The term *Projectiles* is applied to that part of mechanical philosophy which treats of the motion of bodies anyhow projected from the surface of the earth, and influenced by the action of terrestrial gravity.

It is demonstrated in the physical part of astronomy, that a body so projected must describe a conic section, having the centre of the earth in one focus; and that it will describe round that focus areas proportional to the times. It follows also from the principles of that science, that if the velocity of projection exceed 36,700 feet in a second, the body (if not resisted by the air) would describe a hyperbola; if the velocity be just 36,700 feet in a second, the body would describe a parabola; and a body projected with a smaller velocity than this would describe an ellipse. In the first case, if projected directly upwards, it would never return, but proceed for ever, its velocity continually diminishing, but never becoming less than an assignable portion of the excess of the initial velocity above 36,700 feet in a second. In the second case it would never return; its velocity would diminish without end, but never be extinguished. In the third case it would proceed till its velocity was reduced to an assignable portion of the difference between 36,700 feet and its initial velocity, and would then return, regaining its velocity by the same degrees and in the same places as it lost it. These are necessary consequences of a gravity directed to the centre of the earth, and inversely proportional to the square of the distance. But in the greatest projections that we are able to make, the gravitations are so nearly equal, and in directions so nearly parallel, that it would be ridiculous affectation to pay any regard to the deviations from equality and parallelism. A bullet rising a mile above the surface of the earth loses only \( \frac{1}{2000} \) of its weight, and a horizontal range of 4 miles makes only \( \frac{1}{4} \) deviation from parallelism.

Let us therefore assume gravitation as equal and parallel. The errors arising from this assumption are quite insensible in all the uses which can be made of this theory.

Gravity must be considered by us as a constant or uniform accelerating or retarding force, according as it produces the descent, or retards the ascent, of a body. A constant or invariable accelerating force is one which produces an uniform acceleration; that is, which in equal times produces equal increments of velocity, and therefore produces increments of velocity proportional to the times in which they are produced. Forces are of themselves imperceptible, and are only sensible in their effects; and they have no measure but the effect, or what measures the effect; and every thing which we can discover with regard to those measures, we may affirm with regard to the things of which we assume them as the measures. Hence this proposition:

The motion of a falling body, or of a body projected directly downwards, is uniformly accelerated; and that of a body projected directly upwards is uniformly retarded; that is, the acquired velocities are as the times in which they are acquired by falling, and the extinguished velocities are as the times in which they are extinguished. The following consequences are readily deduced from the proposition now enunciated:

1. If bodies simply fall, not being projected downwards by an external force, the times of the falls are proportional to the final velocities; and the times of ascents, which terminate by the action of gravity alone, are proportional to the initial velocities.

2. The spaces described by a heavy body falling from rest are as the squares of the acquired velocities; and the differences of these spaces are as the differences of the squares of the acquired velocities; and, on the other hand, the heights to which bodies projected upwards will rise, before their motions are extinguished, are as the squares of the initial velocities.

3. The spaces described by falling bodies are proportional to the squares of the times from the beginning of the fall; and the spaces described by bodies projected directly upwards are as the squares of the times of the ascents.

4. The space described by a body falling from rest is one-half of the space which the body would have uniformly described in the same time, with the velocity acquired by the fall. And the height to which a body will rise, in opposition to the action of gravity, is one-half of the space which it would uniformly describe in the same time with the initial velocity.

In like manner, the difference of the spaces which a falling or rising body describes in any equal successive parts of its fall or rise, is one-half of the space which it would uniformly describe in the same time with the difference of the initial and final velocities.

The last proposition will be more conveniently expressed for our purpose as follows:

A body moving uniformly during the time of any fall with the velocity acquired thereby, will in that time describe a space double of that fall; and a body projected directly upwards will rise to a height which is one-half of the space which it would describe in the time of its ascent with the initial velocity of projection, if the motion continued uniform.

It is a matter of observation and experience, that a heavy body falls 16 feet and one inch English measure in a second of time; and therefore acquires the velocity of 32 feet 2 inches per second. This cannot be ascertained directly, with the precision that is necessary. A second is too small a portion of time to be exactly measured and compared with the space described; but the observation is made with the greatest accuracy by comparing the motion of a falling body with that of a pendulum. The time of a vibration is to the time of falling through half the length of the pendulum, as the circumference of a circle is to its diameter. The length of a pendulum can be ascertained with great precision; and in this way the space fallen through in a second has been accurately measured.

As all forces are ascertained by the accelerations which they produce, they are conveniently measured by comparing their accelerations with the acceleration of gravity. This therefore has been assumed by all the later and best writers on mechanical philosophy as the unit by which every other force is measured. It gives us a perfectly distinct notion of the force which retains the moon in its orbit, when we say it is the 3600th part of the weight of the moon at the surface of the earth. We mean by this, that if a bullet were weighed at the surface of the earth by a spring steelyard, and pulled the spring out to the mark 3600; if the body were then taken to the distance of the moon, it would pull the spring out only to the mark 1. And we make this assertion on the authority of our having observed that a body at the distance of the moon falls from that distance \( \frac{1}{3600} \) part of 16 feet in a second. We do not therefore compare the forces, which are imperceptible things; we compare the accelerations, which are at the same time their effects and their measures. Philosophers have therefore been at great pains to determine with precision the fall of heavy bodies, in order to have an exact measure of the accelerating power of terrestrial gravity. Now we must here observe, that this measure may be taken in two ways; we may take the space through which the heavy body falls in a second; or we may take the velocity which it acquires in consequence of gravity having acted on it during a second. The last is the proper measure, for it is the immediate effect on the body. The action of gravity has changed the state of the body by giving it a determination to motion downwards; and this both points out the nature, and the degree or intensity, of the force of gravity. The space described in a second by falling, is not an invariable measure; for, in the successive seconds, the body falls through 16, 48, 80, 112, &c. feet, but the changes of the body's state in each second is the same. At the beginning it had no determination to move with any appreciable velocity; at the end of the first second it had a determination by which it would have gone on for ever (had no subsequent force acted on it) at the rate of 32 feet per second. At the end of the next second, it had a determination by which it would have moved for ever, at the rate of 64 feet per second. At the end of the third second, it had a determination by which it would have moved for ever, at the rate of 96 feet per second, &c. The differences of these determinations are constant, and amount to 32 feet per second. This is therefore the indication and proper measure of the constant or invariable force of gravity. The space fallen through in the first second is of use only as it is one-half of the measure of this determination; and as halves have the proportion of their wholes, different accelerating forces may be safely affirmed to be in the proportion of the spaces through which they uniformly impel bodies in the same time. But we should always recollect, that this is but one-half of the true measure of the accelerating force.

We now proceed to investigate the formulae of rectilinear motion.

Let \( x \) be the space described by a moving body in a certain time \( t \), and \( v \) the velocity acquired at the end of the time \( t \). When the motion is uniform, the space described is in the compound ratio of the velocity and the time, therefore \( x = vt \). In the case of accelerated or retarded motion, the velocity varies by infinitely small degrees, and we may therefore assume it to be uniform during an infinitely small portion of time \( dt \). But the space over which the body moves in the time \( dt \), with the uniform velocity \( v \), is \( vdt \); therefore the increment of the space in the time \( dt \) being \( dx \), we have \( dx = vdt \), whence \( v = \frac{dx}{dt} \).

Let \( \phi \) be the force which accelerates or retards the moving body at the end of the time \( t \). The force \( \phi \) is supposed to act in the straight line in which the body is moving at the instant; and as an accelerating force becomes a retarding force by changing the sign from positive to negative, it is usually called, in both cases, the accelerating force. In order to obtain the measure of \( \phi \), we must consider the effect it produces. Since the motion may be regarded as uniform during an infinitely small portion of time \( dt \), if \( v \) be the velocity at the end of the time \( t \), then \( v + dv \) will be the velocity at the end of the time \( t + dt \), so that \( dv \) is the augmentation of velocity produced by the accelerating force \( \phi \) in the time \( dt \). Assume \( F \) to be a constant force, which produces a velocity \( V \) in the portion of time which is taken as unit; then the velocity produced by the force \( F \) in the time \( dt \) is \( Vdt \), and we have this proportion,

\[ \phi : F :: dv : Vdt, \]

the forces being measured by the velocities which they communicate to a given body in the same time. Now, sup-

pose \( F \) to be the unit of force, and \( V \) the unit of velocity, the proportion becomes \( \phi : 1 :: dv : dt \), whence \( \phi = \frac{dv}{dt} \).

On differentiating the equation \( v = \frac{dx}{dt} \), we get \( \frac{dv}{dt} = \frac{d^2x}{dt^2} \),

whence \( \phi = \frac{d^2x}{dt^2} \). The relations between the velocity, the space described, the time, and the accelerating force, are consequently given by the three following equations, of which the last is a consequence of the other two, viz.

\[ v = \frac{dx}{dt}, \quad \phi = \frac{dv}{dt}, \quad \phi = \frac{d^2x}{dt^2}. \]

(1)

Suppose the accelerating force to be terrestrial gravity, and let it be denoted by \( g \). We have then, from the third of the above equations \( \frac{d^2x}{dt^2} = g \); the integral of which is

\[ \frac{dx}{dt} = gt, \quad \text{or} \quad v = gt \quad \text{(which requires no correction, since } v \text{ and } t \text{ vanish together)}. \]

We have also \( dx = gtdt \); whence by second integration \( x = \frac{1}{2}gt^2 \), which is also the complete integral, since \( x \) and \( t \) vanish together. This last equation gives \( 2gx = gt^2 = v^2 \); whence \( v = \sqrt{2gh} \). We have therefore the following expressions for the velocity, the time, and the space described, or height from which a body must fall in vacuo to acquire a given velocity, and which we shall denote by \( h \), viz.

I. \( v = gt = \sqrt{2gh} \), II. \( t = \frac{v}{g} = \sqrt{\frac{2h}{g}} \), III. \( h = \frac{v^2}{2g} = \frac{1}{2}gt^2 \).

It is found by experiment that the height through which a heavy body descends by the force of gravity in a vacuum, in a second of time, is 16 English feet. For convenience, we shall call this 16 feet. Assuming, therefore, one second as the unit of time, the last equation gives \( g = 2h = 32 \) feet; and the three equations become, in respect of any time \( t \), and height \( h \),

I. \( v = 32t = \sqrt{64h} = 8\sqrt{h} \), II. \( t = \frac{v}{32} = \frac{1}{4}\sqrt{h} \), III. \( h = \frac{v^2}{64} = 16t^2 \).

To give some examples of the use of these formulae, let it be required,

1. To find the time of falling through 256 feet. Here \( h = 256, \sqrt{256} = 16, \) and \( 16 \div 4 = 4 \). Therefore \( t = 4 \) seconds. 2. To find the velocity acquired by falling four seconds. Here \( t = 4, 32 \times 4 = 128 \) feet per second. 3. To find the velocity acquired by falling 625 feet. Here \( h = 625, \sqrt{625} = 25, 8\sqrt{h} = 200 \) feet per second. 4. To find the height to which a body will rise when projected with the velocity of 56 feet per second, or the height through which a body must fall to acquire this velocity. In this case we have

\[ v = 56, \quad 56 \div 8 = 7 = \sqrt{h}, \quad 7^2 = h = 49 \text{ feet.} \]

or \( 56^2 = 3136, 3136 \div 64 = 49 \) feet.

5. Suppose a body projected directly downwards with the velocity of 10 feet per second; what will be its velocity after four seconds? In four seconds it will have acquired, by the action of gravity, the velocity of \( 4 \times 32 \), or 128 feet, and therefore its whole velocity will be 138 feet per second.

6. To find how far it will have moved, compound its mo- Suppose the body projected, as in the 5th example, and that it is required to determine the time it will take to pass through 296 feet, and the velocity it will have acquired.

Find the height \( h \), through which it must fall to acquire the velocity of projection, 10 feet, and the time \( t \) of falling from this height. Then find the time \( t' \) of falling through the height \( 296 + h \), and the velocity \( v \) acquired by this fall. The time of describing the 296 feet will be \( t' - t \), and \( v \) is the velocity required.

These examples will suffice for the perpendicular ascents or descents of heavy bodies, abstracting from the resistance of the air. We now proceed to consider the motions of bodies projected obliquely. The subject is one of considerable interest, as it forms the foundation of the theory of gunnery.

Let a body be projected from the point A (fig. 1.) in any direction AK not perpendicular to the horizon, and with a given velocity \( = a \); it is required to determine the path of the body, and all the circumstances of its motion.

