Solving for theta at different time steps

2017 7 5
2 个回答
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更新时间：2017 7 5
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I have the following equation:
47.9618*theta + 15.9809*sin(2*theta) = 20.780136*t
Given, at time t = 0, theta = 0 radians. I want to find theta value at t = 1 second, 2 seconds, etc.
Could this be automated on matlab?

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James Tursa
James Tursa 2017-7-5
编辑：James Tursa 2017-7-5

1 个投票

E.g.,
>> t = 0;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
0
>> t = 1;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
0.2649
>> t = 2;
>> f = @(theta) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
>> fzero(f,0)
ans =
0.5651

3 个评论
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Lee Quan
Lee Quan 2017-7-5
编辑：Lee Quan 2017-7-5
So for getting results from t = 0 to 10 seconds at once, I could do this:
for t=0:10
f = @(theta)47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
fzero(f,0)
end
Awesome! Thanks James!
Lee Quan
Lee Quan 2017-7-5
Could you point me to some resources where the '@' notation is used so that I can familiarize myself with it? The @ part is something I can't find in matlab documentation online.
James Tursa
James Tursa 2017-7-5
编辑：James Tursa 2017-7-5
https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html?searchHighlight=anonymous%20function&s_tid=doc_srchtitle

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更多回答（1 个）

Walter Roberson
Walter Roberson 2017-7-5
编辑：Walter Roberson 2017-7-5

1 个投票

There is no closed form solution for that. Use fzero or fsolve:
eqn = @(theta,t) 47.9618*theta + 15.9809*sin(2*theta) - 20.780136*t;
for t = 1 : 5
theta0 = randn()
fzero( @(theta) eqn(theta,t), theta0)
end

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