How to plot the point where two complicated graphs cross?

Hello! I want to plot blue points where the red circle graph meets the dark blue graph, but I'm having trouble. I'd like to get this figured out before my presentation tomorrow afternoon.

Here's what my code plots out:

Here's my current code:

clear; clc; close all;

% variables used

jd = juliandate(2020,10,13); %launch date; first opposition

jd_1 = juliandate(2022, 12,8); % next opposition

jd_2 = juliandate(2025,1,24); %end of time date

t = linspace (jd, jd_2, 4*366+10);

syms tt ttt

transfer = double(solve(149.6+160*(tt-jd)/687-227.9,tt)); %moment spaceship makes it to Mars

i=3;

transfer4 = transfer - jd

transfer2 = transfer + 883; %Spaceship starts return to earth

transfer3 = transfer2 + transfer4 %time spaceship completed return to earth

% creating arrays to plot the earth, moon, mars, and my spaceship

x_earth=149.6*cos(2*pi*(t-jd)/365.2);

y_earth=149.6*sin(2*pi*(t-jd)/365.2);

x_moon=20*cos(2*pi*(t-jd)/29.53) + x_earth ;

y_moon=20*sin(2*pi*(t-jd)/29.53) + y_earth;

x_mars=227.9*cos(2*pi*(t-jd)/687);

y_mars=227.9*sin(2*pi*(t-jd)/687);

x_earthtoMars=(149.6+160*(t-jd)/687).*cos(2*pi*(t-jd)/687);

y_earthtoMars=(149.6+160*(t-jd)/687).*sin(2*pi*(t-jd)/687);

x_MarsSatellite = 10*cos(2*pi*(t-transfer+2)/10.23) + x_mars;

y_MarsSatellite = 10*sin(2*pi*(t-transfer+2)/10.23) + y_mars;

x_MarstoEarth=(227.9-160*(t-transfer2)/687).*cos(2*pi*(t-jd)/687);

y_MarstoEarth = (227.9-160*(t-transfer2)/687).*sin(2*pi*(t-jd)/687);

%while ((double(x_MarstoEarth(ttt)) == double(x_earth(ttt)))) && ((double(y_MarstoEarth(ttt)) == double(y_earth(ttt))))

% transfer3 = double(solve(227.9-160*(ttt-transfer2)/687 - 149.6, ttt));

%end

msinitial_velocity = 9.9814681210623879384561120582467; %miles per second

initial_velocity = 1.3878963905307975340027267293406; %million km / day

audience_intial_velocity_ms = input ("Pick a number between 8 and 11.") %miles per second

audience_intial_velocity_kms = audience_intial_velocity_ms/0.6213712 %km per second

audience_intial_velocity_mkms = (audience_intial_velocity_kms/1000000) %million km per second

audience_initial_velocity_mkmh = audience_intial_velocity_mkms*3600 %million km per hour

audience_intial_velocity_mkmday = audience_initial_velocity_mkmh*24 %million km per day

x_earthtoMarsalt=(149.6+(230*audience_intial_velocity_mkmday)*(t-jd)/687).*cos(2*pi*(t-jd)/687);

y_earthtoMarsalt=(149.6+(230*audience_intial_velocity_mkmday)*(t-jd)/687).*sin(2*pi*(t-jd)/687);

x_MarstoEarthalt=(227.9-(230*audience_intial_velocity_mkmday)*(t-transfer2)/687).*cos(2*pi*(t-jd)/687);

y_MarstoEarthalt = (227.9-(230*audience_intial_velocity_mkmday)*(t-transfer2)/687).*sin(2*pi*(t-jd)/687);

%while ((double(x_MarstoEarth(ttt)) == double(x_earth(ttt)))) && ((double(y_MarstoEarth(ttt)) == double(y_earth(ttt))))

% transfer3 = double(solve(227.9-160*(ttt-transfer2)/687 - 149.6, ttt));

%end

%{

i3=0

for i=1:length(t)

if (t(i)> transfer3)

i3 = i

i = length(t)

end

end

%}

% plotting the sun

fhandle = figure

fhandle.Position(3) = 2 * fhandle.Position(3)

