Re: [libreoffice-users] To make a graph

Hi Brian:

        As I said to Regina, thank you very much for your explanation. 
I could reproduce the graph as you instructed to me.

        I found both graph clear and making what I need.

        Yes, I need the graph and the coordinates of the interception 
both lines too. I follow your equation to find U, and I reproduced it 
well with the same result. But I tried to do the same with A coordinate 
interception: 14000 * ( 1 - U / 1.31 ) and I got : 5813 not 9126. As you 
said, 9126 is the correct point where intercept but I couldn't get in my 
calculation. Would you please, tell me what would I do wrong ?

Regards,

Jorge Rodríguez


El 29/08/2017 a las 11:31, Brian Barker escribió:
> At 09:45 29/08/2017 -0600, Jorge Rodríguez wrote:
> > Below you can find a link to access a file were I draw manually the 
> > graph that I need to do:
> > https://www.dropbox.com/s/7oqptiz8e6r9gxi/Graph.pdf?dl=0
>
> So your data set is actually:
> A       U1      U2
> 0       1.37    1.31
> 13680   0
> 14000           0
> (with some missing values).
>
> You will have seen Regina Henschel's recipe for a graph. Here is another:
> o Select the data block as above and click the Chart button.
> o Select "X-Y (Scatter)" and leave all other options as default.
> o Select the chart (grey border, not coloured handles).
> o Right-click a point and select Insert Trend Line... and OK.
> o Repeat for a point on the other line.
>
> But do you really want to see the graph, or perhaps just find the 
> coordinates of the point of intersection, as was perhaps hinted at by 
> your original request? Application of similar triangles shows that the 
> U coordinate of the point of intersection is given by
> 1.31 * 1.37 * (14000 - 13680) / (1.37 * 14000 - 13680 * 1.31)
> and that the A coordinate is
> 14000 * ( 1 - U / 1.31)
> - where U is the U coordinate calculated above.
>
> You will see that these give the intersection point as A=9126, U=0.456.
>
> I trust this helps.
>
> Brian Barker


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