Example: (1.) To reduce 832 galls. of spirits at 32.4 per cent. O. P. to an equivalent quantity at 28 per cent. O. P. and 28 per cent. U. P. respectively. Multiplier for given strength 132.4 A proof of the correctness of these results may be thus exhibited; 832 galls. at 32-4 per cent. O. P. are equivalent to 1101 5 galls. at proof, for 832 × 1·324=1101.5; and 860 6 galls. at 28 per cent. O. P. are also equivalent to 1101 5 galls. at proof, for 860.6 x 1.28=1101.5. Similarly, 1529.9 galls. at 28 per cent. U. P. are equivalent to 1101 5 proof galls. for 1529.9 x 72=1101.5. As to the reason of the preceding general rule for reducing spirits from any one strength to an equivalent at any other, it will be obvious that when the given quantity is multiplied by the factor for the given strength, it is thus reduced to its equivalent at proof, and in order to find what quantity at the required strength corresponds to these gallons at proof, our process must be the reverse of that which is used in reducing to proof, that is, we must divide the proof quantity by the factor for the required strength.* Such a number must be obtained as will, when multiplied by the factor for the required strength, give that proof quantity; so that in the latter part of the operation we have to solve the question—Given a quantity at proof, to find its equivalent at a certain specified strength higher or lower than proof, and the rule for this is evidently the converse of that for reducing spirits at a stated strength to proof. It will be often of advantage in reductions of this kind when a division is involved, to place the numbers in the form of a fraction, and to cancel every factor obviously common to numerator and denominator as a preparatory step: thus in the first of the examples above given, *It is unnecessary in this or any similar case to divide the sum or difference of the degrees of strength and 100 by 100, as in the reduction of spirits to proof, although to do so would give no trouble beyond the shifting of the decimal point. Example: (2.) What is the equivalent quantity at 5.8 per cent. U. P., of 765 gallons of spirits at 11.5 per cent. U. P.? Example: (3.) If an officer has occasion to seize 17.5 proof gallons of spirits from a rectifier's stock, what bulk quantity should he take as an equivalent from spirits at 36 6 per cent. O. P. ? This question amounts to asking, what quantity at 36.6 per cent. O. P. will give 17.5 proof gallons, or in other words, what number multiplied by 1.366 will produce 17.5. The answer is evidently the quotient of 17·5 divided by 1·366. 1-366)17.5 (12 8 Galls. Answer. 3840 152 To find the quantity of water requisite for reducing spirits from any given strength, to any other lower strength. GENERAL RULE.-Multiply the given quantity by the difference between the higher and lower strengths, and divide the product by the lower strength-first having added rates per cent. O. P. to 100, and subtracted rates per cent. U. P. from 100. Example (1.) How much water is needed to reduce 154 gallons of spirits from 24.8 per cent. O. P. to 11.6 per cent., O. P. proof; and 31 per cent. U. P. respectively? When the higher and lower strengths are both on the same side of proof, that is, both O. P. or U. P. there is no occasion in taking their difference, to add them to or subtract them from 100, since, for example, 124-8-111-6 is the same as 24-8-11-6, but the lower strength when used as a divisor must always be the sum or difference of the rate per cent. and 100. The principle of the foregoing rule and operations is simply this: As 100 gallons at 24-8 per cent. O. P. are equal in value to 124-8 gallons at proof (see page 98,) it follows, that if the bulk of 100 gallons at 24.8 O. P., be made up with water, syrup, or any other liquid devoid of spirit, to 124-8 gallons, the mixture will have the strength of proof,-it will contain all the original spirit diluted to the extent of 124-8 gallons, and where the strength was 24-8 O. P. it will now be only Proof. But the difference between the present and the previous bulk-24-8 gallons-does not represent the whole of the water necessary to effect the reduction of the spirit in question to proof strength. When spirits and water are mixed, a condensation or lessening of bulk takes place, that is, the bulk of the mixture is always less than the sum of the bulks of the ingredients when separate, and the amount of this condensation can be determined only by experiment for each varying proportion of spirits and water.* All that may properly be inferred, therefore, from any expression of strength, such as 24-8 per cent. O. P., with regard to the quantity of water, &c., required to reduce a spirit of that strength to proof, is that 100 gallons at 24.8 per cent. O. P., if increased in bulk by the addition of water, &c., to 124-8 gallons will then have the strength of proof; or generally, that the water which must be used to effect a given reduction of strength exceeds the quantity assigned by calculation. In the case of a spirit having the strength of 24-8 per cent. O. P., about 26 gallons of water are really required to lower 100 gallons to proof, the condensation which occurs diminishing the bulk of the whole from 126 to 124.8 gallons. It is customary, however, in the business of reduction, to add water in the proportion suggested by the strength of the spirit, and if an accurate result be desired, to dilute the mixture a little further, until a sample tested by the hydrometer shows that the right degree of strength has been reached; but for the ordinary purposes of trade a simple arithmetical process, such as is here exemplified, with a small practical allowance for condensa. tion, is all that is adopted.