Karl Schröder asked [191021 MMDigest]: "Is there a mathematical
equation for "stepping" rolls, i.e., tempo compensation as a spooled
roll fills the take-up spool?"

Of course there is.

Assume that you have a take-up spool of diameter d.  Then, when the
paper has wound up at the take-up spool to diameter D, the speed of the
paper will have increased (linearly) by D/d, and hence all notes at that
point must be lengthened also by D/d so that they take the same time as
at the beginning of the roll.

However, we (typically) want to get an equation where this "speedup"
is computed from the roll position, i.e., the length of paper that is
already wound up on the take-up spool.

To find this, we ask ourselves: How far are we into the roll when the
paper has wound up to diameter D?

The cross section of the paper on the take-up spool is a ring surface
of outer diameter D and inner diameter d (remember, d is the take-up
spool diameter), so it has an area of (D - d ) * 3.14159/4.  On the
other hand, this area is covered by a length L of paper with thickness
t, i.e., it also has area L*t.  We equate these expressions and solve
for D, i.e.

Lt = (D - d ) * 3.14159/4
4Lt/3.14159 = D - d
D = square root of (4Lt/3.14159 + d )

Dividing by d gives us the speedup at position L:

speedup = D/d = square root of (4Lt/3.14159 + d )/d = square root of
(1 + 4Lt/3.14159d )

where t is the paper thickness (when wound up on the take-up spool,
i.e., including some slack), d is the diameter of the empty take-up
spool, and L is the length of the paper currently wound onto the
take-up spool.

For our practical purposes, pi is roughly equal to 4, so the formula
can be simplified to

speedup is about square root of (1 + Lt/d )

Finally, because the Taylor series of the square root of 1 + x starts
with 1 + x/2, we can rewrite this as

speedup is approximately 1 + Lt/2d

Example: A crank organ has a take-up spool of 50mm diameter, and the
paper thickness is about 0.1 mm (0.0039 inch) (500 sheets of standard
80g paper are about 4 cm high; adding some 25% for the not too tightly
wound paper gives us 5cm / 500 = 0.1 mm for one layer of paper).
30 meters (100 feet) into the roll (about at the roll end for a 20er
organ), the speedup, according to the three formulas above, is

square root of (1 + 4 * 0.1 * 30000 / (3.14159 * 50 )) = 1.59

or

square root of (1 + 0.1 * 30000 / 50 ) = 1.48

or

1 + 0.1 * 30000 / (2 * 50 ) = 1.6

The speedup is therefore 50...&0%.  One can see that a take-up spool
of 50mm diameter definitely needs a roll that is punched with speed
compensation -- which I hate, because it makes it impossible to cut
rolls and rearrange them differently.  With a take-up spool of 100mm,
on the other hand, we get a speedup of about

1 + 0.1 * 30000 / (2 * 100 ) = 1.15

or some 15%, which is in the range where the organ grinder can adjust
the cranking speed manually.  With a motor-driven organ with its
constant rpm of the take-up spool, the speedup might be heard, but
there's nothing wrong with a little speedup towards the end of a roll
so that the music gets somewhat happier.

Harald Mueller