Yahoo Tuning Groups Ultimate Backup tuning Test of ascii diagrams, and dictionary selection

hi Robert,

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:

> http://www.robertinventor.com/test_overlib_tool_tips/index.htm
>
> Test of ascii diagrams and a drop list to select the
> dictionary.
>
> Robert

this is looking very nice. it will be great to
read the tuning-math archives and have all those
definitions right to hand.

one thing about your "diesis" definition:

you wrote:

>> "If you play C E G# C in succesion using pure ratios
>> for the intervals, the last C and the first differ by
>> a diesis up to octave equivalence"

that really should say this:

>> "If you play C E G# B# in succesion using pure ratios
>> for the intervals, the B# and the C differ by
>> a diesis up to octave equivalence"

because of their total equivalence in 12edo, people
are used to thinking of C and B# as the same, but anyone
who works in JI will make a difference in the notation,
and that last note is really B# and not C.

-monz

🔗monz <monz@attglobal.net>

🔗Robert Walker <robertwalker@ntlworld.com>

Hi Monz,

> because of their total equivalence in 12edo, people
> are used to thinking of C and B# as the same, but anyone
> who works in JI will make a difference in the notation,
> and that last note is really B# and not C.

Rightio, yes of course, fixed.

BTW the dictionary is beginning to seem somewhat
less daunting a project, because what I hadn't
realised is that once you have made one definition
e.g. for a diesis, you can then go on and
make lots of other definitions using the same
model, just changing the name and the numbers.
- at least for the ones I understand :-).

Let's see, it would be nice to have newbie type
definition of monzo, val and wedgie as
they get used so often in tuning-math - well I can make
a start anyway:

monzo :
a ratio expressed as a row vector
which shows its prime factors.

Example, to find the monzo for 81/80,
factorise it first 81/80 = 3^4/(5*2^2)
where 3^4 is short for 3*3*3*3 (4 times).
Now you use negative numbers for the numbers
on the bottom, e.g. 1/2^2 is shown as 2^-2
81/80 = 3^4 * 5^-1 * 2^-2
Now you arrange the factors in increasing order starting from 2:
81/80 = 2^-2 * 3^4 * 5^-1
Then write these numbers all out as a row vector
to get its monzo {-2,4,-1}.
If a factor doesn't occur, show it as 0,
so 5/4 = {-2,0,1}

wedge product
A way of expressing an area
in algebraic form - a kind of 2D generalisation
of a vector. It is an area with a sense of direction,
just as a vector is a line with a sense of direction
to it. So if a and b are vectors, a^b is
the area swept out from a to b.
The wedge product is often expressed
as a sum of wedge products of unit
vectors. So for instance in 3D space
you can reduce any wedge product
to the form A * u2^u3 + B * u3^u1 + C*u1^u2
by using the rule that ui^ui=0
and ui^uj = - uj^ui.

But I think I'll give up for now on wedgie
and val - I've never studied Clifford Algebra
which would help. I haven't yet
understood how a wedge product
is used in tuning theory, but
look forward to finding out
as I read more.

Robert

🔗Robert Walker <robertwalker@ntlworld.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...> wrote:

> wedge product
> A way of expressing an area
> in algebraic form - a kind of 2D generalisation
> of a vector. It is an area with a sense of direction,
> just as a vector is a line with a sense of direction
> to it. So if a and b are vectors, a^b is
> the area swept out from a to b.
> The wedge product is often expressed
> as a sum of wedge products of unit
> vectors. So for instance in 3D space
> you can reduce any wedge product
> to the form A * u2^u3 + B * u3^u1 + C*u1^u2
> by using the rule that ui^ui=0
> and ui^uj = - uj^ui.
>
> But I think I'll give up for now on wedgie
> and val - I've never studied Clifford Algebra
> which would help. I haven't yet
> understood how a wedge product
> is used in tuning theory, but
> look forward to finding out
> as I read more.

