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How do I know where to place holes on an instrument?
Tutorials or advice on layering synthsHow to connect a line signal to a guitar amp?Note Accentuation (Dynamic) - When should/shouldn't you apply accentuation?Reed of the Duduk —how to get a sound?Is a wind-string instrument (such as the one I will explain) acoustically possible?A seriously difficult question about mistakes and intepretation of musicMy guitar instrument produced a perfect sine wave?How do uncovered tone holes in middle of a flute work?Is there some means of expanding the range of a capped reed instrument?How do you ream a conical bore when making a woodwind instrument?
I've been trying to build a double reeded instrument out of plastic straws. I've run into a problem though, when I place fingering holes, the instrument doesn't seem to follow the $$f=/fracnv4L$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds instrument-making
edited Apr 4 at 16:53
Tim H
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
add a comment |
I've been trying to build a double reeded instrument out of plastic straws. I've run into a problem though, when I place fingering holes, the instrument doesn't seem to follow the $$f=/fracnv4L$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds instrument-making
edited Apr 4 at 16:53
Tim H
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
3
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36
add a comment |
I've been trying to build a double reeded instrument out of plastic straws. I've run into a problem though, when I place fingering holes, the instrument doesn't seem to follow the $$f=/fracnv4L$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds instrument-making
edited Apr 4 at 16:53
Tim H
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
I've been trying to build a double reeded instrument out of plastic straws. I've run into a problem though, when I place fingering holes, the instrument doesn't seem to follow the $$f=/fracnv4L$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds instrument-making
acoustics construction reeds instrument-making
edited Apr 4 at 16:53
Tim H
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
edited Apr 4 at 16:53
Tim H
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
edited Apr 4 at 16:53
Tim H
3,07321944
edited Apr 4 at 16:53
Tim H
3,07321944
edited Apr 4 at 16:53
Tim H
3,07321944
3,07321944
asked Apr 3 at 23:42
tox123tox123
1558
asked Apr 3 at 23:42
tox123tox123
1558
asked Apr 3 at 23:42
tox123tox123
1558
1558
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
3
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36
add a comment |
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
3
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
3
3
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36
add a comment |
2 Answers
2
active
oldest
votes
For a pipe without any holes, the fundamental pitch of a pipe is determined by f = v/2L, where v is the speed of sound and L is the length of the pipe.
But when you start putting holes in, you get a mix of pipe lengths. It gets messy (mathematically) in a big hurry.
Placing a hole in the pipe shortens its length, but the new length - the effective length - isn't the distance to the hole, because that isn't the end of the pipe. The larger the hole, the more it will behave like the ideal. The smaller the hole, the longer the effective length will be.
Because the hole size is a variable, you're not going to find a formula that's going to fit every situation. That's because the hole size is a variable in relation to the diameter of the tube: a 1cm hole in a 10cm tube will have a different effective length than a 1cm hole in a 9cm tube.
Since no formula is going to work for all situations, you have do some trial and error. If the pitch is flat, you can enlarge the hole to shorten the effective length. If the pitch is sharp, you'll have to figure out a way to make that hole smaller (or make the whole tube longer - there's a reason woodwinds have multiple pieces!)
There are other variables, too... conical bores behave differently than cylindrical ones. But I'm assuming you're using a cylindrical tube.
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
add a comment |
The main problem is that it's an oversimplified assumption to consider an open hole as a perfect open boundary condition for the air column. In fact such a hole still has a significant impedance. On the other side, the mouthpiece is not a perfect closed (reeds) or open (flutes) boundary condition, and also a closed hole still affects the column somewhat. A pitch formula would need to take all those factors into account, which depend on hole size and bore. Doing this accurately would require a big CFD model.
In practice, probably almost every wind instrument maker has instead used empirical models, i.e. basically trial and error. It should certainly be possible to fit a formula to that which is more accurate than the overidealisation but still reasonably accurate, but whether one is available publicly I don't know.
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
add a comment |
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2 Answers
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active
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2 Answers
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oldest
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votes
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oldest
votes
For a pipe without any holes, the fundamental pitch of a pipe is determined by f = v/2L, where v is the speed of sound and L is the length of the pipe.
But when you start putting holes in, you get a mix of pipe lengths. It gets messy (mathematically) in a big hurry.
Placing a hole in the pipe shortens its length, but the new length - the effective length - isn't the distance to the hole, because that isn't the end of the pipe. The larger the hole, the more it will behave like the ideal. The smaller the hole, the longer the effective length will be.
