How to find angle in equation?

I want to find omega value in the equation.

can i anybody tell me how to solve this equation.

Answer 1

[Walter Roberson]

The search turns out to take over 20 seconds for each eload value, and even then it seems to fail.

Notice I ended at 0.002, which is the most I could do with this online version without timing out.

tic
syms omega
a=3.666;
k=0.467;
s=100;
r=1.4;
ef=0.08483;
N=(a*k)/(1+(a*k));
eloadvals = 0:0.001:0.002;
for eloadidx = 1 : length(eloadvals)
    eload = eloadvals(eloadidx);
    c1=(1-N)*(((r*a)/2)+(r/2)-1)-(r-1);
    c2=(N*(r/(2*a))+(r/2)-1);
    Ac2=((r*a)*(1-N))/2;
    Ah1=(r*N)/(2*a);
    A=-8*(1-N)*c2*sin(pi-atan(s/((1-N)*omega)))+8*N*c1*sin(2*(pi-atan(1/(N*omega))))+(32*N*ef+32*N*eload); 
    B=16*(1-N)*(-1*(((1-N)*omega)/(2*s))*(((Ah1*c1*sin(pi-atan(1/(N*omega)))*cos(pi-atan(1/(N*omega))))+((((N*s)/((1-N)*omega))+(N*(r/2-1)))*c2*cos((atan((((1-N)*((r/2)-1))/(s))*omega))+(pi-atan(s/((1-N)*omega))))*cos(pi-atan(s/((1-N)*omega)))))))*c2*sin(2*(pi-atan(s/((1-N)*omega))))+(16*(((N*omega)/2)*((((((1-N)/(N*omega))+((1-N)*(((r/2)-1)-(r-1))))*c1*cos((atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega)))+(pi-atan(1/(N*omega))))*cos(atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega))))+(Ac2*c2*sin(pi-atan(s/((1-N)*omega)))*cos(pi-atan(s/((1-N)*omega)))))))+2*c1)*N*c1*sin(2*(pi-atan(1/(N*omega))))+N*c1*c1*sin(4*(pi-atan(1/(N*omega))));
    thissol = vpasolve(A==B);
    if length(thissol) >= 1
       sol(eloadidx) = thissol(1);
    else
        sol(eloadidx) = nan;
    end
end
toc

Elapsed time is 54.511198 seconds.

plot(eloadvals, sol)

Nothing visibile in the plot because vpasolve() failed.

[Walter Roberson]

My tests show that if the question is to find the omega for which A == B, then in order for there to be solutions, eload must be negative, below approximately -.014

You can derive this by taking symbolic eload, and calculate A-B and solving for eload, and start plotting. You will see a discontinuity, and that it never crosses 0; the closest to zero is at approximately omega = 1.42.

[Ali]

right sir can we find eload (positve or negative) at which omega is only one root?

[Walter Roberson]

If I recall correctly, the plot of eload in terms of omega has a discontinuity at 0, and rises from there, then falls to a minima at about omega = 1.42, and then increases again. The places where omega has only one root would then be the point of the minima (at about 1.42), together with the points that are above the local maxima that it rises to after the discontinuity at 0... as a quick mental memory, it was somewhere around omega = 3

I would need to recheck what happened on the other side of the discontinuity.

We could probably identify that singular point, and the range where it goes above the local maxima. But I have to ask why you would want to know those ??

[Walter Roberson]

I cannot quite get three plots without timing out, so the plots will be split between two posts.

syms omega
Q = @(v) sym(v);
a = Q(3.666);
k = Q(0.467);
s = Q(100);
r = Q(1.4);
ef = Q(0.08483);
N = (a*k)/(1+(a*k));
syms eload
Pi = Q(pi);

c1 = (1-N)*(((r*a)/2)+(r/2)-1)-(r-1);
c2 = (N*(r/(2*a))+(r/2)-1);
Ac2 = ((r*a)*(1-N))/2;
Ah1 = (r*N)/(2*a);
A = -8*(1-N)*c2*sin(Pi-atan(s/((1-N)*omega)))+8*N*c1*sin(2*(Pi-atan(1/(N*omega))))+(32*N*ef+32*N*eload); 
B = 16*(1-N)*(-1*(((1-N)*omega)/(2*s))*(((Ah1*c1*sin(Pi-atan(1/(N*omega)))*cos(Pi-atan(1/(N*omega))))+((((N*s)/((1-N)*omega))+(N*(r/2-1)))*c2*cos((atan((((1-N)*((r/2)-1))/(s))*omega))+(Pi-atan(s/((1-N)*omega))))*cos(Pi-atan(s/((1-N)*omega)))))))*c2*sin(2*(Pi-atan(s/((1-N)*omega))))+(16*(((N*omega)/2)*((((((1-N)/(N*omega))+((1-N)*(((r/2)-1)-(r-1))))*c1*cos((atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega)))+(Pi-atan(1/(N*omega))))*cos(atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega))))+(Ac2*c2*sin(Pi-atan(s/((1-N)*omega)))*cos(Pi-atan(s/((1-N)*omega)))))))+2*c1)*N*c1*sin(2*(Pi-atan(1/(N*omega))))+N*c1*c1*sin(4*(Pi-atan(1/(N*omega))));

sol = solve(A-B, eload)

sol = 

double(limit(sol, omega, -inf))

ans = -0.0848

double(limit(sol, omega, inf))

ans = -0.0848

fplot(sol, [-10 10]); xlabel('omega'); ylabel('eload')

