Solving for angle (Beta) using vpasolve

Question

Benneth Perez 2021-10-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1570313-solving-for-angle-beta-using-vpasolve

评论： Star Strider 2021-10-24

Hello I'm trying to see why I'm not getting the correct angle when i Solve for the following equation. I'm not sure if this is due to a coding error that I dont see or if has to do with an angle property like the refence angle.

The equation I'm solving for is the following the resul i get is

beta =

-196.3722

However if im not mistaken the answer i should get is 23.13 deg.

M1 = 3;
P1 = 1;
theta2 = 5;
thetha3 = 20;
gamma = 1.4;
syms beta
tantheta = tand(theta2);
x = 2*cotd(beta)*((M1^2*((sind(beta))^2)-1));
y = M1^2*(gamma+cosd(2*beta))+2;
eqn = x/y;
result = vpasolve(eqn == tantheta,beta);
beta = result

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Answer 1

Star Strider 2021-10-24

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https://ww2.mathworks.cn/matlabcentral/answers/1570313-solving-for-angle-beta-using-vpasolve#answer_815008

在 MATLAB Online 中打开

That’s the result I get when I solve it (and then correct it by

) —

M1 = 3;
P1 = 1;
theta2 = 5;
thetha3 = 20;
gamma = 1.4;
syms beta
tantheta = tand(theta2);
x = 2*cotd(beta)*((M1^2*((sind(beta))^2)-1));
y = M1^2*(gamma+cosd(2*beta))+2;
eqn = x/y;
result = solve(tantheta == eqn, beta);
beta = vpa(result) + 180
beta = 23.133257450785606135761291258498

.

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Star Strider 2021-10-24

在 MATLAB Online 中打开

As always, my pleasure!

The reason is that solve and vpasolve are different functions, and work differently.

The solve function solves the equation symbolically, and then the vpa call returns a numeric equivalent (although off from the desired result by 180°).

The vpasolve function is a numeric solver, and apparently does not solve symbolically first and then createes a numeric representation of that solution. There may be several different results, and since the function is not a poplynomial, only solves for one of them. Giving it different initial estimates for β may result in different results. See the Tips section for a discussion.

There are actually six solutions, so a different starting point would be necessary for each of them in order to return all of them with vpasolve —

M1 = 3;

P1 = 1;

theta2 = 5;

thetha3 = 20;

gamma = 1.4;

syms beta

tantheta = tand(theta2);

x = 2*cotd(beta)*((M1^2*((sind(beta))^2)-1));

y = M1^2*(gamma+cosd(2*beta))+2;

eqn = x/y - tantheta;

figure

hp = fplot(eqn, [0 359]);

grid

xv = hp.XData;

yv = hp.YData;

idx = find(diff(sign(yv)));

idx = idx([1:3 5:7]);

xv(idx);

for k = 1:numel(idx)

idxrng = (-1:1)+idx(k);

xexact(k) = interp1(yv(idxrng), xv(idxrng), 0);

end

xexact

xexact = 1×6

23.3458 88.1438 163.5677 203.2841 268.1513 343.5227

hold on

plot(xexact, zeros(size(xexact)), '+r')

hold off

I did this numerically because that is simply more efficient than making numerical guesses.

So vpasolve returned a ‘correct’ solution, simply not the desired solution!

.

Walter Roberson 2021-10-24

vpasolve() uses a modified Newton-Raphson algorithm. It might need to take the derivative of the function internally first.

Star Strider 2021-10-24

As always, my pleasure!

I looked at the linked Question, and posted an Answer (of sorts) to it, although I could not find a symbolic solution for it.

.

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