Julia D. Posted February 26, 2008 Share Posted February 26, 2008 Hello. I'd like to create a calculation field that converts centimeters to inches with a maximum of 3 decimal places. I set up the field like this: Round (Centimeter_Field * 0.393700787;3) It would be very helpful if I could limit the results to standard inch increments, like: 1/2 inch (.5) 1/4 inch (.25) 3/8 inch (.375) 5/8 inch (.625) And make the results round up OR down to the closest standard increment, so that: 12.376 rounded to 12.375 12.374 rounded to 12.375 12.555 rounded to 12.5 12.430 rounded to 12.5 Is there a way to accomplish this? Thanks for your help! Julia Link to comment Share on other sites More sharing options...

Maarten Witberg Posted February 26, 2008 Share Posted February 26, 2008 try (result is text) Let ( [ inchfull = Round (centi * ,393700787;4) ; whole = Truncate ( inchfull ; 0 ) ; fraction = inchfull - whole ; precision = Case ( fraction ≤ ,0625 ; "" ; fraction ≤ ,1875 ; "1/8" ; fraction ≤ ,3125 ; "1/4" ; fraction ≤ ,4375 ; "3/8" ; fraction ≤ ,5625 ; "1/2" ; fraction ≤ ,6875 ; "5/8" ; fraction ≤ ,8125 ; "3/4" ; fraction ≤ ,9375 ; "7/8" ; "8/8" ) ]; whole + (precision = "8/8") & Case ( precision ≠ "8/8" ; " " & precision ) ) or, result is number Let ( [ inchfull = Round (centi * ,393700787;4) ; whole = Truncate ( inchfull ; 0 ) ; fraction = inchfull - whole ; precision = Case ( fraction ≤ ,0625 ; 0 ; fraction ≤ ,1875 ; ,125 ; fraction ≤ ,3125 ; ,25 ; fraction ≤ ,4375 ; ,375 ; fraction ≤ ,5625 ; ,5 ; fraction ≤ ,6875 ; ,625 ; fraction ≤ ,8125 ; ,75 ; fraction ≤ ,9375 ; ,875 ; 1 ) ]; whole + precision ) please note that these calcs only give an estimate of the original SI centimeter value. Link to comment Share on other sites More sharing options...

Julia D. Posted February 26, 2008 Author Share Posted February 26, 2008 kjoe, You are brilliant and fast! It works like a dream (I used the result is a number version). Beautiful! Thank you so much Julia Link to comment Share on other sites More sharing options...

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