# Rayleigh method Dimensional Analysis

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1. Aug 16, 2016

### williamcarter

1. The problem statement, all variables and given/known data

ii)Rayleigh method gives me inverse of expected result.Would really appreciate your help.

2. Relevant equations
show that $f(gamma)$=$(\frac {ro*R^5} {E*t^2})$
where ro=density
E=energy
3. The attempt at a solution
We will do dimensional analysis on the elements ro,R,E
ro=kg/m^3=M*L-3
R=m=L
E=J=N*m=$\frac {kg*m} {s^2}*m$=M*L2*T-2

Now we say that gamma=C*roa*Rb*Ec*td (eq1)
where C=constant, a,b,c,d=yet other constants
Now we do dimensional analysis on eq(1)
M0*L0*T0=C*(M*L-3)a*(L)b*(M*L2*T-2)c*(T)d
now M:0=a+c =>a= -c
L:0=-3a+b+2c =>b= -5c
T:0= -2c+d =>d=2c
c=c
We wrote everything in terms of constant c.
Now we substitute back those constant in eq(q)
=>gamma=ro-c*R-5c*Ec*t2c

We rearrange => $gamma$=$(\frac {E*t^2} {ro*R^5})$c

I would really appreciate it if you could tell me what exactly I did wrong because it seems I get 1/expected result.

2. Aug 16, 2016

### Staff: Mentor

Who says it's 1/expected result?

3. Aug 16, 2016

### williamcarter

Well if c= -1 then I would get $gamma$=$(\frac {ro*R^5} {E*t^2})$ which they asked me to show.

4. Aug 16, 2016

### Staff: Mentor

c doesn't have to be -1. You showed that $\left(\frac {Et^2} {\rho R^5}\right)^c=g(\gamma)$, not necessarily $\gamma$. So, $$\frac{\rho R^5}{Et^2}=[g(\gamma )]^{-1/c}=f(\gamma )$$which is all you were asked to show.

5. Aug 16, 2016

### williamcarter

Thank you, now is clear