out the rectangle B DOG. And now, by the help of this curve, we have an easy way of conceiving and computing the motion of a body through the air. For the subtangent of our curve now presents twice the height through which the ball must fall in vacuo, in order to acquire the terminal velocity; and therefore serves for a scale on which to measure all the other representatives of the motion. It remains to make another observation on the curve Lp K, which will save us all the trouble of geographical operations, and reduce the whole to a very simple arithmetical computation. In constructing this curve we were limited to no particular length of the line LR, which represented the space A CDB; and all that we had to take care of was, that when O C, OK, O g, were taken in geometrical progression, Ms, M t, should be in arithmetical progression. The abscissæ having ordinates equal to ps, nt, &c., might have been twice as long as is shown in the dotted curve which is drawn through L. All the lines which serve to measure the hyperbolic spaces would then have been doubled. But NI would also have been doubled, and our proportions would have still held good; because this sub-tangent is the scale of measurement of our figure, as E or 2 a is the scale of measurement for the motions. called logarithms are just the lengths of the different parts of this line measured on a scale of equal parts. 42129 Reasons of convenience have given rise to another set of logarithms: these are suited to a logistic curve whose subtangent is only of the ordinate rv, which is equal to the side of the hyperbolic square, and which is assumed for the unit of number. We shall suit our applications of the preceding investigation to both these, and shall first use the common logarithms whose subtangent is 0·43429. The whole subject will be best illustrated by taking an example of the different questions which may be proposed: Recol= 2a, or, lect that the rectangle ACO F is u2 I. It may be required to determine what will be the space described in a given time t by a ball setting out with a given velocity V, and what will be its velocity v at the end of that time. Here we have NI: MI=ACOF: BDCA; now NI is the subtangent of the logistic curve; MI is the difference between the logarithms of OD and OC; that is, the difference between the logarithms of e+t and e; ACOF is 2 ad, or or Ed. Therefore by common logag rithms 0-43429: log. e+t-log.e=2ad: S,= space described, u'd Since then we have tables of logarithms calculated for every number, we may make use of them instead of this geometrical figure, which still requires considerable trouble to suit it to There are two sets of logarithmic every case. tables in common use. One is called a table of hyperbolic or natural logarithms. It is suited to such a curve as is drawn in the figure, where the subtangent is equal to that ordinate which and S.= corresponds to the side of the square π e+t e or 0.43429: log. =2 ad: S, 2 ad x log. ett 0.43429 e inserted between the hyperbola and its assymp- by hyperbolic logarithms S=2 ad × log. ‡t totes. This square is the unit of surface, by which the hyperbolic areas are expressed; its side is the unit of length, by which the lines beis= longing to the hyperbola are expressed; 7 v 1, or the unit of numbers to which the logarithms Now the are suited, and then IN is also 1. square Ox being unity, the area BACD will be some number; # O being also unity, OD is some number: call it x. Then, by the nature of the hyperbola, О В : 0π = 0; DB; that 1 is Now, And e+t is 23",03, of which the log. is 1.36229 from which take the log. of e 0:48145 is, x1=1:- so that D B calling D d the area, B D d b, which is the T fluxion (ultimately) of the hyperbolic area, is Now in the curve Lp K, MI has the same ratio to NI that BACD has to @XO. Therefore, if there be a scale of which N I is the unit, the number on this scale corresponding to MI has the same ratio to 1 which the number measuring BACD has to 1; and I i, which corresponds to BD db, is the fluxion (ultimately) of MI; Therefore, if MI be called the logarithm of a,is properly represented by the fluxion of MI. In short, the line M I is divided precisely as the line of numbers on a Gunter's scale, which is therefore a line of logarithms; and the numbers = 62 1000 0-763, and its logarithm 9.88252, which, log, of e+t, = 0·825, from which take e, and added to 0.03408, gives 9.91660, which is the there remains t = 0′′·062, or of a second, for the time of passage. Now, to find the remaining velocity, say 825: 763=1670 : 1544, = v. But in Mr. Robins's experiment the remaining velocity was only 1425, the ball having have lost only 126. It appears, therefore, that lost 245; whereas by this computation it should the resistance is double of what it would have proportion of the velocity. been if the resistance increased in the duplicate Mr. Robins says it to slow motions much smaller than his own exis nearly triple. But he supposes the resistance periment, so often mentioned, fully warrants. 