An Example in Numbers, and first of the first.

CK, 34: KN, 26 :: GO, 68: OZ, 52; then 68 more 52=120, GZ: then AC 3600; in GZ, 120; is equal to 432000, a third part is 144000, ABCNZ, FE, 676 in OZ, 52 is equal to 35152.

A third part is 11717⅓, HFDEZ, 144000 less 11717⅓ is equal to 132282 [...] the solidity of the frustum ACEF.

The Second way.

AC, 3600 more FE, 676, more AK, 1560 the sum of those are 5836, which multiplyed by GO, 68, is 396848, a third part is 1322 82⅔. These ways Christopher Clavius hath, pag. 208 of his Geometria practica. I shall give a third in conformity to what follows.

A Third way.

Let ACEZHPOF be a frustum pyramide the Bases Squares and Parallel; Let HPOF be projected in the Base ACFZ, as ZVRT; VR and TR be continued to D and B; then will TD be equal to RA, and RC a Square; most manifest it is, that the parallelepipedon ZVRTPOFH, more the prisme VABR OP, more the pyramide RBCDO, more the prisme RDETOF is equal to the said fru­stum pyramide.

[Page 5]For all the parts of any magnitude being taken, together are equal to the whole. Axiom 19. 1.

The prisme V ABROP is e­qual to the prisme RDETFO, be­cause of equal ba­ses and altitude, but the prism VA BROP is half the parallelepipedon, made of the base VABR, and the altitude RO, by 28. 11. therefore the parallelepipe­don made of the base ZABT, that is the Rectangle made of HF and ZA, and the alti­tude RO, more the pyramide RBDCO that is [...] of the square of the difference of the sides of the upper and lower Bases and the altitude RO, shall be equal to the frustum proposed.

Example in Numbers.

ZE, 60; ZT, 26; the product 1560 e­qual to BZ, a third part of RC is 385⅓, which [Page 6] added to 1560, is equal to 1945 [...], this last composed number multiplyed by 68 equal to RO makes 132282⅔ the solidity of the said frustum; then as 14 to 11, or more near the truth, as 452 to 355, so is 132282⅔ to 103894 [...] the frustum cone adscribed within the fru­stum pyramide.

An Example.

CD equal to 25, DE equal to 12, the [...]ike for the other Angles, CE equal to 300, [Page 8] which being multiplyed by 5, that is ⅓ of TP, makes the pyramide CDEPT equal to 1500, which multiplyed by 4 (because there are 4 of them) is 6000; BC, 34; CP, 12; BP, 408; this last being multiplyed by 7 [...] that is [...] the altitude TP, makes 3060 equal to the prisme BCQPTS, this being doubled (because there are two of that magnitude) is 6120; PE, 25; PK, 60; PF, 1500; which be­ing multiplyed by 7 [...] makes 11250, equal to the prisme PEFKZT, this being doubled (because there are two opposites equal) is 22500; PQ, 34; QL, 60; LP, 2040; this last number being multiplyed by 15 the alti­tude of the pyramide TP, produceth 30600; the parallelepipedon QPKLVZTS; these 4 composed numbers being added together, makes the whole frustum pyramide.

A Second way thus.

Let FC be the greater base, RQ the lesser, GO the height; make it as RH: HQ :: FA: AB; draw BE parallel to FA, then will [Page 9] the plane FB be like the plane RQ; Let the Figure be completed, then will the frustum ACDFRHQZ be equal to the frustum ABEFRHQZ more the prisme, BCVT QZ, more the pyramide TVDEZ: here note that TB is made equal to QZ.

The Example in Numbers.

The frustum ABEFRHQZ, is equal to 124501 [...], found by the first Proposition, BC, 16; CV, 26; BV, 416; being multiplyed by ½ the altitude, that is 32, makes 13312, [Page 10] equal to the prisme BCVTQZ; TV, 26; VD, 34; TD, 544; being multiplyed by [...] of GO that is 21 [...], makes 11605⅓ equal to the pyramide TVDEZ; these three com­posed numbers being added together, makes 149418 [...] equal to the frustum pyramide ACD FRHQZ.

A Third way thus.

Let PC the greater base; RF the lesser; Let the upper base be projected in the lower, as RIFV be equal to POHG, most easie it is to apprehend, that the frustum pyramide PACEVFIR, is equal to the parallelepi­pedon POHGVFIR, more the prisme OA BHFI, more the pyramide HBCDF, more the prisme HDEGVF, for the whole is equal to all its parts.

In numbers thus.

Let AC, be 56; AP, 38; RI, 26; RV, 30; RP, 40; RI, 26; VR, 30; IV, 780 equal to PH; which being drawn into RP, 40; makes POHGVFIR, 31200: OA, 12; AB, 30; AH, 360; being multiplyed by 20, that is half the height PR, it makes [Page 11] 7200 equal to OABHFI. BC, 28; HB, 12; BD, 336; being multiplyed by [...] of the altitude that is 13⅓, it produceth 4480, equal to the pyra­mide HBCDF; GH, 26; HD, 28; HE, 728; being multiplyed by 20, the product is 14560 equalto the prisme HDE GVF; these 4 composed num­bers being added together, makes the solidity of the said frustum.

Then,

As 14 is to 11, so is the solidity of any frustum pyramide thus found, to any Ellip­tick frustum, adscribed in that frustum pyra­mide.

An Example in Numbers.

BC, 52; BH, 64; BD, 3328; this last [...]umber being multiplyed by 40, that is half [...]he height, makes 133120 equal to the prisme [...]BCDEF; AB, 32; BH, 64; BG, [...]048: this last being drawn into 26 [...], that [...]s, a third part of the altitude makes 54613 [...] equal to the pyramide GABHF: this prisme [...]nd pyramide being added together, makes 187733 [...], equal to the frustum prisme AGD CEF.

The second Case thus.

Let ZBCDEH be a frustum prisme; Let AC be equal to HE and AO parallel to ZB, the Figure being completed, the prisme AC DOHE less the pyramide ABZOH is equal to the prisme BCDZHE; or thus, the prisme ZBCDFE more the solidity ZBHF, that is, half the pyramide ZBAOH, is equal to the frustum proposed.

Example in Numbers.

AC, 84; AO, 64; AD, 5376; this last drawn into 40, half the height makes 215040; equal to the prisme OACDHE; AC, 32; BZ, 64; AZ, 2048 being multiplyed by 26 [...], that is, one third of the height, it makes [Page 14]

54613⅓ equal to the pyramide ABZOH this pyramide taken from the prisme leaves 160426 [...], equal to the frustum prisme proposed.

