Re: [libreoffice-users] Re: New User: Three Questions

On 19 Jul 2011 at 12:26, Rich Shepard wrote:

Date sent:              Tue, 19 Jul 2011 12:26:12 -0700 (PDT)
From:                   Rich Shepard <rshepard@appl-ecosys.com>
To:                     users@global.libreoffice.org
Subject:                Re: [libreoffice-users] Re: New User: Three 
Questions
Send reply to:          users@global.libreoffice.org

> On Tue, 19 Jul 2011, Andreas Säger wrote:
>
> > =X1*24 returns the time number in cell X1 in unit "Hours".
> > =X1*24*Y1 multiplies the hours with an hourly wage in Y1.
>
> Andreas,
>
>    It's not working for me here. Cell D3 contains '01:00:00 PM' and cell E3
> contains '07:00:00 PM'. In cell F3 I have '=(E3*24)-(D3*24)' and the display
> is 0.00. If cell F3 contains '=E3-D3' then F3 shows '06:00'. The display for
> 1-1/4 hours is '01:15' rather than 1.25. As soon as I put parentheses around
> the cells the resulting display is '0.00'. Strange.

=(E3-D3)/TIME(1,0,0)

The above formula seems to work for my test?

01:00:00 PM     07:00:00 PM     6
01:00:00 PM     04:15:00 PM     3.25

Note that with dates have two parts. The part to the left of the 
decimal is the day, and the right is the time. If both dates are the 
same day, there isn't a problem. One can always use the mod 
function to get rid of the day  portion if.

 =(mod(e3,1)-mod(d3,1)) /time(1,0,0)

The division by time(1,0,0) converts the hours into a numeric 
value.







> Rich
>
>
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