Iterations until condition is met

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aregr8
aregr8 2023-4-7
评论： aregr8 2023-4-7
clear all; clc
ER = 50;
gamma = 1.4;
for M=1:0.01:10
if (1/M) * ((2/(gamma+1))*(1+((gamma-1)/2)*(M^2)))^((gamma+1)/(2*gamma-2)) == ER
mach = M
break
end
end
I am trying to iterate M from 1 to 10 in intervals of 0.01 until the equation value matches my value for ER. Can you please help me figure out where I went wrong?

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采纳的回答

Torsten
Torsten 2023-4-7
Ran in:
ER = 50;
gamma = 1.4;
M = 1:0.00001:10;
expr = (1./M) .* ((2/(gamma+1))*(1+((gamma-1)/2)*(M.^2))).^((gamma+1)./(2*gamma-2)) - ER;
[error,index] = min(abs(expr))
error = 1.6950e-04
index = 491378
Mstar = M(index)
Mstar = 5.9138

更多回答（1 个）

the cyclist
the cyclist 2023-4-7
It is because of the accuracy of floating-point operations.
Instead of checking
if (1/M) * ((2/(gamma+1))*(1+((gamma-1)/2)*(M^2)))^((gamma+1)/(2*gamma-2)) == ER
you could do something like
tol = 1.e-6; % Choose a suitable tolerance here
if abs(((1/M) * ((2/(gamma+1))*(1+((gamma-1)/2)*(M^2)))^((gamma+1)/(2*gamma-2))) - ER) < tol
Probably even better would be to use a while loop instead of a for loop, which is the more natural construct for looping until a condition no longer holds. (But you may still need to be careful about floating point precision.)

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