Since gravity acts only in the vertical direction, and the body, after being projected, is acted on by no other accelerating force, it is evident, in the first place, that the whole of the trajectory, or path described by the projectile, must be contained in the vertical plane which passes through AK. In the second place, since the action of gravity is constant, the projectile must describe a continuous curve, to which AK is a tangent at A. Through A draw AB horizontal, and AH vertical, and let these lines be taken as the axes of the rectangular co-ordinates to which the different points of the curve are referred. Let P be the place of the projectile at the end of the time \( t \); draw PQ perpendicular to AB, and make AQ = \( x \), PQ = \( y \). The object is now to find the equation of the trajectory by establishing a relation between the co-ordinates \( x \) and \( y \).

Let the angle KAB, which the initial direction of the projectile makes with the horizon, be denoted by \( A \). As we abstract from the resistance of the air, the motion in the direction AK is uniform, and proportional to the time \( t \); and the velocity of projection being \( a \), the velocity in the direction AB (which is not affected by gravity) is also uniform, and equal to \( a \cos A \), therefore the space passed over being equal to the product of the time by the velocity, we have \( x = a \cos A \cdot t \). To find \( y \), we must consider that the motion in the vertical direction is made up of two parts; first, that which is due to the velocity of projection; and, second, that which arises from the action of gravity. The velocity in the direction AH, independent of gravity, is \( a \sin A \); and therefore the space described in that direction in the time \( t \), is \( ta \sin A \). But by the formula III. the space through which the body is carried in the time \( t \) by the accelerating force of gravity is \( \frac{1}{2} gt^2 \); and as this is in the direction opposite to the former, it must be taken with the negative sign. We have therefore \( y = ta \sin A - \frac{1}{2} gt^2 \). The values of \( x \) and \( y \) are thus given in terms of the time; and on eliminating \( t \) between the two equations, viz.

\[ x = ta \cos A, \quad y = ta \sin A - \frac{1}{2} gt^2, \]

we shall have the equation of the trajectory. The first gives \( t = \frac{x}{a \cos A} \), on substituting which in the second, there results

\[ y = x \tan A - \frac{1}{2} g \frac{x^2}{a^2 \cos^2 A}, \]

which is an equation to the Apollonian parabola, having its axis parallel to the ordinates \( y \), and its parameter equal to \( \frac{1}{2} g \frac{a^2}{a^2 \cos^2 A} \).

By the formula I. we have \( a = \sqrt{2gh} \), therefore \( a^2 = 2gh \); and on substituting this value of \( a^2 \), the equation becomes

\[ y = x \tan A - \frac{x^2}{4h \cos^2 A}. \]

Having found the equation of the curve, it is easy to assign its position in the plane of the co-ordinates. Let B be the point in which the parabola meets the horizontal line drawn through A, C the vertex of the parabola, and let CD be perpendicular to AB. The point C is thus found. Let \( \theta \) be the angle which the tangent to the curve at any point \( P \) makes with the axis AB; we have then for any curve whatever, \( \tan \theta \ dx = dy \). At the vertex, the tangent becomes parallel to AB, consequently \( \theta \) vanishes, and we have \( \frac{dy}{dx} = 0 \).

Now on differentiating the equation (2), we find

\[ \frac{dy}{dx} = \tan A - \frac{x}{2h \cos^2 A}; \]

therefore at the vertex, \( 0 = \tan A - \frac{x}{2h \cos^2 A} \), whence

\[ x = 2h \cos A \sin A. \]

Substituting this value of \( x \) in equation (3), we get, after reduction, \( y = h \sin^2 A \). Hence denoting the co-ordinates of the vertex by \( x_1 \) and \( y_1 \); that is, making \( AD = x_1 \), \( DC = y_1 \); we have the two equations to determine \( x_1 \) and \( y_1 \), namely

\[ x_1 = 2h \cos A \sin A, \quad y_1 = h \sin^2 A. \]

To find the point B in which the projectile meets the horizontal line AB, it is only necessary to make \( y = 0 \) in equation (3). This gives \( 0 = x \tan A - \frac{x^2}{4h \cos^2 A} \); whence

\[ 4h \cos^2 A \tan A = x, \quad \text{and therefore} \]

\[ x = AB = 4h \cos A \sin A = 2h \sin 2A. \]

From this equation it appears that when the velocity of projection \( a \), or the height \( h \), corresponding to that velocity, is given, the line AB, which is called the amplitude or horizontal range, is a maximum when the angle of elevation \( A \) (that is, KAB) is equal to 45°, for then \( \sin 2A = 1 \). In this case we have \( AB = 2h \); that is to say, when a ball is projected with an angle of elevation equal to 45 degrees, the amplitude or range on the horizontal plane is equal to twice the height from which a body must fall in order to acquire a velocity equal to the velocity of projection.

To find the velocity at any point \( P \) of the trajectory, in the direction of the tangent at that point, let \( z \) denote the arc AP, then the velocity being equal to the increment of the space passed over, divided by the increment of the time, we have \( \frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt} \). Now, on differentiating the two equations (2), we get

\[ \frac{dx}{dt} = a \cos A, \quad \frac{dy}{dt} = a \sin A - gt, \]

the squares of which being added together, give the following relation between the velocity and the time, viz.

\[ v^2 = a^2 - 2agt \sin A + gt^2. \]

At the point B we have \( x = AB = 4h \cos A \sin A \) by equation (5), and \( y = 0 \). But from the equation \( x = ta \cos A \), we have also \( t = \frac{x}{a \cos A} \); hence at the point B, \( t = \frac{4h \sin A}{a \cos A} \); therefore, since by formula I. \( a^2 = 2gh \), \( gt = \) PROJECTILES

Projectiles. \(2a \sin A\). Substituting this value of \(gt\) in the above equation, we get

\[v^2 = a^2 - 4a^2 \sin^2 A + 4a^2 \sin^2 A = a^2;\]

so that the velocity in the parabola at the point B is equal to the initial velocity at A.

The angle under which the projectile meets the horizontal line AB, is also equal to the angle of elevation A. For since the tangent of the inclination of the curve to the axis of \(x\) is given in general by the equation

\[\frac{dy}{dx} = \tan A - \frac{x}{2h \cos^2 A},\]

we have at the point B, where \(x = 4h \cos A \sin A\),

\[\frac{dy}{dx} = \tan A - \frac{2 \sin A}{\cos A} = \tan A - 2 \tan A = -\tan A.\]

From the equation (3) we readily deduce the solution of the following problem, which contains the most important practical application of the theory. Given the velocity of projection \(=a\), it is required to determine A, the angle of elevation, in order that the projectile may hit a given mark, or that the trajectory may pass through a given point E.

Let \(\beta, \gamma\) be the co-ordinates of E, namely \(AF = \beta, FE = \gamma\).

From equation (3) we have

\[\gamma = \beta \tan A - \beta^2 - 4h \cos^2 A,\]

and the object is to find from this equation the angle of elevation, or the value of A. Transposing, and observing that \(1 + \cos^2 A = 1 + \tan^2 A\), the equation becomes

\[\tan^2 A - \frac{4h}{\beta} \tan A = -\frac{4h \gamma}{\beta^2} - 1,\]

which being solved in respect of \(\tan A\), gives

\[\tan A = \frac{2h}{\beta} \pm \frac{1}{\beta} \sqrt{(4h^2 - 4h \gamma - \beta^2)}.\]

When the quantity under the radical sign is positive, that is, when \(4h^2\) is greater than \(4h \gamma + \beta^2\), then \(\tan A\) has two values, and there are consequently two angles of elevation, by either of which the mark may be hit. If \(4h^2 = 4h \gamma + \beta^2\), then \(\tan A = 2h \pm \beta\), and there is consequently only one angle of elevation; and if \(4h^2\) be less than \(4h \gamma + \beta^2\), then the quantity under the radical sign is negative, and the problem becomes impossible. It may be remarked that, supposing \(\gamma\) and \(\beta\) variable, the equation \(4h^2 - 4h \gamma - \beta^2 = 0\), or \(\beta^2 = 4h(h - \gamma)\) is the equation of a parabola, the axis of which is the vertical line AH, the focus at A, and the vertex at a point H, such that \(AH = h\), the height corresponding to the given velocity of projection. For all points within this parabola \(\beta^2\) is less than \(4h^2 - 4h \gamma\), and consequently \(4h^2\) greater than \(4h \gamma + \beta^2\); and for all points without it \(\beta^2\) is greater than \(4h^2 - 4h \gamma\), or \(4h^2\) less than \(4h \gamma + \beta^2\), so that the parabola, whose equation is \(\beta^2 = 4h(h - \gamma)\) forms the line of separation between the points within the range of a body projected from the point A with the velocity \(a\), and which may be each hit by projecting the body under two different angles of elevation, and the points which the projectile cannot reach at all.

The different expressions which have been found in the preceding solution, may be represented by a very simple geometrical construction. On the vertical line passing through A, (fig. 2,) take \(AH = h\) height due to the velocity of projection \(a\), and let AK be the initial direction. On AH describe a semicircle, meeting AK in L, make LM perpendicular to AH, and join HL. Since ALH is a right angle, and therefore AHL = LAB = A, we have \(AL = h \sin A\), \(AM = AL \sin A = h \sin^2 A\); Projectiles, and \(LM = AL \cos A = h \cos A \sin A\). Comparing these expressions with the equations (4), we have \(x_1 = 2LM\), and \(y_1 = AM\), \(x_1\) and \(y_1\) being the co-ordinates of the vertex. Hence the vertex of the parabola lies in the straight line ML, and is obtained by making LC = ML. It follows also from the equation (5), that AB, the horizontal range, is equal to \(4ML\).

Through L conceive a straight line to be drawn perpendicular to AB, or parallel to AH. This line will either be a tangent to the semicircle, or will meet it in another point \(L'\). In the former case the angle of elevation is \(45^\circ\), and there is only one parabola corresponding to the given velocity \(a\), through which a projectile can pass to B. But when the horizontal range AB is less than twice the height AH due to the velocity, the vertical drawn through L again meets the circle in \(L'\), and \(AL'\) is the tangent to another parabola which passes through B, and of which the vertex is situated at the intersection of DC with \(M'L'\) parallel to AB.

From the equation (3) the parameter of the parabola is \(4h \cos^2 A\). But \(DC = \sin^2 A\); whence \(DC + \frac{1}{2}\) parameter \(= h\), and consequently \(MH = \frac{1}{2}\) parameter. But (see Conic Sections, Part I. prop. xiv.) the distance of the directrix of a parabola from its vertex is equal to one-fourth of the parameter; and hence a straight line drawn through H parallel to AB is the common directrix of all the parabolas which can be described by a body projected from A with the velocity due to the height AH.

The focus of the parabola ACB is at the point F where HL meets the axis CD, and the focus of AC'B at the point \(F'\) where \(HL'\) meets the axis; and if a circle be described about A as a centre, with a radius \(= AH\), the foci of all the parabolas which can be described with the given velocity of projection, will be in the circumference of that circle.

These few simple propositions contain the whole theory of the motion of projectiles in vacuo, or independent on the resistance of the air; and being easily investigated, and connected with very interesting practice, they have been much commented on, and have furnished matter for many splendid volumes. But the air's resistance occasions such a prodigious diminution of motion in the great velocities of military projectiles, that this parabolic theory, as it is called, is hardly of any use. Mr. Robins found that a musket-ball, discharged with the usual allotment of powder, had the velocity of 1700 feet in a second. This requires a fall of 45,156 feet, and the range therefore should be 90,312, or 17½ miles; whereas it does not much exceed half a mile. A twenty-four pound ball, discharged with sixteen pounds of powder, should range about sixteen miles; whereas it is generally short of three miles. It is only in the cases of slow motions, not exceeding 300 or 400 feet in a second, that the theory can be of any use, for then the path of the projectile differs little from a parabola.

In such experiments as can be performed with great accuracy in a chamber, the coincidence is as great as can be wished. A jet of water or mercury, gives us the finest example, because we have the whole parabola exhibited in the simultaneous places of the succeeding particles. Yet even in these experiments a deviation can be observed. When the jet is made on a horizontal plane, and the curve carefully traced on a perpendicular plane held close by it, it is found that the distance between the highest point of the curve and the mark is less than the distance between it and the spout, and that the descending branch of the curve is shorter than the ascending branch. And this difference is more remarkable as the jet is made with greater velocity, and reaches to a greater distance; and it is evidently produced by the resistance of the air, which diminishes the velocity, without affecting the gravity of a projectile. It is Projectiles still more sensible in the motion of bombs. These can be traced through the air by the light of their fuses; and we see that their highest point is always much nearer to the mark than to the mortar on a horizontal plane.

