27 C.F.R. Subpart E—Prescribed Tables


Title 27 - Alcohol, Tobacco Products and Firearms


Title 27: Alcohol, Tobacco and Firearms
PART 30—GAUGING MANUAL

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Subpart E—Prescribed Tables

Note. The tables referred to in this subpart appear in their entirety in the “Gauging Manual Embracing Instructions and Tables for Determining Quantity of Distilled Spirits by Proof and Weight” which is incorporated by reference in this part (see §30.1).

§ 30.61   Table 1, showing the true percent of proof spirit for any indication of the hydrometer at temperatures between zero and 100 degrees Fahrenheit.

This table shows the true percent of proof of distilled spirits for indications of the hydrometer likely to occur in practice at temperatures between zero and 100 degrees Fahrenheit and shall be used in determining the proof of spirits. The left-hand column contains the reading of the hydrometer and on the same horizontal line, in the body of the table, in the “Temperature” column corresponding to the reading of the thermometer is the corrected reading or “true percent of proof.” The table is computed for tenths of a percent.

Example.    

   Temperature, °F............................................       75Hydrometer reading.............................................      193True percent of proof..........................................    189.5 

Where fractional readings are ascertained, the proper interpolations will be made (see §30.23). If the distilled spirits contain dissolved solids, temperature-correction of the hydrometer reading by the use of this table would result in apparent proof rather than true proof.

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended (26 U.S.C. 5204))

§ 30.62   Table 2, showing wine gallons and proof gallons by weight.

The wine and proof gallon content by weight and proof of packages of distilled spirits usually found in actual practice will be ascertained from this table. The left-hand column contains the weights. The true percent of proof is shown on the heading of each page in a range from 90 degrees to 200 degrees. Under the true percent of proof and on the same horizontal line with the weight will be found the wine gallons (at 60 degrees Fahrenheit) and the proof gallons respectively. Where either the weight or the proof of a quantity of spirits is beyond the limitations of this table, the number of proof gallons may be ascertained by reference to Table 3. This table may also be used to ascertain the wine gallons (at 60 degrees Fahrenheit) and proof gallons of spirituous liquor containing dissolved solids where the weight, apparent proof (hydrometer indication corrected to 60 degrees Fahrenheit), and obscuration factor have been determined.

Example.  334 lbs. of distilled spirits.

Apparent proof—96.0°.

Obscuration—0.8°.

True Proof 96.0°+0.8°=96.8°.

334 lbs. at 96.0° apparent proof=42.8 wine gallons.

42.8 wine gallons×96.8°=41.4 proof gallons.

  In addition this table may be used to obtain the wine gallons, at the prevailing temperature, of most liquids within the range of the table, from the weight of the liquid and the uncorrected reading of the hydrometer stem. An application of this would be in determining the capacity of a package.

Example.  It is desired to determine, or to check the rated capacity of a package having a net weight of 395 pounds when completely filled with spirits having an uncorrected hydrometer reading of 113.0°. The full capacity of the package, 51.5 wine gallons, would be found by referring to the table at 395 pounds and 113° proof (hydrometer reading).

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended, 1362, as amended (26 U.S.C. 5204, 5211))

§ 30.63   Table 3, for determining the number of proof gallons from the weight and proof of spirituous liquor.

When the weight or proof of a quantity of distilled spirits is not found in Table 2, the proof gallons may be ascertained from Table 3. The wine gallons (at 60 degrees Fahrenheit) may be ascertained by dividing the proof gallons by the proof.

Example.  A tank car of spirits of 190 degrees of proof weighed 60,378 pounds net. We find—

 ------------------------------------------------------------------------                                                                 Proof                                                                gallons------------------------------------------------------------------------60,000 pounds equal to.......................................   16,778.4300 pounds equal to..........................................       83.970 pounds equal to...........................................       19.68 pounds equal to............................................        2.2                                                              ----------      Total..................................................   16,884.1------------------------------------------------------------------------

That is, the total weight of 60,378 pounds of spirits at 190 proof is equal to 16,884.1 proof gallons. The equivalent gallonage for 70 pounds is found from the column 700 pounds by moving the decimal point one place to the left; that for 8 pounds from the column 800 pounds by moving the decimal point two places to the left.

Example.  A package of spirits at 86 proof weighed 3211/2 pounds net. We find—

 ------------------------------------------------------------------------                                                                 Proof                                                                gallons------------------------------------------------------------------------300 pounds equal to..........................................       32.720 pounds equal to...........................................        2.21 pound equal to.............................................         .1\1/2\ pound equal to.........................................         .1                                                              ----------      Total..................................................       35.1------------------------------------------------------------------------

That is, 3211/2 pounds of spirits at 86 proof is equal to 35.1 proof gallons. The equivalent gallonage for 20 pounds is found from the column 200 pounds by moving the decimal point one place to the left; that for 1 pound from the column 100 pounds by moving the decimal point two places to the left; that for the 1/2 pound from the column 500 pounds by moving the decimal point three places to the left.

