Erik Lane wrote: > Having the natural log of 0 might be an issue. I'm pretty sure that > the natural log of 0 would be negative infinity. That would likely not > produce a straight line graph.... > You can take care of that with one or another expediency. One thing you can do is calculate using the 'if': if(that cell == 0; 0; ln(that cell)). The other thing you can do is to just don't graph the first point. > Just have two extra rows where you do your computations. ( =LN(A1) > and =1/B1 ) for example. Then just copy and paste and it will give you > your whole series. > > Then make the chart from those pre-computed values. > > (But isn't a vapor pressure of 0 a vacuum? Is it not 1 for something > at equilibrium in a lab environment? I could be really misremembering, > of course.) > It wouldn't be 1 for something in a sealed container. Perfectly 0 is unachievable, too, but "down in the noise", certainly.
Perhaps he's introducing ethanol into an evacuated flask. > On Sun, Feb 14, 2010 at 4:32 PM, Michael Robinson > <plu...@robinson-west.com> wrote: > >> I have temperature values in Kelvin corresponding to vapor pressure >> values in kPa's. I need to graph natural_log(vapor pressure) as Y >> and (1/temperature) for the X axis. How do I do this in Openoffice >> Calc??? I should end up with a straight line, but I probably won't. >> >> A problem is that my initial vapor pressure is 0, no ethanol in the >> sealed container. I don't think the natural_log(0) is defined. >> >> _______________________________________________ >> PLUG mailing list >> PLUG@lists.pdxlinux.org >> http://lists.pdxlinux.org/mailman/listinfo/plug >> >> > _______________________________________________ > PLUG mailing list > PLUG@lists.pdxlinux.org > http://lists.pdxlinux.org/mailman/listinfo/plug > > -- Tim Wescott Wescott Design Services Voice: 503-631-7815 Cell: 503-349-8432 http://www.wescottdesign.com _______________________________________________ PLUG mailing list PLUG@lists.pdxlinux.org http://lists.pdxlinux.org/mailman/listinfo/plug
