To HIT a MARK, As well upon Ascents and Descents, As upon the Plain of the Horizon: Experimentally and Mathematically DEMONSTRATED. BY ROBERT ANDERSON. LONDON: Printed for Robert Morden, at the Atlas near the Royal Exchange in Cornhill. MDCXC. To the Right Honourable Sir Henry Goodrick, Knt and Baront, Lieutenant General of their MAJESTY'S Ordnance, And One of His MAJESTY'S Most Honourable Privy-Council, AND To the Honourable JOHN CHARLETON, Esq Master-Surveyor of Their MAJESTY'S Ordnance, AND ALSO, To the rest of the Principal Officers of the same: This New Work Is Humbly Dedicated By Your most Observant, ROB. ANDERSON. To all Ingenious Persons that delight in the TRUE USE OF Great Artillery. IN the Year 1673. I published my Genuine Use of the Gun, viz. To strike a Mark at demand, within the reach of the Piece Analytically, upon the Principles of Galileus. In the Year 75. Monsieur Blondell propounded that Problem to the Royal Academy of Sciences of France to be resolved Geometrically, Mons. Buot, Romer the Sweed, and Monsieur de la Hire set upon the Mathematical part, whilst Monsieur Mariote and Perrault, with others, employed themselves about Experiments of several sorts, yet not of the Gun, which is the chief end of that Problem, Monsigneur the Dauphin being present. In the Year 86. Mr. Halley, in Transaction Number 179. gives a Resolution of the same Problem. All these purely Mathematical, without the consideration of any Accidental Impediments. Galileus tells us the beginning of the Parabola will be deformed, by reason of the Impulse of Fire: And the latter end, by reason of the Resistance of the Air, which amounts to very little, saith he. Cavalierus bids us begin the Parabola where the Force leaves the thing projected. And this Line I call the Line of Impulse of Fire, and take it for a right Line for ease of Calculation; although I believe the thing projected moves as it can so far as the Impulse of Fire, or violent shake of the Engine is upon it, and the more irregular the thing projected is, the longer the Line will be before it passeth into its Parabola. In these last 15 Years I have made more than a Thousand Experiments (and am ready to make as many more to illustrate these Truths) although I set down so few; by which I find the Impulse of Fire, and so carry the Work on upon the Principles of Galileus, viz. To hit a Mark not only upon the Plain of the Horizon, but upon Ascents and Descents in the Parabola, according to Mathematical Principles. For ease of the Mortar-Piece, I have made a Table of Horizontal Ranges, and the like may be done for Ascents and Descents, if it were necessary; Also, by the Experiments of Mr. Eldred and my own compared together, I have made Tables of Ranges for Long Guns to 8 degrees of Elevation, which I am apt to believe comes near the matter; that so when proved and approved, the said Ranges may be graved upon the Guns; so their Use will become very easy and exact. Who ever go about to prove or disprove these things, I would have them qualified with these three Qualifications, viz. Art, Care, and Honesty, and not to stand a tiptoe to see farther than Nature and Art have made them capable. To make the Work more perfect, I have given a short Discourse of Granades, Fusees, Carcases, and Fire-Balls. The service of War being ended, we turn our Guns into Pipes of Music, playing the Charms of a Soldiers delight, in Consorts of Musical Instruments, where Sound and Time are actually employed. Last of all, Guns in Geometrical Proportion, beautiful in shape, and delightful in sound, which might delight the Minds of noble Souls. These things I have done for the Truth, which has been my Recreation, and at my own Charge, and now may be useful, if faithfully put in practice, which I am apt to believe will not be corrected in this Age, nor perhaps in the next. In the Mortar-Piece, I verily believe, there cannot be three Minutes got; nor in the Long Guns ten, perhaps not five Minutes, taking every thing in its right sense, which I shall ever endeavour to do, whilst I am R. A. ON THE Most Ingenious AUTHOR, Mr. ROB. ANDERSON, and his New Work. AMongst the several Parts of Mathematical Elements, the Sections of a Cone, and Algebra, have ever been accounted the most abstruse and curious: But the adapting these to explicate the Theories of heavy Bodies forced from Engines, and the like, is that wherein the Age preceding ours, exceeded all that went before it, by the almost immense and universal Wit of Galileus: His Scholar Torricellio succeeded him, and with most admirable Felicity and Learning rendered those matters easy, which his Master had made Mathematical. But, To hit a Mark (within the Pieces reach) in any Angle above or under the Horizontal level of the Piece, was what they could not overcome. And Torricellius does professedly avow the Difficulties thereof to be almost insuperable. But this, our Author, by his extraordinary Abilities in Mathematics, happy Address, and indefatigable Pains and Diligence, has not only overcome, but further added many useful things, to render this New Science profitable. Those who have written on this Subject since the Author's Printing of his Genuine Use of the Gun, etc. 1673. have all confirmed what he first published to the World. Nor can this Doctrine meet with any Opposition, except from those who neither understand the Laws of Motion, nor the Nature of that Line wherein a Bullet moves. His long and great Experince (having in Twenty Years made many Thousand Shots, with greater Accuracy each, perhaps, than Men of common Practice ever have made one); his great Curiosity, Expense, and Labour; of all which this present Piece of his is an undoubted Instance; but most of all his Zeal to benefit the Public, without regard of profiting himself, renders him meritorious, not only of the Thanks and Praises of his Friends, but as he truly is an Honour to our Age and Nation, the same is due to him from all Mankind. The matter of this present Tract being wholly new, it could not reach that Length at which Collectors aim, who write the Histories (as I may say, the Life and Death) of such or such a Science: But it truly seems to be all that was wanting to complete the Art. Now having its Foundation laid in Scientifique Certitude, confirmed by much and long Experience, and thus explained by Numbers; it neither needs nor can have greater Commendation, other than reading of the Book itself, from which I shall no longer keep you. B. M. TO Hit a MARK. PROPOSITION I. GALILEUS Dialogo Quarto, Prop. I. Projectum dum fertur motu composito ex horizontali aequabili, & ex naturaliter accelerato deorsum, lineam Semiperabolicam deseribit in sua latione. If a thing be projected in a Motion compounded of an equal Horizontal, and naturally accelerated motion downward, it describeth in its passage a Semiparabolical Line. And from Page 180 to 190, Discourses of the Accidental Impediments, viz. Of the Roundness of the Earth; The Impediment of the Air; and The Impulse of Fire. Here I shall endeavour to inquire what the Quality and Quantity of these accidental Impediments are. PROP. II. EXperiments made with a Mortar-piece well fixed, the Diameter 4 Inches, the length of the Chase 2 Diameters, the Chase and Ball turned very fit. On Wimbleton Heath, August 24. 1676. ½ at 15 deg. of Elevation Ranged the Shot 1336 1357 1378 ¾ at 75 deg. of Elevation Ranged the Shot 1232 1337 1442 ½ at 10 deg. of Elevation Ranged the Shot 539 538 537 ¾ at 80 deg. of Elevation Ranged the Shot 514 507 500 Septemb. 17. 1677. 