We assume the product on the bobbin does not stretch and the product is tightly wound (no extra (air)space between each layer)

Let :

Do = outside diameter of the bobbin

Ro = (Do/2) = outside radius of the bobbin

Di = inside diameter bobbin = outside diameter of the cardboard residual roll of the bobbin

Ri = (Di/2) = inside radius of the bobbin = outside radius of the cardboard residual roll of the bobbin

n = number of completed windings (the drawing above shows n=3)

t = thickness of the wound product

L = total length of the wound product on the bobbin

Then :

Do = Di + 2.n.t

L = π.(Di+t) + π.(Di+2t+t) +π.(Di+4t+t)+π.(Di+6t+t)+......(applying the neutral fiber of the product)

L = n.π.Di + n^{2}.π.t

L = n.π.(Di + n.t) where n.t =(Do-Di)/2

L = n.π.(Di + (Do-Di)/2)

L = n.π.((Do+Di)/2)

L = n.π.(Ro+Ri) (formula based on the number of windings and the bobbin radii) where n = (Ro-Ri)/t

L = π.((Ro-Ri)/t).(Ro+Ri)

L =
(π/t).(Ro^{2}-Ri^{2}) (formula based on the thickness of the wound product and the bobbin
radii)

Example question 1:

A bobbin of tape has an outside diameter of 60 mm and an inside diameter of 30 mm. The thickness of the tape is 25 micron. What is the length of the tape on the bobbin and how many windings are there?

Given:

Do = 60 mm or Ro = (60/2) = 30 mm

Di = 30 mm or Ri = (30/2) = 15 mm

t = 0,025 mm

Requested:

L ? n ?

Solution:

L =
(π/0,025).(30^{2}-15^{2}) mm

L = 84823 mm

L = 84,823 m

n = (Ro-Ri)/t

n = (30-15)/0,025

n = 600

Check of L with the first formula

L = 600.π. (30+15) = 84823 mm = 84,823 m

Answer:

There is 84,823 m tape on the bobbin and 600 windings.

Example question 2:

At what outside diameter of the bobbin with an outside diameter 600 mm and inside diameter 80 mm shall the length be decreased to 50% of its original total length?

Given:

Do = 600 mm or Ro = (600/2) = 300 mm

Di = 80 mm or Ri = (80/2) = 40 mm

Requested:

Find Do so that L = Loriginal/2 ?

Solution:

Let the requested Do = 2.x

(π/t).(x^{2}-Ri^{2})
= (π/2t).(Ro^{2}-Ri^{2})

2.x^{2}-2.Ri^{2} = Ro^{2}-Ri^{2}

x^{2} = (Ro^{2}+ Ri^{2})/2

x = sqrt((Ro^{2}+ Ri^{2})/2)

x = sqrt((300^{2}+ 40^{2})/2)

x = 214 mm

Do = 2.214 mm = 428 mm

Answer:

Notice that the thickness of the product or the number of windings is not required to solve this question.

At an outside diameter of 428 mm the bobbin has already decreased to 50% of it's original starting length or otherwise formulated : if the net wound radius has decreased with approximately 33%, already 50% of the total length is gone.

## Calculate the length of a bobbin based on outside diameter Do, inside diameter Di and thickness of wound product t

## The calculation result is:

Ro = | 0 | mm |

Ri = | 0 | mm |

Length L on the bobbin = | 0 | m |

number of windings n = | 0 |