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#82
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 06:04:34 -0500, "Arny Krueger"
wrote: I'm just sitting here mostly lurking, watching people argue over the meanings of words. It seems quite clear that something bad is happening in rooms and something might be done about it. How long does this haggling over words go on before someone actually says something about improving sound quality? I've done that in the past, and directed people to BBC technical papers and journals that are an incredible source of free ideas and solutions to bad rooms. This just happens to be a technical thread about what goes on in rooms, and I thought it looked like a good idea to draw attention to the fact that travelling wave and standing wave problems are not the same thing - and hence do not admit of the same solutions. d _____________________________ http://www.pearce.uk.com |
#83
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 06:04:34 -0500, "Arny Krueger"
wrote: I'm just sitting here mostly lurking, watching people argue over the meanings of words. It seems quite clear that something bad is happening in rooms and something might be done about it. How long does this haggling over words go on before someone actually says something about improving sound quality? I've done that in the past, and directed people to BBC technical papers and journals that are an incredible source of free ideas and solutions to bad rooms. This just happens to be a technical thread about what goes on in rooms, and I thought it looked like a good idea to draw attention to the fact that travelling wave and standing wave problems are not the same thing - and hence do not admit of the same solutions. d _____________________________ http://www.pearce.uk.com |
#84
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 06:04:34 -0500, "Arny Krueger"
wrote: I'm just sitting here mostly lurking, watching people argue over the meanings of words. It seems quite clear that something bad is happening in rooms and something might be done about it. How long does this haggling over words go on before someone actually says something about improving sound quality? I've done that in the past, and directed people to BBC technical papers and journals that are an incredible source of free ideas and solutions to bad rooms. This just happens to be a technical thread about what goes on in rooms, and I thought it looked like a good idea to draw attention to the fact that travelling wave and standing wave problems are not the same thing - and hence do not admit of the same solutions. d _____________________________ http://www.pearce.uk.com |
#85
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 06:04:34 -0500, "Arny Krueger"
wrote: I'm just sitting here mostly lurking, watching people argue over the meanings of words. It seems quite clear that something bad is happening in rooms and something might be done about it. How long does this haggling over words go on before someone actually says something about improving sound quality? I've done that in the past, and directed people to BBC technical papers and journals that are an incredible source of free ideas and solutions to bad rooms. This just happens to be a technical thread about what goes on in rooms, and I thought it looked like a good idea to draw attention to the fact that travelling wave and standing wave problems are not the same thing - and hence do not admit of the same solutions. d _____________________________ http://www.pearce.uk.com |
#86
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Non-modal peaks and nulls - here's the proof!
Don Pearce writes:
On Tue, 20 Jan 2004 01:26:05 GMT, Randy Yates wrote: Don Pearce writes: On Mon, 19 Jan 2004 11:31:15 -0500, "Ethan Winer" ethanw at ethanwiner dot com wrote: Don, This is what I would expect at any frequency ... this has nothing to do with room modes or standing waves. Yes, exactly. This is the precise point I have made repeatedly dozens upon dozens of times here and in many other audio newsgroups. But nobody believed me, so I made the video to prove the point. I will mention that standing waves are standing still whether or not they relate to a room's dimensions. As long as a "forcing" tone is present the waves will stabilize into a static pattern. But I accept that many people reserve the term "standing wave" for modal frequencies only. And they'd be wrong. --Ethan OK - I see he problem - one of terminology. What is going on here is most definitely NOT a standing wave phenomenon, but a free wave one. And this is most definitely BULL****. The waves Ethan is describing are precisely "standing waves." Go look it up in a physics book. No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% % 'One Summer Dream', *Face The Music*, ELO |
#87
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Non-modal peaks and nulls - here's the proof!
Don Pearce writes:
On Tue, 20 Jan 2004 01:26:05 GMT, Randy Yates wrote: Don Pearce writes: On Mon, 19 Jan 2004 11:31:15 -0500, "Ethan Winer" ethanw at ethanwiner dot com wrote: Don, This is what I would expect at any frequency ... this has nothing to do with room modes or standing waves. Yes, exactly. This is the precise point I have made repeatedly dozens upon dozens of times here and in many other audio newsgroups. But nobody believed me, so I made the video to prove the point. I will mention that standing waves are standing still whether or not they relate to a room's dimensions. As long as a "forcing" tone is present the waves will stabilize into a static pattern. But I accept that many people reserve the term "standing wave" for modal frequencies only. And they'd be wrong. --Ethan OK - I see he problem - one of terminology. What is going on here is most definitely NOT a standing wave phenomenon, but a free wave one. And this is most definitely BULL****. The waves Ethan is describing are precisely "standing waves." Go look it up in a physics book. No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% % 'One Summer Dream', *Face The Music*, ELO |
#88
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Non-modal peaks and nulls - here's the proof!
Don Pearce writes:
On Tue, 20 Jan 2004 01:26:05 GMT, Randy Yates wrote: Don Pearce writes: On Mon, 19 Jan 2004 11:31:15 -0500, "Ethan Winer" ethanw at ethanwiner dot com wrote: Don, This is what I would expect at any frequency ... this has nothing to do with room modes or standing waves. Yes, exactly. This is the precise point I have made repeatedly dozens upon dozens of times here and in many other audio newsgroups. But nobody believed me, so I made the video to prove the point. I will mention that standing waves are standing still whether or not they relate to a room's dimensions. As long as a "forcing" tone is present the waves will stabilize into a static pattern. But I accept that many people reserve the term "standing wave" for modal frequencies only. And they'd be wrong. --Ethan OK - I see he problem - one of terminology. What is going on here is most definitely NOT a standing wave phenomenon, but a free wave one. And this is most definitely BULL****. The waves Ethan is describing are precisely "standing waves." Go look it up in a physics book. No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% % 'One Summer Dream', *Face The Music*, ELO |
#89
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Non-modal peaks and nulls - here's the proof!