Then as 14 is to 11, so are such prismes to Elliptick solids adscribed in those prismes.

Hence follows a Fourth way for finding the solidity of these irregular frustums.

Suppose ACDFGNIH be the frustum proposed, because FA is equal to DC, and AC equal to DF, and GN equal to IH, and [Page 15] NI to GH; and because GN is less than FA, and HI than DC; therefore if AN, CI, DH, FG, be continued to­wards Z and V, they will meet, suppose at Z and V, then AC DFVZ, less NI HGVZ, there remains ACDF GNIH the fru­stum proposed.

It also follows, That such prismes have such propor­tions one to ano­ther, as Squares and Cubes of their corres­ponding sides, disjoyned, thus.

The pyramide BCDEZ, is to the pyramide RQIHZ, as the cube of BE to the cube of RQ; 8, 12. The prisme FABEVZ, is to the prisme GNQRVZ, as the square of FA, is to the square of GN; the reason why these prismes are not in a triplecate ratio of their homologal sides, as well as the pyra­mides; is because FE, GR, and VZ are [Page 16] equal, and the ratio riseth but from VG [...] GN and VF, and FA; whereas in th [...] pyramide the ratio riseth from three, that is from ZR, RQ, and RH; and from ZE [...] ED, and EB.

In Numbers thus.

Let VZ be 9, ZR, 3; ZE, 6; RQ, [...] EB, 8; RH, 3 ½; ED, 7; Therefore G [...] QRZV is equal to 54, and RQIHZ [...] equal to 14; then as the cube of 3, that [...] 27, to the cube of 6, that is 216; so is th [...] pyramide RQIHZ, that is 14, to the pyr [...] ­mide EBCDZ, that is 112. Then 112 le [...] 14 is equal to 98, equal to RQIHDCB [...] Again, as the square of 3, that is 9, to th [...] square of 6, that is, 36; so is the prisme G [...] QRZV; that is, 54 to the prisme FABE [...] V, that is, 216. Then 216 less, 54 is equ [...] to 162, equal to GNQREBAF. This la [...] frustum, more the frustum RQIHDBC [...] is equal to the frustum GNIHDCA [...] equal to 260.

Or thus.

As 27: 189 :: (that is the cube of ZE le [...] the cube of ZR) 14: 98, equal to the fr [...] ­stum RQIHDCBE. Again, as 9: 27▪ (that is the square of ZE less the square o [...] ZR) 54: 162, equal to the frustum GN [...] REBAF.

[Page 17]What ever is said of these prismes and pyra­mides, the same is to be understood in Cones and Ecliptick prismes;

For they are in duplicate and triplicate ratio of their homologal sides, further for as much as that, the two first terms in each pro­portion may be fixt; there may by the help of a Table of Squares and Cubes, the solidity of such solids easily be calculated gradually, that is, inch by inch, or foot by foot, the two fixed numbers in the pyramide are the Cube of the continuation, or the side RZ; and the pyramide RQIH, the like in the prisme, and with the square of the continuation RZ.

Thus.

27: 14 :: 37: 19 [...], 27: 14 :: 98: 60 [...], 27: 14 :: 189: 98, these three num­bers, namely, 19 5/27, 50 [...], and 98, are the solidity of the frustum pyramide RQIHDC [...]E, the first number is the solidity of one [...]nch, the second number of two inches, the [...]hird of three inches; for the prisme; as 9: 54 :: 7: 42, 9: 54 :: 16: 96, 9: 54 :: 27: 162; these 3 numbers, viz. 42, 96, and 162 are the solidity of the frustum prisme GNQREBAF, taken [...]nch by inch; therefore 42 more 19 5/27 equal [...]o 61 [...], 96 more 50 [...] equal 146 [...], 162 more [...]6 equal to 260, are equal to the solidity of [...]he whole frustum GNIHDCAF, taken inch [...]y inch; the like for any Elliptick frustum.

II. To find the solidity of a triangular frustum pyramide, whose bases are not parallel.

[Page 20]Let CABDHF be a frustum pyramide, CAB not parallel to HDF, let QDE be parallel to CAB; the triangular frustum Q DEBAC is half a quadrangular frustum py­ramide, found by the 2 Proposion, DZ the height of the pyramide QHFED; There­fore the frustum pyramide QDEBAC more the pyramide QHFED will be equal to the frustum pyramide FHDABC.

III. To find the solidity of the frustum pyramide, ACFEDZ.

Here the base AFC is not parallel to the base ZDE but to the plane ZOE; the frustum py­ramide ZOE FCA may be found by the 2 Proposition, the part ZOE D will be a py­ramide, its al­titude may be found, thus, As OF: OG :: OD: DR, [Page 21] the frustum pyramide ZOEFCA more the pyramide ZOED will be equal to the solid ACFEOZD.

IV. To find the solidity of a frustum pyramide, whose bases are not parallel, and whose inclina­tion is from Angle to Angle.

Let CABDFXKQ be a frustum pyramide, the base CABD not parallel to the base FXKQ, but parallel to the plane FEH G; then the fru­stum pyramide C ABDHEFG found by the 2 Propos. more the pyramide EQXG F, found by the 2 case of this Propos. more the frustum pyramide GEHZXQ, found by the 2 Propos. more the pyramide XKZQ, found by the 3 case of this Propos. will be e­qual to the whole frustum pyramide CABDFXKQ.

The second Case.

Let ROBC be ¼ of a Circle, and RSQC ¼ of an Ellipsis complete the Diagram, then will the square of QC, that is IV less the

[Page 24] square of LV be equal to the square of SV. For AR: RD :: TV: VL, there fore the squares of them, and AR: RD :: OV: VS, therefore the squares of them; therefore, as the square of TV, to the square of VL, so the square of OV, to the square of VS; Therefore, as the square of EV, to the square of EV less the square of TV. So the square of IV, to the square of IV less the square of LV, but the square of EV less the square of TV is equal to the square of OV; therefore the square of IV less the square of LV equal to the square of SV.

In Numbers thus.

Let IP be 20; QV, 16; CB, 10; the square of IP is 400, the square of QV 256; the difference of the squares is 144, a third part of that number is 48, which substracted from the greater square 400, leaves 352: or if two thirds of the dif­ference, that is, 96 be added to the lesser square, that is, to 256, the sum will be 352 the true mean square; but if it be as 14: 11 :: 352: 276 4/7 the mean circle, which being mul­tiplyed by the length ZS, 20; the Product is 5531 3/7 the solidity of the portion ZISG PX.

