While-loop only iterates a single time

I'm trying to find a value for h by first guessing a value (0.00001<h<0.003) and then using a while loop to find the correct value.
W = 0.3; %Channel width (m).
S = 0.006; %Channel slope (m/m).
ks = 0.0013; %Nikuradse roughness length (m).
Q = 1e-3; %Water flow (m^3/s)
h = 0.0004; %Initial water depth to start while-loop.
htemp = .9*h; %Controlling value to determine if we are getting a better value for h.
while (h-htemp)>(1e-6)
R = (W * h) / (W + 2*h); %Calculate hydraulic radius.
C = 18*log10((12*R)/ks); %Calculate grain-related Chezy number.
u = C*(h*S)^(.5); %Calculate flow velocity.
htemp = Q / (W*u); %Calculate depth of the water.
h = (h+htemp)/2
end
However, I only seem to get the while loop to do a single iteration, even though the end result does not satisfy the initial condition. How could I get this to work?

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Image Analyst
Image Analyst 2018년 9월 25일
편집: Image Analyst 2018년 9월 25일

0 개 추천

Simple debugging would have let you know. But to be more explicit, try this where I added fprintf's to your code:
W = 0.3; %Channel width (m).
S = 0.006; %Channel slope (m/m).
ks = 0.0013; %Nikuradse roughness length (m).
Q = 1e-3; %Water flow (m^3/s)
h = 0.0004; %Initial water depth to start while-loop.
htemp = .9*h; %Controlling value to determine if we are getting a better value for h.
while (h-htemp)>(1e-6)
R = (W * h) / (W + 2*h); %Calculate hydraulic radius.
C = 18*log10((12*R)/ks); %Calculate grain-related Chezy number.
u = C*(h*S)^(.5); %Calculate flow velocity.
htemp = Q / (W*u); %Calculate depth of the water.
h = (h+htemp)/2;
fprintf('h = %f\n(h-htemp) = %f\n', h, (h-htemp));
if (h-htemp)>(1e-6)
fprintf('It is greater than 1e-6 so the loop will iterate again.\n');
else
fprintf('It is NOT greater than 1e-6 so the loop will not iterate again.\n');
end
end
You will see
h = 0.105771
(h-htemp) = -0.105371
It is NOT greater than 1e-6 so the loop will not iterate again.

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Thom21
Thom21 2018년 9월 25일
A friend also kindly pointed me to this obvious flaw in my code. I changed it to abs(h-htemp), which consistently gives me the same answer :)

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