Balloon theory of yarn during unwinding from packages

Abstract

Yarn unwinding from a package is a key problem in many textile processes, such as weft insertion and warping. Stability of the unwinding has a direct influence on the efficiency of the entire textile process and the quality of the final product. The quality of the yarn is numerically expressed mainly in terms of mechanical quantities. In the unwinding process, viscoelastic properties are the most important. They depend on how the yarn is stressed. The quality of the yarn that is being unwound should not be reduced, unless this reduction does not significantly lower the quality of the fabric. We strive to achieve as large warping and weaving speeds as possible; therefore, our aim is to improve the theory of cross-wound package unwinding and to find the necessary modifications of the yarn unwinding process. The goal of our contribution is to state the equations of motion that describe the unwinding yarn.

Keywords

unwinding yarn movement non-inertial rotating systems balloon theory inertial forces

The theory of yarn unwinding off a package and the balloon theory had a quick development in the 1950s because of Padfield’s work.^1,2 She fixed Mack equations for the balloon³ so they take into account the Coriolis system force. She found the results for a single balloon as it unwinds from a cylindrical package. The same theory was later used to calculate the parameters for multiple consecutive balloons with a nonzero unwinding angle and a cylindrical, conical or empty package.¹ Kothari and Leaf^4,5 derived motion equations that include the effect of the gravity force and air resistance force tangential component. Using extensive numerical methods for cylindrical and conical packages showed these effects can be ignored. Recently, Fraser and Batra⁶ used the motion theory to show that the time dependence can be excluded from motion equations in a mathematical correct way. He derived movable boundary conditions for packages with a small winding angle. Fraser⁷ also determined that the tension inside and the radius of a balloon are smaller for an elastic yarn.

Equations for unwinding yarn from packages

The coordinate system and the mathematical description of yarn position

The chosen coordinate system for the unwinding of the yarn from a cylindrical package is shown in Figure 1.

The yarn is unwinding with speed V through a guide O that is also the origin of the coordinate system. The rise point Lp is the point where the yarn leaves the package and makes the balloon. At the point Up the yarn starts to unwind and slide on the package. The angle of the yarn winding on the cylindrical package is φ. Vector t is the tangential vector to the yarn at the unwinding point.

For the description of the unwinding yarn a rotating coordinate system is used. Base vectors e_i are time dependent¹⁰ as follows:

\frac{\partial e i (t)}{\partial t} = ω \times e i (t)

(1)

where ω is the angular velocity vector that is equal to ω = ωe _z and ω is the angular velocity of unwinding.

From a mathematical point of view the yarn is a curve in space that can be described in a parametric form as r = r(s). The arc length s¹¹ is the length of the yarn from the origin O inside the guide to the chosen point P on the yarn determined by the radius vector r (Figure 1). The position of the point P is time dependent. With the addition of the time parameter t we write the position of P as r = r(s,t) = x(s,t)e _x (t) + y(s,t)e _y (t) + z(s,t)e _z (t).

Because of the symmetry around the z-axis it is better to select the cylindrical coordinate system instead of the Cartesian. As shown in Figure 1, a point in the cylindrical coordinate system is described by the coordinates r (distance from the z-axis), the polar angle θ and the point height z.

The point position in vector form is

r (s, t) = r (s, t) e r (θ (s, t), t) + z (s, t) e z (t)

(2)

Here we need to emphasize that the point with coordinates r,θ,z has its own triplet of base vectors e _r , e _θ , e _z (Figure 1).

Now we must take into account that the base vectors e _r and e _θ explicitly depend on time t because they rotate with the coordinate system (see Figure 2). We can derive the following equations (see Figure 3):

e r (t, θ) = e x (t) \cos (ω t + θ) + e y (t) \sin (ω t + θ) e θ (t, θ) = - e x (t) \sin (ω t + θ) + e y (t) \cos (ω t + θ)

(3)

Here e _x and e _y are the base vectors of the non-rotating coordinate system. The time derivatives of e _r and e _θ are

\overset{\cdot}{e} r (t) = - e x ω \sin (ω t) + e y ω \cos (ω t)

(4)

\overset{\cdot}{e} θ (t) = - e x ω \cos (ω t) - e y ω \sin (ω t)

(5)

By introducing the angular velocity vector ω (and using e _z × e _x = e _y and e _z × e _y = −e _x ), we can rewrite Equations (4) and (5) as

\overset{\cdot}{e} r (t) = ω \times e r (t)

(6)

