Home |
Search |
Today's Posts |
#1
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
I need to enter 10 chars to left of decimal and 5 to right.
I need to enter more than 10 numbers to left of the decimal and more than 5
to the right of the decimal. Right now if I enter more than 10 numbers to the left of the decimal, the number to the right of will start to round automatically. |
#2
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
I need to enter 10 chars to left of decimal and 5 to right.
If you need to do math with these numbers, you'll have to find a
workaround - XL only retains 15 decimal digits of precision. If it's a part number or some other number that you won't do math on, preformat the cell as Text, or prefix the number with an apostrophe ('). In article , "Andre" wrote: I need to enter more than 10 numbers to left of the decimal and more than 5 to the right of the decimal. Right now if I enter more than 10 numbers to the left of the decimal, the number to the right of will start to round automatically. |
#3
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
I need to enter 10 chars to left of decimal and 5 to right.
Hi Andre,
Excel only allows 15 significant digits, which is the quantity of digits from the left-most non-zero digit to the right-most non-zero digit. This limitation is based on an industry standard called "IEEE 754", which increases calculation speed by sacrificing precision. Other spreadsheet programs have the same or very similar limitation. If you don't need to do any math with these numbers of yours that have more than 15 significant digits, you can enter them as text, by either pre-formatting the cell as Text, or by entering a leading apostrophe in the cell. If you do want to do any math with these numbers, you might want to try my Excel add-in xlPrecision 2.0, which allows doing math on numbers with as many as 32,767 significant digits. You can download the free edition of xlPrecision 2.0 here and use it as long as you wish: http://PrecisionCalc.com Thanks, Greg Lovern http://PrecisionCalc.com Get Your Numbers Right "Andre" wrote in message ... I need to enter more than 10 numbers to left of the decimal and more than 5 to the right of the decimal. Right now if I enter more than 10 numbers to the left of the decimal, the number to the right of will start to round automatically. |
#4
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
I need to enter 10 chars to left of decimal and 5 to right.
Greg Lovern wrote...
.... If you do want to do any math with these numbers, you might want to try my Excel add-in xlPrecision 2.0, which allows doing math on numbers with as many as 32,767 significant digits. You can download the free edition of xlPrecision 2.0 here and use it as long as you wish: .... Excel 10 (2002) SP-2. I downloaded and installed the free version. A few quirks (read: NASTY bugs). The formula =xlpROOT(2,3) works as expected, but =xlpROOT(2,1/3) hangs Excel, requiring Task Manager to kill the Excel process. Exception handling doesn't appear to be a strength of this product. Note that in real math, ROOT(x,y) = POWER(x,1/y), but that doesn't seem to be how xlpROOT works. Note that =xlpROOT(2,xlpDIVIDE(2,3)) didn't do any better. |
#5
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
I need to enter 10 chars to left of decimal and 5 to right.
Hi Harlan,
Thanks for catching that bug! It's the first one reported. I've fixed the bug, and I'll get the fixed version up on the website ASAP, after I've done more testing. "Harlan Grove" wrote in message ups.com... I downloaded and installed the free version. A few quirks (read: NASTY bugs). The formula =xlpROOT(2,1/3) hangs Excel, requiring Task Manager to kill the Excel process. Exception handling doesn't appear to be a strength of this product. It was an infinite loop. Note that in real math, ROOT(x,y) = POWER(x,1/y), but that doesn't seem to be how xlpROOT works. As I'm sure you know, Excel turns this: =xlpROOT(2, 1/3) into this, before sending it to xlPrecision: =xlpROOT(2, 0.333333333333333) So the xlpPOWER equivalent would look like this: =xlpPOWER(2, 1/0.333333333333333) However, Excel rounds 1/0.333333333333333 to 8 before sending anything to xlPrecision. So, it isn't equivalent after all. An exact or nearly exact xlpPOWER equivalent would require 1/0.333333333333333 taken to infinite precision. You could get close to the equivalent with this: =xlpPOWER(2,xlpDIVIDE(1,0.333333333333333)) Except that this is affected by the same underlying bug (and fixed by the same bugfix). I'd offer you a free copy of a paid version of xlPrecision for finding this bug, except that I know from previous posts of yours that you'd have no use for it, and might even prefer a root canal. Still, it's yours for the asking if you're interested. Thanks again, Greg Lovern http://PrecisionCalc.com Get Your Numbers Right |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|