In Numbers thus.

Let AG be 150; GN, 54; GE 90; the paralellepipedon made of the square of BE and Altitude GN, in 1749600; then, as AG, 150; to [...] AG, more ⅙ GN; that is 75 more 9; that is 84; so is 1749600, to 979776, that is all the squares in the portion of the Sphere BNRE; then as 14 : 11 :: 979776 : 769824, the portion of the Sphere BNRE.

Thus in Numbers.

Let AB, 10; BC, 10; Therefore AC, 20; Let BE, 30; the square of AC, 20; is 400, which being multiplyed by ½ EB, that is by 15; the Product is 6000, equal to all the squares in the conoide AEC; but if it be made as 14 : 11 :: 6000 : 4714 [...] the solidity of the conoide AEC.

In Numbers thus.

Let AX be 150; RX, 54; ZX, 90; AR, 96; BR, 48; the parallelepipedon whose Base is the Rectangle of RXZ and the Altitude AX is 729000.

The prisme whose Base is the Rectangle ARX and Altitude RD is 233280.

The pyramide whose Base is the square of RX and Altitude RD is 87480.

The parallelepipedon whose Base is the square of ZX and Altitude XR is 437400.

Then 729000 : 233280 more 87480, that is, 320760 :: AX, 150 : BR, 48, more ⅓ of RX, 18; that is 66 :: 437400 : 192456 equal to all the squares in the ¼ of the conoide ZH RK; but if that last number be multiplyed by 4, the Product will be 769824 equal to all the squares in the conoide ZHRK, then 14 : 11 :: 769824 : 533433 [...] the solidity of the Hyperbolick conoide ZHRK.

In Numbers thus.

Let AB, AF, FE, AD, DC, BC each of these be equal to DE, and DE be equal to 204: from E to the first R be 54, and from that first R to D be 150; KR, 204 mul­tiplyed by RX, 54; is XK, 11016; which being multiplyed by ½ RE 27; the Product will be 297232, equal to the prisme KHXR EF; FK, KH, HO, and IK, every of them equal to 54; Therefore the pyramide KHO IF will be 52488; then KHXREF, 297 232; less KHOIF, 52488 there will remain IOXREF, 244944; equal to all the squares in the portion QER; that is, all the squares in the portion of ¼ of a Sphere,

In Numbers Thus.

Let DE the longest Diameter be 36, the Conjugate or shortest 18, therefore FE will be 12, for it is 36 : 18 :: 18 : 12. Let from E to the first R be 9, then FE, 12, multiplyed by R X, 9; XK will be 108; which being multiplyed by ½ of ER, that is 4 ½ the Product will be 486, equal to KHXREF; for the finding of KI, work thus, AF, 36; to AD, 12; so is FK, 9; to IK, 3; then IK, 3, multiplyed by KH 9; the Product is 27, that is, HI; this last number being multiply­ed by ⅓ of KF, that is by 3; the pyramide KHIOF will be equal to 81, which being substracted from the prisme KHXREF there will remain 405 the prisme IOXREF equal to all the squares in the portion of the spheroide QER.

In Numbers thus.

Let GM the Transverse Diameter be 18; MA, 12; AE, 10; ML, 6; AB equal to AM multiplyed by AE, 16; the Product is 72 equal to ABCF, which being multiplyed by ½ AM, that is, 6, the Product is 432 equal to the prisme ABCFLM; FE, 4; [Page 47]

[Page 48] multiplyed by FC, 12; the Product is 48, equal to FCDE, which being multiplyed by ⅓ of FL, that is, 4; the Product is 192, equal to the pyramide FCDEL; this pyra­mide being added to the prisme before found, the whole prisme ABDELM, will be 634, equal to all the squares in the conoide AZM.

In Numbers thus.

Let AP, be 10; AV, 8; Therefore AC will be 80; which being multiplyed by ½ AP, 10; that is, by 5, the Product will be 400, equal to the prisme ABCVHP; that is, all the squares in ¼ of a parabolical conoide whose Altitude is AP, 10; and semidiameter of the Base is AR.

In these 4 last Propositions it ought to be [...]ade, as 14: 11; so are all the squares in the [...]ortion to the portion it self.

1.

The Areas ZXD and ZXA may be found by the 19th. Proposition.

2.

The Spherical Superficie of BDZQ is equal to a Surface whose Base is equal to the Line FLN and Altitude RB.

Or,

The Spherical Superficie of RDZQ is equal to the Area of a Circle whose Diameter is equal to a Line drawn from B to D.

The like for the Spherical Superficie of NA ZT. Archimedes 36. 1. of the Sphere and Cylinder.

3.

Let RDGZQ and MAPZT be Qua­drants of lesser Circles of the same Sphere, RD and MA their Semidiameters; CBZL and CNZE Quadrants of great Circles of the Sphere; AOZ and DIZ Arches of great Circles of that Sphere, the Arch NZ equal to the Arch NA, and the Arch BZ equal to the Arch BD; the right lined Angle DRZ equal to the Spherical Angle DBZ, the Angle ZMA equal to the Angle ANZ.

4.

As 90 Degrees, is to the degrees and parts of a Degree in the Angle DBZ; so is the Spherical Superficie BADGZQ, to the Sphe­rical Superficie of the Triangle BADGZ.

5.

In the Triangle IZBDI; there is given the Arches BD and BZ, and the Angle D BZ; Therefore there are given the Angles BDZ and BZD. The like in the Triangle OZNAO for the Angles NAZ and NZA; by the third Case of Oblique angled Spherical Triangles.

6.

In the Triangle AOZID, there is given the Arch AD and the Angles ZAD and ZDA; Therefore there is given the Angle AOZID, by the 8th. Case of Oblique angled Spherical Triangles.

7.

In the Triangle AOZID there are given the three Angles, Therefore the Area may be found, thus. As 180 Degrees, is to the ex­cess of the three Angles over and above 180 Degrees; so is the Area of a great Circle of that Sphere to the Area of that Triangle, Foster, Miscel. page 21.

Or,

It may be found by that method which is delivered by that Learned Mathematician Mr. Iohn Leek, in page, 116, of Mr. Gibsons Syntaxis Mathematica, Which is this. If the excess of the three Angles above 180 Degrees be mul­tiplyed by half the Diameter of the Sphere, the Superficies of any Spherical Triangle is thereby produced.

