plot graph with value when finding fzero

Dear all,

I have question how to let it plot the graph while finding fzero

frad = 220;
rpm = 20000;
m =3;
alpha = 0.1;
R0 = 0.015;
R1 = 0.020;

rpmtoms = rpm*pi*(R0+R1)/(60);
B=pi*(R0+R1)/(m);
L=R1-R0;
ita=0.01; %from prof.
h0 = 0.00002;
k   =  B*tan(alpha*pi/180)/h0;
% fu  =  -(6*ita*rpmtoms*L*B^2)*(-log(k+1)+(2*k)/(k+2))/((k^2)*(h0^2));
f   =  @(h0) (-(6*ita*rpmtoms*L*B^2)*(-log(k+1)+(2*k)/(k+2))/((k^2)*(h0^2))) - frad;
h0 = fzero(f,h0)

I would like to see the graph because it seems like it depends on the initial value i set so i am not sure how much should i set? and my answer would be close to zero like in 1e-6 but cannot set it to zero. i read in some matlab questions i need to set the range for my h0 but in this case since the result close to zero but cannot be zero.

Thank you

[madhan ravi]

What value are you trying to find from fzero()? , it’s a bit confusing.

[Paranee Sawad]

Dear Madhan Ravi,

I would like to find value of h0 which makes

(-(6*ita*rpmtoms*L*B^2)*(-log(k+1)+(2*k)/(k+2))/((k^2)*(h0^2))) - frad =0

[Walter Roberson]

There are two solutions. Your equation is a quadratic of the form

constant/h0^2 - another_constant == 0

which can be solved by multiplying through by h0^2 to get

constant - another_constant*h0^2 == 0

which has solution

h0 = +/- sqrt(constant / another_constant)

[Paranee Sawad]

Dear Walter Roberson,

Ahhhh so it doesn't correct to use fzero right? because fzero cannot find a root of a function such as x^2.

[Walter Roberson]

It just means that you should provide bounds,

fzero(f, [sqrt(realmin), realmax])

to ensure that it does not accidentally try negative values.

[Paranee Sawad]

Dear Walter Roberson,

I wonder about the value of h0 when it computes k

k   =  B*tan(alpha*pi/180)/h0;

since i set the initial for h0

h0 = 0.00002;

when it tries to find h0 that makes fzero will it also change the value of h0 when it compute the next k or k now be a constant? but i want k changing the value related to h0

frad = 220;
rpm = 20000;
m =7.5;
alpha = 1;
R0 = 0.015;
R1 = 0.020;

rpmtoms = rpm*2*pi*(R0+R1)/(2*60);
B=2*pi*(R0+R1)/(2*m);
L=R1-R0;
ita=0.01; 
h0 = 0.00002;
k   =  B*tan(alpha*pi/180)/h0;
f = @(h0) (-(6*ita*rpmtoms*L*B^2)*(-log(k+1)+(2*k)/(k+2))/((k^2)*(h0^2)))-frad;

[h0,fval] = fzero(f, [sqrt(realmin), realmax])

and i also substitute k in f

h0 = 0.00002;
% k   =  B*tan(alpha*pi/180)/h0;
f = @(h0) (-(6*ita*rpmtoms*L*B^2)*(-log((B*tan(alpha*pi/180)/h0)+1)+(2*(B*tan(alpha*pi/180)/h0))/((B*tan(alpha*pi/180)/h0)+2))/(((B*tan(alpha*pi/180)/h0)^2)*(h0^2)))-frad;

but there is an error said

Error using fzero (line 257)
Function values at interval endpoints must be finite and real.
Error in testhfzero (line 16)
[h0,fval] = fzero(f, [sqrt(realmin), realmax]) 

[Walter Roberson]

Use 1E150 as the upper bound.

[Paranee Sawad]

Dear Walter Roberson,

I do not know how to set the upper bound 1e150 together with the range of h0

[h0,fval] = fzero(f, [sqrt(realmin), realmax])

[Walter Roberson]

[h0, fval] = fzero(f, [sqrt(realmin), 1E150])

[Paranee Sawad]

Dear Walter Roberson,

i used to try this but it does not work but now it works. Thank you very much ^__^

Answer 1

[Andrea Monfredini]

You can substitute the equation expression of the operators * / and ^ with the respective dot operators

f   =  @(h0) (-(6.*ita.*rpmtoms.*L.*B.^2).*(-log(k+1)+(2.*k)./(k+2))./((k.^2).*(h0.^2))) - frad;

and then use semilogx to see a clearer representation of your function. Adjust the arguments of t to your needs.

t=linspace(0,h0*100,1000);
figure()
semilogx(t,f(t))
hold on
yline(0)