Yahoo Tuning Groups Ultimate Backup tuning Calculating the tension of a string

Then let me answer my own question as best I can:

My Bechstein Grand has three A4 strings each about 40 cms long and 1 mms thick all of which are roughly tuned to 440Hz in room temperature and in mild spring season humidity.

The tension equation is given as:

4 (L^2 * f^2 ) M
W = ___________

n^2

Where W is the weight measured in Newtons by which a single string is tightened, L is the lenght of the string, F is the frequency, M is the mass of the string per unit lenght, and N is the harmonic number.

According to the MKS system, the tension of a piano string may vary from 102 kgs in the bass to 82 kgs in the treble regions, with a total of 20-30 tons of pressure exerted to the frame.

The middle strings are manufactured from stainless steel such that each weigh about 20 grams per unit meter, or 8 grams (0.008 kgs) per 0.4 meters.

Thus, for the fundamental tone A4=440Hz, the solution to the above-given equation per string would be:

W= 4 (0.16 * 193 600) * 0.008 Newtons

W= 4 * 30.976 * 0.008 Newtons

W= 990 Newtons

Since 1 kg on Earth weighs 10 Newtons, the total amount of force required to produce a tension that will sound 440 Hz for a single A4 string 0.4 meters long equals:

W= 99 kgs.

For three A4 strings together:

W= 300 kgs.

Delta * W 2 (Delta * F)
Since: _______ = __________

W F

Every 1% change in tension will produce 2% change in frequency.

Now, in mina system, the octave is divided into 2640 equal steps, each 0.4545E cents wide. For practical purposes, let us refer to this as `half a cent`.

If we want to raise 440Hz by a mina, the resulting frequency will be 440.11554Hz according to the conversion calculator here:

http://www.sengpielaudio.com/calculator-centsratio.htm

Remember that every 1% change in the frequency will result in 0.5% change in tension. Since the change in frequency is 1 in 4000, the change in tension will be 1 in 2000.

Thus, the total tension of all three strings should be increased as much as 2000th of a unit of weight, giving us:

Delta W=300kg/2000

Delta W=150 grams.

The Ultratonal Piano (tm) utilizes a technology which can effectively change the string tensions by unit grams. This yields divisions of the octave in the order of +200.000 EDO.

Cordially,
Ozan

----- Original Message -----
From: George D. Secor
To: tuning-math@yahoogroups.com
Sent: 09 Mayıs 2005 Pazartesi 20:43
Subject: [tuning-math] The best ETs and 12276 (was Re: Meantone diesis)

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...>
wrote:
> The technology in question is the one I plan to implement with my
Ultratonal Piano (tm) design. It will hopefully be possible (I dare
presume) to adjust each string tension by about half a cent at minimum
precision. Could you re-affirm this by showing me how much tension
ought to be delivered to the A4 strings in order to raise La one degree
of 2640tET?
>
> Cordially,
> Ozan

Sorry, I have no idea.

--George

🔗Yahya Abdal-Aziz <yahya@melbpc.org.au>

🔗Ozan Yarman <ozanyarman@superonline.com>

Brother, thanks for the encouraging words! But 1/150 mina system equals:

150*2640= 396 000-EDO

I think such a precision with which each of the 88 tones of a piano can be calibrated is fantastic in proportions. With this setup, and with proper tensions applied, practically any scale can be mapped to the Halberstad keyboard with the touch of a button. I am, of course, assuming that the operable range of the strings below the `snap-limit` permit a myriad of tunings. The resulting acoustical phenomena will no doubt be fascinating. Imagine spending hours in front of your house piano equipped with this device. The Ultratonal Piano (tm) is matchless for unbounded acoustical tuning. I hope that I will be able to organize a commitee for the complete technical and financial requirements for the production of this portable machine.

Cordially,
Ozan
----- Original Message -----
From: Yahya Abdal-Aziz
To: tuning@yahoogroups.com
Sent: 12 Mayıs 2005 Perşembe 9:38
Subject: [tuning] RE: Calculating the tension of a string

Ozan,

You wrote:
> Then let me answer my own question as best I can:
...
> Now, in mina system, the octave is divided into 2640 equal steps, each
0.4545E cents wide. For practical
> purposes, let us refer to this as `half a cent`.

