# math ? for BackGear ratio

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I need to know the exact gearing ratio for my mill in backgear. This is for a gear hobbing project I hope to finish within the next X years (X has been less than one for a couple years now)

Anyway, I have an encoder on my drive spindle that does 4000 counts per turn. I ran the machine in backgear exactly 100 revolutions and got

2,451,765 pulses. I didn't use an indicator for the begin or end stop point (mark at end of 4" fly cutter lined up) so the true total could be +/- 250 pulses

This gives a gear ratio of 6.1294. I need a really good guess on the number of teeth in the drive and driven gear. For example, if the drive gear has 10 teeth and the driven gear has 60 teeth, the ratio would be exactly 6.00. There's a zillion possibles, I'm looking for one that give a number within a couple tenths(.0002) of this ratio.

If I have too, I'll tear the machine down. But, I'm hoping there's a math genious out there. If needed, I could put a counter on the spindle and do

1000 turns.

Karl

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49:8? Unlikely, but that gives 6.125, error of only .016% .. - GWE
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Karl,

Since there aren't many choices of small gear size lets do this the brute force way and just bang the first few reasonabley sized gears by the ratio on a calculator

The * indicates it a prime number

6.129 times 8 =49.032 = 8x49 = 6.125 *9 =55.161 = 9x55 = 6.1111111 10=61.29 = 10x61 = 6.1 *11=67.419 = 11x67 = 6.0909090 12=73.548 = 12x73 = 6.0833333 *13=79.677 = 13x80 = 6.1538462 14=85.806 = 14x85 = 6.0714286 15=91.935 = 15x91 = 6.0666666 16=98.064 = 16x98 = 6.125 *17=104.193 = 17x104= 6.1176471 18=110.322 = 18x110= 6.1111111 *19=116.451 = 19x116= 6.1052632

My bet is that it's a 6.125 ratio.. Note it's NOT prime but neither is my Bridgeport (it's 8.25) And I also bet it's the 16x98 gearset and not the 8x49... just a hunch.... I doubt the pinion is much bigger than 19 or 20 teeth because of the physical size that the big get winds up being. Good luck ( and I'd dissassemble the machine and count teeth) Oh and what make of machine is this, I bet some google time would find the 'right' answer...

Dave

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,,,

I'm certain its slightly different from 6.125, that was my first estimate. This is (6.125*4000*100) = 2,450,000 counts. I had 1765 counts more than this in 100 turns, or nearly 1/2 a revolution of the spindle.

The machine is an Excello model 602. An Excello looks like a bridgeport J head on steroids. Made in Canada. I've never found a manual for it.

Karl

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Don't want to get into 'pissing on trees' but an excello 602 is a BP clone and the stock model weighs in at about #300 less than a BP.. I've seen a 602 run and .. no big deal.. (not to say I haven't seen some kickass Excello machines)

But back to your problem... You have my table of likely gear ratios.. pick which one suites you, next closest is 13x80, which *could* be in the ball park, but just as far off on the other end... There isn't any magic here, you see the ratios that are generated,... "ya pays yer money and ya takes yer chances"

I also have trouble with your statment " I had 1765 counts more than this in 100 turns, or nearly 1/2 a revolution of the spindle" If I read you origonal post correctly thats 1765 counts in 100 turns or

17.65 counts in ONE turn.. Hmmm that sure sounds like 6.125 to me.. but hey I just own a wimpy Bridgeport....

Dissasemble and count teeth......... and let Grant and I know when it comes up 6.125 :-)

Dave

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Sorry to reply to my own post. I used Excel to follow this idea. 31*6.1294 =

190.01 or almost identical to my measuremnets.

I've never had a 1J style head apart. Any chance they use gears with this many teeth, 31 driving 190?

Karl

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What you need to do is a continued fraction expansion of 6.1294. Using the program on my site, I get:

RATIONAL FRACTION APPROXIMATION Number to approximate ? 6.1294 Desired approximation accuracy [0.01 %] ? Number = 6.1294

6 & 11/85 = 521/85 = 6.129411765 Approximation Error = 0.00909174 % A 521:85 gear ratio sounds mighty unlikely. I suspect you need to determine the numerical value more accurately.

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things

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The J head has a double reduction. The first stage is a timing belt, the second is a pinion driven by the larger timing belt sprocket which drives the bull gear on the spindle.

The total ratio for a 1J is 8-1/4:1 - I don't know whether that's exact or not. I do have a couple exploded views and photographs of the parts in question from old BP manuals, so it would probably be possible to count teeth on the various pulleys and gears, but I'm not sure it would be much help in figuring out your Excello.

