Mixing
Find three points on a scale using golden ratio
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
My question came from my musicial part:
The note A4 on a piano is 440hz and the note
A5 (one octave higher) is 880hz.
On the piano there are 12 notes
between A4 and A5 (include).
Im trying to find three notes between A4 and A5 (hence on the scale of 440 to 880)
Using the golden ratio: 1.618 (accuracy of three points is enough here).
More specific:
There are 5 points on an axis from 440 to 880:
A4=440 note1=?, note2=?, note3=?, A5=880
I want to find (note1, note2, note3)
Where note2 is: “1.618 closer to note1 than note1 to A4â€Â,
note3 is “1.618 closer to note2 than note2 to note1†and
A5 is “1.618 closer to note3 than note3 to note2.
Im not sure if one can prove that 3 points like this exist,
but I guess we can prove that there are n points (n is natural) like that that exist, right?
How can I find these n’s? And when finding one, how can I find its n notes?
ratio music-theory
asked Jul 30 at 14:01
user1901968
152
add a comment |Â
up vote
1
down vote
favorite
My question came from my musicial part:
The note A4 on a piano is 440hz and the note
A5 (one octave higher) is 880hz.
On the piano there are 12 notes
between A4 and A5 (include).
Im trying to find three notes between A4 and A5 (hence on the scale of 440 to 880)
Using the golden ratio: 1.618 (accuracy of three points is enough here).
More specific:
There are 5 points on an axis from 440 to 880:
A4=440 note1=?, note2=?, note3=?, A5=880
I want to find (note1, note2, note3)
Where note2 is: “1.618 closer to note1 than note1 to A4â€Â,
note3 is “1.618 closer to note2 than note2 to note1†and
A5 is “1.618 closer to note3 than note3 to note2.
Im not sure if one can prove that 3 points like this exist,
but I guess we can prove that there are n points (n is natural) like that that exist, right?
How can I find these n’s? And when finding one, how can I find its n notes?
ratio music-theory
asked Jul 30 at 14:01
user1901968
152
1
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My question came from my musicial part:
The note A4 on a piano is 440hz and the note
A5 (one octave higher) is 880hz.
On the piano there are 12 notes
between A4 and A5 (include).
Im trying to find three notes between A4 and A5 (hence on the scale of 440 to 880)
Using the golden ratio: 1.618 (accuracy of three points is enough here).
More specific:
There are 5 points on an axis from 440 to 880:
A4=440 note1=?, note2=?, note3=?, A5=880
I want to find (note1, note2, note3)
Where note2 is: “1.618 closer to note1 than note1 to A4â€Â,
note3 is “1.618 closer to note2 than note2 to note1†and
A5 is “1.618 closer to note3 than note3 to note2.
Im not sure if one can prove that 3 points like this exist,
but I guess we can prove that there are n points (n is natural) like that that exist, right?
How can I find these n’s? And when finding one, how can I find its n notes?
ratio music-theory
asked Jul 30 at 14:01
user1901968
152
My question came from my musicial part:
The note A4 on a piano is 440hz and the note
A5 (one octave higher) is 880hz.
On the piano there are 12 notes
between A4 and A5 (include).
Im trying to find three notes between A4 and A5 (hence on the scale of 440 to 880)
Using the golden ratio: 1.618 (accuracy of three points is enough here).
More specific:
There are 5 points on an axis from 440 to 880:
A4=440 note1=?, note2=?, note3=?, A5=880
I want to find (note1, note2, note3)
Where note2 is: “1.618 closer to note1 than note1 to A4â€Â,
note3 is “1.618 closer to note2 than note2 to note1†and
A5 is “1.618 closer to note3 than note3 to note2.
Im not sure if one can prove that 3 points like this exist,
but I guess we can prove that there are n points (n is natural) like that that exist, right?
How can I find these n’s? And when finding one, how can I find its n notes?
ratio music-theory
asked Jul 30 at 14:01
user1901968
152
asked Jul 30 at 14:01
user1901968
152
asked Jul 30 at 14:01
user1901968
152
asked Jul 30 at 14:01
user1901968
152
152
1
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28
add a comment |Â
1
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28
1
1
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:
Take:
$$d_1=n_1-A_4$$
$$d_2=n_2-n_1$$
$$d_3=n_3-n_2$$
$$d_4=A_5-n_3$$
Where $n_k$ is the $k_th$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_k-1$ and $n_0=A_4$ and $n_4=A_5$.