+ As 100 gallons of spirits at 24 8 per cent. O. P. need the addition of 24 8 (calculated) gallons of water, or as 1 gallon such spirit needs 248 gallons of water, to render the whole of proof strength, 154 gallons of the spirit require evidently, 154x248=38.2 gallons of water, the result arrived at in the second of the examples immediately preceding. The reason for taking the difference between the numbers for the higher and lower strengths, and for the rest of the rule given on page (102), may be easily understood from the process of finding the quantities of spirits equivalent to each other at different strengths, as exhibited on page (100.) It is there seen, that 832 gallons of spirits at 32-4 per cent. O. P. correspond in value to 860-6 gallons at 28 per cent. O. P. If we subtract 832 from 860-6 we have the computed quantity of water necessary to reduce the spirits from the higher to the A Table showing the degree of condensation of all spirits within the range of Sykes's Hydrometer, will be found in the Appendix. + The subject of the strength of spirits will be more fully treated of in the chapter on "Specific Gravity." lower strength. But this quantity is more readily obtained by the operation directed at page (102,) for, and the latter is the formula on which the rule is founded. By adding to the water thus found, the original quantity of spirits, we have the equivalent at 28 per cent O. P., as otherwise determined on page (101.) Example (2.) How much water will reduce 200 gallons of spirits at proof to 22 per cent. U. P. ? Example (3.) A rectifier has 345 gallons of spirits of the strength of 66 per cent. U. P., which he desires to raise to 45 per cent. O. P. by redistillation: how much spirit of the latter strength will be produced, and what quantity of water will remain after the operation? In this example it is supposed, that the condensing power of the still can be regulated so as to yield a spirit of the desired strength, in which case nothing but water will be left in the still when 80.9 gallons have distilled over. If a spirit stronger than 45 per cent. O. P. were produced, less water would distil over with the alcohol, and a proportionally greater quantity than 264.1 gallons of spiritless liquid would remain as "spent lees." Reduction of Worts.-Wort or worts, is a general name given to any saccharine liquid, such as an infusion of malt, a solution of cane sugar, &c., from which beer, spirits, &c., is prepared. Wort of every description, when unfermented, is heavier, bulk for bulk, than water, and the greater its relative weight, the more solid matter is there dissolved in it, and the more valuable is it to the brewer, &c. In the practice of the Excise, the strength or saccharine richness of wort estimated and expressed in « degrees of gravity," as determined by the action of an instrument called the Saccharometer, each of these degrees being legally equivalent to the one-thousandth part of the density of water assumed as 1 at the temperature of 60° Fah. At that temperature a gallon of pure water weighs almost exactly 10lbs., and as the one-thousandth of 10lbs. is 01lb. it follows that a wort of one degree of gravity weighs 10-01lbs. per gallon at 60° Fah. ; a wort of two degrees gravity weighs 10-02lbs. per gallon, and so on. For Questions in the reduction of worts from one gravity to another, are solved by the same method as questions in the reduction of spirits, but as water itself has no saccharine value, it is in this respect considered to have no gravity, and may, therefore, be numerically represented as 0. The strength of spirits, on the other hand, is estimated with reference to a standard spirit called proof, which possesses a given alcoholic value denoted by the number 100. A spirit stronger than proof contains all the alcohol of the standard spirit and a certain quantity more; a spirit weaker than proof contains so much alcohol less than that of the standard. this reason, when spirits of different strengths are reduced to proof, or compared together, it is necessary to add degrees O. P. to 100, and subtract degrees U. P. from 100, or to perform a process which has the same effect. (See pages 98, 99.) In the case of worts, water as above stated, having no numerical value, there is nothing to add to the degrees of gravity before they can properly be compared with one another, and accordingly, every question of reduction is here solved on the principle, that the values of different worts are in the direct ratio of the numbers ezpressing their gravities-that a wort of 20° gravity has twice the value of a wort of 10° gravity, &c. The rule in every instance is, multiply the quantity of worts by the given gravity, and divide the product by the required gravity. An example or two will be sufficient to illustrate the general mode of operation. Example (1.) Reduce 1000 gallons of worts at 65° gravity to an equivalent 45°. The meaning of this result is, that 1444-4 gallons at 45° gravity, are equal in value to 1000 gallons at a gravity of 65°, or simply that 1444.4 × 45=1000 × 65. The effect of the condensation or lessening of bulk which takes place when worts of different gravities are mixed, is neglected in these questions, as of no practical moment. Example (2.) Reduce 3588 gallons of worts from 34° to 51° gravity. 3588 x 34 51 = 1196 × 2=2392 gallons. Answer. Example (3.) What is the quantity of water required to reduce 567 gallons of worts from 93° to 63° gravity ? The principle of these operations is the same as that which governs the corresponding operations in the reduction of spirits. (See the last article.) 5. PROPORTION.-The term ratio in arithmetic is usually defined to be, the measure of the relation which exists between two numbers, or between two quantities H |