We now proceed to the problem of determining the path of a projectile in a resisting medium, and as the solution is attended with considerable difficulty, and cannot be obtained without the aid of the integral calculus, we shall endeavour to render it more perspicuous, by considering separately, 1st, the motion of a body allowed to fall from rest; 2d, of a body projected vertically upwards; and, lastly, the general case of a body projected with a given velocity in any direction oblique to the horizon. In order to simplify the subject, we shall suppose the density of the medium to be uniform, which may be considered to be the case of the atmosphere at all altitudes which can be reached by a projectile from the ground; and that the projected body is a sphere, either homogeneous, or composed of spherical strata, for each of which the density is uniform.

Suppose then the body to be let fall, or submitted to the action of gravity, and let \( R \) denote the force with which it is resisted at any given instant. By the principles of dynamics the corresponding accelerating force is equal to the moving force \( R \) divided by the mass of the body, (or the moving force referred to the unit of mass,) therefore the accelerating force due to the resistance is \( R - m \). As this force acts in the direction opposite to gravity, it must be taken with the negative sign, or deducted from \( g \), and the whole accelerating force, causing the descent of the body being represented by \( \varphi \), we have

\[ \varphi = g - R + m. \tag{6} \]

It is necessary to remark, for the sake of preserving accurate principles, that a body descends in air, not by the whole of its weight, but by the excess of its weight above that of the air which it displaces. It descends by its specific gravity only as a stone descends in water. Suppose a body 32 times heavier than an equal bulk of air, it will be buoyed up by a force equal to \( 1 - 32d \) part of its weight; and, instead of acquiring the velocity of 32 feet in a second, it will only acquire a velocity of 31, even though it sustained no resistance from the inertia of the air. Let \( w \) be the weight of the body, and \( w' \) the weight of an equal bulk of air, then the relative acceleration of gravity will be correctly expressed by \( g \left( 1 - \frac{w'}{w} \right) \). But in all practical cases, and particularly in the case of military projectiles, which have a great density, the fraction which enters into this expression is so small that it is not worth attending to.

In order to reduce the force \( R - m \) to its simplest expression, let \( r \) denote the radius of the sphere, and \( \delta \) its mean density. Its volume is then \( \frac{4}{3} \pi r^3 \), (\( \pi \) being as usual the ratio of the circumference of a circle to its diameter,) and we have for the mass \( m = \frac{4}{3} \pi \delta r^3 \). With respect to the force \( R \), the usual hypothesis is to suppose it to be proportional to the square of the velocity of the moving body, to the surface of the projectile, and also to the density of the air. If then we take \( \rho \) to represent the density of the air, and assume \( n = \) a numerical factor to be found by experiment, we shall have \( R = n \rho v^2 \), and consequently

\[ R = \frac{n \rho v^2}{m}. \tag{7} \]

Since \( R - m \) is an accelerating force of the same nature as gravity, and may consequently be represented by a quantity of the same kind as \( g \), we shall have a homogeneous expression by making \( \frac{np}{dr} = \frac{g}{b^2} \), \( b \) being a constant. The expression thus becomes \( \frac{R}{m} = \frac{v^2}{b^2} \); and it is necessary to observe that the constant \( b \) denotes the terminal velocity, or uniform final velocity which the body attains when the resistance becomes equal to gravity, or when \( R = m = g \), and it is consequently no longer urged by an accelerating or retarding force. For the resistance being assumed proportional to the square of the velocity, and \( R - m \) being the retarding force due to the velocity \( v \), if \( b \) = the terminal velocity, the retarding force due to which is \( g \), we shall have the proportion \( b^2 : v^2 :: g : \frac{R}{m} \); whence \( \frac{R}{m} = \frac{g}{b^2} \).

Thus the retarding force due to any velocity \( v \) is expressed in terms of the terminal velocity, but the great difficulty is to procure an absolute measure of this quantity. Sir Isaac Newton has attempted to determine the resistance by theory, and employs a great part of the second book of the Principia in demonstrating, that the resistance to a sphere moving with any velocity, is to the force which would generate or destroy its whole motion in the time that it would uniformly move over \( \frac{1}{2} \) of its diameter with this velocity, as the density of the air is to the density of the sphere. This is equivalent to demonstrating that the resistance of the air to a sphere moving through it with any velocity, is equal to half the weight of a column of air having a great circle of the sphere for its base, and for its altitude the height from which a body must fall in vacuo to acquire this velocity. This appears from Newton's demonstration; for, let the specific gravity of the air be to that of the ball as 1 to \( \beta \); then, because the times in which the same velocity will be extinguished by the uniform action of different forces are inversely as the forces, the resistance to this velocity would extinguish it in the time of describing \( \frac{1}{2} \beta d \), \( d \) being the diameter of the ball. Now \( I \) is to \( \beta \) as the weight of the displaced air to the weight of the ball, or as \( \frac{1}{2} \beta \) of the ball to the length of a column of air of equal weight. Call this length \( k \); \( k \) is therefore equal to \( \frac{1}{2} \beta d \). Suppose the ball to fall from the height \( k \) in the time \( t \), and acquire the velocity \( u \). If it moved uniformly with this velocity \( u \) during the time \( t \), it would describe a space \( = 2k \), or \( \frac{1}{2} \beta d \). Now its weight would extinguish this velocity, or destroy this motion, in the same time, that is, in twice the time of falling. The resistance therefore is equal to half the weight of the ball, or to half the weight of the column of air whose height is the height producing the velocity. But the resistances to different velocities are as the squares of the velocities, and therefore, as their producing heights; and, in general, the resistance of the air to a sphere moving with any velocity, is equal to the half weight of a column of air of equal section, and whose altitude is the height producing the velocity. The result of this investigation has been acquiesced in by all Sir Isaac Newton's commentators. Many faults have indeed been found with his reasoning, and even with his principles; and it must be acknowledged that his reasoning is liable to serious objections, which his most ingenious commentators have not completely removed. We must add, however, that all the causes of deviation from the duplicate ratio of the velocities, and the causes of increased resistance, which the latter authors have valued themselves for discovering and introducing into their investigations, were pointed out by Sir Isaac Newton, but purposely omitted by him, in order to facilitate the discussion in re difficillima. (See Schol. prop. 37. book ii.)

It is known that the weight of a cubic foot of water is 62½ pounds, and that the medium density of the air is \( \frac{1}{10} \) of water; therefore, let \( h \) be the height producing the velocity (in feet), and \( d \) the diameter of the ball (in inches), and \( \pi \) the periphery of a circle whose diameter is 1; the resistance of the air will be \( \frac{62\frac{1}{2}}{840} \times \frac{\pi}{4} \times \frac{d^2}{144} \times \frac{h}{2} \). Projectiles.

\[ \frac{ha^2}{4928} \text{ pounds, very nearly, or, since by formula III. } h = \frac{v^2d^2}{4928} \times 64 = \frac{v^2d^2}{315417} \text{ pounds.} \]

We may take an example. A ball of cast iron weighing 12 pounds, is \(4\frac{1}{2}\) inches in diameter. Suppose this ball to move at the rate of \(25\frac{1}{2}\) feet in a second (the reason of this choice will appear afterwards). The height which will produce this velocity in a falling body is \(97\) feet. The area of its great circle is \(0.11044\) feet, or \(11.848\) inches of one foot. Suppose water to be 840 times heavier than air, the weight of the air incumbent on this great circle, and 97 feet high, is \(0.081151\) pounds; half of this is \(0.0405755\), or nearly \(3\frac{1}{2}\) of a pound. This should be the resistance of the air to this motion of the ball.

Newton himself attempted to compare his propositions with experiment. Some were made by dropping balls from the dome of St. Paul's cathedral; and all these showed as great a coincidence with his theory as they did with each other; but the irregularities were too great to allow him to say with precision what was the resistance. It appeared to follow the proportion of the squares of the velocities with sufficient exactness; and though he could not say that the resistance was equal to the weight of the column of air having the height necessary for communicating the velocity, it was always equal to a determinate part of it; and might be stated \(=n h\), \(n\) being a number to be fixed by numerous experiments.

Newton made another set of experiments with pendulums; and has pointed out some very curious and unexpected circumstances of their motions in a resisting medium. Their results were much more uniform, and confirmed his general theory; which, as we have said above, has been acquiesced in by the first mathematicians.

But the deductions from this theory were so inconsistent with the observed motions of military projectiles, when the velocities are prodigious, that no application could be made which could be of any service for determining the path and motion of cannon shot and bombs; and although John Bernoulli gave, in 1718, a most elegant determination of the trajectory and motion of a body projected in a fluid which resists in the duplicate ratio of the velocities, it has remained a dead letter. Mr. Benjamin Robins was the first who suspected the true cause of the imperfection of the usually received theories; and in 1737 he published a small tract, in which he showed clearly, that even the Newtonian theory of resistance must cause a cannon ball, discharged with a full allotment of powder, to deviate farther from the parabola, in which it would move in vacuo, than the parabola deviates from a straight line. But he farther asserted, on the authority of good reasoning, that in such great velocities the resistance must be much greater than this theory assigns; because, besides the resistance arising from the inertia of the air which is put in motion by the ball, there must be a resistance arising from a condensation of the air on the anterior surface of the ball, and a rarefaction behind it; and there must be a third resistance, arising from the statical pressure of the air on its anterior part, when the motion is so swift that there is a vacuum behind. Even these causes of disagreement with the theory had been foreseen and mentioned by Newton (see the Scholium to prop. 37, book ii. Principi); but the subject seems to have been little attended to.

Two or three years after this first publication, Mr. Robins invented that ingenious method of measuring the great velocities of military projectiles, which has handed down his name to posterity, and discovered the prodigious resistance of the air by observing the diminution of velocity which it occasioned. This made him anxious to examine what was the real resistance to any velocity whatever, in order to ascertain what was the law of its variation; and he was equally fortunate in this attempt. His method of measuring the resistance has been fully described in the article GUNNERY.

It appears (Robins's Math. Works, vol. i. p. 205,) that a sphere of \(4\frac{1}{2}\) inches in diameter, moving at the rate of \(25\frac{1}{2}\) feet in a second, sustained a resistance of \(0.04914\) pounds, or \(4\frac{1}{2}\) lbs. of a pound. This is a greater resistance than that of the Newtonian theory, which gave \(0.0405755\), in the proportion of 1000 to 1211, or very nearly in the proportion of five to six in small numbers. And we may adopt as a rule in all moderate velocities, that the resistance to a sphere is equal to \(3\frac{1}{2}\) of the weight of a column of air having the great circle of the sphere for its base, and for its altitude the height through which a heavy body must fall in vacuo to acquire the actual velocity.

This experiment is peculiarly valuable, because the ball is precisely the size of a 12-pound shot of cast iron; and its accuracy may be depended on. There is but one source of error. The whirling motion must have occasioned some whirl in the air, which would continue till the ball again passed through the same point of its revolution. The resistance observed is therefore probably somewhat less than the true resistance to the velocity of \(25\frac{1}{2}\) feet, because it was exerted in a relative velocity which was less than this, and is, in fact, the resistance due to this relative and smaller velocity. The Chevalier de Borda made experiments similar to those of Mr. Robins, and his results exceed those of Robins in the proportion of 5 to 6. It is much to be regretted, that in a subject so interesting both to the philosopher and the engineer, experiments have not been multiplied. Nothing would tend so much to perfect the science of gunnery, and indeed until this be done, no practical advantage can be derived from the theory.

Adopting this experimental measure of Robins, we have the product \(np\) of equation (7) \(=0.04914\) pounds in respect of a sphere of \(4\frac{1}{2}\) inches diameter, or \(2\frac{1}{2}\) inches radius, moving at the rate of \(25\cdot2\) feet in a second. In order to determine from this experiment the resistance on a sphere of a different size, and moving with a different velocity, it will be convenient to reduce it to the case corresponding to the units of magnitude and velocity; therefore, since the resistance is assumed proportional to the square of the velocity and the surface of the projectile, or \(R=npv^2r^2\), if we denote by \(R_1\) the corresponding resistance, in respect of a sphere whose radius is 1 inch, moving with the velocity of 1 foot per second, we shall have

\[ R_1 = np = \frac{0.04914}{(2\cdot25)^2 \times (2\cdot25)} = 0.00015285 \text{ pounds.} \]

The logarithm of this number is 5.18427. The resistance here determined is independent of the density of the ball; but the retardation occasioned by it being in proportion to the quantity of motion, will therefore be proportional to the density of the ball, or its specific gravity. Military projectiles are of cast iron or lead, the specific gravities of which may be taken at 7.207, and 11.37 respectively, that of water being 1.

The following examples will illustrate the application of the above formula.

