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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