Hornresp

[Sabbelbacke]

Power PC is "too far away" from the x86 world, even an old outdated P4 from the trash will easily outrun every emulation of x86 on any commercialy available power PC. With x86-MACs, it´s easy... Why not just get one old Windows-PC, put XP on it?

[Don Hills]

Originally Posted by David McBean

...

I'll summarise it like this, then.

When Hornresp exports a vented enclosure design in AkAbak format, it models it as a large Duct (the enclosure) attached to a small Duct (the port). Therefore, end correction needs to be added to the internal end of the small Duct in the AkAbak model to simulate the same as the Hornresp model. The standard "add 0.425 * the port diameter" is suggested.

For a long version of this, see here:

Originally Posted by Don Hills

...

[David McBean]

Originally Posted by Don Hills

When Hornresp exports a vented enclosure design in AkAbak format, it models it as a large Duct (the enclosure) attached to a small Duct (the port). Therefore, end correction needs to be added to the internal end of the small Duct in the AkAbak model to simulate the same as the Hornresp model. The standard "add 0.425 * the port diameter" is suggested.

Hi Don,

Even though the Hornresp schematic diagram shows the port tube located external to the enclosure for the sake of simplicity, the simulation model actually assumes that the tube is positioned inside the enclosure, which is normally the case in practice.

Because the "inner" end of the tube does not terminate in a baffle, the end correction required is 0.306 * port diameter not 0.425 * port diameter.

Hornresp actually uses 4 / (3 * Pi * (2 ^ 0.5)) * port diameter, which is slightly more accurate than the 0.306 value suggested by AkAbak.

Kind regards,

David

[Don Hills]

Originally Posted by David McBean

Hi Don,

Even though the Hornresp schematic diagram shows the port tube located external to the enclosure for the sake of simplicity, the simulation model actually assumes that the tube is positioned inside the enclosure, which is normally the case in practice.

David,

Do you allow for the port tube volume in the enclosed volume? In other words, if someone builds an enclosure to the Hornresp calculated enclosure volume and port dimensions, will it resonate at the right frequency, or at a higher frequency due to the physical enclosure volume being smaller by the amount of the port volume?

Originally Posted by David McBean

Because the "inner" end of the tube does not terminate in a baffle, the end correction required is 0.306 * port diameter not 0.425 * port diameter.

The topology (port internal versus external) makes no difference for the total end correction. Either way, you have one end terminated in 2Pi space and the other terminated in 4Pi space. I believe that's why most programs use a lumped end correction of 0.732 * diameter. Actually, the AkAbak manual seems to "fudge factor" this. When talking about calculating the vent for a reflex form of the Enclosure element, it says:

If you specify the Helmholtz resonance fb= the program will evaluate the vent length Len=. In this case the length is evaluated with the Helmholtz-formula with a one-sided end correction of a duct mounted in an infinite baffle. The other side is loaded by the enclosure. Usually there is a good approximation of the inner loading due to the fact that standing waves are calculated (Lb=).

Originally Posted by David McBean

Hornresp actually uses 4 / (3 * Pi * (2 ^ 0.5)) * port diameter, which is slightly more accurate than the 0.306 value suggested by AkAbak. ...

It seems to me it should be possible to also account for the end topology (number of steradians) in this formula?

[David McBean]

Hi Don,

Originally Posted by Don Hills

Do you allow for the port tube volume in the enclosed volume? In other words, if someone builds an enclosure to the Hornresp calculated enclosure volume and port dimensions, will it resonate at the right frequency, or at a higher frequency due to the physical enclosure volume being smaller by the amount of the port volume?

The enclosure volume is given by Vrc and the port tube volume by Ap * Lpt, as shown in the schematic diagram. If the port tube is fully inside the enclosure then the total cabinet volume is actually Vrc + (Ap * Lpt). The Helmholtz resonance frequency will be correct.

Originally Posted by Don Hills

The topology (port internal versus external) makes no difference for the total end correction. Either way, you have one end terminated in 2Pi space and the other terminated in 4Pi space.

As far as exporting to AkAbak is concerned, it is necessary to specify only the 4Pi value (0.306 * port diameter) because AkAbak will automatically include the 2Pi value (0.425 * port diameter) when either the enclosure element or a terminating radiator is used.

Originally Posted by Don Hills

It seems to me it should be possible to also account for the end topology (number of steradians) in this formula?

Hornresp already does this :-).

Kind regards,

David

[Don Hills]

Originally Posted by David McBean

Hi Don,

The enclosure volume is given by Vrc and the port tube volume by Ap * Lpt, as shown in the schematic diagram. If the port tube is fully inside the enclosure then the total cabinet volume is actually Vrc + (Ap * Lpt). The Helmholtz resonance frequency will be correct. ...

Oh... yeah. The schematic shows the port as external, but the system volume at the top includes the port volume. I should be more observant.

[David McBean]

Hi Everyone,

Hornresp Version 28.50 has just been released. Changes are:

1. An option has been included to generate the "perfect axisymmetric profile" for a Le Cléac'h horn. See Jean Michel on LeCleac'h horns for further details.

2. Port tube end corrections are now included in exported AkAbak scripts, when appropriate.

Kind regards,

David

[blo06]

Hi Everyone,

I'm beginning with Hornresp, trying different things.

When simulating a Bass Reflex and comparing the result with WinISD's (to be sure I'm doing ok), I don't get the same response... Could you tell me why ?

I model the same box with the same vent and get two different Helmoltz freq and two different response curves...

In HR, I did it with the Loudspeaker Wizard (wonderful feature by the way !!)

Perhaps all of this is already explain in this tread, but there's # 2000 messages ! If so tell me where !

Regards.

[Sabbelbacke]

Originally Posted by blo06

When simulating a Bass Reflex and comparing the result with WinISD's (to be sure I'm doing ok), I don't get the same response... Could you tell me why ?

There can be many reasons - but the most important one: hornresp uses a completely different formula to calculate response and everything else.. hornresp´s model is way better than WinISD, it`s much more accurate.

[mashaffer]

I did some searching through this thread (that was fun) and have found some reference to this idea but am not really finding what I want to know.

What I am looking at is a quarter wave pipe (TL) with several identical drivers distributed along a portion of the pipe, not for smoothing necessarily but for convenience sake. I realize that HR can not simulate this directly (nor would it be reasonable to rewrite it to handle it) but what I am wondering is if one could not get a close enough approximation of the total response by simulating each position and averaging the results.

If the number of drivers is "n" and each segment area for the total pipe is S(x)total would one simulate each driver's contribution using segment areas of S(x)total/n (i.e. for 4 drivers pipe areas are 1/4 of what they will be in the final cabinet) or do each drivers contribution in the total pipe size?

Sorry if I missed this in earlier posts.