52' 0.763 The time e in which the resistance of the air would extinguish the velocity is 0"-763. Gravity, or the weight of the bullet, would have done it 1670 in or 52′; therefore the resistance is 32 times, or nearly sixty-eight times its weight, by this theory, or 5.97 pounds. If we calculate from Mr. Robins's experiment, we must say log. : 043429 100: e V, which will be 630-23, V v and e= -9-63778 3.02009 This reduction will be produced in about seveneighths of a second. III. To determine the time which a ball, beginning to move with a certain velocity, employs in passing over a given space, and the diminution of velocity which it sustains from the resistance of the air; proceed thus:e+t 2ad: S 0.43429: log e+t e e =t. Then to log. add log. e, and we obtain log. e+t, and e+t; from which if we take e we have t. Then, to find v, say e+t:e=V: v. These examples may be concluded by applying this last rule to Mr. Robins's experiments on a musket bullet of three-fourths of an inch in diameter, which had its velocity reduced from 1670 to 1425 by passing through 100 feet of air. This we do to discover the resistance which it sustained, and compare it with the resistance to a velocity of one foot per second. We must first ascertain the first term of our analogy. The ball was of lead, and therefore 2 a must be multiplied by d and by m, which expresses the ratio of the density of lead to that of cast iron, d is 630.23 1670 52 0.3774 =0".3774, and gives 138 for the proportion of the resistance to the weight, and makes the resistance 12.07 pounds, fully double of the other. With this velocity, which greatly exceeds that with which the air can rush into a void, there must be a statical pressure of the atmosphere equal to six pounds and a half. This will make up the difference; and allows us to conclude that the resistance, arising solely from the motion communicated to the air, follows very nearly the duplicate proportion of the velocity. The next experiment, with a velocity of 1690 feet, gives a resistance equal to 157 times the weight of the bullet, and this bears a much greater proportion to the former than 1690 does to 16702; which shows that, although these experiments clearly demonstrate a prodigious augmentation of resistance, yet they are by no means susceptible of the precision which is necessary for discovering the law of this augmentation, or for a good foundation of practical rules; and it is still greatly to be wished that a more accurate mode of investigation could be discovered. We have thus explained, in detail, the principles and the process of calculation for the simple case of the motion of projectiles through the air. The learned reader will think that we have been the hyperbolic logarithm of the quantity annexea to it, and A may be used as to express its common logarithm. See article FLUXIONS. The constant quantity C for completing the fluent is determined from this consideration, that the space described is o, when the velocity is o : us u2 C- — — × L√w2 = o, and C = "—-— × L√®, ·× g and the complete fluent S= X L√w2=L√w2—v2==× unreasonably prolix, and that the whole might SECT. V. OF THE PERPENDICULAR ASCENTS g Let u, as before, be the terminal velocity, and the accelerating power of gravity: when the body moves with the velocity u, the resistance is equal to g; and in every other velocity v, we g v2 must have u2: v2 = g : = r, for the resistu2 ance to that velocity. In the descent the body is urged by gravity g, and opposed by the resist therefore the remaining accelerating force, which we shall call f, is g or The fundamental theorem for varied motion is v i 2 i v fsuv, and &= ~ X and s u2 0.43429 g g u2 2 g or (putting M for 0-43429, the modulus or subtangent of the common logistic curve) = u2 Mg therefore, considering the above fraction as a logarithmic secant, look for it in the tables, and then take the sine of the arc of which this is the secant, and multiply it by u; the product is the velocity required. An example may be given of a ball whose terminal velocity