1
$\begingroup$

The mouth of the horn is circular and there are many models in order to approximate the radiation impedance, such as:

  1. massless circular piston in an infinite baffle,
  2. unbaffled massless piston,(planar piston in end of a long tube,)
  3. spherical cap vibrating in a sphere, and so on.

What's the most appropriate model thatcan be used to calculate the radiation impedance of the horn?

Obviously, piston in an infinite baffle is the simplest one, but the radius of the piston seems to be some fractions of the mouth radius. (I don't know exactly,what's the ratio between the massless-piston radius and the mouth radius)

The unbaffled massless piston, or a planar piston at the end of a long tube, is very hard to calculate. (H. Levine and J. Schwinger: "On the Radiation Sound from an Unflanged Circular Pipe" . Honestly, I cannot understand this one at all, is there any simpler explanation or calculation for this model?

is there any paper discussing or comparing these models? does anyone have any good articles?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yeah, these theories aren't a piece of cake at all. In his well-known article dealing with horn modeling (reading that might be useful!) Dan Mapes-Riordan used the first of your possibilities, i.e. circular piston in an infinite baffle:

$$ Z = \rho_0 c_0\pi R^2 \left(1 - \frac{J_1(2kR)}{kR} - i\frac{H_1(2kR)}{kR} \right) $$

where J1 is the 1st order 1st kind Bessel function and H1 1st order Struve function, R radius of the piston and the other variables denoted as usual in acoustics. Both the functions can be expressed as series (see the links) so you can easily obtain a numerical solution.

Maybe useful for you would be the solution for the low frequencies:

$$ Z \approx \rho_0c_0\pi R^2 \frac{(kR)^2}{2}+i\omega\rho_0\pi R^2\frac{8R}{3\pi} \ , \ for\ kR<<1 $$

The key problem in validity of this entire model should be rate of the horn flaring (the lower, the better for this model).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.