Don Pearce writes:
On Tue, 20 Jan 2004 01:26:05 GMT, Randy Yates wrote: Don Pearce writes: On Mon, 19 Jan 2004 11:31:15 -0500, "Ethan Winer" ethanw at ethanwiner dot com wrote: Don, This is what I would expect at any frequency ... this has nothing to do with room modes or standing waves. Yes, exactly. This is the precise point I have made repeatedly dozens upon dozens of times here and in many other audio newsgroups. But nobody believed me, so I made the video to prove the point. I will mention that standing waves are standing still whether or not they relate to a room's dimensions. As long as a "forcing" tone is present the waves will stabilize into a static pattern. But I accept that many people reserve the term "standing wave" for modal frequencies only. And they'd be wrong. --Ethan OK - I see he problem - one of terminology. What is going on here is most definitely NOT a standing wave phenomenon, but a free wave one. And this is most definitely BULL****. The waves Ethan is describing are precisely "standing waves." Go look it up in a physics book. No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% % 'One Summer Dream', *Face The Music*, ELO |
#90
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote:
No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. So the condition you need for standing waves is the simultaneous transfer of energy in opposite directions, resulting in no (or at least reduced) net energy flow in either direction. Failing that condition, all you have is superposed TRAVELLING waves. Now that, unless you have something more useful to add than your last effort, can be the end of the discussion. d _____________________________ http://www.pearce.uk.com |
#91
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote:
No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. So the condition you need for standing waves is the simultaneous transfer of energy in opposite directions, resulting in no (or at least reduced) net energy flow in either direction. Failing that condition, all you have is superposed TRAVELLING waves. Now that, unless you have something more useful to add than your last effort, can be the end of the discussion. d _____________________________ http://www.pearce.uk.com |
#92
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote:
No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. So the condition you need for standing waves is the simultaneous transfer of energy in opposite directions, resulting in no (or at least reduced) net energy flow in either direction. Failing that condition, all you have is superposed TRAVELLING waves. Now that, unless you have something more useful to add than your last effort, can be the end of the discussion. d _____________________________ http://www.pearce.uk.com |
#93
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Non-modal peaks and nulls - here's the proof!
On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote:
No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. So the condition you need for standing waves is the simultaneous transfer of energy in opposite directions, resulting in no (or at least reduced) net energy flow in either direction. Failing that condition, all you have is superposed TRAVELLING waves. Now that, unless you have something more useful to add than your last effort, can be the end of the discussion. d _____________________________ http://www.pearce.uk.com |
#94
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Non-modal peaks and nulls - here's the proof!
Svante,
There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Thanks. --Ethan |
#95
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Non-modal peaks and nulls - here's the proof!
Svante,
There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Thanks. --Ethan |
#96
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Non-modal peaks and nulls - here's the proof!
Svante,
There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Thanks. --Ethan |
#97
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Non-modal peaks and nulls - here's the proof!
Svante,
There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Thanks. --Ethan |
#98
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
Svante, There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. I am not arguing that you are wrong in that there might be zeroes at other frequencies than at those of the room modes. I argue that your calculation of the room modes does not take into account that the room is not "perfect". This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Yes, it probably is. But if we go back to your test, and its validity, I also have doubts about the mode frequencies that you show on http://www.realtraps.com/quarter_wave.htm I made my own spreadsheet calculation using the formula c nx ny nz f(x,y,z)= - sqrt((--)^2+(--)^2+(--)^2) 2 lx ly lz where nx, ny, and xz are integers 0,1,2,3,4... corresponding to the mode number, and lx, ly, lz are the dimensions of the room. I used velocity of sound=344.5 m/s and converted the feet and inches to metres (gosh, how DO you enter feet and inches into a spreadsheet ?!?!) and got a room size of 4.953 x 2.9845 x 2.3368 metres. I ended up with these mode frequencies: x y z c 4.953 2.9845 2.3368 344.5 Mode number Mode frequency [Hz] 0 0 0 0 1 0 0 34.7769028871391 0 1 0 57.7148601105713 1 1 0 67.3827726648608 2 0 0 69.5538057742782 0 0 1 73.7119137281753 1 0 1 81.5038600306235 ---- Near 85,23 Hz 2 1 0 90.3810653581204 0 1 1 93.6186482654646 1 1 1 99.8693360219956 2 0 1 101.346820981992 3 0 0 104.330708661417 0 2 0 115.429720221143 2 1 1 116.628397917226 3 1 0 119.230456878167 1 2 0 120.554772965456 ----- Almost 121.7 Hz 3 0 1 127.743269862891 2 2 0 134.765545329722 0 2 1 136.957900596502 4 0 0 139.107611548556 3 1 1 140.176132322326 1 2 1 141.304279872277 0 0 2 147.423827456351 4 1 0 150.605221251877 1 0 2 151.470188077725 2 2 1 153.607286394517 3 2 0 155.592149802375 4 0 1 157.430536479471 0 1 2 158.318634340569 1 1 2 162.093253881475 2 0 2 163.007720061247 4 1 1 167.676411262279 3 2 1 172.169577177836 2 1 2 172.923456700208 0 3 0 173.144580331714 5 0 0 173.884514435696 1 3 0 176.602600979337 ---- Almost 177.9 Hz 3 0 2 180.606427548062 4 2 0 