you did not take into account the internal gear, then I found your last posts. I took your word "EXACT" too exactly. But I agree that the solution is "good enough".
If I had to work on similar problem I would probably use the vectors locating the centers of gears and solve/approximate it in Mathcad.
I think we did enough for GFinCA, I hope that it was not his college project.
This is my last post in this thread.
gearguru

gearguru :
Mine too I expect (is GFinCA still reading this I wonder?)- sorry about the "exact" comment - thought you might pick up on that after I wrote it - have to be on your toes all the time with you guys! Just possible that the equations might actually be simpler or more elegant if you used a true curved sector gear. Glad the equations weren't screwed up - hope solution converges for CFinCA - probably requires double precision. Unless it's come up before, I think I'm going to post a question about the pros and cons of mathcad, matlab, mathematica etc - don't know much about any of them.

Wow!  I get back to work today and find my mailbox full of new replies to my post!  Thanks for all the interest.  I think EnglishMuffin has given me what I was looking for, a way to determine the locations through mathematics.  We would prefer that the backlash be equal, but didn't want to complicate the problem too much, so thanks for bringing that assumption in.  As to the application, the geartrain will be a high-reliability redundant drive system for an aircraft door.  Both pinions are required to be capable of driving the internal gear independent of the other, hence the need not to have pinion 1 drive idler 1.  As mentioned previously, the gears are intermeshed to keep idler 1 in synch with idler 2 when it leaves the internal gear.  Of course pinion 1 can drive idler 1 when idler 1 leaves the internal gear, but this only happens during maintenance.

Thanks for all the help!

G

EnglishMuffin,
That's right, the elastic deflection was caused by dynamics, but the drawing helped to get a grip of the rotation recuired to get the CLUNCK in other words, to get an idea when things would go wrong, and what we could do to prevent is. Changing the first few teeth of the rack, giving it more 'backlash' could help (we ended up not trying sadly), because the pinion would then not collide head-on with the rack, but at an angle to the flank of a tooth from the pinion. To get this, we would have to grind the head of the teeth further backwards, reducing the amount further op the rack. Because of the elasticity in the system, the rack will cause both pinions to get back in mesh. (We hoped, but never proved!).

Regards,

Pekelder

Pekelder:
So hopefully this won't be a problem for CFinCA, since his system is relatively stiff torsionally. - don't want to be responsible for any aircraft crashes, do we?

pekelder, EnglishMuffin,

EM is right that our torsional stiffness is high, not to mention our speed is extremely low (18 rpm at the idlers), so I can't see there will be any issue.

Thanks for the reply,

G