Then, by the golden ratio constraints you've asked for, we end up with:
$$n_3-n_2=(A_5-n_3)cdot1.618$$
$$n_2-n_1=(n_3-n_2)cdot1.618$$
$$n_1-A_4=(n_2-n_1)cdot1.618$$
Solving these taking $A_4=440$ and $A_5=880$, we end up with
$$A_4=440$$
$$n_1=636.7701$$
$$n_2=758.3833$$
$$n_3=833.5459$$
$$A_5=880$$
All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
add a comment |Â
up vote
0
down vote
If what you want is the ratio of the frequencies to be $varphi$ for each of the four ratios then this is impossible.
The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is
$1.05946309436 approx 1.06$. Four your five note scale you want
$$
2^1/5 = 1.148698355 approx 1.15 .
$$
That's pretty far from $varphi approx 1.618$.
Even just a two note scale gives you a ratio pf $sqrt2 approx 1.4 < varphi$ .
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:
Take:
$$d_1=n_1-A_4$$
$$d_2=n_2-n_1$$
$$d_3=n_3-n_2$$
$$d_4=A_5-n_3$$
Where $n_k$ is the $k_th$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_k-1$ and $n_0=A_4$ and $n_4=A_5$.
Then, by the golden ratio constraints you've asked for, we end up with:
$$n_3-n_2=(A_5-n_3)cdot1.618$$
$$n_2-n_1=(n_3-n_2)cdot1.618$$
$$n_1-A_4=(n_2-n_1)cdot1.618$$
Solving these taking $A_4=440$ and $A_5=880$, we end up with
$$A_4=440$$
$$n_1=636.7701$$
$$n_2=758.3833$$
$$n_3=833.5459$$
$$A_5=880$$
All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
add a comment |Â
up vote
2
down vote
accepted
From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:
Take:
$$d_1=n_1-A_4$$
$$d_2=n_2-n_1$$
$$d_3=n_3-n_2$$
$$d_4=A_5-n_3$$
Where $n_k$ is the $k_th$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_k-1$ and $n_0=A_4$ and $n_4=A_5$.
Then, by the golden ratio constraints you've asked for, we end up with:
$$n_3-n_2=(A_5-n_3)cdot1.618$$
$$n_2-n_1=(n_3-n_2)cdot1.618$$
$$n_1-A_4=(n_2-n_1)cdot1.618$$
Solving these taking $A_4=440$ and $A_5=880$, we end up with
$$A_4=440$$
$$n_1=636.7701$$
$$n_2=758.3833$$
$$n_3=833.5459$$
$$A_5=880$$
All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:
Take:
$$d_1=n_1-A_4$$
$$d_2=n_2-n_1$$
$$d_3=n_3-n_2$$
$$d_4=A_5-n_3$$
Where $n_k$ is the $k_th$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_k-1$ and $n_0=A_4$ and $n_4=A_5$.
Then, by the golden ratio constraints you've asked for, we end up with:
$$n_3-n_2=(A_5-n_3)cdot1.618$$
$$n_2-n_1=(n_3-n_2)cdot1.618$$
$$n_1-A_4=(n_2-n_1)cdot1.618$$
Solving these taking $A_4=440$ and $A_5=880$, we end up with
$$A_4=440$$
$$n_1=636.7701$$
$$n_2=758.3833$$
$$n_3=833.5459$$
$$A_5=880$$
All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
From what I'm understanding, you want the ratio between $d_2$ and $d_1$ to be 1.618, and so on for $d_3$ to $d_2$, and $d_4$ to $d_3$. You basically end up with a system of equations:
Take:
$$d_1=n_1-A_4$$
$$d_2=n_2-n_1$$
$$d_3=n_3-n_2$$
$$d_4=A_5-n_3$$
Where $n_k$ is the $k_th$ note and $A_4<n_1<n_2<n_3<n_4<A_5$ and $d_k$ is the distance between $n_k$ and $n_k-1$ and $n_0=A_4$ and $n_4=A_5$.