Actually many people on this list who have not been following the
jargon explosion should be familiar with area as a concept in tuning
theory. Suppose we have a Fokker block using the commas 128/125 and
81/80. If we look at the 5-limit lattice of octave classes, 128/125 a
|0 -3>, and 81/80 is a |4 -1>, and the area of the parallogram they
define is 12, which we can get by evaluating the 2x2 matrix they
determine. We don't need to just look at Fokker blocks with octave
periods; if we look at what we get for 3 ("tritrave") periods, we
would take the determinant of [|7 -3>, |-4 1>] and get an area of
19--there are 19 notes in the Fokker block they determine using
tritrave equivalence. Finally, using 5-equivalence, we would have a
Fokker block the number of whose notes would be defined by the
determinant of [|7 0>, |-4 4>], namely 28. Putting these together we
get the val for 12-equal, though technically we have to take a
complement; but the numbers are <12 19 28|, which is 12-equal, and
which was obtained by looking at areas.

🔗monz <monz@attglobal.net>

hi Gene,

--- In tuning@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...> wrote:

> Actually many people on this list who have not been
> following the jargon explosion should be familiar with
> area as a concept in tuning theory. Suppose we have a
> Fokker block using the commas 128/125 and 81/80. If we
> look at the 5-limit lattice of octave classes, 128/125
> [is] a |0 -3>, and 81/80 is a |4 -1>, and the area of the
> parallogram they define is 12, which we can get by evaluating
> the 2x2 matrix they determine. We don't need to just look at
> Fokker blocks with octave periods; if we look at what we get
> for 3 ("tritrave") periods, we would take the determinant
> of [|7 -3>, |-4 1>] and get an area of 19--there are 19 notes
> in the Fokker block they determine using tritrave equivalence.
> Finally, using 5-equivalence, we would have a Fokker block
> the number of whose notes would be defined by the determinant
> of [|7 0>, |-4 4>], namely 28. Putting these together we
> get the val for 12-equal, though technically we have to take
> a complement; but the numbers are <12 19 28|, which is
> 12-equal, and which was obtained by looking at areas.

now *that* is the most comprehensible thing i've ever
seen from you!

thanks ... things are really beginning to make sense now ...

-monz

🔗monz <monz@attglobal.net>

hi Robert,

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:

> Let's see, it would be nice to have newbie type
> definition of monzo, val and wedgie as
> they get used so often in tuning-math - well I can make
> a start anyway:
>
> monzo :
> a ratio expressed as a row vector
> which shows its prime factors.

it should say "a row-vector which shows the
exponents of its prime-factors".

(i like hyphenation more than most people ... so that's
optional ... but get "exponent" in there)

> Example, to find the monzo for 81/80,
> factorise it first 81/80 = 3^4/(5*2^2)

oops ... that should be 81/80 = 3^4/(5*2^4)

> where 3^4 is short for 3*3*3*3 (4 times).
> Now you use negative numbers for the numbers
> on the bottom, e.g. 1/2^2 is shown as 2^-2

so use:

1/2^4 is shown as 2^-4

> 81/80 = 3^4 * 5^-1 * 2^-2

81/80 = 3^4 * 5^-1 * 2^-4

> Now you arrange the factors in increasing order starting from 2:
> 81/80 = 2^-2 * 3^4 * 5^-1

81/80 = 2^-4 * 3^4 * 5^-1

> Then write these numbers all out as a row vector
> to get its monzo {-2,4,-1}.

instead:

>> Then write these exponents all out as a row-vector
>> to get its monzo [-4 4, -1>.

many of us have endorsed the bra-ket notation, and
also have agreed that it makes sense to put comma-marks
after the exponent of 3 and then subsequently after
every third exponent.

-monz

🔗Kurt Bigler <kkb@breathsense.com>

on 7/27/04 12:41 AM, monz <monz@attglobal.net> wrote:

> --- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
> wrote:
>
>> Let's see, it would be nice to have newbie type
>> definition of monzo, val and wedgie as
>> they get used so often in tuning-math - well I can make
>> a start anyway:
>>
>> monzo :
>> a ratio expressed as a row vector
>> which shows its prime factors.
>
>
> it should say "a row-vector which shows the
> exponents of its prime-factors".