1. Let it be proposed to find the resistance to a 24-pound iron ball (whose diameter is 5.603 inches), moving with the velocity of 1670 feet in a second, which is nearly the velocity communicated by 16 lbs. of gunpowder.

Here \(v=1670\), \(r=2\cdot8015\), and the formula \(R=npv^2r^2\) gives

\[ R=0.00015285 \times (1670)^2 \times (2\cdot8015)^2 = 334\cdot57 \text{ lbs.} \]

But it is found, by unequivocal experiments, that the retardation of such a motion, is equivalent to 504 lbs. This is owing to the causes often mentioned, the additional resist-

Projectiles, once to great velocities, arising from the condensation of the air, and from its pressure into the vacuum left by the ball.

2. Let the terminal velocity of the same ball be required.

Here the equation (7) gives \( \frac{R}{m} = g \frac{v^2}{b^2} \); therefore in respect of the ball whose velocity is 1 and radius 1, we have \( R_1 = mg - b^2 \), \( b \) being the terminal velocity. Now it is to be observed that the product of the mass of a body into the force of gravity expresses its weight, which we may denote by \( w \); we have then \( R_1 = w + b^2 \), and consequently \( b^2 = w + R_1 \). Now in respect of another body of the same specific gravity, whose radius is \( r \), we have \( R : R_1 :: r^2 : 1 \), whence \( R = R_1 r^2 \), and the equation becomes \( b^2 = w + R_1 r^2 \). Substituting in this equation the numbers given in the question, we get

\[ b^2 = \frac{24}{-000015285 \times (28015)^2} = 200060, \]

whence \( b = 447.3 \) feet.

As the terminal velocity \( b \), and its producing height \( h \), enter into all computations of military projectiles, we have inserted the following Table for the usual sizes of cannon-shot, computed both by the Newtonian theory of resistance, and by the resistances observed in Robins's experiments. A similar table, computed from Dr. Hutton's experiments, and not materially different, is given in Hutton's Tracts, vol. iii. p. 247.

| Ball | Term Vel. | 24. | Term Vel. | 24. | Diam. Inch. | |------|-----------|-----|-----------|-----|-------------| | 1 | 289.9 | 2626.4 | 263.4 | 2168.6 | 1.94 | | 2 | 324.9 | 2929.5 | 295.2 | 2723.6 | 2.45 | | 3 | 348.2 | 3788.2 | 316.4 | 3127.9 | 2.80 | | 4 | 365.3 | 4170.3 | 331.9 | 3442.6 | 3.08 | | 5 | 390.8 | 4472.7 | 355.1 | 3940.7 | 3.52 | | 6 | 418.1 | 5463.5 | 379.9 | 4511.2 | 4.04 | | 7 | 438.6 | 6010.6 | 398.5 | 4962.9 | 4.45 | | 8 | 469.3 | 6883.3 | 425.5 | 5683.5 | 5.09 | | 9 | 492.4 | 7576.3 | 447.4 | 6265.7 | 5.61 | | 10 | 512.6 | 8024.8 | 465.8 | 6780.4 | 6.21 | | 11 | 540.5 | 9129.9 | 491.5 | 7538.3 | 6.75 |

Having thus explained the nature of the quantities employed in the investigation, we now proceed with the solution of the problem of determining the circumstances of the motion of a body falling vertically in the atmosphere. In consequence of the equation \( \frac{R}{m} = g \frac{v^2}{b^2} \), the relation between the accelerating force and the resistance given by equation (6), becomes \( \phi = g - g \frac{v^2}{b^2} = g \frac{b^2 - v^2}{b^2} \). Hence from the second of equations (1), \( \frac{dv}{dt} = g \frac{b^2 - v^2}{b^2} \), whence \( gdt = \frac{b^2 dv}{b^2 - v^2} \), or, which is the same, \( gdt = \frac{b}{2} \left( \frac{dv}{b+v} + \frac{dv}{b-v} \right) \).

On integrating this equation, and observing that as \( t \) and \( v \) commence together no correction is required, we get \( gt = \frac{b}{2} \log (b+v) - \log (b-v) \), whence

\[ t = \frac{b}{2g} \log \frac{b+v}{b-v}. \]

From this equation the time \( t \) during which a body must fall in a resisting medium in order to acquire the velocity \( v \), is readily computed, supposing the terminal velocity \( b \) to be known. It will be observed that the logarithm is hyperbolic, and therefore, if the common tables are used, the tabular logarithm of the number in question must be divided by the modulus of the system \( M = 43429 \) (see Logarithms, vol. xiii. p. 429). The terminal velocity \( b \), as appears by equation (7), varies with the radius of the ball, and the ratio of its density to that of the air, and must be determined by experiment.

As an example of the formula, assume the terminal velocity \( b = 689.5 \) feet in a second, and let it be proposed to determine the time in which the body will acquire a velocity of 323.62 feet per second.

Here \( b = 689.33 \), \( v = 323.62 \), \( b + v = 1012.95 \), \( b - v = 365.71 \), whence the common logarithm of \( \frac{b+v}{b-v} \) is 0.44245 and hyp. log \( \frac{b+v}{b-v} = 44245 \). Now \( b + 2g = 689.33 + 64 = 10.77 \), therefore \( t = 10.77 \times \frac{44245}{43429} = 10.97 \) seconds, which is the time required.

Assume \( r = gt + b \), and let \( e \) denote the base of the Napierian or hyperbolic logarithms, that is, let \( e = 2.71828 \); the equation (8) gives, on passing to numbers \( e^{2r} = \frac{b+v}{b-v} \), or

\[ e^{2r} = \frac{b-v}{b+v}, \quad \text{whence} \quad e = \frac{b(1-e^{-2r})}{1+e^{-2r}}. \]

Let the terms of this expression for \( e \) be multiplied by \( e \) and the equation becomes

\[ v = \frac{b(e^r - e^{-r})}{e^r + e^{-r}}. \]

which gives the velocity in terms of \( r \), that is, in terms of the time \( t \), and the terminal velocity \( b \).

To find the space described, the equations (1) give \( vdt = dx \). Now we have assumed \( gt + b = r \), therefore \( gdt = bdv \), whence and from equation (9)

\[ gdx = \frac{b^2(e^r - e^{-r})}{e^r + e^{-r}} dt. \]

The integral of this is \( gx = b^2 \log (e^r + e^{-r}) + C \), \( C \) being a constant. When \( x = 0 \), then \( v = 0 \), and the equation becomes \( 0 = b^2 \log 2 + C \); therefore the complete integral is

\[ x = \frac{b^2}{g} \log \frac{1}{e^r + e^{-r}}. \]

which gives the space in terms of the time and final velocity.

The space may also be expressed in terms of the velocity. From equations (1) we have \( vdt = dx \) and \( gdt = dv \). Eliminating \( dt \) between these two equations we get \( \phi dt = vde \). But \( \phi = \frac{dv}{dt} = g \frac{b^2 - v^2}{b^2} \), therefore \( gdx = b^2 vdv \).

The integral of this gives \( gx = \frac{1}{2} b^2 \log (b^2 - v^2) + C \). To determine the constant, we have \( x = 0 \), when \( v = 0 \), and therefore \( 0 = \frac{1}{2} b^2 \log (b^2 + C) \); whence the complete integral is \( gx = \frac{1}{2} b^2 \log \left( \frac{b^2}{b^2 - v^2} \right) \), or

\[ x = \frac{b^2}{2g} \log \frac{b^2}{b^2 - v^2}. \]

As an example of this formula, suppose, as above, \( b = 689.5 \) feet, and let it be required to find the space through which the body has descended when the acquired velocity \( v = 323.62 \) feet. Here the common logarithm of \( b^2 \) is 5.67686, and that of \( b^2 - v^2 \) is 5.56872; therefore the common logarithm of \( b^2 + (b^2 - v^2) \) is 0.10814, and its hyperbolic logarithm is \( \frac{1}{2} \log \frac{b^2}{b^2 - v^2} \). We have also \( b^2 + 2g = 7429.6 \). The formula may also be computed by means of the trigonometrical tables; for since \( \frac{1}{2} \log \frac{b^2}{b^2-v^2} = \log \sqrt{\frac{b^2}{b^2-v^2}} \)

\( = \) logarithmic secant of an arc whose radius is \( b \) and sine \( v \), or of an arc whose radius is \( 1 \) and sine \( = \frac{v}{b} \); if we seek in the table the arc whose logarithmic sine \( = \log (v + b) \), its logarithmic secant, or the logarithm of its cosine subtracted from unity, and divided by the modulus \( M \)

\( = 43429 \), will be the value of \( \frac{1}{2} \log \frac{b^2}{b^2-v^2} \). Thus in the last example we have the common logarithm of \( 1 + b = 9.67160 \), which corresponds to an arc of \( 28^\circ \). Now log \( \cos 28^\circ = 9.94593 \), whence log secant \( 28^\circ = 0.05407 \). But \( b^2 + g = 148592 \) (the double of \( b^2 + 2g \) or \( 7429.6 \) found above), therefore \( x = 148592 \times \frac{43429}{43429} = 1850 \) feet, as before.

In this manner the equation (11) will also give the velocity when the space \( x \) is known. For let \( \psi \) denote the arc whose logarithmic sine in the common tables is \( \log (v + b) \), and the equation becomes \( x = \frac{b^2}{gM} \log \secant \psi \), whence

\( \log \secant \psi = \frac{gMx}{b^2} \). Having therefore found \( \psi \) corresponding to this secant, we shall have \( v + b = \sin \psi \), and consequently \( v = b \sin \psi \). Suppose, as before, \( b = 689.4 \) feet, and let it be required to find the velocity when the body has fallen 1850 feet. Here \( \frac{gMx}{b^2} = \frac{32 \times 43429 \times 1850}{(689.4)^2} = 0.05407 \), which is the logarithmic secant of \( \psi \); but \( 1 - 0.05407 = 0.94593 = \log \cos \psi \), whence \( \psi \) is found in the tables \( = 28^\circ \), and \( \log \sin \psi = 9.67160 \). Adding log \( b = 2.83343 \), we have \( \log b \sin \psi = 2.51003 \), whence \( v = b \sin \psi = 32362 \) feet.

We come now to consider the case of a body, supposed, as before, to be a sphere of given magnitude and density, projected vertically upwards. In this case the resistance of the air and the force of gravity act both in the same direction, and the accelerating force becomes \( \phi = -g \frac{v^2}{b^2} \); whence, since \( \phi = dv + dt \), we obtain the equation

\[ \frac{bdv}{b^2 + v^2} = -\frac{gd}{b}. \]

To integrate the first member of this equation, let \( \psi \) denote an arc of a circle, then \( d \tan \psi = d \psi + \cos^2 \psi \). Now let \( \tan \psi = v + b \), then \( d \tan \psi = dv + b \), and \( 1 + \cos^2 \psi = 1 + \tan^2 \psi = \frac{b^2 + v^2}{b^2} \), and we get \( d \psi = \frac{bdv}{b^2 + v^2} \), whence

\[ \int \frac{bdv}{b^2 + v^2} = \psi = \arctan \left( \frac{v}{b} \right) \]

and the integral of the above equation becomes

\[ \arctan \left( \frac{v}{b} \right) = -\frac{gt}{b} + C. \]

To determine the constant, let \( a \) as before denote the initial velocity. When \( v = a \), then \( t = 0 \), and we have

\[ \arctan \left( \frac{a}{b} \right) = +C, \]

therefore the complete integral is \( \arctan \left( \frac{v}{b} \right) = \arctan \left( \frac{a}{b} \right) - \frac{gt}{b} \), whence

\[ t = \frac{b}{g} \left\{ \arctan \left( \frac{a}{b} \right) - \arctan \left( \frac{v}{b} \right) \right\}, \quad (12) \]

which gives the relation between the velocity and time. The quantities within the brackets express a portion of the arc of a circle whose radius is unit, and are therefore abstract numbers, multiplying \( b + g \), which is the number of units of time in which a heavy body falling in vacuo, acquires the velocity \( b \), by formula 11.