[Page 57]By this Rule and by what Mr. Gibson delive­red in that 116 page, I found this way to resolve this Proposition.

8.

By this last Rule we are to find the Areas of the Triangles BDIZ and NAOZ: the difference between the Areas of these, and those found by the 4. of this sheweth the Areas of the two Figures, viz. GZID and OZPA, which added to the Area DIZOA gives the Superficie of the mixt Lined Triangle APZ GD.

9.

In the 9th. Chapter of the forementioned Syntaxis Mathematica there is given some Rules about some solids; but when it was time to treat of these second fragments, that Author breaks off thus; There may be other parts of a Sphere besides those which are here called Fragments, (not to speak of those which are irregular and multiform) which are either Cones or Pyramides, whose Bases lie in the su­perficies of the Sphere, and their vertices at the Center, the solidity of one of these is found by multiplying the third part of the Base by the Altitude (which here is the semiaxis) the pro­duct is the solidity: these Fragments are those which are usually called Solid Angles. Thus fa [...] Mr. Gibson.

In the Diagram, Let the Triangle AZF, be equal to the Triangle ZPADG in the last Diagram, the Spherical Superficie of that Triangle multiplyed by one third of the Semi­diameter CA, CF or CZ produceth the Spherical pyramide ZFAC.

This pyramide may be divided into three solids, viz. a pyramide whose Base is the seg­ment ZFX and Altitude ZB, the pyramide whose Base is the Area ZAX and Altitude XE; and the solid AZXF the solid requi­red. But the pyramides are given, because [Page 59] the Areas ZXF and ZXA may be found by the I of this, and the Altitudes XB and XE are given to limit the Proposition; There­fore the pyramides ZXFC and ZXAC taken from the pyramide ZFAC leaves the solid FXAZ.

10.

Let ♄ ♃ ♂ ☉ ♀ ☿ ☽ ABD ♄ HFCT be [...] of a Spheroide, ♄ ☉ OXVIABD ♄ HF [...]T, be ⅛ of a Sphere. Unto the Planes ♄ ♃ ♂ ☉ ♀ ☽ and ♄ LO ☉ XC, Let there be Planes at right Angles as ♂ QDH, ☉ QBF and ♀ ☿ ☽ AC. OIDH, ☉ IBF and XIAC. From the points L parallel to ♀ C draw the Line ♃ L E, because the Line ♃ LE is parallel to ♀ C, Therefore the Line ♃ T is equal to LT, and the Line T ☽ is equal to TI, Therefore the Quadrants of the Circle T ♃ ☽ and TLI are equal. In these Planes, viz. ♂ QDH, ☉ QBF and ♀ ☽ AC, Let there be Lines drawn, viz. P comma, and QN, parallel to ♂ H. P comma, and QN parallel to ☉ F. ☿ comma, and ☽ N parallel to ♀ C. The points ♂ PQD, ☉ PQD and ♀ ☿ ☽ A in those El­lipses, the points OVID, ☉ VIB and XV IA in those Circles.

[Page 60]

♀ C : XC :: ♂ H : OH, by the 10. Prop.

♀ C : XC :: ♃ E : LE,

♂ H : OH :: ♃ E : LE, 11. 5.

♂ H-OH: OH :: ♃ E-LE:LE, 17. 5.

♂ O : OH :: ♃ L : LE, 19. 5.

☉ F : ♃ E :: ☉ F : LE, 10th. Prop.

☉ F-♃E: ♃ E :: ☉ F-LE: LE, 17. 5.

☉ R : ♃ E :: ☉ Z : LE, 19. 5.

☉R♂O: ♃ E :: ☉ZOZ: LE, 12. 5.

♃ E : LE :: ♀ C : XC,

♀ C : XC :: ☉R♂O : ☉ZOZ, 11. 5.

♀ C : XC :: Area R☉♀♃ : Area Z☉OL,

[Page 61]Again.

♂ H : ♃ E :: (QN) OH: IN, 10. Prop.

♂ H-QN: QN :: OH-IN (ZH) IN, 17. 5.

♂ O : OH :: OZ : ZH, 19. 5.

P, : QN :: V, : IN,

P,-QN : QN :: V,-IN: IN, 17. 5.

P. : QN :: V, : IN, 19. 5.

♂OP. : QN :: (♃E)OZVI:IN; (LE,) 12. 5.

♂OP. : ♃ E :: OZ + VI : LE,

♀C : XC :: ♃ E : LE,

♀C : XC :: ♂OP. : OZVI, that is,

♀C:XC :: Area OQP♂: Area ZIVO. the Plane.

☉ PQBF or any other drawn parallel to these, shall have the same qualifications; which makes us conclude that the whole Section ILTX shall be to the whole Section ☽ ♃ T ♀ as XC is to ♀ C, and as XC to ♀ C so the segment ILZO to the segment Q ♃ O ♂, the second Section in the Spheroide.

In Numbers thus.

4 : 9 :: 16 : 36.

16 : 36 :: 16 : 36.

16 : 36 :: 256 : 576.

4 : 6 :: 16 : 24.

1.

KC : HC :: GF : FE

KF : HE :: GF : FE, 19. 5.

KC : HC :: KF : HE, 11. 5.

GC : FC :: GF : FE, 11. 5.

KCG : HCF :: KFG : HEF, 23. 6.

KCG : KFG :: HCF : HEF, 16. 5.

HCF : HEF :: CBC : ELE, 11 Pro.

KCG : KFG :: CBC : FMF=ELE, 11. 5. 11. Prop.

[Page 64]Therefore FM is equal to the Line LE, and for as much as GC is divided in F, in the same Ratio as FC is in E, Therefore,

GG : FC :: GF : FE, 17. 7.

GC : FC :: FC : EC, 19. 5.

GC : FC :: MD : LD,

GC : FC :: area CBG : area CBF, 12. 5.

2.

From the Vertex of the Hyperbola CBF, Let there be a Semihyperbola as CAF; FH its Transverse Diameter.

Then as,

HCF : HEF :: CBC : ELE, 11. Prop.

HCF : HEF :: CAC : ENE, 11. Prop.

CBC : CAC :: ELE : ENE, 11. 5.

CB : CA :: EL : EN 22. 6.

CB : CA :: area CBF : area CAF, 12. 5.

3.

Let CVQ be ¼ of an Ellipsis, its Trans­verse Diameter FV, its semiconjugate CQ, Let CGQ be ¼ of another Ellipsis its Trans­verse Diameter EG, its conjugate CQ.