Well, if you're going to get technical ;) ...

"its prime factors" is incorrect because the referent of "its" in this
context is "row vector". The row vector is not showing the prime factor of
the row vector. How about:

"monzo:
a row vector which expresses a rational number in terms of the integral (not
necessarily zero or positive) exponents of its prime factors"

In that definition the referent of "its" is the rational number, no longer
the row vector, so it is correct. And a couple of other useful facts are
included. I don't think you want non-integral exponents.

>> Then write these numbers all out as a row vector
>> to get its monzo {-2,4,-1}.
>
> instead:
>
>>> Then write these exponents all out as a row-vector
>>> to get its monzo [-4 4, -1>.
>
>
> many of us have endorsed the bra-ket notation, and
> also have agreed that it makes sense to put comma-marks
> after the exponent of 3 and then subsequently after
> every third exponent.

Where did that notation come from? Is it specific to tunings-list
developments or is it used in math somewhere?

-Kurt

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Robert Walker <robertwalker@ntlworld.com>

Hi Gene,

> Actually many people on this list who have not been following the
> jargon explosion should be familiar with area as a concept in tuning
> theory. Suppose we have a Fokker block using the commas 128/125 and
> 81/80. If we look at the 5-limit lattice of octave classes, 128/125 a
> |0 -3>, and 81/80 is a |4 -1>, and the area of the parallogram they
> define is 12, which we can get by evaluating the 2x2 matrix they
> determine. We don't need to just look at Fokker blocks with octave
> periods; if we look at what we get for 3 ("tritrave") periods, we
> would take the determinant of [|7 -3>, |-4 1>] and get an area of
> 19--there are 19 notes in the Fokker block they determine using
> tritrave equivalence. Finally, using 5-equivalence, we would have a
> Fokker block the number of whose notes would be defined by the
> determinant of [|7 0>, |-4 4>], namely 28. Putting these together we
> get the val for 12-equal, though technically we have to take a
> complement; but the numbers are <12 19 28|, which is 12-equal, and
> which was obtained by looking at areas.

Great thanks - that is all very clear.
I'll see if I can make a newbie type definition of val.

What is a wedgie?

Also a notational question - the bra-ket notation | , > here seems
to indicate a missing exponent, e.g. |7 , -3> = 2^7 * 5^-3
with the missing exponent understood by context.

Does it always mean that - perhaps you can explain a bit more about
how the notation works.

I read a few chapters of Dirac's classic book on quantum mechanics
once - enjoyed the beauty of his style of maths writing,
but it was quite slow going - and I can't remember how his bra-ket
notation worked now.

Robert

🔗monz <monz@attglobal.net>

--- In tuning@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...> wrote:

> --- In tuning@yahoogroups.com, Kurt Bigler <kkb@b...> wrote:
>
> > > many of us have endorsed the bra-ket notation, and
> > > also have agreed that it makes sense to put comma-marks
> > > after the exponent of 3 and then subsequently after
> > > every third exponent.
> >
> > Where did that notation come from? Is it specific to
> > tunings-list developments or is it used in math somewhere?
>
> Dirac invented bra-ket notation; it is used a lot in physics
> and less often by mathematicians. The comma-marks business
> of course is purely a tuning thing.

but is a useful idea, especially when the monzo is
very long, representing a lot of prime-factors.

using the prime-factors themselves to illustrate
their monzo exponents, the punctuation goes like this:

[2 3, 5 7 11, 13 17 19, 23 29 31, 37 41 43, 47 53 59, ...>

several of us like that because:

- grouping more than one exponent between comma-marks
makes the monzo easier to read, similar to the way
people commonly use comma-marks in regular numbers
above 1,000; and

- for one reason or another the exponents which
happen to fall before each comma-mark (at least
going as far as prime 31) can be thought of as
historically important prime-limits.

don't read too much into that last statement ... it
was just fortuitous that things worked out this way.
but that reasoning was partly behind the decision
to put the first comma-mark after 3 instead of after 5.

and everyone please notice how careful i was to
write "comma-mark" instead of simply "comma" everywhere
that i referred to the puncutation-mark ... to help
avoid confusion with the tuning term "comma".