1 at 15 deg. of Elevat. Ranged the Shot 901 2 at 75 deg. of Elevat. Ranged the Shot 848 3 at 00 deg. of Elevat. Ranged the Shot 125 The Shot fell lower than the Centre of the Mouth of the Piece 2.35 Lo Specchio ustorio, Bonaventura Cavasieri, Pag. 99 Dico adunque, che I gravi spinti dal proiciente à qualsivoglia banda, fuorche per la perpendicolare all' Orizonte, separati che siano da quello, & esclusa l'impedimento dell' ambiente, descrivono una linea curva, insensibilinente different dalla Parabola. I say then, that heavy Bodies being projected by any force, except it be perpendicular to the Horizon, and excluding the Impediment of the Medium, do after their separation from that force describe a curve Line insensibly differing from a Parabola. PROP. III. PROB. I. TWO Ranges made upon the plain of the Horizon, equi-distant above and below 45 deg. of Elevation, with the same Piece, Ball, and quantity of Powder being given. To find the Line of impulse of Fire, and greatest Range in the Parabola. Let R be the Axis on which the Piece moves, in Fig. 1. The Range RI at 15 deg. of Elevation 901 the half 450.5 = B The Range RX at 75 deg. of Elevat. 848 the half 424 = C The sine of the Angle RFZ = S The sine of the Angle R nr = s The Line of Impulse of Fire RF the thing sought = Z Then as Radius: Z:: S: RZ in the Diagoras. that is in Diagoras. and in the Diagoras. then , Galileus Dialog. quart. Prop. oct. that is, RB − ZS = RC − Z s, that is , that is, Half the difference of the Ranges multiplied by Radius, divided by the difference of the sine Compliment of Elevations, the Quote is the Radius RF = Z the thing sought, the Line of Impulse of Fire. Here we take RO to be equal to HEY, which is not exactly so; yet nevertheless it is sufficient for the purpose. Example. ½ The difference of the Ranges in Radius 2650000 6.423246 The difference of the sins of 15 and 75 deg. 70711 sub 4.849487 The Radius of the Circle RF = Z the Line of Impulse of Fire 37.48 1.573759 To find the greatest Range in the Parabola. As the Radius 90 10.000000 is to the sine of 15 9.412996 so is the Radius RF 37.48 1.573759 to R r 9.7 0.986755 As the Radius 90 10.000000 is to the sine of 75. 9.984944 so is the Radius RF 37.48 1.573759 to RZ 36.2 1.558703 ½ the Range on the Horiz. at 75° 424 ½ the Range on the Horizon at 15° 450.5 R r sub 9.7 RZ subst. 36.2 ND 414.3 FV 414.3 The Range at 75° NE, in Parab. 828.6 equ. to the Ran. FG at 15° in Parab. 828.6 Then the double of NE or FG 1657.2 the greatest Rang in the Parabola. PROP. IU. PROB. II. THE Piece put parallel to the Horizon at R, Ranged the shot RMC, that is, the horizontal distance AC, the descent RAMIRES equal to MB, with the Range at 15 deg. of Elevation upon the plain of the Horizon RI being given. To find RM the Line of Impulse of Fire. The Range RI at 15 deg. of Elevat. 901 the half 450.5 = B The Piece put parallel to the Horizon Ranged the shot 125 AC = A RA the height of the Piece above where the shot fell 2.35 = C The sine of the Angle RFZ = S Radius = R RM = AB the Line of impulse of Fire, the thing sought = Z Then in the Diagoras. in the Diagoras. Then A − Z its square A ² − 2AZ+Z ², that square divided by 8C is in Diagr. so then , that is RAMIRES ² − 2RAZ+RZ ² = 8CBR − 8CSZ, that is, . In Numbers thus, Z ² = 231.84 Z = 7155.6 it's square Root is 36.67 = Z = RF the Line of Impulse. Here note, if A ² − 2AZ+Z ² be divided by C, the Quote will be the double of the greatest Range in the Parabola; if by 2C the greatest Range, if by 4C the half of the greatest Range, if by 8C a quarter of the greatest Range in the Parabola FU. PROP. V. PROB. III. THE horizontal distance, and the descent being given. To find the double of the greatest Range in the Parabola. BM, 2.35. AC, 125 less AB, 3667. Their remains BC, 88.33 being given. To find the greatest Range in the parabola. BC 88.33 1.946108 squared 3.892216 BM 2.35 sub 0.371068 The double of the greatest Range which we call G 3320 3.521148 The half thereof is 1060 the greatest Range. PROP. VI PROB. IU. THE horizontal distance, the descent, and the double of the greatest Range in the Parabola being given. To find the Line of impulse of Fire. AC 125 = A RA equal to MB 2.35 = C G the double of the greatest Range 3320 being given. To find RM the Line of impulse = Z. Then A − Z, squared is A ² − 2AZ+Z ² = GC, that is, In Numbers thus Z ² = 250Z − 7823. Whose square Root is 36.68 = Z = RF the Line of Impulse. A method to extract the Roots of square Equations. Take half the number of the coefficient. To the square of that half, add or subtract the absolute number, according to the Sign + or −, then extract the square Root of that sum or difference, which Root added to or substracted from the half coefficient, the sum or difference will be the Root of the Equation. Example. Take half of 250 (which is called the Coefficient) 125 it's square is 15625 from that square subtract the absolute Number 7823 by the sign −, the remainder is 7802, the Root thereof 88.32, substracted from 125 the remainder is 36.68 the Root required. In these Experiments at 0 degree of Elevation, and at 15 deg. of Elevation there could be very little resistance of the Medium; therefore it is the impulse of Fire carrying the Ball in a right line, so far as the blast of the Piece is upon it, which after its separation from that force, passeth into the Curve of a Parabola. By this means the point N is removed from the point F, which is one, if not the chief reason, that RX and RI are not equal upon the plain of the Horizon. PROP. VII. PROB. V. THE three last Propositions, viz. the 4th. 5th. and 6th. may be done Geometrically. Thus, if (in the 4th.) BC in the Diagr. be made equal to 8C in the equation, and CD in the Diagoras. be equal to B in the equation, than CE is a mean proportion. If EA in the Diagoras. be equal to A in the equation, and OF in the Diagoras. be equal to half the Coefficient in the equation, then HA and GA' will be the Roots of the equation, and GA' equal to 36.67. If we take mean proportions between 2A and R, and 8C and S in the Coefficient part of the equation, the construction will be more completed. In the 5th. If BC in the Diagr. be made equal to BM in the 5th. Prop. and CE in the Diagr. be equal to BC in the 5th. Prop. Then CD in the Diagr. a 3d. proportion will be equal to G the double of the greatest Range in the Parabola. In the 6th. If BC in the Diagr. be made equal to C in the equation, and CD in the Diagoras. be equal to G in the equation, than CE is a mean proportion, make EA in the Diagoras. equal to A in the equation, than IE, and KE are the Roots of the equation, and KE = 36.67. PROP. VIII. PROB. VI IF two shot be made at the same degree of elevation, with different quantities of Powder, the corresponding parts of the Ranges will be in proportion, sufficiently near the truth. And to find the parts simile. june the 5th. 1677. On Wimbleton-Heath, a Mortar-piece whose diameter of bore 4 inches, and the length of the Chase 2 diameters, charged with 4 ounces of Powder, and laid to 15 deg. of elevation, Ranged its shot to the horizontal distance of 659 paces. By this, and what is done by the Experiments before cited, the qualifications of all Mortar-pieces are known, being of the same kind. Example. As RI is to RF fig. 1. the Range at 15 deg. of elevation 901 2.954725 the Line of impulse 37.48 1.573759 so is AB to AD fig. 3. the Range at 15 deg. of elevation 659 2.818885 the Line of impulse 27.41 1.437919 As RI is to FG fig. 1. the Rang. at 15 deg. of elevat. 901 on the Hor. 2.954725 the Rang. at 15 deg. of elevat. 8286 in the Parab. 2.918345 so is AB to DE fig. 3. the Rang. at 15 deg. of elevat. 659 on the Horiz. 2.818885 the Rang. at 15 deg. of elevat. 606 in the Parab. 