Valve Lengths

I'm trying to solve this problem:

How can I most efficiently build a brass instrument with valves that give me pipe length discrimination to a certain scale? That is, trying to keep the number of valves low, what lengths should I give each valve so that combinations of them give me discrimination of pipe length down to a given scale like, say, quarters of a centimeter?

I think I found a system that works but you all here might be able to figure out something that works better. If I have the length of the valves following the fibonacci sequence, I end up needing relatively very few valves for a very accurate length discrimination. That is, if the first valve is 1 unit long, the second 2, the third 3, then 5, 8, 13, etc., using combinations thereof I can reach any total length up to two units less than the fibonacci number after the number used as the length of the last valve.

For instance, with the valve lengths 1, 2, 3, 5, 8, 13, 21, and 34, I could get any integer unit length up to 53 (two less than the next fibonacci number, 55). If I used two more valves (with lengths 55 and 89) I could get any integer unit length up to 142 (two less than 144). That is, with only 10 valves (for ten fingers?) I could get 142 discrete lengths, which doesn't sound half bad.

Is there any way to prove whether or not this is the most efficient use of valves and piping?

John

--- In tuning-math@yahoogroups.com, "jlmoriart" <JlMoriart@...> wrote:
> I think I found a system that works but you all here might be able to figure out something that works better. If I have the length of the valves following the fibonacci sequence, I end up needing relatively very few valves for a very accurate length discrimination. That is, if the first valve is 1 unit long, the second 2, the third 3, then 5, 8, 13, etc., using combinations thereof I can reach any total length up to two units less than the fibonacci number after the number used as the length of the last valve.

Why wouldn't powers of two be better than fibonacci numbers?

> For instance, with the valve lengths 1, 2, 3, 5, 8, 13, 21, and 34, I could get any integer unit length up to 53 (two less than the next fibonacci number, 55). If I used two more valves (with lengths 55 and 89) I could get any integer unit length up to 142 (two less than 144). That is, with only 10 valves (for ten fingers?) I could get 142 discrete lengths, which doesn't sound half bad.

For example, with 8 valves, the same number as your first example, why can't I make them 1, 2, 4, 8, 16, 32, 64, and 128, and then get any integer length up to 255?

It seems to me like Fibonacci numbers could only be optimal if there were some restriction on which valves you could use together (specifically, if for some reason you couldn't use two adjacent valves in combination, then Fibonacci numbers would be the answer). But as I remember from playing my friend's brass instruments in high school, you can use any combination of valve states and they all work fine, so there are 2^n different states for n valves. If these don't yield 2^n different pipe lengths, then your system is inefficient.

I said "my friend's" instruments rather than my own, because I played trombone, which leads my to my more flippant answer to this question: Get a damn slide!! =)

Keenan

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "jlmoriart" <JlMoriart@> wrote:
> > I think I found a system that works but you all here might be able to figure out something that works better. If I have the length of the valves following the fibonacci sequence, I end up needing relatively very few valves for a very accurate length discrimination. That is, if the first valve is 1 unit long, the second 2, the third 3, then 5, 8, 13, etc., using combinations thereof I can reach any total length up to two units less than the fibonacci number after the number used as the length of the last valve.
>
> Why wouldn't powers of two be better than fibonacci numbers?

I don't think I'm explaining the problem correctly. I want to be able to add any pipe length (down to say, a centimeter) using combinations of valves.

Using powers of two would give me valve lengths of 2, 4, 8, 16 etc, but no combination of those numbers would allow me to add 1, 3, 5, 7, etc.

I could use one valve per integer I want to add, with valves lengths of 1, 2, 3, 4, 5, etc, but then I'd be wasting a lot of piping because I could use the second and third valves to give me 2+3=5. Also, there would be gaps between the min and max combinations.

Fibonacci numbers work out so that with combinations I can get every integer between 1 and the total of all numbers in the series. Is there any other series with this property?

John

One is a power of two. If you factor one into everything, you will be able
to get to any integer, just like how a computer can represent every number
in binary.

Sent from my iPhone

On Jun 21, 2011, at 1:17 AM, jlmoriart <JlMoriart@gmail.com> wrote:

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "jlmoriart" <JlMoriart@> wrote:
> > I think I found a system that works but you all here might be able to
figure out something that works better. If I have the length of the valves
following the fibonacci sequence, I end up needing relatively very few
valves for a very accurate length discrimination. That is, if the first
valve is 1 unit long, the second 2, the third 3, then 5, 8, 13, etc., using
combinations thereof I can reach any total length up to two units less than
the fibonacci number after the number used as the length of the last valve.
>
> Why wouldn't powers of two be better than fibonacci numbers?

