Table of the 19 epacts, according to the Julian and Gregorian account, by the golden number. Golden Julian Greg: Golden Julian Greg: Golden Jalian Greg. Number. Epact. Epact. Number. Epact. Epact. Number. Epact. Epact. 1 29 7 17 0 13 23 12 2 22 11 28 17 14 4 23 3 22 28 15 15 4 14 3 10 20 16 26 15 25 14 11 17 26 6 25 12 18 18 18 19 29 18 A Table of the dominical letters for the new stile, according to the solar cycle, or cycle of the sun. Cycle. 22 IS 2 Dom. Dom. Letter, Letter. E 17 S E DE 24 с E 18 с 25 Ş BAŞ 27 F 14 A G FS | 28 E 5$ 6 7 21 % NOTE. If you want the epact, first find the golden number for the given year, which you will see in the column headed golden number; and on the same line, to the right hand, you will find either the Julian or Gregorian epact. To find the dominical letter for any given year, first find the cycle of the sun for the required year, which you will find in the column headed cycle; and in the next column, towards the right hand, you will see the dominical letters, or letter, answering thereto. The leap-years are marked & under the cycle and dominical letters. To find the length of the day and night by the sun's rising and setting Rule. Multiply the time the sun rises by 2, and you have the length of the night. Multiply the time the sun sets by 2, and you have the length of the day. EXAMPLE. Required the length of the day and night on Sept. 28th, 1827, the sun rising on that day at 7 minutes past 6, and setting at 53 minutes past 5 o'clock ? Rises 6 7x2=12 14 length of the night. Day and night 24 00 hours. Ans. USEFUL TABLES. 28 = 1 gt. Cubic inches contained in gallons, pecks and bushels of dry measure. Cub.irc. gal. 2684 1 pk. Inches contained in links, poles, chains, furlongs and miles of long measure. Inches. link. 1 pole. 7,92 TO yard. Square inches contained in feet, yards, poles, roods, acres and miles of land measure. Sq. inches. feet. 144= 1 1296= 9 1 pole. 40= 1 627 2640= 43560- 4840= 160= 4014489600=2787400=3097600=102400=2560=640=1 acre. 1 m. Cubic inches contained in pints, quarts and gallons of beer aud ale measure. Cub. inc. pt. 351 el gt. 1 2 = OF GAUGING. I shall not here give the whole art of gauging, there being several books of that art already extant, but will shew how to find the quantity of liquor in any cask, upon occasion. How to measure any cask. In order to perform this part of gauging, the three follow ing dimensions of the cask must be truly taken, viz. The bung diameter, the head diameter, and the length of the cask, all within the cask. The bunghole must be in the middle of the cask, and the bungstave, and the stave opposite to the bunghole, both regular and even: the heads of the cask equal and truly circular. If so, the distance between the inside of the chine, to the outside of its opposite stave, will be the head diame. ter within the cask very near. Rule. Multiply the difference between the bung and head diameter by ,67 or ,64 or ,6 according to the greater or less curvature of the staves, between the head and bung. To the product add the head diameter, and the sum will be a mean diameter; then find the contents of the mean diameter in gallons, which multiply by the length of the cask, and the product is the capacity in wine gallons. EXAMPLES. 1. Suppose the bung diameter 32 inches, the head diame. ter 24 inc. and the length 40 inc. required the contents ? Bung diameter 32 -8 diff. which multiplied by ,6 4,8 + 24, the head diameter, equal 28,8, the mean diameter. Now find the area of the mean diameter by dividing its square by 359,05 for ale, or by 294,12 for wine. 28,8 X 28,8=829,44-7-359,05=2,31 gallons of ale. 829,44;-294,12=2,71, area in wine. 2,71 X 40, the length, equal 108,4 gallons wine, the contents. 2. Suppose the bung diameter 20 inches, the head diame. ter 16 inches, and the length 32 inches, required the capacity in wine gallons ? Bung diameter 20 4 diff. then 4 X,6= 2,4 + 16, the head diameter, equal 18,4 X 18,4 = 338,56 294,12 = 1,15, area of the mean diameter. Then 1,15 X 32 = 36,8 gallons, the contents. 222 CONNOLLY'S ARITHMETIC. ULLAGE. To find the ullage of a cask lying with its axis parallel to the horizon. RULE. Divide the dry inches of the bung diameter by the bung diameter, to two decimal places, and multiply the quotient thus found by the correspondent multiplier in the following table, and the product, multiplied by the whole contents of the cask, will give the ullage required. Quot. ,04 ,07 ,09 ,14 ,20 ,24 ,31 ,40 ,51 EXAMPLE. Suppose the bung diameter 20 inches, of which 5 inches are dry, and the whole contents of the cask 36,8 gallons, required the ullage ? 5+20=,25x,7 --,175x36,8-6,44, ullage in gallong. TO MEASURE A STILL. Required the contents of a still, the diameter of the great. est section inside being 25 inches, mouth 6 inches, bottom 18 inches, middle section, between the greatest and mouth, 20 inches, middle section, between the greatest and bottom, 24 inches, depth from mouth to the greatest section, 12 inc. and depth from greatest section to the bottom, 18 inches? Greatest section 25X25= 625 Mouth 6x6= 36 Middle, between great- } 20x20x4=1600 est and mouth 2261 - 6x12=4522 Greatest section 25 x 25= 625 Bottom 18*18= 324 Middle, between greatest and bottom 24 x 24x4=2304;6X18=9759 14281 14281 X ,00344 = 484 gallons wine measure. } |