180.762130716241 5 1 0 183.212525330886 2 3 0 186.592544320322 0 2 2 187.237296530929 0 3 1 188.182071206891 5 0 1 188.863100117539 3 1 2 189.604026194741 1 2 2 190.439591961946 ....which is quite a few more modes than those you listed. (Pardon me for the many decimals on the frequencies) Modes 1,2,0 and 1,3,0 are near the frequencies you used, however the nearest frequency to 85,23 Hz is mode 1,0,1 at 81.5 Hz. According to my list above, there ARE room resonances near at least two of the three frequencies you used. I hope that I am wrong, could others please check if these modes exist and/or if my calculations are right? Add to this that the modes might actually be shifted a bit from these theoretical values, and the validity of the test is seriously threatened. Again I am not disputing that there could be zeroes for other reasons than room resonances, but rather your statement that you DON'T have room resonances at the frequencies you used. That is still to be proven. Perhaps a frequency sweep with the microphone in one corner could indicate the frequencies of the actual room modes, as they are in the actual room. (That was the software I was talking about, to make and record sinusoidal sweeps. Tell me if you want it.) |
#99
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
Svante, There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. I am not arguing that you are wrong in that there might be zeroes at other frequencies than at those of the room modes. I argue that your calculation of the room modes does not take into account that the room is not "perfect". This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Yes, it probably is. But if we go back to your test, and its validity, I also have doubts about the mode frequencies that you show on http://www.realtraps.com/quarter_wave.htm I made my own spreadsheet calculation using the formula c nx ny nz f(x,y,z)= - sqrt((--)^2+(--)^2+(--)^2) 2 lx ly lz where nx, ny, and xz are integers 0,1,2,3,4... corresponding to the mode number, and lx, ly, lz are the dimensions of the room. I used velocity of sound=344.5 m/s and converted the feet and inches to metres (gosh, how DO you enter feet and inches into a spreadsheet ?!?!) and got a room size of 4.953 x 2.9845 x 2.3368 metres. I ended up with these mode frequencies: x y z c 4.953 2.9845 2.3368 344.5 Mode number Mode frequency [Hz] 0 0 0 0 1 0 0 34.7769028871391 0 1 0 57.7148601105713 1 1 0 67.3827726648608 2 0 0 69.5538057742782 0 0 1 73.7119137281753 1 0 1 81.5038600306235 ---- Near 85,23 Hz 2 1 0 90.3810653581204 0 1 1 93.6186482654646 1 1 1 99.8693360219956 2 0 1 101.346820981992 3 0 0 104.330708661417 0 2 0 115.429720221143 2 1 1 116.628397917226 3 1 0 119.230456878167 1 2 0 120.554772965456 ----- Almost 121.7 Hz 3 0 1 127.743269862891 2 2 0 134.765545329722 0 2 1 136.957900596502 4 0 0 139.107611548556 3 1 1 140.176132322326 1 2 1 141.304279872277 0 0 2 147.423827456351 4 1 0 150.605221251877 1 0 2 151.470188077725 2 2 1 153.607286394517 3 2 0 155.592149802375 4 0 1 157.430536479471 0 1 2 158.318634340569 1 1 2 162.093253881475 2 0 2 163.007720061247 4 1 1 167.676411262279 3 2 1 172.169577177836 2 1 2 172.923456700208 0 3 0 173.144580331714 5 0 0 173.884514435696 1 3 0 176.602600979337 ---- Almost 177.9 Hz 3 0 2 180.606427548062 4 2 0 180.762130716241 5 1 0 183.212525330886 2 3 0 186.592544320322 0 2 2 187.237296530929 0 3 1 188.182071206891 5 0 1 188.863100117539 3 1 2 189.604026194741 1 2 2 190.439591961946 ....which is quite a few more modes than those you listed. (Pardon me for the many decimals on the frequencies) Modes 1,2,0 and 1,3,0 are near the frequencies you used, however the nearest frequency to 85,23 Hz is mode 1,0,1 at 81.5 Hz. According to my list above, there ARE room resonances near at least two of the three frequencies you used. I hope that I am wrong, could others please check if these modes exist and/or if my calculations are right? Add to this that the modes might actually be shifted a bit from these theoretical values, and the validity of the test is seriously threatened. Again I am not disputing that there could be zeroes for other reasons than room resonances, but rather your statement that you DON'T have room resonances at the frequencies you used. That is still to be proven. Perhaps a frequency sweep with the microphone in one corner could indicate the frequencies of the actual room modes, as they are in the actual room. (That was the software I was talking about, to make and record sinusoidal sweeps. Tell me if you want it.) |
#100
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
Svante, There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. I am not arguing that you are wrong in that there might be zeroes at other frequencies than at those of the room modes. I argue that your calculation of the room modes does not take into account that the room is not "perfect". This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Yes, it probably is. But if we go back to your test, and its validity, I also have doubts about the mode frequencies that you show on http://www.realtraps.com/quarter_wave.htm I made my own spreadsheet calculation using the formula c nx ny nz f(x,y,z)= - sqrt((--)^2+(--)^2+(--)^2) 2 lx ly lz where nx, ny, and xz are integers 0,1,2,3,4... corresponding to the mode number, and lx, ly, lz are the dimensions of the room. I used velocity of sound=344.5 m/s and converted the feet and inches to metres (gosh, how DO you enter feet and inches into a spreadsheet ?!?!) and got a room size of 4.953 x 2.9845 x 2.3368 metres. I ended up with these mode frequencies: x y z c 4.953 2.9845 2.3368 344.5 Mode number Mode frequency [Hz] 0 0 0 0 1 0 0 34.7769028871391 0 1 0 57.7148601105713 1 1 0 67.3827726648608 2 0 0 69.5538057742782 0 0 1 73.7119137281753 1 0 1 81.5038600306235 ---- Near 85,23 Hz 2 1 0 90.3810653581204 0 1 1 93.6186482654646 1 1 1 99.8693360219956 2 0 1 101.346820981992 3 0 0 104.330708661417 0 2 0 115.429720221143 2 1 1 116.628397917226 3 1 0 119.230456878167 1 2 0 120.554772965456 ----- Almost 121.7 Hz 3 0 1 127.743269862891 2 2 0 134.765545329722 0 2 1 136.957900596502 4 0 0 139.107611548556 3 1 1 140.176132322326 1 2 1 141.304279872277 0 0 2 147.423827456351 4 1 0 