Assume, as above, \( \frac{gt}{b} = \tau \), and make \( \frac{a}{b} = \tan \alpha, \frac{v}{b} = \tan \beta \); the equation (12) then becomes \( \beta = \alpha - \tau \), whence by trigonometry

\[ \tan \beta = \frac{\tan \alpha - \tan \tau}{1 + \tan \alpha \tan \tau}. \]

On substituting for \( \alpha \) and \( \beta \) their values as above, this gives

\[ \frac{v}{b} = \frac{a}{b} \frac{\sin \tau}{\cos \tau} = \frac{a}{b} \frac{\sin \tau}{\cos \tau} = \frac{b(a \cos \tau - b \sin \tau)}{a \sin \tau + b \cos \tau}. \]

Now, since \( gt = br \), we have \( gdt = bdr \), and consequently

\[ gdt = \frac{b^2(a \cos \tau - b \sin \tau)}{a \sin \tau + b \cos \tau} dr; \]

therefore, integrating, and observing that \( fdt = x \), we have

\[ gx = b^2 \log (a \sin \tau + b \cos \tau) + C. \]

When \( x = 0 \), then \( t = 0, \tau = 0, \sin \tau = 0, \cos \tau = 1 \); therefore \( 0 = b^2 \log b + C \), and the complete integral becomes

\[ gx = b^2 \log (a \sin \tau + b \cos \tau) - \log b, \]

or

\[ gx = b^2 \left( \log \frac{a}{b} \sin \tau + \cos \tau \right); \]

whence on restoring the value of \( \tau \),

\[ x = \frac{b^2}{g} \log \left( \frac{a}{b} \sin \frac{gt}{b} + \cos \frac{gt}{b} \right), \quad (13) \]

which gives the space described in a function of the time.

To express the space in terms of the velocity, we have

\[ \phi dx = edv. \]

But in this case \( \phi = -g \frac{(b^2 + v^2)}{b^2} \), therefore,

\[ dx = -\frac{b^2}{g} \cdot \frac{v}{b^2 + v^2}. \]

The integral of this gives \( 2gx = -b^2 \log (b^2 + v^2) + C \); and when \( x = 0 \), then \( v = a \); therefore \( 0 = -b^2 (b^2 + a^2) + C \), and the complete integral becomes

\[ x = \frac{b^2}{2g} \log \frac{b^2 + a^2}{b^2}. \quad (14) \]

Let \( h \) denote the greatest altitude. When the projectile reaches this altitude, \( v = 0 \), and the last equation gives

\[ h = \frac{b^2}{2g} \log \frac{b^2 + a^2}{b^2}. \quad (15) \]

This formula may be computed from the trigonometrical tables, by observing that \( \frac{1}{2} \log \frac{b^2 + a^2}{b^2} = \log \sqrt{\frac{b^2 + a^2}{b^2}} \).

Logarithmic secant of an arc \( \psi \) of which the tangent is \( a + b \). Suppose the velocity of projection to be 411 feet per se- Projectiles.

cond, and \( b = 689 \frac{1}{2} \), the calculation of the height \( h \) will be as follows:

\[ \log a = \log 411 = 2.61384 \\ \log b = \log 689 \frac{1}{2} = 2.83843 \\ \log \tan \psi = 9.77541 \]

From the table of logarithmic tangents we find \( \psi = 30^\circ 48' 20'' \), the logarithmic cosine of which is 9.93395, whence \( \log \sec \psi = 0.06605 \). We have therefore

\[ h = \frac{b^2 \times 0.06605}{gM} = \frac{(689 \frac{1}{2})^2 \times 0.06605}{32 \times 43429} = 2259 \text{ feet}. \]

Having reached the greatest altitude \( h \), the body will immediately begin to fall. From equation (11) we have \( x = \frac{b^2}{2g} \log \frac{b^2}{b^2 - v^2} \) for the space through which the body must fall to acquire the velocity \( v \). Now, if we make \( x = h \), we shall have, on comparing this with the equation (13)

\[ \log \frac{b^2 + a^2}{b^2} = \log \frac{b^2}{b^2 - v^2} \quad \text{whence} \quad \sqrt{\frac{b^2}{b^2 - v^2}} = \sqrt{\frac{b^2}{b^2 - v^2}}. \]

But the first number of this equation is the secant of an arc, whose tangent is \( a + b \); and the second the secant of an arc whose sine is \( a + b \); and since the secants are equal, the arcs must be equal; 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 an arc. Suppose the body projected with a velocity equal to the terminal velocity, that is, suppose \( a = b \); the above equation will then become \( 2(b^2 - v^2) = b^2 \), or \( b^2 = 2v^2 \), whence \( v = b \sqrt{2} \). If \( b = 689 \frac{1}{2} \), then \( v = 487 \), or the body will reach with a velocity of 487 feet per second.

The time required for reaching the greatest altitude is found by making \( v = 0 \) in equation (12). In this case the equation becomes \( t = \frac{b}{g} \arctan \left( \frac{a}{b} \right) \). We learn from this expression of the time, that however great the velocity of projection, and the height to which the body will rise, may be, the time of its ascent is limited. It never can exceed the time of falling in vacuo from the height due to the terminal velocity \( b \), namely \( t = b + g \), in a greater proportion than that of a quadrantial arc to the radius, nearly the proportion of 8 to 5. A 24-pound iron ball cannot continue rising above 14 seconds, even if the resistance to quick motions did not increase faster than the square of the velocity. It probably will attain its greatest height in less than 12 seconds, let its velocity be ever so great.

Having thus determined the relations subsisting between the velocity, the space described, and the time, in the cases of vertical ascent and descent in a resisting medium, we come now to consider the general problem, and to determine the motion of a body projected in any direction whatever. Our readers will readily conceive that this must be a difficult subject, when they see the simplest cases of rectilinear motion abundantly abstruse. In the infancy of the fluxionary calculus, the difficulty was indeed such that Newton himself did not succeed in obtaining a complete solution of the problem. In the tenth and subsequent propositions of the second book of the Principia, he shows what state of density in the air will comport with the motion of a body in any curve whatever; and then, by applying this result to several curves which have some similarity to the path of a projectile, he found one which is not very different from what may be supposed to obtain in the atmosphere. But even the approximation was involved in calculations of such intricacy, that it seemed impossible to make any use of it. In the second edition of the Principia, published in 1713, he corrects some mistakes into which he had fallen in the first, and carries the approximation much farther, but still does not attempt a direct investigation of the trajectory. In proposition 14, &c., he shows how a body, actuated by a centripetal force, in a medium of a density varying according to certain laws, will describe an eccentric spiral, of which he assigns the properties, and the law of description. Had he supposed the density constant, and the difference between the greatest and least distances from the centre of centripetal force exceedingly small in comparison with the distances themselves, his spiral would have coincided with the path of a projectile in the air of uniform density, and the steps of his investigation would have led him immediately to the complete solution of the problem. For this is the real state of the case. A heavy body is not acted on by equal and parallel gravity, but by a gravity inversely proportional to the square of the distance from the centre of the earth, and in lines tending to that centre nearly; and it has been only with the view of simplifying the investigation, that mathematicians have adopted the other hypothesis.

Soon after the publication of this second edition of the Principia, the dispute about the invention of the fluxionary calculus became very violent, and the great promoters of that calculus upon the continent were in the habit of proposing difficult problems as trials of skill to each other. Challenges of this kind frequently passed between the British and foreigners. Dr. Keill of Oxford had keenly espoused the claim of Sir Isaac Newton to this invention, and had engaged in a very acrimonious altercation with the celebrated John Bernoulli of Basle. Bernoulli had published in the Acta Eruditorum of Leipzig an investigation of the law of the forces, by which a body moving in a resisting medium might describe any proposed curve, reducing the whole to the simplest geometry; and Dr. Keill proposed to him the particular problem of the trajectory and motion of a body moving through the air. Bernoulli very soon solved the problem in a way much more general than it had been proposed, viz. without any limitation either of the law of resistance, the law of the centripetal force, or the law of density, provided only that they were regular, and capable of being expressed algebraically. Dr. Brook Taylor, the celebrated author of The Method of Increments, solved it at the same time, in the limited form in which it was proposed. Other authors since that time have given other solutions. But they are all (as indeed they must be) the same in substance with Bernoulli's. Indeed they are all (Bernoulli's not excepted) the same with Newton's first approximations, modified by the steps introduced into the investigation of the spiral motions mentioned above.

We now proceed to the problem itself, restricted to the case of uniform density and a resistance proportional to the square of the velocity.

**Problem.** To determine the trajectory, and all the circumstances of the motion of a body projected through the air, from a given point, and in a given direction, and resisted in the duplicate ratio of the velocity.

Suppose the body to be projected from \( A \), in the direction \( AK \), with a velocity \( = a \), and let \( AB \) and \( AH \), horizontal and vertical lines drawn through \( A \), be assumed as the axes of the rectangular co-ordinates \( x \) and \( y \). Let \( P \) be the place of the projectile at the end of the time \( t \), make the arc \( AP = z \), and let \( Pp = dz \), be an infinitely small portion of the arc, which may be regarded as coinciding with \( PM \), the tangent to the curve at \( P \); draw \( PQ \) and \( pg \) parallel to \( AH \), and \( Pn \) parallel to \( AB \), and let the angle \( pPn (= \text{PMA}) \), which the tangent makes with the horizontal...

Projectile's axis), be denoted by \( t \); we have then \( Pn = dx \), \( np = dy \), and \( dy = dx \tan \theta \).

The intensity of the force of resistance being denoted as before by \( R \), and the mass of the projectile by \( m \), the accelerating force due to the resistance is \( R + m \). This force is exerted in the direction of the tangent at \( P \), and as it tends to diminish the velocity in the direction \( Pp \), it must be taken with the negative sign. Now, the accelerating force in the direction \( Pp \) is to the accelerating force in the direction \( Pn \) as \( 1 : \cos \theta \), and to the force in the direction \( np \) as \( 1 : \sin \theta \); therefore in the direction of the axis \( x \) the accelerating force is \( (R + m) \cos \theta \), and in the direction of the axis \( y \), \( (R + m) \sin \theta \). But \( \cos \theta = \frac{dx}{dz} \), and \( \sin \theta = \frac{dy}{dz} \), therefore the accelerating forces due to the resistance, in the direction of the co-ordinates, are \( -\frac{R}{m} \cdot \frac{dx}{dz} \) and \( -\frac{R}{m} \cdot \frac{dy}{dz} \) respectively. In the horizontal direction, the sole force acting on the projectile is that which is due to the resistance; in the vertical direction the body is also acted on by the accelerating force of gravity, \( g \), which, when the body is ascending, acts in the same direction with the resisting force of the atmosphere, and must therefore be taken with the same sign. Now, the accelerating forces in the directions \( x \) and \( y \), are respectively \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \) by equations (1); we have therefore for the equations of motion,

\[ \frac{d^2x}{dt^2} = -\frac{R}{m} \cdot \frac{dx}{dz}, \quad \frac{d^2y}{dt^2} = -g - \frac{R}{m} \cdot \frac{dy}{dz} \]

Assuming \( r = \text{radius of the projectile}, \delta = \text{its density}, \rho = \text{density of atmosphere}, v = \text{velocity at the point } P, \text{ and } n = \text{a constant co-efficient}, \text{we have}, \frac{R}{m} = \frac{np}{\delta r} v^2 \text{by equation (7). In the case of the vertical motion, we represented the co-efficient of } v^2 \text{ by the fraction } b + b' \text{ (} b \text{ being the terminal velocity)}; \text{it will now be more convenient to denote it by a single letter}; \text{we shall therefore assume } c = n\rho + b + b'; \text{and consequently, as } n \text{ and the ratio } \rho + \delta \text{ are both abstract numbers, the reciprocal of the constant } c, \text{or } 1 + c, \text{will represent a straight line}; \text{in fact, the reciprocal of } c \text{ is twice the altitude producing the terminal velocity } b \text{ in vacuo, for calling } h' \text{ this altitude, we have by formula (III)} b' = 2gh', \text{therefore } g + b' = 1 + 2h' = c. \text{We have then } R + m = cv^2; \text{whence, since } v = \frac{dz}{dt} \frac{R}{m} = \frac{dz}{dt}. \text{Substituting this in the above equations of motion (16) we get}

\[ \frac{d^2x}{dt^2} + c \frac{dz}{dt} \frac{dx}{dt} = 0 \]

\[ \frac{d^2y}{dt^2} + c \frac{dz}{dt} \frac{dy}{dt} + g = 0, \]

from which the equation of the trajectory must be deduced by integration.

To integrate the first of these equations, suppose \( \frac{dx}{dt} = ke^{-z} \), where \( k \) is an arbitrary constant, and \( e \) the base of the hyperbolic logarithms. The differential of this equation is

\[ \frac{d^2x}{dt^2} = -kce^{-z} dz, \text{or, dividing by } dt, \text{and substituting for } ke^{-z} \text{ its value, } \frac{d^2x}{dt^2} = -c \frac{dz}{dt} \frac{dx}{dt}, \]

which is the first of the above equations. Hence the integral of that equation is \( \frac{dx}{dt} = ke^{-z} \).

