Please help me with this as soon as possible

Why do I get the same answer as 733 for all columns. The code I want is divide every single a with b so I use a./b but I end up getting all same 733. Why??? Please help .

[KSSV]

Copy the code here.....image snippet will not work.

You are not supposed to mention ASAP.

[carson yeoh]

wab =733;r=0.0405;L=0.13;
theta = 0:0.1:2*pi;
x = (r*sin(theta))/L;
phi = asin(x);
a = wab*r*sin(theta);
b = L*sin(phi);
wbc = a./b

[KSSV]

Then what you expect..? You are effectively finding:

wab*r/L*L/r

You will get wab...

[carson yeoh]

theta is not equal to phi

why will I get wab???

[carson yeoh]

I tried to calculate separetly and I got different answer for each column

but then I tried every way to combine it, i will get wab

[Walter Roberson]

x = (r*sin(theta))/L;

Ok.

phi = asin(x);

asin() of a constant times sin(theta)... perhaps there is a trig identity that could be used, but it is not one of the common ones.

b = L*sin(phi);

phi is asin(x) and x = (r*sin(theta))/L so provided that r/L*sin(theta) happens to fail within the first period then sin(asin(x)) would be x, so for at least part of the theta range, L*sin(phi) -> L*sin(asin(x)) -> L*x -> L * r*sin(theta) / L -> r * sin(theta)

a = wab*r*sin(theta);

Look, there is an r*sin(theta) in there...

wbc = a./b

So wab*r*sin(theta) / (r * sin(theta)) -> wab

r/L is about 0.3 and 0.3*sin(theta) is going to be within the primary range for real-valued theta that is in the primary range.

Answer 1

[Rasul Khan]

wab =733;r=0.0405;L=0.13;
theta = 0:0.1:2*pi;
x = (r*sin(theta))/L;
phi = asin(x);
a = wab*r*sin(theta);
b = L*sin(phi);
wbc = a./b

If you look at your code and replace the value of phi you will get

 b = L * x

And calculating a./b wil give you 'wab' like @KSSV mentioned here

https://www.mathworks.com/matlabcentral/answers/554386-please-help-me-with-this-as-soon-as-possible#comment_912661