2.782505 DE 606 the Range at 15 deg. of elevat. in the Parabola, the double of it is 1212 Paces the greatest Range at 45 deg. of elevat. in the Parabola. PROP. IX. PROB. VII. THE Line of impulse, the greatest Range in the Parabola, and the Angle of elevation being given. To find the Range upon the plain of the Horizon. In the right angled Triangle ABC, there is given the Angle BAC, AD the line of impulse, and G the double of the greatest Range in the Parabola, with this qualification, as G: DC:: DC: CB. To find AB the horizontal Range. BAC = S sine 70 93969 R = 100000 G = 2424 AD = r 27.41 DC = Z 2304.89 Then, as R: r+Z:: S: BC in diag. that is in diag. and in diag. by Lemma to the 23 Prop. De motu projectorum Liber Secundus, Torricel. Therefore That is Z1R = ZGS+G rS. In Numbers thus, Z ² = 2277. 80856Z+62434.7326296, the Root is 2304.89 = DC. Then DC 2304. 89+DA27.41S = AC2332. 3 3.367784 sine 70 9.534052 AB 797.69 2.901836 A Second way by Lines thus, Geometrically. In Fig. X. If AC be made Radius, and CB the sine of the Angle of elevation, and AD the Line of impulse, draw DE parallel to CB, then DE will be a fourth proportion by 12.6 Euclid. Then make VL in Fig. Y. equal to DE in Fig. X, and LE in Fig. Y, equal to the double of the greatest Range; the Line LN will be a mean proportion, 13.6. Euclid. Further, if AC in Fig. X. be made Radius, and CB the sine of the Angle of elevation, and AD the greatest Range from the point D draw DE parallel to BC, DE will be a fourth proportion; Lastly, make OL in Fig. Y. equal to DE in Fig. X. then draw the Line NOM, and it will be the Line required, viz. = Z in the equation, and DC in the first Fig. of this Proposition. A Third way by Logarithms. G = 2424 3.384533 G = 2424 3.384533 S s 93969 4.972985 S s 93969 4.972985 r r 27.41 1.437909 8.357518 from 9.795427 from 2R = 200000 5.301030 subst. R = 100000 5.000000 subst. OL 1138.9 3.056488 subst. 4.795427 rest LN+Radius 12.397713 from LN = 2.397713 half LON tang. 12.22.30 9.341225 rest LON sine 12.22.30 9.331041 subst. LN+Radius 12.397713 from ON 1165.9 3.066672 rest LO+ON 2304.8 viz. = Z in the Equation, and DC in the first Fig. of this Proposition. A Fourth Way shall be showed, where we come to calculate the horizontal Ranges. PROP. X. PROB. VIII. THE Horizontal distance, the Line of Impulse, the double of the greatest Range in the Parabola being given. To find the Angle of Elevation, to hit an Object at the Horizontal distance given. In the Right angled triangle ABC, AB the Horizontal distance, AD the Line of Impulse, the Line G the double of the greatest Range, with this qualification, as G: DC:: DC: CB being given. To find the Angle BAC, the angle of Elevation, to hit the Object at B, at the Horizontal distance AB. AB = H = 264 G = 568.867 To find the Roots Z = 487.3 = DC Paces. AD = r 6.534 Z = 302.8 = DC Then in the Diagr. r+Z = AC in Diagr. squared is I. Euclid. that is That Equation in Numbers. − Z4+323609. 663689Z2+4228931. 085087852Z = 22540483202.613561987516 The Roots are had, by the usual method of extracting mixed Powers, which will be very easy to any ingenious Person, with a little practice; the lesser Root is greater than EBB, and lesser than DC at 45 deg. of Elevation. The greater Root DC, is greater than DC at 45 deg. of Elevation, and lesser than DC, at the Compliment of the lower Elevation from the Zenith. DC 302.8 DC 487.3 AD 6.534 AD 6.534 AC 309.334 AC 493.834 As AC 309.33 2.490422 is to the Radius 90 10.000000 so is AB 264 2.421604 to the sine of the angle BAC 31° 24′ 30″ 9.931182 As AC 493.83 2.490422 is to the Radius 90 10.000000 so is AB 264 2.421604 to the sine of the angle BAC 57° 41′ 9.728027 On Wimbleton-Heath, 1689. june 17. 1. at 10 deg. of Elevation the shot Ranged 610 2. at 78.56 of Elevation the shot Ranged 616 3. at 15 deg. of Elevation the shot Ranged 680 4. at 73.57 of Elevation the shot Ranged 724 August 13. 1. at 15 deg. of Elevation the shot Ranged 694 2. at 15 deg. of Elevation the shot Ranged 680 3. at 73.57 of Elevation the shot Ranged 711 4. at 73.57 of Elevation the shot Ranged 682 1. at 31.24 of Elevation the shot Ranged 1953 2. at 31.24 of Elevation the shot Ranged 1932 3. at 57.41 of Elevation the shot Ranged 1975 4. at 57.41 of Elevation the shot Ranged 2020 1. at 31.24 of Elevation the shot Ranged 1077 2. at 31.24 of Elevation the shot Ranged 1105 3. at 57.41 of Elevation the shot Ranged 1077 4. at 57.41 of Elevation the shot Ranged 1135 Here note, If the Horizontal distance be greater than the greatest Range of the Piece; the Object is beyond the reach of the Piece. On Wimbleton-Heath, June 26. 1689. 1. at 15 deg. of Elevation the shot Ranged 1149 2. at 15 deg. of Elevation the shot Ranged 1197 − 1173 3. at 75 deg. of Elevation the shot Ranged 1104 4. at 75 deg. of Elevation the shot Ranged 1102 − 1103 In Fig. I. ½ the difference of the Ranges in Radi 3500000 6.544068 the difference of the Sins of 15 and 75 deg 70711 subst. 4.849487 the Radius of the Circle RF the line of Impulse 49.5 1.694581 As the Radi 90 10.000000 is to the sine of 15° 9.412996 so is the Radi RF 49.5 1.694581 to R r 12.8 1.107577 As the Radi 90 10.000000 is to the sine of 75° 9.984944 so is the Radi RF 49.5 1.694581 to RZ 47.8 1.679525 ½ the Rang. on the Hor. at 75° 551.5 ½ the Rang. on the Horiz. at 15° 586.5 R r subst. 12.8 RZ subst. 478 ND 538.7 FV 538.7 the Ran. at 75° NE in the Par. 1077.4 equ. to the Ran. FG at 15° in Par. 1077.4 As RI is to RF Fig. I. the Range at 15 deg. of Elevation 901 2.954725 the Line of Impulse 37.48 1.573759 In this Experiment the Range at 15 deg. of Elevation 1173 3.069298 the Line of Impulse 48.8 1.688332 Here may be seen, the fair agreement of the Parabola, with Experiments: For 49.5 less 48.8 is but seven Tenths. Upon Ascents and Descents. PROP. XI. PROB. IX. THE angle of Elevation, the double of the greatest Range, the Ascent or Descent being given. To find the Horizontal distance. In the right angled Triangle ABC, the Angle BAC, the double of the greatest Range G, and the Ascent BE, with this qualification; as G: AC:: AC: CE being given: To find AC = Z. Then, as R: Z:: S: BC in Diagr. that is in Diagr. Then in Diagoras. Then in Diagoras. that is Z2R = ZG − GPR. G = 2424 BE = P = 120 the sine of 80 = S = 98581 Radius = R = 100000 AC = Z = 2258.37 In Numbers thus, Z ² = 2387. 17944Z − 290880 whose Roots are 2258.37 = AC = Z, and 128.8 = AC = Z. By Logarithms thus, G = 2424 3.384533 G = 2424 3.384533 P = 120 2.079181 S s 98481 4.993352 ½ 5.463714 8.377885 from LO+Radius is 12.731857 2R = 200000 5.301030 subst. HO = 1193.6 3.076855 subst. LO+Radius 12.731857 from sine LHO 9.655002 rest co-sine LOH 9.950405 HO = 1193.6 3.076855 HL = 1064.8 3.027260 HO+HL = 2258.4 = AC = Z HO − HL = 128.8 = AC = Z Radius 10.000000 AC 2258.4 3.353801 BCA sine 10° 9.239670 AB 392.17 2.593471 AB the Horizontal distance. In Fig. Z, if VL be made equal to P, and LE equal to G, then LO will be a mean Proportion. Further, in Fig. X, if AC be made equal to Radius, and BC equal to the sine of the Angle of Elevation; as also, AD be made equal to ½ G, that is equal to the greatest Range in the Parabola, then DE will be a fourth Proportion. Further, from the point O in Fig. Z, make OH equal to DE in Fig. X, continue the Line OH to I. Lastly, set one point of the Compasses in H, and at the distance HL, sweep the Arch ILR, then are IO and RO the Lines required. PROP. XII. PROB. X. THE Angle of Elevation, the double of the greatest Range, and the Descent being given. To find the Horizontal distance. In the right angled Triangle ABC, the Angle BAC, the double of the greatest Range G, and the descent BE, with this qualification as G: AC:: AC: CE being given. To find AC = Z. Then, as R: Z:: S: BC in Diagram that is in Diagr. Then in Diagoras. Then in Diagr. that is Z2R = ZGS+GPR. G = 2424 BE = P = 120 The sine of 80 = S = 98481 Radius = R = 100000 AC = Z = 2503.37 In Numbers thus, Z ² = 2387. 