To determine the constant \( k \), suppose \( z = 0 \). When \( z = 0 \), the velocity in the direction of the tangent \( AK \) is the initial velocity \( a \), and in the direction \( AB \) it is \( a \cos A \), assuming \( A \) as before to represent the angle \( KAB \). But \( \frac{dx}{dt} = v \), the velocity at \( P \), therefore when \( z = 0 \), the equation \( \frac{dx}{dt} = ke^{-z} \) gives \( a \cos A = k \). The complete integral of the first of the equations (17) is therefore

\[ \frac{dx}{dt} = a \cos A e^{-z}. \]

(18)

The integral of the second equation (17) is found as follows. Since \( dy = \tan \theta dx \), we have, on supposing \( x \) and \( y \) to be functions of \( t \), \( \frac{dy}{dt} = \tan \theta \frac{dx}{dt} \). Differentiating this equation, and observing that \( d \tan \theta = \frac{d}{\cos^2 \theta} d \theta \), we get

\[ \frac{d^2y}{dt^2} = \frac{d}{\cos^2 \theta} \frac{dx}{dt} + \tan \theta \frac{d^2x}{dt^2}. \]

Substituting these values of \( \frac{d^2y}{dt^2} \) and \( \frac{dy}{dt} \) in the equation in question, it becomes

\[ \frac{d}{\cos^2 \theta} \frac{dx}{dt} + \tan \theta \left\{ \frac{d^2x}{dt^2} + c \frac{dz}{dt} \frac{dx}{dt} \right\} + g = 0; \]

whence, by reason of the first of the equations (17)

\[ \frac{d}{\cos^2 \theta} \frac{dx}{dt} + g = 0; \]

(19)

and on dividing by \( \frac{d^2x}{dt^2} = a^2 \cos^2 A e^{-2z} \) (18), we get

\[ \frac{d}{\cos^2 \theta} \frac{dx}{dt} = -\frac{g}{a^2 \cos^2 A e^{-2z}}. \]

Now if \( \theta \) be regarded as a function of \( x \), then \( \frac{d}{dt} \frac{dx}{dt} = \frac{d}{dx} \frac{dx}{dt} \); and from the formula (III) we have also \( a^2 = 2gh \) (\( h \) being the height due to the velocity \( a \)); therefore by substitution

\[ \frac{d}{\cos^2 \theta} \frac{dx}{dt} = \frac{1}{2h \cos^2 A e^{2z}}; \]

But we have \( dx = dz \cos \theta \), therefore this equation becomes

\[ \frac{d}{\cos^2 \theta} \frac{dz}{dt} = \frac{1}{2h \cos^2 A e^{2z}}; \]

(20)

which is the differential equation of the trajectory.

The integral of this equation is found as follows. By Fluxions (Art. 153),

\[ \int \frac{d}{\cos^2 \theta} = \frac{\sin \theta}{2 \cos^2 \theta} + \frac{1}{2} \int \frac{d \theta}{\cos \theta}; \]

(Art. 155),

\[ \int \frac{d}{\cos^2 \theta} = \log \tan (45^\circ + \frac{1}{2} \theta) \]

We have also

\[ \int e^{2z} dz = \frac{1}{2e} e^{2z}; \]

therefore assuming \( K \) to denote an arbitrary constant, the general integral becomes

\[ \frac{\sin \theta}{\cos^2 \theta} + \log \tan (45^\circ + \frac{1}{2} \theta) = K - \frac{1}{2ch \cos^2 A e^{2z}}. \]

To determine the constant, suppose \( z = 0 \). When \( z = 0 \) then \( \theta = A \), and the last equation gives

\[ K = \frac{1}{2ch \cos^2 A} + \frac{\sin A}{\cos^2 A} + \log \tan (45^\circ + \frac{1}{2} A). \]

(21)

For the sake of abridging, let us assume

\[ F(\theta) = \frac{\sin \theta}{\cos^2 \theta} + \log \tan (45^\circ + \frac{1}{2} \theta), \]

(22) then the complete integral of equation (20) becomes \( F(t) \)

\[ F(A) = \frac{\sin A}{\cos^2 A} + \log \tan (45^\circ + \frac{1}{2}A), \]

an equation which gives the length \( z \) of the arc \( AP \) in terms of the angle which the tangent at \( P \) makes with the horizontal line \( AB \).

We have now to find expressions for the co-ordinates \( x \) and \( y \). In consequence of the equation (21), the last may be otherwise written

\[ \frac{1}{2h \cos^2 A} e^{2z} = c \left[ K - F(\theta) \right]. \]

Substituting this in (20) we get

\[ \frac{d\theta}{\cos^2 \theta} = -dzc \left[ K - F(\theta) \right], \]

whence \( dz = \frac{d\theta}{c \cos^2 \theta \left[ F(\theta) - K \right]} \). Now, we have also \( dx = dz \cos \theta \), \( dy = dz \sin \theta \), therefore by substitution,

\[ dx = \frac{c \cos^2 \theta \left[ F(\theta) - K \right]}{\cos^2 \theta \left[ F(\theta) - K \right]}, \]

\[ dy = \tan \theta d\theta. \]

These expressions are not susceptible of integration in finite terms; but as \( x \) and \( y \) have the forms \( x = f(\psi(t)) \) and \( y = f(\psi(t)) \) denoting known functions of \( t \) and as \( \theta \) varies within given limits, from \( \theta = A \) to \( \theta = 0 \) in the ascending branch of the curve, and from \( \theta = 0 \) to \( \theta = 90^\circ \) in the descending branch, the problem is thus brought within the method of quadratures; that is to say, it depends on finding the areas of two curves, of which the ordinates are respectively \( \psi(\theta) \) and \( \psi(\theta) \), corresponding to the abscissas \( t \). Hence \( x \) and \( y \) may be found for given values of \( \theta \), and consequently any number of points may be found in the trajectory.

Instead of employing the equations (25), the curve may be constructed by means of equation (24), which is the method followed by Euler. That equation gives

\[ e^{2z} = 2ch \cos^2 A \left[ K - F(\theta) \right], \]

whence we have

\[ \log z = \frac{1}{2c} \log \left[ 2ch \cos^2 A \left[ K - F(\theta) \right] \right]. \]

Now, if \( \theta \) be the angle which the tangent makes with the horizon at the point \( P \) of the curve, and \( z \) the length of the arc \( AP \), and if at the point \( p \) at a small but finite distance from \( P \), the angle becomes \( \theta - \Delta \theta \), and the arc \( \Delta z \), then if the formula be computed for a given value of \( \theta \), (for example \( 30^\circ \)), and also for \( \theta - \Delta \theta \) (suppose \( 25^\circ \)), the difference between the two resulting numbers will give the length of the arc \( PP' = \Delta z \). Thus we can find the values of a great number of small portions, and the inclinations of the two tangents at their extremities. We can also assign to each of these portions its proportion of the abscissa and ordinate, without having recourse to the values of \( x \) and \( y \); for the portion of the abscissa corresponding to \( PP' \) or \( \Delta z \) being \( \Delta x \), and \( \theta - \Delta \theta \) being the inclination of its middle point to the horizon, we have \( \Delta x = \Delta z \cos (\theta - \Delta \theta) \); and in like manner \( \Delta y = \Delta z \sin (\theta - \Delta \theta) \). By adding all these portions together, the co-ordinates will be obtained, and consequently the points of the curve from \( \theta = A \) to \( \theta = 0 \), or through the ascending branch from \( A \) to the vertex. Near the point of projection, the variation of the length of the arc is much greater than that of the angle \( \theta \), and therefore at the commencement it may be necessary to compute \( \theta \) for each degree, that is, to assume \( \Delta \theta = 1^\circ \); but a sufficiently near approximation will be obtained by taking intervals of \( 5^\circ \), or \( \Delta \theta = 5^\circ \) throughout the rest of the curve. It will be observed that the logarithms of the formulae are hyperbolic; and therefore the logarithms of the common tables must be divided by the modulus \( M \).

We have still to find expressions for the time and velocity. From equation (19) \( \frac{d\theta}{\cos^2 \theta} \cdot dx = -gdt^2 \). Multiplying by \( c \), and substituting for \( dx \), its value given by the first of the equations (25), this becomes

\[ \frac{d\theta}{\cos^2 \theta \left[ F(\theta) - K \right]} = -cgdt^2, \]

whence, transposing and extracting the root,

\[ \sqrt{\frac{c}{g}} \cdot dt = \frac{d\theta}{\cos^2 \theta \left[ F(\theta) - K \right]}. \]

In this equation the radical must be regarded as a positive quantity, because when the time \( t \) increases, the angle \( \theta \) diminishes.

For the velocity we have from (19) \( \frac{dx}{dt} = -g \cos^2 \theta \frac{dt}{d\theta} \).

But \( dx = dz \cos \theta \), therefore \( v = \frac{dz}{dt} = -g \cos \theta \frac{dt}{d\theta} \), and

\[ v^2 = g^2 \cos^2 \theta \frac{d\theta}{d\theta}. \]

Substituting in this equation the value of \( \frac{d\theta}{d\theta} \) given by (26), we have

\[ v^2 = \frac{g}{c \cos^2 \theta \left[ K - F(\theta) \right]}. \]

As the vertex, \( \theta = 0 \), and \( \cos^2 \theta = 1 \); therefore, assuming \( u \) to denote the velocity at the vertex, the equation becomes

\[ u^2 = g + bK; \quad u^2 = b^2 + K; \]

whence it appears that \( K \) is the ratio of the square of the terminal velocity to the square of the velocity in the vertex.

If there were no resistance, the smallest velocity would be at the vertex, and beyond the vertex it would immediately increase by the action of gravity conspiring (in a small degree, however) with the motion of the body. But in a resisting medium the velocity at the vertex is diminished by a quantity to which the acceleration of gravity in that point bears no assignable proportion. It is therefore diminished upon the whole, and the point of smallest velocity is a little way beyond the vertex. It is not material for our present purpose to determine the exact position of that point.

By means of the expressions which have now been found, we are enabled to compute for every value of \( \theta \), from the point of projection where \( \theta = A \), to the vertex where \( \theta = 0 \), the corresponding point of the trajectory, the time which the projectile takes to arrive at that point, and the velocity with which it is then moving. We now proceed to consider the motion in the descending branch.

Let the origin of the co-ordinates (fig. 4.) be transferred to the vertex \( C \); let \( CE \) be horizontal, \( CD \) vertical, \( S \) a point in the descending branch, \( SN \) a tangent at \( S \), meeting \( CE \) in \( N \), and \( ST \) parallel to \( CD \). Also let \( x' \) and \( y' \) be the co-ordinates of \( S \) referred to the vertex, that is,

Projectiles make \( x = CT \), \( y = ST \), and suppose the co-ordinates of the vertex to be \( p \) and \( q \), namely \( AD = p \), \( CD = q \). Then the co-ordinates of \( S \) referred to \( A \) being always represented by \( x \) and \( y \), we have

\[ x = p + x', \quad y = q - y', \quad dx = dx', \quad dy = -dy'. \]

Now it will be observed, that in the descending branch the angle \( \theta \) is negative, or that \( ENS = -\theta \); whence the first of the two equations in (25) becomes in the present case

\[ dx' = \frac{-d\theta}{c \cos^2 \theta \{ F(\theta) - K \}}. \]

But from (23), \( F(-\theta) = \frac{\sin \theta}{\cos^2 \theta} + \log \tan (45^\circ - \frac{1}{2}\theta) \); and

\[ \tan (45^\circ - \frac{1}{2}\theta) = \cot (45^\circ + \frac{1}{2}\theta) = \frac{1}{\tan (45^\circ + \frac{1}{2}\theta)}; \quad \text{whence} \]

\[ \log \tan (45^\circ - \frac{1}{2}\theta) = \log \tan (45^\circ + \frac{1}{2}\theta), \quad \text{and consequently} \]

\( F(-\theta) = F(\theta) \). Equations (25) therefore become

\[ dx' = \frac{d\theta}{c \cos^2 \theta \{ K + F(\theta) \}}, \quad dy' = \frac{-\tan \theta d\theta}{c \cos^2 \theta \{ K + F(\theta) \}}. \tag{28} \]

In like manner the equation (24) becomes in this case

\[ e^{2\theta} = 2ch \cos^2 \theta \{ K + F(\theta) \}, \]

\( z \) being the arc counted from the vertex; and the expressions for the time and velocity become respectively

\[ \sqrt{cg} \cdot dt = \frac{d\theta}{\cos^2 \theta \{ K + F(\theta) \}}, \quad v^2 = \frac{g}{c \cos^2 \theta \{ K + F(\theta) \}}. \tag{29} \]

All these expressions are computed for successive values of \( \theta \) from \( \theta = 0 \); and a sufficient approximation will be obtained by assuming \( \theta \) to vary by intervals of \( 5^\circ \). The ascending and descending branches of the trajectory are thus computed independently, as if they belonged to different curves.