Ned Simmons

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Karl, can you contact Excello?

Are you positive you have just one set of gears operating here?

Grant

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Looks like it may be time for me to find them and buy that manual. I've been lucky and never needed special parts. But, I bet the parts manual won't show enough detail to count teeth. By chance, does anyone know where they are located?

I'll guess that its a compound reduction. After all, you don't have to reverse the motor when you go to back gear, like on the older J head Bridgeport machine. BTW, I just noticed an error in my post a ways back. This machine looks a lot like a Series 1 CNC Bridgeport with a 2J style head.

Right now, I'm a cogitatin' on how to mount an encoder in a tool holder. Then run the machine and get the ratio out to the sixth decimal place. Seems easier than tearing it down. Then use that real nice gear calculator Marvin offered.

Karl

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The Bridgeport 1J and 2J use 2 stages. There is a toothed belt with a small sprocket on the driven pulley and a great big sprocket behind it that turns the back gear's pinion. Then, the pinion turns the big bull gear below the driven pulley. The center of the bull gear is the female spline that drives the spindle spline.

Jon

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Karl I have a manual for my 602 and it will be no help in determining gear tooth counts. Just has a cut away view to identify parts. The machine uses a double reduction gearsets.

Be careful if you disassemble the head. I had mine apart and replaced all the bearings and seals. When putting it back together the meshing of the plastic spindle feed gear is critical. If not meshed with the mating steel gear, guess what happens to the little \$350 gear. The "second time' I used spacers to slowly lower the castings together while meshing the gears.

Good luck

John Normile

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John,

Cam I email you a pic of my machine? Based on another poster, I may have the model number wrong. My CNC machine has a Taper 40 spindle, is a bit larger than a 2J Bridgeport, and has a 2J style head. Sound like yours? My machine has a serial no. but can't find a model number on it.

BTW, I need to do the same job. My machine drips oil on the table. Maybe 1 drip per hour, but its a mess after a couple of weeks with no use. Thanks for the heads up.

Karl

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My model 602 is in exactly the same condition. (1 drip / hour plus or minus). So I've been all ears lurking on this thread. If you find the source for the manual, please mail me or publish it here.

Thanks,

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Someone said it was probably two sets of gears. I wrote a simple program having 4 gears, with net ratio of a/b * c/d. I ran this for all combinations of teeth with the following ranges:

a 8 to 100 b 8 to 50 c 8 to 50 d 8 to 100

as a more-or-less arbitrary set of values. I had it save all sets with total ratio within 0.0001 of 6.1294. There were 24 successes. The one with the smallest max number of teeth on any gear was:

65, 8, 43 and 57 teeth, ratio of 6.12939.

These are all prime numbers in the sense that they are divisible by no number less than 8. (I figure 8 is about the minimum number of teeth on a real gear)

There were two other sets with same ratio:

65, 12,43, 38 65, 24 43, 19

When I did it with a single pair of gears I got no hits even with tolerance loosened to .001.

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Meant to say no number greater than 8 but smaller than the number itself. A nit, doesn't matter.

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There's one more piece of information you could throw in to narrow things down. Presumably the input and output of the gear train are coaxial as on a BP, so if the pitch on both gear sets is the same

a+b = c+d

If the pitches are different

(a+b)/dp1 = (c+d)/dp2

where dp1 and dp2 are standard diametral pitches in a reasonable range, say 8,10,12,16,20. An estimate of the center distance between the two shafts would limit the range of possibilities further.

Ned Simmons

Ned Simmons

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I still get some minor dripping from the quill/housing area, and from the gearbox for the quill feed. When it gets to be a problem I will get a cookie sheet and use it to catch the oil.

Excello of Canada will sell you a manual for about \$60 (1997 price)

Ex-Cell_O Machine Tools

6015 Center Drive Sterling Heights MI 48312 810 / 939-1330

Email if I can be of help

John Normile

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Interesting observation, Ned!

Assuming that the ratio for such a speed reduction is somewhat arbitrary, (so why not 6.25: 1?) and they wouldn't use overly-large gears if they could help it, I widened the tolerance to .001 and limited the tooth counts to 50,30, 50, 30, then computed the difference you mention (a+b -(c+d)) for each set that made the cut.

Results: 39, 20, 44 and 14 give a ratio of 6.128571 with a difference of 1 tooth in the sums, while 44, 19, 45 and 17 give a ratio of

6.130031 with a difference of 1 tooth.

The best I could do with DP variation was 22,10, 39,14 with DP's of

12 and 20, ratio of 6.128571 -- not

Now, Karl, it's time to count teeth and see what the *real* story is!

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