mir
Guest
Hi.
Sorry but the quota on the diameter
+0.5
168.3
+0.6
What actual diameter would it give me?
Thank you.
♪
Sorry but the quota on the diameter
+0.5
168.3
+0.6
What actual diameter would it give me?
Thank you.
♪
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zeigs
Guest
that I know is as to say 168.8 (+0.1;0). that in parentheses seems more logical
Marco74
Guest
"o pienamente.that I know is as to say 168.8 (+0.1;0). that in parentheses seems more logical
ciao.
mir
Guest
I just imagined that I wanted to hold the correct nominal quota then I tried in cad to make this notice. I thought it didn't work a tolerance range that didn't include the nominal fee.
Thank you.
♪
Thank you.
♪
zeigs
Guest
apart h, j, js and k are all deviations that do not include the nominal diameter. But usually we talk about cents of deviation from the nominal
radio
Guest
if you want to use unified tolerances you may consider your share as:I just imagined that I wanted to hold the correct nominal quota then I tried in cad to make this notice. I thought it didn't work a tolerance range that didn't include the nominal fee.
Thank you.
♪
if it is a tree
168.3 z9 +0.465/+0.565 or
168.3 z10 +0.465/+0.625
if it is a hole you may consider an a9 or a10 or a b9 or b10 but in this case you are less close to the tolerance you require at first.
said this, I also agree with zeigs, 168.8 +0.1/-0 in general seems more logical.
Bye.
zeffiro62
Guest
notes the greater tolerance should be placed above the lower tolerances should be expressed in microns
e.g. 8 +0.010
-0.010
you have done exactly the contractor
e.g. 8 +0.010
-0.010
you have done exactly the contractor
radio
Guest
you are right for positions, I reversed them; but the values are correct.notes the greater tolerance should be placed above the lower tolerances should be expressed in microns
e.g. 8 +0.010
-0.010
you have done exactly the contractor
Bye.
zeigs
Guest
only one note, I don't want it zeffiro62: tolerances should be expressed in mm. then if you tell me that you have to put 3 decimals to show that precision in defining tolerance is the micron, therefore also the last zero is significant, then I can agree. after this statement you can give me some mathematician:biggrin:
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ciao