Horn experts - Please help to determine optimal throat size for low-mid horn

[pk]

I want to build a pair of rectangular horns for the 160-600 Hz range for my home horn rig (not PA). The driver will probably be the 10” Beyma 10MI100 driver (see specs in the link):

http://www.usspeaker.com/beyma 10MI100-1.htm

However, I am bit confused about how to determine the optimal throat size for this application.

Often the rule of thumb is to use a compression ratio between 2:1 and 3:1, which with the Beyma driver SD=380 cm2 leads to a throat size somewhere between 190 cm2 and app. 127 cm2. This, however, correponds badly with the throat sizes optained if I use the formula from Melhuish.org horn website where the throat size for given driver can be calculated with the fomula:

St= [2 x Pi x Fs x Qt x Vas]/c which lead to a throat size = 81 cm2 for the Beyma driver.

And if I “ask” McBeans excellent HornResp for the optimal throat size given the driver parameters for the 160-600 Hz horn, then the program suggests an even smaller throat size= 57 cm2.

If I understand it correctly then McBeans HornResp calculations is probably optimized for max efficiency, while I go for best home hifi sound quality where a few dB lesser efficiency is less important.

In HornResp I can achive fairly flat responses using quite different throat sizes, but I simply do not know which throat size is the best sounding.

If I want the driver/horn combo to sound as clear and natural as possible in a home environment, then what throat size should I head for with the Beyma driver?


Thanks a lot!

Kind regards
Peter

 

[WithTarragon]

Good Luck on your project!

One calculation (from Keele) is

S sub t (throat area) = (2*pi)*f sub s * Q sub ts * V sub as) / 343

An alternative is to use Q sub es rather than Q sub ts

These two may not differ that much.

The suggested size can be fiddled with. The trade off is that if the throat size is made smaller then the SPL increases and Bandwidth decreases.

-Tom

 

[MaVo]

rules of thump say 2-3 for a basshorn and up to 10 for a compression driver. so your low mid horn is inbetween. i dont like rules of thump, but there is no real formula to figure it out. two options remain: play safe and use the rule of thump or build a horn with a higher ratio and drive it to death, while measuring where your cone fails and after that build an optimum one.

if you dont drive your horn into high levels in a home listening setup, i suppose the compression ratio can be higher, because the force on the cone will be less.

i would simply use the compression ratio which gives the best frequency response and hope for good

 

[ingvar ahlberg]

Hi PK
As You`re aiming for a hifi system go for a 1:2 throat ratio, You don´t need any extra efficiency, instead ponder on horngeometry. If possible go biradial, gives all that You wanted in the first place but with the drawbacks the least present. Also before choosing driver take a look at Beymas 102ND/n, designed for hornloading.

 

[tinitus]

I want to build a pair of rectangular horns for the 160-600 Hz range

Take a look at this nice 12"...seems to be made fore what you want

http://www.precisiondevices.co.uk/showdetails.asp?id=70

 

[bear]

For a given expansion of a given type (exponential, conical, hyperbolic, tractrix, etc...) the throat size/area is going to be a cross section of that expansion at a given distance from the mouth.

So, what you probably want to do is to build a horn with a removable tail section, such that you can merely bolt on a tail that is the proper throat area (length) for a given experimental set up that you want to try.

That's how I'd do it.



_-_-bear

 

[pk]

Hi all,

Thanks a lot for all your constructive suggestions! I really appreciate it!

Kind regards
Peter

 

[sauuuuuce]

excellent dialogue. I am in the "development" stages of a similar project and enjoy seeing all the input. Thanks to the OP for starting this.

 

[alenordio]

excuse me all for late response,
the optimal dimension for maximum efficiency is the db keekle formula.
you can make a throat big than the area resulting by db keele formula or small than it.
if you make a big throat the efficiency go down but you have a low 2 harmony distortion tipically of horn loaded.
if you make a throat small than db keele formula,you add the bandwidht in the upper frequency but you loose efficiency.
The maximum efficiency is maximum only with dbkeele formula....if you move the dimension upper or down from the result,you loose always efficency.
For build a midrange horn i think you have to calculate the acoustic power at the throat for know the value of distortion that you listen at maximum power of your amplifier because the midrange is placed in a frequency-range where the distortion is very audible.
Best regards to all for the interesting discussion

 

[whgeiger]

Also Geometry

More important than the compression ratio is the geometry used to transition from a large circular driver diaphragm to a small rectangular (or square) throat. Consider using a phase plug (Smith) to achieve this transition. Example of this design may be found in the JBL CMCD & PTW Tech Notes [1] & [2].

References:
[1] JBL :: Technical Library
[2] JBL :: Technical Library

Regards,
WHG