A technician draws every test specimen of yarn from the outer layers of packages that are easiest to reach. Which statement is correct?

The sample is biased, and testing more specimens in the same way does not remove the bias

The sample is random, because every package in the lot was sampled at least once by the technician

The sample is biased, and testing more specimens in the same way does not remove the bias

The sampling error vanishes, because outer layers are the least variable

The sample is stratified, because the layers of a package differ from one another in their properties

Why is a specimen conditioned from the dry side before testing in the standard atmosphere?

A fibre holds more moisture at a given humidity when approached from the wet side, so the route must be fixed

Dry fibres cannot absorb any moisture until they have first been heated well above room temperature

Wet fibres lose weight when conditioned, which would make the test result come out too low to be useful

The humidity of the standard atmosphere is then reached faster by the room and its air-conditioning plant

An airflow (micronaire) test gives a low reading for a cotton sample. What can be concluded?

The fibres are either fine or immature, because the airflow reading combines fineness and maturity and cannot separate them

The sample is too moist, because micronaire measures the moisture regain of the sample directly

The fibres must be coarse and fully mature, because the airflow resistance of the plug falls with increasing fibre diameter

The fibres are either fine or immature, because the airflow reading combines fineness and maturity and cannot separate them

The fibres are long and strong, because micronaire is really a measure of the staple length of the cotton

The same fibre is tested in single-fibre CRE tests at gauge lengths of 10 mm and 50 mm. What is the usual outcome for the mean breaking strength?

It is higher at 10 mm, because a short gauge is less likely to contain a weak place

It is higher at 50 mm, because the fibre is then stretched further before it finally breaks

It is the same, because the cross-section of the fibre is the same at both gauge lengths

It is lower at 10 mm, because a short fibre cannot develop its full strength between the clamps

It is higher at 10 mm, because a short gauge is less likely to contain a weak place

A 20 tex yarn and an 80 tex yarn of the same fibre are spun to the same tex twist factor (α = turns per cm × √tex). How do their twists in turns per cm compare?

The 20 tex yarn has twice the twist of the 80 tex yarn

The 20 tex yarn has four times the twist of the 80 tex yarn

The 20 tex yarn has twice the twist of the 80 tex yarn

The 80 tex yarn has twice the twist of the 20 tex yarn

Cotton yarn A is Ne 30 with a lea strength of 60 lbf, and yarn B is Ne 40 with a lea strength of 45 lbf. Which statement about their count-strength products is correct?

The two have equal CSP of 1800, so the yarns are of equal quality on this measure

B has the higher CSP because its count is the finer of the two yarns

The two have equal CSP of 1800, so the yarns are of equal quality on this measure

CSP cannot be computed because the two counts of the yarns differ from each other

A has the higher CSP because its lea strength of 60 lbf is the larger of the two values

Which of the following statements about sampling and sample size are correct?

The number of observations needed for a given precision rises in proportion to the square of the coefficient of variation

Repeating a test on the same unit reduces the testing error, but not the variation between units

The number of observations needed for a given precision rises in proportion to the square of the coefficient of variation

Repeating a test on the same unit reduces the testing error, but not the variation between units

Systematic sampling of every k-th unit is always unbiased, whatever pattern the lot has

Halving the standard error of the mean needs only twice as many observations

Which of the following statements about fibre testing instruments are correct?

A Baer sorter arranges fibres into length groups that are weighed to give the length distribution

HVI readings are comparable between laboratories only when the instrument is calibrated against reference cottons

A Baer sorter arranges fibres into length groups that are weighed to give the length distribution

HVI readings are comparable between laboratories only when the instrument is calibrated against reference cottons

The micronaire value alone gives the maturity ratio of the cotton

A Fibrograph measures the tenacity of single fibres

Which of the following statements about yarn evenness testing are correct?

