Since there are three simples, there are three unknown quantities, any two of which may be assumed at pleasure, and the value of the third deduced from equation (3). If, now, we divide 200 by 10, the whole these restrictions greatly limit the generality number of ounces in the mixture, we shall of the problem. find 20 carats for the quality of the mixture. The principle in the last example is in no wise different from that in the former, the apparent difference lying entirely in the language employed in stating the proposition. We may in this example regard 24 as the value of pure gold per ounce; then 22 and 17 will be the respective values of each specimen mixed, and we shall find, as before, 20 for the value of an ounce of the mixture, that is, an ounce will contain ths of pure gold. We may then write this rule for solving all questions in alligation medial : RULE.-Multiply the price or quality of a unit of each simple by the number of such units; take the sum of their products, and divide it by the whole number of units; the quotient will be the price or quality of a unit of the mixture. ALLIGATION ALTERNATE, as may be seen from the definition, gives rise to the solution of an indeterminate problem in Algebra. According as fewer or more restrictions are imposed, the solutions will be more or less numerous. There will be several cases. We shall first discuss the general one, in which it is required to find the amount of each simple of known value, which must be mixed so that each unit of the mixture shall have a given value. If there are n simples, the equation of condition will contain n unknown quantities, and (n - 1) of them may be assumed at pleasure. In order to deduce a practical rule for solving questions in alligation alternate, let us begin with the case where there are but two simples. Denote the price or quality of a unit of the mixture by a; let a + b and a — c - c be the respective values of a unit of each simple, and let x and y denote, as before, the number of units of these simples that are taken. We shall have, as before, (a + b) x + (a− c) y = a (x + y) (1). Whence, by reduction, we find ō'a relation which shows that any two values of y and x which are to each From a consideration of the notation em Let there be three simples of the respective values of a, b, and c; let x, y, and z denote ployed, it appears that b is the excess of the the number of units taken from the respect-of a unit of the mixture, and c is the excess value of a unit of the first simple over that ive simples to form m units of the mixture; and let d denote the price or quality of a unit of the mixture. Then, from the conditions of the question, we shall have of value of a unit of the mixture over that of a unit of the second simple. The above discussion indicates the following rule, when there are but two simples: Write down the values of a unit of each simple beginning with the greatest, and link them together by a bracket; write on their left the value of a unit of the mixture; subtract the last value from the first value given, and set the difference opposite the second; subtract the second value from the last, and set the difference opposite the first: these differences, or any equi-multiples of them, will be answers to the question proposed. 1. Required the number of bushels of oats at 50 cents per bushel, and of wheat at 120 cents per bushel, that must be mixed, so that The result may be easily verified: Taking 5 4 25 bushels of wheat at 120 cents, gives 3000 cents, and 45 bushels of oats at 50 cents, 2250 cents. Hence we see that 70 bushels of the mixture is worth 5250 cents, and one bushel 75 cents, as was required. By an analagous train of reasoning, we may, when there is any number of simples of different prices, establish the following general RULE. Write down the prices of the simples under each other, and the price of the mixture at the left-hand; link the prices of the simples two and two, so that each price greater than that of the mixture may be linked with one less than it, and the reverse. ...3lb. 3×2= 6 3d. Method of Linking. 13 )65(5. Verification: Ans. 1+2...3lb. 3x8=24 3 ..3lb. 3×6=18 And so on, many other ways of linking can readily be conceived, and the number of ways becomes greater as the number of simples is increased. The following are cases that may arise: 1. When the amount of one simple is given. Solve the general problem by the rule already given; divide the amount of the given simple by the amount opposite to its price found by the rule; multiply the amount opposite the price of each of the other simples by this ratio and the products will be the respective amounts required. Subtract the price of the mixture from each greater price of the simples, and write the difference opposite the price or prices with which it is linked; subtract each less price of the For example, in the last case, solved by the simples from that of the mixture, and write the first method of linking, let it be required to difference opposite the price or prices with which form a mixture which shall contain 4 pounds it is linked; then, the number or sum of the num- of tea at 8s. The ratio found is ; there bers written opposite each price, will express will, therefore, be the amount of that simple which is to be taken. Any equi-multiples of these numbers will also satisfy the conditions of the problem. It is to be observed that the quantities may be linked in many different ways, but the answers in all cases will be true. 1. Required the number of pounds of tea of the respective values of 2s., 38., 4s., 6s., and 88. per pound, which must be taken so that the mixture may be worth 5s. Verification 4 X 8=32 4lbs. at 8s. at 68. 4 × 6=24 2. When the amount of the mixture is given. Solve the general problem as before, and take the sum of the results; divide the amount of the mixture given by this amount, and multiply the amount opposite the price of each simple by the ratio found; the products will be the respective amounts required. For example, take the case already considered, and let it be required to form a mixture )55(5. of 65 pounds. Then the ratio will be 65 Ꮐ A result which may be verified as before. It is evident, from a review of the preceding GHD, are alternate exterior angles. discussion, that we may assume the amount The angles IFB and CHG, also AFI and of all the simples except one, and the amount ALTERNATE PROPORTION. Quantities are in of that one can then be found from the equa-proportion alternately or by alternation, when tion of condition. If the amount so found is antecedent is compared with antecedent and positive, the answer will be true; if it is consequent with consequent. Thus, negative, we must vary our assumption till a positive result is found. a : b :: C: d. if we change the order of the terms so as