The Motion of Projectiles Near the Earth's Surface Consider'd, Independent of the Properties of the Conic Sections; In a Letter to Martin Folkes Esquire, Pr. R. S. by Mr. Tho. Simpson F. R. S.

happen'd about the Time of the natural Birth, the Child then must have continued alive some considerable Time afterwards, during which these bony Excrescences were formed; there being a perfect Ossification, as performed by the Laws of Circulation, and not by any vegetative or petrifying Power, as in inanimate Bodies. Two or three of the lateral Processes of the Spine were what first passed thro' the little Ulcer; the rest of the Bones (except a few that were lost in cleaning) were presented by the Doctor to the Museum of the Royal Society. They retain a very strong and singular Smell, though they were immediately cleansed from the rotten Flesh, and well washed. The Woman came by Sea to Stockholm above a Year after this Cure, and was presented to the Academy in good Health; and the Doctor believes she is still alive and well. II. The Motion of Projectiles near the Earth's Surface consider'd, independent of the Properties of the Conic Sections; in a Letter to Martin Folkes Esquire, Pr. R. S. by Mr. Tho. Simpson F. R. S.. Read Feb. 4. AFTER so much as has been already said upon the Motion of Projectiles in vacuo, it may seem needless to attempt any thing further on that Head; nevertheless, as a thorough Knowledge in the Art of Gunnery is become more than than ever necessary, and as Gentlemen employ'd in the Practice of that Art are (I am sensible) too often deterr'd from applying themselves to the Theory, by the Difficulties they imagine they shall meet with in the Conic Sections, you will, I hope, pardon the Liberty I have taken, in troubling you with my Thoughts on a Subject, in which little or nothing new is to be expected besides the Method. When I first drew up this Paper (which was about two Years ago) I did intend, had Health permitted me to make the proper Experiments, to have also attempted something with respect to the Resistance of the Atmosphere, whereof the Effects are indeed too considerable to be intirely disregarded: But if the Amplitude of the Projection, answering to one given Elevation, be first determined by Experiment (which our Method supposes) the Amplitudes in all other Cases, where the Elevations and Velocities do not very much differ from the first, may be determined, by the Proportions here laid down, to a sufficient Degree of Exactness: Because, in all such Cases, the Effects of the Resistance will be nearly as the Amplitudes themselves; and were they accurately so, the Proportions of the Amplitudes, at different Elevations, would be exactly the same as in vacuo; which Proportions I now proceed to determine. **Problem I.** Let two Balls be projected with the same Celerity, at different, but given Elevations, 'tis proposed to determine the Ratio of the Times of their Flight, Flight, of their greatest Altitudes, and of their horizontal Amplitudes. Let $Pq$ (Fig. 1.) represent the Plane of the Horizon, $PEQ$ and $peq$ the Paths of the Projectiles, described in the Flight; moreover let $QPT$ and $qpt$ be the given Angles of Elevation, and let $PQ$ and $pq$ be bisected in $H$ and $h$; drawing $HE$, $QT$, $he$ and $qt$, all perpendicular to $Pq$; and making the Sine of $QPT = s$, its Co-sine $= c$, the Sine $qpt = s$, its Co-sine $= c$, and Radius $= r$. Therefore, since the Distances descended by heavy Bodies (whether from a Point at Rest, or from the right Lines in which they would move, if not acted upon by Gravity) are known to be as the Squares of the Times, $QT$ will be to $qt$, as the Square of the Time of describing $PEQ$ (or of that wherein the Ball would move uniformly over the Space $PT$ with its first Velocity at $P$) is to the Square of the Time of describing $peq$ (or of that wherein the other Ball would move uniformly thro' the Length $pt$). But the Celerities at $P$ and $p$ being equal, by Hypothesis, the Times in which the said Lines $PT$ and $pt$ would be uniformly described, are manifestly, as the Lines themselves: Whence the Squares of those Lines must, also, be as the Squares of the Times, and, consequently, as the Distances descended: that is, $Pt^2 : pt^2 :: TQ : tq$. Now, by Plane Trigonometry $TQ = \frac{s \times PT}{r}$ and $tq = \frac{s \times pt}{r}$; therefore $PT^2 : pt^2 (\because \frac{s \times PT}{r} : \frac{s \times pt}{r}) :: S \times PT : s \times pt$; whence, by dividing the Antecedents by $T$ by \( PT \), and the Consequents by \( pt \), we have \( PT \). \( pt :: S : s \); from which it appears, that the Times of Flight are directly as the Sines of Elevation. Again, the Times of describing \( EQ \) and \( eq \) (which are the Halves of the Wholes) being also to one another as \( S : s \), and the Distances \( EH \), \( eb \) descended in them, as the Squares of the Times, it likewise follows, that \( S^2 : s^2 :: EH : eb \); or that the greatest Altitudes are as the Squares of the Sines of Elevation. Moreover, because (by Trigonometry) \( PT = \frac{r \times PQ}{C} \) and \( pt = \frac{r \times pq}{c} \); and it has been already proved, that, \( S : s :: PT : pt \), it follows, that \( S : s :: \frac{r \times PQ}{C} : \frac{r \times pq}{c} \); whence, by multiplying the Antecedents by \( \frac{2C}{r} \) and the Consequents by \( \frac{2c}{r} \), it will be \( \frac{2SC}{r} : \frac{2cs}{r} \) (\( :: 2PQ : 2pq \)) :: \( PQ : pq \). But \( \frac{2SC}{r} \) is known to be the Sine of double the Angle whose Sine is \( S \), and Co-sine \( C \), &c. Therefore the horizontal Amplitudes are to one another, as the Sines of the double Elevations. Corol. 1. Hence it follows, that the greatest Amplitude possible will be, when the Elevation is half a Right Angle, or 45 Degrees (because the Sine of \( 90^\circ \) is the greatest of all others). Corol. Corol. 2. Therefore, if the greatest Amplitude be given (from Experiment) the Amplitude answering to any proposed Elevation, above, or below, 45 Degrees, may from hence be found: For it will be as the Radius, to the Sine of double the given Elevation, so is the greatest, to the required, Amplitude. Corol. 3. Hence, also, the Altitude of the Projection may be known; for QT, when the Angle QPT is half a Right Angle, will be = PQ; and therefore HE \( \frac{1}{4} TQ \) = \( \frac{1}{4} PQ \); also, in this Case, \( S^2 = \frac{1}{2} r^2 \); whence our Proportion \( S^2 : s^2 :: HE : he \) will here become \( \frac{1}{2} r^2 : s^2 :: \frac{1}{4} PQ : he \); from whence it appears, that, as the Square of the Radius is to the Square of the Sine of any given Elevation, so is half the greatest horizontal Amplitude, to the Altitude of the Projection. Hence it also follows, that the Height to which the Ball would ascend, if projected directly upwards, is just half the greatest Amplitude. Corol. 4. Therefore, since it is well known, that a Body in vacuo ascends and descends with the same Velocity; and that the Distances descended are as the Squares of the Velocities; it follows, that the Amplitudes, at the same Elevation, with different Velocities, will also be to one another as the Squares of the Velocities; because they are as the greatest Amplitudes, with the same Velocities (by Corol. 2.) and these are as the Distances perpendicularly descended (by the precedent). Whence, universally, if both the Elevations and the Velocities differ, the Amplitudes will be to each other in a Ratio compounded of the Ratio's of the Sines of double the Angles of Elevation, and of the duplicate Ratio's of the Velocities, or impelling Forces. Problem II. The Angle of Elevation, and the greatest horizontal Amplitude, being given, to find at what Distance the Piece ought to be planted, to hit an Object, whose Distance, above or below the Plane of the Horizon, is also given. Let \( AB \) (Fig. 2 and 3.) be the Plane of the Horizon, \( BC \) the perpendicular Height or Depression of the Object, and \( AB \) the required Distance: Also let \( BC \) be produced to meet the Line