For a normal distribution of mass per unit length, CV is about 1.25 times U

The index of irregularity is the measured CV divided by the limit CV = 100 ÷ √n for a random arrangement of n fibres

For a normal distribution of mass per unit length, CV is about 1.25 times U

The index of irregularity is the measured CV divided by the limit CV = 100 ÷ √n for a random arrangement of n fibres

Thin places are defined as those 50% above the mean mass per unit length

A capacitance sensor measures the colour of the yarn

Textile Testing of Fibres and Yarns: Sampling, Moisture, Fibre, Yarn and Evenness Tests

This chapter answers the first half of Section XE9.4 of XE9 Textile Engineering and Fibre Science, “Textile Testing”, in the Engineering Sciences (XE) paper. XE9 is one of the nine optional sections, any two of which a candidate chooses in the exam. The syllabus lines covered are: “Sampling techniques for fibres, yarns and fabrics; Sample size and sampling errors. Moisture in textiles; Fibre length, fineness, crimp, maturity and trash content; Tensile testing of fibres; High volume fibre testing. Linear density of sliver, roving and yarn; Twist and hairiness of yarn; Tensile testing of yarns; Evenness testing; Fault measurement and analysis of yarns.” It teaches how to draw a sample and size it, moisture regain and conditioning, the measurement of fibre length, fineness, crimp, maturity and trash, fibre and yarn tensile testing, linear density and twist, and evenness and imperfection analysis. The numericals are sample size, commercial mass at standard regain, length averages, fibre fineness by vibration, twist contraction and the index of irregularity.

1. Sampling, sample size and sampling error

A test measures only a tiny part of a lot, so the result is useful only if the sample represents the lot. The chain is lot → gross (lot) sample (units taken from the lot) → laboratory sample → test specimens. Sampling must be random (every unit has an equal chance), and it is often multistage: bales, then tufts within each bale for fibre; packages, then lengths within each package for yarn; rolls, then positions for fabric. Stratified sampling draws from each known sub-group (shift, machine, bale grade), and systematic sampling takes every k-th unit, which is safe only if the lot has no periodic pattern of period k. A biased sample (the easily reached top layer, the outer wraps of a package, the fabric near the selvedge or roll end) is wrong in a way that no sample size can repair.

For fibre, tufts are taken from many places in many bales, then blended; for yarn, packages are drawn at random and the outer layers discarded before testing; for fabric, specimens are cut away from selvedge and fold, spread across the width and length, and cut on a diagonal so that no two specimens share the same warp or weft yarns.

Sampling error. For n independent observations of coefficient of variation CV (per cent), the standard error of the mean, as a percentage of the mean, is CV ÷ √n. The ±limit at a confidence level is E = t × CV ÷ √n, so the number of observations needed for a stated precision E (per cent of the mean) is n = (t CV ÷ E)². For 95% confidence take t ≈ 1.96 when n is large (use the t-distribution for small n). Because n grows as CV², a variable material needs far more tests, and halving the error needs four times as many.

Worked example. Yarn strength has CV = 15%. For the mean to be within ±5% at 95% confidence: n = (1.96 × 15 ÷ 5)² = 5.88² = 34.6, so test 35 specimens. With 36 tests the 95% limit is 1.96 × 15 ÷ 6 = ±4.9%.

⚠️ Round up, and test different units

Always round n up, since 34.6 tests cannot be done and 34 would miss the target. Ten readings from one package are not ten independent samples of the lot: the CV that matters is the one between the units sampled, and repeating a test on the same unit only reduces the testing error.

2. Moisture in textiles: regain, conditioning and measurement

Fibres take up water vapour from the air, and most properties (mass, strength, extension, electrical resistance) change with it, so moisture is defined and controlled. With W the mass of the specimen and D its oven-dry mass: moisture regain R = 100 (W − D) ÷ D, and moisture content M = 100 (W − D) ÷ W. They convert by R = 100 M ÷ (100 − M) and M = 100 R ÷ (100 + R), and R is always larger than M. Trade uses regain: the commercial (conditioned) mass is D (1 + R_c ÷ 100), where R_c is the agreed commercial regain, so a seller is paid for dry fibre plus an allowance, whatever the weather.

Standard atmosphere. Specimens are conditioned to equilibrium in a standard atmosphere, 20 ± 2 °C and 65 ± 4% relative humidity in the international standard (27 ± 2 °C and the same humidity is used in tropical countries). Equilibrium is approached from the dry side, because a fibre holds more water when approached from the wet side at the same humidity (sorption hysteresis), so the route must be fixed. Equilibrium is confirmed when successive weighings differ by less than a stated small percentage.

