# Cutting Involute Gears

• posted

I have ivan Law's book, "Gears and Gear Cutting" and a copy of John Stevenson's paper, "Cutting involute Gears with Form Tools". In each are detailed instructions for making "button type" form tools for the fabrication of gear cutters. Choose any specific gear and pressure angle and the 2 sets of instructions show different sizes of buttons, pin centers, and etc. Can anyone elaborate on why this is so? I would think the specifications for making precision gear cutters would be consistent among "authorities".

Bob Swinney

• posted

I guess you gotta choose your authority. I've made gears with Ivan Law's instruction and it worked perfectly, the second time. (The first time was for seed)

Karl

• posted

If I remember correctly it is impossible to cut perfectly accurate involute gears except by generating them. Thus any calculation to make a "gear cutter" would be, at best, an approximation so a difference in calculations might well be the norm.

• posted

I don't think that's the case, Bruce. It depends on how much undercutting is required at the base of the teeth. And that's a function of the pressure angle and the number of teeth per gear, plus the geometry of the tips of the opposing teeth.

That is, for straight-cut gears. The geometry of most such gears is perfectly achievable with a milling (or straight shaping) cutter. But there are some extreme ones that require generating.

-- Ed Huntress

• posted

I can't quote an authority as it is something I remember (I think) from my apprentice days - that the only method of producing a perfectly accurate involute gear was to generate it. Now there is a possibility I "learned" this little fact when visiting Fellows Gear Shapers, who would be somewhat prejudiced, being in that business, but it is stuck in my memory from somewhere.

Having said that we habitually cut perfectly acceptable gears using a set of gear cutters and In fact I have ground single point tools using the gear itself as a template and then used a fly cutter type device to cut the gear with the single point cutter.

It was a one off for a large computer installation in the A.F. with no part numbers, drawings or data available about the "special gear". As far as I know the thing was still running some years later when I left the Base.

I throw in the "war story" to illustrate the fact that sometimes "close enough" works and you don't need perfect parts. Thus my remarks about two different calculations for "close enough" being possible.

• posted

I think that's true, and you can make a small compromise in the tooth tip shape to avoid undercuts, which, IIRC, is done with some fairly high-performance gears that are production-cut with hollow broaches.

But gear geometry is something I've long forgotten. I do remember reading a paper on broach-cut gears that showed how to adjust simple gear shapes to avoid undercuts, with no loss of performance.

On the other hand, the most sophisticated paper on gears I ever read was for some work I did for Hoescht Celenese a few years back, about plastic gears, which showed how even hobbed gears were really compromised from ideal shapes -- and which you could achieve with relative ease in plastic. The key example was an injection-molded plastic planetary set that would transmit something like 150 ft-lb of torque at high speeds. Pretty impressive.

-- Ed Huntress

• posted

Stevenson's paper, "Cutting

"button type" form

pressure angle and the 2

Can anyone elaborate on

cutters would be

ISTR seeing a mention of the discrepancy in Model Engineer a while back. The answer had something to do with differences in the reference points that the measurements were based upon. I was interested enough to read the stuff, but was not taking notes, if you know what I mean.

Same John Stevensen that hangs out on uk.rec.model.engineering ? Ask there, if you can access it.

IIRC, George H Thomas and the crew he conspired with, came up with a form releiving tool called the "Eureka", that was to be used with such cutters. It was in relation to one of these form releiving tools that I recall reading about the different sets of formulae for coming up with the buttons.

Cheers Trevor Jones

• posted

Years back, as a mental exercise I worked out the math to duplicate Stevenson's chart for 14 1/2 deg pa.

When I cross checked my math, I found some discrepancies with his 20 deg pa chart. It wasn't much, but I lost interest and never followed it up. It might have been a clearance issue or simply which tooth count is really the average of the cutter range.

Now I'll have to get a copy of Law's book and see if my formulas match his numbers.

Paul K. Dickman

• posted

Thanks Trevor,

The Eureka is mentioned in Ivan Laws book.

Stevenson's paper, "Cutting

"button type" form

pressure angle and the 2

Can anyone elaborate

cutters would be

ISTR seeing a mention of the discrepancy in Model Engineer a while back. The answer had something to do with differences in the reference points that the measurements were based upon. I was interested enough to read the stuff, but was not taking notes, if you know what I mean.

Same John Stevensen that hangs out on uk.rec.model.engineering ? Ask there, if you can access it.

IIRC, George H Thomas and the crew he conspired with, came up with a form releiving tool called the "Eureka", that was to be used with such cutters. It was in relation to one of these form releiving tools that I recall reading about the different sets of formulae for coming up with the buttons.

Cheers Trevor Jones

• posted

Paul sez: Now I'll have to get a copy of Law's book and see if my formulas match his numbers. Good go, Paul. While you're at it, see if you can dupe the math that correlates the radii of standard end mills with that of gear cutters. I haven't tried to reconcile the math yet but I recently made some 45 DP gear cutters (form) with the radius of a 3/32" end mill. This was according to a Jerry Keiffer article in Home Shop Machinist. I spoke briefly with Jerry at the Sherline booth at N.A.M.E.S. He put me onto the idea that most gear DPs can be matched with the radii of standard end mills. Thus far, I've only tried 56DP with a 3/32 end mill. It runs perfectly wtih the gears of my Sherline lathe.

Bob Swinney

Years back, as a mental exercise I worked out the math to duplicate Stevenson's chart for 14 1/2 deg pa.

When I cross checked my math, I found some discrepancies with his 20 deg pa chart. It wasn't much, but I lost interest and never followed it up. It might have been a clearance issue or simply which tooth count is really the average of the cutter range.