In the descending branch the angle \( \theta = ENS \) continually increases with the distance of the point \( S \) from \( C \), until it becomes a right angle, which is its limiting value. But \( \theta \) can only increase to a certain magnitude, which it would attain when \( S \) is at an infinite distance, and consequently this branch of the curve has an asymptote, which is perpendicular to the horizon. To prove this, let the factor \( K + F(\theta) \) be developed, and let us consider what it becomes when \( \theta \) approaches its limit. We have \( F(\theta) = \frac{\sin \theta}{\cos^2 \theta} + \log \tan (45^\circ + \frac{1}{2}\theta) \). Now when \( \theta \) approaches very nearly to a right angle, \( \sin \theta = 1 \), and \( \frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta \)

\[ = \tan^2 \theta; \quad \text{therefore} \quad \frac{\sin \theta}{\cos^2 \theta} = \tan^2 \theta. \quad \text{With respect to the second term, when } \theta \text{ is nearly } 90^\circ, \text{ then } \tan (45^\circ + \frac{1}{2}\theta) \text{ becomes very large, and nearly equal to } \tan \theta; \text{ but the logarithm of a very large number may be neglected as insensible in comparison of the square of the number, we have therefore simply } F(\theta) = \tan^2 \theta. \quad \text{In like manner, } K, \text{ which is a finite quantity, may be neglected in comparison of } \tan^2 \theta; \text{ hence the first of equations (28) becomes } dx' = \frac{d\theta}{c \sin^2 \theta}. \quad \text{The integral of this (Fluxions, art. 153) is } x' = C - \frac{1}{\tan^2 \theta}, \text{ where } C \text{ is an arbitrary constant. Hence } x \text{ does not increase indefinitely, but as } \theta \text{ increases, it ap-}

proaches nearer and nearer to a constant } C, \text{ with which it Projectiles coincides when } \theta \text{ is infinite. The constant } C \text{ cannot be determined from this equation; but if we assume } k = \frac{1}{c} \int_0^\infty \cos^2 \theta \{ K + F(\theta) \} d\theta, \text{ the value of } k \text{ may be approximate-ly computed by the method of quadratures. Suppose } k \text{ to be determined, make } CE = k, \text{ and draw through } E \text{ the vertical } EF, \text{ then } EF \text{ is the asymptote of the descending branch of the curve.}

The terminal velocity \( b \) may be determined from equation (27). On changing \( \theta \) into \(-\theta\) in that equation, and supposing \( \theta \) to become a right angle, we have \( b^2 = \frac{g}{c \sin^2 \theta} = \frac{g}{c}; \quad \text{whence} \quad b = \sqrt{\frac{g}{c}}, \text{ which agrees with the assumed value of } c, \text{ namely } c = g + b^2.

Although the motion commences at the point } A, \text{ the curve may be supposed to be continued infinitely below the line } AB. \text{ This branch of the curve has also an asymptote, the position of which it is important to determine. If in the equation (23)

\[ e^{2\theta} - 1 = 2ch \cos^2 \theta \{ F(A) - F(\theta) \}, \]

we suppose \( z \) to be negative and infinite, and assume } B \text{ to denote the value of } \theta \text{ when } z \text{ is infinite, we shall have

\[ F(B) = \frac{1}{2ch \cos^2 \theta} + F(A), \tag{30} \]

whence from equation (22), \( F(B) = K \). Substituting this value of } K \text{ in the two equations (25), and observing, that for any point in this branch of the curve below } AB, x \text{ and } y \text{ are both negative, we have}

\[ dx = \frac{d\theta}{c \cos^2 \theta \{ F(B) - F(\theta) \}}, \quad dy = \frac{-\tan \theta d\theta}{c \cos^2 \theta \{ F(B) - F(\theta) \}}. \]

Now let } P \text{ (fig. 4.) be a point in the curve, } PM \text{ a tangent at } P, \text{ meeting } AB \text{ in } M, \text{ let } AQ \text{ and } PQ \text{ be the co-ordinates of } P, \text{ and make } AM = l. \text{ We have then } l = AQ - MQ = x - \frac{y}{\tan \theta}. \quad \text{Suppose now the point } P \text{ at an infinite distance, then } \theta \text{ becomes } B, \text{ the tangent } MP \text{ becomes an asymptote, and } l = x - \frac{y}{\tan B}. \quad \text{Hence } dl = dx - \frac{dy}{\tan B}; \quad \text{and therefore, on substituting the above values of } dx \text{ and } dy, \text{ we get}

\[ dl = \frac{(\tan B - \tan \theta) d\theta}{c \tan B \cos^2 \theta \{ F(B) - F(\theta) \}}. \]

The integral of this expression from } \theta = A \text{ to } \theta = B \text{ will give the distance of the asymptote from the point. Let the definite integral be denoted by } \lambda, \text{ that is, let

\[ \lambda = \frac{1}{c \tan B} \int_A^B \frac{(\tan B - \tan \theta) d\theta}{\cos^2 \theta \{ F(B) - F(\theta) \}}. \tag{31} \]

then } \lambda \text{ may be computed by the method of quadratures. If we then take on the prolongation of } AB, AL \text{ equal to } \lambda, \text{ and draw through } L \text{ the line } LG, \text{ making with } AB \text{ an angle } = B, \text{ then } LG \text{ will be the asymptote of the branch } AC. \text{ The whole curve will then lie between the two asymptotes, which are inclined to each other in an angle } = 90^\circ - B.

Such, then, is the process by which the form and magnitude of the trajectory, and the motion in it may be determined. But it does not yet appear how this is to be appli- Projectiles.

By means of the preceding formulae, we are enabled to compute the motion from any point in the ascending branch to the vertex; where the motion becomes horizontal, and the motion from the vertex to a point in the descending branch, where the motion has acquired another determinate direction; the constants which enter into the formulae being $c$ and $K$, of which $c = \frac{g}{\sqrt{b}}$ depends on the terminal velocity, and therefore on the nature of the projectile and the ratio of its density to that of the air; and $K$ on the initial velocity, the terminal velocity, and the angle of elevation. We have also shown that $K$ denotes the ratio of the square of the terminal velocity to the square of the velocity in the vertex; therefore the terminal velocity and velocity in the vertex may be regarded as the two constants of the formula. The mode in which the formulae are applied to the solution of the usual question, namely, to find the motion of a ball projected in a certain direction, with a certain velocity, is as follows: Suppose a trajectory, computed for a particular terminal velocity $b$, and for a particular velocity at the vertex, and that the velocity at that point of the ascending branch where the inclination of the tangent is $30^\circ$ is 900 feet per second. Then, we are certain, that if a ball, whose terminal velocity is $b$, be projected with the velocity of 900 feet per second, and an elevation of $30^\circ$, it will describe this very trajectory, and the velocity and time corresponding to every point will be such as is here determined.

Now this trajectory will, in respect to form, answer an infinity of cases: for its characteristic is the proportion of the velocity in the vertex to the terminal velocity. If, therefore, we compute the trajectories for a sufficient variety of these proportions, we shall find a trajectory that will nearly correspond to any case that can be proposed: and an approximation sufficiently exact will be had by taking a proportional medium between the two trajectories which come nearest to the case proposed.

Accordingly, a set of tables or trajectories have been computed by the English translator of Euler's Commentary on Robins's Gunnery. They are in number 18, distinguished by the position of the asymptote of the ascending branch. This is given for $5^\circ$, $10^\circ$, $15^\circ$, &c., to $85^\circ$, and the whole trajectory is computed as far as it can ever be supposed to extend in practice.

Since the path of a projectile is much less incurvated, and more rapid in the ascending than in the descending branch, and the difference is so much the more remarkable in great velocities, it must follow, that the range on a horizontal or inclined plane depends most on the ascending branch; therefore the greatest range will not be made with that elevation which bisects the angle of position, but with a lower elevation; and the deviation from the bisecting elevation will be greater as the initial velocities are greater. It is difficult to frame a rule for determining with much precision, the elevation which gives the greatest range.

When the angle of elevation is very small, and consequently the path of the projectile nearly horizontal, the formulae admit of simplifications which it is important to consider. Suppose KAB (fig. 5) the angle of elevation to be small, then $\theta$, the angle which the tangent at any other point of the trajectory makes with AB, will be very small, and $\sin \theta$ and $\tan \theta$ will be very small fractions. Suppose the angle so small that $\tan \theta$ may be neglected as insensible; then $\cos \theta$ may be regarded as unit, and the equation $dx = dz \cos \theta$ gives $dx = dz$. Hence $x = z$ and equation (20) becomes

$$\frac{d\theta}{\cos^2 \theta} = -\frac{dx}{2h \cos^2 A e^{2\theta}},$$

the integral of which, since $d.e^{2\theta} = 2c e^{2\theta} dx$, is

$$\tan \theta = \frac{1}{4ch \cos^2 A e^{2\theta}} + C.$$

To determine the constant we suppose $x = 0$, in which case $\theta = A$, and we have

$$\tan A = \frac{1}{4ch \cos^2 A} + C,$$

whence the complete integral is

$$\tan \theta = \tan A - \frac{1}{4ch \cos^2 A} \left( e^{2\theta} - 1 \right).$$

Multiplying both sides of this equation by $dx$, observing that $dy = dx \tan \theta$, and integrating, we get

$$y = \tan A x - \frac{1}{4ch \cos^2 A} \left( \frac{1}{2c} e^{2\theta} - x \right) + C.$$

But when $x = 0$, $y = 0$, and this equation becomes

$$0 = \frac{1}{8c^2 h \cos^2 A} + C;$$

therefore the complete integral is

$$y = x \tan A - \frac{1}{8c^2 h \cos^2 A} \left( e^{2\theta} - 2c \theta - 1 \right),$$

which gives an approximate equation of the trajectory between the co-ordinates $x$ and $y$.

To determine the time corresponding to the horizontal distance $x$, we have from equation (19)

$$\frac{d\theta}{\cos^2 \theta} = -gd\theta,$$

whence by reason of (32) $dt = \frac{1}{2gh \cos^2 A e^{2\theta} dx}$, and consequently

$$dt = \frac{1}{\sqrt{(2gh) \cos A}} e^{2\theta} dx.$$

Integrating this expression, and determining the constant so that $x = 0$ when $t = 0$, we get

$$t = \frac{1}{c \cos A \sqrt{(2gh)}} \left( e^{2\theta} - 1 \right).$$

The two equations (33) and (34) may be employed for the purpose of determining the coefficient $c$ of the resistance on the terminal velocity which depends on it, from experiments. Suppose the trajectory to be continued to D, a small distance below the horizontal line AB. For this point of the curve the approximation may still be regarded as sufficiently accurate. Let DE be vertical, make AE = $a$, ED = $\beta$, and the time in which the projectile passes from A to D = $r$. Substituting these values of $x$, $y$, and $t$ in the equations (33) and (34), making $\cos A = 1$, because A is by supposition a very small angle, and observing that since D is below AB, $y$ must be taken as negative, the two equations give

$$8h c^2 (\beta + a \tan A) = e^{2\theta} - 2ca - 1,$$

$$r c \sqrt{(2gh)} = e^{2\theta} - 1.$$

Now, of the quantities involved in these equations, $g$ and $\tan A$ are known, and $a$, $\beta$, $r$, can be determined by direct experiments, leaving only $c$ and $h$ to be computed from the equations. On squaring the second, and substituting the resulting value of $h$ in the first, we obtain the equation

$$\frac{(4e^{2\theta} - 1)^2}{r^2 g} (\beta + a \tan A) = e^{2\theta} - 2ca - 1,$$

from which $c$ may be computed, and thence $h$ by means of one of the equations (35). Of the quantities which enter into the equations (35) and which are to be found from experiment, that which can be determined with the least certainty is \( t \), the time of flight. This, however, may be eliminated by varying the experiment. Suppose the projectile to be discharged at a different height above the ground, for instance the top of a tower, (the angle of elevation \( A \), and the initial velocity remaining the same,) and let \( a, b \) become respectively \( a', b' \); we have then, from the first of equations (35),

\[ 8ac^2(b + a\tan A) = e^{2ca} - 2ca - 1 \]

whence

\[ (b + a\tan A)(e^{2ca} - 2ca - 1) = (b + a\tan A)(e^{2ca'} - 2ca' - 1) \]

an equation from which the value of \( c \) may be computed when \( a, b \) and also \( a', b' \) are determined from experiments.