\overset{\cdot}{e} θ (t) = ω \times e θ (t)

(7)

The base vector e _z around which the coordinate system rotates does not depend on time. Thus, it can be written as $\overset{\cdot}{e} z = 0 = ω \times e z .$ This relation will be used in the following. The base vectors e _r and e _θ do not depend only on time t but also on the angle θ = θ(t) (see Figure 4). We have

e r (θ) = \cos θ e x + \sin θ e y e θ (θ) = - \sin θ e x + \cos θ e y

(8)

We differentiate with respect to θ and obtain

\frac{\partial e r}{\partial θ} = - \sin θ e x + \cos θ e y = e θ

(9)

\frac{\partial e θ}{\partial θ} = - \cos θ e x - \sin θ e y = - e r

(10)

Assumptions

While deriving the equations of motion of yarn we will apply the following four assumptions.

The yarn has no bending stiffness and has constant density and diameter.

The yarn is unstretchable.^6,7

Gravity is ignored.^4,5

The yarn is moving in the air. We take into account air drag with a quadratic dependence on the velocity.^8,9

Kinematics: yarn point velocity and acceleration

Let us consider the motion of a point defined by the vector r(s, t). The velocity is the total derivative of the position vector with respect to time:¹⁰

v (s, t) = \frac{d r (s, t)}{dt} = \frac{\partial r}{\partial t} + \frac{\partial r}{\partial s} \frac{\partial s}{\partial t}

(11)

The yarn is dragged with speed V towards the guide, which implies that ∂s/∂t = −V. The above equation can then be simplified to

v = \overset{\cdot}{r} - V \frac{\partial r}{\partial s}

(12)

The base vectors e _r (θ(s,t),t) and $e θ$ (θ(s,t),t) are time dependent because of the rotation of the coordinate system around the axis z, and because of the movement of the selected point P inside this coordinate system. Hence,

\overset{\cdot}{e} r = \frac{\partial e r}{\partial t} + \frac{\partial e r}{\partial θ} \frac{\partial θ}{\partial t} = ω \times e r + \overset{\cdot}{θ} e θ \overset{\cdot}{e} θ = \frac{\partial e θ}{\partial t} + \frac{\partial e θ}{\partial θ} \frac{\partial θ}{\partial t} = ω \times e θ - \overset{\cdot}{θ} e r

(13)

Here we considered Equation (1), which applies to every rotating coordinate system, Cartesian or cylindrical. We also used Equations (9) and (10). They represent a basic property of a cylindrical coordinate system. The time derivative of position can be then written as

\overset{\cdot}{r} = \overset{\cdot}{r} e r + r \overset{\cdot}{e} r + \overset{\cdot}{z} e z + z \overset{\cdot}{e} z = \overset{\cdot}{r} e r + r \overset{\cdot}{θ} e θ + \overset{\cdot}{z} e z + ω \times (r e r + z e z) = v rel + ω \times r

(14)

We introduced the relative velocity $v re l = \overset{\cdot}{r} e r + r \overset{\cdot}{θ} e θ + \overset{\cdot}{z} e z$ , the velocity of a selected point according to the rotating coordinate system. The total velocity is

v = v rel + ω \times r - V \frac{\partial r}{\partial s}

(15)

where ω × r is the angular velocity of the point P around the axis z and

- V \frac{\partial r}{\partial s}

is the velocity of dragging the yarn through the guide. This component of the velocity points in the direction of the tangent of the infinitesimal yarn section with the orientation towards the origin (i.e. in the direction of dragging).

Since r was parametrized by the natural parameter, the tangent vector is

t = \frac{\partial r}{\partial s}

(16)

The acceleration is the total derivative of velocity with respect to time:¹⁰

a = \frac{d v}{dt} = \frac{d}{dt} (\frac{\partial r}{\partial t} - V \frac{\partial r}{\partial s}) = \frac{\partial}{\partial t} (\frac{d r}{dt}) - V \frac{\partial}{\partial s} (\frac{d r}{dt}) = \frac{\partial}{\partial t} (\frac{\partial r}{\partial t} - V \frac{\partial r}{\partial s}) - V \frac{\partial}{\partial s} (\frac{\partial r}{\partial t} - V \frac{\partial r}{\partial s}) = \frac{\partial 2 r}{\partial t 2} - 2 V \frac{\partial 2 r}{\partial t \partial s} + V 2 \frac{\partial 2 r}{\partial s 2}

(17)

