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Let us continue our theme about a yarn retraction,
based on the ideal helical model. We spoke
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about the starting position of fiber bundle,
parallel fiber bundle and the twisted yarn.
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In idea of Braschler, we assume that the total
fiber volume in yarn and fiber cross-sectional
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area, means fiber diameter do not change due
to twist.
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May be, that on the periphery fiber is a little
longer than the volume of such fiber is a
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little, may be small, may be in the center
is a little pressed. So, it can be opposite,
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but these differences are so small that Braschler
mentioned that, this assumption is possible
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to use.
When we use this, we can write V 0 is equal
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V, so that it is a constant yeah and also
fiber cross section S is constant, now change
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through the process of twisting. Fiber volume
in non-twisted V 0 is n times because in bundle
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is in fibers, n times S fiber cross section
times lengths. Length is zeta 0 fiber cross
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section is S times zeta 0 is volume of parallel
fiber times, numbers of fibers. In twisted
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form, the volume of fibers, all fibers material
in this portion is S times zeta. S is the
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substance cross section of our yarn times,
final zeta because V equal V 0 equal. We using
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this formula, we obtain this here. Then, we
obtain zeta by zeta 0, which is S by S by
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n and capital S by n substance cross section
by n. It is in the bean area, sectional area
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pair one fiber s star bar and it was definition
of K n. It was K n. So, zeta by zeta 0 is
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equal to K n. How it is now in ideal helical
model with yarn retraction? Yarn retraction
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delta is 1 minus zeta by zeta 0. It was starting
slight to retraction problem, but this is
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K n, so it is 1 minus K n.
We know K n from earlier analysis of ideal
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helical model number of fibers in cross section
yarn corresponding to ideal helical model.
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Isn’t it? So, we can use our equation derived
on the place of K n and we obtained that delta
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is here. A lot is only rearranging. Let us
rearrange this equation using this way to
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this equation rearranging only common denominator
and so on and so forth.
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Well, I think you can make it own self. No
special routine is here. Well, so on the end
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we obtain delta square root from one plus
pi DZ square minus 1 by 1 plus pi DZ square
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plus 1 or because pi DZ is tangents beta D.
We can write it also in that form.
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Second expression for K n, second version
of rearranging of K n for helical model was
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this here. K n as a function of, it is this
here. So, that after rearranging, we can write
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the same delta also in such form. The fact
is rearranging according Braschler, when delta
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is this here, we write it here. Then, after
rearranging, we obtain delta is equal tangent
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square beta D by 2. I think it is this. This
is the easiest formally easiest. All is identical
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all is the same it is only all different shapes
of interpretation.
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Well, now we want also to know this delta
yarn retraction as a function of this coefficient
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alpha. We know that intensity of this intensity
is tangents beta BD is pi DZ and it was derived
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that it is also this here. Using this, we
obtain for delta on the place of tangents
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beta D here or pi DZ here.
We obtain this equation on so that after small
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rearranging this. This may be fourth or fifth
identical equation for yarn retraction. More
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interesting will be to express yarn retraction
as a function of latent twist coefficient
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related to starting lengths where Z twist
and the count, I relate it to starting values.
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Let us remember it.
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We know that it was shown lot pages back.
It was shown that alpha is alpha 0 by 1 minus
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delta power 3 by 2. Sorry, I was here. So,
we use this. We use this equation on the place
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of alpha here. We obtain this equation. The
same is the here, this equation. The problem
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is that, now we have delta on left hand side
as earlier, but also here and here. So, we
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must rearrange; its only mathematical operation.
We must rearrange this equation because to
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obtain the delta explicitly, it means in the
form beta delta is equals some function on
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right hand side must not be delta. It is possible
the way it is shown here. What is you can
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quietly home step by step to see it. It is
good in first step. For first step, call this
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ratio as a, then delta is this here rearranging.
Then, I make square of left hand side as well
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as right hand side rearranging here. Yes,
this, this, this and here. I give back to
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the helping quantity a, this ratio here. After
rearranging, we obtain such resulting equation.
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It is permanently same delta, but now the
delta is function is explained as a function
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of latent twist coefficient alpha 0. See it
from a one point of view. It is interesting.
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Why? Under square root must not be a value.
