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Math problem - gear jammer - 11-29-2016
I am looking at on old delta lathe. It has a 1 hp at 2450 rpm motor with a 2 inch pulley to a 3 pulley driven on the shaft. 2,3and 4 inch. I would add a jack shaft to the machine as I would like one with some power at 30 to 40 rpms on the pieces I want to turn. Old root balls with a lot of dirt and very odd shapes.
How do you figure the math for getting the rpm's I want other than trial and error.
RE: Math problem - EricU - 11-29-2016
Is your current low speed really 1225 rpm? If so, it looks like it's not going to happen with one jackshaft. Maybe look for a gearbox on ebay. Of course, I violated my rule about doing algebra in my head, so I could be wrong
output speed would be = input speed* (motor pulley diameter * jackshaft output pulley diameter)/(jackshaft input pulley diameter * existing shaft pulley diameter)
If you want to know the jackshaft ratio, that is JR = desired output speed * shaft pulley /(input speed * motor pulley)
So JR = 40 * 4/(2450 * 2) = 1/30
JR = Jackshaft output pulley dia/ Jackshaft input pulley dia
RE: Math problem - BloomingtonMike - 11-29-2016
An optical non-contact tachometer for $20 or less on ebay or amazon will tell you rpm.
RE: Math problem - Ohio Mike - 11-30-2016
If the motor pulley is 2 inches and the jackshaft input pulley is 48 inches, you would get a 24x reduction. So the jackshaft would be spinning at about 102 RPM.
If you put a 2 inch pulley on the output of the shaft and connect to the 4 inch on the lathe, you would get an additional 2x reduction. Final spindle speed = 51 RPM.
Is replacing the motor a possibility? Maybe with a 3 phase motor connected to a single phase variable frequency drive (VFD)??
Mike
RE: Math problem - EricU - 11-30-2016
vfd is a good idea, you could run the motor at low speed. They start out at 1750 anyway, but a vfd would easily get it down to half that much.
My lathe used to have a 3 phase, but the previous owner replaced it with a single phase motor
RE: Math problem - Leo Van Der Loo - 11-30-2016
I don’t know where you live, but your motor speed RPM is not what you are posting, the N. American 60 cycle net gives you these speeds as indicated in the table.
The typical RPM is either 3450 or 1725 rpm, so I assume you did mis-reed it and it is 3450 rpm, and too fast for pretty well any lathe setup, going to a 1725 RPM motor would halve your starting off speed.
If you then go to a 2” pulley on your motor and a 6” speed on the jackshaft input side you are down to a jackshaft turning at approx. 575 RPM.
A 2” pulley on the output side of the jackshaft that goes to the 4” pulley on the lathe will bring that speed down to 287 RPM.
That would be a good speed for roughing out a larger blank
Ok I can’t load a picture here, anyway the speeds I gave are what we all are working with id we use a 2 or 4 pole AC motor.
You can find the table and more info here.
http://www.engineeringtoolbox.com/speed-electrical-motors-d_738.html
RE: Math problem - gear jammer - 11-30-2016
Thank you. My apologies I did misread the plate on the motor. It is 3450. It is an old lathe with the old style oil cup bearings and a sturdy frame. He wants $40 for it. It would be easy to add a jack shaft even a second one if I want. I have power washed the root balls and cut some into pieces. The figure in the wood is great on what I hand planed to get a look at. I will set the lathe up outside I don't need all that dirt still left on inside the shop I have some old tools to use to get them round then take them in on better light and better machine. I can't see putting much money into a vfd motor on and leave it outside under a tarp.