Fractional gallons beyond the first decimal ascertained through use of this table will be dropped if less than 0.05 or will be added as 0.1 if 0.05 or more. The wine gallons (at 60 degrees Fahrenheit) may be determined by dividing the proof gallons by the proof. For example: 35.1 divided by 0.86 equals 40.8 wine gallons.

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended (26 U.S.C.5204))

§ 30.64   Table 4, showing the fractional part of a gallon per pound at each percent and each tenth percent of proof of spirituous liquor.

This table provides a method for use in ascertaining the wine gallon (at 60 degrees Fahrenheit) and/or proof gallon contents of containers of spirits by multiplying the net weight of the spirits by the fractional part of a gallon per pound shown in the table for spirits of the same proof. Fractional gallons beyond the first decimal will be dropped if less than 0.05 or will be added as 0.1 if 0.05 or more.

Example.  It is desired to ascertain the wine gallons and proof gallons of a tank of 190 proof spirits weighing 81,000 pounds.

81,000×0.14718=11,921.58=11,921.6 wine gallons.

81,000×0.27964=22,650.84=22,650.8 proof gallons.

This table may also be used for ascertaining the quantity of water required to reduce to a given proof. To do this, divide the proof gallons of spirits to be reduced by the fractional part of a proof gallon per pound of spirits at the proof to which the spirits are to be reduced, and subtract from the quotient the net weight of the spirits before reduction. The remainder will be the pounds of water needed to reduce the spirits to the desired proof.

Example.  It is desired to ascertain the quantity of water needed to reduce 1,000 pounds of 200 proof spirits, 302.58 proof gallons, to 190 proof:

302.58 divided by 0.27964 equals 1,082.03 pounds, weight of spirits after reduction.

1.082.03 minus 1,000 equals 82.03 pounds, weight of water required to reduce to desired proof.

The slight variation between this table and Tables 2, 3, and 5 on some calculations is due to the dropping or adding of fractions beyond the first decimal in those tables. This table may also be used to determine the wine gallons (at 60 degrees Fahrenheit) of distilled spirits containing dissolved solids from the total weight of the liquid and its apparent proof (hydrometer indication, corrected to 60 degrees Fahrenheit). The proof gallons may then be found by multiplying the wine gallons by the true proof.

Example.  5,350 pounds of blended whisky containing added solidsTemperature °F75.0°Hydrometer reading92.0°Apparent proof85.5°Obscuration0.5°True proof86.0°

5,350.0 lbs.×0.12676 (W.G. per pound factor for apparent proof of 85.5°)=678.2 wine gallons

678.2 W.G.×0.86=583.3 proof gallons

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended 1362, as amended (26 U.S.C. 5204, 5211))

§ 30.65   Table 5, showing the weight per wine gallon (at 60 degrees Fahrenheit) and proof gallon at each percent of proof of spirituous liquor.

This table may be used to ascertain the weight of any given number of wine gallons (at 60 degrees Fahrenheit) or proof gallons of spirits by multiplying the pounds per gallon by the given number of gallons of the spirits. The table should be especially useful where it is desired to weigh a precise quantity of spirits.

Example.  It is desired to ascertain the weight of 100 wine gallons of 190 proof spirits:

6.79434×100 equals 679.43 pounds, net weight of 100 wine gallons of 190 proofs spirits.

Example.  It is desired to ascertain the weight of 100 proof gallons of 190 proof spirits.

3.57597×100 equals 357.60 pounds, net weight of 100 proof gallons of 190 proof spirits.

The slight variation between this table and Tables 2 and 3 on some calculations is due to dropping or adding of fractions beyond the first decimal on those tables. This table also shows the weight per wine gallon (at the prevailing temperature) corresponding to each uncorrected reading of a proof hydrometer.

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended (26 U.S.C. 5204))

§ 30.66   Table 6, showing respective volumes of alcohol and water and the specific gravity in both air and vacuum of spirituous liquor.

This table provides an alternate method for use in ascertaining the quantity of water needed to reduce the strength of distilled spirits by a definite amount. To do this, divide the alcohol in the given strength by the alcohol in the required strength, multiply the quotient by the water in the required strength, and subtract the water in the given strength from the product. The remainder is the number of gallons of water to be added to 100 gallons of spirits of the given strength to produce a spirit of a required strength.

Example.  It is desired to reduce spirits of 191 proof to 188 proof. We find that 191 proof spirits contains 95.5 parts alcohol and 5.59 parts water, and 188 proof spirits contains 94.0 parts alcohol and 7.36 parts water.

95.5 (the strength of 100 wine gallons of spirits at 191 proof) divided by 94.0 (the strength of 100 wine gallons of spirits at 188 proof) equals 1.01.

7.36 (the water in 188 proof) multiplied by 1.01 equals 7.43.