% Plot both versions simultaneously

subplot(1,2,1)

plot (0,0, '*', 'color', 'yellow', 'linewidth', 8)

hold on

xlabel('horizontal movement in space')

ylabel('vertical movement in space')

title('Planets around the Sun')

subplot(1,2,2)

plot (0,0, '*', 'color', 'yellow', 'linewidth', 8)

xlabel('horizontal movement in space')

ylabel('vertical movement in space')

title('Planets around the Sun')

hold on

while i<=4*368

subplot(1,2,1)

hold on

xlim([-300 300])

ylim([-300 300])

plot([x_moon(i) x_moon(i+2)], [y_moon(i) y_moon(i + 2)], 'k')

plot([x_moon(i-2) x_moon(i)], [y_moon(i-2) y_moon(i)], 'Color', 'white')

if t(i) <= transfer

plot([x_earth(i) x_earth(i+2)], [y_earth(i) y_earth(i + 2)], 'g')

plot([x_earthtoMars(i) x_earthtoMars(i+2)], [y_earthtoMars(i) y_earthtoMars(i + 2)], 'b')

plot([x_mars(i) x_mars(i+2)], [y_mars(i) y_mars(i + 2)], 'r')

else if ((transfer < t(i))) && ((t(i) <= transfer2))

plot([x_MarsSatellite(i) x_MarsSatellite(i+2)], [y_MarsSatellite(i) y_MarsSatellite(i + 2)], 'b')

plot([x_MarsSatellite(i-2) x_MarsSatellite(i)], [y_MarsSatellite(i-2) y_MarsSatellite(i)], 'Color', 'white')

plot([x_earth(i) x_earth(i+2)], [y_earth(i) y_earth(i + 2)], 'Color', 'cyan')

plot([x_mars(i) x_mars(i+2)], [y_mars(i) y_mars(i + 2)], 'Color', 'magenta')

else if ((transfer2 < t(i))) && ((t(i) <= transfer3))

plot([x_MarstoEarth(i) x_MarstoEarth(i+2)], [y_MarstoEarth(i) y_MarstoEarth(i + 2)], 'b')

plot([x_earth(i) x_earth(i+2)], [y_earth(i) y_earth(i + 2)], 'g')

plot([x_mars(i) x_mars(i+2)], [y_mars(i) y_mars(i + 2)], 'r')

end

end

end

subplot(1,2,2)

plot([x_moon(i) x_moon(i+2)], [y_moon(i) y_moon(i + 2)], 'k')

plot([x_moon(i-2) x_moon(i)], [y_moon(i-2) y_moon(i)], 'Color', 'white')

if t(i) <= transfer3

plot([x_earth(i) x_earth(i+2)], [y_earth(i) y_earth(i + 2)], 'g')

plot([x_earthtoMarsalt(i) x_earthtoMarsalt(i+2)], [y_earthtoMarsalt(i) y_earthtoMarsalt(i + 2)], 'b')

plot([x_mars(i) x_mars(i+2)], [y_mars(i) y_mars(i + 2)], 'r')

end

hold on

xlim([-300 300])

ylim([-300 300])

i = i + 2;

pause(0.0000000001)

end

subplot(1,2,1)

plot (x_earthtoMars(fix(transfer-jd)),y_earthtoMars(fix(transfer-jd)), '*', 'color', 'blue', 'linewidth', 5)

plot (x_MarstoEarth(fix(transfer2-jd)),y_MarstoEarth(fix(transfer2-jd)), '*', 'color', 'blue', 'linewidth', 5)

subplot(1,2,2)

plot (x_earthtoMarsalt(fix(transfer-jd)),y_earthtoMarsalt(fix(transfer-jd)), '*', 'color', 'blue', 'linewidth', 5)

% position is in millions of kilometers

% time is in days

Answer 1

[Chad Greene]

Does this do it for ya? https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections

[Rebekah Ericson]

How do I run it?

Answer 2

[Star Strider]

Clever animation!

Since I assume you wnat to know where the ‘x_eartoMars’ and ‘x_mars’ intersect, the easiest way is to simply find where those two curves cross:

int_idx = find(diff(sign(x_earthtoMars-x_mars)))

Plotting them as:

figure
plot(t,x_earthtoMars,'-b', t,x_mars,'-r')
hold on
plot(t(int_idx),x_earthtoMars(int_idx),'sb', t(int_idx),x_mars(int_idx),'dr')
hold off
grid

demonstrates that the indices are likely precise enough that it is not necessary to interpolate them to get more precise values. (In any event, when I attempted that it was possible to get accurate x-intercept values, however not y-intercept values, so using the indices is likely all that is necessary.)

To get other intersections, use this idea with other vector differences.