Let it be made, as,

FC : EC :: CG : CD, 17. 7.

FC : CG :: EC : CD, 16. 5.

FC : FCCG :: EC : ECCD, 18. 5.

FC : FG :: EC : ED, 18. 5.

CV : GV :: CG : DG, 17. 7.

FCV : FGV :: ECG : EDG, 23. 6.

FCV : FGV :: CQC : GLG, 10. Prop.

ECG : EDG :: CQC : DHD=GLG, 10. Prop.

[Page 66]Therefore DH and GL are equal, Be­cause the Line CV is divided in G in the same ratio as CG is in D.

Therefore, as,

CV: VG :: CG: GD,

CV: CG :: CG: DC, 19. 5.

CV: CG :: IL: IH,

CV: CG :: area CQV: area CQG,

4.

From the vertex of the Quarter of the Ellipsis CGQ; Let there be ¼ of another Ellipsis, as CGR, its Transverse Diameter EG the semiconjugate CR,

Then, as,

ECG: EDG :: CQC: DHD, 10. Pr.

ECG: EDG :: CRC: DKD, 10. Pr.

CQC: DHD :: CRC: DKD, 11. 5.

CQ: CR :: DH: DK, 22. 6.

CQ: CR :: area CDQ: area CGR, 12. 5.

5.

Let CGB be half a parabola, its Diame­ter CG, its Ordinate CB. Let CVB be half of another parabola, its Diameter CV, its Ordinate CB.

[Page 67]

Let it be,

CV: CG :: GV: GD,

CG: GD :: CBC: DND, 9. Prop.

CV: GV :: CBC: GTG=DND, 9. Prop.

[Page 68]Therefore GT is equal to DN; because the Line CV is divided in G in the same ratio as CG is in D.

Therefore,

CV: CG :: CG: CD,

CV: CG :: AT: AN,

CV: CG :: area CVB: area CGB, 12. 5.

6.

From the vertex of the semiparabola CGB; Let there be another semiparabola, as CGZ,

Then, as,

GC: GD :: CBC: DND, 9. Pr.

GC: GD :: CZC: DOD, 9. Pr.

CZC: CBC :: DOD: DND, 11. 5.

ZC: BC :: OD: ND, 22. 6.

ZC: BC :: area ZGC: area BGC, 12. 5.

7.

Hence,

It follows that the areas of hyperbolas, Ellip­ses and parabolas may be increased or decreased in any proportion assigned; either according to their Transverse diameters or Ordinates: or both together.

[Page 69]It follows also,

That the areas of hyperbolas, ellipses and pa­rabolas of the same bases are as their Alti­tudes.

And also,

That the areas of hyperbolas, ellipses and parabolas of the same Altitudes are as their bases.

Further it follows,

If there be two hyperbolas, namely, A and B; if the Transverse diameter of A, be to the Transverse diameter of B; as the conjugate diameter of A, is to the conjugate diameter of B; if the Transverse diameter of A, is to the Transverse diameter of B; as the inter­preted diameter of A, is to the interpreted diameter of B: if the conjugate diameter of A, is to the conjugate diameter of B; as the Ordinate of A, is to the Ordinate of B. Then such hyperbolas are called like hyperbolas.

And the area of A, will be to the area of B; as the Rectangled Figure of the Transverse and conjugate diameters of A, to the Rectan­gled Figure of the Transverse and conjugate diameters of B.

Or,

The area of A, will be to the area of B; as the Rectangled Figure of the interpreted di­ameter, and Ordinate of A, is to the Rectan­gled Figure of the interpreted diameter and Ordinate of B.

8.

Here note,

In the byperbolas CFB, CFA and CGB the Lines CB, EL; and CA, EN; and CB, FM, are called Ordinates.

In the Ellipses, CGQ, CGR and CVQ; the Lines DH, and DK and GL are Ordinates; the Lines CQ and CR conjugate diameters.

In the parabolas.

CGB, CGZ and CV. B; the Lines CB, DN; and CZ, DO; and CB, GT are called Ordinates.

9.

In the hyperbola CFB; CF, EF, CB and EL being given, to find the Transverse dameter FH.

Let FE=A. FC=B. CB=D. EL=E. Z=HF. Therefore HE=ZA. HC=ZB.

Therefore,

DD: EE :: ZBBB: ZAAA, 11. Prop.

ZADDAADD=ZBEEBBEE, 16. 6.

-ZBEE -AADD.

In Words thus.

The difference between the square of B in the square of E, and the square of A in the square of D; being divided by the difference between A in the square of D and B in the square of E; the Quotient is the value of Z.

10.

In the Ellipsis CQG; CQ, DH, and CD being given, to find the Transverse dia­meter EG.

Let CQ=A. DH=B. CD=D, GC=Z. Therefore ZD=ED. Z-D=DG.

ZZ: ZD in Z-D :: AA: BB, 10. Prop.

ZZBB=ZZAA-DDAA. 16. 6.

DDAA=ZZAA -ZZBB

In Words thus.

As the square Root, of the difference be­tween the square of A and the square of B, is to A, so is D, to Z, the semitransverse diame­ter. 22. 6.

11.

In the parabola CGB; CB, CD and DN being known, to find the diameter DG.

Let CB=A. DN=B. CD=C. DG=Z. Then

CZ=CG. AA: BB :: CZ: Z, 9. Prop.

AAZ=BBCBBZ 16. 6.

-BBZ

In Words thus.

As the difference betwixt the square of A and B, is to the square of B; so is C, to Z,

Or thus.

AA: BB :: CZ: Z, 9. Prop.

AA-BB: BB :: CZ-Z: Z, 17. 5.

12.

Yet further, it may be made manifest, that by the points of B and F there may be infinite hyperbolick lines pass: the area of the least, shall not be so little as half the parallelogram, whose base is BC, and Altitude CE: nor the area of the greatest, so great, as three fourths of the said parallelogram.

1.