17944Z+290880, whose Root is 2503.37. By Logarithms thus, Radi 90 10.000000 AC 25034 3.398530 BCA sine 10 9.239670 AB 434.7 2.638200 AB the Horizontal distance. By Lines thus. In Fig. X and Z, if VL be made equal to P, and LE equal to G, then LO will be a mean Proportion. Further, in Fig. X, if AC be made equal to Radius, and CB equal to the sine of the Angle of Elevation; as also, AD be be made equal to ½G than DE will be a fourth Proportion. Again make LH in Fig. Z, equal to DE in Fig. X, draw the Line OHI, set one foot of the Compasses in H, and at the distance HL strike the Arch LI, then IO will be the Line required. PROP. XIII. PROB. XI. THE Line of Impulse, the double of the greatest Range in the Parabola, the Ascent or Descent, and the Angle of Elevation being given. To find the Horizontal distance. Then R: r+Z:: S: BC, that is in Diagoras. in Diagr. then in Diagoras. that is i. e. G = 2424 BE = P = 120 AD = r = 2741 sine of 80 = S = 98481 Radius = R = 100000 DC = Z = 2288.67 In Numbers, thus, Z ² = 2387. 17944Z − 225447.4115496 whose Roots are 2288.67 and 98.51. Upon an Ascent. In Fig. Y, if AC be made Radius, and CB the sine of the Angle of Elevation, and AD the Line of impulse, and DE drawn parallel to CB, then DE will be a fourth Proportion. In Fig. Z, if ODD be made equal to DE in Fig. Y, and DE in Fig. Z be made equal to the double of the greatest Range in the Parabola, then DC will be a mean Proportion. Further, if RC be made equal to P the perpendicular Ascent, and CB equal to the double of the greatest Range, then CA will be a mean Proportion. Again, in Fig. Y, if AC be made equal to the Radius, and CB equal to the sine of the angle of Elevation, and AD the greatest Range, then DE will be a fourth Proportion, make A in Fig. Z, equal to DE in Fig. Y, then LA and HA are the Lines desired. By Logarithms thus, G Log. 3.384533 G Log. 3.384533 G Log. 3.384533 r 1.437909 S 4.993352 P 2.079181 S 4.993352 8.377885 ½ 5.463714 9.815794 2R 5301030 AC is 2.731857 subst. R sub. 5.000000 OF 1193.6 3.076855 DC+Ra. 12.407897 from 4.815794 9.676040 si. of DAC DC 2.407897 ½ 9.944661 si. of DCA Lastly. AC 2.731857 AD 27.41 AD from 12.676518 DC 2288.7 OF subst. 3.076855 1193.6 AC 2316.1 3.364757 rest 9.599663 si. of AFD ABC 10 sine. 9.239670 add OF 9.962590 si. of DAF AB 402.2 2.604427 3039445 1095.1 the Horizontal distance desired. ALL the Number 2288.7 de. PROP. XIV. PROB. XII. THE Line of Impulse, the Descent, the greatest Range in the Parabola; the Angle of Elevation being given. To find the Horizontal distance. In the right angled Triangle ABC, the Angle BAC, AD the Line of Impulse, and G the double of the greatest Range in the Parabola being given, with this qualification G: DC:: DC: CE. To find AB the Horizontal distance. Then R: r+Z:: S: BC that is in Diagr. in Diagram. Then in Diagr. that is that is . G = 2424 BE = P = 120 the Sine of 80 = S = 98481 Radius = R = 100000 AD = r = 27.41 DC = Z = 2528.11 In Numbers thus, Z ² = 2387. 17944Z+356312.5884504 the Root is 2528.11. In Fig. Y, if AC be made Radius, and CB, the sine of the Angle of Elevation; and AD the Line of impulse; then DE will be a fourth Proportion. In Fig. X. If ODD be made equal to DE in Fig. Y, and DE in Fig. X, be made equal to the double of the greatest Range, than DC will be a mean Proportion. In Fig. X, if DR be equal to the Descent, and DB equal to the double of the greatest Range, DA will be a mean proportion. Then draw the Line AC. Further, in Fig. Y. If AC be Radius, and DC the sine of the Angle of Elevation, and AD the greatest Range, then DE will be a fourth proportion. In Fig. X, draw CF (at right Angle to AC) equal to DE in Fig. Y, then set one point of the Compasses in the point F, and take the distance FC, sweep the Arch CL, then LA will be the Line desired. By Logarithms thus, G Log. 3.384533 G Log. 3.384533 G Log. 3.384533 S 4993352 S 4.903352 P 2.079181 r 1.437909 8. 3●7885 ½ 5.463714 9.815794 2R sub. 5.301030 DAMN is 2.731857 R subst. 5.000000 GF 1193.6 3.076855 DC ½ 4.815794 DC+Radi. is 12.407897 from DA 2.731857 subst. Tang. DAC 9676040. sine DAC 9631992 subst. AD 27.41 DC+Radi. 12.407897 from DC 2528.1 AC 2.775905 subst. AC 2555.5 3.407476 FC+Radi. 13.676855 from ACB sine 10 9.239670 Tang. CAF 10.300950 AB 443.8 2.647146 sine CAF 9.951528 subst. OF 1334.5 3.125327 CF 1193.6 ALL 2528.1 the Numb. desired. Ascents. Descents. AB 402.2 with the Line of Impulse. AB 443.8 with the Line of Impulse. AB 392.2 without the Line of Impul. AB 434.7 without the Line of Imp. 10.0 the Horizontal difference. 9.1 the Horizontal difference. PROP. XV. PROB. XIII. THE greatest Range, the Horizontal distance, Perpendicular Ascent, and Line of Impulse being given. To find the Angle of Elevation; to hit the said Ascent at the Horizontal distance given. Then in Diagoras. Further, in Diagr. that squared is that +H ² is equal to the square of r+Z, that is r2+2Z r+Z ² that is . 47. I. Euclid. that is G = 2424 AB = H = 392.2 BE = P = 120 AD = R = 27.41 Radius = R = 100000 DC = Z = 2290.6 In Numbers thus, Z4 − 5294016Z ² − 322110040. 32Z = − 984013457269.2544. Whose Root is 2290.6 = DC for the upper Elevation. Then, DC 2290.6 AD 27.41 AC 2318.01 subst. 3.365113 Radius 90 10.000000 AB 392.17 2.593474 Co-sine of the Angle ACB 80ᵒ 15′ 40″ 9.228361 The Angle of Elevation to hit the Object at E. PROP. XVI. PROB. XIV. THE double of the greatest Range in the Parabola, the Horizontal distance, the Perpendicular Ascent or Descent being given. To find whether the Object be within the Reach of the Piece or not. Then AY equal to 2AF, or 4AE in EO equal to the square of OL. Appol. Coni. Lib. 1. Prop. 11. Therefore if the Horizontal Distance AGNOSTUS be equal to OL, it is in the farthest extent of the Projection; if it be less as AB, the Point C is within; but if greater, as AI, the Point N is without, therefore beyond the Reach of the Piece, that is, if half the greatest Range, less the Altitude of the Object in Ascents, more in Descents, be multiplied by the Parameter (equal to the double of the greatest Range) whose square Root, if less than the Horizontal distance, the Object is out of the Reach of the Piece. AB = 392.17 OE = 486 2.686636 EY = 2424 EY = 2424 3.384533 OF = 1212 6.071169 A = 606 OL = 1085.4 3.035584 AO = 120 AB = 392.17 less than OL = 1085.4 therefore the Object at C is within the Reach of the Piece. If any find the XV. Proposition tedious, they may make use of the Tables in my genuine Use of the Gun; for the difference is not much, as appears in the Close of the XIV. Proposition; the difference is 10 in the Ascent and 9 in the Descent, in that place; and in the XV. Proposition it is 16 minutes. By the Tables in my genuine Use of the Gun. Greatest Rang. G 1212 sub. 3.083503 greatest Range G 1212 sub. 3083503 Greatest Rang. G 2000 3.301030 greatest Range G 2000 3.301030 Horizont. dist. H 392.17 2.593474 Perpend. Ascent P 120 2. 