%fplot(sol, [1.35 1.5]); xlabel('omega'); ylabel('eload')
%fplot(sol, [-1.5 -1.35]); xlabel('omega'); ylabel('eload')

[Walter Roberson]

syms omega
Q = @(v) sym(v);
a = Q(3.666);
k = Q(0.467);
s = Q(100);
r = Q(1.4);
ef = Q(0.08483);
N = (a*k)/(1+(a*k));
syms eload
Pi = Q(pi);

c1 = (1-N)*(((r*a)/2)+(r/2)-1)-(r-1);
c2 = (N*(r/(2*a))+(r/2)-1);
Ac2 = ((r*a)*(1-N))/2;
Ah1 = (r*N)/(2*a);
A = -8*(1-N)*c2*sin(Pi-atan(s/((1-N)*omega)))+8*N*c1*sin(2*(Pi-atan(1/(N*omega))))+(32*N*ef+32*N*eload); 
B = 16*(1-N)*(-1*(((1-N)*omega)/(2*s))*(((Ah1*c1*sin(Pi-atan(1/(N*omega)))*cos(Pi-atan(1/(N*omega))))+((((N*s)/((1-N)*omega))+(N*(r/2-1)))*c2*cos((atan((((1-N)*((r/2)-1))/(s))*omega))+(Pi-atan(s/((1-N)*omega))))*cos(Pi-atan(s/((1-N)*omega)))))))*c2*sin(2*(Pi-atan(s/((1-N)*omega))))+(16*(((N*omega)/2)*((((((1-N)/(N*omega))+((1-N)*(((r/2)-1)-(r-1))))*c1*cos((atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega)))+(Pi-atan(1/(N*omega))))*cos(atan((((1-N)*((r/2)-1)-(r-1))/(1-N))*(N*omega))))+(Ac2*c2*sin(Pi-atan(s/((1-N)*omega)))*cos(Pi-atan(s/((1-N)*omega)))))))+2*c1)*N*c1*sin(2*(Pi-atan(1/(N*omega))))+N*c1*c1*sin(4*(Pi-atan(1/(N*omega))));

sol = solve(A-B, eload)

sol = 

double(limit(sol, omega, -inf))

ans = -0.0848

double(limit(sol, omega, inf))

ans = -0.0848

%fplot(sol, [-10 10]); xlabel('omega'); ylabel('eload')

fplot(sol, [1.35 1.5]); xlabel('omega'); ylabel('eload')

fplot(sol, [-1.5 -1.35]); xlabel('omega'); ylabel('eload')

[Walter Roberson]

So look at the top plot. The left and right halfs are symmetric around eload = -0.0848

You can see near the discontinuity at 0 that the left side close to 0 shares values with the right side close to 0, so that section is not unique.

If we look down the discontinuity on the left side close to 0, to the point where the one on the right close to 0 gives out, and then mentally follow that level near -0.1 to the left, we can see that we are not unique, there is the other side of the valley that rises up again. How high does the left valley rise? Well it rises to double(limit(sol, omega, -inf)) = -0.0848, and by following across towards the discontinuity at 0, we can see that that entire area gets us to a matching edge just to the left of 0.

What point does not have a duplicate? Only the lowest part of the valley, at approximately omega = -1.42 corresponding to eload of about -0.15342 : if you follow that one point left and right, there is no corresponding point on either side of the discontinuity.

Likewise, when we look on the right side of the discontinuity, we have a balance point at about 1.42 (same distance away from 0 as the other one), corresponding to eload of about -0.01403 .

Those are the only two omega that have unique solutions: +/- 1.42 (approximately) corresponding to eload of about -0.01403 and eload of about -0.15342 .

If there are any stable points, it would be only those two.

If omega were being controlled by some force pushing towards a balance, then if you were to disturb omega a little, it should rebalance itself. But if not... well, if there is no restoring pressure on omega, then if the system gets knocked slightly so that omega varies, then you would start getting two eload solutions.

[Ali]

Thanks a lot sir

Answer 2

[Reshma Nerella]

Hi,

You can use solve function from symbolic math toolbox to solve equations and obtain the values of variables.

Refer to the documentation page for more informataion and examples: Solve.