"Don Pearce" wrote in message

On Mon, 19 Jan 2004 20:08:16 GMT, ow
(Goofball_star_dot_etal) wrote:

On Mon, 19 Jan 2004 19:19:51 +0000, Don Pearce
wrote:

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'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?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

"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.)

"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.)

"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.)

"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.)

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.

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.

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.

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.

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

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

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

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

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

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

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

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

"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.

"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.

"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.

"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.