Isn’t it? So, this value must be higher
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equal 0.
So, that this value must be higher equal 0
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because under square root is not possible.
We speak about the real values, not complex
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values. Well, it must be this here. So, Z
is here. So, that this here is very important.
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It is very important because it say mu is
some quantity from 0 to 1. Some think it is
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rho specific mass density of fibers. It is
some value, some constant. It shown here that
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this alpha 0 by this constant square root
must be smaller than some parameter 1 by square
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root of 4 pi.
So, that alpha 0 must have some border, must
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be smaller than some border value. Isn’t
it? It is not possible to give number of horse
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to run unit for starting such a starting fiber
bundle. How you want? It exist some moment
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in which is not possible more to give the
coil inside. When you prove it, then you obtain
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something. Other, I will show you later. Do
understand what represent this equation here?
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Let us calculate the parameters by maximum
possible twist of the yarn. The maximum twist
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is, if this equivalency here is valid. So,
that its valid alpha 0 by square root of mu,
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rho is equal 1 by square root 4 times pi,
by the way 0.28 and something. Using this
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to the equation for retraction yarn retraction
on this place, we obtain 0. We obtain delta
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is 1 half 50 percentage of this theoretical
maximum retraction of the yarn.
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Since, delta equal tangent beta square by
2, hence 1 half 13 peripheral angle in this
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point of maximum twist beta D is 70 degree.
Now, using equations derived, we obtain also
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tangents beta D, which is 2 times square root
of 2, roughly 2.8. So, other equations or
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results, the border results for maximum twist.
As a root from quadratic equation which was
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here, we obtain plus minus. The question is
what is real? Automatically, we have two roots.
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Physically, only one is real, one is not.
The second is not possible in reality to obtain.
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It is plus or minus C.
Plus somebody said plus. Plus do not be possible.
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Yes. You are right, you are right because
let us imagine the structure without twist,
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then alpha 0 is equal 0 square root, this
0 square root from 1 is 1. So, it is 1 half
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plus minus 1 half. You can obtain 0 or 1.
When you have parallel fiber bundle without
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twist, you cannot have the value of retraction
1 to 0. You must have the value to 0 once.
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I mean, you must be having this value; this
retraction must be equal to 0. Therefore,
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this minus is real, real symbol here.
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The third idea based on axial forces is also
now important for us. This version according
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to Braschler is enough good. Yes and please
any minutes for change of yarn 2. Let us continue
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our theme about a yarn retraction.
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It is a summary for all three ideas. We analyze
only the idea number 2, according to Braschler
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and we obtained this expression for yarn retraction.
How it is graphically interpreted? Here is
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alpha 0 latent twist co-efficient by square
root of mu rho in relation to yarn retraction
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delta. Our curve is the curve number 2. It
is here. What we obtained by first course?
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Now, significant change in lengths, then the
speed with increasing, with shortening increase
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and increase and here is the maximum value.
Here is the maximum possible value, the border
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value and now more is possible.
Theoretically, when we use minus than plus
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in our equation, you obtain the second thin
curve. It is only for completeness. The real
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is this thin line. Real is the thick line
here. Usually, we have the chance to go to
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this theoretically derived end point and practical
maximum twist is coming earlier, may be in
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this moment. How is this graph, this part
of our graph is shown here?
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The second curve is the middle curve among
these three. So, you see that all three hypotheses
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are very similar. Is it well? Is not well?
In relation to the reality because we apply
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ideal helical model, evidently to this model
nearest structure is the structure of twisted
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filament yarn. Lot years ago, we measured
together with my older colleague, Professor
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Marco, we measured lot of filament yarns on
a special instrument which we created and
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we measured the yarn retraction.
You can see that all different counts, all
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different twists together are lying on only
one curve. Using peck intensity 0.75 in our
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equation, this continual line show the curve
in our, according our theoretical model. The
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curve Perkin is excellent. It is the best
in my life. The comparison between experimental
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research and theoretical curve, so perfect
it is, but around 0.25, around this point,
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stop the possibility in practice in laboratory,
stop the possibility to test filament yarn
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was not more possible. In relation to the
theoretical model, it is roughly here in this
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moment, now here.