LEt MACFTP be half a parabolick Cylinder, the base MAC parallel, equal and a like to the base FTP. Let C and F be the vertices of the parabolas ACM and TFP, CM and FP their diameters. Let this cylinder be cut by the Plane HVKB parallel to the Plane FPMC, the Line HB in the superficie of the cylinder. Let it be cut by another Plane, as HGRB parallel to the Plane PTAM, the line HB in the superficie of the said cylinder. Let it be cut by the Plane OADQM, the line AOD in the super­ficie of the cylinder, and the line DQM in the Plane CFPM; the Ordinate AM the common Section of the Planes ABCM and AODQM. Further, Let it be cut by the Plane AIELM, the line. AIE in the superficie of the cylinder and the line ELM in the Plane CFPM. From K, to O and I let lines be drawn, and also from M to D and E: from the intersection of GR and ME, that is from L, to the intersection of the lines HB and AE, that is to I, let there be a line drawn, as LI; [Page 74] and likewise the line QO; then the line LI and QO will be equal and parallel to the line RB. Because the Planes AODQM and AIELM, cut the Plane GFPM at right Angles▪ and the points B, O, I, H, are in the superficie of the cylinder and in the lin [...] BH.

Betwixt the Planes ACM and AODQM there is a solid made, as ABCMQDOA; also betwixt the said base ABCM and the Plane AIELM there is another solid made as AB [...] MLEIA: such solids are called cylindric [...] hoofs, and they take their particular name [...] from such cylinders as they are part of; viz. if the base ABCM be half, or a quarter of a circle, it may be called a circular cylindrick hoof; If half or a quarter of an ellipsis, then an elliptick cylindrick hoof. If the base be half, or a whole parabola, then a parabolick cylin­drick hoof. If half, or a whole hyperbola, then a hyperbolick cylindrick hoof; the like for any other.

[Page 75]

2.

To prove that the Section AODQM is of the same kind and degree as the base ABCM.

The Angle BKO is equal to the Angle RMQ; because BK and RM are parallel, equal and in the Plane ABCM, the Lines OK and QM are parallel, equal and in the Plane AODQM, also the Lines BO and RQ are parallel, equal and in the Plane BHGR; Therefore the Triangles BKO and RMQ are equal and a like.

Because it is, as, CM: CR :: MAM: RBR. 9th. Prop. as MC: MR :: MD: MQ. 4. 6. but AM is common to both and OQ equal to BR; Therefore by the 5th. of the 23. Propos. as, DM: DQ :: MAM: QOQ. Therefore the Section AODQM is a para­bola by the 9th. Proposition.

3. To find the relation of one hoof to another.

The Triangles MCD and MCE are upon the same base MC, Therefore they are as their Altitudes CD and CE. 1. 6. Again, the Triangles KBO and KBI are upon the same base KB, Therefore they are as their Altitudes BO and BI, 1. 6.

[Page 77]Further, the Triangles KBO and MCD are alike, and also the Triangles KBI and MCE are alike, therefore their sides are in proportion. 4. 6.

The Triangle MCD: Triangle MCE :: CD: CE. 1. 6.

The Triangle KBO: Triangle KBI :: BO: BI. 1. 6.

But BO: BI :: CD: CE. 4. 6.

The Triangle KBO: Triangle KBI :: CD: CE. 11. 5.

KBOMCD: KBIMCE :: CD: CE. 12. 5.

Hence it follows,

That the solidities of hoofs upon the same base are as their Altitudes, that is, the hoof ABCMQDOA, is to the hoof ABCMLEIA; as CD, is to CE.

Further it follows.

Because it is BO: BI :: CD: CE, There­fore the superficies of hoofs, upon the same base are as their Altitudes, that is, the super­ficie CBAOD, is to the superficie CBAIE; as CD, is to CE.

4. To find the solidity of cylindrick hoofs.

The Triangles MCD and KBO a [...] alike, Therefore as, as KB: BO :: MC CD, 4. 6. and KB: BO / 2 :: MC: CD / 2, by the converse of 17. 7. then as, KB, to a Geome­trical mean proportion between KB and ha [...] BO, so is MC, to a Geometrical mean pro­portion between MC and half CD, by th [...] 22. Proposition. Let MX be made equal to the last term in the last Proportion, viz. the Geometrical mean proportion between MC and half CD. Then will KZ be equal to a Geometrical mean proportion between K [...] and half BO, and AZXM will be a semi­parabola by the 5th. of the 23: Propsition: that is, as CM: XM :: BK: ZK. Further▪ the area of the Triangle MCD is equal to the Product of MC in half CD, that is, a Square whose side is a Geometrical mean pro­portion between MC and half CD, that is, equal to the Square of MX. Also the area of the Triangle KBO is equal to the Product of KB in half BO, that is, equal to a square whose side is a Geometrical mean proportion between KB and half BO; that is equal to the Square of KZ.

5.

Hence it follows.

That the solidity of the hoof MQDOA BC is equal to all the squares in one eighth of a parabolick Spindle whose semiaxis is AM and semidiameter is MX; but all the squares in a parabolick Spindle is to a parallelepipedon of the same Base and Altitude; as 8, is to 15. Bonav. Caval. Exerc. quer. page 282.

6.

It further follows.

As all the squares in the whole solid AZXM are equal to the whole hoof MQDOAC, so are their parts equal, if the axis be cut by a Plane at right Angle.

Example.

The Plane BKO cuts the axis AM in K at right Augle, that is, the Plane KBO is parallel to the Plane MCD, the part of the Spindle AKZ, is equal to the part of the hoof AKOB.

7.

If the hoof ABCMQDOA be cut by a Plane QON parallel to the Base ABCM, the part QOND may be found thus.

First, take away the part KABO equal to the part of the Spindle KAZ: then take away [Page 80] the prisme KBRMQO; Lastly, take away the parabolick Semicylinder RBCNOQ there will remain the little hoof QOND.

8.

If ACM be a semihyperbola its Transver [...] diameter C (rum), the hyperbolick cylinder may be MABCFHTP: this cylinder being c [...] by Planes according to the First of this Pro­position, the Sections AODQM and AI [...] LM will be semihyperbolas, DS and EW their Transverse diameters by the First of the 23d. Proposition. If between M (rum) and ha [...] (rum)S there be a Geometrical mean proportio [...] found, suppose it M♄, and if between M [...] and half DC there be taken a Geometrica [...] mean proportion, suppose it MX: And also if a Geometrical mean proportion be taken between KB and half BO, suppose it KZ; the Plane AZXM will be the area of a hyper­bola its Transverse diameter X♄; by the 22d. Proposition and by the First of the 23d. Pro­position all the Squares in one Fourth of the hyperbolick Spindle AZXM taking AM for its axis, will be equal to the hyperbolick hoof ABCMQDOA, by the 4th. and 5th. of this Proposition. The relations of the soli­dities and supersicies, by the Third of this.

9.