Paul K. Dickman

• posted

Involute gears have the tooth flank form required to produce constant velocity transmission i.e. constant ratio. Any deviation from this results in slight variations in instantaneous ratio (causing noise and roughness) as individual teeth move in and out or mesh.

Generation methods are necessary if the"perfect" constant velocity tooth form is needed. However the flank form of involute gears with large numbers of teeth is very close to a single radius curve. Acceptable gears can be cut by a cutter formed with a simple cylindrical cutting edge.

The best centre position and the curve radius changes with the number of teeth in the gear. This change is small for gears with a large number of teeth but rapidly increases for gears with smaller numbers of teeth.

Purpose made gear cutters for milling teeth come as set of 8. No 2 covers the range 55 to 134 teeth but No 8 only covers the range 12 to 13 teeth.

Jim

• posted

Am I right in assuming that you are using the side of the endmill to cut the concavity on the flank of the form tool?

While some may work perfectly, the main reason it would be useful for other ranges, is that for most of what you or I would use gears for, close enough is good enough.

Heck, the gears in windmills were made out of logs and sticks. If backlash, perfectly smooth power transmission, and high speed operation are taken out of the requirements, all gears have to do is mesh.

The formulas I came up with are these:

First, you needed to figure the pitch radius of the gear from the standard formulas. The base circle radius (from which the arc of the involutes is defined) BCR = PR*cos(PA) Then the tooth radius (radius of the form cutter) TR = tan(PA)*BCR

This is based on 1DP so you have to divide by the DP you are using.

Also, remember those numbers are radii not diameters. I have a formula for the center to center spacing of the circular pins, but it contains some scrawled additions, that I don't remember the logic behind, and this is where the discrepancy showed up

You could take the numbers on Stevenson's chart and divide them by standard cutter diameters, the ones with quotients close to whole numbers, would be the DPs they would work for.

Paul K. Dickman

• posted

Sorry,

Make that "recently made some 56 DP gear cutters with the radius of a 3/32 end mill. . ."

Bob Swinney

Bob Swinney

Years back, as a mental exercise I worked out the math to duplicate Stevenson's chart for 14 1/2 deg pa.

When I cross checked my math, I found some discrepancies with his 20 deg pa chart. It wasn't much, but I lost interest and never followed it up. It might have been a clearance issue or simply which tooth count is really the average of the cutter range.

Paul K. Dickman

• posted

Paul sez:

"Am I right in assuming that you are using the side of the endmill to cut the concavity on the flank of the form tool?"

Precisely. Unlike Ivan Law's and John Stevenson's method, the form tool is not used to make a gear cutter. The form tool unto itself is the gear cutter, used in fly-cutter fashion.

Paul, please elaborate a bit on your equations: BCR = PR*cos(PA) and TR = tan(PA)*BCR In each, please explain what the term (PA) represents.

Thank you,

Bob Swinney

While some may work perfectly, the main reason it would be useful for other ranges, is that for most of what you or I would use gears for, close enough is good enough.

Heck, the gears in windmills were made out of logs and sticks. If backlash, perfectly smooth power transmission, and high speed operation are taken out of the requirements, all gears have to do is mesh.

The formulas I came up with are these:

First, you needed to figure the pitch radius of the gear from the standard formulas. The base circle radius (from which the arc of the involutes is defined) Then the tooth radius (radius of the form cutter)

This is based on 1DP so you have to divide by the DP you are using.

Also, remember those numbers are radii not diameters. I have a formula for the center to center spacing of the circular pins, but it contains some scrawled additions, that I don't remember the logic behind, and this is where the discrepancy showed up

You could take the numbers on Stevenson's chart and divide them by standard cutter diameters, the ones with quotients close to whole numbers, would be the DPs they would work for.

Paul K. Dickman

• posted

PA stands for pressure angle.

Let us say that you have decided to make a 37-47 metric transposition set of gears in 56DP 20deg PA. For that you would need a #3 cutter which will cut 35 to 54 tooth gears. Cutters are sized for the lowest in the range, so we design it for 35 tooth gears.

First work your calculations like you were making a 1DP gear to simplify the calculations.

35 tooth gear 1dp has a pitch diameter of 35 inches or a pitch radius of 17.5. Base circle radius or BCR = PR*cos(PA) = 17.5 * cos(20deg) = 17.5 * .9396 = 16.443

Tooth radius (the radius of the pins in Stevenson's method) or TR = tan(PA)*BCR = tan(20deg) * 16.443 = 5.9848 Diameter if the pins is 2* TR or 11.9696. Stevenson's chart shows 11.97.

Divide that by your desired DP 11.9696/56 and you need pins of .2137 diameter

For 14.5 deg PA gears, substitute the cosine and tangent of 14.5 des for those of 20 deg.

The formula I had for the pin spacing was 2(cos(PA+(1/4 the angular spread of the teeth))*TR)+ the chordal thickness from tables.

This is where it fell apart and I can't find the drawings I used to work out the formula, so I can't really explain the logic I used to derive it.

Paul K. Dickman

• posted

tan(PA)*BCR In each,

I'll bet PA represents pressure angle.

• posted

Right on, Don. I should have known that but for some reason I couldn't get my mind around it. Paul answered it is in fact, pressure angle. If I have been reading everything right and you make a rack from the appropriate cutter, the sides of the teeth would be perfectly straight and inclined on each side at the pressure angle.

tan(PA)*BCR In each,

I'll bet PA represents pressure angle.

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