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Converting Centimeters to Inches


Julia D.

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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

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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.

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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

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