John, Nicholas, all --

John and Nicholas, thanks for your measurements on the Triola rolls.
In the following, I'd like to give the mathematical formulas for the
spiral correction (without their derivation; it is not too hard) and
check them against your numbers; and then tell you what we (Rudolf
Klomfar + I) decided for the punching of my Triola arrangements.

The following symbols are used:

d ... diameter of empty take-up spool
D ... the diameter of the paper of length L wound onto the take-up spool
L ... a given length of paper wound onto the take-up spool
n ... number of turns
p ... paper thickness including the "air gap" resulting from non-tight
      winding
pi .. the constant 3.1415...
t ... total playing time for length L
v ... initial speed
V ... average speed

Then (ignoring terms of "second order") we have

(1a) nd + n^2p = L/pi
(1b) d + 2np = D

or

(2a) n = (L/pi) / ((D+d)/2)
(2b) p = (D-d) / 2n

(3) V = L/t
(4) v = dn pi/t

John's measurements are redundant enough to serve as "test data".
He measured (_j is for "John"):

 d_j = 25.4 mm (= 1")
 L_j = 6300 mm
 D_j = 41.9 mm (= 1.65")
 n_j = 58
 p_j = 0.1422 mm (= 0.0056")
 V_j = 42 mm/s
 v_j = 31 mm/s
 t_j = 150 s

The formulas yield ("=" here is a "technical equals" or "approximately
equals"; _c is for "computed"):

 n_c = (6300/3.1416) / ((25.4+41.9)/2) = 59.6;
for a length L_j' = 6130 mm, we get n_c' = 58 = n_j.

John, is the length from start of first to end of last note really
6300 mm, or rather a little less?

 p_c = (41.9-25.4) / (2*58) mm = 0.1422 mm = p_j.
 v_c = 25.4*58*3.1416 / 150 mm/s = 30.85 mm/s = v_j.

So it seems we are all "d'accord!"

For the Triola, we will now punch a few test arrangements with the
following data:

(a) initial speed v_h = 38 mm/s. This is quite a bit faster than
the v_j = 31 mm/s reported by John Wolff and also faster than the
v_n = 33..36 mm/s reported by Nicholas Simons. I risk this because
otherwise I'd have to rewrite a few arrangements which have a somewhat
higher repetition than is possible with the lower rates (in the first
few bars, at least).

(b) minimal note length 6.5 mm

(c) minimal distance between repeating notes 4.3 mm

(d) maximum length of slot in paper 110 mm

Reducing (b) and/or (c) would of course allow to reduce (a)
proportionally to the sum of (b) and (c).  The lengths we use for
(b) and (c) are derived from a single original Triola roll we got
from our customer.  I hope they are conservative enough, but not
too conservative.  (d) is, in my opinion, quite uncritical -- it only
serves to limit the possibility of parts of the paper wandering around.

If possible (= if we have enough empty spools), I would like to send
a roll or two to you, John and Nicholas, so that you could critically
check whether these assumptions are acceptable (and of course also
whether the arrangements "make sense"). I'll let you know when we have
the rolls!

Nicholas, you corrected my "wild guess" how the chords of the Triola
are struck.  I shouldn't have written this, as I have never seen a
Triola up to now.  Thanks for pointing out how it really works.

And, on a different note, thanks for your compliments on my
arrangements, although some of them (like Shreveport Stomp) are
certainly taking the 20er quite (probably too) far ... but I have
to say (also) those were fun to write!

Best regards and many thanks again for the ample input!

Harald M. Mueller
Grafing b. Muenchen, Germany
http://www.haraldmmueller.de/