My Ideal lathe
My Old Iron
Old Machine Pic's
This appears to be a subject that gives a lot of people a lot of problems, I hope I can provide a little light here.
I firmly believe that a picture is worth a thousand words, and as I can I will use them. But a lot of this will require your reading and understanding. If you dont understand fractions take a detour now.
Firstly "idle" gears, a lot of people think every gear in a train of gears changes the result, this is not the case. Take two gears the first having 20 teeth, the second having 40, if we mesh these together with the 20 driving then the result is 20/40 which equals 0.5 or half the speed you started with, so if the starting speed was 500rpm then the speed would now be 500 x 0.5 = 250rpm
But what if we couldn't mesh these two together because machine wouldn't allow it. Well we could place an idle gear between them, let's assume we have one with 30 teeth, this will give a train of 20 - 30 - 40, so what happens. Well the speed change is exactly the same, lets see why. You see a lot of explanations as just accept it, but I think if you understand it properly it will stick, so....
we still have 500 rpm and the first two gears
20/30 = 0.6r times 500 = 333.3r rpm
so now we have the idler running at 333.3r rpm and the second two gears
30/40 = 0.75 times 333.3r = 249.975
I chose an awkward one to prove a point, ignore the repeating part of each, what you see is the same result.
So the idle wheel is revolving at a different speed, but the end result is the same, for example 4 divided by 2 equals 2 and multiplied by 2 equals 4 again, it is that simple, honestly.....
But, the one thing the idle wheel has done is to change the direction of rotation, every addition to a train of gears changes the rotation, each even number is the opposite, each odd number is the same direction, which of course we need that to happen to cut left and right handed threads.
So you see it was easy, it's not rocket science. Let's move on one more step.
We want to cut a thread, so what do we need to know ? Come to that what don't we need to know.........
Well we don't need to know the speed of the mandrel for a start, because whatever the speed is the result will be the same, if that's confused you let me explain. What's a good analogy, how about stacking a pile of bricks one at a time, it doesnt matter how fast you do it, the pile still only increases one brick at a time doesnt it.
All we need to know is how many threads per inch are cut on the lead screw of your lathe and also how many threads per inch we want to cut. If your playing around in metric, it is still the same story okay.
I dont wish to insult anyones intelligence here, so forgive me when I say, if you dont know how many threads per inch (TPI) are on your lead screw, this is how you find out. Measure from the end of one tooth for an inch and count complete teeth over that distance. Or if metric over a suitable equivalent, like this (example). Which is 8 tpi.
Okay, so lets assume we wish to cut 16tpi, a nice round number and we have a lead screw of 8tpi. If the lead screw revolved at the same speed as the mandrel then we would cut 8 tpi.
Express this as a vulgar fraction as 16/8 this is the ratio and if we had these gears ie one with 16 teeth and the other with 8 all would be well, but we dont, so let's change it and it is very simple, anything you do to the top you must also do to the bottom, thus the ratio doesnt change. for example by 2 = 32/16 or by 5 = 80/40 what if we now divide the latter by 2 we get 40/20, it's exactly the same ratio ie; 2 to 1, but we might just have 2 gears of 40 and 20 teeth, so we are done. Now it is the lead screw that we want to slow down so the 40 tooth goes on there and the 20 on the mandrel, or we will be cutting twice as fast ie instead of 16 tpi it will cut 4 tpi, got it......not that difficult.
Now we shall go on to something that is or at least appears to be difficult, but it is exactly the same formula, we just need to shake it around a bit. Lets assume we want to cut 40 tpi. So 40/8 and the answer is of course 5. We multiply up that fraction so that the denominator is equal to one of our change wheels, so by 5 = 200/40, well we most likely will have the 40, but not the 200, so what to do next. Well what will make up 200/40, how about 20/8 x 10/5, we now have two fractions that still make up the same answer, check it yourself. We still dont have gears that size, so now we can once again change the numerators and denominators to suit. lets take the first one 20/8 x 3 = 60/24 these two gears we have, now for the second set 10/5 x 5 = 50/25 and we shall assume we have these too. So the final result is 60/24 x 50/25 = 5 which is of course what we started with, and the gears we need to achieve this are those four 60,24,50,25 and we make up a gear train to suit. Just one more thing to say here and that is that the driven wheels are the 60 and the 50 the driving wheels the 24 and 25, so all you now need do is compound together one driving and one driven the other two taking up their respective places, one on the mandrel and the other on the leadscrew, hey presto. And thats about as hard as it gets with one exception.
What if the thread we want to cut doesnt work out to (x)tpi, lets suppose like my Fenn mandrel it is something weird, and for our example we will choose 6.25 tpi. Okay complex, no, no its not. Just look at it in a different light what can you do to it to make it a whole number, in this particular case multiply it by 4 and you get 25, so all you now need do is allow for your leadscrew, so the result is in four inches there are 25 threads, and in four inches of leadscrew there is 32 threads, so the ratio is 25/32 and off we go again to find some suitable gears, just as we did above.
And just to put it in perspective what if we cant make it an exact amount over a set amount of inches, well if your only cutting a thread half an inch long and your error is out by ten thousandths of an inch in four inches, what is the problem, the nut wont know and that's a fact.