Hi,
I'm going through the formulas for a closed box design found in Leach's Intro to Electroacoustics and I find a huge discrepency between his design formulas and those of Dickason, Weems & others.
Firstly in Section 7.5 he states: "The equations below... should never be used for design purposes." These are the equations that everyone else seems to be using to model a closed box system.... (?)
In sectin 7.7 - "System Design with a Given Driver", Example 3, with a 12" driver having Fs=19Hz, Qms=3.7, Qes=0.35, & Vas= 19 cubic ft., using his method we end up with an unstuffed box volume of 2.78 cubic ft.
Using Dickason, Weems, etc., the Vb comes out to 4.9 cubic feet.
What gives? Am I doing something wrong here?
Would someone care to elaborate?
Thanx,
fred p.
I'm going through the formulas for a closed box design found in Leach's Intro to Electroacoustics and I find a huge discrepency between his design formulas and those of Dickason, Weems & others.
Firstly in Section 7.5 he states: "The equations below... should never be used for design purposes." These are the equations that everyone else seems to be using to model a closed box system.... (?)
In sectin 7.7 - "System Design with a Given Driver", Example 3, with a 12" driver having Fs=19Hz, Qms=3.7, Qes=0.35, & Vas= 19 cubic ft., using his method we end up with an unstuffed box volume of 2.78 cubic ft.
Using Dickason, Weems, etc., the Vb comes out to 4.9 cubic feet.
What gives? Am I doing something wrong here?
Would someone care to elaborate?
Thanx,
fred p.
Not familiar with Leach's book.
I can only ask which Qtc-final Q in the box-you are aiming for.
According to the old formula listed below, a 2.78 cu ft box yields a Atc of 0.9, approximately.
Also according to it, a 4.9 cu ft box yields a Qtc of 0.7.
Qtc = 0.7 once was considered the ideal, but many box builders aim for various Qtc's between 0.5 and 1.0, depending on the application.
So I can only ask-did Leach say that with a box volume of 2.78, you will end up with a Qtc of 0.7? If he did, we definitely have a conflict between Leach and Small, Weems, etc.
Or did Leach merely say that with a box volume of 2.78 cu ft, you will have an acceptable speaker? Because many people will consider a sealed box with a Qtc = 0.9 as being a well-designed box.
I can only ask which Qtc-final Q in the box-you are aiming for.
According to the old formula listed below, a 2.78 cu ft box yields a Atc of 0.9, approximately.
Also according to it, a 4.9 cu ft box yields a Qtc of 0.7.
Qtc = 0.7 once was considered the ideal, but many box builders aim for various Qtc's between 0.5 and 1.0, depending on the application.
So I can only ask-did Leach say that with a box volume of 2.78, you will end up with a Qtc of 0.7? If he did, we definitely have a conflict between Leach and Small, Weems, etc.
Or did Leach merely say that with a box volume of 2.78 cu ft, you will have an acceptable speaker? Because many people will consider a sealed box with a Qtc = 0.9 as being a well-designed box.
Attachments
This is a link to an exerpt from his book which has the formulas he uses and works out the model with the listed driver parameters.
http://users.ece.gatech.edu/~mleach/ece4445/downloads/closedbox.pdf
You'll notice that he specifies a Qtc of .7.
The method he uses gives an equivalent stuffed box volume and then the actual volume is derived from Vab. i.e.,
Vab=3.47 cu. ft.
Vb = (3.47/1.25) = 2.78 cu. ft.
This of course is the reverse methology used by others. That is using the "classic" formulas we arrive at Vb and to get the stuffed equivalent Vb is multiplied by 1.25 to get Vab.
I read through the Leach material. I see how he derives the values but it looks optimistic to me. You will just not see that much improvement in volume from stuffing except perhaps in small boxes (a ft^3 or less). I'm sure he has a rational explanation(the man is nothing if not thorough), but I would be inclined to agree with the other formulas.
Just for the heck of it, I used winISD pro to figure the box, and there is agreement with Weems/Dickason. A box of 3.47 ft^3 will yield a Q of ~.8, an frc of 50.7 Hz, leaving the stuffing deduction out of it. You would need a box of ~4.9 ft^3 to yield a .707 Q and 44 Hz frc.
Not sure of the date of the material, but it may have been updated. I have seen responses from Georgia Tech on other audio matters, so I would email him there. A grad student or someone will reply, if they all haven't left for the holidays.
Tim
PS We just got about 4" of snow overnite. Still snowing.
Just for the heck of it, I used winISD pro to figure the box, and there is agreement with Weems/Dickason. A box of 3.47 ft^3 will yield a Q of ~.8, an frc of 50.7 Hz, leaving the stuffing deduction out of it. You would need a box of ~4.9 ft^3 to yield a .707 Q and 44 Hz frc.
Not sure of the date of the material, but it may have been updated. I have seen responses from Georgia Tech on other audio matters, so I would email him there. A grad student or someone will reply, if they all haven't left for the holidays.
Tim
PS We just got about 4" of snow overnite. Still snowing.
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