-monz

🔗monz <monz@attglobal.net>

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:
> Hi Monz,
>
> Thanks, I've done those fixes for 81/80
> with 2^-4 instead of 2^-2, and I've
> changed monzos in the dictionary to the bra-ket
> notation. Yes commas every three numbers help to
> make it easier to read the complex ones doesn't it,
> like: 33/32 = [-5 1, 0 0 1>
>
> Robert

exactly ...

i'm guessing that as more and more people
get interested in microtonality and actually make
microtonal music, most of them are who work in JI are
going to find, like Partch, that 11-limit gives more
than enough harmonic variety to their music, and that
they really don't need the numerical complexity of
going to the 13-limit or beyond.

so that's a really excellent example. that one little
comma-mark makes the 11-limit monzo much easier to read
than it would be without it.

-monz

🔗monz <monz@attglobal.net>

hi Kurt (and Gene and others)

--- In tuning@yahoogroups.com, Kurt Bigler <kkb@b...> wrote:

> "monzo:
> a row vector which expresses a rational number in terms
> of the integral (not necessarily zero or positive) exponents
> of its prime factors"
>
> In that definition the referent of "its" is the rational
> number, no longer the row vector, so it is correct. And
> a couple of other useful facts are included. I don't think
> you want non-integral exponents.

this came up in discussion about a week ago and
i asked about it, but got conclusive response.

i *do* find it useful to include rational exponents
in a monzo, because they map linear temperaments across
the JI lattice *as* a line, which is what i think a
"linear temperament" should look like.

i proposed that if Gene and others think "monzo"
should be limited to integers (and i haven't really
seen the argument yet why it should), then we need
a name for the rational version. any ideas?

-monz

🔗Robert Walker <robertwalker@ntlworld.com>

Hi there,

Attempt again at wedge product and val, though they
are rather long still:

wedge product:
Required: understanding of monzo and periodicity block.

Used to work out the number of notes in a periodicity block
scale. It is an area with a sense of direction
to it, just as a vectpr is a line with a sense
of direction. So a^b is the parallelogram swept out from
a to b. Because the direction counts, then a^b = - b^a.
Also a^a = 0.

Vectors are often expressed coordinate style in terms
of unit vectors in two or more directions of travel.
A wedge product is expressed similarly in terms of wedge products
of unit vectors - which are used as the units of directed
area.

The wedge product of [a1 b1> with [a2 b2>
is (a1*b2 - a2*b1) * u1^u2
where u1^u2 has magnitude 1.

So, to find the number of notes in a five
limit scale with the diesis and syntonic comma as unison
vectors, the wedge product of 81/80 = [4,-1> and
128/125 = [ 0,-3> has magnitude:
abs( 4*-3 - (-1 * 0) ) = abs(-12) = 12
which shows that there are twelve notes in the scale.
It is normaly written as [|4 -1> , |3 0>] = 12

val:
Required: understanding of monzo, wedge product, and periodicity block.

Used for scales defined using unison vectors.
val2 is the number of notes in the periodicty block if you assume
octave equivalence.

Example: with the syntonic comma 81/80 = [-4 4,-1> and
the diesis 125/128 = [-7 3, 0> as your unison vectors,
assuming octave equivalence, you have a twelve note scale:
val2 = [ |4, -1>, |3, 0> ] = 12

However if you drop octave equivalence and instead consider notes at an
interval of 3/1 as equivalent, then you find there are
[|-7 -3>, |-4 1>] = (-4*-3) - (-7*1) = 19 notes so val3 is 19.
Since this periodicity block scale has 19 notes which is a prime
number, the periodicity block has to have one side of length 19.

An example 19 note scale would be (under equivalence of multiples of 3):
1/1 2/1 4/3 8/3 16/9 32/27 64/27 128/81 ... 2^19/3^5
The way it works is that using |-7,-3> as one unison vector,
2^7 ~= 5^-3 and using |-4 1>, 2^4 ~= 5
so 2^19 ~= 2^4 * 2^4 * 2^4 * 2^7
~= 5^-1 * 5^-1 * 5^-1 * 5^3 ~= 1.

