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Commission Regulation (EEC) No 000/90 of 17 September 1990 determining Community methods for the analysis of wines (repealed)
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The chromatic properties of a wine are defined as its luminosity and its chromaticity.
The luminosity is represented by the transmittance and it varies inversely with the colour intensity of the wine.
The chromaticity is represented by the dominant wavelength (which characterizes the tint) an the purity of the colour.
By convention, and for reasons of convenience, the chromatic properties of red and rosé wines are given as the colour intensity and the tint, in keeping with a procedure adopted as the usual method.
This is a spectrophotometric method which makes it possible to determine the tristimulus values and three chromaticity coordinates necessary for the specification of the colour as laid down by the International Commission on Illumination (CIE).
This is a spectrophotometric method by which the chromatic properties are expressed by convention as follows:
The colour intensity is given by the sum of the absorbences at wavelengths af 420, 520 and 620 nm for radiation traversing a 1 cm optical path in the sample.
The tint is expressed by the ratio of the absorbences at 420 nm and 520 nm.
Cloudy wine must be clarified by centrifugation. The bulk of the carbon dioxide in young and sparkling wines must be removed by shaking under vacuum.
The optical path, b, in the glass cell should be so chosen that the measured absorbence lies between 0,3 and 0,7.
The following guidance is given for the appropriate choice of the optical path: use cells of 2 (or 4) cm optical path for white wines, with 1 cm for rosé wines and with 0,1 cm (or 0,2 cm) for red wines.
The spectrophotometric measurements should be made using distilled water, in a cell with the same optical path, b, as reference liquid to establish the zero of the absorbence scale at wavelengths 445, 495, 550 and 625 nm.
The four corresponding absorbences for the wine should then be measured to three decimal places for the optical path, b. Let these be A 445, A 495, A 550, A 625.
Together with Table I, use these values of the absorbences for the optical path, b cm to obtain the corresponding transmittances (T%). Let these be T 445, T 495, T 550 and T 625.
Calculate the tristimulus values X, Y and Z expressed as decimal fractions from the following expressions:
X = 0,42 T 625 + 0,35 T 550 + 0,21 T 445
Y = 0,20 T 625 + 0,63 T 550 + 0,17 T 495
Z = 0,24 T 495 + 0,94 T 445
Calculate the chromaticity coordinates x and y from:
The determination of these two quantities makes use of the chromaticity diagram bounded by the spectral locus as given in Figure 1. The point O plotted in this diagram represents the white light source used and has the coordinates of a standard source, C, xo = 0,3101 and yo = 0,3163, representing daylight of average brightness.
Dominant wavelength
Plot the point C with coordinates x, y on the chromaticity diagram.
If C is outside the triangle AOB, draw the straight line joining O to C and extend it to cut the spectral locus at the point S, which corresponds to the dominant wavelength.
If C is inside the triangle AOB, draw the straight line from C to O and extend it to intersect the spectral locus at a point corresponding to the wavelength of the colour complementary to that of the wine. This wavelength is denoted by its value followed by the letter C.
Purity
If the point C is outside the triangle AOB, the purity is given as a percentage by the ratio:
If the point C is inside the triangle AOB, the purity is given as a percentage by the ratio:
(1)
where P is the point where the straight line OC cuts the line of purples (line AB).
Purity is also given directly by chromaticity diagrams from the known values of x and y (Figures 2, 3, 4, 5 and 6).
The colour of a wine is completely defined by its luminosity, its chromaticity (expressed by the dominant wavelength) and its purity.
These should be indicated in the analysis report with the value of the optical path in which the measurements were made.
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Textual Amendments
Find the first decimal figure of the absorbence in the first vertical column, and call its row R. Find the second decimal figure of the absorbence in the top horizontal row and call its column C. Read the figure in the box at the intersection of the row R and the column C. To calculate the transmittance, divide this figure by 10 if the absorbence is less than 1, by 100 if it lies between 1 and 2, and by 1 000 if it lies between 2 and 3.
The figure in the top right-hand corner of each box enables the third decimal figure of the absorbence to be taken into account by interpolation.U.K.
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
23 | 22 | 22 | 21 | 21 | 20 | 20 | 19 | 19 | 19 | |||||||||||
0 | 1000 | 977 | 955 | 933 | 912 | 891 | 871 | 851 | 832 | 813 | ||||||||||
18 | 18 | 17 | 17 | 16 | 16 | 16 | 15 | 15 | 15 | |||||||||||
1 | 794 | 776 | 759 | 741 | 724 | 708 | 692 | 676 | 661 | 646 | ||||||||||
14 | 14 | 14 | 14 | 13 | 13 | 13 | 12 | 12 | 12 | |||||||||||
2 | 631 | 617 | 603 | 589 | 575 | 562 | 549 | 537 | 525 | 513 | ||||||||||
11 | 11 | 11 | 11 | 10 | 9 | 9 | 10 | 10 | 9 | |||||||||||
3 | 501 | 490 | 479 | 468 | 457 | 447 | 436 | 427 | 417 | 407 | ||||||||||
9 | 9 | 9 | 8 | 8 | 8 | 8 | 8 | 7 | 8 | |||||||||||
4 | 398 | 389 | 380 | 371 | 363 | 355 | 347 | 339 | 331 | 324 | ||||||||||
7 | 7 | 7 | 7 | 6 | 7 | 6 | 6 | 6 | 6 | |||||||||||
5 | 316 | 309 | 302 | 295 | 288 | 282 | 275 | 269 | 263 | 257 | ||||||||||
6 | 5 | 6 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | |||||||||||
6 | 251 | 245 | 240 | 234 | 229 | 224 | 219 | 214 | 209 | 204 | ||||||||||
4 | 5 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | |||||||||||
7 | 199 | 195 | 190 | 186 | 182 | 178 | 174 | 170 | 166 | 162 | ||||||||||
3 | 4 | 3 | 4 | 4 | 3 | 3 | 3 | 3 | 3 | |||||||||||
8 | 158 | 155 | 151 | 148 | 144 | 141 | 138 | 135 | 132 | 129 | ||||||||||
3 | 3 | 3 | 2 | 3 | 2 | 3 | 2 | 3 | 2 | |||||||||||
9 | 126 | 123 | 120 | 117 | 115 | 112 | 110 | 107 | 105 | 102 |
Absorbence | 0,47 | 1,47 | 2,47 | 3,47 |
T % | 33,9 | 3,4 | 0,3 | 0 |
Transmittances T % are to be expressed to the nearest 0,1 %.
The absorbence of the rectified concentrated must is measured at 425 nm through a thickness of 1 cm after dilution to bring the sugar concentration to 25 % (m/m) (25° Brix).
Use the solution with a sugar concentration of 25 % (m/m) (25° Brix) prepared as described in the chapter ‘pH’, section 4.1.2. Filter is through a membrane filter of pore diameter 0,45 µm.
Zero the absorbence scale at a wavelength of 425 nm using a cell with an optical path of 1 cm containing distilled water.
Measure the absorbence A at the same wavelength of the solution containing 25 % sugar (25° Brix) prepared as in 2.3.1 and placed in a cell with an optical path of 1 cm.
The absorbence at 425 nm of the rectified concentrated must in a solution with 25 % sugar (25° Brix) is quoted to two decimal places.
This distance must be given in a direction going from O to C.
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