hornresp excursion plots |
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roborg
Registered User Joined: 05 July 2004 Location: United Kingdom Status: Offline Points: 1026 |
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My mid-tops use two sn12-b's in a dual siamesed porn-horn sized horn & about 5 litres rear chamber (actually one hasnt been restricted yet hehe!) I modelled them in akabak, hornresp is good for bass-horns but not for mid-range.
For the sn12-b the response starts to become nicely flat 120-1.2Khz without EQ when using 2-4litres rear volume per driver. Larger volumes allow for slightly higher low-output, at the cost of slight loss of higher frequency sensitivity.
At 3litres rear volume the simulated excursion is less than 3.5mm when being fed a 1.2Kw sine down to 100hz. So basically around 3-5litres seems pretty optimal. To be honest it's hard to get down to such small volumes without using a specifically moulded back chamber (i used polyurethane foam to restrict the rear chamber.)
The smaller it is the more distortion from the rear chamber & the hotter it'll run too.
cheers,
Rob.
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What I cannot create, I do not understand
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roborg
Registered User Joined: 05 July 2004 Location: United Kingdom Status: Offline Points: 1026 |
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The front chamber volume is almost not worth modelling as the comp-ratio in the porn horn is pretty low. I've modelled 1 litre with my horns to represent the volume 'trapped' behind the phase-plug.
The effect of modelling front-chambers is irrelevent for maximum excursion calculations, smaller front chambers only really give you more higher bandwidth.
A simple akabak script (google for akabak, its a free download) i used to model my mid-range horns to get parameters about right is this:
| 2x sn12-b DDR mid range horn with phase plugs
Def_Driver 'sn-12b'
Meas_DoNotModify SD=530cm2 dD1=9.6cm tD1=5cm |Cone fs=50Hz Mms=47.83g Qms=4.32 Qes=0.17 Re=5.4ohm Le=0.79mH ExpoLe=0.618 System 'DDR mid horn' | the two speakers, defined as drivers Driver 'D1' Def='sn-12b' Node=1=0=25=5 Driver 'D2' Def='sn-12b' Node=1=0=26=6 | the rear chambers of the speakers
Enclosure 'rearchamber1' Node=5
Vb=3L Qb/fo=10e-3 Sb=500cm2 fb=100Hz dD=2cm QD/fo=.0001 x=0 y=0 z=0 HAngle=0 VAngle=0 Enclosure 'rearchamber2' Node=6
Vb=3L Qb/fo=10e-3 Sb=500cm2 fb=100Hz dD=2cm QD/fo=.0001 x=0 y=0 z=0 HAngle=0 VAngle=0 Horn 'top half' Node=25 STh=232cm2 WMo=60cm HMo=41cm Vf=1L Len=70cm T=1 x=0 y=25cm z=0 HAngle=0 VAngle=0 Horn 'bottom half' Node=26
STh=232cm2 WMo=60cm HMo=41cm Vf=1L Len=70cm T=1 x=0 y=-25cm z=0 HAngle=0 VAngle=0 Cheers,
Rob
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What I cannot create, I do not understand
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tommysb
Registered User Joined: 24 April 2006 Location: Finland Status: Offline Points: 1036 |
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Hey Rob, how is the phase plug defined here? Could you shed a bit more light/a bit more explanation please?
Tom |
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roborg
Registered User Joined: 05 July 2004 Location: United Kingdom Status: Offline Points: 1026 |
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Hi Tommy, The phase plug is implied by the area of the throat Sth being less than the area of the driver. This is only a really simple script to get an idea of performance (IE low cutoff & bandwidth) of the horn.
Any directivity information & a more accurate high frequency response would need the throat area & different parts of the horn to be modelled individually. I've done this a few times, but get fed up with having to change the akabak scripts when i change my mind about the design.
As i rarely array my mid-top designs, usually cos i only build two & want a stereo system! (they quite often get arrayed when in storage hehe!) I'm not too fussy about constant directivity, more so with efficiency & wide dispersion. I try to match up the -6db points with the lower & higher frequency sections tho, so that the change in directivity varies smoothly down the stack.
I'm a keen fan of keeping only 1 mid-top per side to keep to the ideal of a point source (or cylindrical source in this case) as faithfully as poss.
You can really go to work with ducts in akabak, but it very quickly becomes hard work to make sure all the dimensions & areas are correct.
cheers,
Rob.
PS I dont have a mid modelled as ducts & waveguides on the pc i've got with me here but i did have this approximation to the folding of an 1850 here. This was a model i based the simulation of my smooth contoured 1850It uses ducts & waveguides to approximate the geometry in the 1850 horn:
Def_Driver 'sub800'
Meas_DoNotModify dD=41cm dD1=8.5cm tD1=4cm |Cone fs=30Hz Mms=201g Qms=10 Qes=0.23 Re=5ohm Le=1.5uH ExpoLe=0.618 System 'better 1850'
Driver 'D1' Def='sub800' Node=1=0=20=3 Enclosure 'rearchamber1' Node=3
Vb=45L Sb=2400cm2 | simulate v wedge with few ducts Duct 'Du1' Node=20=21 WD=60cm HD=3mm Len=4cm Visc=0 Duct 'Du2' Node=20=21 WD=60cm HD=0.5cm Len=8cm Duct 'Du3' Node=20=21 Visc=0 WD=60cm HD=0.7cm Len=12cm Duct 'Du4' Node=20=21 Visc=0 WD=60cm HD=1cm Len=16cm Duct 'Du5' Node=20=21 Visc=0 WD=60cm HD=1.5cm Len=20cm Duct 'Du6' Node=20=21 Visc=0 WD=60cm HD=3cm Len=24cm Duct 'Du7' Node=20=21 Visc=0 WD=60cm HD=1.5cm Len=28cm Duct 'Du8' Node=20=21 Visc=0 WD=60cm HD=1cm Len=32cm Duct 'Du9' Node=20=21 Visc=0 WD=60cm HD=0.7cm Len=36cm Duct 'Du0' Node=20=21 Visc=0 WD=60cm HD=0.5cm Len=40cm Duct 'Du0' Node=20=21 Visc=0 WD=60cm HD=0.3cm Len=44cm | first bend
Waveguide 'bend1' Node=21=22
WTh=60cm HTh=3.5cm WMo=60cm HMo=4.66cm Len=15.5cm T=1 Waveguide 'bend1' Node=21=22 WTh=60cm HTh=3.5cm WMo=60cm HMo=4.66cm Len=22cm T=1 Waveguide 'bend1' Node=21=22 WTh=60cm HTh=3.5cm WMo=60cm HMo=4.66cm Len=28.5cm T=1 | first expo section
Waveguide 'downpipe' Node=22=23
WTh=60cm HTh=14cm WMo=60cm HMo=21.5cm Len=44cm T=1 | second bend (three waveguides) Waveguide 'bend1' Node=23=24 WTh=60cm HTh=7cm WMo=60cm HMo=11cm Len=30cm T=1 Waveguide 'bend1' Node=23=24 WTh=60cm HTh=7cm WMo=60cm HMo=11cm Len=45cm T=1 Waveguide 'bend1' Node=23=24 WTh=60cm HTh=7cm WMo=60cm HMo=11cm Len=75cm T=1 | exit horn
Horn 'top half' Node=24 WTh=60cm HTh=32cm WMo=60cm HMo=60cm Vf=0L Len=65cm T=1 x=0 y=0 z=0 HAngle=0 VAngle=0 |
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What I cannot create, I do not understand
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