Moisture measurement methods
Method	Principle	Remark
Oven drying	Dry to constant mass in an oven at about 105 °C and weigh	Reference method; destructive and slow; other volatiles are counted as water
Electrical resistance or capacitance	Resistance falls and dielectric constant rises steeply with moisture	Fast, for bales and packages; needs calibration per fibre and affected by temperature
Humidity-sensing and chemical titration	Equilibrium humidity above the sample, or a water-specific titration reaction	Used for small samples or very low moisture

Worked example. A lot weighs 1000 kg when its regain is 6.0%. Oven-dry mass D = 1000 ÷ 1.06 = 943.4 kg. At a commercial regain of 8.5%, the commercial mass = 943.4 × 1.085 = 1023.6 kg. A moisture content of 7.5% corresponds to a regain of 100 × 7.5 ÷ 92.5 = 8.11%.

3. Fibre length, fineness, crimp, maturity and trash

Fibre properties and common measurements
Property	Measures and instruments
Length	Single fibres straightened and measured; Baer sorter arranges fibres by length into groups that are weighed; Fibrograph scans a combed beard to give span lengths (2.5% and 50%) and uniformity ratio = 50% span ÷ 2.5% span × 100; HVI reports upper-half mean length, mean length and uniformity index; AFIS individualises fibres and gives number- and weight-based length and short-fibre content
Fineness	Linear density in tex or dtex (1 dtex = 0.1 tex) by weighing a measured length, or by vibroscope; airflow instruments (micronaire) give a combined fineness-maturity value in µg/inch (1 µg/inch = 0.0394 tex)
Crimp	Crimp % = (straightened length − crimped length) ÷ crimped length × 100 (convention to be stated), and crimps per cm
Maturity	Cotton wall thickness, by swelling in caustic soda and counting normal, thin-walled and dead fibres; maturity ratio M = (N − D) ÷ 200 + 0.70 with N and D the percentages of normal and dead fibres (Lord’s relation)
Trash	Trash % = mass of trash separated ÷ mass of sample × 100, in a cleaning analyser (Shirley analyser) or by image analysis in HVI

Fineness by vibroscope. A fibre of length L under tension T vibrates like a string, f = (1 ÷ 2L) √(T ÷ μ), so the mass per unit length is μ = T ÷ (4 L² f²). For L = 20 mm, T = 6.0 mN and f = 5000 Hz: μ = 6.0 × 10⁻³ ÷ (4 × 4 × 10⁻⁴ × 2.5 × 10⁷) = 1.5 × 10⁻⁷ kg/m = 0.15 tex.

Length averages. If n fibres of length L are present, the number-average length is L_n = Σ n L ÷ Σ n, and the length-weighted (mass-weighted) average for equal fineness is L_w = Σ n L² ÷ Σ n L, which is always at least L_n. Fibre counts of 20, 30 and 50 at 10, 20 and 30 mm give L_n = 2300 ÷ 100 = 23.0 mm and L_w = 59,000 ÷ 2300 = 25.7 mm. Instruments that sample by weight or from a beard are biased toward long fibres, so state which average is quoted.

4. Tensile testing of fibres and high-volume fibre testing

A tensile test stretches the specimen between clamps over a gauge length and records force against extension. In a constant-rate-of-extension (CRE) tester the clamp moves at a fixed speed and the force is measured, while constant-rate-of-loading (CRL) and constant-rate-of-traverse (CRT) testers use other drives, so results from different types are not interchangeable. Results are breaking force, tenacity (breaking force per linear density, cN/tex), breaking extension (per cent of gauge length), initial modulus (slope of the first, steepest part of the curve) and work of rupture (area under the curve). Because tenacity uses linear density, it compares fibres of different fineness.

Conversions. Breaking stress (MPa) = tenacity (cN/tex) × density (g/cm³) × 10, because the cross-section is tex ÷ (1000 ρ) mm². A fibre of 1.2 dtex (0.12 tex) that breaks at 3.6 cN has tenacity 3.6 ÷ 0.12 = 30 cN/tex, and with ρ = 1.40 g/cm³ the stress is 30 × 1.40 × 10 = 420 MPa. Also 1 gf/den = 8.83 cN/tex. Gauge length matters: a short gauge excludes weak places, so strength rises as the gauge falls (weakest-link effect), and extension is a percentage of that gauge, so quote it.

