Danger, Will Rogers!

John deLaubenfels wrote:

> [Robert Walker wrote:]
> >I wonder if you could also try, for fun, making the step size
> >distinctions even larger by reversing the sign for the melodic springs,
> >if this makes sense, and see what happens, might be interesting in
> >context of adaptive j.i.
>
> Robert, you're a "badder boy" than I even realized! I think that
> negative spring coefficients would make the matrix unstable, however,
> because a node becomes pushed to one side or the other rather than being
> pulled toward the best compromise. I guess you could get away with them
> if they were used carefully, but there'd be a danger of the matrix going
> unstable - more push away the farther one gets.

Yeah, but we should start a tuning rodeo. Who can stay on the bucking matrix the longest? Yeeeeee
haww! Livin' on the edge, baby.

Or you could go with the Star Trek analogy, if that suits you: Bravely retuning where no man has
tuned before.

Go for it!

--
David J. Finnamore
Nashville, TN, USA
http://personal.bna.bellsouth.net/bna/d/f/dfin/index.html
--

[Robert Walker wrote:]
>>>I wonder if you could also try, for fun, making the step size
>>>distinctions even larger by reversing the sign for the melodic
>>>springs, if this makes sense, and see what happens, might be
>>>interesting in context of adaptive j.i.

[I wrote:]
>>Robert, you're a "badder boy" than I even realized! I think that
>>negative spring coefficients would make the matrix unstable, however,
>>because a node becomes pushed to one side or the other rather than
>>being pulled toward the best compromise. I guess you could get away
>>with them if they were used carefully, but there'd be a danger of the
>>matrix going unstable - more push away the farther one gets.

[David Finnamore:]
>Yeah, but we should start a tuning rodeo. Who can stay on the bucking
>matrix the longest? Yeeeeee haww! Livin' on the edge, baby.

>Or you could go with the Star Trek analogy, if that suits you: Bravely
>retuning where no man has tuned before.

Har! I think that perhaps I qualify for that description even without
negative spring coefficients! ;->

If the matrix goes unstable, it wouldn't be a pretty, or good-sounding,
thing. In fact, it'd be impossible even to play, 'cause the pitches
would slide off to infinity. But, at some low level of negative
spring constant, it would work, simply because the other springs present
would keep the matrix stable. Each sequence would have a stability
limit, the finding of which would be a matter of experimentation. When
I get the melodic springs wired up, perhaps I'll give it a try...

BTW, wasn't that "Danger, Will Robinson!"?

JdL

--- In tuning@y..., "John A. deLaubenfels" <jdl@a...> wrote:
> If the matrix goes unstable, it wouldn't be a pretty, or good-
sounding,
> thing. In fact, it'd be impossible even to play, 'cause the pitches
> would slide off to infinity. But, at some low level of negative
> spring constant, it would work, simply because the other springs
present
> would keep the matrix stable. Each sequence would have a stability
> limit, the finding of which would be a matter of experimentation.

Exactly. Thus the rodeo analogy, riding that fine line between
stable and unstable. See how far you can push it without going over
the edge.

> BTW, wasn't that "Danger, Will Robinson!"?

Doh! And I call myself an educated American!

David F.

This reminds me of Buckminster Fuller's "tensegrity" idea, although that applies to three dimensional structures and your matrix is, I believe, two-dimensional. However, some have spoken of three dimensional lattices. Perhaps this might be easier to imagine as a structure held together with rubber bands, which can be stretched but tending to return to its original form.
I recently found (second-hand) a modeling set (not really a game) based on this, using rods connected by elastic bands at their ends.

>>> daeron@bellsouth.net 05/15/01 07:23PM >>>
--- In tuning@y..., "John A. deLaubenfels" <jdl@a...> wrote:
> If the matrix goes unstable, it wouldn't be a pretty, or good-
sounding,
> thing. In fact, it'd be impossible even to play, 'cause the pitches
> would slide off to infinity. But, at some low level of negative
> spring constant, it would work, simply because the other springs
present
> would keep the matrix stable. Each sequence would have a stability
> limit, the finding of which would be a matter of experimentation.

Exactly. Thus the rodeo analogy, riding that fine line between
stable and unstable. See how far you can push it without going over
the edge.
Yes, it was Robinson, from the TV sci-fi series, "Lost in Space", recently used as the basis for a feature length movie. I believe it was the robot's saying.

> BTW, wasn't that "Danger, Will Robinson!"?

Doh! And I call myself an educated American!

David F.

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[John F. Sprague wrote:]
>This reminds me of Buckminster Fuller's "tensegrity" idea, although
>that applies to three dimensional structures and your matrix is, I
>believe, two-dimensional.

I guess it is: all the displacements to be solved for are
one-dimensional (tuning, up or down), but the links between pairs of
nodes make it a 2-D matrix.

>However, some have spoken of three dimensional lattices.

Yes, and lattices can go into much higher dimensional space, but the
tuning targets projected from them would still form the same matrix I'm
making already (though with different rest points for each spring).

>Perhaps this might be easier to imagine as a structure held together
>with rubber bands, which can be stretched but tending to return to its
>original form.

To me, there is no difference between a rubber band and a spring,
except that a spring resists displacement in both directions, whereas
a rubber band goes limp under compression.

>I recently found (second-hand) a modeling set (not really a game) based
>on this, using rods connected by elastic bands at their ends.

Kyool!

JdL