150.605221251877 1 0 2 151.470188077725 2 2 1 153.607286394517 3 2 0 155.592149802375 4 0 1 157.430536479471 0 1 2 158.318634340569 1 1 2 162.093253881475 2 0 2 163.007720061247 4 1 1 167.676411262279 3 2 1 172.169577177836 2 1 2 172.923456700208 0 3 0 173.144580331714 5 0 0 173.884514435696 1 3 0 176.602600979337 ---- Almost 177.9 Hz 3 0 2 180.606427548062 4 2 0 180.762130716241 5 1 0 183.212525330886 2 3 0 186.592544320322 0 2 2 187.237296530929 0 3 1 188.182071206891 5 0 1 188.863100117539 3 1 2 189.604026194741 1 2 2 190.439591961946 ....which is quite a few more modes than those you listed. (Pardon me for the many decimals on the frequencies) Modes 1,2,0 and 1,3,0 are near the frequencies you used, however the nearest frequency to 85,23 Hz is mode 1,0,1 at 81.5 Hz. According to my list above, there ARE room resonances near at least two of the three frequencies you used. I hope that I am wrong, could others please check if these modes exist and/or if my calculations are right? Add to this that the modes might actually be shifted a bit from these theoretical values, and the validity of the test is seriously threatened. Again I am not disputing that there could be zeroes for other reasons than room resonances, but rather your statement that you DON'T have room resonances at the frequencies you used. That is still to be proven. Perhaps a frequency sweep with the microphone in one corner could indicate the frequencies of the actual room modes, as they are in the actual room. (That was the software I was talking about, to make and record sinusoidal sweeps. Tell me if you want it.) |
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
Svante, There were, as I understand some small deviations from the "straight angled, empty" room, like the ceiling, and the fact that there were three persons and some equipment inside the room. You are correct about the beams, anyway, and the floor is 4-1/2 inches lower for the last 4'1" of the room. But you know what? It doesn't matter. You can play pretty much any tone in any room and you'll be able to find a null due to the 1/4 wavelength boundary interference shown in the video. I say "pretty much any tone" to ward off the wise guys who inevitably pipe up that 14 Hz is too low for a room that small. Okay, so sue me. :-) The basic phenomenon is valid, and it occurs exactly as described. I am not arguing that you are wrong in that there might be zeroes at other frequencies than at those of the room modes. I argue that your calculation of the room modes does not take into account that the room is not "perfect". This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. Maybe this could shed some light on what the ACTUAL modes frequencies are. I have software for this if you are interested. I wrote my own program for computing axial modes, and have spreadsheets for the others. One problem is no room is perfectly square. And there's always a door that's set into a frame a few inches. Or someone could complain that the walls or ceiling aren't rigid enough to define the modal response. Naysayers will always find something to object to. The fact that 1/4 wavelength cancellations occur outdoors is enough to prove the concept to my satisfaction. Yes, it probably is. But if we go back to your test, and its validity, I also have doubts about the mode frequencies that you show on http://www.realtraps.com/quarter_wave.htm I made my own spreadsheet calculation using the formula c nx ny nz f(x,y,z)= - sqrt((--)^2+(--)^2+(--)^2) 2 lx ly lz where nx, ny, and xz are integers 0,1,2,3,4... corresponding to the mode number, and lx, ly, lz are the dimensions of the room. I used velocity of sound=344.5 m/s and converted the feet and inches to metres (gosh, how DO you enter feet and inches into a spreadsheet ?!?!) and got a room size of 4.953 x 2.9845 x 2.3368 metres. I ended up with these mode frequencies: x y z c 4.953 2.9845 2.3368 344.5 Mode number Mode frequency [Hz] 0 0 0 0 1 0 0 34.7769028871391 0 1 0 57.7148601105713 1 1 0 67.3827726648608 2 0 0 69.5538057742782 0 0 1 73.7119137281753 1 0 1 81.5038600306235 ---- Near 85,23 Hz 2 1 0 90.3810653581204 0 1 1 93.6186482654646 1 1 1 99.8693360219956 2 0 1 101.346820981992 3 0 0 104.330708661417 0 2 0 115.429720221143 2 1 1 116.628397917226 3 1 0 119.230456878167 1 2 0 120.554772965456 ----- Almost 121.7 Hz 3 0 1 127.743269862891 2 2 0 134.765545329722 0 2 1 136.957900596502 4 0 0 139.107611548556 3 1 1 140.176132322326 1 2 1 141.304279872277 0 0 2 147.423827456351 4 1 0 150.605221251877 1 0 2 151.470188077725 2 2 1 153.607286394517 3 2 0 155.592149802375 4 0 1 157.430536479471 0 1 2 158.318634340569 1 1 2 162.093253881475 2 0 2 163.007720061247 4 1 1 167.676411262279 3 2 1 172.169577177836 2 1 2 172.923456700208 0 3 0 173.144580331714 5 0 0 173.884514435696 1 3 0 176.602600979337 ---- Almost 177.9 Hz 3 0 2 180.606427548062 4 2 0 180.762130716241 5 1 0 183.212525330886 2 3 0 186.592544320322 0 2 2 187.237296530929 0 3 1 188.182071206891 5 0 1 188.863100117539 3 1 2 189.604026194741 1 2 2 190.439591961946 ....which is quite a few more modes than those you listed. (Pardon me for the many decimals on the frequencies) Modes 1,2,0 and 1,3,0 are near the frequencies you used, however the nearest frequency to 85,23 Hz is mode 1,0,1 at 81.5 Hz. According to my list above, there ARE room resonances near at least two of the three frequencies you used. I hope that I am wrong, could others please check if these modes exist and/or if my calculations are right? Add to this that the modes might actually be shifted a bit from these theoretical values, and the validity of the test is seriously threatened. Again I am not disputing that there could be zeroes for other reasons than room resonances, but rather your statement that you DON'T have room resonances at the frequencies you used. That is still to be proven. Perhaps a frequency sweep with the microphone in one corner could indicate the frequencies of the actual room modes, as they are in the actual room. (That was the software I was talking about, to make and record sinusoidal sweeps. Tell me if you want it.) |
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Non-modal peaks and nulls - here's the proof!
Don Pearce wrote in message . ..