By using Equations (1, 9 and 10) we write ∂²r/∂t² as

\frac{\partial}{\partial t} \overset{\cdot}{r} = \frac{\partial}{\partial t} (\overset{\cdot}{r} e r + r \overset{\cdot}{θ} e θ + \overset{\cdot}{z} e z + ω \times r) = \overset{··}{r} e r + \overset{\cdot}{r} \frac{\partial e r}{\partial θ} \frac{\partial θ}{\partial t} + \overset{\cdot}{r} \overset{\cdot}{θ} e θ + r \overset{··}{θ} e θ + r \overset{\cdot}{θ} \frac{\partial e θ}{\partial θ} \frac{\partial θ}{\partial t} + r \overset{\cdot}{θ} \frac{\partial e θ}{\partial t} + \overset{··}{z} e z + \overset{\cdot}{z} \overset{\cdot}{e} z + \overset{\cdot}{ω} \times r + ω \times \overset{\cdot}{r} = \overset{··}{r} e r + \overset{\cdot}{r} \overset{\cdot}{θ} e θ + \overset{\cdot}{r} ω \times e r + \overset{\cdot}{r} \overset{\cdot}{θ} e θ + r \overset{··}{θ} e θ - r \overset{\cdot}{θ} 2 e r + r \overset{\cdot}{θ} ω \times e θ + \overset{··}{z} e z + \overset{\cdot}{z} \overset{\cdot}{e} z + \overset{\cdot}{ω} \times r + ω \times \overset{\cdot}{r}

= (\overset{··}{r} - r \overset{\cdot}{θ} 2) e r + (2 \overset{\cdot}{r} \overset{\cdot}{θ} + r \overset{··}{θ}) e θ + \overset{··}{z} e z + ω \times (\overset{\cdot}{r} e r + r \overset{\cdot}{θ} e θ + \overset{\cdot}{z} e z) + ω \times (v rel + ω \times r) + \overset{\cdot}{ω} \times r = a rel + 2 ω \times v rel + ω \times (ω \times r) + \overset{\cdot}{ω} \times r

(18)

We introduced the relative acceleration $a rel = (\overset{··}{r} - r \overset{\cdot}{θ} 2) e r + (2 \overset{\cdot}{r} \overset{\cdot}{θ} + r \overset{··}{θ}) e θ + \overset{··}{z} e z .$ The part of the equation with mixed derivatives can be expressed as

\frac{\partial 2 r}{\partial s \partial t} = \frac{\partial}{\partial s} (\frac{\partial r}{\partial t}) = \frac{\partial}{\partial s} (v rel + ω \times r) = \frac{\partial v rel}{\partial s} + ω \times \frac{\partial r}{\partial s}

(19)

The total acceleration is then equal to

a = a re l + 2 ω \times v rel - 2 V ω \times \frac{\partial r}{\partial s} + ω \times (ω \times r) + \overset{\cdot}{ω} \times r - 2 V \frac{\partial v rel}{\partial s} + V 2 \frac{\partial 2 r}{\partial s 2}

(20)

We introduce the operator D for the total derivative with respect to time that follows the motion of a selected point inside a rotating coordinate system:⁶

D = \frac{\partial}{\partial t} | r, θ, z - V \frac{\partial}{\partial s}

(21)

We partially differentiate only the cylindrical coordinates r, θ and z and not the base vectors e _r , eθ and e _z . This way we achieve that operator D only follows the points inside the rotating coordinate system. We get

D r = v rel - V \frac{\partial r}{\partial s}, D 2 r = a rel - 2 V \frac{\partial v rel}{\partial s} + V 2 \frac{\partial 2 r}{\partial s 2}

(22)

With the help of D we can write the acceleration in a simpler form:

a = D 2 r + 2 ω \times (D r) + ω \times (ω \times r) + \overset{\cdot}{ω} \times r

(23)

Dynamics: forces on an infinitesimally small yarn section

The forces that act on an infinitesimal section of a yarn with length δs are the tension T and air resistance (Figure 5). The force on the first edge of the section within lengths s and s+δs is

- (T \frac{\partial r}{\partial s}) (s)

(24)

The reason for the negative sign is the direction of the yarn tension. At point r(s) the tension generates a force in the opposite direction to the tangential vector δr/δs. The force on the other end of the infinitesimal section equals

(T \frac{\partial r}{\partial s}) (s + δ s)

(25)

The mass of a small section is m = ρδs, where ρ is the linear mass density. The second Newton’s law states that