Similarly using 5/1 as your equivalence vector, val5 is 28.
Let's leave that one as an "exercise for the reader" to figure it out.
Putting these together, the val for the periodicity blocks
marked out using the syntonic comma and diesis is <12, 19, 28]
which is the val for twelve equal. You use < ... ] for vals.

Robert

🔗Robert Walker <robertwalker@ntlworld.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...> wrote:

> What is a wedgie?

A wedge product reduced to a standard form which uniquely denotes a
regular temperament.

> Also a notational question - the bra-ket notation | , > here seems
> to indicate a missing exponent, e.g. |7 , -3> = 2^7 * 5^-3
> with the missing exponent understood by context.

> Does it always mean that - perhaps you can explain a bit more about
> how the notation works.

People standardly leave off the 2 exponent when looking at octave
classes, so I left off the 3 for tritave classes. It makes the
matricies automatically square, but you can take 2x3 matricies and
minors and it might be clearer to you.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗monz <monz@attglobal.net>

--- In tuning@yahoogroups.com, "monz" <monz@a...> wrote:
> hi Kurt (and Gene and others)
>
>
> --- In tuning@yahoogroups.com, Kurt Bigler <kkb@b...> wrote:
>
> > "monzo:
> > a row vector which expresses a rational number in terms
> > of the integral (not necessarily zero or positive) exponents
> > of its prime factors"
> >
> > In that definition the referent of "its" is the rational
> > number, no longer the row vector, so it is correct. And
> > a couple of other useful facts are included. I don't think
> > you want non-integral exponents.
>
>
>
> this came up in discussion about a week ago and
> i asked about it, but got conclusive response.

oops ... i got *no* conclusive response.

> i *do* find it useful to include rational exponents
> in a monzo, because they map linear temperaments across
> the JI lattice *as* a line, which is what i think a
> "linear temperament" should look like.
>
> i proposed that if Gene and others think "monzo"
> should be limited to integers (and i haven't really
> seen the argument yet why it should), then we need
> a name for the rational version. any ideas?
>
>
>
> -monz

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...> wrote:
> Hi there,
>
> Attempt again at wedge product and val, though they
> are rather long still:
>
> wedge product:
> Required: understanding of monzo and periodicity block.
>
> Used to work out the number of notes in a periodicity block
> scale. It is an area with a sense of direction
> to it, just as a vectpr is a line with a sense
> of direction. So a^b is the parallelogram swept out from
> a to b. Because the direction counts, then a^b = - b^a.
> Also a^a = 0.

This won't do at all, since wedge products can be products of vals,
and even if they are products of monzos it may have nothing whatever
to do with periodicty blocks.

I thought we already had a definition where Graham had explained how
to compute a wedge product in reasonably simple and explicit language?

> val:
> Required: understanding of monzo, wedge product, and periodicity block.
>
> Used for scales defined using unison vectors.
> val2 is the number of notes in the periodicty block if you assume
> octave equivalence.

Again, a val need have nothing to do with a scale. The most typical
vals are either the ones we find in prime mappings for temperaments
(most typically the period and generator maps) or equal tenperament vals.

A val takes a monzo and maps it to the integers; it's a group
homorphism but a nontechnical definition hardly needs to babble about
Hom(*, Z).

A definition of vals and monzos probably should not place much
emphasis on Fokker blocks, I fear.

🔗Robert Walker <robertwalker@ntlworld.com>

Hi Gene,

Yes, I understand that wedge products have many other
uses, though I don't know how they work.

A dictionary type definition needn't go into
all the details, it can't. Just enough to
help give some understanding so reader can
have a first hint about how it is understood.

But I see indeed, that as a dictionary
definition, it is going to confuse the reader
if they look at a tip for wedge product
and it describes it as a way of finding numbers
of notes, when the concept is something quite different.

