An alternate complete and correct string tension equation

>W = (m^2 * 1/s^2) * (kg * m)/s^2
>His last /s^2 does not correspond to anything.

Dear Ozan,

Exactly right. Congratulations!

On the right hand side of your following equation, and I quote:

******************************

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

******************************

.008 Newtons, or

.008 (kg * m)/s^2

corresponds to zip.

However, an alternate complete and correct string tension
equation that includes "4" on the right hand side would be:

T = 4 * F^2 * L^2 * Pi * r^2 * rho

In all cases, except weighing unit-length string segments --
where for obvious reasons the string radius remains shrouded
in a profound and even sarcastic and belligerent mystery --
the diameter or the radius of a string must be included
in all complete and dimensionally consistent string tension equations.

Enthralling, indeed! But only for skilled craftsmen who
know that if string tension calculations are not correct,
the whole piano could explode through the roof.

But then again, you are probably not familiar with the concept
of "potential energy," or could care less.

Very nice chatting with you, and Hamlet's ghost.

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

Dear Cris,

There is no ghost in this story. The person in question is my father whom I found the need to consult and benefit from his opinion. You might still remember that he is an MIT Ph. D. in Nuclear Engineering, and might yourself benefit from his input.

Moreover, I didn't present any material that eludes me. Let me clarify a misunderstanding first:

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

The Newtons here obviously correspond to the entire RHS of the equation, whereas you somehow assume that I refer to 0.008 only. 0.008 is the `mass per unit lenght data in kilograms`, which should be understood as 0.008 kilograms per meter (not 0.008 kg per 0.4 meters as previously stated).

There is no mystery at all in determining the diameter of the strings when it is just a question of specifying mass per unit lenght. It is clear from the relationships I provided that the diameter is inescapably bound to mass per unit lenght, or vice versa, once the density is fixed. It is not as if one can play with the thickness parameter in the equation I gave without affecting M:

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

which is once again exactly identical to and as complete as the equation you gave:

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

merely because,

Pi * D^2 * Rho
M= ___________

4

There is no information lost when taking M into account alone, and nothing is missing or elusive when I want to describe tension the way I did above. Likewise, there is no additional data to be gained from your equation as compared to mine, since they are one and the same.

Look, you can either take 1 meter of wire and weigh it to determine M, or you can calculate M by measuring the diameter of the cross section area of that string and knowing the density of the material. The results will be the same no matter what.

In other words, for a given M, for a given material, there is only one D, or vice versa.

I imagine that this is clear at last. There is nothing etherial to this at all.

If the problem is calculating the limit before a string made of a certain material of a certain thickness breaks, the equation for that is not the one I or you gave (which should equally be dismissed as incomplete following your logic). "Hamlet's Ghost" tells me that, such a problem is much more complicated than we would imagine.

Anyhow, that particular piece of information is irrelavent at this point when it is only a matter of raising a certain piano string by a mina.

It is indeed very nice to chat with you, but it would be even nicer to prolong our conversation if you improved your geometry and physics a little.

Cordially,
Ozan

----- Original Message -----
From: Cris Forster
To: Moderator
Sent: 14 Mayıs 2005 Cumartesi 23:16
Subject: [tuning] An alternate complete and correct string tension equation

>W = (m^2 * 1/s^2) * (kg * m)/s^2
>His last /s^2 does not correspond to anything.

Dear Ozan,

Exactly right. Congratulations!

On the right hand side of your following equation, and I quote:

******************************

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

******************************

.008 Newtons, or

.008 (kg * m)/s^2

corresponds to zip.

However, an alternate complete and correct string tension
equation that includes "4" on the right hand side would be:

T = 4 * F^2 * L^2 * Pi * r^2 * rho

In all cases, except weighing unit-length string segments --
where for obvious reasons the string radius remains shrouded
in a profound and even sarcastic and belligerent mystery --
the diameter or the radius of a string must be included
in all complete and dimensionally consistent string tension equations.

Enthralling, indeed! But only for skilled craftsmen who
know that if string tension calculations are not correct,
the whole piano could explode through the roof.

But then again, you are probably not familiar with the concept
of "potential energy," or could care less.

Very nice chatting with you, and Hamlet's ghost.

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