Because the hole size is a variable, you're not going to find a formula that's going to fit every situation. That's because the hole size is a variable in relation to the diameter of the tube: a 1cm hole in a 10cm tube will have a different effective length than a 1cm hole in a 9cm tube.
Since no formula is going to work for all situations, you have do some trial and error. If the pitch is flat, you can enlarge the hole to shorten the effective length. If the pitch is sharp, you'll have to figure out a way to make that hole smaller (or make the whole tube longer - there's a reason woodwinds have multiple pieces!)
There are other variables, too... conical bores behave differently than cylindrical ones. But I'm assuming you're using a cylindrical tube.
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
add a comment |
For a pipe without any holes, the fundamental pitch of a pipe is determined by f = v/2L, where v is the speed of sound and L is the length of the pipe.
But when you start putting holes in, you get a mix of pipe lengths. It gets messy (mathematically) in a big hurry.
Placing a hole in the pipe shortens its length, but the new length - the effective length - isn't the distance to the hole, because that isn't the end of the pipe. The larger the hole, the more it will behave like the ideal. The smaller the hole, the longer the effective length will be.
Because the hole size is a variable, you're not going to find a formula that's going to fit every situation. That's because the hole size is a variable in relation to the diameter of the tube: a 1cm hole in a 10cm tube will have a different effective length than a 1cm hole in a 9cm tube.
Since no formula is going to work for all situations, you have do some trial and error. If the pitch is flat, you can enlarge the hole to shorten the effective length. If the pitch is sharp, you'll have to figure out a way to make that hole smaller (or make the whole tube longer - there's a reason woodwinds have multiple pieces!)
There are other variables, too... conical bores behave differently than cylindrical ones. But I'm assuming you're using a cylindrical tube.
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
add a comment |
For a pipe without any holes, the fundamental pitch of a pipe is determined by f = v/2L, where v is the speed of sound and L is the length of the pipe.
But when you start putting holes in, you get a mix of pipe lengths. It gets messy (mathematically) in a big hurry.
Placing a hole in the pipe shortens its length, but the new length - the effective length - isn't the distance to the hole, because that isn't the end of the pipe. The larger the hole, the more it will behave like the ideal. The smaller the hole, the longer the effective length will be.
Because the hole size is a variable, you're not going to find a formula that's going to fit every situation. That's because the hole size is a variable in relation to the diameter of the tube: a 1cm hole in a 10cm tube will have a different effective length than a 1cm hole in a 9cm tube.
Since no formula is going to work for all situations, you have do some trial and error. If the pitch is flat, you can enlarge the hole to shorten the effective length. If the pitch is sharp, you'll have to figure out a way to make that hole smaller (or make the whole tube longer - there's a reason woodwinds have multiple pieces!)
There are other variables, too... conical bores behave differently than cylindrical ones. But I'm assuming you're using a cylindrical tube.
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
For a pipe without any holes, the fundamental pitch of a pipe is determined by f = v/2L, where v is the speed of sound and L is the length of the pipe.
But when you start putting holes in, you get a mix of pipe lengths. It gets messy (mathematically) in a big hurry.
Placing a hole in the pipe shortens its length, but the new length - the effective length - isn't the distance to the hole, because that isn't the end of the pipe. The larger the hole, the more it will behave like the ideal. The smaller the hole, the longer the effective length will be.
Because the hole size is a variable, you're not going to find a formula that's going to fit every situation. That's because the hole size is a variable in relation to the diameter of the tube: a 1cm hole in a 10cm tube will have a different effective length than a 1cm hole in a 9cm tube.
Since no formula is going to work for all situations, you have do some trial and error. If the pitch is flat, you can enlarge the hole to shorten the effective length. If the pitch is sharp, you'll have to figure out a way to make that hole smaller (or make the whole tube longer - there's a reason woodwinds have multiple pieces!)
There are other variables, too... conical bores behave differently than cylindrical ones. But I'm assuming you're using a cylindrical tube.
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
edited Apr 4 at 13:41
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
answered Apr 4 at 3:49
Tom SerbTom Serb
1,267110
1,267110
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
add a comment |
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
“In a perfect world” – that's a bit of a silly way to put it. You might also say in a perfect world the air flow is perfectly linear and laminar – however, in that case the mechanism by which flutes bring the air to vibrate in the first place also wouldn't work! And even assuming you get the air to vibrate in the perfect cylinder: nobody would be able to hear the sound, because all the energy is trapped! I wouldn't call that perfect. No, the world is exactly as good as is actually happens to be – what's imperfect are the models with which we describe certain phenomena in the world.
