How to plot implicit function in an iteration ?
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for b= 5:1:40
f = @(o,i) sind(b+i) + sind(b+o)-( 1140/0.04 + sqrt ((1140/0.04 -2*sind(b)).^2 - (cosd(b- o)- cos(b +i)).^2));
fimplicit(f,[0 90 0 90])
hold on
end
1 个评论
Anshuman S
2017-12-20
编辑：Anshuman S
2017-12-20
I wanted to plot δi vs δo for different β values ranging from 5 to 40. and the δi & δo values ranging from 0 to 90 degrees. values for w = 2.4 m and d = 0.4 m .
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Walter Roberson
2017-12-20
1 个投票
You accidentally used cos() in one place instead of cosd()
But even fixing that, there are no solutions over that range of values. The function is quadratic-like, and has a minima at [85, 85] (to within round-off) where it is roughly -57000. The maxima would be somewhere on the boundary of the [0 90 0 90] region, with a value that would be barely different.
2 个评论
Anshuman S
2017-12-20
编辑：Anshuman S
2017-12-20
I wanted to plot δi vs δo subjected to this equation for different β values ranging from 5 to 40. and the δi & δo values ranging from 0 to 90 degrees. values for w = 2.4 m and d = 0.4 m .
Walter Roberson
2017-12-20
Look at your right hand side. The square root part is never negative (by convention), so the smallest that can contribute is 0. The w/d part is 2.4/0.4 = 6 . So in the case where the square root happens to come out as 0 you would have
sin(beta + delta_i) + sin(beta - delta_o) = 6
which is never possible for real-valued beta and delta_* .
If you reverse the values of w and d then there are solutions. Those solutions happen to lie along a slightly curved line, and the solutions for the various beta values overlap each other so it may be difficult to tell them apart with the plot you are doing.
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