is 6893 feet, and ascertain its velocity after a fall of 1848 feet. Here, u2 = 475200, and its log. u = 689} g= 32 S= 1848 5.67688 2-83844 1.50515 3.26670 Then log. 27.794 +1.44396 log. S +3.26670 -5.67688 0-10809 is the logarithm of 1.2826 =n, and n— 2. Required the velocity acquired by the body by falling 1848 feet. Say 14850 1848 = 0-43429 0-05407. Look for this number among the logarithmic secants. It will be found at 28°, of which the logarithmic sine is Add to this the log. of u The sum 9.67161 2-83844 2.51005 is the logarithm of 323-62, the velocity required. From these solutions we see that the acquired velocity continually approaches to, but never equals, the terminal velocity. For it is always expressed by the sine of an arch of which the terminal velocity is the radius. The motion of a body projected downwards next merits consideration. While the velocity of projection is less than the terminal velocity, the motion is determined by what we have already said; for we must compute the height necessary for acquiring this velocity in the air, and suppose the motion to have begun there. But, if the velocity of projection be greater, this method fails. We pass it over (though not in the least more difficult than what has gone before) because it is of mere curiosity, and never occurs in any interesting case. We may just observe that, since the motion is swifter than the terminal velocity, the resistance must be greater than the weight, and the motion will be retarded. The very same process will give us for the space u2 described S= ·× L√ g V being the འ —tua| འ༠velocity of projection greater than u. Now as this space evidently increases continually (because the body always falls, but does not become inV2_u3 finite in any finite time), the fraction བཻ—u? does not become infinite; that is, v2 does not become equal to u3: therefore, although the velocity V is continually diminished, it never becomes so small as u. Therefore u is a limit of diminution as well as of augmentation. The relation between the time of the descent u g Mg This fluent needs no constant quantity to complete it, or rather Co; for t must beo when vo. This will evidently be the case; for then "+" is L√, = L 1, = 0. Lvu+v น---บ ample. In what time will the body acquire the น Mg น t, and we have 0-43429 : 0-22122=21′′-542: 10" -973, the time required. U-v g We should remember that the numbers or symbols which we call logarithms are really parts of the line MI in the figure of the logistic curve, and that the motion of a point in this line is precisely similar to that of the body. The marquis Poleni, in a dissertation published at Padua in 1725, has with great ingenuity constructed logarithmics suited to all the cases which can occur. By reversing this proportion we get the velocity (+3)=2Mg. Therefore let & be the corresponding to a given time. To compare this descent of 1848 feet in the air with the fall of the body in vacuo during the same time, say 21" 5422 10" 973 1848 1926 6, which makes a difference of seventy-nine' feet. COR. 1. The time in which the body acquires the velocity u by falling through the air is, to the time of acquiring the same velocity by falling to v; for it would in vacuo, as ú L√u+v acquire this velocity in vacuo during the time-, g 2. The velocity which the body acquires by u+v falling through the air in the time u-v is, to the velocity which it would acquire in vacuo during the same time as v to u L√u+v for the velocity would acquire in vacuo during the time Lv 'u + v g (because in any time quired). ข must be u U-v u + v L√"+ น-บ the velocity w is ac g Next, let a body whose terminal velocity is u be projected perpendicularly upwards, with any velocity V. It is required to determine the height to which it ascends, so as to have any remaining velocity v, and the time of its ascent; as also the height and time in which its whole motion will g (u2 + v2) be extinguished. We have now for u2 the expression of f; for both gravity and resistance act now in the same direction and retard the motion of the ascending body; therefore in the ascent, is equal to u2. ” plier in the descent. The first is the secant of an arch whose tangent is V; the other is the secant of an arch whose sine is v. These secants are equal, or the arches are the same; therefore the velocity of projection is to the final returning velocity as the tangent to the sine, or as the radius to the cosine of the arch. Thus suppose the body projected with the terminal velocity, or V = u; then v If V=689, v=487. |