0●9181 Horizont. dist. H 647.15 2.811001 Perpend. Ascent P 198 2.296708 Then entering the Tables with the Horizontal distance 647.15 and Perpendicular Ascent 198, and you will find the Elevations 80 deg. and 27. deg. Thus have I given a Method to strike any place at demand within the Reach of the Peice, as well upon Ascents and Descents, as upon the plain of the Horizon, by Experimental and Mathematical demonstration. It remains now to say something of Mortarpieces, Granadaes, Fusees, Bombs, Carcases, and Fire-Balls, etc. About 70 or 80 degrees of Elevation is the best place to work a Mortar-Piece for service: Therefore I have given Rules from the Elevation given, to find the Horizontal distance, which is as necessary as the Converse. These Examples are of both kinds, viz. as well upon Descents as Ascents: As also, with the Line of Impulse and without it, that the difference may be seen: But this is to be sure, that, with the Line of Impulse, will come nearest the Mark. In these Examples I have made use of the Ascent of Edinburgh Castle, which is the highest Ascent, if below, or lowest Descent if above, of any Garrison in this part of the World. If Mortarpieces were all similis, and their requisits of Powder as the Cube of the Diameter of their Boars, and their Granades, Bombs, Carcases, and Fire-Balls were similis, as they ought to be, their Ranges upon the plain of the Horizon under the same deg. of Elevation would be equal, comparing like with like: So one Piece being well proved, that is, the Range of the Granada, Bomb, Carcase, and Fireball being found to any degree of Elevation, the whole Work of a Mortar-Piece would become very easy and exact. Considering they are not similis, there is required the Range of the Piece, at any convenient deg. of Elevation, with its requisite of Powder, then proceed by the Tables. A Table of Horizontal-Distances. de. ●i nu. dif. de. m● nu. dif. de. mi. nu. dif. de mi. nu. dif. de. mi. nu. dif. de. mi. nu. dif. 521 145 82 4 89 151 30 521 30 5406 30 8851 30 9996 30 8523 30 4808 190 144 79 7 92 152 711 16 5550 8930 46 9989 61 8431 76 4656 181 142 76 10 94 154 30 892 30 169● 30 9006 30 9979 30 8337 30 4502 177 149 73 13 97 155 2 1069 17 5832 32 9079 47 9966 62 8240 77 4347 175 138 71 16 99 157 30 1244 30 5970 30 9150 30 9950 30 8141 30 4190 173 132 69 19 101 157 1417 18 6107 9219 48 9931 63 8040 78 4033 172 135 65 22 103 159 30 1589 3● 6242 30 9284 30 9909 30 7937 30 4874 170 133 62 25 107 160 4 1759 19 375 34 9346 49 9884 64 7830 79 3714 170 131 61 28 110 161 30 2929 30 6506 30 9407 30 9856 30 7720 30 3553 168 129 57 31 111 162 5 2097 20 663● 35 9464 50 9825 65 7609 80 3391 16● 128 54 34 113 16● 30 2264 30 6763 30 9518 30 9791 30 7496 30 3228 167 126 51 36 116 165 6 2431 21 6889 36 9569 51 9755 66 7380 81 3063 166 123 49 40 118 165 30 2597 30 7012 30 9618 30 9715 30 7262 30 2898 166 122 46 42 120 166 7 2763 22 7134 37 9664 52 9673 67 7142 82 2732 164 119 43 45 121 166 30 2927 30 7253 30 9707 30 9628 30 7021 30 2566 163 117 40 49 125 168 8 3090 23 7370 38 9747 53 9579 58 6896 83 2398 193 115 37 51 126 168 30 3253 30 ●485 30 9784 30 9528 30 6770 30 2230 162 113 34 53 128 16● 9 3415 24 ●598 39 9818 54 9475 69 6642 84 2061 160 111 32 58 131 17● 30 3575 30 ●709 30 9850 30 9417 30 6511 30 1891 159 108 28 59 132 17● 10 3734 25 ●817 40 9878 55 ●358 70 6379 85 1721 158 106 26 6● 134 17● 30 3892 30 ●923 30 9904 30 9295 30 6245 30 1550 158 104 22 65 136 171 11 4050 26 8027 41 9926 56 9230 71 6109 86 1379 156 102 20 6● 138 17● 30 ●206 30 8129 30 9946 30 ●162 80 5971 30 1208 155 98 16 71 139 172 12 1361 27 8227 42 9962 57 9091 72 5832 87 1036 153 96 14 73 141 17● 30 ●514 30 8323 30 9976 30 9018 30 5691 30 864 151 95 11 76 144 173 13 ●665 28 8418 43 9987 58 8942 73 5547 88 691 151 92 8 7● 144 17● 30 ●816 30 8510 30 9995 30 8864 30 5403 30 519 150 89 4 8● 146 17● 14 ●966 29 8599 44 9999 59 8782 ●4 5257 89 346 148 87 2 8● 148 17● 30 5114 30 8686 30 10001 30 8698 30 5109 30 173 147 83 1 8● 150 17● 15 5261 30 8769 45 10000 60 861● ●5 4959 ●● ●000 The Use Table of Horizontal Distances. I. Any degree of Elevation under 45 being given, to find at what degree above 45 will hit the same horizontal distance; suppose 12 degrees, I look against 12 degrees in the Table, and find 4●61, which I look for beyond 45 degrees, and find it against 76 deg. 57 min. so I conclude, a Piece charged with the same quantity of the same Powder, and the same Ball put to 12 deg. and 76 deg. 57 min. of Elevation, will Range the Shot to the same Horizontal distance. Here note, Suppose a Piece be charged with 1, 2, 3, and 4 parts of Powder and the same Ball, and put to those degrees of Elevation, if the upper and lower Ranges be equal, there is no sensible resistance of the Medium. II. june the 5th. 1●77. on Wimbleton Heath I charged the Mortar Piece with 4 ounces of Powder, and put it to 15 deg. of Elevation, it ranged the Ball to the Horizontal distance of 659 paces; with that, I would hit a Mark with the same Piece, Ball, and quantity of Powder, at the Horizontal distance of a 1000 paces. Then as 659, is to a 1000, so is 5261 the tabular Number of 15 deg. to 7983, which gives in the Table 25 deg. 47 min. and 63 deg. 16 min. to hit a Mark at the Horizontal distance of a 1000 paces. III. Feb. the 12th 1677/8. on Wimbleton Heath, a Piece whose Length of its Chase is 18 Inches and Diameter of Boar 3 Inches, charged with 8 ounces of Powder, and laid to 10 deg. of Elevation, Ranged its shot to the Horizontal distance of 1805 paces; with that I would hit a Mark at the Horizontal distance of 2112 paces, that is 2 English Miles: Then as 805, is to 2112, so is 3734 the Tabular Number at 10 deg. to 9797, which gives in the Table 38 deg. 41 min. and 50 deg. 25 min. to hit a Mark at the Horizontal distance of 2112 paces, viz. 2 English Miles. The Making of the Table. In Fig. to the III. Proposition, the Angle MRF being given, RZ, FV and HEY being found; that is RI the Horizontal distance to that degree of Elevation; and then reduced, as in the Table. A Table of the Requisites of Powder of like Mortarpieces, from 6 Inches to 20 Inches Diameter; and they are as the Cubes of the Diameters of their Boars. In. Dec. Pou. Ou. In. Dec. Pou. Ou. A B A B 6 13 13 8 09 6 5 1 01 13 5 9 10 7 1 05 14 10 11½ 7 5 1 10 14 5 11 14 8 2 00 15 13 03 8 5 2 06 15 5 14 09 9 2 14 16 16 00 9 5 3 06 16 5 17 09 10 3 14½ 17 19 03 10 5 4 08 17 5 20 15 11 5 03 18 22 12½ 11 5 5 15 18 5 24 11 12 6 12 19 26 13 12 5 7 10 19 5 28 14 20 31 04 The Use of the Table. The Diameter of a Mortar Piece 9 Inches and ½ being given. To find the Requisite of Powder. Look in Column A, and there find 9 Inches and 5 Tenths, and against it in Column B there is 3 Pound 6 Ounces, its Requisite of Powder. The like in the rest. OF GRANADES, CARCASES, And FIRE-BALLS. OF GRANADES. AS a Granarium keeps Corn for the Preservation of the Life of Man, so these Granariums (corruptly called Granades) are filled with Corns of Fire for the Destruction of Mankind. A Granade is a hallow Sphere of Iron (as we may so call it) filled with Corn-Powder, with a Fusee to fire the dry Powder to break the Shell when it arrives to the designed Object (as in Fig. A) the Fusee B, the Vents R, ZR equal to two Thirds of the Diameter of the Granado. If the Granado be too little for the Chase of the Mortar, marl it with slack twisted Thread, brushing it all over with hot Pitch, throwing Brick-dust thereon, till it be fit for the Piece, by its Gage C, charge the Fusees with one of these Compositions following, 1. Of Powder 1 ℥ Of Saltpetre 1 ℥ Of Sulphur 1 ℥ 2. Of Powder 3 ℥ Of Saltpetre 2 ℥ Of Sulphur 1 ℥ 3. Of Powder 4 ℥ Of Saltpetre 3 ℥ Of Sulphur 2 ℥ 4. Of Powder 4 ℥ Of Saltpetre 3 ℥ Of Sulphur 1 ℥ Casimire Simienowiez, Page 206. Drive it into the Granade to the Shoulder Z, being well glued. Mortars usually Range the Granads with their due Requisite of Powder about 1250 paces upon the plain of the Horizon; the best place generally to work a Mortar, is from 70 to 80 deg. of Elevation, the greatest Altitude of a Projection at 75 deg. of Elevation whose Horizontal distance is 1250 paces, is about 583 paces, and the time of the Flight of the Granad will be about 27 Seconds of time, which time the Fusee ought to continue burning before it fires the Granad; but if upon an Ascent or Descent, less or more time is to be allowed, according as the Object is situated. And this in the general. Example. The greatest Range in the Parabola 1250 A ●/4 part equal to ND in Fig. I 312.5 In the Right Angled Triangle NDL. As Radius 90 10.000000 is to ND 312.5 2.494850 so is Tang. of the Angle DNL 75 10.571948 to DL 1166.2 3.066798 the half DQ 583.1 Then, As 20.2 Paces 1.305351 is to the square of 2″. 