> If we want to raise 440Hz by a mina, ...
> ..., the total tension of all three strings should be increased as much as
2000th of a unit of weight, giving > us:
> ...
> Delta W=150 grams.

[YA] I'm impressed! Sounds like good solid applied maths to me :-)

> The Ultratonal Piano (tm) utilizes a technology which can effectively
change the string tensions by unit
> grams. This yields divisions of the octave in the order of +200.000 EDO.
[YA] Ummm ... How can that be? If you mean that your
Ultratonal Piano can change tension by one gram at a time,
then according to your calculations, it's changing the note
by 1/150 mina at a time. This is equivalent to selecting notes
from a 150*2640 = 39600-EDO scale.

Regards,
Yahya

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🔗chrysalis2grow <76153.763@compuserve.com>

🔗Ozan Yarman <ozanyarman@superonline.com>

Dear Cris,

You presume to know better when it comes to the scientific input of my MIT graduate father (one of the first in his class) who is a renown professor of nuclear engineering and an expert in physics, relativity and quantum mechanics?

Surely the calculations I gave in this regard are no jest. I forwarded your response to him, and this is what he has to say about your "correction":

--------------------------------------------------------------------------------

My Dear Ozan,

1) The equation he gives and your equation are identical. He did not understand this.
The simple reason is that your M (mass per unit length of the string), at the RHS of your equation, is equal to D^2 x pi x rho/4, in his terms (for a cylindrical string).

More specifically, by definition the density rho is

rho = mass/volume.

Thus

mass = rho x volume.

The volume for a cylinder of height L and base of radius R is the surface area of the base, i.e. pi x R^2, times its height, i.e. L.

So,

volume = pi x R^2 x L.

Then,

mass = rho x pi x R^2 x L.

But by definition of the diameter:

D (his diameter) = 2 R,

R =D /2.

Thus the mass of the string of "mass density" rho, lenght L, and base radius R is

mass = rho x pi x [(D^2)/ 4] x L.

This mass, again is the mass of the string of length L.

Thus the lenght of the string of a "unit lenght", should be L times less (since to obtain 1 unit length, we have to reduce L unit lenghts by L). This makes that the M that you used in your relationship becomes

M = {rho x pi x [(D^2)/ 4] x L} / L,

M = rho x pi x [(D^2)/ 4] .

In other terms,

4 M = rho x pi x D^2 .

So the two formulae are identical...

Now as to what he says about the RHS and the LHS units of your equation, this is just nonsense...

Sorry but this is the way it is...

Pap T.

M is the mass of the string of a unit lenght.

----- Original Message -----
From: chrysalis2grow
To: tuning@yahoogroups.com
Sent: 13 Mayıs 2005 Cuma 19:19
Subject: [tuning] Calculating the tension of a string

Ozan,

The correct string tension equation is:

T = F^2 * L^2 * D^2 * Pi * rho

where T is tension, F is frequency, L is string length, D is string
diameter, and rho is the mass density of the stringing material.

Please note that your equation:

W= 4 (0.16 * 193 600) * 0.008 Newtons

is dimensionally incorrect. Your W (my T) is a force unit on the
left side of the equation; however, you also have a force unit on
the right side of the equation (Newtons). When you solve for a
given variable, you cannot have the same dimensions on both sides of
an equation.

Cris Forster, Music Director
www.Chrysalis-Foundation.org

🔗Ozan Yarman <ozanyarman@superonline.com>

Dear Cris,

The subject is so enthralling that I cannot restrict our discussion to
personal correspondence. I find it prudent to deliver my father's answer to
the tuning list.

Cordially,
Ozan

----------------------------------------------------------------------------
---

My dear Ozan,

I would like to communicate the following, if appropriate, to your
correspondent.

First of all, his relationship and yours, as I pointed out previously, are
identical. Simply because, as I have previously shown, the mass per unit
lenght (your M) is equal to (Pi) times (diameter square) times (the density
of the string material) divided by 4.

