Room Modes 101 in 14 Minutes

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  • Опубліковано 21 жов 2024

КОМЕНТАРІ • 18

  • @JoseGarcia-yu9nd
    @JoseGarcia-yu9nd Рік тому +1

    Thanks for the information.

  • @tritonedelta3464
    @tritonedelta3464 2 роки тому +1

    Thumbs up for knowledge! Good videos.

  • @Canadian_Eh_I
    @Canadian_Eh_I 3 роки тому

    Phenominal series, thankyou. You have a gift for explaining

  • @Rene_Christensen
    @Rene_Christensen 2 роки тому

    In a mode null you do have sound, it is just that the mode does not support that pressure. The dips in the frequency response are not from being in mode nulls. Look up Boundary Interference.

  • @AsifKhan-gb2qu
    @AsifKhan-gb2qu 3 роки тому

    Could you please share some insights into , where to place membrane absorbers to treat modal issues, taking any room as an example. At what frequencies must they all be tuned?

    • @AVRoomService
      @AVRoomService  Рік тому

      Corners would likely be good because all the modes congregate there. Half way point is often good for treatment. Problem modes can be calculated or discovered with an SPL or RTA meter.

  • @ahmedzengy
    @ahmedzengy 2 роки тому +1

    How i can calculate an L shape room ?!

    • @manamimnm
      @manamimnm Рік тому

      Use software like COMSOL.

  • @Rene_Christensen
    @Rene_Christensen 2 роки тому

    Why would sound take longer to travel at different frequencies? The sound speed is the same at all frequencies (there is not dispersion in air).

    • @AVRoomService
      @AVRoomService  Рік тому

      The speed of sound is the same, but higher frequencies take less time to develop than lower frequencies.

  • @DmitryMyadzelets
    @DmitryMyadzelets 4 роки тому

    Norman, thank you for explanation. Could you explain the following, please?
    Let's take the ceiling height 10'8" for instance. A wave length equal to this size has frequency 105.7 Hz. This wave would have a mode exactly in the middle between the ceiling and floor, forming a so called standing wave. A longer wave with frequency 52.8 Hz has no ability to form standing wave, to my understanding. So, why do you have 52.8 Hz in the table?

    • @AVRoomService
      @AVRoomService  4 роки тому

      You need to multiply your numbers by two in order to account for the wave to reflect back, creating the standing wave. Rounding the numbers, 53 would be f1, 106 would be f2, 158 f3, etc.

    • @AlexLapugean
      @AlexLapugean 4 роки тому

      That's actually the issue with small rooms. All those waves that do not fully fit in the room dimension can still create standing waves. But they do it at half the wave length (wave gets reflected and completes the cycle at twice the room length), or quarter the wave length (4 room lengths for a cycle), etc. All those then have a much higher chance to overlap with other higher room modes, to create stronger high / low pressure points.

    • @DmitryMyadzelets
      @DmitryMyadzelets 4 роки тому

      @@AlexLapugean Thank you for the participation. Indeed! I've taken a pencil and have drown a wave in the wave-length room, then in the half-length room etc. The wave perfectly stands! (if I can say so :)
      However, it's still not clear for me why Norman wants me to multiply the frequency by two. Is it because it has most of energy when overlaps?

    • @AlexLapugean
      @AlexLapugean 4 роки тому

      @@DmitryMyadzelets Well he did not say to multiply frequency by 2, but "your numbers". The only thing making sense in that context is the room dimension to be multiplied by 2, in order to also find the waves with twice the wavelength of your room dimension, that would still create a standing wave due to reflection at half the cycle.

    • @DmitryMyadzelets
      @DmitryMyadzelets 4 роки тому

      @@AlexLapugean Correct. Thank you.