30‴ that is 150‴ that is 22500 4.352183 so is DQ 583.1 2.765743 to the square of the Time 5.812575 The Root 805.9 i e. 13.43 2.906287 The Double, that is near 27″, 26.86 To give a Fusee its just time of burning, viz. 27 Seconds. Take a Bullet of any convenient weight, fasten a String to it, measure from the Centre of the Bullet, 39 Inces 2 Tenths, make a Loop there, hang it upon a Pin, move the Bullet with your Hand to a convenient distance out of its Perpendicularity, then let it lose, these Swings or Vibrations will be Seconds of Time sufficiently near the truth. Drive your Fusee with one of the former Compositions, than fire it, if it continues burning 27 Seconds of Time, you have hit right, if the Fusee continued not burning the full Time, give it a slower Composition, or the contrary. Few Trials make Perfectness. A Bomb has the same Office as a Granade, only it is in the form of a Spheriod, that is, longer one way than the other, therefore it's Range will be more uncertain. There are also Granads to be cast with the hand; their Office is the same as the former; they are of Iron, Brass, Glass, and other mixed Metals that are brittle. Thomas Malthus, an Engineer of England, was the first that exercised the Mortar-Piece, and taught the Use of Granades in the 17 Provinces in Germany and Poland. The French King hearing of his good success, sent for him, to instruct the French as he had done the Dutch, and employed him in several Sieges, and at the Siege of Graveline, where he was slain by a Musquet-shot in the Head. William Eldred, 60 Years a Gunner, sometime Master-Gunner of Dover-Castle, an honest industrious Person, gives us many good Experiments of Long Guns, for the space of thirty Years; a good Work: At the latter end of his days he eat the Charity of charitable Sutton, and not the Bounty of his Prince; for he died in the Charter-House. OF CARCASES. TAke two Armillas of Iron of a convenient breadth, a little less than the Diameter of the Mortar-Piece to which they belong, put them at right Angles, rivet them fast at each Pole, at a and b in Fig. D. Rivet a Basin of Iron at the bottom, as c, a and d, to bear the Impulse of Powder and fall of the Carcase, a lesser Basin at the top, as gbh, in which there ought to be Holes to fire the Carcase, take an Armilla fit to the Diameter of the Piece, and place it equidistant from each Pole, as ef. Fill it with this Composition. 12 ℥ of Pitch 12 ℥ of Colophone. 4 ℥ of Oil of Turpentine. 1 ℥ of Hemp cut short. Or, 2 lb of Pitch. 4 ℥ of Oil of Turpentine. 1 ℥ of Hemp cut short. All these melted together, take fit Marlin, and from the two Shoulders d and c marl it fast, fill it as you marel, till you come to gh, fasten the Marlin to the Armillas with fine Marlin, if need be. Coat it with this Receipt. Pitch 4 ℥ Rosin 3 ℥ Wax 2 ℥ Turpentine 1 ℥ Fit it to its Gage C, make Holes in the top of the Carcase about 3 Inches deep, more or less, according to the magnitude of the Carcase. Use for the Priming the same Composition as for Fusees of Granades, with quick Match moderately droven into the Holes. If it be designed to place Petards in the Carcase, let there be Holes made in the Armillas to fix them in, or they may be placed in the Marlin, as the Carcase is a filling: These are things of Common Sense. There is another kind of Carcase, namely, in form of an Egg, as E, whose length is a Diameter and half of the Piece to which it belongs, composed of two Hoops of Iron and two Basins, as in the former, with a Zone of Iron in the middle, as also a Trunk of Iron E, with two Shoulders, as Z and X, to be placed in the Carcase, as AB, with Holes to place the ends of the Petards in, the Trunk filled with a slow Composition to fire the Petards as the Carcase consumes; these Petards, as F, are to be charged with dry Powder and Bullets, to destroy those that come to quench the Carcase. The Range of this Carcase will be very uncertain, by reason of the irregular form. OF FIRE-BALLS, IN Fig. G, Take the Truncheon H, wind Marlin about it in form of a Sphere, as AB, leaving the first end of the said Marlin out, as I, about this Sphere of Marlin put Canvas several thicknesses, dipping the Sphere of Marlin in Glue betwixt every thickness of Cloth. Then coat it with the Coating before cited, till it fits its Gage C, pluck out the Truncheon H, then draw the Marlin I till it all comes out, there will remain a Concave Sphere AB, which being filled with this Composition, viz. Powder-dust 12 ℥ Saltpetre 6 ℥ Sulphur 3 ℥ Colophone 2 ℥ more or less, as necessity requires. Driving it hard as you fill, it being finished, it's fit for use. Lastly, place a Basin of Lead at E, as CED, to bear the blast of the Piece, and carry that side of the Ball E foreright, that the mouth of the Ball may be uppermost, when it arrives to its assigned place. Fill the Concave Sphere AB with either of the Compositions made for the Carcases, and that will be another sort of Fireball. Much more might be said of this Subject, as well as of the whole, but this may suffice at the present. To HIT a MARK WITH Long Guns. PROPOSITION I. THE Length of the Chase, Diameter of the Boar, and Requisite of Powder of any Piece; with the Length of the Chase, Diameter of the Boar of any other Piece being given. To find its Requisite of Powder. For if, as A BD:: OF: BL, then as the Cylinder GE, is to the Cylinders ID; so is the Requisite of Powder in GF, to the Requisite of Powder in IL. But as quantity is to quantity, so is weight to weight being of the same kind. Therefore as the square of AGNOSTUS in A is to the square of BY in BD; so is the weight of the Powder GF, to the weight of Powder in IL. Here we compare the Basilisk with the Culverin, EAG the Basilisk, and DBI the Culverin. L the Length of the Culverin 11 Foot. D the Diameter of the Boar 5 Inch. P the Requisite of Powder 9.977 R the greatest Range 4.837 T the Time of the Flight of the shot. F the comparative Force. L the Length of the Basilisk 23.5 Foot. D the Diameter of the Boar 46 Inch. P the Requisite of Powder. R the greatest Range. T the Time of the Flight of the shot. F the comparative Force. 1 the Diameter of the Boar of the Culverin BY 5 Inches 0.698970 2 the Square of the first 1.397940 3 the Length of the Chase of the Culverin BD 11 Foot 1.041393 4 the Log of the second and third 2.439333 5 the Diameter of the Boar of the Basilisk AGNOSTUS 4.6 0.662758 6 the Square of the fifth 1.325516 7 the length of the Chase of the Basilisk A 23.5 1.371068 8 the Log of the sixth and seventh 2.696584 9 the Requisite of Powder of the Culverin 9 l. 977 3.998999 10 Log of the 8 and 9 6.695583 11 the Log of the 4 subst. 2.439333 12 the Requisite of Powder of the Basilisk AGFH 18 l. 041 4.256250 That is, LD ²: LD2:: P: P, That is, LD ² P = LD2 P, so any five of these six being given, the sixth is also given. 1. LD ²: LD2:: P: P 2. D ² D2P:: L: L 3. L LP:: D2: D ² 4. LD2: LD ²:: P: P 5. DP: D ² P:: L: L 6. LP: L P:: D ²: D2 PROP. II. THE Length of the Chase, the Diameter of the Boar and greatest Range of any Piece, with the Length of the Chase and Diameter of the Boar of any other Piece being given, to find the greatest Range. Let XY = ZR; then as A to BD, so is the Range of the Cylinder HV, to the Range of the Cylinder KN, but as XC is to XY; so the Range of the Cylinder KN to the Range of the Cylinder KM. That is as A in XC, is to BD in ZR, so is the Range of the Cylinder HV, to the Range of the Cylinder KM. But a Sphere is two Thirds of its circumscribed Cylinder, therefore the Ranges of the Bullets are in the same proportion. 1 the length of the Culverin BD 11 Foot 1.041393 2 the Diameter of the Boar of the Basilisk 45 Inches 0.662758 3 the Log of the 1 and 2 1.704151 4 the length of the Basilisk A 235 Foot 1.371068 5 the Diameter of the Boar of the Culverin BY 5 Inches 0.698970 ● the Log of the 4 and 5 2.070038 7 the greatest Range of the Culverin 4837 3.684576 8 the Log of the 6 and 7 5.754614 9 the 3 subst. 