In other words:

Pi * D^2 * Rho
M= ___________

Therefore, you see that the mass per unit lenght and the diameter of the
cross sectional area of the string are inherently linked, so that for a
given density:

If the diameter is changed %1, then the mass per unit lenght is changed by
%2. In mathematical words:

Delta M 2 Delta D
______ = ________

M D

Delta M being the change in the unit mass, and Delta D being the
corresponding change in the diameter.

Thus, your relationship:

W (tension of the string) = 4 (L^2) * (F^2) * M

is perfectly valid for the first harmonic.

Here, L is the lenght of the string, F is the frequency achieved applying
the weight W to the string, and M again is the mass per unit lenght.

Thus, this relationship is not approximate at all; it is identical to the
relationship given by your correspondent:

T = F^2 * L^2 * D^2 * Pi * Rho,

assuming that his T and your W mean the same quantity.

On the other hand, I'm afraid he's mistaken in his dimension analysis when
he writes for the dimension analysis of your W:

W = (m^2 * 1/s^2) * (kg * m)/s^2

His last /s^2 does not correspond to anything.

So you have no problems dimension-wise.

(Anyway, since your W and his T are identical, there should not be any
dimension problem in the first place.)

Let me mention though, that esentially it is M (mass per unit lenght) which
is the basic quantity that comes into the equation behind the derivation of
the tension quantity, for, the velocity (V) of the soundwave in the string
is expressed as:

V = sqr (Tension / M)

Thus, it is not only that M embodies the diameter in your relationship
(since the mass per unit lenght and the diameter are linked as I pointed
out), but also the mass per unit lenght occurs to be an initial input to the
setup of the equations in question.

Pap T.

----- Original Message -----
From: "Cris Forster" <76153.763@compuserve.com>
To: "Ozan Yarman" <ozanyarman@superonline.com>
Sent: 14 Mayï¿½s 2005 Cumartesi 19:55
Subject: Re: Calculating the tension of a string

Dear Ozan,

Newton's second law of motion states:

Force (or Newtons) = Mass (or kilograms) * Acceleration (or
meters/second^2).

In this context: Newtons = (kg * m)/s^2

For strings, mass per unit length, or M/u.l. = Pi * r^2 * rho,
where r is the radius of the string, in meters,
and rho is the mass density of the stringing material,
in kg/m^3.

For a simplified correct string tension equation that gives

T = F^2 * L^2 * D^2 * Pi * rho,

a dimensional analysis (without dimensionless Pi) shows that

N = 1/s^2 * m^2 * m^2 * kg/m^3.

A cancellation of dimensions gives the correct unit for Newtons:

N = (kg * m)/s^2

However, for your equation below

W = 4 (.16 m^2 * 193600/s^2) * .008 N

or (without numerical values)

W = (m^2 * 1/s^2) * (kg * m)/s^2

a dimensional analysis gives an incorrect unit for Newtons
or for your tension variable (W):

W = (kg * m^3)/s^4

Your string tension equation is not a correct equation,
but an approximate equation based on taking a unit length
of steel wire and weighing it on a scale, or calculating its
mass per unit length from a table of properties for steel, around
7850 kg/m^3.

All competent builders of stringed instruments - especially of
pianos -
know that mass per unit length varies with string diameter. String
diameters are not only critical in calculating tension, but they are
extremely important in calculating inharmonicity. A beautifully
built
piano can be rendered useless if string diameters - tension AND
inharmonicity -
are not controlled.