– leftaroundabout
Apr 4 at 11:45
3
3
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
"In a perfect world" is a common idiom that means a world that conforms to the model in use. You're taking it too literally.
– chepner
Apr 4 at 11:56
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@leftaroundabout I changed the answer -- see if it is more readable now.
– Carl Witthoft
Apr 4 at 13:04
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@CarlWitthoft - I altered your edit. Helmholtz resonance is the frequency created by air "bouncing" at the mouth hole. It contributes to the overall timbre whether there are holes or not, and introducing holes does not change the Helmholtz resonance - that's fixed by the size of the mouth hole and the thickness of the tube material. Open holes don't have Helmholtz resonance frequencies, because the air pressure in the tube is always higher than that outside; the air simply escapes without additional vibration.
– Tom Serb
Apr 4 at 13:47
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
@TomSerb thanks for doing that. I should have emphasized where Helmholtz was applicable (see "Ocarina" :-) )
– Carl Witthoft
Apr 4 at 14:34
add a comment |
The main problem is that it's an oversimplified assumption to consider an open hole as a perfect open boundary condition for the air column. In fact such a hole still has a significant impedance. On the other side, the mouthpiece is not a perfect closed (reeds) or open (flutes) boundary condition, and also a closed hole still affects the column somewhat. A pitch formula would need to take all those factors into account, which depend on hole size and bore. Doing this accurately would require a big CFD model.
In practice, probably almost every wind instrument maker has instead used empirical models, i.e. basically trial and error. It should certainly be possible to fit a formula to that which is more accurate than the overidealisation but still reasonably accurate, but whether one is available publicly I don't know.
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
add a comment |
The main problem is that it's an oversimplified assumption to consider an open hole as a perfect open boundary condition for the air column. In fact such a hole still has a significant impedance. On the other side, the mouthpiece is not a perfect closed (reeds) or open (flutes) boundary condition, and also a closed hole still affects the column somewhat. A pitch formula would need to take all those factors into account, which depend on hole size and bore. Doing this accurately would require a big CFD model.
In practice, probably almost every wind instrument maker has instead used empirical models, i.e. basically trial and error. It should certainly be possible to fit a formula to that which is more accurate than the overidealisation but still reasonably accurate, but whether one is available publicly I don't know.
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
add a comment |
The main problem is that it's an oversimplified assumption to consider an open hole as a perfect open boundary condition for the air column. In fact such a hole still has a significant impedance. On the other side, the mouthpiece is not a perfect closed (reeds) or open (flutes) boundary condition, and also a closed hole still affects the column somewhat. A pitch formula would need to take all those factors into account, which depend on hole size and bore. Doing this accurately would require a big CFD model.
In practice, probably almost every wind instrument maker has instead used empirical models, i.e. basically trial and error. It should certainly be possible to fit a formula to that which is more accurate than the overidealisation but still reasonably accurate, but whether one is available publicly I don't know.
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
The main problem is that it's an oversimplified assumption to consider an open hole as a perfect open boundary condition for the air column. In fact such a hole still has a significant impedance. On the other side, the mouthpiece is not a perfect closed (reeds) or open (flutes) boundary condition, and also a closed hole still affects the column somewhat. A pitch formula would need to take all those factors into account, which depend on hole size and bore. Doing this accurately would require a big CFD model.
In practice, probably almost every wind instrument maker has instead used empirical models, i.e. basically trial and error. It should certainly be possible to fit a formula to that which is more accurate than the overidealisation but still reasonably accurate, but whether one is available publicly I don't know.
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
answered Apr 4 at 1:40
leftaroundaboutleftaroundabout
20.8k3690
20.8k3690
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
add a comment |
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
I suspect that Adolphe Sax used the same spacing ratios in (soprano, alto, tenor, bari) at least as a starting point, given the similarity in the bore shapes.
– Carl Witthoft
Apr 4 at 13:06
add a comment |
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I guess the answer is guess! :D It may also be trial and error.
– Xilpex
Apr 3 at 23:44
3
It seems that MathJax (the fraction code) isn't supported on Music SE.
– Bladewood
Apr 4 at 10:36