solve trigonometrical equations with input range
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How to solve on matlab this system trignometric? I need the code
1.42*cos(alfa)+3.3*cos(beta)+3.33*cos(gamma)-6= 0
1.42*sin(alfa)+3.3*sin(beta)+3.33*sin(gamma)=0
with alfa=linspace(19.7, -103, 14)
(19.7 and 103 are in degrees)
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Stephan
2019-2-2
2 个投票
23 个评论
pablolama
2019-2-2
Stephan
2019-2-2
This is not the way things work here. Provide your attempts so far an ask a specific question to the problems you have by trying on your own.
John D'Errico
2019-2-2
+1. Stephan is right on the money here. You don't learn MATLAB when someone gives you a complete answer that solves your problem. The main thing you do learn then is how to ask a question so that someone will do your thinking for you, the next time you find yourself even slightly over your head.
John D'Errico
2019-2-2
@Pablolama - Please don't add an answer just to ask a question. Moved to a comment:
Why it doesn't work?
OA=1.42
AB=4.3
BO=3.33
oo=6
alfa=linspace(19.7, -103, 14)
syms OA AB alfa BO beta gamma
Result=solve(OA*cosd(alfa)+AB*cosd(beta)+BO*cosd(gamma)-OO==0),...
OA*sind(alfa)+AB*sind(beta)+BO*sind(gamma)==0,beta,gamma
Result.beta
Result.gamma
madhan ravi
2019-2-2
Why ? see stephens answer carefully.
pablolama
2019-2-2
John D'Errico
2019-2-2
编辑：John D'Errico
2019-2-2
It fails first, because you overwrote all of your variables.
OA=1.42
AB=4.3
BO=3.33
oo=6
alfa=linspace(19.7, -103, 14)
syms OA AB alfa BO beta gamma
so you created OA, AB, BO as double precision numbers. Then you re-defined them, when you created them as syms. The values they originally had when you created them? Those numbers diasppeared.
OA
OA =
OA
So now, OA, AB, and BO are all gone, turned into purely symbolic unknowns.
Next, you defined the variable oo. But then you use the variable OO.
And finally, you had a right parens in thewrong place.
OO= 6;
Result=solve(OA*cosd(alfa)+AB*cosd(beta)+BO*cosd(gamma)-OO==0,...
OA*sind(alfa)+AB*sind(beta)+BO*sind(gamma)==0,beta,gamma);
Basically, you need to learn to be more careful in your typing.
pablolama
2019-2-2
pablolama
2019-2-2
it is the counter variable of my for loop. Since you did not define a for loop there is an error message.
With every call of fsolve (or solve) i can get a solution for ONE discrete value of alfa. So the for loop should call fsolve (or solve) as often as there are different values for alfa to solve for.
Stephan
2019-2-2
This will only use alfa at the index 14. You get one result. Where is the for loop?
pablolama
2019-2-2
Stephan
2019-2-2
it depends on how you saved them.
gamma = gamma(:,2)
for example if you saved gamma as a 28x2 array.
syms beta gamma
OA=1.42;
AB=4.3;
BO=3.33;
OO=6;
alfa=linspace(19.7, -103, 14);
% preallocate solution arrays
result_beta = nan(numel(alfa),2);
result_gamma = nan(numel(alfa),2);
% solve and save the results
for k = 1:numel(alfa)
[res_beta, res_gamma]=solve([OA*cosd(alfa(k))+AB*cosd(beta)+BO*cosd(gamma)-OO==0,...
OA*sind(alfa(k))+AB*sind(beta)+BO*sind(gamma)==0],beta,gamma);
result_beta(k,:)=double(res_beta);
result_gamma(k,:)=double(res_gamma);
end
madhan ravi
2019-2-2
In your mind you know what is beta but matlab doesn’t know that so you have to tell it to matlab in the proper manner.
pablolama
2019-2-2
Stephan
2019-2-2
quick and dirty:
gamma(1:3,:)=-gamma(1:3,:)
pablolama
2019-2-2
pablolama
2019-2-3
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