It is sometimes necessary to compound together more than two steps. Thus, in the example just used, the train is moving relatively to the Earth, the Earth is moving round the Sun, and the Sun is moving on his own account through space,-or rather, for this is all we can be sure about, he is moving relatively to certain stars. So that to get the actual motion of the man in the train relative to these stars, we must compound all these motions together. The rule for this is very easily found when the straight lines representing the steps to be compounded are so arranged that the end of each is the beginning of the next. Then the resultant is the step from the beginning of all to the end of all. Thus the steps ab, bc, cd, de have the resultant ae; for ab and be give ac, then ac and cd give ad, and finally ad and de give ae. But when the lines are all arranged so as to have a common beginning, the rule is rather more complex, and will be examined after we have found a shorter way of writing about the composition of steps. What is true of two steps, that their resultant is independent of the order in which they are taken, is true of any number of steps. This we shall now prove. First, the resultant is unaltered by the interchange of two successive steps. For to interchange the steps bc, cd, that is, to take cd before bc, we must draw bc' equal and parallel to cd, and then from c' a line equal and parallel to bc. But this line will end precisely at d, because bcdc' is a parallelogram. Nothing after the point d will be altered, and consequently the resultant ae will be the same as before. Next, any change whatever in the order can be produced by a sufficient number of interchanges of successive steps. This statement clearly does not apply to steps only, but to any things whatever that can be arranged in order; for example, letters or figures. The truth of the statement will be made clear by an example of the process to be used. Thus, let it be required to change COMPOSITION OF STEPS. 7 the order 123456 into the order 314625. Bring 3 to the first place by successively interchanging it with 2 and 1. Then 1 will be in the second place as required. Bring 4 to the third place by interchanging it with 2, and then bring 6 to the fourth place by interchanging it with 5 and 2; lastly, interchange 5 and 2, and the required transformation is complete. As no one of these six interchanges has altered the resultant, it remains the same as at first. Thus the proposition is proved. When we have to deal with steps. which are all in the same straight line, as ab, bc, cd, we may describe each of them as a step of so many inches to the right or to the left. To find the resultant we must add together the lengths of all the steps to the right, and also the lengths of all the steps to the left. The resultant is a step whose length is the difference between these two sums, and it is to the right if the former is greater, to the left if the latter is greater. Thus the resultant of the steps ab, bc, cd is, as we know, ad; and the length of ad is ab+cd-cb. The resultant is a step to the right because the sum abcd is greater than cb. It is convenient to regard a step to the left as a negative quantity, the addition of which is equivalent to the subtraction of its length from that of a step to the right. Thus bc is taken to be the same as - cb. And thus we may write either or else ad = ab+cd - cb, ad = ab+cd+bc. The symbol +, placed between two steps, is thus made to mean that their resultant is to be found, regard being had to their directions. The resultant ab+be is always ac, no matter how the points are situated; but the length ac is a sum or a difference of the lengths ab and bc, according as they are in the same direction or not. We shall extend this meaning of the symbol + to cases in which the component steps are not in the same straight line; that is to say, ab+cd shall always mean the resultant of the steps ab and cd, not the sum of their lengths unless this is expressly mentioned. Similarly ab - cd will mean the resultant of ab and a step the reverse of cd, namely dc. After a little practice, the student will find that this extension of the meaning of the signs +, -, and = does not cause any confusion, but on the contrary enables us to reason more clearly because more compactly. We