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?

 채택된 답변

Torsten
Torsten 2023년 4월 7일

1 개 추천

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

추가 답변 (1개)

the cyclist
the cyclist 2023년 4월 7일

0 개 추천

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.)

카테고리

도움말 센터File Exchange에서 Loops and Conditional Statements에 대해 자세히 알아보기

제품

릴리스

R2020b

질문:

2023년 4월 7일

댓글:

2023년 4월 7일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by