of Direction \( AD \) in \( D \), and let \( P \) be the Place where the Path of the Projectile would meet the Horizon; moreover, let \( PQ \) be perpendicular to \( AP \), and \( CN \) parallel to \( AD \). Then, by the preceding Problem, it will be as Radius: the Sine of \( 2BAD \) :: the given (or greatest) Amplitude: \( AP \); which therefore, is known. Moreover, the Areas of similar Triangles being as the Squares of their homologous Sides, we have \( AP \times PQ : AB \times BD :: AQ^2 : AD^2 \). But \( AQ^2 : AD^2 :: AB \times BD : QP : DC \) (from Principles already explained) therefore, by Equality, \( AP \times PQ : AB \times BD :: QP : DC \); and consequently \( AP : AB :: BD : CD \); but (because of the parallel Lines \( CN \) and \( AD \)) \( BD : CD :: AB : AN \); whence, again again by Equality, \( AP : AB :: AB : AN \); therefore, by Division, \( AP : BP :: AB : BN \); and, consequently \( AP \times BN = BP \times AB \). Let \( AP \) be now bisected in \( O \); then \( BP \times AB \) being \( AO^2 - OB^2 \) (in the first Case) and \( OB^2 - AO^2 \) (in the second Case), we shall therefore have \( OB^2 = AO^2 + AP \times BN = AO \times AO + 2BN \): whence the Distance \( AB \) is likewise known. Q.E.I. Corollary. Hence, if the Elevation, and the greatest Amplitude, together with the Distance \( AB \) of the Object be given, the Height or Depression of the Ball in the Perpendicular \( BCD \) will be known: For it is proved, that \( AP : BP :: BA : BN \); whence \( BN \) is known: But, as the Radius to the Tangent of \( BNC \) (\( BAD \)): so is \( BN \) to \( BC \). Problem III. The greatest horizontal Amplitudes of the Piece, together with the Distance and Height (or Depression) of the Object being given, to find the Direction or Angle of Elevation. Let \( BC \) (Fig. 4 and 5.) be the perpendicular Height or Depression of the Object, \( AB \) its given horizontal Distance, and \( AH \) the required Direction; Also let \( PQ \) (Fig. 6.) be the greatest Amplitude (answering to 45° of Elevation); draw \( AC \), in which produced (if need be) take \( AG = PQ \); make \( MGO \) perpendicular to \( AG \), meeting \( AB \) produced (if need be) in \( O \); and from the Centre \( O \), with the Interval \( OA \), let a Circle be described, intersecting \( AG \), produced in \( E \), and the Line of Direction \( AD \) in \( H \); join \( E, H \), and let \( HI, AN \) and \( QR \), be perpendicular to \( AE, AO \), and \( PQ \) respectively, and let \( BC \), produced, meet \( AH \) in \( D \). It will appear, from what has been said above, that \( AD^2 : PR^2 :: DC : RQ \); therefore \( PR^2 \) being \( = 2PQ^2 = 2AG^2 = \frac{1}{2}AE^2 \), and \( RQ = PQ = \frac{1}{2}AE \) (by Construction), we have \( AD^2 : \frac{1}{2}AE :: DC : \frac{1}{2}AE \), and therefore \( AD^2 = AE \times DC \). Now, the Triangles \( ADC, AEH \), being equiangular (because \( ADC = DAN = A EH \), and \( DAC \) common to both) we likewise have \( AD : DC :: AE : EH \), and consequently \( AE \times DC = AD \times EH = AD^2 \) (per above); whence \( EH = AD \). Therefore, as the Triangles \( ADB \) and \( EHI \) are equiangular, they are equal in all respects; and so \( HI = AB \): Whence follows this easy Construction. Having described the Circle \( AEF \), as above directed, and drawn \( MG \) perpendicular to \( AE \), take \( Gn \) equal to \( AB \), and thro’ \( n \), parallel to \( AE \), draw \( Hb \), cutting the Circle in \( H \) and \( b \); join \( A, H \), and \( A, b \); then either of the Directions \( AH \) or \( Ab \), will answer the Conditions of the Problem. From this Construction we have the following Calculation, viz. As \( AB \) is to \( BC \), so is \( AG \) to \( OG \); which added to, or subtracted from, \( Gn \) (\( AR \)) gives \( On \): Then, it will be, as \( AG : On :: \) the Co-sine of \( OAG : \) Co-sine of \( HOn (= HA_b) \) the Difference of the two required required Elevations; whence the Elevations themselves are known. Q.E.I. Corol. 