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#1
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Converting time to Minutes
Converting 5:08:00 using =(HOUR(D2) *60)+MINUTE(D2) works perfect.
That turns out 308 minutes. However, converting 119:38:00 the same way doesn't work. As a matter of fact, HOUR(XN) where the value of the cell XN = 119:38:00 returns 23. It should be 119. How can I convert 119:38:00 to 7178 minutes? |
#2
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Converting time to Minutes
Hi,
Am Thu, 21 Jul 2011 05:32:48 -0700 (PDT) schrieb rlm: Converting 5:08:00 using =(HOUR(D2) *60)+MINUTE(D2) works perfect. That turns out 308 minutes. =D2*1440 and format General Regards Claus Busch -- Win XP PRof SP2 / Vista Ultimate SP2 Office 2003 SP2 /2007 Ultimate SP2 |
#3
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Converting time to Minutes
"rlm" wrote:
Converting 5:08:00 using =(HOUR(D2) *60)+MINUTE(D2) works perfect. That turns out 308 minutes. However, converting 119:38:00 the same way doesn't work. "Claus Busch" wrote: Am Thu, 21 Jul 2011 05:32:48 -0700 (PDT) schrieb rlm: =D2*1440 and format General Although that does work by coincidence for D2 equal to 5:08:00 and 119:38:00, I would suggest the following instead: =ROUND(D2*1440,0) Try putting Claus's formula into D3, and put =D3-ROUND(D3,0)=0 into D4. Then put 1:03:00 into D2. You will see that D4 is FALSE (!). The point is: Excel time is stored as decimal fraction of a day. As with all non-integers, the representation is usually not exact. See #2 below. Some other details.... 1. Note that I write =D3-ROUND(D3,0)=0 instead =D3=ROUND(D3,0). The latter returns TRUE, but it is only an illusion due to the dubious heuristic poorly described under the misleading title "Example When a Value Reaches Zero" at http://support.microsoft.com/kb/78113. In other words, Excel forces the latter difference to zero artificially. We cannot always count on that. 2. I write ROUND(D3,0) instead of INT(D3). This is because the floating-point representation of D3 when D2 is 1:03:00 is actually less than 63, namely 62.9999999999999,928945726423989981412887573242187 5. As it happens, Excel INT(D3) does return 63 for that representation. But I argue that it is a defect. Note that VBA Int(Range("d3")) returns 62, as I would expect. |
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