The solution which we have now given of this celebrated problem is exact in principle, and the application of it by no means difficult or even opose. But let us see what advantage we are likely to derive from it.

In the first place, it is very limited in its application. There are few circumstances of general coincidence, and almost every case requires an appropriate calculus. Perhaps the only general rules are the two following:

1. Balls of equal density, projected with the same elevation, and with velocities which are as the square roots of their diameters, will describe similar curves. This is evident, because, in this case, the resistance will be in the ratio of their quantities of motion. Therefore all the homologous lines of the motion will be in the proportion of the diameters.

2. If the initial velocities of balls projected with the same elevation are in the inverse subduplicate ratio of the whole resistances, the ranges, and all the homologous lines of their track, will be inversely as those resistances.

These theorems are of considerable use; for by means of a proper series of experiments on one ball projected with different elevations and velocities, tables may be constructed which give the motions of an infinity of others.

But when we take a retrospective view of what we have done, and consider the conditions which were assumed in the solution of the problem, we shall find that much yet remains before it can be rendered of great practical use. The resistance is all along supposed to be in the duplicate ratio of the velocity; but even theory points out many causes of deviation from this law, such as the pressure and condensation of the air, in the case of very swift motions; and Mr. Robins's experiments are sufficient to show us that the deviations must be exceedingly great in such cases. Euler and all subsequent writers have allowed that it may be three times greater, even in cases which frequently occur; and Euler gives a rule for ascertaining with tolerable accuracy what this increase and the whole resistance may amount to.

Let \( H \) be the height of a column of air whose weight is equivalent to the resistance taken, in the duplicate ratio of the velocity. The whole resistance will be expressed by \( H + \frac{H^2}{28845} \). This number 28845 is the height in feet of a column of air whose weight balances its elasticity. We shall not at present call in question his reason for assigning this precise addition. They are rather reasons of arithmetical convenience than of physical import. It is enough to observe, that if this measure of the resistance is introduced into the process of investigation, it is totally changed; and it is not too much to say, that with this complication it requires the knowledge and address of a Euler to make even a partial and very limited approximation to a solution. Any law of the resistance, therefore, which is more complicated than what Bernoulli has assumed, namely, that of a simple power of the velocity, must be abandoned on account of the intricacy of the Projectiles calculations; and mathematicians have attempted to avoid the errorising from the assumption of the duplicate ratio of the velocity, either by supposing the resistance throughout the whole trajectory to be greater than what it is in general, or they have divided the trajectory into different portions, and assigned different resistances to each, which vary, through the whole of that portion, in the duplicate ratio of the velocities. In this manner they make up a trajectory and motion which correspond, in some tolerable degree, not with an accurate theory, but with a series of experiments. For, in the first place, every theoretical computation that we make, proceeds on a supposed initial velocity; and this cannot be ascertained with any thing approaching to precision, by any theory of the action of gunpowder that we are yet possessed of. In the next place, our theories of the resisting power of the air are entirely established on the experiments on the flights of shot and shells, and are corrected and amended till they tally with the most approved experiments we can find. We do not learn the ranges of a gun by theory, but the theory by the range of the gun. Now the variety and irregularity of all the experiments which are appealed to are so great, and the acknowledged difference between the resistance to slow and swift motions is also so great, that there is hardly any supposition which can be made concerning the resistance, that will not agree in its results with many of those experiments. It appears from the experiments of Dr. Hutton of Woolwich, in 1784, 1785, and 1786, that the shots frequently deviated to the right or left of their intended track 200, 300, and sometimes 400 yards. This deviation was quite accidental and anomalous, and there can be no doubt but that the shot deviated from its intended and supposed elevation as much as it deviated from the intended vertical plane, and this without any opportunity of measuring or discovering the deviation. Now, when we have the whole ranges from one to three to choose among for our measure of resistance, it is evident that the confirmations which have been drawn from the ranges of shot are but feeble arguments for the truth of any opinion. Mr. Robins finds his measures fully confirmed by experiments made at Metz and Minorca. Mr. Muller finds the same. Yet Mr. Robins's measures, both of the initial velocity and of the resistance, are at least treble of Mr. Muller's; but by compensation they give the same results. The Chevalier Borda has added the very same experiments in support of his theory, in which he abides by the Newtonian measure of the resistance, which is about \( \frac{1}{2} \) of Mr. Robins's, and about \( \frac{3}{4} \) of Muller's. From all this we may conclude, that we have as yet a very uncertain and imperfect knowledge of the law and the measure of the air's resistance.

There is another essential defect in the conditions assumed in the solution. The density of the air is supposed uniform; whereas we are certain that it is less by one-fifth or one-sixth towards the vertex of the curve, in many cases which frequently occur, than it is at the beginning and end of the flight. This is another latitude given to authors in their assumptions of the air's resistance. Borda has, with considerable ingenuity, accommodated his investigation to this circumstance, by dividing the trajectory into portions, and, without much trouble, has made one equation answer them all. We are disposed to think that his solution of the problem (in the Memoire of the Academy of Paris for 1769) corresponds better with the physical circumstances of the case than any other. But this process is there delivered in too concise a manner to be intelligible to a person not perfectly familiar with all the resources of modern analysis.

After all, the practical artillerist must rely chiefly on the records of experiments contained in the books of practice at the academies, or those made in a more public manner. Projectiles. Even a perfect theory of the air's resistance can do him little service, unless the force of gunpowder were uniform. This is far from being the case even in the same powder. A few hours of a damp day will make a greater difference than occurs in any theory; and, in service, it is only by trial that everything is performed. If the first shell fall very much short of the mark, a little more powder is added; and in cannonading, the correction is made by varying the elevation.

The experiments of Mr. Robins and Dr. Hutton show, in the most incontrovertible manner, that the resistance to a motion exceeding 1100 feet in a second, is almost three times greater than in the duplicate ratio of the resistance to moderate velocities. Euler's translator, in his comparison of the author's trajectories with experiment, supposes it to be no greater. Yet the coincidence is very great. The same may be said of Borda's. Nay, the same may be said of Mr. Robins's own practical rules: for he assumes the quantity which represents the height due to the terminal velocity, almost double of what these authors do, and yet his rules are confirmed by practice.

But it must not be inferred from all this, that the physical theory is of no use to the practical artillerist. It plainly shows him the impropriety of giving the projectile an enormous velocity. This velocity is of no effect after 200 or 300 yards at farthest, because it is so rapidly reduced by the prodigious resistance of the air. Mr. Robins has deduced several practical maxims of the greatest importance from what we already know of this subject, and which could hardly have been even conjectured without this knowledge. See Gunnery.

It must also be acknowledged, that this branch of physical science is highly interesting to the philosopher; nor should he despair of carrying it to a greater perfection. The defects arise almost entirely from our ignorance of the law of variation of the air's resistance. Experiments may be contrived much more conducive to our information here than those commonly resorted to. The oblique flights of projectiles are, as we have seen, of very complicated investigation, and ill fitted for instructing us; but numerous and well contrived experiments on the perpendicular ascents are of great simplicity, being affected by nothing but the air's resistance.

We shall conclude this article, by giving two tables, computed from the principles established above, and which serve to bring into one point of view the chief circumstances of the motion in a resisting medium. Although the result of much calculation, as any person who considers the subject will readily see, they must not be considered as offering any very accurate results; or that, in comparison with one or two experiments, the differences shall not be considerable. Let any person peruse the published registers of experiments which have been made with every attention, and he will see such enormous irregularities, that all expectations of perfect agreement with them must cease. In the experiments at Woolwich in 1785, which were continued for several days, not only do the experiments of one day differ among themselves, but the mean of all the experiments of one day differs from the mean of all the experiments of another no less than one-fourth of the whole. The experiments in which the greatest regularity may be expected, are those made with great elevations. When the elevation is small, the range is more affected by a change of velocity, and still more by any deviation from the supposed or intended direction of the shot.

The first table shows the distance in yards to which a ball projected with the velocity of 1600 feet in a second will go, before its velocity is reduced one-tenth, and the distance at which it drops 16 feet from the line of its direction. This table is calculated by the resistance observed in Mr. Robins's experiments. The first column is the weight of the ball in pounds. The second column remains the same whatever be the initial velocity; but the third column depends on the velocity. It is here given for the velocity which is very usual in military service, and its use is to assist us in directing the gun to the mark. If the mark at which a ball of 24 pounds is directed is 474 yards distant, the axis of the piece must be pointed 16 feet higher than the mark. These deflections from the line of direction are nearly as the squares of the distances.

| L | II | III | |---|----|-----| | 2 | 92 | 420 | | 4 | 121 | 428 | | 9 | 159 | 456 | | 18 | 200 | 470 | | 32 | 272 | 479 |

The next table contains the ranges in yards of a 24-pound shot, projected at an elevation of 45°, with the different velocities in feet per second, expressed in the first column. The second column contains the distances to which the ball would go in vacuo in a horizontal plane; and

| L | II | III | IV | V | |---|----|-----|----|---| | 200 | 416 | 349 | 106 | 360 | | 400 | 1664 | 1121 | 338 | 1150 | | 600 | 3740 | 1812 | 606 | 1859 | | 800 | 6649 | 2373 | 806 | 2435 | | 1000 | 10300 | 2845 | 1138 | 2919 | | 1200 | 14961 | 3259 | 1378 | 3343 | | 1400 | 20364 | 3640 | 1606 | 3734 | | 1600 | 26597 | 3950 | 1814 | 4050 | | 1800 | 33963 | 4235 | 1992 | 4345 | | 2000 | 41559 | 4494 | 2168 | 4610 | | 2200 | 50286 | 4720 | 2348 | 4842 | | 2400 | 59846 | 4917 | 2460 | 5044 | | 2600 | 70235 | 5106 | 2630 | 5238 | | 2800 | 81456 | 5293 | 2760 | 5430 | | 3000 | 93465 | 5455 | 2862 | 5596 |

the third contains the distances to which it will go through the air. The fourth column is added, to shew the height to which it rises in the air; and the fifth shows the ranges corrected for the diminution of the air's density as the bullet ascends, and may therefore be called the corrected range.

The initial velocities can never be pushed as far as they have been calculated for in this table; but we mean it for a table of more extensive use than appears at first sight. Recollect, that while the proportion of the velocity at the vertex to the terminal velocity remains the same, the curves will be similar; therefore, if the initial velocities are as the square roots of the diameters of the balls, they will describe similar curves, and the ranges will be as the diameters of the balls.

Let it be required, for example, to find the range of a 12-pound shot, projected at an elevation of 45°, with the velocity 1500; suppose the diameter of the twelve-pounder to be d, and that of the twenty-four-pounder D; and let the velocities be v and V. Then say, \( \sqrt{d} : \sqrt{D} = 1500 : \) a fourth proportional V. If the twenty-four-pounder be projected with the velocity V, it will describe a curve similar to that described by the twelve-pounder, having the initial velocity 1500. Therefore find (by interpolation) the range of the twenty-four-pounder, having the initial velocity V. Call this R. Then \( D : d = R : r \), the range of the twelve-pounder which was wanted, and which, assuming the diameters of Projection: the balls to be 4-45 and 5-61 inches, respectively, will be found 3312.

This table shews the immense difference between the motions through the air and in a void. It shews also that the ranges through the air, instead of increasing in the duplicate ratio of the initial velocities, really increase slower than those velocities in all cases of military service; and in the most usual cases, viz. from 800 to 1600, they increase nearly as the square roots of the velocities.

A set of similar tables, made for different elevations, would almost complete what can be done by theory, and would be much more expeditious in their use than Mr. Euler's Trajectories, computed with great labour by his English translator.

The same table may also serve for computing the ranges of bomb-shells. We have only to find what must be the initial velocity of the twenty-four-pound shot which corresponds to the proposed velocity of the shell. This must be deduced from the diameter and weight of the shell, by making the velocity of the twenty-four pounder such, that the ratio of its weight to the resistance may be the same as in the shell.

Dr. Hutton, in his Mathematical Tracts (vol. iii. p. 276), has given a similar table, deduced partly from theory and partly from experiment, but the numbers in the three last columns differ greatly from the above, particularly for the higher velocities. Thus, for the velocity of 1600 feet, Dr. Hutton's corrected range is 2264 yards, and the height to which the ball rises 650 yards, instead of 4050 and 1814, as in our table.

For further information on this subject, see Gunnery; Robin's Mathematical Tracts; Brown's True Principles of Gunnery; Hutton's Mathematical Tracts; and for an example of the computation of a trajectory by the method of quadratures, Legendre's Exercices de Calcul, tome i. p. 390.