Absolutely: 

The warpage in either axis is a direct linear %. For example if the width is 10.00 in and the allowance is 1% ( .010 in/in) , then the maximum allowable warp displacement for that axis would be 0.100".

However, twist is a different problem. If you are trying to determine the allowance for displacement at opposite diagonal ends then you must first determine the diagonal length.
The formula for that is: (Diagonal) = the square root of the total of the (length squared + width squared). Therefore the diagonal of a 10.00" x 5.00 in board would be 11.180" rounded to 3 places. Now the allowable displacement would be whatever the allowable % was for the 11.180 dimension. This will depend an what class you are building to. 

Note: if you are checking twist and using the method by which you hold down one corner while measuring the opposite corner, then don't forget to divide your displacement by 2.  

Hope this helps,

Ed Cosper
Director Quality Assurance and Engineering
Graphic Electronics Inc.

----------
From:  MIKE J. LOPEZ[SMTP:[log in to unmask]]
Sent:  Tuesday, March 04, 1997 9:55 AM
To:  [log in to unmask]
Subject:  BOARD WARPAGE

can anyone tell me what the formula is for calculating acceptable board
warpage per ipc specifications.

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