127 hole plate for B&S BS0

[Cuebuilder]

Hi, i'm new to this forum. I want to know if I build a 127 hole plate for my B&S BS0 dividing head if I will end up with 127 teeth on a gear? I need to make a 127 tooth gear for my Birmingham lathe so I can cut Diametral pitch and Module pitch worms. I will have to cut other gears also, but this is the only one I'm currently having trouble with. By my math if I rotate the plate 40 holes and make a cut and then rotate another 40 holes, when I get to the end I will have 127 teeth. Is this right? Thanks for any information or experience you have in this matter. Cuebuilder

 

[Bartonius]

So in your Case: 127teeth=(90/T) >>> T=90/127 >>>> T=0.70866

Where 90 is the worm ratio substitute your ratio if different

If you had a 127 hole plate you would need to cover .70866 handle rev*127=89.999 or 90 holes to move 1/127th part rev

I think a 24 hole plate would get you where you need to be as well depending on accuracy (17/24 = 0.7083333333) >>> 90 / 0.7083333333 = 127.058

 

[Cuebuilder]

Sorry I should have mentioned that the ratio is 40:1 for the Brown and Sharp BS0. So, my thinking is that I need 3 cuts per turn of the crank handle, or so. With the 127 hole plate that is 40+40+40=120 with 7 holes left over. 7 holes X 40 revolutions= 280 holes. 280 holes /40 =7. The 7 divisions that I still need will come from the left over 280 holes.
So what I'm really wanting to know is my math correct? And will this work to cut my 127 tooth Gear?

Thanks for any and all help. Cuebuilder

 

[Bartonius]

So with a goal of a 127 tooth gear and a 40:1 Worm, you need 0.315 turns of the handle between teeth.

If you use a 127 hole plate you need to go 40 holes on the plate to get .315 revolutions of the handle.

If you already have plates available, a 19 hole plate covering 6 holes will get you very close. (6/19=.3157)

 

[rgray]

Sorry I should have mentioned that the ratio is 40:1 for the Brown and Sharp BS0. So, my thinking is that I need 3 cuts per turn of the crank handle, or so. With the 127 hole plate that is 40+40+40=120 with 7 holes left over. 7 holes X 40 revolutions= 280 holes. 280 holes /40 =7. The 7 divisions that I still need will come from the left over 280 holes.
So what I'm really wanting to know is my math correct? And will this work to cut my 127 tooth Gear?

Thanks for any and all help. Cuebuilder

The info that came with my 40:1 dividing head says use a 16 hole plate and advance 5 holes per tooth....Not counting the hole the pin is in.
That comes out to 5/16 = .3125 not near as close as the 19 hole plate ???
My dividing head came with a 19 hole plate. is it just a typo on the included chart? I have found other typos on it. It shows the 16 hole plate and 5 hole advancement for 126, 127, and 128 tooth gears.

 

[Cuebuilder]

I'm pretty sure the 16 hole plate is only accurate to cut 128 teeth. If you look at the chart it shows 4 change gears 24,64,40, and 48 to be used with the B&S BS2 to cut 127 teeth. I don't have the BS2, so I can't use the change gears as mine didn't come with them nor is there a place to mount them. So, a 16 hole plate moving 5 holes will only get you close. that is why I'm planning on making the 127 tooth hole on my taig cnc mill.

Cuebuilder

 

[benmychree]

The problem that you guys neglect is that that "tiny" error accumulates and you end up with a substantial error when you get to the last tooth to be cut. I do have a B&S #2 universal dividing head with change gears, and did the differential indexing to make a master change gear to fit my automatic gear cutter so now I can produce 127 tooth change gears of any common pitch. If you would like help with this, and you make the blank, I can cut the teeth for a nominal fee.
John York york@napanet.net

 

[eightball]

I agree with John. If memory serves me correctly 127 is a prime number. I think you will have to use differential indexing to get an accurate plate. Out of curiosity, are you indexing something metric?

 

[Cuebuilder]

I will be making a 127 tooth gear for my Birminghm lathe. I need to cut a new worm and worm gear for my Zubal Lathe. I will have to cut some other gears also as it is a module 2 worm, the worm gear is module 2 with 25 teeth. I have the idea that if I make a new index plate for my bs0 indexing head with 127 holes in it that I will end up with 127 teeth on the gear that I need to make. I should just need to rotate 40 holes between the sector arms and that should give me 127 teeth. I started this thread so I could get someone with more experience cutting gears to check my math before I make the plate. The plate itself should be straight forward to make on my taig cnc mill. I've already drawn up the plate in bobcad but for some reason I can't post a picture of it on here. It's a dwg file. I'm going to make the plate out of 1/4 inch lexan as this will be for a one off job and won't see hardly any use, I think it should hold up fine. Any comments or suggestions would be greatly appreiciated. Also does anyone know how I can get the drawing on here, it might be of use to someone else with the same problem.

Thanks
Cuebuilder

 

[benmychree]

I very much doubt that you will be able to fit 127 holes on a plate that will fit your dividing head, likely, they would even overlap one another. Look at the highest number plate you have, the holes are pretty close to each other on that (39) hole plate, could you fit that many more?