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Calculating number of teeth for a gear

jarhead jim

Aluminum
Joined
Apr 28, 2009
Location
Bartley,NE
Needing to make a large spur gear. 12 dp 14-1/2 degree. 8.750 to 9" outside diameter. My machinery's handbook goes up to 8.50 but want to make one a tad larger. My math shows a 9" gear with 106 teeth, using either a #53 plate with a 20/53 advance or the 43 plate with a 16/43 advance. Am I anywhere close?

Thanks, Jim
 
Needing to make a large spur gear. 12 dp 14-1/2 degree. 8.750 to 9" outside diameter. My machinery's handbook goes up to 8.50 but want to make one a tad larger.
106 teeth if you can get a clean 9".

(Add 2 teeth for the o.d., divide by the DP. 106+2=108/12=9)

(Avoid big primes, like 103. Makes it a pita to cut because of the change gears. On the other hand, if you're into that hunting tooth stuff, have at it :))
 
Pitch dia controls number of teeth - OD on a 12DP standard depth will be .167" larger (because addendum on a 12DP is .0833")

As above 12 into a 9" PD is 108 - and OD will be 9.167

A 106 will have a 8.833 PD and a 9" OD
 
IMG_3019.jpgFrom what I came up with..... with a 106 tooth gear on the lead screw and an 18 tooth gear on the other shaft, and with a 6tpi lead screw, I should get roughly 70 tpi. I'm doing this mainly to use the half nut to engage the automatic feed to do the finer turning/cutting.... Using my chart in the photo down at the bottom, if I divide 60 by 18 and multiply that by 6 and again multiply that by 2 I get 40tpi. If I use 72 and divide that by 18 and multiply that times 6 and again multiply that times 2 I get 48tpi. Soooooooooo if I divide 106 by 18 and multiply that times 6 and again by 2 I get 70..... But I suck at math so this may just be coincidence. LMFAO!

Jim
 
From what I came up with..... with a 106 tooth gear on the lead screw and an 18 tooth gear on the other shaft, and with a 6tpi lead screw, I should get roughly 70 tpi. I'm doing this mainly to use the half nut to engage the automatic feed to do the finer turning/cutting.... Using my chart in the photo down at the bottom, if I divide 60 by 18 and multiply that by 6 and again multiply that by 2 I get 40tpi. If I use 72 and divide that by 18 and multiply that times 6 and again multiply that times 2 I get 48tpi. Soooooooooo if I divide 106 by 18 and multiply that times 6 and again by 2 I get 70..... But I suck at math so this may just be coincidence. LMFAO!

Jim



Do you know about how things work on the Le Blond?

Info may help.

Info on feeds.jpg

If you use the feed rod and its feed clutch, you get to take advantage of the 3 to 1 ratio in the apron

As to the selector deal, you generally put it in the middle (for neutral) when using the half nuts (and lead screw) to cut a thread.

Suggestion on slowing down feed rod for fine feeds

IMG_3019.jpg
Here is a shot showing the spindle cone pulley missing from yours

01515_iP4thq4qpIrz_0CI0t2_1200x900.jpg
 
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More and related to end gearing. IF the stud gear went the same speed as the spindle AND the screw gear had half as many teeth as the stud the lathe with 6 TPI lead screw would cut 3 threads per inch. IF the stud was swapped with the screw gear it would cut 12 TPI. If both screw and stud had same tooth count, it would cut 6 TPI

and so on

Scan from 1911

End Gears.jpg
 
IMG_3139.jpgIMG_3142.jpgIMG_3143.jpgIMG_3144.jpgF... me... I cut the blank to an exact 9.00. Set the plate on the 43 hole and indexed 16. And ended up with 108 or so teeth in what shoulda been a 106 tooth gear. The earlier teeth were narrower on the outter edge than the latter ones. In some areas the edges of the teeth weren't even the same. Gear blank didn't move in the chuck as I marked it this time. The gear cutter looked great and was still sharp. Correct gear cutter too, 55-135 or so. Depth of cut .180 like the others.

First picture is the earlier cuts. The second picture is the latter ones. The 3rd is the area where teeth were thick and thin. 4th is compared to a 64 tooth gear I made last week.

Jim
 
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(43×40)/16 = 107.5
Use a 53 and index 20
(53×40)/106 = 20
The prime factorization of 106 is 2 and 53, so you need a 53 to make it work.
 
(43×40)/16 = 107.5
Use a 53 and index 20
(53×40)/106 = 20
The prime factorization of 106 is 2 and 53, so you need a 53 to make it work.

And still go with a 9" gear blank? And ugh, I was mistaken on my numbers on the plate. I don't have a 53, rather a 54. What other combo can I use to get 106 teeth on a 9" gear blank. The 43 hole and 16 index was a failure. I have numbers 16, 18, 24, 28, 30, 34, 37, 38, 39, 41, 42, 43, 46, 47, 49, 54, 58, 62 and 66.

Jim
 
As far as "how to index", see bottom of page 1502 for 106 and top of page 1503 for 108 in my 21st edition of Machinery's Handbook - getting rid of guess work

106 by the way involves Differential Indexing - meaning extra gearing helping out the dividing head
 
As far as "how to index", see bottom of page 1502 for 106 and top of page 1503 for 108 in my 21st edition of Machinery's Handbook - getting rid of guess work

106 by the way involves Differential Indexing - meaning extra gearing helping out the dividing head

I settled for the 104 tooth gear using the #39 hole, indexing 15. About done. Hopefully this one will turn out alright. Thanks again.

Jim
 








 
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