e-side,
Take a look at figure 3 at LOW-FREQUENCY HORN DESIGN USING THIELE/SMALL DRIVER PARAMETERS by D.B Keele, Jr. http://www.dbkeele.com/papers.htm Paper no. 11 and first decide what you want from your horn.
If you want to maximize the mid band for your horn, your choice for k is 4 and Q= 0.5 for Fh(s) as in Leach example and Vf is Rho (0)*c^2*0.686*Qtc^2*Cat: Cat = Cab//(Sd^2*Cmd).
Now for the choice of driver: (In Keele’s fig.3) fhm = 2*fs/Qts is important because for maximum efficiency, fhm gives the upper highest possible cut off.
But if you want to trade efficiency for more bandwidth for an unknown reason, then another Q for Fh(s) must be chosen and you have to decide a new upper corner closer to fhvc or fhc.
This results in that you cannot use Q=0,5 for Fh(s) and you have to step back to eq. 34 (Leach) and consider a new value for Caf.
For fhvc you must pay attention for your driver’s voice coil inductance and for fhc you must decide the size for Vf.
Or the easy way...get Hornresp and start fiddle with tedious simulations using known good drivers for the right purpose!
B
Take a look at figure 3 at LOW-FREQUENCY HORN DESIGN USING THIELE/SMALL DRIVER PARAMETERS by D.B Keele, Jr. http://www.dbkeele.com/papers.htm Paper no. 11 and first decide what you want from your horn.
If you want to maximize the mid band for your horn, your choice for k is 4 and Q= 0.5 for Fh(s) as in Leach example and Vf is Rho (0)*c^2*0.686*Qtc^2*Cat: Cat = Cab//(Sd^2*Cmd).
Now for the choice of driver: (In Keele’s fig.3) fhm = 2*fs/Qts is important because for maximum efficiency, fhm gives the upper highest possible cut off.
But if you want to trade efficiency for more bandwidth for an unknown reason, then another Q for Fh(s) must be chosen and you have to decide a new upper corner closer to fhvc or fhc.
This results in that you cannot use Q=0,5 for Fh(s) and you have to step back to eq. 34 (Leach) and consider a new value for Caf.
For fhvc you must pay attention for your driver’s voice coil inductance and for fhc you must decide the size for Vf.
Or the easy way...get Hornresp and start fiddle with tedious simulations using known good drivers for the right purpose!
B
Thank you for your answer bjorno
I experimented somewhat and came up with the following.
In case of a maximum sensitivity (Qlc = 2Qtc) design, k = 4 for maximum w3. It ran a couple of calculations and it appeared that Qlc isn't always 2Qtc , so k, Caf and Vaf will change. Therefore a new value of k (to maximize w3) should be determined.
In my case Qtc = 0,35 and Qlc = 0,61
This gives Ral/Rat = 1,35 and Rat/Ral = 0,74
Solving eq. 36 gives a maximum value of w3 for k = 3,48
Caf should be calculated, so we need a value for Cat.
This can be done by solving eq. 25, the other variables in this equation are known (or can be calculated).
For my design Cat ~ 1,3547E-8
Then Caf and Vaf can be calculated, this results in a throat chamber of 0,159 litres.
regards
Erwin
I experimented somewhat and came up with the following.
In case of a maximum sensitivity (Qlc = 2Qtc) design, k = 4 for maximum w3. It ran a couple of calculations and it appeared that Qlc isn't always 2Qtc , so k, Caf and Vaf will change. Therefore a new value of k (to maximize w3) should be determined.
In my case Qtc = 0,35 and Qlc = 0,61
This gives Ral/Rat = 1,35 and Rat/Ral = 0,74
Solving eq. 36 gives a maximum value of w3 for k = 3,48
Caf should be calculated, so we need a value for Cat.
This can be done by solving eq. 25, the other variables in this equation are known (or can be calculated).
For my design Cat ~ 1,3547E-8
Then Caf and Vaf can be calculated, this results in a throat chamber of 0,159 litres.
regards
Erwin
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.