δ s ρ a = (T \frac{\partial r}{\partial s}) (s + δ s) - (\frac{\partial r}{\partial s}) (s) + δ F

(26)

where δF is the air resistance force. We divide the above equation by δs. The first two terms on the right-hand side can be simplified by using the definition of the derivative to ∂/∂s (T∂r /∂s). We also introduce the linear density of the air resistance force f = F/δs. The second Newton’s law can then be rewritten as

ρ a = \frac{\partial}{\partial s} (T \frac{\partial r}{\partial s}) + f

(27)

If we use the derived acceleration in Equation (27), we get

ρ (D 2 r + 2 ω \times D r + ω \times (ω \times r) + \overset{\cdot}{ω} \times r) = \frac{\partial}{\partial s} (T \frac{\partial r}{\partial s}) + f

(28)

This is the equation of motion for a yarn that we have been looking for. The first term on the left represents the relative acceleration, the same as the simple second derivative with respect to time in the inertial coordinate system. The next three terms are the system forces that appear in non-inertial rotating coordinate systems.

–ρ 2 ω × Dr is the Coriolis force.

−ρ ω × ω × r) points in a radial direction towards the outside. It is the centrifugal force.

$- ρ \overset{\cdot}{ω} \times r$ creates a system force inside the systems where the angular velocity changes with time. It is the Euler's force.

Our Equation (28) is more general than the similar equations in the literature.^{1–7,12–21} Namely, we obtain an additional term $- ρ \overset{\cdot}{ω} \times r$ that corresponds to the nonconstant angular velocity of the rotating coordinate system. This term is almost as important as the Coriolis force term in the Padfield’s correction of Mack’s equations.

Condition for inextensibility

We assume that the yarn is inextensible. We will show what this means from a mathematical point of view. We select two nearby points on the yarn: A at length s and B at s + δs. Because the arc length s is measured from the origin of the coordinate system, the length between A and B is δs. We define the vector from A to B as δr = r _A – r _B with length |δr|. In the limit δs→0 the distance between A and B equals

| δ r | = δ s

(29)

\frac{\partial r}{\partial s} \cdot \frac{\partial r}{\partial s} = 1

(30)

By differentiating (2) with respect to s (here we use relations (9) and (10)),

\frac{\partial r}{\partial s} = \frac{\partial r}{\partial s} e r + r \frac{\partial e r}{\partial s} + \frac{\partial z}{\partial s} e z = \frac{\partial r}{\partial s} e r + r \frac{\partial e r \partial θ}{\partial θ \partial s} + \frac{\partial z}{\partial s} e z = \frac{\partial r}{\partial s} e r + r \frac{\partial θ}{\partial s} e θ + \frac{\partial z}{\partial s} e z

(31)

and calculating the dot product in Equation (30), we get the condition for inextensibility:

\frac{\partial r}{\partial s} \cdot \frac{\partial r}{\partial s} = (\frac{\partial r}{\partial s}) 2 + r 2 (\frac{\partial θ}{\partial s}) 2 + (\frac{\partial z}{\partial s}) 2 = 1

(32)

Guide boundary conditions

Figure 1 shows the coordinate systems and points O, Lp and Up that we will use to determine the yarn unwinding boundary conditions.

The boundary condition for the guide in the coordinate system base where the arc length is s = 0 is

r (s = 0, t) = 0

(33)

or, written componentwise,

r (s = 0, t) = 0, z (s = 0, t) = 0

(34)

The boundary condition for θ (s = 0,t) with the condition r (s = 0,t)=0 is not fixed because any θ (s = 0,t) satisfies this equation. For a quasi-stationary state, when variables r, θ, z and T are time independent, we can choose any boundary condition for θ (s = 0), for example

θ (s = 0) = 0

(35)

Boundary condition at the point where the yarn creates the balloon

At the point where the yarn raises from the package and makes the balloon (point Lp in Figure 1), the solution has to satisfy three conditions as follows.

The yarn must not be broken: the spatial vector r(s,t) is a continuous function.

The yarn must not be bent: the tangent vector ∂r/∂s on the rope is continuous. This condition can be justified as follows: on the boundary between the yarn and the package (point Lp) there is no cohesive force that would push the yarn on to the package.^12–14

The velocity v before and the velocity after the point Lp must be equal. If this is not true the yarn would break immediately. This is in conflict with the first condition.