to read a : C :: b : d, Many other problems than those already discussed may arise; in fact their number is infinite, but an attentive consideration of the principles discussed will readily present the comparison is said to be made by alterthe proper mode of procedure for their solution. AL'MA-CAN-TAR. See Almucantar. AL'MA-GEST. A collection of problems in astronomy and geometry, drawn up by Ptolemy. The same name has been given to other works of a like kind AL'MU-CAN-TAR [Arabic]. A circle of the celestial sphere, whose plane is parallel to the horizon. Since every point of an almucantar has the same altitude, it is often called a circle of equal altitude. AL'MU-CAN-TAR STAFF. An instrument having an arc of about 15°, used for observing the sun or a star when near the horizon, to find the amplitude or the variation of the needle. nation. AL'TERN-A'TION. Sometimes used in Algebra and Arithmetic for permutation, to express the changes in the order of the quantities considered. AL-TIM'E-TER. [L. altus, high, and Gr. HETрOV, measure]. An instrument for measuring altitudes, as a quadrant, sextant, or theodolite. AL-TIM'E-TRY. The art of measuring altitudes by means of an altimeter, and the application of geometrical principles. ALTI-TUDE. [L. altus, high]. The third dimension of a body, or its height. ALTITUDE OF A TRIANGLE. The perpendicular distance from the vertex of the triangle to the base, or base produced. Either side ALTERN-ATE [L. alternatus, by turns]. of a triangle may be regarded as a base, and Succeeding each other by turns. ALTERNATE ALLIGATION In arithmetic. See Alligation. ALTERNATE ANGLES. In elementary geometry, if two parallel straight lines are intersected by a third, the two inner angles, on opposite side of the cutting line, are called alternate interior angles: also, the two outer angles, on opposite sides of the third line, are called alternate exterior angles. If two parallel planes are intersected by a third, the analogous angles are called by the same names. then the vertex of the opposite angle is the vertex of the triangle. The side which appears horizontal, in viewing the figure, is generally considered as the base, unless some other side is especially pointed out. In the right angled triangle, the base is always one of the sides about the right angle, the other one being the measure of the altitude. If either angle at the base is obtuse, the line on which the altitude is measured will fall upon the base produced in the direction of the obtuse angle. ALTITUDE OF A TRAPEZOID. The perpen dicular distance between its parallel sides, which are then called bases. ALTITUDE OF A PARALLELOGRAM is the perendicular distance between any two parallel sides taken as bases. ALTITUDE OF A CONE OR PYRAMID is the perpendicular distance of the vertex from the plane of the base. ALTITUDE OF A FRUSTRUM of either a cone or pyramid, is the perpendicular distance between the planes of its bases. ALTITUDE OF A PARALLELOPIPEDON is the distance between the planes of any two parallel faces taken as bases. ALTITUDE OF A SPHERICAL SEGMENT, OR ZONE, is the distance between the planes of the circles which constitutes its bases. If the segment or zone has but one circular base, the altitude is the distance between the plane of that base and a plane drawn parallel to it and tangent to the surface of the segment or zone. In Leveling, the altitude of one point above another, is the difference of their distances from the centre of the earth. In Surveying, the altitude of an object is the distance between two horizontal planes, drawn one through the highest, and the other through the lowest point of the object, or through the position of the observer. Such altitudes are divided into two classes, accessible and inaccessible. ACCESSIBLE ALTITUDE is the altitude of an object whose base is accessible, so that the surveyor may measure the distance from his station to it. INACCESSIBLE ALTITUDE is the altitude of an object such, that the surveyor cannot measure the distance from his station to its base, by direct measurement, on account of some intervening obstacle. I. To measure an accessible altitude. There are three principal methods: set up a vertical stake; from C, sight to the top of the object, and note the point D where this line of sight cuts the stake, and then measure DE. We shall have, from similar triangles, CE: DE :: CA : BA; Let AB represent the object whose altitude 1. Let it be required to determine the alti- is to be determined, and let CD represent a tude of an object AB. staff planted vertically, whose length above the ground is known. At any moment of time note the point E where the shadow of the point B falls, and also the point F where the shadow of the point D falls. Measure FC and AE; then from the similar triangles FCD and EAB, we shall have FC: CD: EA: AB, whence, CD × EA If we now conceive a straight line DG to be drawn from D perpendicular to AC, we shall This method does not require that the plane have, from a known principle of geometry, AF should be horizontal. II. To measure an inaccessible altitude. 1. Let it be required to determine the altitude of the point D above the horizontal plane AC. = AD2 AB2 + BD + 2AB BG and BC + BD2. - CD2 whence, by substituting the expressions for the several distances aiready deduced, and the values of AB and BC, we have x2 cot a = a + x2 cotß + 2a BG, x2 cota y = b2 + x2 cot2 ß - 26 · BG. Eliminating BG from these equations, and finding the value of x, we have Select two points A and B in a vertical x plane through D, and measure the distance AB between them. At the points A and B, with some suitable instrument. measure the angles of elevation DAC and DBC. Then, in the triangle DAB, since the sine of DBC is equal to the sine of DBA, we have Denote the distances measured by a and b, and the angles of elevation by a, ẞ and y, and the required altitude DE, above the horizontal plane through AC, by x. From the right-angled triangles, in the figure, we have AD = x cot a, BD = x cot ß, CD = x cot v. x= This method enables us at the same time to determine the distances from the stations to the object, by simply multiplying the altitude formed by the cotangents of the angles of elevation. 3. When no suitable instrument can be had for measuring angles, the altitude may be determined as follows: let AB represent the required altitude, and denote it by z. Select two points C and D in a vertical plane through B, and measure the distance between them and call it c. At some point E, between C and A, plant a vertical staff, and having measured the distance CE, call it f. From C sight to B, and note the point F; also from D sight to B, and note the point G: Denote the distance AC by y. the distance EF by a, and the distance FG by b. From the similar triangles CEF and CAB we have |