7.43 less 5.59 (the water in 191 proof spirits) equal 1.84 gallons of water to be added to each 100 wine gallons of 191 proof spirits to be reduced.

This rule is applicable for reducing to any proof; but when it is desired to reduce to 100 proof, it is sufficient to point off two decimals in the given proof, multiply by 53.73, and deduct the water in the given strength. Thus, to reduce 112 proof spirits to 100 proof:

1.12×53.73–47.75 equals 12.42 gallons of water to be added to each 100 wine gallons of spirits to be reduced.

This table may also be used to obtain the proof gallonage of spirituous liquor according to weight and percent of proof.

Example.  It is desired to determine the number of gallons in 400 pounds of spirits of 141 percent of proof. Multiply the weight of one gallon of water in air by the specific gravity in air of the spirits—8.32823 by 0.88862—the product (7.40063) divided into 400 gives 54.049 wine gallons, which rounded to the nearest hundredth is 54.05 and multiplied by 1.41 gives 76.2 proof gallons. In rounding off where the decimal is less than five, it will be dropped; if it is five or over a unit will be added.

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended (26 U.S.C. 5204))

§ 30.67   Table 7, for correction of volume of spirituous liquors to 60 degrees Fahrenheit.

This table is prescribed for use in correcting spirits to volume at 60 degrees Fahrenheit. To do this, multiply the wine gallons of spirits which it is desired to correct to volume at 60 degrees Fahrenheit by the factor shown in the table at the percent of proof and temperature of the spirits. The product will be the corrected gallonage at 60 degrees Fahrenheit. This table is also prescribed for use in ascertaining the true capacity of containers where the wine gallon contents at 60 degrees Fahrenheit have been determined by weight in accordance with Tables 2, 3, 4, or 5. This is accomplished by dividing the wine gallons at 60 degrees Fahrenheit by the factor shown in the table at the percent of proof and temperature of the spirits. The quotient will be the true capacity of the container.

Example.  It is desired to ascertain the volume at 60 degrees Fahrenheit of 1,000 wine gallons of 190 proof spirits at 76 degrees Fahrenheit:

1,000×0.991 equals 991 wine gallons, the corrected gallonage at 60 degrees Fahrenheit.

Example.  It is desired to ascertain the capacity of a container of 190 proof spirits at 76 degrees Fahrenheit, shown by Table 2 to contain 55.1 wine gallons at 60 degrees Fahrenheit:

55.1 divided by 0.991 equals 55.6 wine gallons, the true capacity of the container when filled with spirits of 60 degrees temperature.

It will be noted the table is prepared in multiples of 5 percent of proof and 2 degrees temperature. Where the spirits to be corrected are of an odd temperature, one-half of the difference, if any, between the factors for the next higher and lower temperature, should be added to the factor for the next higher temperature.

Example.  It is desired to correct spirits of 180 proof at 51 degrees temperature:

1.006 (50°)−1.005 (52°)=0.001 divided by 2=0.0005

0.0005+1.005=1.0055 correction factor at 51 °F.

Example.  It is desired to correct spirits of 180 proof at 53 degrees temperature:

1.005 (52°)−1.003 (54°)=0.002 divided by 2=0.001

0.001+1.003=1.004 correction factor at 53 °F.

Where the percent of proof is other than a multiple of five, the difference, if any, between the factors for the next higher and lower proofs should be divided by five and multiplied by the degrees of proof beyond the next lower proof, and the fractional product so obtained should be added to the factor for the next lower proof (if the temperature is above 60 degrees Fahrenheit, the fractional product so obtained must be subtracted from the factor for next lower proof), or if it is also necessary to correct the factor because of odd temperature, to the temperature corrected factor for the next lower proof.

Example.  It is desired to ascertain the correction factor for spirits of 112 proof at 47 degrees temperature:

1.006 (46°)−1.005 (48°)=0.001 divided by 2=0.0005

0.0005+1.005=1.0055 corrected factor at 47 °F.

1.007 (115 proof)−1.006 (110 proof)=0.001

0.001 divided by 5=0.0002 (for each percent of proof)×2 (for 112 proof)=0.0001

0.0004=1.0055 (corrected factor at 47 °F.)=1.0059 correction factor to be used for 112 proof at 47 °F

Example.  It is desired to ascertain the correction factor for spirits of 97 proof at 93 degrees temperature:

0.986 (92°)−0.985 (94°)=0.001 divided by 2=0.0005

0.0005+0.985=0.9855 corrected factor at 93 °F.

0.986 (95 proof)−0.985 (100 proof)=0.001

0.001 divided by 5=0.0002 (for each percent of proof)×2 (for 97 proof)=0.0004

0.9855 (corrected factor at 93 °F.)=0.0005=0.9851 correction factor to be used for 97 proof at 93 °F.

(Sec. 201, Pub. L. 85–859, 72 Stat. 1358, as amended (26 U.S.C. 5204))

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