Maybe just need to keep to the mathematical sense:

wedge product:

Used in many ways, for instance, to find out the area of a periodicity
block.

A wedge product is an area with a sense of direction
to it, just as a vector is a line with a sense
of direction. So a /\ b is the parallelogram swept out from
a to b. Since the direction counts, a /\ b = - b /\ a.
However if one is interested in the area, all one cares
about is the magnitude of the result.

The two dimensional case is the simplest - the
magnitude of the wedge product of [a b> and [c d>
is a*c - b*d

The idea of a wedge product can be extended to 3D and higher dimensions.
There one needs to go into more detail about how
it is undersood. Just as a vector
is considered to be made up of unit vectors,
for two or more directions of travel
so a wedge product is made up of wedge products
of unit vectors which are our units of directed area.

Then, the 3D wedge product can be obtained by simply multiplying
the two vectors together using ui/\uj = - uj/\ui and
ui/\ui = 0 to end up with an expression in the form
A * u2/\u3 + B* u3/\u1 + C * u1/\u2. See the encyclopedia
for details of how to calculate a 3D wedge product.

Yes I saw the explanation in the encyclopedia.
It explains how to do the calculation.
Here, tried to give some motivation for it
and to show that if you spent timeon it you
could figure out the 3D wedge product for yourself,
not that you need to if it is explaiend how to
do it - maybe that is partly personal inclination
- I've never been happy with mathematical rules for
doing things if the reason for doing things that
way isn't explained, so want to supply that
but others find such explanations only
get in the way - so put it at the end
now where it is more easily ignored.

It's too long anyway for a dictionary
definition still. Don't know what
the solution is - but perhaps
the best solution is just to cut out the
entire bit about 3D wedge products
as more appropriate for the encyclopedia
than the dictionary, leaving:

wedge product:

One example of its use is to find out the
area of a periodicity block.

The two dimensional case is the simplest - the
magnitude of the wedge product of [a b> and [c d>
is a*c - b*d

For example, the wedge product of 81/80 = [4,-1> and
128/125 = [ 0,-3> has magnitude:
abs( 4*-3 - -1*0 ) = abs(-12) = 12
which shows that there are twelve notes in the
scales constructed using these as unison vectors.
The idea can also be extended to 3D and higher
dimensions - look up in Tonalsoft Encyclopedia.

Robert

🔗Robert Walker <robertwalker@ntlworld.com>

Hi Gene,

Since there is no notation yet to show whether
one includes the exponent of 2 in a monzo or is assuming
octave equivalence, or tritave equivalence
or whatever, how about introducing one?

I've been adding monzo's to the dictionary
definitions and sometimes the power of 2
isn't relevant to the definition if one is
talking about everything up to octave
equivalence, and it would be nice to have a
way to leave it out without having to say that the
exponent of two is left out in every definition
- so that is the immediate motivation for it.

How about using a wild card * - or is that
confusing for some reason
- so
81/80 = 3^4/(4^4*5) = [4 -4, -1>
= [* -4, -1> assuming octave equivalence
= [4 *, -1> assuming tritave equivalence
= [4 -4, *> assuming pentave equivalence

Example of use in a definition:

skhisma:
schisma:
The interval between 8/5 and (3/2)^8 up to octave equivalence.
For instance if you play C G D A E B F# G#
using pure ratio 3/2s for the intervals, then go by 5/4 from the G#
to B#, the B# and C differ by a schisma up to
octave equivalence.
Ratio: 32805/32768 Monzo: [* 8, 5> Cents: 1.954

where I think it adds clarity to leave out the
exponent of 2 - it isn't particularly useful to know
that the exponent of 2 is -15.

We could use this format for all these various named
intervals in the dictinoary. Those named intervals
can all now be done very quickly - a few minutes each.