shall now use this method to investigate the resultant of several steps the lines representing which are so placed as all to have the same beginning. In the case of two steps oa and ob, the rule is to complete the parallelogram oapb, and then the diagonal op is the resultant. But if we join the points ab by a straight line meeting Thus op in c, both op and ab are bisected at the point c. op is twice oc, which may be written op=2oc. Observe that 2oc means a step in the direction of oc, of twice its length. We may now state our rule as follows:-find c the middle point of ab, then the resultant of the steps oa and ob is twice oc; or, more shortly, oa + ob = 2oc. We may extend this result. Let ab be divided in c so that ac is to cb as m to l, where 7 and m are any two numbers. Then 1. ac m.cb and (l+m). ac = m. ab. = a Now oc=oa+ac; that is to say, the (l+m) oc = (l + m) oa + (l + m) ac. m But (1+m) ac = m. ab, and ab =ao + ob = ob — oɑ. Substituting this value, we find (l +m) oc = (1+m) oa+m. ab That is, if ab be divided in the ratio m: 1 at the point c, then the resultant of l times oa and m times ob is l+m times oc. have COMPOSITION OF STEPS. 9 We shall now write this proof in a shorter form. We therefore oa oc + ca, ob= =oc + cb; l.oa+m.ob = (1 + m). oc + 1. ca+m.cb = = (2 + m).ỌC, because the point c was so chosen that l.acm.cb, or (which is the same thing), l.ca+m.cb= 0. The former investigation exhibits the process of finding oc in terms of oa and ob; the latter is a shorter and more symmetrical proof of the result when it is known. We proceed now to the case of three steps, oa, ob, oc. Bisect ab in f, then 2ofoa + ob, so that it remains 0. f to find the resultant of oc and twice of. This is a case of the last proposition, in which 7=1 and m = 2; we must therefore divide ef in the ratio of 2: 1. Taking then a point g at two-thirds of the way from c to f, we find 3og for the resultant of oa, ob, oc; or, more shortly, oa + ob+ oc= 30g. This result is true wherever the point o is: whether in the plane abc or out of it. And the method of determining g is quite independent of the position of o. By making o coincide with g, so that og is zero, we find that ga+gb+gc = 0. This is independently clear, because ga+gb=2gf, and 2gf+ge=0 by construction. Hence also we see that g is of the way from a to d, and from b to e, if d, e are the middle points of bc, ca. Or the lines joining the angles of a triangle to the middle points of its sides meet in a point which divides each of them in the ratio of 2 to 1. To find the resultant of l times oa, m times ob, and n times oc, we must observe that whatever the point g is, oa = og + ga, ob = og + gb, oc = og + gc, and therefore l.oa+m.ob+n.oc = (1+m+ n). og +l.ga+m.gb+n.gc. f 19 If therefore we can find a point g such that l.ga+m.gb+n. gc = 0, we shall have l. oa + m. ob+n.oc = (l+m+n).og. Now 1. ga+m.gb = (1+m). gf, if ƒ is the point dividing ab in the ratio ml. Hence (l+m).gf+n.gc = 0, or g is the point dividing cf in the ratio l+m: n. We might equally well have found g by dividing be in the ratio n: m at d, or ca in the ratio : n at e. That is, we have the equations l.fam.bf, n.ec=l.ae, and m.db=n.cd, between the lengths of the six segments into which the sides are divided. Multiplying these equations together, we find that the product Imn divides out, and that fa.ec. db = bf. ae. cd. Hence if ad, be, cf meet in a point, then af.ce.bd ea. dc.fb. This theorem is a = useful criterion for the concurrence of three lines drawn through the vertices of a triangle. A similar set of theorems belongs to the composition of four steps. If f, g, h are the middle points of bc, ca, ab, and f', g', h' of da, db, dc, then ff', gg', hh' bisect one another at a point k, such that a с also if k be taken at the middle point of hh', oh + oh'= 2ok; therefore oa+ob + oc + od = 2 (oh +oh') = 4ok. And the symmetry shews that this k is also the middle point of gg' and of ff'. Moreover, if we take a of the way from f to d, then k is of the way from a to a. For we know that ob+oc + od 30%, and therefore oa + 30x = 4ok, wherever 0 is or taking o to coincide with k, ka + 3ka = 0, which shews that k divides aa in the ratio of 3: 1. = Observe that the points abcd may either be in the same plane, or form a triangular pyramid, or tetrahedron. |