1.704151 10 the greatest Range of the Basilisk 11.232 4.050403 1 L LD:: R: R 2 DR: DR:: L: L 3 L LR:: D: D 4 LD: L D:: R: R 5 DR: DR:: L: L 6 LR: L R:: D: D PROP. III. THE length of the Chase, Diameter of the Boar and force of any Piece; with the length of the Chase, and Diameter of the Boar of any other Piece being given. To find the force. 1 the greatest Range of the Culverin AB 4.837 3.684576 2 the square Root of the 1 1.842288 3 the greatest Range of the Basilisk 11.232 4.050457 4 the square Root of the 3 2.025228 5 the Diameter of the Boar of the Basilisk 4.6 0.662758 6 the Cube of the 5 1.988274 7 the Log of the 4 and 6 4.013502 8 the 2 subst. 1.842288 9 the force of the Basilisk 148 2.171214 1 L LD:: D6: F2 2 DF2: D D6:: L: L 3 L F2: LD6:: D: D 4 LD: L D:: F2: D6 5 D D6: DF2:: L: L 6 LD6: L F2:: D: D PROP. IU. THE Length of the Chase, Diameter of the Boar of any Piece, with the Time of the flight of the shot; with the length of the Chase, Diameter of the Boar of any other Piece being given. To find the Time or Duration of the shot in its flight. Let AB be the greatest Range in the Culverin, CD = OF the greatest Range of the Basilisk, the time of the falling of the Ball from A to B, is to the time of the falling of the Ball from A to F; as the square Root of AB, is to the square Root AF. Or as AB to OF, so the square of the time of AB, to the square of the time AF. Torricel. de Motu Proj. Lib. II. Prop. XIX. june 24. 1686. I made an Experiment of the falling of heavy Bodies from the top of Cripplegate-Steeple, which is 101 Foot higher than the place where they fell, viz. Three Iron Balls, one of 5 Inches Diameter, another of 3 Inches, and another of 2 Inches and an half Diameter, which constantly passed that space in 2′ 30‴ of Time, with such exactness, that not any difference could be discerned; Mr. Leake, Mr. Tompion, our famous Watchmaker, who kept time with a Watch that moved ¼ seconds, Mr. Norris, Mr. Morden, with several other Persons there present. If any heavy Body fall 101 foot, or 20.2 paces in 150 thirds of time, what time shall it require to fall 4837 paces, the greatest Range of the Culverin. As 20.2 is to the squarè of 150, so is 4837 to the square of the time. 1 the Time 150 thirds 2.176091 2 the square of the 1 4.352182 3 the greatest Range of the Culverin 4837 3.684576 4 the Log of the 2 and 3 8.036758 5 the first space 20.2 subst. 1.305351 6 the Log of the square of the time 6.731407 7 the time 2321‴, that is, 38″ 41‴ 3.365703 The time of the flight of the shot of any Piece at 45 degrees of Elevation, is the time of the falling of an heavy Body the greatest Range of the same Piece. 1 the Log of the square of the time of the Culverin 6.731407 2 the greatest Range of the Basilisk 11.232 4.050463 3 the Log of the 1 and 2 10.781870 4 the greatest Range of the Culverin Log. subst. 4837 3 684576 5 the Log of the square of the time of the Basilisk 7.097294 6 the time of the Basilisk 3537‴ that is 58″ 57‴ 3.548647 Or, 1 L LD:: T ²: T2 2 L T2: LT ²:: D: D 3 DT2: DT ²:: L: L 4 LD: L D:: T2: T ² 5 DT ²: DT2:: L: L 6 LT ²: L T2:: D: D PROP. V. The greatest Range in the Parabola 1657.2 DE, the Line of Impulse AD, the Angle KAD 45 deg. of Elevation. To find AB the Range upon the plain of the Horizon. As Rad. 90 10.000000 is to AD 37.48 1.573800 so is the sine of the Angle KAD 45 deg. 9.849485 to KD or AK 26.5 1.423285 DAK being 45 deg. therefore ADK is 45 deg. therefore AK is equal to KD, and by the same reason, GP and DG are equal, therefore GH is equal to half DG. HG 414.3 DK 26.5 HN 440.8 As HG 414.3 2.617315 is to HN 440.8 2.644242 so is the square of DG or GE 8286 5.836690 to the square of NB 5.863617 ND 854.7 2.931808 NK 828.6 AK 26.5 AB 1709.8 Hear note, AK and FB differs but 4 Tenths of a Unit, therefore in the following Work, I take the double of AK for them both. PROP. VI TO make a Table of Ranges, of any Gun; for the plain of the Horizon to every 30 Minutes and single degree of Elevation. And here remember, that the Ranges upon the plain of the Horizon, are as the sins of the double of the Angle of Elevation. For Example sake, the Cannon Royal whose greatest Range is 3298 paces. As AB 1709.8 3.232945 is to AD 37.48 1.573800 so is the greatest Range of the Cannon 3298 3.518251 to the Line of Impulse 72 paces 1.859106 If we make use of the last Diagram, and suppose AB to be the Horizontal Range at 45 deg. of Elevation of the Cannon Royal, and AD the Line of Impulse 72 paces. Then, As the Rad. 90 10.000000 is to the sine of the Angle ADK 45 deg. 9.849485 so is the Line of Impulse AD 72 1.859106 to AK 51 1.708591 The greatest Range of the Cannon Royal AB 3298 subst. FB+AK = 102 The greatest Range in the Parabola DE 3196 As the Rad. 90 subst. 10.000000 is to DE 3196 3.504607 so is the sine of 1 deg. 8.241855 to 56 1.746462 L: P 72 2 deg. 8.542819 56 112 2.047426 30 128 3 deg. 8.718800 112 167 2.223407 1 184 4 8.843585 167 223 2.348192 30 239 etc. 223 2 295 etc. These Calculations may suffice, till such time as we have more Experiments to confirm these; or to make Tables of Ranges more exactly. The greatest Ranges of Long Guns, are deduced from many Experiments made by Eldred in his Gunner's Glass, especially those in Pag. 74. 1611. july 2. and Pag. 75. 1636. August 30. The manner how they are deduced may be seen in my Genuine Use of the Gun, Prop. 1, 2, 3, 4. Printed for Robert Morden, at the Atlas in Cornhill, near the Royal Exchange. These Ranges for single degrees, are deduced from those Ranges, and what was done on Wimbleton-Heath, Septemb. 17. 1677. A Table of the Names, Diameters of the Boars, and Length of the Chases of Ten several Pieces of Canon; with their Requisites of Powder, greatest Ranges, Comparative Force, with their Ranges to 8 deg. of Elevation, Experimentally and Mathematically demonstrated. The Names of each Piece. Lengths of the Chases. Diamet. of the Boars. Requisits of Powder. Greatest Ranges. Comparative Force. Feet. Inches. lb ℥ Paces. A Rabbinet 1 3 1.75 5 3769 38 A Falconet 2 4 2 9 4398 61 A Falcon 3 6 2.75 1 10 4797 166 A Minion 4 8 3 2 10 5864 238 A Saker 5 9 3.5 4 5654 371 A Demi Culverin 6 10 4.5 7 5 4886 733 A Culverin 7 11 5 10 4837 1000 A Demi Cannon 8 11 6 14 6 4031 1575. A Whole Cannon 9 12 7 21 5 3769 2422 A Cannon Royal 10 12 8 27 14 3298 3382 The USE of these Ranges. Suppose an Engineer be commanded to batter a Bastion, or a Curtain, at the Horizontal distance of 900 paces, with a whole Cannon, a Demi Cannon, and a Culverin; I look in the Table, and find 900 against the Demi Cannon, and at the head 6°, for the whole Cannon 6° 27′ and for the Culverin 4° 52′ to batter the said Bastion or Curtain, taking the proportional parts; thus, in the Culverin, 80: 30′: 60: 22′, whole Cannon 63: 30′: 58: 27′. 0′ 30′ 1° 30′ 2° 30′ 3° 30′ 4° L.P. D. R. D. R. D. R. D. R. D. R. D. R. D. R. D. R. D. 1 83 64 147 63 210 64 274 64 338 63 401 64 465 63 528 63 591 63 2 96 74 170 75 245 74 219 74 393 74 467 74 541 74 615 74 689 74 3 105 81 186 81 267 81 348 81 429 81 510 81 591 80 671 81 752 80 4 128 99 227 99 326 99 425 93 524 99 623 99 722 98 820 99 919 98 5 124 96 220 95 315 96 411 95 506 95 601 9 697 95 792 94 886 95 6 107 83 190 82 272 83 355 82 437 83 520 82 602 82 644 82 766 82 7 106 83 189 81 270 82 352 81 433 82 515 81 596 82 678 80 758 82 8 88 68 156 68 224 68 292 69 361 68 429 67 496 68 564 68 632 67 9 83 64 147 63 210 64 274 64 338 69 401 64 465 63 528 63 591 63 10 72 56 128 56 184 55 239 56 295 56 351 55 406 55 461 56 517 55 Here note well also; The Object that the Ball is to hit, aught to be as high above the Horizon, as the Mouth of the Piece; if not, the Ranges at 0, 1, 2, and 3 degrees of Elevation will sensibly differ, those Ranges at 6, 7, and 8 degrees of Elevation not so much as the former. 