Bundle testing breaks a combed bundle of many fibres at once, at zero gauge (Pressley type) or a short gauge, and divides by the bundle’s mass; it is quick and averages many fibres but gives strength lower than the mean single-fibre value because fibres break at different strains, and the weakest do not share load evenly.

High-volume testing. A high-volume instrument (HVI) tests a sample from each bale in about a minute, with length, strength, micronaire, colour and trash from one sample, so every bale of a lot can be graded for purchasing and laydown. Its readings are only comparable if calibrated to reference cottons. Single-fibre systems such as AFIS are slower but give the distributions of length, neps and trash.

5. Linear density, twist and hairiness

Linear density. The direct systems give mass per length: tex (g per 1000 m) and denier (g per 9000 m), so denier = 9 × tex. The indirect systems give length per mass: English cotton count Ne (number of 840-yard hanks per lb), with Ne × tex = 590.5, and metric count Nm = 1000 ÷ tex. Sliver and roving are expressed in ktex or grains per yard (1 grain = 64.8 mg), with 1 ktex = 1 g per m. A wrap reel makes a skein of a stated number of wraps of known circumference; it is weighed on a balance, and tex = mass (g) × 1000 ÷ length (m). For a 125 m skein weighing 3.0 g, tex = 3.0 × 1000 ÷ 125 = 24, so Ne = 590.5 ÷ 24 = 24.6. The CV of count over many skeins measures count variation.

Twist. Twist is turns per unit length (turns/cm, tpcm). Twist factor (tex system) α = tpcm × √tex, and for cotton counts α_e = tpi ÷ √Ne with α_tex ≈ 9.57 α_e. Equal twist factor means equal yarn character across counts. Measurement by the direct method holds a length of yarn, untwists it until the fibres lie parallel (a pointer or needle can pass between them) and counts the turns; the untwisting also lengthens the specimen. Twist contraction = (L_u − L_t) ÷ L_u × 100, where L_t is the twisted length and L_u the length after untwisting: a 50 cm specimen that needs 400 turns, giving 8 tpcm, and lengthens to 51.5 cm has a contraction 1.5 ÷ 51.5 × 100 = 2.9%.

Hairiness is the protrusion of fibre ends from the yarn body. Optical instruments count hairs longer than set lengths per unit length of yarn, or give a hairiness index H, the total length of protruding fibre in cm per cm of yarn. Hairiness raises with yarn twist falling and with fibre length falling, and a hairy yarn has a fuzzier surface and sheds more fly in weaving.

6. Yarn tensile testing, evenness and fault analysis

Yarn tensile. A single-yarn CRE tester gives force, tenacity and elongation, with a stated gauge length, strain rate and pre-tension, with the CV of strength as the quality measure; a lea (skein) test breaks a 120-yard lea and gives a single large figure. The count-strength product (CSP) = Ne × lea strength (lbf) makes strength comparable across counts. A yarn is weaker than the sum of its fibres because fibres slip, break at unequal strains and are inclined by twist, so tenacity first rises with twist (more grip) and then falls (more obliquity).

Evenness. Capacitance testers pass the yarn between condenser plates, where the signal tracks mass per unit length; optical testers sense diameter. The variation is reported as U% = mean deviation ÷ mean × 100 and CV% = standard deviation ÷ mean × 100. For a normal distribution the mean deviation is 0.798 σ, so CV = 1.25 U. The limit for a random arrangement of n fibres in the cross-section is CV_lim = 100 ÷ √n, and the index of irregularity is I = CV ÷ CV_lim; a perfect random sliver has I = 1, and real yarns are above it. With fibres of fineness 0.20 tex in a 30 tex yarn, n = 150, CV_lim = 8.16%, and a measured CV of 12.25% gives I = 1.50.

Faults and periodicity. Imperfections are counted per km at stated sensitivities: thin places (usually 50% below mean), thick places (50% above mean) and neps (200% above, over a short length). A spectrogram plots variation against wavelength; a peak means a periodic fault from a mechanical source, such as a damaged drafting roller. A faulty roller of circumference C gives period C in the product at that roller’s delivery, and a wavelength C × (draft from that roller to the delivery) in the final strand: a back roller of 25 mm diameter (C = 78.5 mm) with a draft of 20 beyond it gives 1571 mm.

