Nate. If the globe be so light as to ascend in the fluid; it is only necessary to change the signs of the first two terms in the value of ƒ, or the accelerating force, by which it be To compare this theory, contained in the last four problems, with experiment, the few following numbers are here extracted from extensive tables of velocities and resistances, resulting from a course of many hundred very accurate experiments, made in the course of the year 1786. In the first column are contained the mean uniform or greatest velocities acquired in air, by globes, hemispheres, cylinders, and cones, all of the same diameter, and the altitude of the cone nearly equal to the diameter also, when urged by the several weights expressed in avoirdupois ounces, and standing on the same line with the velocities, each in their proper column. So, in the first line, the numbers show, that, when the greatest or uniform velocity was accurately 3 feet per second, the bodies were urged by these weights, according as their different ends went foremost; namely, by 028 oz. when the vertex of the cone went foremost; by 064 oz. when the base of the cone went foremost; by 027 oz. for a whole sphere; by 050 oz. for a cylinder; by '051 oz. for the flat side of the hemisphere ; and by ·020 oz. for the round or convex side of the hemisphere. Also, at the bottom of all, are placed the mean proportions of the resistances of these figures in the nearest whole numbers. Note, the common diameter of all the figures, was 6.375, or 63 inches; so that the area of the circle of that diameter is just 32 square inches, or of a square foot; and the altitude of the cone was 6 inches. Also, the diameter of the small hemisphere was 43 inches, and consequently the area of its base 17 square inches, or of a square foot nearly. From the given dimensions of the cone, it appears, that the angle made by its side and axis, or direction of the path, is 26 degrees, very nearly. The The mean height of the barometer at the times of making the experiments, was nearly 30'1 inches, and of the thermometer 62°; consequently the weight of a cubic foot of air was equal to 1 oz. nearly, in those circumstances. From this table of resistances, several practical inferences may be drawn. As, 1. That the resistance is nearly as the surface; the resistance increasing but a very little above that proportion in the greater surfaces. Thus, by comparing together the numbers in the 6th and last columns, for the bases of the two hemispheres, the areas of which are in the proportion of 174 to 32, or as to 9 very nearly; it appears that the numbers in those two columns, expressing the resistances, are nearly as 1 to 2, or as 5 to 10, as far as to the velocity of 12 feet; after which the resistances on the greater surface increase gradually more and more above that proportion. And the mean resistances are as 140 to 288, or as 5 to to 10%. This circumstance therefore agrees nearly with the theory. 2. The resistance to the same surface, is nearly as the square of the velocity; but gradually increasing more and more above that proportion, as the velocity increases. This is manifest from all the columns. And therefore this circumstance also differs but little from the theory, in small velocities. 3. When the hinder parts of bodies are of different forms, the resistances are different, though the fore parts be alike; owing to the different pressures of the air on the hinder parts. Thus, the resistance to the fore part of the cylinder, is less than that on the flat base of the hemisphere, or of the cone; because the hinder part of the cylinder is more pressed or pushed, by the following air, than those of the other two figures. 4. The resistance on the base of the hemisphere, is to that on the convex side, nearly as 2 to 1, instead of 2 to 1, as the theory assigns the proportion. And the experimented resistance, in each of these, is nearly part more than that which is assigned by the theory. 5. The resistance on the base of the cone is to that on the vertex, nearly as 23 to 1. And in the same ratio is the inclination of the side radius to the sine of the angle of of the cone, to its path or axis. So that, in this instance, the resistance is directly as the sine of the angle of incidence, the transverse section being the same, instead of the square of the sine. 6. Hence we can find the altitude of a column of air, whose pressure shall be equal to the resistance of a body, moving through it with any velocity. Thus, Let a = the area of the section of the body, similar to any of those in the table, perpendicular to the direction of motion; the resistance to the velocity, in the table; and the altitude sought, of a column of air, whose base is a, and its pressure r. Then at the content of the column in feet, and far or far its weight in ounces; r therefore far = r, and x = 응 X is the altitude sought in feet, feet, namely, of the quotient of