If ABCM be a quarter of a Circle, the circular cylinder will be ABCMTHFP; this cylinder being cut by Planes, according to the First of this, the Sections AODQM and AIELM will be quarters of Ellipses by the Third of the 23d. Proposition. If a Geometrical mean proportion be taken be­tween MC and half CD, suppose it MX, and also a Geometrical mean proportion be taken between KB and half BO, suppose it KZ. The area AZXM will be one quarter of an Ellipsis by the 22d. Proposition, and by the Third of the 23d. Proposition.

All the squares in one fourth of the semi­spheroide, taking AM for its Semiaxis, and MX for its semidiameter, will be equal to the circular cylindrick hoof AODQMCBA, by the Fourth and Fifth of this Propo­sition. As the whole, so the parts, according to the Sixth and Seventh of this Proposition. How to find all the squares in any Sphere or spheroide is taught Proposition the 15th. and 16th.

Here Note,

If the Altitude of the hoof be equal to four diameters of the base, that hoof will be equal to all the squares in a Sphere adscribed in that cylinder.

[Page 82]If the Altitude be less than Four diameters of the base, that hoof will be equal to all the squares in a spheroide whose longest diameter shall be the axis.

But if the Altitude be greater than Four diameters of the base, then that hoof will be equal to all the squares in a spheroide whose shortest diameter shall be the axis.

10.

If ABCM be a quarter of an Ellipsis, the Elliptick cylinder will be ABCMTHFP; this cylinder being cut by Planes, according to the First of this, the Sections AODQM and AIELM will be quarters of Ellipses by the 3d. of the 23d. Proposition.

If Geometrical means be taken between MC and half CD, and also between KB and half BO, suppose them to be MX and KZ, Then will AZXM be a quarter of an Ellipsis by the 22d. Proposition, and by the 3d. of the 23d. Prop. all the squares in one Fourth of a semispheroide taking AM for its semiaxis and MX for its semidiameter, will be equal to the Elliptick cylindrick hoof AODQMCBA by the Fourth and Fifth of this Proposition. The parts KABOBCRQON and QOND are found as in the Sixth and Seventh of this.

2.

Let AKD be a Hemisphere; the Plane AKD and the Plane L ♄ F ♂ cutting one ano­ther at right Angles their common section the line LF. From the points L and F draw the Lines as LG and FM parallel to the line AD; also from the points L and G, to the points F

[Page 91] and M let there be lines drawn as GM and LF their common section the point Z. From the point L, to the line MF let there be a perpendicular line as LE. Let YP be equal to ♄ ♂; then will the Spheroide whose axis is EL and its conjugate diameter YP be equal to the Excavatus part MZLGZF of the Zone MLGF; or the Zone MLGF less the cones GZL and FZM will be equal to the Sphe­roide EYLP. By the 47. of the 1. of Euclid, and by the 10th. 6th. of Euclid, and by the 3d. of the 23d. Propos. Here note, Z is the vertex of the two cones. Here also note, that L ♃ is equal to ♃ F. Further note, the lines IO, ♄ ♃ and BO are supposed to be at right Angles with the line LF.

3.

Let AGE be a Hemisphere; the planes AGE and GIE cutting one another at right Angles; their common section the line GE. Let GZ be equal to ZE; and also RL e­qual to ZI; then will the Hemisphere AGE less the cone whose diameter of the Base is AE and Altitude CG, that is the Cone AGE, be equal to a Spheroide whose axis is CG and conjugate semidiameter is RL. By the 47th. 1. and 10th. and 6th. and by the third of the 23d. Prop.

4.

Let AGE be a Hemisphere; the Planes AGE and G ♂ Q cutting one another at right Angles; their common section the line GB. Let G ♀ be equal to ♀ B being conti­nued till it meets with the Circle GYA be­ing continued; Let RO be equal to ♂ ♀. Then will the Hemisphere AGE, less the cone whose diameter of the Base is BD and alti­tude CG, that is the cone BGD, be equal to the frustum spheroide whose axis is BP and conjugate semidiameter is OR. By the 47. 1. and 10. 6. and by the third of the 23d. Proposition.

This holds true not only in the Sphere, but also in both the Spheroides; as well in the cone as also in both the Conoides, not only when [Page 93] the cutting plane cuts the axis, but when it is parallel thereunto; but when it is parallel to the axis, there will be a cylinder instead of these cones. The Excavatus parts in the sphere and both the spheroides, will always be spheres or spheroides, or parts thereof their demon­strations by the 47. 1. 10. 6. and by the third of the 23d. Prop.

5.

The Excavatus parts of a frustum cone, will have relation to the hyperbola, ellipsis and pa­rabola; Their demonstrations from the 47. 1. 10. 6. and from the first, third and fifth of the 23. Prop.

6.

The Excavatus parts of a hyperbolick conoide will have relation to all the three sections, viz. the hyperbola, ellipsis and parabola; their de­monstrations by the 47. 1. 10. 6. and by the first, third and fifth of the 23. Prop.

7.

The Excavatus parts of a parabolick conoide, will have relation but to the ellipsis and para­bola, their demonstrations from the 47. 1. and 10. 6. and from the third and fifth of the 23. Proposition.

8.

The solidity of every frustum cone, is equal to a cylinder whose Base is the lesser Base of the frustum, and its altitude the altitude of the frustum, more a hyperbolick conoide whose Base is equal to the difference of the Bases of the said frustum, and its altitude the altitude of the frustum; the Transverse Diameter of the hyperbolick conoide will be a line intercep­ted, betwixt the continuation of the other side of the frustum cone, and the intercepted Diameter, being continued.

9.

The solidity of every frustum parabolick conoide, is equal to a cylinder whose Base is the lesser Base of the frustum, and altitude the altitude of the said frustum, more a parabolick conoide whose Base is equal to the difference betwixt the Bases of the said frustum, and its altitude the altitude of the frustum.

10.

The solidity of every frustum hyperbolick conoide, is equal to a cylinder whose Base is the lesser Base of the frustum, and altitude the altitude of the said frustum; more a hyperbo­lick conoide whose Base is equal to the diffe­rence betwixt the Bases of the said frustum; [Page 95] and its altitude the altitude of the frustum.

This latter hyperbolick conoide, is like to that hyperbolick conoide, of which the frustum is a part.

11.

A Sphere being cut by two parallel planes, both of them equidistant & parallel to a plane passing through the Center of the Sphere; in­cludes a part of the Sphere, which for distin­ction sake may be called a middle Zone.

