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| DavidA:
The new lathe (new to me that is) came without change wheels. I want to cut an 8tpi thread. The gear on the back of the spindle is 34 tooth. It drives a 21 tooth reverser set up that drives a 54 tooth 'stud' gear' that will drive the required gear train. My leadscrew is 6 tpi. I will need to obtain the gears, but what gears do I need ? I can't get my head around the selection due to the odd stud gear that means I don't have a one to one connection to the spindle. I think it may be easier to change the stud gear for a 34 tooth gear than it will turn at the same speed as the spindle; I can deal with the maths for that assembly. Ideas, anyone ? Dave |
| andyf:
Hi Dave, There's a formula for this: Product of all the Drivers = leadscrew tpi Product of all the Drivens desired tpi Your leadscrew is 6tpi, and you require 8 tpi. Your spindle gear is 34 teeth, and that is the first of the drivers. The reverser set can be ignored; they are just idlers, and each idler acts as both a driver and a driven, so comes on the top and bottom of the equation and so can be cancelled out. The gear below the reversing set is 54T, and that is your first driven gear. So we have: 34 x A = 6 54 x B 8 , where A is an extra driver and B an extra driven This can be switched around to A = 54 x 6 B 34 x 8 So a 324T driver rotating with the 54T stud gear, and a 272T driven on the leadscrew, would do the trick, if they weren't such impractical sizes. Dividing each by 4 gives an 81T driver and a 68T driven, which might fit with another idler of suitable size between them to fill up the gap. That is as low as you can go; there's nothing else which can be divided into both of them. 81T and 68T gears aren't common sizes, and you will need a third gear to act as an idler between them. If you are going to do a lot of screwcutting of different pitches, it might be a good idea to equalise the spindle and stud gears and use 44T for each. That would mean repositioning the tumbler reverse a bit lower down, but it would enable more common gears to be used to cut different threads per inch. I suggest 44T for each in case the position of your stud gear is fixed; the gap between two 44Ts would be the same as that between the 34T and the 54T. If you can move the stud gear closer to the spindle, your suggestion of using another 34T for it would be fine. Apologies in advance if I have erred in the arithmetic; if so, I'm sure some kind soul will correct me. Andy |
| Raggle:
If a 34t gear was on the leadscrew and all intervening gears were a simple train you would cut a thread equal to the leadscrew 8tpi. So you have a constant: desired tpi x 34/8 or 17/4 = leadscrew gear teeth 6 tpi required = 6 x 17/4 = 25.5 Can't get a gear with 25.5 teeth, so a 51t gear on the leadscrew and a 2:1 compound (say 20 and 40) in the train would do it. However, a 34t gear on the spindle seems very constricting to me with 17 and 2 as the prime numbers. Is it possible to change this gear for, say 32t? In that case all simple gear trains would have a leadscrew carrying a gear equal to 4x the desired tpi. In the case of your 6 tpi this would be a 24t gear with no compounding. Simples :) I hope you can make sense of my rambling. I don't know how to do andyf's delightful fractions. Ray |
| andyf:
Your way is much simpler than mine, Ray - it uses the awkward 54T as an idler, so it takes no part in determining the overall ratio. But you have transposed things: Dave's leadscrew is 6tpi, not 8 tpi. It is his desired thread which is 8tpi. So using your method: If 34T was on the leadscrew, as well as the spindle, and they were connected by a simple train of idlers, Dave would cut a thread equal to the leadscrew: 6 tpi. So he has a constant: desired tpi x 34/6 (or 17/3) = leadscrew gear teeth. 8tpi required = 8x17/3 = 45.3333 teeth. Can't get a gear with 45.3333 teeth, so one three times the size, being 136T, on the leadscrew and a 1:3 compound (say 60 and 20) in the train would do it. To check, apply those figures to the formula I quoted, which was Product of all the Drivens = leadscrew tpi Product of all the Drivens desired tpi Then 34x60 = 6 136x20 8 Both sides of that equation work out to 0.75, so that’s OK. The problem is that a 136T gear for the leadscrew will be hard to find, and even if one were bought or made, there might not be enough distance between Dave’s spindle and leadscrew for a train which included 54T as an idler and 136T on the leadscrew. If there is room, the position of Dave’s 54T stud may be fixed, so he can’t adjust it get the 54T into mesh with the 136T, rendering another idler necessary to bridge the gap. Andy |
| Fergus OMore:
I think that this is where 'number crunching' as per Martin Cleeve in Screwcutting in the Lathe comes in. I'd say that a 127 (or a 63) transposing gear would be more available- and give pretty well the same pitch Always remember that Professor Dennis Chaddock said 'If you have 25 gears, you can have a choice of 750,000 different pitches- provided it was not all tied up in a gear box' It's a bit like my dividing head that has ONE hole. |
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