Calculating the mechanical index of applied ultrasound

[rwooduk]

On the surface, the equation is simply the peak rarefractional pressure divided by the root of the applied frequency:

$MI = \frac{P_{ra}}{\sqrt{f}}$

But the pressure is reduced/derated by an attenuation factor/coefficient that is dependent on depth and frequency e.g. $0.3 \ dB / (cm \cdot MHz)$

An example calculation is given in the paper "Tutorial paper: thermal and mechanical indices". Where it takes a 4 MHz pulse, at 6cm, 2 Mpa, $0.3 \ dB / (cm \cdot MHz)$ and says the MI would be (2*0.44)/SQRT(4) = 0.44. I can't get the 0.44 because the dimensional analysis with units doesn't seem right.

$MI = \frac{MPa}{\sqrt{MHz}}\cdot \frac{dB}{cm\cdot MHz}$

I can easily calculate the attenuation, which is simply the attenuation coefficient multiplied by distance and frequency. Which for his example is 7.2 dB, I can also calculate intensity from this from I = I * 10^(-dB/10), but still don't get the 0.44.

Am I overcomplicating things?

 

[jim mcnamara]

This is probably some help - note their approach seems a little different from yours-
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5336845/

 

[rwooduk]

This is probably some help - note their approach seems a little different from yours-
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5336845/

Thanks, I did see that paper but it seems they do not factor in the attenuation due to tissue / depth at all. I should have probably put this in the physics sections above as I think I'm having a calculation problem, rather than a definition problem. But I really appreciate the reply, thanks.
[Thread moved from the Medical forum to the Physics forums by a Mentor]

 

[nasu]

There is a normalization factor in the formula. If the factor is written in the denominator the units are $\frac{MPa}{MHz^{1/2}}$. The formula I use is $MI=\frac{p_{r.03}}{C_{MI}\sqrt{f}}$ where $C_{MI}$ is this normalization constant. This is concerning the units.
For the value, the attenuation refers to the intensity. The intensity is proportional to the square of the acoustic pressure. So you have $10 log(I_1/I_2)= 7.2 dB$ and $20 log(p_1/p_2)= 7.2 dB$ so the pressure at 6 cm is derated to $\frac{2 Mpa}{10^{0.36}}$ which is about 0.87 MPa. Divide this by $\sqrt{4}$ and you get about 0.44.

Edit.
In some papers they don't put the normaliation factor in the formula but it is assumed to be there.
It does not change the values.

 

[rwooduk]

There is a normalization factor in the formula. If the factor is written in the denominator the units are $\frac{MPa}{MHz^{1/2}}$. The formula I use is $MI=\frac{p_{r.03}}{C_{MI}\sqrt{f}}$ where $C_{MI}$ is this normalization constant. This is concerning the units.
For the value, the attenuation refers to the intensity. The intensity is proportional to the square of the acoustic pressure. So you have $10 log(I_1/I_2)= 7.2 dB$ and $20 log(p_1/p_2)= 7.2 dB$ so the pressure at 6 cm is derated to $\frac{2 Mpa}{10^{0.36}}$ which is about 0.87 MPa. Divide this by $\sqrt{4}$ and you get about 0.44.

Edit.
In some papers they don't put the normaliation factor in the formula but it is assumed to be there.
It does not change the values.

Thank you very much nasu, I had completely messed up my calculation. Extremely helpful!