The solidity of every middle Zone of any Sphere, is equal to a cylinder of the same Base and altitude as the Zone; more a Sphere whose diameter is equal to the altitude of the Zone.

12.

A Spheroide being cut by two parallel planes, both of them equidistant and parallel to a plane passing through the Center of the Sphe­roide and cutting the Axis of the said spheroide at right Angles, includes a part of the sphe­roide, which part for distinction sake may be called the middle Zone of a spheroide.

The solidity of the middle Zone of any spheroide, is equal to a cylinder of the same Base, and altitude as the Zone; more a sphe­roide whose Axis is equal to the altitude of the Zone.

[Page 96]The ellipsis which generates this last sphe­roide is like to that ellipsis which generated that spheroide of which this middle Zone is a part.

13.

The solidities of hyperbolick conoides upon the same Base, are as their altitudes. Here the Transverse Diameter is increased or de­creased in the same proportion as the inter­cepted Diameter. By the first of the 23d Proposition.

And also,

The solidities of hyperbolick conoides under the same altitude, are as their Bases. Here the conjugate Diameter is increased or de­creased in the same proportion as the ordi­dinate. By the second of the 23d. Proposi­tion.

The solidities of like hyperbolick Conoides, are in a triplicate ratio of their corresponding terms, that is, their Transverse Diameters, or their intercepted Diameters, their con­jugate Diameters, or the Ordinates.

14.

The solidities of hemispheroides upon the same Base, are as their Altitudes. By the 3d. of the 23d. Prop.

The solidities of hemipheroides under the same Altitude, are as their Bases, by the 4th. of the 23d. Prop.

[Page 97]The solidities of like spheroides, are in a triplicate ratio of their corresponding terms.

15.

The solidities of parabolick conoides upon the same Base, are as their Altitudes. By the 5th. of the 23d. Prop.

The solidity of parabolick conoides under the same Altitude, are as their Bases. By the 6th. of the 23d. Prop.

The solidities of like parabolick conoides, are in the triplicate ratio of their correspon­ding terms.

An Example thus.

Suppose there be two parabolick conoides, A and B, the solidity of A, will be to the so­lidity of B; as the Cube of the axis of A, is to the Cube of the axis of B.

Further, as the parabolick conoide of A, is to the parabolick conoide of B; so is the Cube of latus rectum of A, to the Cube of latus rectum of B.

Yet further, if these conoides both equally incline, it will be;

As the conoide of A, is to the conoide of B; so will the Cube of the Diameter of A, be to the Cube of the Diameter of B. The like in the rest.

If there be two parabolick conoides unlike, suppose A and D. Then,

A will have that proportion to D, as is [Page 98] composed of the Base and altitude of A, to the Base and altitude of D.

In parabolick conoides, as A and B.

If the conoide of A be equal to the conoide of B, then their Bases and altitudes are reci­procal, and if their Bases and altitudes are re­ciprocal; those conoides are equal.

Thus.

As the Base of A, is to the Base of B; so is the altitude of B to the altitude of A.

These are said to be reciprocal, and their magnitudes are equal.

The like in hemispheroides, but not in hy­perbolick conoides; for there may be infinite hyperbolick conoides, yet having the same Base & altitude; the least not so little as a Cone, no [...] the greatest so great as a parabolick conoide of that same Base and altitude. In this condition the semidiameter of the Base, and the altitude are supposed to be equal.

16.

Spheres of equal Diameters may be added together, their sum will be a spheroide, whose axis will be equal to the Diameter of one of the spheres; But the area of that Circle which passeth through the Center of this spheroide, and cutteth the axis at right Angles; will be equal to the areas of so many great Circles of those spheres, as there are spheres in number. Suppose there be six spheres to be added toge­ther, [Page 99] the axis of the spheroide will be equal to one of their Diameters; but the area of that Circle, that passeth through the Center of that spheroide, and cutteth the axis at right Angles, shall be equal to the areas of six Cir­cles whose Diameter is equal to the Diameter of one of the spheres. The areas of the Circles are added, or substracted, by the 47th. of the 1. of Euclid.

Spheres and Spheroides of the same axis, may be added to, or substracted from each other, their sum, and difference will be spheres or spheroides.

17.

Hyperbolick conoides of the same axis may be added too, or substracted from each other, their sum and difference will be hyperbolick co­noides. Here you are to take the sum of the Bases, for the base of the sum, and the diffe­rence of the bases, for the base of the diffe­rence. Further, We are to take both the parameters for the parameter of the sum, and the difference of both the parameters for the parameter of the difference.

18.

Parabolick conoides of the same axis, may be added too, or substracted from each other, the sum and difference will be parabolick co­noides, using the former rules for their bases and parameters,

[Page 100]Here note, the line PZ in the Diagram for the 9th. Propos. the line PN in the Dia­gram for the 10th. Propos. the line NH in the Diagram for the 11th. Propos. are called parameters.

19.

If the axis, Transverse and conjugate Dia­meters of a hyperbolick conoide be equal one to another; and the axis equal to the axis of a Sphere: this hyperbolick conoide and Sphere being added together, they make a parabolick conoide; its Base and altitude equal to the Base and altitude of the hyperbolick conoide, its parameter the double of the Diameter of the Sphere.

If there be a hyperbolick conoide, A; and a Spheroide, B; their axis equal: If the transverse Diameter of A, is to the parameter of B; as the parameter of A, is to the parameter of B. Two such conoides being added together, they will make a parabolick conoide; its Base the same as the hyperbolick conoide; its para­meter equal to the parameters of the other co­noides, being added together.

If there be a Cone, whose axis is equal to the Diameter of the Base; and a Sphere whose axis is equal to the axis of the Cone; this Cone and Sphere being put together, makes a parabolick conoide; its Base equal to the Base [Page 101] of the Cone, its parameter equal to the Dia­meter of the Sphere:

20.

Parabolick Conoides, may always be added to, and sometimes substracted from hyperbolick conoides of the same axis; that sum will be a perbolick conoide, and that difference when a difference may be, will be a hyperbolick conoide, the sum of their Bases for the Base of the sum: and the difference of their Bases for the Base of their difference: the sum of their parameters for the parameter of the sum, and the diffe­rence of the parameters for the parameter of the difference.

Here note, when the parameter of the hy­perbolick conoide is greater than the parameter of the parabolick conoide then a difference may be: but if it be lesser, then no difference can be taken.

The Demonstrations of these are in Prop. 15, 16, 17 and 18, compared with Propos. 9, 10 and 11. of this Book.