How to display Intersection of two graphs

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Austen Thomas
Austen Thomas 2018-3-15
评论： David Lont 2020-3-28
I am writing a code and plotting to set of equations against each other T_R and T_A. How can you make it so it disaplys the point where the two lines intersect?
I have attached the code and a picture of the graph thanks to anyone who can help
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Austen Thomas
Austen Thomas 2018-3-15
编辑：Walter Roberson 2018-3-15
%%Problem 1 a
clc, clear
S = 135.2; % ft^2
b = 35.5; % ft
C_D0 = 0.019;
e = 0.83;
W = 3000; % lbs
P_SSL = 195 * 0.85; %hp
rho_SSL = 0.002377;
AR = 9.321;
K = 1/(pi*e*AR);
hG = 1;
rho = 0.002377;
v = 40:10:750;
for i = 1:length(v)
q(i) = 0.5 * rho * v(i)^2;
C_L(i) = W / (q(i) * S);
Dp(i) = q(i) * S * C_D0;
Di(i) = q(i) * S * K * C_L(i)^2;
T_R(i) = Dp(i) + Di(i);
end
plot(v,T_R);
hold on
for i = 1:length(v)
P_A(i) = P_SSL * (1.132 * (rho / rho_SSL) - 0.132);
T_A(i) = (P_A(i) * 550) / v(i);
end
David Lont
David Lont 2020-3-28
Thank you! This is very useful as I believe it is the same exact assigment I have with Dr. Ro at WMU in spring of 2020.

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采纳的回答

Star Strider
Star Strider 2018-3-15
You need to create ‘T_A’ and ‘T_R’ as anonymous functions. After that, you can use fzero to calculate the intersection.
The Code —
S = 135.2; % ft^2
b = 35.5; % ft
C_D0 = 0.019;
e = 0.83;
W = 3000; % lbs
P_SSL = 195 * 0.85; %hp
rho_SSL = 0.002377;
AR = 9.321;
K = 1/(pi*e*AR);
hG = 1;
rho = 0.002377;
v = 40:10:750;
q = @(v) 0.5 * rho * v.^2;
C_L = @(v) W ./ (q(v) * S);
Dp = @(v) q(v) * S * C_D0;
Di = @(v) q(v) * S * K .* C_L(v).^2;
T_R = @(v) Dp(v) + Di(v);
P_A = P_SSL * (1.132 * (rho / rho_SSL) - 0.132);
T_A = @(v) (P_A * 550) ./ v;
v_int = fzero(@(v) T_R(v) - T_A(v), 300);
plot(v, T_R(v), v, T_A(v))
hold on
plot(v_int, T_R(v_int), 'pg', 'MArkerFaceColor','g', 'MarkerSize',8)
hold off
grid
xlabel('v')
legend('T_R', 'T_A')
text(v_int+20, T_R(v_int)+20, sprintf('\\leftarrow (%.1f, %.1f)', v_int, T_R(v_int)), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

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