arbitrary point with the know slope to another line to specify point on that second line
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I have two lines, one of them is operating line and the other is equilibrium line. I need to specify an arbitary point on a operating line, then with the known slope I need to draw a line that stops on the equilibrium line and obtain the point on equlibrium line. How can I achieve that?
I need it for separation process course to find interface mole fraction
5 件のコメント
Sam Chak
2022 年 6 月 24 日
OK I guess you want to find the intersections between multicolored lines and the thick blue line. Right?
If so, can you provide the parameters of line equation?
y = m*x + c
回答 (1 件)
Sam Chak
2022 年 6 月 24 日
Since you are unable to provide additional info, perhaps you can do something like this:
% Analysis
x1 = linspace(0, 14e-3, 1401);
y1 = 220/7*x1; % blue line
x2 = linspace(0, 5e-3, 501);
y2 = 64*x2 + 0.03; % red line
plot(x1, y1, 'LineWidth', 1.5), hold on,
plot(x2, y2, 'LineWidth', 1.5)
x3 = linspace(2.5e-3, 5e-3, 251);
y3 = 0.28 - 36*x3; % orange line
plot(x3, y3, 'LineWidth', 1.5), hold off, axis([-3/1000 15/1000 -0.25 0.55]), grid on
% find intersection between orange line and blue line
f1 = @(x) 0.28 - 36*x - 220/7*x;
x0 = 4/1000; % initial guess (near the intersection)
interX = fzero(f1, x0)
% New Plot
x4 = linspace(2.5e-3, interX, 251);
y3 = 0.28 - 36*x4;
plot(x1, y1, 'LineWidth', 1.5), hold on,
plot(x2, y2, 'LineWidth', 1.5)
plot(x4, y3, 'LineWidth', 1.5), hold off, axis([-3/1000 15/1000 -0.25 0.55]), grid on
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