1. Hence, if the Elevation of the Piece, with the Distance and the Height (or Depression) of the Object be given, the greatest horizontal Amplitude may be found: For it will be $AB : BC :: \text{Radius} : \text{Tang. of } BAC$; whence $CAD$ is also known. Then, $S.CAD : S.ACD (AHE) :: AD (HE) : AE$. And, $S.ADC : \text{Radius} :: AB : AD$. Therefore, by compounding these Proportions, we have $S.CAD \times S.ADC : \text{Radius} \times S.ACD :: AB : AE$; which is equal to twice the required Amplitude, by Construction. Corol. 2. Moreover, if the Elevation, and the greatest horizontal Amplitude be given, the Amplitude of the Projection on any ascending or descending Plane $AE$, whose Inclination $FAE$ is also given, may from hence be derived. For, $S.AHE (AC'D) : S.EAH (CAD) :: AE (2PQ) : EH (AD)$ and $S.ACD : S.ADC :: AD : AC$; whence, by compounding the two Proportions, $Sq.S.ACD : S.CAD \times S.ADC :: 2PQ : AC$; from which $AC$ is known. Corol. 3. Since it appears, that the Triangles $ADB$ and $EHI$ are equal and alike in all respects, and, therefore, the horizontal Distance $AB$, universally, equal to the Perpendicular $HI$, it is manifest, that, when HI is the greatest possible, AB will also be the greatest possible; in which Circumstance AC (if the Angle FAE be given) will likewise be the greatest possible: And this, it is evident, must be, when HI coincides with MG, or when the Angles HEA and HAE are equal (as in Fig. 7 and 8); at which time the Point D coincides with H; because AD and EH are always equal to each other. Therefore, since, in this Case, HAE (HEA) is = NAH, it follows, that the Amplitude, on any inclined Plane AE, will be the greatest possible, when the Line of Direction AH bisects the Angle made by the Plane and Zenith. Corol. 4 Hence the greatest Amplitude on any inclined Plane may also be known; for the right-angled Triangles AOG and HOB, having AO = HO and the Angle O common, are equal in all respects; and therefore, as Tang. of AHG (BAH the Angle of Elevation): Tang. of CHG (CAB the Plane's Inclination) :: AG : GC; whence AC = AG + GC is also known. Corol. 5. Hence, also, if the greatest Amplitude on an inclin'd Plane be given, the greatest horizontal Amplitude may be determined: For, Radius : S. BAC :: AC : BC = CG = the Difference of the given, and the required, Amplitudes. Corol. 6. But if, instead of the Plane's Inclination, the perpendicular Height, or Depression, of the Object be given; then, AC (AG + BC) being to BC, as Radius to to the Sine of \( BAC \), and Radius : Cotang. \( BAC :: BC : AB \); the greatest Distance \( AB \), at which the Ball can possibly hit the Object, will from hence be given: which Distance (because \( AC = AG + BC \), and \( AB^2 = AC + CB \times AC - BC \)) will also be expressed by \( \sqrt{AG \times AG + 2BC} \). Hence the greatest horizontal Amplitude of a Ball, projected from a given Height above the Plane of the Horizon is known: For \( ST \) (Fig. 8.) may here be supposed to represent the Plane of the Horizon, and \( SA \) the given Height; and then \( SC \), being equal to \( AB \), is given from above \( = \sqrt{AG \times AG + 2BC} \). Corol. 7. But, if the horizontal Distance \( AB \) be given, and it be required to find the greatest Height the Ball can possibly reach in the Perpendicular \( BCD \); we shall have \( HG : AB : AG :: \text{Radius : Tang. of the Elevation } (BAH \text{ or } AHG) \); and Radius : Tang. \( BAC \) (\( 2BAH = 90^\circ \)) :: \( AB : BC \); which therefore is known. But (because \( AC + BC = AG \), and \( AC + CB \times AC - CB = AB^2 \)) the same will also be truly exhibited by \( \frac{AG^2}{2AG} \). Corol. 8. Lastly, let the Height, or Depression, of the Object be given, together with its Distance \( AB \), to determine the Direction, and the least Impetus possible, to hit the Object: Then \( AB : BC :: \text{Radius : Tang. } BAC \); whence the Elevation \( BAH \) is known: And as Radius : Tang. \( AHG \) (\( BAH \)) :: \( MG \) (\( AB \)) : \( AG \); whence the Impetus is also known.