Since the speed v (Equation (12)) and the tangent vector ∂r/∂s are continuous at the lift point, the vector ∂r/∂t is also continuous. We deduce the following:

r (s_{Lp}^{-}, t) = r (s_{Lp}^{+}, t) \partial r (s_{Lp}^{-}, t) / \partial t = \partial r (s_{Lp}^{+}, t) / \partial t \partial r (s_{Lp}^{-}, t) / \partial s = \partial r (s_{Lp}^{+}, t) / \partial s θ (s_{Lp}^{-}, t) = θ (s_{Lp}^{+}, t) \partial θ (s_{Lp}^{-}, t) / \partial t = \partial θ (s_{Lp}^{+}, t) \partial t \partial θ (s_{Lp}^{-}, t) / \partial s = \partial θ (s_{Lp}^{+}, t) / \partial s z (s_{Lp}^{-}, t) = z (s_{Lp}^{+}, t) \partial z (s_{Lp}^{-}, t) / \partial t = \partial z (s_{Lp}^{+}, t) \partial t \partial z (s_{Lp}^{-}, t) / \partial s = \partial z (s_{Lp}^{+}, t) / \partial s

(36)

Here s^-_Lp is the arc length right before the lift point and s⁺_Lp is the arc length right after the lift point.

For the yarn in the currently unwinding layer the vector r is

r = c e r + z e z

(37)

where c is the package radius at the selected point. Compared to other packages the value c for cylindrical packages changes very slowly during the yarn unwinding. In the first approximation we can take a constant value for c. This means that ∂c/∂s = 0 and ∂c/∂t = 0.

We get two more conditions:

\partial r (s_{Lp}^{-}, t) / \partial t = 0, \partial r (s_{Lp}^{-}, t) / \partial s = 0

(38)

We need these conditions if we want to link the balloon and package solutions.

Boundary conditions at the point where the yarn slides on the package

At some point on the package the yarn starts to slide. We call it the starting point of the unwinding, Up. The tangent vector at this point must correspond to the winding angle φ (Figure 6). We write this as

t = \frac{\partial r (s Up)}{\partial s} = \cos φ e θ + \sin φ e z

(39)

Inserting r = c and ∂r/∂s into Equation (31) we get

\frac{\partial r (s Up)}{\partial s} = c \frac{\partial θ}{\partial s} e θ + \frac{\partial z}{\partial s} e z

(40)

Equating expressions in Equations (39) and (40) gives us two boundary conditions:

c \frac{\partial θ (s Up)}{\partial s} = \cos φ, \frac{\partial z (s Up)}{\partial s} = \sin φ

(41)

The system forces in a non-uniformly rotating frame

The term D²r in the equation of motion (28) represents the acceleration of a point in a rotating coordinate system. By moving the remaining three terms from the right-hand to the left-hand side of the equation we can reinterpret them as system forces caused by the non-inertial character of the observation frame: the Coriolis force, the centrifugal force and Euler's force due to changes of the rotational velocity (Figures 7 and 8).

The system forces acting on a short yarn segment are shown schematically in Figures 7 and 8. The centrifugal and the Coriolis force are well known, but we would like to emphasize the presence on additional system Euler's force $- ρ \overset{\cdot}{ω} \times r$ . This force is oriented along the Coriolis force when the lift-off point is near the front edge of the package. At that moment the angular velocity $ω$ is increasing. The angular acceleration vector $\overset{\cdot}{ω}$ is shown by a dashed line in the figure. When, however, the lift-off point is near the rear edge of the package, the force is oriented in the opposite direction to the Coriolis force, since the angular velocity $ω$ is decreasing and the angular acceleration $\overset{\cdot}{ω}$ points in the opposite direction compared to the situation near the front edge of the package. In the center part of the package the conditions are quasi-stationary (they do change very slowly with time): the additional system force is absent. (This is clearly only true for layers with a large number of loops, that is, for precision wound packages.) This additional system force affects the motion of the yarn at the package edges where the winding angle is reversed. Such a rapid change of the winding angle leads to a sudden change of the angular velocity $ω$ , which implies that the angular acceleration $\overset{\cdot}{ω}$ is large. For this reason, the system Euler's force $- ρ \overset{\cdot}{ω} \times r$ is also large and it affects the yarn dynamics. The conditions are then not quasi-stationary but transitory, since the type of motion changes. For this reason there might be instabilities of the balloon shape, and the yarn can break.

