Chuck speed of a lathe

by JamesCalculus
Tags: chuck, lathe, speed
 P: 2 Hi, i would be grateful if anyone could help. When using a lathe to cut a spindle of 10mm diameter at a cutting speed of 45m/min, how would i convert the 45m/min into rev/min and rad/s? Also how would i covert the cutting speed of 45m/min into mm/s for a spinde of 7mm? Thanks
 P: 1,127 So, the spindle is 10 mm diameter. The circumference is thus 31 mm or 0.031m which is 0.031 m/revolution. Then 45 m/min divided by 0.031 m/revolution gives 1450 rpm. Note that, as your tool goes into the spindle, the surface speed will decrease as the diameter.
 Sci Advisor P: 5,095 This question is posed rather strangely. The 45 m/min looks like a feed rate. It appears that the OP needs a depth of cut number. I'm probably misunderstanding the problem.
HW Helper
P: 2,537
Chuck speed of a lathe

 Quote by FredGarvin This question is posed rather strangely. The 45 m/min looks like a feed rate. It appears that the OP needs a depth of cut number. I'm probably misunderstanding the problem.
I think this is just a conversion problem:
1 revolution=2 $\pi$ radians
1 minute = 60 seconds
1 meter = 1000 millimeters

Now, in this problem
1 diameter = 10 mm

So - to get from meters per minute to revolutions per second:
45 meters/minute * (1 minute/60 seconds) * (1000 milimeters/1 meter) * (1 diameter/10 millimeters) * (2 radians/1 diameter) * (1 revolution/ 2 $\pi$ radians)

Each of the fractions is 1, and you should be able to cross off the units. Regrouping gives...
(45 * 1000 * 2) revolution / (60*10*2 $\pi$) seconds.
P: 1,127
 Quote by FredGarvin This question is posed rather strangely. The 45 m/min looks like a feed rate. It appears that the OP needs a depth of cut number. I'm probably misunderstanding the problem.
I think (hope) that he was asking for cutting speed in surface m/min cause that's how I answered. It sounds about right for free machining low carbon steel (~150 fpm). OP, help us out. But, yes, he sorta asked it backwards of the way you normally hear it, so I'm not sure.

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