30′ 5° 30′ 6° 30′ 7° 30′ 8° R. D. R. D. R. D. R. D. R. D R. D. R. D. R. 1 654 3 717 63 780 62 842 63 905 62 967 61 1028 62 1090 2 763 73 836 73 909 73 982 73 1055 72 1127 72 1199 72 1271 3 832 80 912 80 992 79 1071 80 1●51 78 1229 79 1308 78 1386 4 1017 98 1115 97 1212 97 1309 97 1406 97 1503 96 1599 95 1694 5 981 94 1075 94 1169 94 1263 93 135 93 1449 93 1542 92 1634 6 848 81 929 81 ●010 81 1091 81 1172 80 1252 80 1332 80 1412 7 840 80 920 81 1001 79 1080 81 1161 79 1240 80 1320 78 1398 8 699 67 766 67 833 67 900 67 967 66 1033 66 1099 66 1165 9 654 63 717 63 780 62 842 63 905 62 967 61 1028 62 1090 10 572 55 627 55 682 54 736 55 791 54 845 54 899 54 953 Warlike Music ILLUSTRATED, In several Consorts of Phrygian Flutes. Clearly demonstrated by Principles of Music and Mathematics. PROP. I. THE length of the Chase 12 foot of a Cannon Royal being given. To find the length of the Chases of seven lesser Pieces of Cannon that shall be in Musical Proportion. The greater tones 9/8 their Log. 95424251 90308999 The difference of their Log. 5115●52 The lesser tone 10/9 their Log. 100000000 95424251 The difference of their Log. 4575749 The half Note 16/15 their Log. 1.20411998 1. 1●6●9126 The difference of their Log. 2802872 The perfect Eight 2/1 12 1.07918125 4575749 10/9 The middle Seventh 9/5 10.8 1.03342376 5115252 9/8 The lesser Sixth 8/5 9.6 0.98227124 2802872 16/15 The perfect-Fifth 3/2 9 0.95424252 5115252 9/8 The perfect Fourth 4/3 8 0.90309000 4575749 10/9 The lesser Third 6/5 7.2 0.85733251 2802872 16/15 The greater Second 9/8 6.75 0.82930379 5115252 9/8 The Unisound 1/1 6 0.77815127 PROP. II. THE Diameter of the Boar of the Cannon Royal being 8 Inches. To find the Diameter of the Boar of seven lesser Pieces of Cannon, which shall be in Musical Proportion: By the first. PROP. III. THE Requisite of Powder of the Cannon Royal being 28 Pounds. To find the Requisite of Powder of seven lesser Pieces of Cannon in Musical Proportion: By the first. PROP. IU. THE Force of the Cannon Royal being taken to be as the Cube of the Diameter of the Boar, viz. 512. To find the Force of seven lesser of Pieces of Cannon which shall be in Musical Proportion: By the first. PROP. V. THE Time of the Flight of the shot of the Cannon Royal 1000 To find the time of the flight of the shot of seven lesser Pieces of Cannon which shall be in Musical Proportion: By the first. PROP. VI THE Length of the Chase, Diameter of the Boar of the Cannon Royal, as also the Length of the Chase, and Force of any of the aforesaid Pieces being given. To find the Diameter of its Boar. PROP. VII. THE Length of the Chase, Diameter of the Boar, and the Time of the Flight of the shot of the Cannon Royal, as also the Length of the Chase, and time of any of the aforesaid Pieces being given. To find the Diameter of its Boar. PROP. VIII. THE Length of the Chase 12 Foot, the Diameter of the Boar 8 Inches, the Requisite of Powder 28 lb of a Cannon Royal being given. To find the Length of the Chase, Diameter of the Boar, and Requisite of Powder of seven lesser Pieces of Cannon, which shall be in Geometrical Proportion. The Length of the Chase 12 Foot Log. 1.07918125 It's half 6 Foot Log. 0.77815127 The difference of the Log. 30102998 The 7th. part is, add 4300428 Add to the length of the least 1 6 foot 0.77815127 2 6.624 0.82115555 3 7.314 0.86415983 4 8.076 0.90718411 5 8.916 0.95016839 6 9.844 0.99317267 7 10.869 1.03617655 8 12. 1.07918123 The like for the Diameters. The Requisite of Powder 28 lb Log. 1.44715803 The eighth part thereof is 3.5 Log. 0.54406804 The difference of the Log. 0.90308999 The 7th part is 0.12901285 1 3.5 0.54406804 2 4.71 0.67308089 3 6.34 0.80209374 4 8.53 0.93110659 5 11.48 1.06011944 6 15.46 1.18913229 7 20.80 1.31814514 8 28. 1.44715809 I. By the first Proposition for the length of the Chase, and second Proposition for the Diameter of the Boar; we compose the first Consort of Phrygian Flutes, whose visible shapes are in Musical Proportion, the Ranges and Time upon the same degree of Elevation are equal, their Force and Requisite of Powder as the Cube of the Diameters of their Boar. II. By the first Proposition for the Length of the Chase, and third Proposition of the Requisite of Powder, we compose the second sort of Phrygian Flutes, whose Sounds or Reports and Ranges shall be in Musical Proportion, the Diameter of their Boar's equal. III. By the first Proposition for the Length of the Chase, and the fourth Proposition for the Force, and Proposition the 5th. for the Diameter of the Boar, we compose a third sort of Phrygian Flutes whose Force shall be in Musical Proportion. IV. By the first Proposition for the Length of the Chase, and 6th. Proposition for the Time of the Flight of the shot, and Proposition the 7th. for the Diameter of the Boar, we compose a fourth sort of Warlike Flutes, whose Times shall be in Musical Proportion. V. By the eighth Proposition we find the Length of the Chase, Diameter of the Boar, Requisite of Powder of Eight Pieces of Cannon in Geometrical Proportion, their Ranges upon the same degree of Elevation equal. So then, here is nothing wanting in these things that the Heart of Man can in reason desire; for in the first, the Eye is satisfied, their shapes are in Musical Proportion. In the second, the sense of Hearing is delighted, their Sounds are in Musical Proportion. In the third the sense of Feeling may be terrified, the Smart of the Blow in Musical Proportion. In the fourth, that precious thing called Time is employed in delightful Musical Proportion. What can be presented to a Prince more delightful, or what can a Prince more delight in, than to conquer his Enemies with Music and Delight? FRom what has or may be said concerning Guns in Musical and Geometrical Proportion, and what may be said concerning Guns under other Qualifications, it may manifestly appear, that the business of Gun-founding lies as open for Improvements in these days, as their Uses did when Galileus first appeared in the World: For, An Engineer draws the Draught of a Gun, and gives it a certain Fortification, the Founder found'st it according to the Draught or Pattern, a Proof-Master proves it, and gives it its Requisite of Powder. Quere, Whether it be the Engineers or Draught-Man's business to give it its Requisite of Powder, and consequently its Range, or the Proof-Master's? Then, I. In the Diagram, Let K, G, B, and P, be any Piece of Cannon. It being demanded to increase the Metal to I, H, A, and Q; or decrease the Metal to L, F, C, and O; so as the strength of the Metal K, G, B, and P, be to the strength of the Metal I, H, A, and Q; as 3 to 7; and to L, F, C, and O, (being of the same Metal) as 7 to 3. The Requisite of Powder, and Range at 45 degrees of Elevation of the Piece I, H, A, and Q; and the Piece L, F, C, and O, are required. II. If the Lengths of the Chases of two Pieces, and the Diameters of the Boars of the same two Pieces be as 8 to 9 To give them such Metal that the strength of one Piece may be to the strength of the other, as 3 to 7. Their Requisites of Powder and greatest Ranges are required. III. If the Lengths of the Chases of two Pieces be as 9 to 10, and the Diameters of their Boars be as 4 to 5. To give them such Metal, that their strengths may be as 3 to 7. Their Requisites of Powder, and greatest Ranges are required. To Conclude. HERE I might produce a Jury of Mathematicians, famous for their Learning, for the confirmation of that of the Parabola: But it is so evident in itself, that he that is ignorant of, and denies that, viz. The Flight of the Bullet to be in the Curve of a Parabola, deserves not the Name of a Mathematician, Philosopher, Engineer, or Fire-Master, no, not of a private Gunner. FINIS.