On Mon, 19 Jan 2004 19:03:52 GMT, ow (Goofball_star_dot_etal) wrote: On Mon, 19 Jan 2004 09:30:22 +0000, Don Pearce wrote: This is what I would expect at any frequency, and looks a lot like normal comb filtering - the signal arrives at the microphone via two paths, one direct and one reflected. For some positions there is reinforcement and for others cancellation. But this has nothing to do with room modes or standing waves. It is a purely travelling wave phenomenon. Ahem, I thought a standing wave *was* traveling waves (or components thereof) of the same frequency travelling in opposite directions at the same speed. In a standing wave there is no net transfer of energy along the direction of "travel" of the wave. The energy remains constrained between two boundaries. This is why they build up; they can be pumped - and also why they don't die the instant the stimulus is removed. In fact, standing waves make it impossible to measure the reverberation time of a room at very low frequencies - the wave collapses when it will, and the time taken is generally longer than the T60 of the room. Travelling waves possess none of these qualities, and must be handled, both analytically and practically, quite differently. I think there is a slight terminology confusion here. When standing waves are treated mathematically, there is often only ONE boundary, and the standing wave occurs at any frequency. Here, there is no energy buildup over time, there are merely two identical waves traveling in opposite directions, and no energy transportation. On the other hand I often see, and perhaps say myself, that a "standing wave" is used as a synonym to an excited room resonance. Here we have an energy buildup, two or more boundaries, and this phenomenon occurs (of course) only at frequencies near the mode frequency. |
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Non-modal peaks and nulls - here's the proof!
Don Pearce wrote in message . ..
On Mon, 19 Jan 2004 19:03:52 GMT, ow (Goofball_star_dot_etal) wrote: On Mon, 19 Jan 2004 09:30:22 +0000, Don Pearce wrote: This is what I would expect at any frequency, and looks a lot like normal comb filtering - the signal arrives at the microphone via two paths, one direct and one reflected. For some positions there is reinforcement and for others cancellation. But this has nothing to do with room modes or standing waves. It is a purely travelling wave phenomenon. Ahem, I thought a standing wave *was* traveling waves (or components thereof) of the same frequency travelling in opposite directions at the same speed. In a standing wave there is no net transfer of energy along the direction of "travel" of the wave. The energy remains constrained between two boundaries. This is why they build up; they can be pumped - and also why they don't die the instant the stimulus is removed. In fact, standing waves make it impossible to measure the reverberation time of a room at very low frequencies - the wave collapses when it will, and the time taken is generally longer than the T60 of the room. Travelling waves possess none of these qualities, and must be handled, both analytically and practically, quite differently. I think there is a slight terminology confusion here. When standing waves are treated mathematically, there is often only ONE boundary, and the standing wave occurs at any frequency. Here, there is no energy buildup over time, there are merely two identical waves traveling in opposite directions, and no energy transportation. On the other hand I often see, and perhaps say myself, that a "standing wave" is used as a synonym to an excited room resonance. Here we have an energy buildup, two or more boundaries, and this phenomenon occurs (of course) only at frequencies near the mode frequency. |
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Non-modal peaks and nulls - here's the proof!
Don Pearce wrote in message . ..
On Mon, 19 Jan 2004 19:03:52 GMT, ow (Goofball_star_dot_etal) wrote: On Mon, 19 Jan 2004 09:30:22 +0000, Don Pearce wrote: This is what I would expect at any frequency, and looks a lot like normal comb filtering - the signal arrives at the microphone via two paths, one direct and one reflected. For some positions there is reinforcement and for others cancellation. But this has nothing to do with room modes or standing waves. It is a purely travelling wave phenomenon. Ahem, I thought a standing wave *was* traveling waves (or components thereof) of the same frequency travelling in opposite directions at the same speed. In a standing wave there is no net transfer of energy along the direction of "travel" of the wave. The energy remains constrained between two boundaries. This is why they build up; they can be pumped - and also why they don't die the instant the stimulus is removed. In fact, standing waves make it impossible to measure the reverberation time of a room at very low frequencies - the wave collapses when it will, and the time taken is generally longer than the T60 of the room. Travelling waves possess none of these qualities, and must be handled, both analytically and practically, quite differently. I think there is a slight terminology confusion here. When standing waves are treated mathematically, there is often only ONE boundary, and the standing wave occurs at any frequency. Here, there is no energy buildup over time, there are merely two identical waves traveling in opposite directions, and no energy transportation. On the other hand I often see, and perhaps say myself, that a "standing wave" is used as a synonym to an excited room resonance. Here we have an energy buildup, two or more boundaries, and this phenomenon occurs (of course) only at frequencies near the mode frequency. |
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Non-modal peaks and nulls - here's the proof!
Don Pearce wrote in message . ..
On Mon, 19 Jan 2004 19:03:52 GMT, ow (Goofball_star_dot_etal) wrote: On Mon, 19 Jan 2004 09:30:22 +0000, Don Pearce wrote: This is what I would expect at any frequency, and looks a lot like normal comb filtering - the signal arrives at the microphone via two paths, one direct and one reflected. For some positions there is reinforcement and for others cancellation. But this has nothing to do with room modes or standing waves. It is a purely travelling wave phenomenon. Ahem, I thought a standing wave *was* traveling waves (or components thereof) of the same frequency travelling in opposite directions at the same speed. In a standing wave there is no net transfer of energy along the direction of "travel" of the wave. The energy remains constrained between two boundaries. This is why they build up; they can be pumped - and also why they don't die the instant the stimulus is removed. In fact, standing waves make it impossible to measure the reverberation time of a room at very low frequencies - the wave collapses when it will, and the time taken is generally longer than the T60 of the room. Travelling waves possess none of these qualities, and must be handled, both analytically and practically, quite differently. I think there is a slight terminology confusion here. When standing waves are treated mathematically, there is often only ONE boundary, and the standing wave occurs at any frequency. Here, there is no energy buildup over time, there are merely two identical waves traveling in opposite directions, and no energy transportation. On the other hand I often see, and perhaps say myself, that a "standing wave" is used as a synonym to an excited room resonance. Here we have an energy buildup, two or more boundaries, and this phenomenon occurs (of course) only at frequencies near the mode frequency. |
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Non-modal peaks and nulls - here's the proof!
