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How do I know where to place holes on an instrument?
How do I know where to place holes on an instrument?
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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 $$f=/frac{nv}{4L}$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds
asked 5 hours ago
tox123tox123
1187
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 $$f=/frac{nv}{4L}$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds
asked 5 hours ago
tox123tox123
1187
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago
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 $$f=/frac{nv}{4L}$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds
asked 5 hours ago
tox123tox123
1187
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 $$f=/frac{nv}{4L}$$ formula. Is there a formula that would allow me to calculate where along the instrument I should place the holes?
acoustics construction reeds
acoustics construction reeds
asked 5 hours ago
tox123tox123
1187
asked 5 hours ago
tox123tox123
1187
asked 5 hours ago
tox123tox123
1187
asked 5 hours ago
tox123tox123
1187
asked 5 hours ago
tox123tox123
1187
1187
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago
add a comment |
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago
I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
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 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
add a comment |
In a perfect world, 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 we don't live in a perfect world.
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 1 hour ago
Tom SerbTom Serb
1,132110
add a comment |
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2 Answers
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2 Answers
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active
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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 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
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 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
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 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
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 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
answered 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
answered 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
answered 3 hours ago
leftaroundaboutleftaroundabout
20.7k3690
20.7k3690
add a comment |
add a comment |
In a perfect world, 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 we don't live in a perfect world.
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 1 hour ago
Tom SerbTom Serb
1,132110
add a comment |
In a perfect world, 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 we don't live in a perfect world.
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 1 hour ago
Tom SerbTom Serb
1,132110
add a comment |
In a perfect world, 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 we don't live in a perfect world.
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 1 hour ago
Tom SerbTom Serb
1,132110
In a perfect world, 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 we don't live in a perfect world.
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 1 hour ago
Tom SerbTom Serb
1,132110
answered 1 hour ago
Tom SerbTom Serb
1,132110
answered 1 hour ago
Tom SerbTom Serb
1,132110
answered 1 hour ago
Tom SerbTom Serb
1,132110
1,132110
add a comment |
add a comment |
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I guess the answer is guess! :D It may also be trial and error.
– Xilpex
5 hours ago