posted
I have been looking around and cannot seem to find how to calcuate tuning of a port that changes area. Such as one that gets wider as it grows longer, not near as wide as a true horn though. Any ideas or comments would be very appreciated.
posted
Wow.. That would probably be kinda complex, and involve calculus, unless one could solve it as two generic shapes (IE two cones + cylinders or two cylinders and a sphere)
ShadowStar
------------------ Still looking for that CHEAP Thunderdome :D
Got Ears? Get Oz!
You can't build a reputation for what you're GOING to do.. But you can build one for TALKING about it!
-Eclipse-Rockford Fosgate-Oz Audio-Precision Power Inc.-AudioControl-Stinger-Tsunami-Clifford
It's all about knowledge, love and respect.
Posts: 2578 | From: Somewhere In the Northeast | Registered: May 1999
| IP: Logged |
posted
Thanks for the input ShadowStar. I can see how it would be hard to find the area and the mass of the air inside such a shape. But if I could do it then could I tune it like a port? that has the same mass of air as the horn type port? Love to work with fiberglass and not scared of calculations if I know I have the right idea...at least a clue anyways. thanks, Tim.
------------------ StangDb 1989 Mustang 5.0 GT. Currently running 2 Vega 15's with a 200dhc. Subject to change weekly.
posted
you can't do it in two different shapes, as port tuning isn't based on volume, but rather area. There would be alot of calculus involved and it would probably be easier to just test it rather than figure it out....unless someone knows a formula or something.
posted
Ok, so could I use the average area of the port to calculate the length needed? Just to guess at the right frequency. Then test it and make it so that I can change the shape and or area? Good idea? thanks for the info.
------------------ StangDb 1989 Mustang 5.0 GT. Currently running 2 Vega 15's with a 200dhc. Subject to change weekly.
posted
I thought this might entertain some of you on this topic. I done some searching and found alot on home audio waveguides. Thanks to a very helpful individual. Here is what I found.
A waveguide is a structure which forces wave propagation along a path parallel to its longest dimension. Acoustic wavequides are structures with constant cross-sectional area and shape. Simple examples of such structures include hoses, tubes, and pipes, referred to hereafter as ducts. If a duct is excited by a pressure disturbance with a wavelength larger than twice the duct's largest cross-sectional dimension, then only plane waves will propagate down the duct. For a circular duct containing air at room temperature, the highest frequency at which only plane waves will propagate is given by f = 100/a where a is the radius of the duct cross-section. Once plane waves are generated inside the duct, they will propagate down the duct, even if the duct has bends or turns in it. A propagating plane wave may encounter a change in the acoustic impedance of the duct when the duct (i) opens into free space, (ii) is connected to another section of duct with a different cross-section, (iii) branches off into two ducts, or (iv) is terminated in some other way. This impedance change causes partial reflection and partial transmission of the incident plane waves.
Assume that a duct of cross-sectional area S and length L is driven by a piston at x=0. The pipe is terminated at x=L by an acoustic impedance ZL. The input acoustic impedance as seen by the driver (looking into the duct at x=0) may be written in terms of the terminal impedance, (1) Equation (1) is called a transmission line equation; similar equations are used for electromagnetic waveguides (transmission lines), as well as for comparing the input mechanical impedance to the termination impedance for transverse waves on a string or longitudinal waves in a beam.
B. Duct driven at x=0 and open at x=L Let the duct be driven by a rigid piston at x=0 and open-ended at x=L. At first guess one might think that the termination impedance for an open end would be ZL =0, which would reduce Eq. (1) to Z0 = (p c/S) i tan (kL) resulting in resonance frequencies occurring at fn = nc/2L, for n=1,2,3, . . . This is the assumption made in most elementary physics books; it is not, however, correct. The boundary condition is not zero at the open end, because the open end of the duct radiates sound into the surrounding medium. The proper value for the terminating impedance is then ZL= Zr where Zr is the radiation impedance of the open end of the pipe. The radiation impedance is complex; the real part (radiation resistance) represents the energy radiated away from the open end in the form of sound waves, and the imaginary part (radiation reactance) represents the mass loading of the air just outside the open end. For unflanged and flanged open ends, the radiation impedance is unflanged; (2a) flanged. (2b)
The input acoustic impedance for an unflanged, open-ended duct may be obtained by substituting Eq. (2) into the transmission equation line equation Eq. (1). Resonance occurs when the input impedance becomes a minimum; or when 1/Z0 becomes a maximum. Figure 1 shows the input acoustic impedance calculated from Eqs. (1) and (2a) for a duct of length L=1.1 m and radius a= 2.0 cm. The first two resonances occur at approximately f1 = 170 Hz and f2=340 Hz.
Figure 1: Input admittance (inverse of input impedance) for an unflanged, open-ended duct of length L=1.01 m and radius a= 2.0 cm. The resonance frequencies for an unflanged, open duct may be approximated by (3) where n=1,2,3, . . . , c is the speed of sound (343 m/s for air), and the length of the duct includes an "end correction" for the open end.
If that did not give you a headace then let me bow to you. hehehehe
------------------ StangDb 1989 Mustang 5.0 GT. Currently running 2 Vega 15's with a 200dhc. Subject to change weekly.
posted
No headache, but the simple way to do it would be to test the sub in the enclosure before you finish anding/painting etc. then find the tuned frequency by playing tones. if you want to tune it lowere then fibreglass a bit more length onto the port inside it, or cut a bit off it to tune higher. If you make a drawing of roughly what yo're considering it would help as well. i had to consider this problem as well and ended up cutting a bit off the inside of my port. For pictures of my box click on the "spy pics" link in my signature.
Colin
------------------ Spy Shots Bass: Sitting IN a port while playing a 30Hz test tone flat out. See also: Fun; Insane.
posted
That's very interesting. I'm wonderin if you could find any SPL applications for that. If anyone does teh experiment it would eb interesting to see how lout it is compared to just having the speaker itsef there.
Printer-friendly view of this topic