Then, by the golden ratio constraints you've asked for, we end up with:
$$n_3-n_2=(A_5-n_3)cdot1.618$$
$$n_2-n_1=(n_3-n_2)cdot1.618$$
$$n_1-A_4=(n_2-n_1)cdot1.618$$
Solving these taking $A_4=440$ and $A_5=880$, we end up with
$$A_4=440$$
$$n_1=636.7701$$
$$n_2=758.3833$$
$$n_3=833.5459$$
$$A_5=880$$
All in Hz. Each of those are in the golden ratio to the preceding one. You can generalize this as long as you know your initial and final note.
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
edited Jul 30 at 14:46
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
answered Jul 30 at 14:41
Sriram Gopalakrishnan
1564
1564
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
add a comment |Â
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
2
2
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
This is a correct solution to the problem as stated. It seems strange that the small interval is at the top of the range, but that is what OP asked for.
– Ross Millikan
Jul 30 at 14:51
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
Exact solution for my specific question and generalization.
– user1901968
Jul 30 at 20:37
add a comment |Â
up vote
0
down vote
If what you want is the ratio of the frequencies to be $varphi$ for each of the four ratios then this is impossible.
The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is
$1.05946309436 approx 1.06$. Four your five note scale you want
$$
2^1/5 = 1.148698355 approx 1.15 .
$$
That's pretty far from $varphi approx 1.618$.
Even just a two note scale gives you a ratio pf $sqrt2 approx 1.4 < varphi$ .
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
add a comment |Â
up vote
0
down vote
If what you want is the ratio of the frequencies to be $varphi$ for each of the four ratios then this is impossible.
The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is
$1.05946309436 approx 1.06$. Four your five note scale you want
$$
2^1/5 = 1.148698355 approx 1.15 .
$$
That's pretty far from $varphi approx 1.618$.
Even just a two note scale gives you a ratio pf $sqrt2 approx 1.4 < varphi$ .
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If what you want is the ratio of the frequencies to be $varphi$ for each of the four ratios then this is impossible.
The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is
$1.05946309436 approx 1.06$. Four your five note scale you want
$$
2^1/5 = 1.148698355 approx 1.15 .
$$
That's pretty far from $varphi approx 1.618$.
Even just a two note scale gives you a ratio pf $sqrt2 approx 1.4 < varphi$ .
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
If what you want is the ratio of the frequencies to be $varphi$ for each of the four ratios then this is impossible.
The interval on the ordinary $12$ tone scale is the twelfth root of $2$, which is
$1.05946309436 approx 1.06$. Four your five note scale you want
$$
2^1/5 = 1.148698355 approx 1.15 .
$$
That's pretty far from $varphi approx 1.618$.
Even just a two note scale gives you a ratio pf $sqrt2 approx 1.4 < varphi$ .
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
answered Jul 30 at 14:14
Ethan Bolker
35.7k54199
35.7k54199
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
add a comment |Â
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
As I read it, he doesn't want the ratio between the frequencies to be $varphi$, but the ratio between the intervals between the notes. Whether that means frequency interval or tone interval is still unclear.
– Arthur
Jul 30 at 14:20
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
Exactly. And i meant the frequency interval
– user1901968
Jul 30 at 14:34
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
@user1901968 I may come back to this later - but my suspicion is that it's still impossible. If you constrain all the intervals and want a whole octave then the number of intervals determines the scale factor and it probably won't be the golden ratio.
– Ethan Bolker
Jul 30 at 15:38
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867053%2ffind-three-points-on-a-scale-using-golden-ratio%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
First of all, you have to define what "some specific amount closer" means. Do you mean in frequency, so that A4 is as close to A5 as A5 is to E5 at 1320 Hz (same number of Hertz between them), or do you mean in tone, so that A4 is as close to A5 as A5 is to A6 at 1760 Hz (same interval of an octave between them)?
– Arthur
Jul 30 at 14:13
I meant for the first approach in your comment.
– user1901968
Jul 30 at 14:28