Calling all lathe experts!

I'm making two circular parts that fit one inside the other, the inner part is 40mm outside diameter.

How much bigger does the outer part need to be internally so that I get a good sliding fit, without sideways movement.

Millimeters (well fractions of) please, I don't do imperial

Ta.

Chris

A couple of thou? I don't do metric

1 thou = 0.001" = 0.0254 mm

Al

quote:

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Originally posted by rusty nuts
A couple of thou? I don't do metric

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You want a 40 H7 hole I think.

Shaft will be 40 g6.

ROYMECH is good for this sort of thing.

40.000 - 40.025 hole

39.991- 39.975 shaft

This on the 'hole' based system of tolerancing, the hole is generally size to plus. Ground bar normally comes size to minus and reamers will rarely cut less than the nominal size.

[Edited on 26/10/07 by Mansfield]

quote:

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Originally posted by Wadders
1 thou = 0.001" = 0.0254 mm

Al

quote:

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Originally posted by rusty nuts
A couple of thou? I don't do metric

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Of course, it all depends on the degree of surface finish of the sliding surfaces. The better the finish the tighter tolerance that you can have, up to a point.

Depending on the length of the parts unless you have a very good machine you will struggle to bore the internal parallel. If I was doing this I would turn the shaft and then ream the hole to suit using an expanding reamer.

What you are describing sounds really easy....until you try to do it. I really struggle to take a 1 thou cut on my lathe at home.

I,ll second that, just about fits, just about fits, a wafer to go.... start again.

doh

LOL Iv'e spent hours turning stuff 'somewhere near' and then seconds making it fit like a wizards sleeve.
One day i will treat my lathe to a DRO
Al.

Originally posted by mark chandler
I,ll second that, just about fits, just about fits, a wafer to go.... start again.

doh