Key takeaways

Sampling must be random and unbiased; a biased sample cannot be rescued by size. Standard error = CV ÷ √n, and n = (t CV ÷ E)² rounded up, with t ≈ 1.96 for 95%. Halving the error needs four times the tests.
Regain R = 100 (W − D) ÷ D, moisture content M = 100 (W − D) ÷ W, R = 100 M ÷ (100 − M). Commercial mass = D (1 + R_c ÷ 100). Standard atmosphere is 20 ± 2 °C and 65 ± 4% RH, conditioned from the dry side.
Fibre length by Baer sorter, Fibrograph, HVI and AFIS; weight-biased averages exceed number averages. Fineness by vibroscope μ = T ÷ (4 L² f²) or airflow micronaire. Maturity from Lord’s ratio M = (N − D) ÷ 200 + 0.70; trash from a cleaning analyser.
Tenacity = force ÷ tex (cN/tex); stress (MPa) = tenacity × ρ × 10. Shorter gauge gives higher strength by the weakest-link effect, and bundle strength is below the single-fibre mean. HVI gives bulk bale data fast when calibrated.
Tex = g/km, denier = 9 × tex, Ne × tex = 590.5. Twist factor = tpcm × √tex. Twist contraction = (L_u − L_t) ÷ L_u × 100. Hairiness index H is protruding fibre length per cm of yarn.
CV = 1.25 U for a normal distribution; CV_lim = 100 ÷ √n with n = yarn tex ÷ fibre tex; I = CV ÷ CV_lim. CSP = Ne × lea strength. Spectrogram peaks show periodic faults, with wavelength C × draft for a faulty roller.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A technician draws every test specimen of yarn from the outer layers of packages that are easiest to reach. Which statement is correct?
1. The sample is random, because every package in the lot was sampled at least once by the technician
2. The sample is biased, and testing more specimens in the same way does not remove the bias
3. The sampling error vanishes, because outer layers are the least variable
4. The sample is stratified, because the layers of a package differ from one another in their properties
Show answer
Answer: B — The sample is biased, and testing more specimens in the same way does not remove the bias
Outer layers are handled, exposed and often differ from the interior, so a sample taken only from them is systematically unrepresentative: a bias that no sample size can repair. It is not random because not every part of the package had a chance of selection. Stratified sampling would draw from each layer deliberately, not only from the easiest.
A cotton specimen has a moisture content of 10.0% (water as a percentage of the wet mass). What is its moisture regain, to one decimal place?
1. 12.0%
2. 9.0%
3. 10.0%
4. 11.1%
Show answer
Answer: D — 11.1%
R = 100 M ÷ (100 − M) = 1000 ÷ 90 = 11.1%. Regain is based on the smaller dry mass, so it is larger than moisture content. 9.0% would come from subtracting the wrong way, and 10.0% from treating the two as equal.
Why is a specimen conditioned from the dry side before testing in the standard atmosphere?
1. A fibre holds more moisture at a given humidity when approached from the wet side, so the route must be fixed
2. Dry fibres cannot absorb any moisture until they have first been heated well above room temperature
3. Wet fibres lose weight when conditioned, which would make the test result come out too low to be useful
4. The humidity of the standard atmosphere is then reached faster by the room and its air-conditioning plant
Show answer
Answer: A — A fibre holds more moisture at a given humidity when approached from the wet side, so the route must be fixed
Sorption shows hysteresis: at the same humidity the regain reached by desorption is higher than that reached by absorption, so equilibrium is approached from the dry side by convention. Dry fibres absorb readily at room temperature, and the speed of the room’s humidity is unrelated.
A Fibrograph test of a cotton gives a 2.5% span length of 28 mm and a 50% span length of 14 mm. What is its uniformity ratio, defined as 50% span length ÷ 2.5% span length × 100?
1. 50
2. 14
3. 200
4. 42
Show answer
Answer: A — 50
14 ÷ 28 × 100 = 50. Inverting the ratio gives 200, which is the common slip, and 42 is the sum of the two lengths.
An airflow (micronaire) test gives a low reading for a cotton sample. What can be concluded?
1. The sample is too moist, because micronaire measures the moisture regain of the sample directly
2. The fibres must be coarse and fully mature, because the airflow resistance of the plug falls with increasing fibre diameter