From point of view of yarn retraction, the
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difference is high, but from point of view
of alpha 0, the difference is not too high.
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The question is, why coming this border situation
earlier? It is evident, because now our expressions
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are related to the very complicated structure
of the real yarn. Real yarn is not axially
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perfectly symmetric. Therefore, one bending
mechanism also is coming, when the peripheral
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fibers press the central and so on because
the yarn is not absolutely perfect in symmetric
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axial symmetric. Therefore, this critical
moment is coming a little earlier than our
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results from theory, but it is coming. It
is coming.
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What is twisted filament yarn? When we give
more coils inside this one, the maximum of
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possible twist, we call it as a saturated
twist, that terminology saturated twist. Then,
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it is not possible more to give the twist
to our filament yarn. I start it first, and
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then, more and more helicos of twist of second
order. It is important, be under this border
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in the technology of texturising because when
you have in machine, you use too high twist.
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Then, such moments like this or this can break
your yarn and destroy this. The process is
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not continual.
More difficult for filament yarn, this model
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based on ideal helical model can be applied
without problem, nevertheless for staple yarn,
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it brink complicate. It is more complicated.
Why? Staple yarns are from staple fibers,
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so that the ends, fiber ends and something
can slip, especially on the periphery of the
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yarn by twisting. Isn’t it?
Therefore, not so high force for retraction
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as by filament yarn, therefore the yarn retraction
in staple yarn is usually smaller. Not too
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much, but a little smaller than the retraction
by filament yarn. It is difficult to measure
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it, very difficult, retraction by staple yarn.
Now, some experimental results exist. Therefore,
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I can recommend you. When you have not something
better, then I can recommend you, use it.
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To use our equation for, the shape of our
equation for yarn retraction but on the place
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of real yarn diameter D gives some more defined
diameter D delta. Some smaller, a little smaller
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value, then real yarn diameter D and this
D delta is related to D. Using this pure empirical
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form, it is pure empiric expression based
on some set of experimental results.
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So, that D yarn diameter, staple yarn diameter
times this expression, where mu specking density
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and S is substantial cross section of the
yarn. It is better than nothing, built now
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to too good, but it is difficult to say what
is good and what is bad because it is very
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difficult to measure yarn retraction by yarn,
experimentally. When we are speaking about
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different applications of helical model, let
us mention also yarn stress strain relation
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model, according a generalize theory of Gegauff
and others. By the way, Gegauff was one of
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us and of 19th century, second half of 19th
century.
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In the picture, we have two fiber elements,
2 red fiber elements. 1 red fiber element
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is laying on the green surfaces of some cylinder
and elemental part of this cylinder surface
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is shown. This height is D Zeta. This ends
must be this is radius r. So, this ends is
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r times D Zeta. Well, the angle, the element
of angle is D phi. Such element, this green
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element we said red fiber is also here. So,
this length is RD phi. This is D zeta and
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this angle, this angle, this angle beta our
known angle beta.
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Well, now, let us take the yarn and load this
yarn in some breaking machine or so. Then,
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you obtain a new position, a new geometry
of our fiber element. By the way, do you know
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how it is in English? ((FL)). You know it.
Did you use as a yes and when you use it make
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also no.
What did you saw that the thickness of this
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is decreasing, some contraction. Isn’t it?
The same effect is in the yarn by its holding,
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so that our element which started on the radius
R is changed his position to another smaller
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radius r dash. Of course, because we elongate
the yarn, balance the zeta starting length
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D Zeta which also elongate to new length D
Zeta dash. Sorry, it is my mistake here. This
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arrow shall be here. It is my mistake in this
in this picture. Sorry. Nobody is perfect.
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Well, to mu lengths d dash. So, the lengths
of the fiber DL is now longer, have the lengths
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DL dash and the angle beta is changed, B may
be smaller, mu angle is B dash. Isn’t it?
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Let us, now define some relative quantities.
Epsilon L, epsilon in axial direction, epsilon,
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therefore a. It is D Zeta dash minus the zeta
by D Zeta, each strain axial strain. So, that
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it is this one or D Zeta dash is 1 plus epsilon
a times D Zeta. Well, it is a repetition from
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last picture.
