Calculate the difference between minimum values of a parabola and straight line (from a plot)
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Hello everyone, please could you help me with a code.
How I can calculate the values from a plot? I need the difference between straight line (P) and between the minimum value of a parabola (P) for each curve.
So, for example in this curve, the difference between the straight blue curve and the first parabola, then the next straight line and green parabola and so on.
My code is:
clear all; close all
W = 60000;
S = 28.2;
AR=7;
cd0 = 0.02;
k = 0.04;
RC=0.51;
clalpha = 2*pi;
Psl=741000;
hv=0:1:10;
cdminp=4*cd0;
clminp=sqrt(3*cd0/k);
Vmin=sqrt(2*W/(1.225*28.2*clminp));
D=0.5*1.225*Vmin^2*S*cdminp;
Pminreq=D*Vmin;
deltaPgiven=RC*W;
figure(1);hold on; xlabel('V');ylabel('P')
hv=0:1:10;
for k1 = 1:numel(hv)
h = hv(k1);
i=0;
for alpha = 1:0.25:15
i=i+1;
rho(i)=1.225*exp(-h/10.4);
cl(i) = clalpha * alpha * pi/180;
V(i) = sqrt(2*W/rho(i)/S/cl(i));
L(i) = 0.5 * rho(i) * V(i) * V(i) * S * cl(i);
cd(i) = cd0 + k * cl(i) * cl(i);
D(i) = 0.5 * rho(i) * V(i) * V(i) * S * cd(i);
clcd(i) = cl(i)/cd(i);
p(i) = D(i)*V(i);
Ph(i)=Psl*(rho(i)/1.225).^0.75;
end
figure(1); plot(V,p)
hold on
plot(V,Ph);
end
0 个评论
采纳的回答
Les Beckham
2023-2-10
编辑:Les Beckham
2023-2-10
Maybe this?
W = 60000;
S = 28.2;
AR=7;
cd0 = 0.02;
k = 0.04;
RC=0.51;
clalpha = 2*pi;
Psl=741000;
hv=0:1:10;
cdminp=4*cd0;
clminp=sqrt(3*cd0/k);
Vmin=sqrt(2*W/(1.225*28.2*clminp));
D=0.5*1.225*Vmin^2*S*cdminp;
Pminreq=D*Vmin;
deltaPgiven=RC*W;
figure(1);hold on; xlabel('V');ylabel('P')
hv=0:1:10;
for k1 = 1:numel(hv)
h = hv(k1);
i=0;
for alpha = 1:0.25:15
i=i+1;
rho(i)=1.225*exp(-h/10.4);
cl(i) = clalpha * alpha * pi/180;
V(i) = sqrt(2*W/rho(i)/S/cl(i));
L(i) = 0.5 * rho(i) * V(i) * V(i) * S * cl(i);
cd(i) = cd0 + k * cl(i) * cl(i);
D(i) = 0.5 * rho(i) * V(i) * V(i) * S * cd(i);
clcd(i) = cl(i)/cd(i);
p(i) = D(i)*V(i);
Ph(i)=Psl*(rho(i)/1.225).^0.75;
end
figure(1); plot(V,p)
hold on
plot(V,Ph);
[pmin, imin] = min(p); % find the min p
deltas(k1) = Ph(1) - pmin; % calculate the difference
tolerance = 5000; % or whatever you want
if (abs(deltas(k1) - 300000) < tolerance)
fprintf('delta = %8.1f at h = %4.1f, rho = %.5f, V = %.2f, Ph = %.1f, p = %.1f\n', ...
deltas(k1), h, rho(imin), V(imin), Ph(imin), p(imin))
end
end
legend(compose('h = %.1f', hv), 'location', 'northwest')
grid on
8 个评论
Les Beckham
2023-2-10
It just represents how close to 300000 the delta has to be to make the code print the results. Adjust as desired.
更多回答(1 个)
Torsten
2023-2-10
编辑:Torsten
2023-2-10
syms h V W S rho cd0 k cl Psl
eqn = V == sqrt(2*W/rho/S/cl);
cl = solve(eqn,cl);
cd = cd0 + k * cl^2;
D = 0.5 * rho * V * V * S * cd;
p = D*V;
Vmin = solve(diff(p,V)==0,V);
pmin = subs(p,V,Vmin);
Ph = Psl*(rho/1.225)^0.75;
hnum = 0:0.01:10;
Wnum = 60000;
Snum = 28.2;
rhonum = 1.225*exp(-hnum/10.4);
cd0num = 0.02;
knum = 0.04;
Pslnum = 741000;
for i = 1:numel(hnum)
Phnum(i) = double(subs(Ph,[rho Psl],[rhonum(i),Pslnum]));
pm = double(subs(pmin,[W S rho cd0 k],[Wnum Snum rhonum(i) cd0num,knum]));
pminnum(i) = pm(end);
deltaP(i) = Phnum(i)-pminnum(i);
end
format long
deltaP.'
idx = deltaP > 2.8e5 & deltaP < 3.2e5; % select those deltaP with 2.8e5 <= deltaP < = 3.2e5
[hnum(idx).' deltaP(idx).'] % Show these values together with the corresponding h values
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