Robert

🔗monz <monz@attglobal.net>

hi Robert and Gene,

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:
> Hi Gene,
>
> Since there is no notation yet to show whether
> one includes the exponent of 2 in a monzo or is assuming
> octave equivalence, or tritave equivalence
> or whatever, how about introducing one?

there *is* a notation which shows the absence or
presence of 2, and that is the comma-mark, in
the usage i just wrote about in a previous post.

the first comma-mark always comes after prime-factor 3,
so if one sees this:

[a b, c d ...>

then one knows that a is the exponent of 2 and b is
the exponent of 3, c is 5, etc.

if one sees this:

[a, b c d, ...>

then one knows that 2 is being omitted, because
a is the exponent of 3, b is 5, etc.

> I've been adding monzo's to the dictionary
> definitions and sometimes the power of 2
> isn't relevant to the definition if one is
> talking about everything up to octave
> equivalence, and it would be nice to have a
> way to leave it out without having to say that the
> exponent of two is left out in every definition
> - so that is the immediate motivation for it.

use the comma-marks consistently in the way
i present them in the "monzo" definition, and
everything will be concise and readily understood.

> How about using a wild card * - or is that
> confusing for some reason
> - so
> 81/80 = 3^4/(4^4*5) = [4 -4, -1>
> = [* -4, -1> assuming octave equivalence
> = [4 *, -1> assuming tritave equivalence
> = [4 -4, *> assuming pentave equivalence
>
> Example of use in a definition:
>
> skhisma:
> schisma:
> The interval between 8/5 and (3/2)^8 up to octave equivalence.
> For instance if you play C G D A E B F# G#
> using pure ratio 3/2s for the intervals, then go by 5/4 from the G#
> to B#, the B# and C differ by a schisma up to
> octave equivalence.
> Ratio: 32805/32768 Monzo: [* 8, 5> Cents: 1.954
>
> where I think it adds clarity to leave out the
> exponent of 2 - it isn't particularly useful to know
> that the exponent of 2 is -15.
>
> We could use this format for all these various named
> intervals in the dictinoary. Those named intervals
> can all now be done very quickly - a few minutes each.
>
> Robert

i think the wildcard * is a great idea!

there are many times where an equivelence-interval is
invoked, probably more often than not actually, and
this is very useful!

i'm going to put it into the official definition.

-monz

🔗monz <monz@attglobal.net>

hi Robert,

i fully endorse your wildcard * for representing
the equivalence-interval in a monzo ... and have
already put it into the Encyclopaedia.

however, there are some math errors in your examples ...

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:

the exponent of 3 in every case here should be 4, not -4.

> Example of use in a definition:
>
> skhisma:
> schisma:
> The interval between 8/5 and (3/2)^8 up to octave equivalence.
> For instance if you play C G D A E B F# G#
> using pure ratio 3/2s for the intervals, then go by 5/4 from the G#
> to B#, the B# and C differ by a schisma up to
> octave equivalence.
> Ratio: 32805/32768 Monzo: [* 8, 5> Cents: 1.954

the monzo for the skhisma is [* 8, 1> .

it's a little difficult to keep the values of
prime-factors and their exponents straight in
your mind if you only use algebra or numbers
... frequent exposure to the lattice diagrams
helps in this regard.

-monz

🔗Robert Walker <robertwalker@ntlworld.com>

Hi Monz,

Thanks for the corrections. Yes I find indeed that
it is easy to get the sign reversed or do something like
write down the prime itself instead of the exponent
when the exponent is 1, until you are well
familiar with it I suppose.

Anyway the easiest way to avoid them altogether
for now will simply be to get FTS to generate
the line automatically (perhaps as a new calculator
option in the drop list), so it shows e.g.:
Ratio: 531441/524288 Monzo: [* 14, 0 1> Cents: 23.46

So after that, I can just paste that line into
the dictionary, and it should eliminate these errors
in the future.

Anyway be sure to say if you notice anything else
as it helps a lot.

Glad you like the * notation. Yes I can see that
the comma-mark placement helps so [14, 0 1> is unambiguous,
as long as octave equivalence is expected.

A reasonable convention might be that [n, ...>
always means that the 2 exponent is left out,
unless the context makes it clear
that tritave equivalence is intended.

I'll call it a comma-mark too, to avoid confusion
with comma, seems a good convention to me.

Robert