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Radial strain is final radius minus starting
radius by starting radius. So, it is this
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one. So, it is this here. Let us think that
epsilon r must be major difference because
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r dash is smaller than starting r. The chewing
gum effect contraction, isn’t it?
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Contraction ratio you know, it may be under
the Poisson contraction ratio. It is defined
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as a minus epsilon r by epsilon a. So, it
is positive value because minus epsilon r
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is positive value. Well, fiber strain epsilon
L is DL dash minus DL by D L evidently. So,
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it is this one. DL dash by DL is 1 plus epsilon
L.
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Now, based on Pythagorean Theorem, we can
write D square L dash is DL dash. So, square
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of this is equal D Zeta dash square D square
zeta dash this length square plus this length
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square. It is this here. Nevertheless, according
the reality quantities which we derived, we
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can write on the place of this here, such
expression and here, this expression and after
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rearranging using also the quantity contraction
ratio, we obtain D square L dash in such form,
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rearranging trivial rearranging.
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Now, all geometric quantities are changed
by yarn elongation, but the angle D phi must
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stay the same. Why? We have not only one element,
but lot of elements around and sum of all
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D phi around our yarn is cross section give
two phi the 160 degree before as well as after
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elongation. If D phi for example, will be
smaller than by elongation on the yarn, it
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will be some whole some. Something so, therefore
D phi must not be changed.
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Well, now your D square, it is here. From
this triangle, the Pythagorean Theorem is
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easier is D square L is D square zeta plus
R D phi square from this triangle. So, it
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00:33:28,580 --> 00:33:35,580
is d square L and D square L dash. The ratio
D square L dash by D square L is equal to
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1 plus epsilon L. We said we used what? We
derived this and this. We arranged this, we
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00:33:45,630 --> 00:33:52,630
know what that for example, r d phi by D Zeta
is tangents beta and so on and so on.
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00:33:52,639 --> 00:33:59,639
After re-arranging, we obtain this ratio where
the green numbers have linear epsilon a, the
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black this and this, has epsilon r square
and violet are result epsilon a. Some of these
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00:34:24,899 --> 00:34:31,899
violet are same than the denominator here.
So, that we can write our expression which
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00:34:33,750 --> 00:34:40,750
start 1 plus epsilon L square is equal to
such form and because 1 plus tangent beta
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square is on B by square is here too. So,
we obtain this expression.
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00:34:49,700 --> 00:34:56,700
In our lecture, we usually solve the easiest
cases. So, let us imagine because be easier
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00:35:03,320 --> 00:35:10,320
and sometimes, it is real. In lot cases, it
is possible to imagine that the formation,
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the elongation of the yarn is small.
If it is small, then epsilon L square is very
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00:35:25,470 --> 00:35:32,470
small and epsilon a, sorry epsilon a square
is also very small. It is limited to 0, so
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00:35:36,880 --> 00:35:43,880
that 1 plus 2 time epsilon l plus epsilon
l square. We can write roughly 0 on the place
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00:35:50,180 --> 00:35:57,180
of you have 1 plus 2 times epsilon l for epsilon
r. We use those, we derived and then after
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00:35:59,220 --> 00:36:06,220
rearranging formation, only we obtain epsilon
l is epsilon a times square beta minus eta
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00:36:10,780 --> 00:36:17,780
times sine square beta. Eta is a contraction
ratio, Poisson’s contraction ratio. Some
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00:36:21,350 --> 00:36:28,350
think between 0 and 0.5 may be. Without this
member which is the influence of contraction,
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this equation was derived by Gegauff in 1907.
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You see that I do not present you some quite
a new theoretical model. It is 100 year old
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model. Well, now it is which forces? In a
fiber, the axial force is called as F1. F1
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because pair one fiber. The fiber tensile
stress-strain relation, let us imagine I said
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easiest case that its linear, then the stress
sigma is proportional to epsilon l. In Hooks
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00:37:27,750 --> 00:37:34,750
law, then E is young modulus, but it is a
constant in our case. Axial force F1 is to
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00:37:43,700 --> 00:37:50,700
the area perpendicular to fiber axis, of course
is sigma times s. So, an epsilon l times s.
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00:37:55,410 --> 00:38:02,410
The vertical component FA is F1 times cos
beta. So, that it is this here.
