Then why have the EXP, CON, & PAR segment options if 99% of BUILT models are for rectangular enclosures?
Con is used to model a conical flare horn segment. See Attachment 1 for examples of conical horns.
Exp is used to model an exponential flare horn segment. See Attachment 2 for examples of exponential horns.
Par (parabolic flare) is used to model a rectangular cross-sectional area horn segment having two parallel straight sides and two sloping straight sides.
Other available horn flares are:
Bes - Bessel
Hyp - Hyperbolic-exponential
Lec - Le Cléac'h
Obl - Oblate spheroidal
Rad - Radius
Sph - Spherical wave
Tra - Tractrix
Attachments
Could I just expect to use this port in the BP6 enclosure and achieve the same tuning and response as before, just with lower port compression?
The results will be a little different because the simulation uses horn segments to specify the flared port, meaning that the internal end correction is not taken into account.
Do we get a physical description of the flange? Width? Angle? Depth? I have read obout flanges and corrections for 40 years. Never seen numbers. I generally do a large round over or thumbnail profile. I guess a chamfer could also work.Hornresp models the port tube as a cylinder, open at both ends. Each end can be flanged or unflanged.
Flanged end correction = 8 / (3 * Pi) * Rp
Unflanged end correction = 0.1952 * Pi * Rp
Where Rp = port tube radius
In the BP6S example used previously:
Ap1 = 800 cm^2
Ap2 = 200 cm^2
The Ap1/Lp1 port tube is shown highlighted in yellow in Attachment 1. One end is flanged and the other is unflanged. Because the radiation impedance is known at the flanged end, no correction is required at that point.
The correction at the internal unflanged end is:
0.1952 * Pi * (800 / Pi) ^ 0.5 = 9.79 cm
Total end correction for the Ap1/Lp1 port tube is 9.79 cm
The Ap2/Lp2 port tube is shown highlighted in yellow in Attachment 2. One end is flanged and the other is unflanged.
The correction at the flanged end is:
8 / (3 * Pi) * (200 / Pi) ^ 0.5 = 6.773 cm
The correction at the unflanged end is:
0.1952 * Pi * (200 / Pi) ^ 0.5 = 4.893 cm
Total end correction for the Ap2/Lp2 port tube is 6.773 + 4.893 = 11.67 cm
By double-clicking on the Lp2 slider label in the loudspeaker wizard the internal port tube Ap2/Lp2 can be changed from the default flanged/unflanged to unflanged/unflanged as shown in Attachment 3. The end corrections are adjusted accordingly.
As far as i know it's just a flat (no roundover) flange. E.g. a tube flush mounted to enclosure in contrast to a tube protruding into (out of) the box.description of the flange
You didn't post a PAR enclosure because 99% of builds have 2 parallel straight sides. If anything, HR should default to PAR and the modeler would have to switch to the other horn profiles if they are going to build a non-PAR enclosure.Con is used to model a conical flare horn segment. See Attachment 1 for examples of conical horns.
Exp is used to model an exponential flare horn segment. See Attachment 2 for examples of exponential horns.
Par (parabolic flare) is used to model a rectangular cross-sectional area horn segment having two parallel straight sides and two sloping straight sides.
Other available horn flares are:
Bes - Bessel
Hyp - Hyperbolic-exponential
Lec - Le Cléac'h
Obl - Oblate spheroidal
Rad - Radius
Sph - Spherical wave
Tra - Tractrix
Wow, not every day you get a picture ref by David of something you've created🙂Con is used to model a conical flare horn segment. See Attachment 1 for examples of conical horns.
Exp is used to model an exponential flare horn segment. See Attachment 2 for examples of exponential horns.
Par (parabolic flare) is used to model a rectangular cross-sectional area horn segment having two parallel straight sides and two sloping straight sides.
Other available horn flares are:
Bes - Bessel
Hyp - Hyperbolic-exponential
Lec - Le Cléac'h
Obl - Oblate spheroidal
Rad - Radius
Sph - Spherical wave
Tra - Tractrix
I of course used Hornresp👍 to model them and get the curve data etc.