Svante,
I argue that your calculation of the room modes does not take into account that the room is not "perfect". No disagreement there. This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. I also have doubts about the mode frequencies that you show on Jeff Szymanski of Auralex also pointed out to me those missing modes. I used the McSquared web site, listed on the video's web page. Assuming you guys both calculated correctly, I have to agree there are modes very close to the frequencies we tested. I still maintain these nulls exist anyway because of boundary intrerference, whether the frequencies are modal or not. your statement that you DON'T have room resonances at the frequencies you used. Again, I agree in that sense the test is flawed. --Ethan |
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Non-modal peaks and nulls - here's the proof!
Svante,
I argue that your calculation of the room modes does not take into account that the room is not "perfect". No disagreement there. This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. I also have doubts about the mode frequencies that you show on Jeff Szymanski of Auralex also pointed out to me those missing modes. I used the McSquared web site, listed on the video's web page. Assuming you guys both calculated correctly, I have to agree there are modes very close to the frequencies we tested. I still maintain these nulls exist anyway because of boundary intrerference, whether the frequencies are modal or not. your statement that you DON'T have room resonances at the frequencies you used. Again, I agree in that sense the test is flawed. --Ethan |
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Non-modal peaks and nulls - here's the proof!
Svante,
I argue that your calculation of the room modes does not take into account that the room is not "perfect". No disagreement there. This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. I also have doubts about the mode frequencies that you show on Jeff Szymanski of Auralex also pointed out to me those missing modes. I used the McSquared web site, listed on the video's web page. Assuming you guys both calculated correctly, I have to agree there are modes very close to the frequencies we tested. I still maintain these nulls exist anyway because of boundary intrerference, whether the frequencies are modal or not. your statement that you DON'T have room resonances at the frequencies you used. Again, I agree in that sense the test is flawed. --Ethan |
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Non-modal peaks and nulls - here's the proof!
Svante,
I argue that your calculation of the room modes does not take into account that the room is not "perfect". No disagreement there. This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. I also have doubts about the mode frequencies that you show on Jeff Szymanski of Auralex also pointed out to me those missing modes. I used the McSquared web site, listed on the video's web page. Assuming you guys both calculated correctly, I have to agree there are modes very close to the frequencies we tested. I still maintain these nulls exist anyway because of boundary intrerference, whether the frequencies are modal or not. your statement that you DON'T have room resonances at the frequencies you used. Again, I agree in that sense the test is flawed. --Ethan |
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Non-modal peaks and nulls - here's the proof!
Don,
First of all, let me apologize for not getting back sooner. This is really the first chance I've had. Let me respond to you point-by-point below. Don Pearce writes: On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote: No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. When you have two sources, I think the more appropriate term would be "constructive" and/or "destructive" interference. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. Perhaps a "full energy" standing wave does, but standing waves in general do not. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. From Halliday and Resnick's "Fundamentals of Physics": In a one-dimensional body of finite size, such as a taut string held by two clamps a distance l apart, traveling waves in the string are reflected from the boundaries of the body, that is, from the clamps. The reflected waves add to the incident waves according to the principle of superposition. [derivation of the equation for such a wave follows] Equation 16-18b is the equation of a *standing wave.* So according to Halliday and Resnick, a standing wave is what results when we have a single traveling wave reflecting from a boundary and adding to the incident wave according to the principle of superposition. However, you state that there is nothing in a scenario involving the superposition of a direct and reflected wave to suggest that there is a standing wave. Whom shall we believe, you or Halliday and Resnick? Note that they also state: In a traveling wave each particle of the string vibrates with the same amplitude. Characteristic of a standing wave, however, is the fact that the amplitude is not the same for different particles but varies with the location x of the particle. So we can say by this definition alone that the waves Ethan is measuring are standing waves, NOT traveling waves. Ok, Mr. Pearce, let's say you don't believe Halliday and Resnick. Let's crack open Kinsler, Frey, Coppens, and Sander's "Fundamentals of Acoustics" to section 9.7, "The Rectangular Cavity" and see what *they* have to say: Consider a rectangular cavity of dimensions L_x, L_y, L_z, as indicated in Fig. 9.6. This box could represent a living room or auditorium, a simple model of a concert ahll, or any other rectangular space that has few windows or other openings and fairly rigid walls. After some derivations they arrive at the result (9.49) p_{lmn} = A_{lmn} * cos(k_{xl} * x) * cos(k_{ym} * y) * cos(k_{zn} * z) * e^{j * \omega_{lmn} * t}, where the components of (vector) k are k_{xl} = l * pi / L_x, l = 0, 1, 2, ... k_{ym} = m * pi / L_y, m = 0, 1, 2, ... k_{zn} = n * pi / L_z, n = 0, 1, 2, ... and where \omega_{lmn} = c*(k_{xl}^2 + k_{ym}^2 + k_{zn}^2)^{1/2}. They then state: The form (9.49) gives three-dimensional standing waves in the cavity with modal planes parallel to the walls. Ok, so Kinsler et al. say these waves in a room are *standing waves*. But then you probably don't believe them either. You probably just want to make your own definitions. However, this does bring out a good point: the modes above are for values of \omega_{lmn}, which is not continuous but rather a set of discrete frequencies depending on the integers l, m, and n. This comes about when solving the differential wave equation with subject to rigid boundary conditions. The whole point of Ethan's experiment was to show that you get these "modes" for ANY frequency, not just certain discrete values. I suspect we can resolve this discrepency by changing the boundary conditions from "complete rigidity" (\partial p/\partial x) = 0, and similar for y and z) to "partial rigidity" and resolving the differential wave equation. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% % 'Waterfall', *Face The Music*, ELO |
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Non-modal peaks and nulls - here's the proof!