the resistance of any body divided by its transverse section; which is a constant quantity for all similar bodies, however different in magnitude, since the resistance r is as the section a, as was found in art. 1. When a = of a foot, as in all the figures in the forego 5 r ing table, except the small hemisphere: then, a = { x a becomes xr, where r is-the resistance in the table, to the similar body. I 4 If, for example, we take the convex side of the large hemisphere, whose resistance is 634 oz. to a velocity of 16 feet per second, then r = 634, and x = r = 2·3775feet, is the altitude of the column of air whose pressure is equal to the resistance on a spherical surface, with a velocity of 16 feet. And to compare the above altitude with that which is due to the given velocity, it will be 32: 162 :: 16:4, the altitude due to the velocity 16; which is near double the altitude that is equal to the pressure. And as the altitude is proportional to the square of the velocity, therefore, in small velocities, the resistance to any spherical surface, is equal to the pressure of a column of air on its great circle, whose altitude is 2 or 594 of the altitude due to its velocity. But if the cylinder be taken, whose resistance r = 1·526: then x = 5 72; which exceeds the height, 4, due to the velocity in the ratio of 23 to 16 nearly. And the difference would be still greater, if the body were larger; and also if the velocity were more. 7. Also, if it be required to find with what velocity any flat surface must be moved, so as to suffer a resistance just equal to the whole pressure of the atmosphere: The resistance on the whole circle whose area is of a foot, is 051 oz. with the velocity of 3 feet per second; it is of 051, or 0056 oz. only, with a velocity of 1 foot. But 2 × 13600 × 7555 oz. is the whole pressure of the atmosphere. Therefore, as 0056: 7556 :: 1: 1162 nearly, which is the velocity sought. Being almost equal to the velocity with which air rushes into a vacuum, 8. Hence may be inferred the great resistance suffered by military projectiles. For, in the table, it appears, that a globe of 63 inches diameter, which is equal to the size of an iron ball weighing 36lb, moving with a velocity of only 16 feet per second, meets with a resistance equal to the pressure of of an ounce weight; and therefore, comput ing only according to the square of the velocity, the least resistance that such a ball would meet with, when moving with a velocity of 1600 feet, would be equal to the pressure of 417lb, and that independent of the pressure of the atmosphere itself on the fore part of the ball, which would be 487lb more, as there would be no pressure from the atmosphere on the hinder part, in the case of so great a velocity as 1600 feet per second. So that the whole resistance would be more than 900lb to such a velocity. 9. Having said, in the last article, that the pressure of the atmosphere is taken entirely off' the hinder part of the ball moving with a velocity of 1600 feet per second; which must happen when the ball moves faster than the particles of air can follow by rushing into the place quitted and left void by the ball, or when the ball moves faster than the air rushes into a vacuum from the pressure of the incumbent air: let us therefore inquire what this velocity is. Now the velocity with which any fluid issues, depends on its altitude above the orifice, and is indeed equal to the velocity acquired by a heavy body in falling freely through that altitude. But, supposing the height of the barometer to be 30 inches, or 2 feet, the height of a uniform atmosphere, all of the same. density as at the earth's surface, would be 2 × 13.6 × 8331 or 28333 feet; therefore 16 : 28333 :: 32: 8√28333 = 1346 feet, which is the velocity sought. And therefore, with a velocity of 1600 feet per second, or any velocity above 1346 feet, the ball must continually leave a vacuum behind it, and so must sustain the whole pressure of the atmosphere on its fore part, as well as the resistance arising from the vis inertia of the particles of air struck by the ball. 10. On the whole, we find that the resistance of the air, as determined by the experiments, differs very widely, both in respect to its quantity on all figures, and in respect to the proportions of it on oblique surfaces, from the same as determined by the preceding theory; which is the same as that of Sir Isaac Newton, and most modern philosophers. Neither should we succeed better if we have recourse to the theory given by Professor Gravesande, or others, as similar differences and inconsistencies still occur. We conclude therefore, that all the theories of the resistance of the air hitherto given, are very erroneous. And the preceding one is only laid down, till further experiments, on this important subject, shall enable us to deduce from them another, that shall be more consonant to the true phænomena of nature. LOGARITHMS |