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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 Comment

Anshuman S
Anshuman S on 20 Dec 2017
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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Accepted Answer

Walter Roberson
Walter Roberson on 20 Dec 2017
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 Comments

Anshuman S
Anshuman S on 20 Dec 2017
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
Walter Roberson on 20 Dec 2017
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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