Don,
First of all, let me apologize for not getting back sooner. This is really the first chance I've had. Let me respond to you point-by-point below. Don Pearce writes: On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote: No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. When you have two sources, I think the more appropriate term would be "constructive" and/or "destructive" interference. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. Perhaps a "full energy" standing wave does, but standing waves in general do not. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. From Halliday and Resnick's "Fundamentals of Physics": In a one-dimensional body of finite size, such as a taut string held by two clamps a distance l apart, traveling waves in the string are reflected from the boundaries of the body, that is, from the clamps. The reflected waves add to the incident waves according to the principle of superposition. [derivation of the equation for such a wave follows] Equation 16-18b is the equation of a *standing wave.* So according to Halliday and Resnick, a standing wave is what results when we have a single traveling wave reflecting from a boundary and adding to the incident wave according to the principle of superposition. However, you state that there is nothing in a scenario involving the superposition of a direct and reflected wave to suggest that there is a standing wave. Whom shall we believe, you or Halliday and Resnick? Note that they also state: In a traveling wave each particle of the string vibrates with the same amplitude. Characteristic of a standing wave, however, is the fact that the amplitude is not the same for different particles but varies with the location x of the particle. So we can say by this definition alone that the waves Ethan is measuring are standing waves, NOT traveling waves. Ok, Mr. Pearce, let's say you don't believe Halliday and Resnick. Let's crack open Kinsler, Frey, Coppens, and Sander's "Fundamentals of Acoustics" to section 9.7, "The Rectangular Cavity" and see what *they* have to say: Consider a rectangular cavity of dimensions L_x, L_y, L_z, as indicated in Fig. 9.6. This box could represent a living room or auditorium, a simple model of a concert ahll, or any other rectangular space that has few windows or other openings and fairly rigid walls. After some derivations they arrive at the result (9.49) p_{lmn} = A_{lmn} * cos(k_{xl} * x) * cos(k_{ym} * y) * cos(k_{zn} * z) * e^{j * \omega_{lmn} * t}, where the components of (vector) k are k_{xl} = l * pi / L_x, l = 0, 1, 2, ... k_{ym} = m * pi / L_y, m = 0, 1, 2, ... k_{zn} = n * pi / L_z, n = 0, 1, 2, ... and where \omega_{lmn} = c*(k_{xl}^2 + k_{ym}^2 + k_{zn}^2)^{1/2}. They then state: The form (9.49) gives three-dimensional standing waves in the cavity with modal planes parallel to the walls. Ok, so Kinsler et al. say these waves in a room are *standing waves*. But then you probably don't believe them either. You probably just want to make your own definitions. However, this does bring out a good point: the modes above are for values of \omega_{lmn}, which is not continuous but rather a set of discrete frequencies depending on the integers l, m, and n. This comes about when solving the differential wave equation with subject to rigid boundary conditions. The whole point of Ethan's experiment was to show that you get these "modes" for ANY frequency, not just certain discrete values. I suspect we can resolve this discrepency by changing the boundary conditions from "complete rigidity" (\partial p/\partial x) = 0, and similar for y and z) to "partial rigidity" and resolving the differential wave equation. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% % 'Waterfall', *Face The Music*, ELO |
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Non-modal peaks and nulls - here's the proof!
Don,
First of all, let me apologize for not getting back sooner. This is really the first chance I've had. Let me respond to you point-by-point below. Don Pearce writes: On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote: No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. When you have two sources, I think the more appropriate term would be "constructive" and/or "destructive" interference. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. Perhaps a "full energy" standing wave does, but standing waves in general do not. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. From Halliday and Resnick's "Fundamentals of Physics": In a one-dimensional body of finite size, such as a taut string held by two clamps a distance l apart, traveling waves in the string are reflected from the boundaries of the body, that is, from the clamps. The reflected waves add to the incident waves according to the principle of superposition. [derivation of the equation for such a wave follows] Equation 16-18b is the equation of a *standing wave.* So according to Halliday and Resnick, a standing wave is what results when we have a single traveling wave reflecting from a boundary and adding to the incident wave according to the principle of superposition. However, you state that there is nothing in a scenario involving the superposition of a direct and reflected wave to suggest that there is a standing wave. Whom shall we believe, you or Halliday and Resnick? Note that they also state: In a traveling wave each particle of the string vibrates with the same amplitude. Characteristic of a standing wave, however, is the fact that the amplitude is not the same for different particles but varies with the location x of the particle. So we can say by this definition alone that the waves Ethan is measuring are standing waves, NOT traveling waves. Ok, Mr. Pearce, let's say you don't believe Halliday and Resnick. Let's crack open Kinsler, Frey, Coppens, and Sander's "Fundamentals of Acoustics" to section 9.7, "The Rectangular Cavity" and see what *they* have to say: Consider a rectangular cavity of dimensions L_x, L_y, L_z, as indicated in Fig. 9.6. This box could represent a living room or auditorium, a simple model of a concert ahll, or any other rectangular space that has few windows or other openings and fairly rigid walls. After some derivations they arrive at the result (9.49) p_{lmn} = A_{lmn} * cos(k_{xl} * x) * cos(k_{ym} * y) * cos(k_{zn} * z) * e^{j * \omega_{lmn} * t}, where the components of (vector) k are k_{xl} = l * pi / L_x, l = 0, 1, 2, ... k_{ym} = m * pi / L_y, m = 0, 1, 2, ... k_{zn} = n * pi / L_z, n = 0, 1, 2, ... and where \omega_{lmn} = c*(k_{xl}^2 + k_{ym}^2 + k_{zn}^2)^{1/2}. They then state: The form (9.49) gives three-dimensional standing waves in the cavity with modal planes parallel to the walls. Ok, so Kinsler et al. say these waves in a room are *standing waves*. But then you probably don't believe them either. You probably just want to make your own definitions. However, this does bring out a good point: the modes above are for values of \omega_{lmn}, which is not continuous but rather a set of discrete frequencies depending on the integers l, m, and n. This comes about when solving the differential wave equation with subject to rigid boundary conditions. The whole point of Ethan's experiment was to show that you get these "modes" for ANY frequency, not just certain discrete values. I suspect we can resolve this discrepency by changing the boundary conditions from "complete rigidity" (\partial p/\partial x) = 0, and similar for y and z) to "partial rigidity" and resolving the differential wave equation. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% % 'Waterfall', *Face The Music*, ELO |
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Non-modal peaks and nulls - here's the proof!
