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Linear Dimension to GD&T Callout

Linear Dimension to GD&T Callout

Linear Dimension to GD&T Callout

(OP)
Hi All

Used this forums many times but never posted. I've got a query regarding converting a basic linear dimension using a specification to a GD&T callout this is the example below.

If the specification states a hole location is +/-0.3 from a edge what would this equate to for the feature control frame:

POS|dia??|A|B|C

Most drawings I deal with are fully GD&T and its not very often I need to refer back to a specification but I remember there is some form of specific conversion

Thanks in advance

RE: Linear Dimension to GD&T Callout

First, I cannot see the enclosure. The most typical problem is symbol "&" in the file name.

The most typical "conversion" everyone is looking for is square root of 2 or 1.4

So it makes dia 0.85

"For every expert there is an equal and opposite expert"
Arthur C. Clarke Profiles of the future

RE: Linear Dimension to GD&T Callout

CH beat me to it...

a2 + b2 = c2 -> .32 * 2 = .18
√.18 = .42426 = r
2 * r = TP Ø = .8485

"Know the rules well, so you can break them effectively."
-Dalai Lama XIV

RE: Linear Dimension to GD&T Callout

If you want to define position tolerance zone that is cylindrical, you will not find a value of diameter that would result in exactly the same requirement.
Usually in situations like this the approach is to set diameter of the cylindrical tolerance zone at 0.6*sqrt(2), but that gives a tolerance zone that is approximately 57% larger in area in comparison to square +/-0.3 tolerance zone. In most cases this is exactly how position of a hole should be defined, but that is not 1 to 1 transition from square tz.

However, if that is really what function requires (which I doubt), you can define a square tolerance zone of 0.6 width/height using a method shown in fig. 7-28 (Y14.5-2009) or fig. 5-41 (Y14.5M-1994); if you put 0.6 in both feature control frames (with no M modifier) you will define 0.6x0.6 tolerance zone.

RE: Linear Dimension to GD&T Callout

(OP)
Thanks for the quick replies

now its been mentioned i recognise using that equation to calculate before

what a great forum!

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