Cris Forster, Music Director
www.Chrysalis-Foundation.org

--- In tuning@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...> wrote:
> Dear Cris,
>
> You presume to know better when it comes to the scientific input
of my MIT graduate father (one of the first in his class) who is a
renown professor of nuclear engineering and an expert in physics,
relativity and quantum mechanics?
>
> Surely the calculations I gave in this regard are no jest. I
forwarded your response to him, and this is what he has to say about
your "correction":
>
>
> -------------------------------------------------------------------
-------------
>
>
>
> My Dear Ozan,
>
> 1) The equation he gives and your equation are identical. He did
not understand this.
> The simple reason is that your M (mass per unit length of the
string), at the RHS of your equation, is equal to D^2 x pi x rho/4,
in his terms (for a cylindrical string).
>
> More specifically, by definition the density rho is
>
> rho = mass/volume.
>
> Thus
>
> mass = rho x volume.
>
> The volume for a cylinder of height L and base of radius R is the
surface area of the base, i.e. pi x R^2, times its height, i.e. L.
>
> So,
>
> volume = pi x R^2 x L.
>
> Then,
>
> mass = rho x pi x R^2 x L.
>
>
> But by definition of the diameter:
>
> D (his diameter) = 2 R,
>
> or
>
> R =D /2.
>
> Thus the mass of the string of "mass density" rho, lenght L, and
base radius R is
>
> mass = rho x pi x [(D^2)/ 4] x L.
>
> This mass, again is the mass of the string of length L.
>
> Thus the lenght of the string of a "unit lenght", should be L
times less (since to obtain 1 unit length, we have to reduce L unit
lenghts by L). This makes that the M that you used in your
relationship becomes
>
> M = {rho x pi x [(D^2)/ 4] x L} / L,
>
> or
>
> M = rho x pi x [(D^2)/ 4] .
>
> In other terms,
>
> 4 M = rho x pi x D^2 .
>
> So the two formulae are identical...
>
> Now as to what he says about the RHS and the LHS units of your
equation, this is just nonsense...
>
> Sorry but this is the way it is...
>
>
> Pap T.
>
> M is the mass of the string of a unit lenght.
>
>
> ----- Original Message -----
> From: chrysalis2grow
> To: tuning@yahoogroups.com
> Sent: 13 Mayï¿½s 2005 Cuma 19:19
> Subject: [tuning] Calculating the tension of a string
>
>
> Ozan,
>
> The correct string tension equation is:
>
> T = F^2 * L^2 * D^2 * Pi * rho
>
> where T is tension, F is frequency, L is string length, D is
string
> diameter, and rho is the mass density of the stringing material.
>
> Please note that your equation:
>
> W= 4 (0.16 * 193 600) * 0.008 Newtons
>
> is dimensionally incorrect. Your W (my T) is a force unit on
the
> left side of the equation; however, you also have a force unit
on
> the right side of the equation (Newtons). When you solve for a
> given variable, you cannot have the same dimensions on both
sides of
> an equation.
>
>
> Cris Forster, Music Director
> www.Chrysalis-Foundation.org

🔗Yahya Abdal-Aziz <yahya@melbpc.org.au>

Oops! :-)

But 396 000 is still twice the 200 000 divisions of the octave
that you indicated.

Imagine the fun that such an instrument may bring us! I do hope
that it will make it easy to achieve any desired (fixed) tuning,
simply and quickly. Does it have presets? Where can I find out
more about the Ultratonal Piano?

It occurred to me while writing the above that a piano capable of
Adaptive JI would be great ...

Regards,
Yahya

Ozan wrote:
________________________________________________________________________

Brother, thanks for the encouraging words! But 1/150 mina system equals:

150*2640= 396 000-EDO

I think such a precision with which each of the 88 tones of a piano can be
calibrated is fantastic in proportions. With this setup, and with proper
tensions applied, practically any scale can be mapped to the Halberstad
keyboard with the touch of a button. I am, of course, assuming that the
operable range of the strings below the `snap-limit` permit a myriad of
tunings. The resulting acoustical phenomena will no doubt be fascinating.
Imagine spending hours in front of your house piano equipped with this
device. The Ultratonal Piano (tm) is matchless for unbounded acoustical
tuning. I hope that I will be able to organize a commitee for the complete
technical and financial requirements for the production of this portable
machine.

Cordially,
Ozan
----- Original Message -----
From: Yahya Abdal-Aziz
To: tuning@yahoogroups.com
Sent: 12 Mayï¿½s 2005 Perï¿½embe 9:38
Subject: [tuning] RE: Calculating the tension of a string

Ozan,

> If we want to raise 440Hz by a mina, ...
> ..., the total tension of all three strings should be increased as much
as
2000th of a unit of weight, giving > us:
> ...
> Delta W=150 grams.

[YA] I'm impressed! Sounds like good solid applied maths to me :-)

Regards,
Yahya

________________________________________________________________________

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