Don,
First of all, let me apologize for not getting back sooner. This is really the first chance I've had. Let me respond to you point-by-point below. Don Pearce writes: On Tue, 20 Jan 2004 12:55:14 GMT, Randy Yates wrote: No they aren't. They are waves travelling away from a source into space. The difference between a travelling wave and a standing wave is the presence of a boundary. The wall is definitely a boundary. End of discussion. Don't tell me when a discussion is ended. The presence of a boundary does not necessarily generate a standing wave, although a boundary is certainly a necessity if you want standing waves. Actually, let me clarify that. If you point two loudspeakers at each other, both playing the same tone, you will find a very nice standing wave between them, and no need for boundaries. When you have two sources, I think the more appropriate term would be "constructive" and/or "destructive" interference. A boundary at an angle to the incoming wave will reflect energy at some arbitrary angle. A standing wave will only result if that energy is directed back along the original arrival path. A full energy standing wave needs both total reflection and perfect alignment. Perhaps a "full energy" standing wave does, but standing waves in general do not. The phenomenon described by Ethan is simply superposition caused by simultaneous arrival of a direct and reflected wave at a point in space. There is absolutely nothing in this scenario to suggest that there is a standing wave. From Halliday and Resnick's "Fundamentals of Physics": In a one-dimensional body of finite size, such as a taut string held by two clamps a distance l apart, traveling waves in the string are reflected from the boundaries of the body, that is, from the clamps. The reflected waves add to the incident waves according to the principle of superposition. [derivation of the equation for such a wave follows] Equation 16-18b is the equation of a *standing wave.* So according to Halliday and Resnick, a standing wave is what results when we have a single traveling wave reflecting from a boundary and adding to the incident wave according to the principle of superposition. However, you state that there is nothing in a scenario involving the superposition of a direct and reflected wave to suggest that there is a standing wave. Whom shall we believe, you or Halliday and Resnick? Note that they also state: In a traveling wave each particle of the string vibrates with the same amplitude. Characteristic of a standing wave, however, is the fact that the amplitude is not the same for different particles but varies with the location x of the particle. So we can say by this definition alone that the waves Ethan is measuring are standing waves, NOT traveling waves. Ok, Mr. Pearce, let's say you don't believe Halliday and Resnick. Let's crack open Kinsler, Frey, Coppens, and Sander's "Fundamentals of Acoustics" to section 9.7, "The Rectangular Cavity" and see what *they* have to say: Consider a rectangular cavity of dimensions L_x, L_y, L_z, as indicated in Fig. 9.6. This box could represent a living room or auditorium, a simple model of a concert ahll, or any other rectangular space that has few windows or other openings and fairly rigid walls. After some derivations they arrive at the result (9.49) p_{lmn} = A_{lmn} * cos(k_{xl} * x) * cos(k_{ym} * y) * cos(k_{zn} * z) * e^{j * \omega_{lmn} * t}, where the components of (vector) k are k_{xl} = l * pi / L_x, l = 0, 1, 2, ... k_{ym} = m * pi / L_y, m = 0, 1, 2, ... k_{zn} = n * pi / L_z, n = 0, 1, 2, ... and where \omega_{lmn} = c*(k_{xl}^2 + k_{ym}^2 + k_{zn}^2)^{1/2}. They then state: The form (9.49) gives three-dimensional standing waves in the cavity with modal planes parallel to the walls. Ok, so Kinsler et al. say these waves in a room are *standing waves*. But then you probably don't believe them either. You probably just want to make your own definitions. However, this does bring out a good point: the modes above are for values of \omega_{lmn}, which is not continuous but rather a set of discrete frequencies depending on the integers l, m, and n. This comes about when solving the differential wave equation with subject to rigid boundary conditions. The whole point of Ethan's experiment was to show that you get these "modes" for ANY frequency, not just certain discrete values. I suspect we can resolve this discrepency by changing the boundary conditions from "complete rigidity" (\partial p/\partial x) = 0, and similar for y and z) to "partial rigidity" and resolving the differential wave equation. -- % Randy Yates % "So now it's getting late, %% Fuquay-Varina, NC % and those who hesitate %%% 919-577-9882 % got no one..." %%%% % 'Waterfall', *Face The Music*, ELO |
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. Yes, the outdoor test proves that zeroes can occur without modes and I see no reason why this phenomenon would not occur indoors as well. My writing "your test" above referred to the indoor test. |
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. Yes, the outdoor test proves that zeroes can occur without modes and I see no reason why this phenomenon would not occur indoors as well. My writing "your test" above referred to the indoor test. |
#116
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. Yes, the outdoor test proves that zeroes can occur without modes and I see no reason why this phenomenon would not occur indoors as well. My writing "your test" above referred to the indoor test. |
#117
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Non-modal peaks and nulls - here's the proof!
"Ethan Winer" ethanw at ethanwiner dot com wrote in message ...
This means that you could actually have modes at the frequencies you have selected (see below, also) and that this would make your test invalid as proof of zeroes at non-mode frequencies. If the phenomenon didn't happen outdoors I'd say you have a point. But it does happen outdoors. ALL that changes when the one wall is enclosed completely to make a "room" is the addition of yet more reflections. And those make it harder to find nulls because of all the wave interaction. Yes, the outdoor test proves that zeroes can occur without modes and I see no reason why this phenomenon would not occur indoors as well. My writing "your test" above referred to the indoor test. |
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