Gear help needed

[Suzuki4evr]

Just to cover all the basis. Would it be possible to make a 57 hole plate on a rotary table with dividing plates?

 

[Lo-Fi]

If the rotary table is 90:1 you'd be able to use the 19 hole plate.

To go through the maths:

We know 57 is divisible by 19. So let's take a stab at dividing with that plate. 40 turns for a complete revolution of the chuck, so 40 X 19 = 760 holes for one chuck rev. Now let's check if we can divide that by 58 and get a whole number: 760 ÷ 57 = 13.3333333. So you'd need to move 13 and a third holes each time. Not cool.

The same calc for 90:1 is 90 X 19 = 1710. 1710 ÷ 57 = 30. Ah ha! So that's one revolution and 11 holes for each move. 90:1 works here because it's wholely divisible by one of 57's factors (3). Something 30:1 would work for the same reason. It's easy to see why primes require their own plates or some trickery with differential dividing.

Just expanding on Brino's excellent post from earlier really

 

[GunsOfNavarone]

So if my gear works out with 57 teeth,the gear would mesh with the OD of 118?

Oh God man! I have no idea! I just happen to have a program that calcs that stuff. You did some damn nice work in my opinion, so disappointing one gear didn't work out.

 

[GunsOfNavarone]

If the rotary table is 90:1 you'd be able to use the 19 hole plate.

To go through the maths:

We know 57 is divisible by 19. So let's take a stab at dividing with that plate. 40 turns for a complete revolution of the chuck, so 40 X 19 = 760 holes for one chuck rev. Now let's check if we can divide that by 58 and get a whole number: 760 ÷ 57 = 13.3333333. So you'd need to move 13 and a third holes each time. Not cool.

The same calc for 90:1 is 90 X 19 = 1710. 1710 ÷ 57 = 30. Ah ha! So that's one revolution and 11 holes for each move. 90:1 works here because it's wholely divisible by one of 57's factors (3). Something 30:1 would work for the same reason. It's easy to see why primes require their own plates or some trickery with differential dividing.

Just expanding on Brino's excellent post from earlier really

I got a nose bleed trying to figure that out! (or more like trying to follow that.) It's like watching David Blaine levitate....

 

[Suzuki4evr]

If the rotary table is 90:1 you'd be able to use the 19 hole plate.

To go through the maths:

We know 57 is divisible by 19. So let's take a stab at dividing with that plate. 40 turns for a complete revolution of the chuck, so 40 X 19 = 760 holes for one chuck rev. Now let's check if we can divide that by 58 and get a whole number: 760 ÷ 57 = 13.3333333. So you'd need to move 13 and a third holes each time. Not cool.

The same calc for 90:1 is 90 X 19 = 1710. 1710 ÷ 57 = 30. Ah ha! So that's one revolution and 11 holes for each move. 90:1 works here because it's wholely divisible by one of 57's factors (3). Something 30:1 would work for the same reason. It's easy to see why primes require their own plates or some trickery with differential dividing.

Just expanding on Brino's excellent post from earlier really

I made a mistake on my post about the rotary table. I forgot to put the word "NO", infront of dividing plates. There is NO dividing plates for the rotary table. So the same question apply.

 

[Suzuki4evr]

Nope. Sorry.

You need a 57-hole plate and to advance 40 holes for each tooth for a 57-tooth gear.

Otherwise you need to do more complex differential indexing.

If it had been a 90:1 ratio, we could have got there!
(one full turn and 22 holes on a 38 hole plate)




That's probably the way to go.

Note to check before you start how many holes to specify.
Apparently on some DRO's you may need to specify N+1 or 58 holes _IF_ you specify both start and end angles the same......see this thread:
https://www.hobby-machinist.com/threads/do-what-i-want-not-what-i-say.84534/post-748360

-brino

Lo-fi,now that you gave GunsOfNavarone a nose bleed,can you confirm that no matter how I achieve getting a 57 hole plate that my OD of 118mm is correct and that when I sit with a 57 tooth gear in my hand, that it should now mesh with my the 95 tooth gear on the lathe?

 

[jaek]

You can totally do this on a rotary table, it'll just be a minor hassle.

This is a spreadsheet I threw together with the angular offsets for each of the 57 holes. Just print it out, cross of each row once you drill it.

 

[Suzuki4evr]

You can totally do this on a rotary table, it'll just be a minor hassle.

This is a spreadsheet I threw together with the angular offsets for each of the 57 holes. Just print it out, cross of each row once you drill it.

Thank you for this awsome spread sheet. Now there is one more option.

 

[Lo-Fi]

Lol! Yep, spot on with that. Zm + 2m = (57*2)+(2*2) = 114+4 = 118

With the right diameter and wrong tooth count, you've not technically cut a module 2 pitch gear, so the mesh won't be perfect. Is the mesh adjustable by moving the banjo? Or is this a fixed relationship? Difficult to see from the photos. It kinda looks like you've run out of banjo adjustment rather than it being purely a meshing issue?

A pic of the lathe without the gears mounted could be quite handy

 

[Suzuki4evr]

The banjo is not the problem here. As you can see here the pinion gears and the driven gear is on a set distance where as you set the play between the drive and pinion gears.