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00:38:04,600 --> 00:38:11,600
The fiber sectional area, this red is s star
which is s by cos beta. In today, we said
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00:38:18,080 --> 00:38:25,080
it. Normal stress on this area, on this red
area is sigma a, which is normal force FA
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00:38:30,040 --> 00:38:37,040
by area s star. It is this here. Then, epsilon
l, for epsilon l we use the expression for
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00:38:40,170 --> 00:38:47,170
small deformation of course. We obtain sigma
a, is E times epsilon a times this expression
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00:38:50,580 --> 00:38:57,580
for small deformation I said.
Now back to our scheme which we used for number
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00:39:01,060 --> 00:39:08,060
of fibers in yarn cross section. We know that
in elemental annulus, the area of fibers is
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00:39:13,380 --> 00:39:20,380
Ds. You know it was derived. Now, the
sigma a is given by this equation, so that
we can integrate, we can say what is the total
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00:39:34,530 --> 00:39:41,530
axial force as a sum from all areas of fiber
sections. It is shown here.
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00:39:43,690 --> 00:39:50,690
All other substitution, only one substitution
rearranging. Then, it is a way we assume.
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00:39:55,740 --> 00:40:02,740
Yes in one moment, we assume that eta contraction
ratio is constant is same in v in each point
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00:40:06,120 --> 00:40:13,120
in yarn. Hence, we integrate and integrate
and integrate. Nothing more traditional integration
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00:40:14,930 --> 00:40:21,930
which you must absorb in your first semesters
by mathematic, isn’t it?
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Well, then on the final form, we obtain this
equation, nothing more than integration of
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00:40:36,430 --> 00:40:43,430
this one. Logically, it is clear. We obtained
this equation for force P force. P is given
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00:40:43,880 --> 00:40:50,880
by this pi times mu pecking density times
module, may be young modulus times epsilon
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00:40:53,280 --> 00:41:00,280
a axial strain of our yarn 1 half of diameter
square and here is angle beta D peripheral
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00:41:04,050 --> 00:41:11,050
angle of fiber in yarn on the yarn surface
and eta and eta. It is nice, but more interesting
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can be the tensile force utilization coefficient.
It was one side. We know the force which we
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need for elongation of yarn using strain epsilon
a axial strain. In other case, let us imagine
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00:41:42,500 --> 00:41:49,500
another yarn without twist parallel fiber
bundle or something. So, let us ask how we
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00:41:54,850 --> 00:42:01,850
must, how is the force for the same strain
in such parallel fiber bundle strain? Yeah
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00:42:02,360 --> 00:42:09,360
same count.
Well, sigma a, was e times epsilon a, because
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00:42:13,180 --> 00:42:20,180
the fibers are perpendicular to cross section
parallel fiber bundle. So, it is equal and
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00:42:26,500 --> 00:42:33,500
S is equal to PD square by 4 times mu. Isn’t
it? Using this, we obtain such expression
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00:42:36,320 --> 00:42:43,320
for force P star, for force in the same yarn,
but without twist. Then, the effect of twist
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00:42:48,680 --> 00:42:55,680
to mechanical properties, we can do tensile
force. We can express as a ratio axial force
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00:43:00,780 --> 00:43:07,780
in twisted yarn by axial force in non-twisted
yarn.
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00:43:08,720 --> 00:43:14,470
Using our equations, you can obtain this.
Here, you can see that this expression is
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00:43:14,470 --> 00:43:21,470
same than which is before brackets. Therefore,
it is this here. Well, how it is theoretically
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00:43:26,540 --> 00:43:33,540
and how is the comparison to some experiments?
This is the utilization coefficient phi is
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00:43:37,200 --> 00:43:44,200
called phi here. It is utilization coefficient
and it based on angle beta D peripheral angle
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00:43:50,130 --> 00:43:57,130
and also, is playing eta contraction ratio.
This is angle beta d. This is our utilization
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00:44:05,260 --> 00:44:12,260
coefficient and there are functions for different
values of eta. If eta is equal 0, so that
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00:44:15,340 --> 00:44:22,340
the traditional Gegauff’s idea, we obtain
such function. Here higher is eta, then the
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00:44:22,700 --> 00:44:29,700
function is smaller. If smaller and smaller
values of phi, but is same. The theoretical
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00:44:30,270 --> 00:44:35,590
curves are here based on our equation, which
is this here.
