# I want to obtain value of "W" for different combination of values of (psi,r) example. "W" for (psi=0.0001,r = 2.1891) ; (psi=0.001,r = 2.1940); (psi=0.01,r = 2.2373) and so on

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AVINASH SAHU on 22 Jun 2022
Commented: AVINASH SAHU on 22 Jun 2022
psi = 0.001;
alpha = (12*psi)^(1/3);
r = 2.1940;
l1 = (r + alpha)*(r + alpha);
l1a = r*r - r*alpha + alpha*alpha;
L1 = log(l1/l1a);
t1 = (2*r - alpha) / (alpha*sqrt(3));
T1 = atan(t1);
l2 = (1 + alpha)*(1 + alpha);
l2a = (1*1) - (1*alpha) + (alpha*alpha);
L2 = log(l2/l2a);
t2 = (2*1 - alpha) / (alpha*sqrt(3));
T2 = atan(t2);
W1 = 4 * sqrt(3) * (T1 - T2) * log((r+alpha)/(1+alpha));
W2 = (L1 - L2)*log((r*r - r*alpha + alpha*alpha)/(1 - alpha + alpha*alpha));
W3 = 4*(T1 -T2)*(T1 -T2);
W4 = (L1 - L2) + 2 * sqrt(3) * (T1 - T2);
W5 = 3 ./ ((r - 1) * (r - 1));
W = ((W1 + W2 -W3) * W5)/W4
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AVINASH SAHU on 22 Jun 2022
thank you for clarification but how can we use loop so that we don't have to enter manually for large dataset?

Dyuman Joshi on 22 Jun 2022
%loop
psi=[0.0001,0.001,0.01];
r=[2.1891,2.194,2.2373];
for i=1:numel(psi)
W(i)=calculateW(psi(i),r(i));
end
W
W = 1×3
0.1602 0.1595 0.1534
%arrayfun
W1=arrayfun(@calculateW, psi, r)
W1 = 1×3
0.1602 0.1595 0.1534
function W = calculateW(psi,r)
alpha = (12*psi)^(1/3);
l1 = (r + alpha)*(r + alpha);
l1a = r*r - r*alpha + alpha*alpha;
L1 = log(l1/l1a);
t1 = (2*r - alpha) / (alpha*sqrt(3));
T1 = atan(t1);
l2 = (1 + alpha)*(1 + alpha);
l2a = (1*1) - (1*alpha) + (alpha*alpha);
L2 = log(l2/l2a);
t2 = (2*1 - alpha) / (alpha*sqrt(3));
T2 = atan(t2);
W1 = 4 * sqrt(3) * (T1 - T2) * log((r+alpha)/(1+alpha));
W2 = (L1 - L2)*log((r*r - r*alpha + alpha*alpha)/(1 - alpha + alpha*alpha));
W3 = 4*(T1 -T2)*(T1 -T2);
W4 = (L1 - L2) + 2 * sqrt(3) * (T1 - T2);
W5 = 3 ./ ((r - 1) * (r - 1));
W = ((W1 + W2 -W3) * W5)/W4;
end
AVINASH SAHU on 22 Jun 2022
Thank you so much