how to solve an equation by using trial and error method?

์กฐํšŒ ์ˆ˜: 30 (์ตœ๊ทผ 30์ผ)
Ali Deniz
Ali Deniz 2022๋…„ 5์›” 16์ผ
๋Œ“๊ธ€: Ali Deniz 2022๋…„ 5์›” 16์ผ
๐‘‡โ„Ž๐‘Ÿ๐‘œ๐‘Ž๐‘ก (๐ดโˆ—)=647.5 ๐‘š๐‘š2
A=1027.5 mm^2
How can I calculate ๐‘€๐‘ฅ value by using trial and error method or how can I find the value of Mx by another method. Thank you.

์ฑ„ํƒ๋œ ๋‹ต๋ณ€

Sam Chak
Sam Chak 2022๋…„ 5์›” 16์ผ
I'm not sure what the trial-and-error method is for this, but I think that basic Algebraic methods can be used to rearrange the equation. I followed the example in fzero and make an assumption:
% Assume
gamma = 1.55526564459735;
% function to be solved
f = @(M) (1/M)*((2*(1 + (gamma - 1)/2*M^2))/(gamma + 1))^((gamma + 1)/(2*(gamma - 1))) - Ax/Astar;
% initial guess
M0 = 3;
% root-finding solver
Msol = fzero(f, M0)
There should be two real roots.
  ๋Œ“๊ธ€ ์ˆ˜: 1
Ali Deniz
Ali Deniz 2022๋…„ 5์›” 16์ผ
Thank you.
-Ali

๋Œ“๊ธ€์„ ๋‹ฌ๋ ค๋ฉด ๋กœ๊ทธ์ธํ•˜์‹ญ์‹œ์˜ค.

์ถ”๊ฐ€ ๋‹ต๋ณ€ (1๊ฐœ)

Walter Roberson
Walter Roberson 2022๋…„ 5์›” 16์ผ
Multiply both sides by Mx. Subtract the left side from the right. Expand out. You now have a polynomial of degree 2 in Mx.
  ๋Œ“๊ธ€ ์ˆ˜: 2
Ali Deniz
Ali Deniz 2022๋…„ 5์›” 16์ผ
I think you neglect the power of the RHS.
Walter Roberson
Walter Roberson 2022๋…„ 5์›” 16์ผ
Ah, I am on mobile. I had tried scrolling the image but it didn't scroll so I didn't see the power.

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