Perforator Math #1
By Robbie Rhodes
Imagine that a customer asks me to make a music roll for a special nickelodeon. The roll will resemble a long-play 88-note roll. The paper is pulled across the tracker bar by winding it upon a take-up spool turning at constant shaft speed. The paper drive motor is adjustable from Tempo 75 to Tempo 85. His requirements:
tracker bar: 100 channels, 9 channels per inch
paper width: 11.25 inches width
take-up spool: 2.00 inch diameter = D_initial
spool flange: 5.00 inch diameter
playing time: 30 minutes
initial speed: 7.5 to 8.5 feet per minute (T75 to T85)
I talk with Richard Tonnesen at Custom Music Rolls, and he tells me his perforator capabilities:
perf advance: 0.022 inch per step, constant (45 steps per inch)
paper thickness: 0.0027 inch effective, after perforating
Now I have enough independent data to calculate the _dependent_ parameters and answer some important questions.
1. Will it fit on the spool?
I must check that 30 minutes of music roll will not overflow the take-up spool flanges. I _assume_ an initial paper speed of 7.85 feet per minute ("Tempo 78.5"), or 1.57 inches per second. (This is as slow as a piano roll can go and still maintain a nice repetition rate for "marimba effects".) What is the speed of the take-up spool shaft?
The circumference of the spool is pi*Diameter = 3.14 x 2.0 = 6.28 inch circumference. When the spool turns, the circumference moves 6.28 inches per revolution (in/rev). Therefore the shaft speed N (rev/sec) is
N = (in/sec) / (in/rev) = 1.57 / 6.28 = 0.25 rev per second,
or 15 revolutions per minute. In 30 minutes there will be 450 revolutions, and each revolution of the take-up spool adds _two_ thicknesses of paper. The total diameter after playing 30 minutes will be the initial spool diameter (2.0 inches) plus 900 thicknesses of paper.
D_final = 2.0 + 900 * 0.0027 = 2.0 + 2.43 = 4.43 inch diameter
Therefore 30 minutes of music will fit within the 5-inch flanges.
2. How fast is it moving after 30 minutes?
The initial paper speed is 1.57 inches per second when the take-up diameter is 2.00 inches. After thirty minutes the paper speed on the (full) spool is
(0.25 rev/sec) * ( pi*4.43 inch/rev) =
= 0.25 * 3.14 * 4.43 = 3.478 in/sec = T 174 = 17.4 feet/min
I could also get this answer by ratio math, since the paper speed is proportional to the diameter. Thus after thirty minutes the paper speed is the initial speed times the ratio of final diameter over the initial diameter:
final_speed = initial_speed * (D_final / D_initial) =
= 1.57 in/sec * (4.43 / 2.00) = 3.478 in/sec.
3. How much paper is needed for 30 minutes?
Richard Tonnesen uses a simple and elegant approach to this problem. He notes that "it's the same amount of paper whether it's rolled up or if it's laid out flat!" Here's his method:
Find the area of the annular ring occupied by the paper; this is simply the area of the 4.43-inch circle minus the area of the 2-inch spool. Area = pi*r^2 ("pi r-squared"), where r is the radius of the circle.
Paper_area = pi * (4.43/2)^2 - pi * (2.0/2)^2
= 15.41 - 3.14 = 12.27 square inches
This area is also the product of thickness times length. We figuratively lay the paper flat and observe the total area of the edge; solving for length:
Length = Area/Thickness = 12.27/0.0027 = 4545 inches = 379 feet
But how will the Midi music files be converted to control the perforator? This will be discussed in the next article, Perforator Math #2.
Robbie Rhodes
[ Computer Operator's Note:
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[ Author unknown. |
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