Filter löschen
Filter löschen

arbitrary point with the know slope to another line to specify point on that second line

1 Ansicht (letzte 30 Tage)
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 Kommentare
Sam Chak
Sam Chak am 24 Jun. 2022
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
Ceren Ecemsu Varan
Ceren Ecemsu Varan am 24 Jun. 2022
the main red line is operating line
the main blue line is equilibrium line
and I want to know the points that end on the blue line for that 10 interface line

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Sam Chak
Sam Chak am 24 Jun. 2022
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)
interX = 0.0042
% 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

Produkte


Version

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by