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00:44:35,590 --> 00:44:42,590
How this practically? John Heiro, a very known
professor from mist from Manchester, now he
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00:44:45,200 --> 00:44:52,200
is old man, but he was really the top person
among his years. As his PHD student studied
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00:44:57,500 --> 00:45:04,500
the strength of different types of filament
yarns, and they obtained such points for these
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00:45:09,100 --> 00:45:16,100
yarns. It is from the book of Heiro and co-workers.
Now, book on mechanics.
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00:45:16,970 --> 00:45:23,970
Well, this is the axis of our phi. This is
our beta d and using eta equal 0.5, we obtain
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00:45:28,240 --> 00:45:35,240
this function. So, you see that, only here
is not it is little order for, but from this
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00:45:41,270 --> 00:45:48,270
moment, it corresponds to our result. So,
you can say that from filament yarn, this
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00:45:57,380 --> 00:46:04,380
result can be roughly used for estimation
of strength. Having higher twist, this quantity
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00:46:17,620 --> 00:46:24,620
phi is decreasing. It means, when I want elongate,
I do not know 5 percentage parallel fibers
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00:46:28,860 --> 00:46:35,860
bundle and twisted yarns and counts, then
the twisted, both filament yarn. I mean, then
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00:46:38,060 --> 00:46:45,060
in parallel fiber bundle, I need a higher
force than by twisted yarn. Twisted yarn is
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00:46:49,160 --> 00:46:56,160
not, filament yarn is not so tough than parallel
fiber bundles.
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00:47:02,660 --> 00:47:09,660
Why because the forces, it will be a variant
I think from equations which I have and not
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00:47:16,510 --> 00:47:23,510
to the strength of staple yarn. You know that
the staple yarn among others, for strength
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00:47:27,670 --> 00:47:34,670
of staple yarn influences two important phenomenons.
One is twist and second is our friction among
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00:47:38,000 --> 00:47:45,000
fibers, then the internal geometry of the
yarn. The internal geometry brings some decreasing
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00:47:49,920 --> 00:47:56,920
curve, this green decreasing curve for yarn
tenacity or something. In similar way, then
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00:48:06,680 --> 00:48:12,960
our model of course, our model is based on
lot of assumptions, but principally this is
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00:48:12,960 --> 00:48:19,960
same effect.
The second effect is the blue curve here,
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00:48:20,310 --> 00:48:27,310
which shows symbolically the effect of friction.
When we have very small twist or 0 twist,
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00:48:31,920 --> 00:48:38,920
then in filament yarn, for elongation of filament
yarn, we need highest force. Therefore, also
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00:48:43,750 --> 00:48:50,750
easier, say the strength of parallel fiber
bundle is highest, but we cannot use it by
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00:48:55,940 --> 00:49:02,940
staple yarn because friction fiber was so
small as a blast in air. So, the friction
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00:49:10,570 --> 00:49:17,570
effect is increasing principally according
to this blue curve and two such influences
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00:49:19,780 --> 00:49:26,780
together with energy give something like this,
a red curve. Therefore, this curve has some
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00:49:28,220 --> 00:49:35,220
maximum point, point of critical twist point
in which the strength of the yarn is maximum.
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Isn’t it?
Now, to the future of such modeling, I said
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00:49:49,370 --> 00:49:56,370
that this curve, we principally are able to
create. The easiest version of it, I present
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00:49:56,770 --> 00:50:03,770
it here now. From others eye, the blue curve
is in whole world, the secret to these days.
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00:50:08,590 --> 00:50:15,590
The problem is friction among fibers. We only
know that the traditional equations like Coulomb
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00:50:17,730 --> 00:50:24,730
equation and Euler’s equations and so on
are not enough well for use in modeling of
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00:50:30,700 --> 00:50:37,700
yarn internal structure. How it is? That is
a question. Hope, somebody of you in future
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00:50:40,300 --> 00:50:47,300
will be successful and will solve these special
laws which are valid for friction among fibers
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00:50:50,660 --> 00:50:57,660
in yarn and other fibers as well. I think
that is all for today.