Iterations until condition is met

3 ビュー (過去 30 日間)
aregr8
aregr8 2023 年 4 月 7 日
コメント済み: aregr8 2023 年 4 月 7 日
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?

採用された回答

Torsten
Torsten 2023 年 4 月 7 日
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
  1 件のコメント
aregr8
aregr8 2023 年 4 月 7 日
Thank you very much for your help.

サインインしてコメントする。

その他の回答 (1 件)

the cyclist
the cyclist 2023 年 4 月 7 日
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.)
  1 件のコメント
aregr8
aregr8 2023 年 4 月 7 日
Thank you very much for your help.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeLoops and Conditional Statements についてさらに検索

製品


リリース

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by