3. The fibres are either fine or immature, because the airflow reading combines fineness and maturity and cannot separate them
4. The fibres are long and strong, because micronaire is really a measure of the staple length of the cotton
Show answer
Answer: C — The fibres are either fine or immature, because the airflow reading combines fineness and maturity and cannot separate them
The airflow through a plug of fixed mass depends on the specific surface of the fibre, which reflects both fineness and wall thickness, so a low value can mean fine or immature fibre. It is not a length or moisture measure, and low values do not mean coarse fibre.
The same fibre is tested in single-fibre CRE tests at gauge lengths of 10 mm and 50 mm. What is the usual outcome for the mean breaking strength?
1. It is higher at 50 mm, because the fibre is then stretched further before it finally breaks
2. It is the same, because the cross-section of the fibre is the same at both gauge lengths
3. It is lower at 10 mm, because a short fibre cannot develop its full strength between the clamps
4. It is higher at 10 mm, because a short gauge is less likely to contain a weak place
Show answer
Answer: D — It is higher at 10 mm, because a short gauge is less likely to contain a weak place
Failure starts at the weakest place in the gauge length, and a longer gauge samples more places, so mean strength falls as gauge length rises: a weakest-link effect. Cross-section does not control this, and stretching further is not a source of extra strength.
A 20 tex yarn and an 80 tex yarn of the same fibre are spun to the same tex twist factor (α = turns per cm × √tex). How do their twists in turns per cm compare?
1. The 20 tex yarn has four times the twist of the 80 tex yarn
2. The two yarns have equal twist
3. The 20 tex yarn has twice the twist of the 80 tex yarn
4. The 80 tex yarn has twice the twist of the 20 tex yarn
Show answer
Answer: C — The 20 tex yarn has twice the twist of the 80 tex yarn
Turns per cm = α ÷ √tex, so the ratio is √(80 ÷ 20) = 2, and the finer yarn carries twice the twist. A ratio of four would arise if twist were proportional to 1/tex rather than 1/√tex.
Cotton yarn A is Ne 30 with a lea strength of 60 lbf, and yarn B is Ne 40 with a lea strength of 45 lbf. Which statement about their count-strength products is correct?
1. B has the higher CSP because its count is the finer of the two yarns
2. The two have equal CSP of 1800, so the yarns are of equal quality on this measure
3. CSP cannot be computed because the two counts of the yarns differ from each other
4. A has the higher CSP because its lea strength of 60 lbf is the larger of the two values
Show answer
Answer: B — The two have equal CSP of 1800, so the yarns are of equal quality on this measure
CSP = Ne × lea strength, giving 30 × 60 = 1800 for A and 40 × 45 = 1800 for B. A finer yarn is naturally weaker per lea, which is why the product is used to compare across counts. Comparing raw lea strengths or raw counts alone is misleading.
A drafting roller of diameter 30 mm has a damaged surface that gives a periodic mass variation. A draft of 8 is applied between this roller and the delivery. Taking the roller circumference as π × 30 = 94.2 mm, what is the wavelength of the fault in the delivered strand?
1. 2260 mm
2. 11.8 mm
3. 94.2 mm
4. 754 mm
Show answer
Answer: D — 754 mm
The fault repeats once per roller revolution, which delivers 94.2 mm of material at that roller, and the draft of 8 stretches this by a factor of 8 in the delivered strand: 94.2 × 8 = 754 mm. 94.2 mm ignores the draft, and 11.8 mm divides by the draft instead of multiplying.
Which of the following statements about sampling and sample size are correct?
1. The number of observations needed for a given precision rises in proportion to the square of the coefficient of variation
2. Repeating a test on the same unit reduces the testing error, but not the variation between units
3. Systematic sampling of every k-th unit is always unbiased, whatever pattern the lot has
4. Halving the standard error of the mean needs only twice as many observations
Show answer
Answer: A — The number of observations needed for a given precision rises in proportion to the square of the coefficient of variation; B — Repeating a test on the same unit reduces the testing error, but not the variation between units
