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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