Need help with fabric roll - number of max turns calculation

[dgiardina]

Hello all !

I need help with writing a formula - I’m trying to calculate the number of rotations of a spool based upon the following: outside diameter of spool, adding material thickness for each rotation & known final length of material.

To give you an idea of the application, I work in a fabric mill & we need to make a roller shade at a certain length but need to add spring assist because it is so wide. Unfortunately, with the spring assist, you can only turn it 14 times before the spring will be at its tension limit.

Thanks !

[mehmetcik]

Try this as a starting point

[newdoverman]

Add 14 thicknesses of the material to the diameter of the spool (yes there are 28 thicknesses in total for 14 turns) to get the diameter after 7 turns use pi * diameter for the circumference. This will be the average circumference for the roll. Multiply by 14 to get the total circumference (length of material wound on the spool). This will give you an approximation of the amount of material that will be wound by the 14 turns. Tension of the material on the spool can make a big difference in the results.

[dgiardina]

well, here's what I came up with manually....

109" fabric length, .2 mm material thickness, 2.5" outside diameter of center spool = 2.70976" outside diameter of fabric on spool. (I used the following calculator: http://www.handymath.com/cgi-bin/roll4.cgi?submit=Entry

take difference of OD fabric - OD spool = .20976"

.20976" / .0079" (which is .2mm / 25.4mm to convert to inches) = 26.55 / 2 (divided to get number of wraps) = 13.27 rotations

Can anyone confirm if this is correct?

If I run the formula backwards, using 14 as the actual number of rotations, I arrive at 115.1" max fabric length.

[shg]

Here's a little different approach.

The area occupied by fabric is the area of the ring formed between the ODs of the spool (d0) and the fabric (d1):

A = pi/4 * (d1^2 - d0^2) (that's approximate because the OD of the fabric varies about the spool)

The length of the fabric is the area divided by the thickness:

L = A / t = pi/(4t) * (d1^2 - d0^2)

The number of turns is

N = (d1 - d0) / (2*t), so

d1 = d0 + 2*t*N

Solving for L,

L = pi * N * (d0 + t*N^2)

That gives a result pretty close to yours:

Row\Col	A	B	C	D
1	Max turns	14		B1: Input
2	Spool OD	2.50	in	B2: Input
3	Fabric Thickness	0.2	mm	B3: Input
4	Thickness	0.00787	in	B4: =B3/25.4
5	Max Length	114.80	in	B5: =PI() * (B1 * B2 + B4 * B1^2)

[newdoverman]

With fabric the number of revolutions to roll up a certain length will vary with the tension applied. The tension will vary both the length of the fabric a little and will compress the roll to a certain degree also. None of this should practically affect something like a roller blind unless it is very long....close is about as good as it gets