10 milligrams (mg)= 1 centigram (cg). Observe, in the case of water, that 1ml (=1ccm) weighs 18; 11 (=1cdm) weighs 1Kg; 1K (=1cum) weighs 17. 155. Kilogram is called Kilo. Quintal is not often used. The cubic centimeter of water, which is used as the standard unit, must be distilled, must be at a temperature of 39.2° F. (4° C.), and must be weighed in a vacuum at the level of the sea. EXAMPLES LII. Written Exercises. 1. Read 64.958 as dg, cg, mg, and Mg. 2. Read 1256" as Kg, Q, T, and g. 3. What is the weight of 1ml of standard water? Of 10 ml? Of 1cl? Of 10cl? Of 3d? Of 31? Of 1000 ccm? 4. Iron is 7.8 times as heavy as water; what is the volume (in cdm) of 29.25 Kg? What is the weight of 2 cdm? Of 55 ccm? Of 7.2 ccm? Of 1.67 ccm? Of 125 cum? 5. Find the value (in grams) of 4Kg 18 dg +188 + 67.896 mg 126.73% +4T — 11.6 Mg. - 6. Gold is 19.5 times as heavy as water; what is the weight of 1ccm? Of one cubic meter? CHAPTER VI. NON-DECIMAL MEASURES. 156. The simplicity of calculations when using decimal measures is due to the facts that changes can be easily made from one denomination to another by moving the decimal point, and that several denominations can be expressed together in one set of figures. In Non-Decimal measures, called also Denominate numbers and Compound Quantities, a variety of divisors is used in the different tables in order to change from low denominations to higher ones; also, it is unusual to express several denominations together in one set of figures. For example, consider the case of the string mentioned in Art. 138. There, 12 inches equal 1 foot, and 3 feet equal 1 yard; and the length of the string must be expressed, not with the denominations together in one set of figures, but each denomination separately, — 6 yards, 1 foot, 6 inches. To express a compound (Art. 139) quantity, express the number of units of each denomination separately, indicating the denominations, as in the above illustration. To read compound quantities, read them exactly as expressed. 157. Table of Measures of Time. The Standard Unit of Time is the Mean Solar Day; that is, the mean interval between two successive passages of the sun across the meridian of any place. A day is supposed to begin at midnight. The year is divided into 12 months, called Calendar Months, which contain an unequal number of days, namely: January 31, February 28, March 31, April 30, May 31, June 30, July 31, August 31, September 30, October 31, November 30, and December 31. Every fourth year contains 366 days, and is called Leap Year, and in these years February has 29 days. It is a Leap Year when the number of the year is exactly divisible by 4; thus, 1896 we a Leap Year. The Solar Year contains 365 da. 5 hr. 48 min. 46 sec., very nearly. Now it would clearly be very inconvenient to reckon by years which did not contain an exact number of days; hence, as the Solar Year contains very nearly 365 days, we have 3 years (called Civil Years) of 365 days each, and then one year of 366 days. The Solar Year is, however, somewhat less than 365 days, and the necessary correction is made by omitting three Leap Years in every 400 years, the years which are not counted as Leap Years (although divisible by 4) are the years which end the Centuries, and are such that the number of the Century is not divisible by 4. Thus, 1800 was not a Leap Year, and 1900 will not be a Leap Year; the year 2000 will, however, be a Leap Year. 158.* Reduction of Compound Quantities. The method by which a compound quantity can be expressed as a simple quantity will be seen from the following example. * The methods of reductions of compound quantities, also addition, etc., will be illustrated by the use of the above table because the different units are familiar to all. Ex. Reduce 7 da. 3 hr. 12 min. 26 sec. to seconds. 10260 12 Adding the 3 hr., 7 da. 3 hr. 171 hr. 10272 min. 60 Adding 26 sec., Adding 12 min., 171 hr. 12 min. = 10272 min. = - 616320 sec. 616320 26 616346 sec. 159. To reduce a Simple Quantity to a Compound Quantity. Ex. Reduce 14678 sec. to hr., min., and sec. 6,0)1467,8 sec. 4 hr. 4 min. 38 sec. Since 60 sec. make 1 min., if we divide the number of sec. by 60, we shall obtain the number of min. equivalent to 14678 sec., i.e., 244 min., but shall have 38 sec. over. We then divide the number of min. by 60 and obtain the number of hours with 4 min. over. 160. Addition, Subtraction, Multiplication, and Division of Compound Quantities. It will be seen that no new principle is involved. Care, however, must always be taken in regard to the number of units of one denomination required to make one unit of the next higher. (a) Compound Addition [see Art. 142]. Ex. Find the sum of 14 da. 41 min. 11 sec., 121 da. 18 hr. 16 min. 29 sec., 201 da. 13 hr. 4 sec., and 11 hr. 23 min. 30 sec. Here the sum of the seconds equals 74 = 1 min. 14 sec.; write the 14 and carry the 1. The number of min. = 81 = 1 hr. 21 min.; write the 21 and carry the 1. The number of hr. equals 43 = 1 da. 19 hr.; write the 19 and carry the 1. The number of days = 337. (b) Compound Subtraction. Ex. From 16 da. 12 min. and 50 sec. subtract 4 da. 12 hr. 13 min. and 54 sec. Here 54 cannot be subtracted from 50; therefore we take 1 min. from the 12 min., change it to sec., and we have with the 50 sec. 110 sec. in all; subtract 54 sec. from 110 sec., and we have 56 sec. remainder. Now 13 from 11 we cannot take, therefore we take 1 hr. from the next column and proceed as before. (c) Compound Multiplication. CASE I. When the multiplier is not greater than 12. Ex. Multiply 9 da. 10 hr. 31 min. 14 sec. by 7. Here 14 sec. x 7 98 sec. = 1 min. 38 sec.; carry 1. 31 min. × 7 = 217 min.; 217 min. + write the 38 and 1 min. = 218 min. = 10 hr. x 770 hr.; 3 hr. 38 min. ; write the 38 and carry the 3. 70 hr. + 3 hr. = 73 hr. = 3 da. 1 hr.; write the 1 and carry the 3. Finally, 9 da. × 7 = 63 da.; 63 da. + 3 da. 1 hr. 38 min. 38 sec. = 66 da. Ans. = : 66 da. CASE II. When the multiplier can be seen to be the product of factors each not greater than 12. Ex. Multiply 9 da. 10 hr. 31 min. 14 sec. by 35. |