Iterations until condition is met

3 Ansichten (letzte 30 Tage)
aregr8
aregr8 am 7 Apr. 2023
Kommentiert: aregr8 am 7 Apr. 2023
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?

Akzeptierte Antwort

Torsten
Torsten am 7 Apr. 2023
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

Weitere Antworten (1)

the cyclist
the cyclist am 7 Apr. 2023
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.)

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Produkte


Version

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by