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How do I Visualize the Width of an Audio Frequency? . . .

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  • DrewsBrews
    replied
    I use this because, ya know.. laziness


    I currently have a 2-way that I have been using as my main TV/listening speakers for a couple years. Small shallow horn tweeter and an 8" woofer. ~6" c2c. They cross around 3k (4.52" wavelength).

    I figured they would be finicky on the vertical axis, but they don't seem to be. More so on horizontal. And that is probably more due to the woofer dispersion dropping off than anything else.

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  • Steve Lee
    replied
    Thanks Dezzar!

    I'll check it out later when I can focus upon it, Sir.

    Leave a comment:


  • DeZZar
    commented on 's reply
    In terms of visualization, found this recently: https://www.youtube.com/watch?v=kgAQy4a8Q2w

  • Steve Lee
    replied
    dcibel, thanks for your continued support to this hobby - for now I will work with what I have and see what happens - all this measurement stuff is complex and time consuming and I really don't have suitable facilities to take this to the degree suggested - but in the effort I will come to understand what you are proposing.

    The reality is that I don't need these speakers - it is just a WANT and another experiment to satisfy my addiction to speakers so that I stop mentally designing and buying drivers and start building and learning again.

    Best!

    Leave a comment:


  • dcibel
    replied
    The ROT a4e mentions really applies to flat faceplate tweeters and 4th order LR filters. It's a ROT for the most common type of speaker design.

    Best solution is to measure your horn on a baffle full horizontal and vertical response sets, horizontal only is okay if the horn is round or square, but ellipse or rectangles should include both axis of data. You can then observe the balance of driver separation in VituixCAD extremely easily. Just pound the response into shape with active blocks and adjust filter knee and driver spacing y axis dimension until you get a good smooth slope of in-room and power response between the horns and woofer. This is a good exercise to go through with any drivers, it will show easily that placing drivers as close together as possible can possibly provide the worst result.

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  • Steve Lee
    replied
    ^ Thanks for the clarification, Sir!

    I am crossing a 12" woofer at 800 Hz to a 12" x 4" midrange horn and then that to a 4" x 4" Tweeter horn at 4KHz.

    After running the numbers it looks like it should be fine if I overlap the tweeter horn rim over the mid horn.

    Leave a comment:


  • a4eaudio
    replied
    Originally posted by Steve Lee View Post
    C-C spacing --> is it 1/4 wave or less than 1 wave length at the XO Freq?
    There are several threads at Diyaudio.com that will say the same thing....Planet10 will talk about how it SHOULD be 1/4 wave but that is impractical for almost all designs unless you have a coaxial, so a rule of thumb (ROT) is to have C-t-C spacing no more than 1 wavelength at the xo point...then some people will talk about how they have heard speakers that sound great that break the rule.

    More recently Kimmo Saunisto (author of VituixCAD) has suggested that 1.2 x one wavelength often sounds the best to him. (This is an over-generaliztion of what he says.)

    Bottom line...its a rule of thumb, good to think about, but don't take it all too seriously. What I have been doing lately is figure out close to where I think I will xo, keep track of the length of 1 wavelength and 1.2 times that wavelength and then mess around in Sketchup for what will actually look okay.

    One place it becomes restrictive is if you are using a relatively large woofer then some large-faceplate tweeters may get ruled out if you aren't able to cross quite low.

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  • Steve Lee
    commented on 's reply
    Thanks, Bill - this makes good sense.

  • djg
    replied
    Easy.

    Click image for larger version

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  • Steve Lee
    replied
    ^ NO!



    C-C spacing --> less than 1 wave length at the XO Freq is GOOD ENOUGH.
    Last edited by Steve Lee; 09-12-2022, 09:27 PM.

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  • dcibel
    replied
    Sorry, 344.4m/s, I’m off by 0.4%, back to the drawing board!

    Trying to keep it simple for someone who just wants to know how long 100Hz wave is, does it matter much if it’s 3.43meters long or 3.444 meters long?

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  • nigelb
    replied
    Pressure variables such as temperature, humidity affect the speed of sound. The other variable is altitude.

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  • JRT
    replied
    Originally posted by dcibel View Post
    The math here is super basic. Speed of sound is 343m/s…Frequency is the number of cycles per second so the math is simply (1/F)*343 for wave length in meters.
    343 m/s (1125.3 ft/sec) is true for dry air at 18.85C (65.93_degF). Temperature and humidity change that.

    Here in western Massachusetts, the air temperature outside of my backdoor is cureently 78 degF, and indoor temp is 72 degF with windows open (no A/C). Not sure about the humidity, but it does feel more humid than usual (sky is overcast and it is supposed to rain later this afternoon).

    At 72 degF (22.22C) in dry air, speed of sound increases to 1130.6 ft/sec (344.6 m/s), and 50% relative humidity increases that another 1 ft/sec.

    12*1131.6= 13579.2 inches per second

    I more usually use 69 degF and 50% relative humidity, which gets to
    1130 ft/sec, 13560 in/sec, 344.4 m/s.






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  • dcibel
    replied
    The math here is super basic. Speed of sound is 343m/s…Frequency is the number of cycles per second so the math is simply (1/F)*343 for wave length in meters.

    Leave a comment:


  • Paulryun
    replied
    270 cm(32hz) overlaps perfectly with90 cm (96hz ) in parallel ‘TL’s converging to same exit . (neat stuff to play with

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