Converting port dimensions.

[haffe]

Hi all. I am in process of building an elf 1.0 creativesound.ca/pdf/SuperELF.PDF However basports with a diameter of 1" are not easily found here i Europe. So I want to convert this into a port with a diameter of 35 mm. Should I just take the total volume of the first port and convert it to corresponding volume with the 35 mm port?

[Svante]

If you increase the cross-sectional area by a factor x, so should the effective length of the tube. The helmholtz resonance frequency is

fh=c/(2*pi)*sqrt(S/LV)

where c=345 m/s, S is the cross-sectional area of the port, L is the effective port length, and V is the box volume. As you can see; if S/L is constant, so is fh.

The effective length of the tube is the physical length plus ~1.7 times the radius of the tube. This little addition takes care of the co-oscillating air outside the tube.

In your case, if the physical length was 10 cm and the diameter 2.5 cm, the effective length was 10+1.7*2.5/2=12.125 cm. Increasing the diameter to 35 mm increases the area by a factor (35/25)^2=1.96 times. So the effective length should now be 12.125*1.96= 23.765 cm. Subtract end corrections from this:
23.765 - 1.7*3.5/2 = 20.79 cm

The factor 1.7 could be discussed depending on if the port ends are seen as flanged or not, and wether the space inside the box is seen as a free half space or not. Anyway, this correction is small, so an error in it is not very harmful.

HTH

[haffe]

Ok. Thanks.

[Dave Jones]

Can you estimate the effective length of a tube with a typical flare on one end or both?

[Svante]

Originally posted by Dave Jones
Can you estimate the effective length of a tube with a typical flare on one end or both?

Not as easy as the straight tube. I think I have seen tables for this, but the easiest way is probably to build an approximation, measure and adjust the length of the cylidrical part until fh is right.

If you are really into simulating it, you could model it by dividing the tube in several coaxial parts with different lengths and diameters. Each would have an acoustic mass of

MA=rho0*L/S

where rho0 is 1.2 kg/m3, L is the length of that section, and S is the section's cross sectional area. Add an end correction of 0.85 r at each end. All these acoustic masses add up to one big acoustic mass, which should be the same as the acoustic mass of the original tube (also including its end corrections of 0.85 r).

Note that acoustic mass is something different than mechanic mass (in case you found the equation for MA strange )

HTH