I don't think I'm explaining the problem correctly. I want to be able to add
any pipe length (down to say, a centimeter) using combinations of valves.

Using powers of two would give me valve lengths of 2, 4, 8, 16 etc, but no
combination of those numbers would allow me to add 1, 3, 5, 7, etc.

I could use one valve per integer I want to add, with valves lengths of 1,
2, 3, 4, 5, etc, but then I'd be wasting a lot of piping because I could use
the second and third valves to give me 2+3=5. Also, there would be gaps
between the min and max combinations.

Fibonacci numbers work out so that with combinations I can get every integer
between 1 and the total of all numbers in the series. Is there any other
series with this property?

John

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> One is a power of two. If you factor one into everything, you will be able
> to get to any integer, just like how a computer can represent every number
> in binary.

Right, what Mike said. Very explicitly, if my valve lengths are 1, 2, 4, and 8, then I can get total lengths of

1 = 1
2 = 2
1 + 2 = 3
4 = 4
1 + 4 = 5
2 + 4 = 6
1 + 2 + 4 = 7
8 = 8
1 + 8 = 9
2 + 8 = 10
1 + 2 + 8 = 11
4 + 8 = 12
1 + 4 + 8 = 13
2 + 4 + 8 = 14
1 + 2 + 4 + 8 = 15

Is there some reason this is unacceptable for what you have in mind? Are we still misunderstanding each other?

> --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "jlmoriart" <JlMoriart@> wrote:
> > Why wouldn't powers of two be better than fibonacci numbers?
>
> I don't think I'm explaining the problem correctly. I want to be able to add
> any pipe length (down to say, a centimeter) using combinations of valves.
>
> Using powers of two would give me valve lengths of 2, 4, 8, 16 etc, but no
> combination of those numbers would allow me to add 1, 3, 5, 7, etc.
>
> I could use one valve per integer I want to add, with valves lengths of 1,
> 2, 3, 4, 5, etc, but then I'd be wasting a lot of piping because I could use
> the second and third valves to give me 2+3=5. Also, there would be gaps
> between the min and max combinations.
>
> Fibonacci numbers work out so that with combinations I can get every integer
> between 1 and the total of all numbers in the series. Is there any other
> series with this property?

Yes, the powers of two (including 2^0 = 1) have this property and they're more efficient. That's what we're trying to tell you.

Keenan

Gooootcha. Thanks for being willing to point out the obvious...

John

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
> >
> > One is a power of two. If you factor one into everything, you will be able
> > to get to any integer, just like how a computer can represent every number
> > in binary.
>
> Right, what Mike said. Very explicitly, if my valve lengths are 1, 2, 4, and 8, then I can get total lengths of
>
> 1 = 1
> 2 = 2
> 1 + 2 = 3
> 4 = 4
> 1 + 4 = 5
> 2 + 4 = 6
> 1 + 2 + 4 = 7
> 8 = 8
> 1 + 8 = 9
> 2 + 8 = 10
> 1 + 2 + 8 = 11
> 4 + 8 = 12
> 1 + 4 + 8 = 13
> 2 + 4 + 8 = 14
> 1 + 2 + 4 + 8 = 15
>
> Is there some reason this is unacceptable for what you have in mind? Are we still misunderstanding each other?
>
> > --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@>
> > wrote:
> > >
> > > --- In tuning-math@yahoogroups.com, "jlmoriart" <JlMoriart@> wrote:
> > > Why wouldn't powers of two be better than fibonacci numbers?
> >
> > I don't think I'm explaining the problem correctly. I want to be able to add
> > any pipe length (down to say, a centimeter) using combinations of valves.
> >
> > Using powers of two would give me valve lengths of 2, 4, 8, 16 etc, but no
> > combination of those numbers would allow me to add 1, 3, 5, 7, etc.
> >
> > I could use one valve per integer I want to add, with valves lengths of 1,
> > 2, 3, 4, 5, etc, but then I'd be wasting a lot of piping because I could use
> > the second and third valves to give me 2+3=5. Also, there would be gaps
> > between the min and max combinations.
> >
> > Fibonacci numbers work out so that with combinations I can get every integer
> > between 1 and the total of all numbers in the series. Is there any other
> > series with this property?
>
> Yes, the powers of two (including 2^0 = 1) have this property and they're more efficient. That's what we're trying to tell you.
>
> Keenan
>