n = (t CV ÷ E)² grows as CV², and repeated tests on one unit tell nothing new about the lot. Systematic sampling can be biased if the lot has a periodic pattern matching k. The standard error falls as 1/√n, so halving it needs four times as many observations, not two.
Which of the following statements about fibre testing instruments are correct?
1. A Baer sorter arranges fibres into length groups that are weighed to give the length distribution
2. HVI readings are comparable between laboratories only when the instrument is calibrated against reference cottons
3. The micronaire value alone gives the maturity ratio of the cotton
4. A Fibrograph measures the tenacity of single fibres
Show answer
Answer: A — A Baer sorter arranges fibres into length groups that are weighed to give the length distribution; B — HVI readings are comparable between laboratories only when the instrument is calibrated against reference cottons
The Baer sorter and calibrated HVI are described correctly. Micronaire combines fineness and maturity, so it cannot by itself yield the maturity ratio. The Fibrograph scans a combed beard to give span lengths, not strength.
Which of the following statements about yarn evenness testing are correct?
1. For a normal distribution of mass per unit length, CV is about 1.25 times U
2. The index of irregularity is the measured CV divided by the limit CV = 100 ÷ √n for a random arrangement of n fibres
3. Thin places are defined as those 50% above the mean mass per unit length
4. A capacitance sensor measures the colour of the yarn
Show answer
Answer: A — For a normal distribution of mass per unit length, CV is about 1.25 times U; B — The index of irregularity is the measured CV divided by the limit CV = 100 ÷ √n for a random arrangement of n fibres
The factor 1.25 follows from the mean deviation 0.798 σ of a normal distribution, and I compares real yarn with the random limit. Thin places are below the mean (about 50% below), not above. A capacitance sensor senses mass, as the dielectric between its plates, and not colour.
The coefficient of variation of fibre-bundle strength between bales is 10%. The mean is to be estimated within ±2% of its value at 95% confidence using n = (t CV ÷ E)² with t = 1.96. The smallest whole number of bales that must be tested is ____.
Numerical answer — type the value.
Show answer
Answer: 97
n = (1.96 × 10 ÷ 2)² = 9.8² = 96.04. The number of observations must be a whole number at least this large, so 97. Rounding down to 96 would leave the precision slightly worse than required.
A lot of cotton weighs 2400 kg at its actual regain of 5.0%. The agreed commercial regain is 8.5%. Taking regain as a percentage of the oven-dry mass, the commercial mass of the lot in kg, to the nearest whole number, is ____.
Numerical answer — type the value.
Show answer
Answer: 2480
Oven-dry mass D = 2400 ÷ 1.05 = 2285.7 kg. Commercial mass = D × (1 + 0.085) = 2285.7 × 1.085 = 2480 kg. Adding 8.5% to the actual mass, or subtracting 5% and adding 8.5%, gives wrong values.
A sample has 10 fibres of length 15 mm, 30 fibres of 25 mm and 10 fibres of 35 mm, all of equal fineness. The length-weighted average length L_w = Σ n L² ÷ Σ n L, in mm to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 26.6
Σ n L = 150 + 750 + 350 = 1250. Σ n L² = 10 × 225 + 30 × 625 + 10 × 1225 = 2250 + 18,750 + 12,250 = 33,250. L_w = 33,250 ÷ 1250 = 26.6 mm. The number average is 1250 ÷ 50 = 25.0 mm, which the weighted value exceeds.
A 25.0 cm length of twisted yarn is untwisted in a twist tester until the fibres are parallel, and its length after untwisting is 25.8 cm. Using twist contraction = (L_u − L_t) ÷ L_u × 100, where L_u is the untwisted length and L_t the twisted length, the contraction in per cent, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 3.1
Contraction = (25.8 − 25.0) ÷ 25.8 × 100 = 0.8 ÷ 25.8 × 100 = 3.10%, so 3.1. Dividing by the twisted length 25.0 would give 3.2%, a different definition.
A 20 tex yarn is spun from fibres of 0.16 tex. Its measured evenness is U = 8.94%, and the mass variation is normally distributed so that CV = 1.25 U. Taking n = yarn tex ÷ fibre tex and CV_lim = 100 ÷ √n, the index of irregularity I = CV ÷ CV_lim, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.25
n = 20 ÷ 0.16 = 125 fibres in the cross-section, so CV_lim = 100 ÷ √125 = 8.944%. CV = 1.25 × 8.94 = 11.175%. I = 11.175 ÷ 8.944 = 1.249, so 1.25. Using U in place of CV gives I = 1.00, which would wrongly suggest a perfect yarn.