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Place Dimensions

02/11/2010 1:42 AM

How to interpret the sentences in bold?

1.All dimensions are important and all parts should be supplied to print.

2.This system is intended to identify those features that are more important (on relative basis) than other dimensions on any print.

1 - 1 place dimensions are ± .052 -

2- 2 place dimensions are ± .013 -

3- 3 place dimensions are ± .004,

how to interpret these sentences in bold?

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#1

Re: how to interpret these sentences in bold?

02/11/2010 1:47 AM

They look like tolerances to me. Machining of the parts in question can be plus or minus the amount given for each.

For example, the reason might be a rod inserted into a socket. If they were machined exactly the same size you might never push the rod into the socket. By designing in tolerances you leave some room for the part to move. Too much tolerance and it will be loose, too little and it will be tight.

Drew

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#2

Re: how to interpret these sentences in bold?

02/11/2010 2:03 AM

It is rather surprising that 1- and 2-place dimensions would have tolerances to 3 places.

Customers/specifiers can sak for whatever they want, I suppose. They may or may not get it (at an economical price).

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#3

Re: how to interpret these sentences in bold?

02/11/2010 4:14 AM

OD = 200.0 is 199.948 to 200.052

OD = 200.00 is 199.199.987 to 200.013

OD = 200.000 is 199.996 to 200.004

Actually I have no idea, but that is how I would interpret it.

Then I would go and ask the client what it was that he thought he had said.

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#4

Re: how to interpret these sentences in bold?

02/11/2010 4:16 AM

Presumably your drawing has dimensions like 12.3, 12.34 and 12.345. These would be 1, 2 and 3 place dimensions respectively, and the specified tolerances would apply.

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#5

Re: How to Interpret These Sentences in Bold?

02/11/2010 11:13 AM

Looks like somebody converted imperial to metric without using any common sense.

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#6

Re: Place Dimensions

02/12/2010 8:30 AM

like this....

eg: 65.2" will have the ±0.52. 65.20" will be ±0.013. 65.200" will be ±0.004.

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