Air resistance force

For a numerical calculation of the balloon shape we need to find the linear density of the air resistance force first shown in Equation (28). The air resistance force of a moving object through air or liquid can be described by laws where the velocity term is either linear or quadratic.⁸ The correct law is determined by the Reynolds number:

Re = \frac{l ρ v}{η}

(42)

where l is the yarn diameter, ρ is the air density, v is the normal component of the velocity and η is the dynamic viscosity of air. Approximately we can say that the linear law is used for Re < 0.5 and the square law for Re > 1000. In the area in between, Tables 1 and 2 can be used to determine the appropriate law. Also, we need to be careful because the square law does not apply for every speed.

Table 1.

Yarn unwinding parameters.

Parameter	Value
Air density, ρ	1.22 kg/m³
Yarn diameter, l = d	10^–4m
Dynamic viscosity of air, η	1.7·10^–5 kgm/s
Unwinding speed, V	1000–2000 m/min

Table 2.

Reynolds numbers.

Unwinding speed	Re	c_u
1000 m/min	7100	0.95
1500 m/min	10,700	1.05
2000 m/min	14,000	1.1

Table 1 shows some typical yarn unwinding parameters. The normal component of the yarn velocity v_n is estimated by the unwinding speed V. The resulting Reynolds numbers are given in Table 2.

For these typical velocities the use of the quadratic drag law is fully justified. The effective drag coefficients c_u were taken from tables.⁹ If we would want to be more accurate we would need to calculate the coefficient c_u for every point on the yarn. We will use the approximation that the velocities are always the same.

The drag force is written as⁸

F = \frac{1}{2} c u ρ v_{n}^{2} S

(43)

where c_u is the effective drag coefficient described above, ρ is the air density, v is the normal component of the velocity and S is the area of the body cross-section that is normal to the direction of velocity vector. For a small yarn section with length δs, the front section is dδs where d is the yarn diameter. The linear density of the force, f = δF/δs, is then

f = \frac{1}{2} c u ρ v_{n}^{2} d

(44)

We introduce the abbreviation D_n = 1/2c_uρd and we take into account that the drag force points in the opposite direction than the normal component of the velocity. In vector form we write the air resistance force as

f = - D n | v n | v n

(45)

The normal component of the velocity is obtained by subtracting the tangential component t(t·v) from the total velocity. The tangent vector is t = ∂ r /∂s. We get

v n = v - (v \frac{\partial r}{\partial s}) \frac{\partial r}{\partial s}

(46)

We simplify the above equation by using the vector identity a × (b × c) = b(a · c) – c(a · b):

v n = v (\frac{\partial r}{\partial s} \frac{\partial r}{\partial s}) - \frac{\partial r}{\partial s} (v \frac{\partial r}{\partial s}) = \frac{\partial r}{\partial s} \times (v \times \frac{\partial r}{\partial s})

(47)

where the velocity v is defined by Equation (15).

Let us calculate an approximation of the air resistance force. For the unwinding speed of V = 2000 m/min we get

f u = \frac{1}{2} c u ρ v_{n}^{2} d \approx 0.5 \times 1.3 \times 1.22 \times (2000 / 60) 2 \times 10 - 4 N / m \approx 0.1 N / m

(48)

The typical strain tension is around⁶

T \approx 10 ρ l V 2 \approx 10 \times 27.3 \times 10 - 6 (2000 / 60) 2 \approx 0.3

(49)

In the calculation we have used the value ρ_l = 27.3 tex. The air resistance force is of the same order of magnitude as the tension, which means that the air resistance is crucially important for the balloon shape and the yarn tension inside the balloon.

We can also approximate the force of gravity pushing on the yarn. The gravitational force density is f_g = ρg, where g = 9.8 m/s². We get

f g \approx 10 - 4 N / m

(50)

which is insignificant compared to the air resistance force. This is why we ignore the gravitational force during theoretical calculations.

Conclusion

We derived a system of differential equations that describes the yarn movement in a rotating cylindrical coordinate system. Equations of this form are very useful when we are addressing the yarn unwinding from a package in various textile processes. The derived system is entirely general. We have emphasized the commonly neglected Euler's force and we have shown how it can affect the yarn dynamics at the edges of the package. The problem is mathematically well defined because we have four unknown variables: spatial curve r(s,t) and tension T(s,t) and four equations: the equation of motion for the yarn (Equation (28)) and the condition for inextensibility (Equation (32)). We derived also the appropriate boundary conditions that gave us the full mathematical description of the yarn unwinding.

Footnotes

Declaration of conflicting interests

The authors declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The authors received no financial support for the research, authorship and/or publication of this article.

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