STEREOMETRICAL Propositions:  
VARIOUSLY APPLICABLE;  
But PARTICULARLY INTENDED For GAGEING.  
By ROBERT ANDERSON.  
LONDON,  
Printed by William Godbid for Joshuah Conniers at the Sign of the Black Raven in Duck-lane, 1668.   

To the READER.  
ABout seven years since I resolved these following propositions, with others of the like nature; but did not intend ever to have published them: But being over-persuaded by some of my Friends, I have here made so many of them public, as I thought would be convenient to what here is intended.  
Here I ought to acknowledge that great Respect which I owe to my Worthy Friend Mr. john Collins, one of the Royal Society; which propounded to me about 6 years since, that proposition of the second Sections of the Sphere and Spheroide; and also in giving me Information, and helping me to such Books as might be most advandageous: Further, not only I, but all Lovers of the Mathematics are very much obliged to this our Worthy Friend for his good Intelligence, great care in sending for Books from beyond the Seas, and his continual love in promoting the Mathematics, and Mathematical Men.  
Forasmuch as throughout this little Tract, here is much use made of Parallelepipedons, Prismes and Pyramids: I think fit, a little to insist thereon; chiefly for the young Learners sakes, for which this is only intended.  

In the Diagram,  
Let there be a Solid, as POHG VFIR; the plane POHG parallel equal and alike to the plane RIFV; Further, POIR is parallel equal and like to the plane GHFV: Yet further, the plane OHFI is parallel equal and like the plane PRUG: such a Solid is called a Parallelepipedon.  
Let there be a Solid as GHDEV F, the plane GEV is parallel equal and like the plane HDF; Further, the planes HGFV and DEUF are alike and the line FV common to both: such a Solid is called a Prism.  
Let there be a Solid as HBCDF, the planes BCF, CDF, DHF and HBF all meet at the point F, such a Solid is called a Pyramid.  
Further,  
Pyramids, Prismes and Parallelepipedons upon the same Base and Altitude, are as, one, one and a half, and three, that is, if the Pyramid be 4, the prism will be 6, and the parallelepipedon will be 12.   

An Example in Numbers.  
If PG, 6; PO, 8; and HF the Altitude be 9; Then multiply 6 by 8, and the product will be 48, the Area POHG, this Area multiplied by 9 the Altitude, the product is 432, the parallelepipedon POHGUFIR, if the Area 48 be multiplied by half 9 that is 4●… the product will be 216, equal to a prism of the Base POHG and altitude HF.  
If the Area POHG, 48, be multiplied by one third of 9, that is by 3, the product will be 144 equal to a pyramid whose Base is POHG and altitude HF.   

To find the solidity of the prism ●…HDEVF.  
Let GE, 10; GH, 8; HF 9; multiply 10 by 8, and the product will be 80 the Area GHDE, this 80 being multiplied by half the altitude HF, 9; that is, 4½, the product will be 360, the prism GHDEUF.   

To find the solidity of the pyramid HBCDF.  
Let HD, 10; HB, 4; multiply 10 by 4 the product will be 40, the Area HBCD, this Area multiplied by 3, that is one third of the Altitude HF, the product will be 120, the pyramid HBCDF.  
  

To find the solidity of a Solid, that hath one of its Bases an Ellipsis and the other Base a Circle, these two Bases are supposed to be parallel.  
In such a Solid, the sides being continued will never meet at a point as in circular and Elliptic cones, and therefore the general Rule will not resolve these kind of Solids: therefore consider it thus;  
Let PE be equal to the Transverse diameter in an Ellipsis, PA its Conjugate diameter. Let RV and RI be the diameters of a Circle. First, Let there be made a Rectangled figure of the Transverse and Conjugate diameters of the Ellipsis, as PACE. Let there be a square made of the diameter RV, as RIFV. Let the altitude of the Solid be HF, find the solidity of the Solid PACEUFIR, thus, Let it be cut into as many parallelepipedons, prismes, & pyramids as are necessary: then find the solidity of those parallelepipedons, prismes and pyramids, as before is taught, those parallelepipedons, prismes & pyramids being added together will be equal to the whole Solid, thus, the parallepipedon POHG VFIR, more the prism GEDHFV, more the pyramid DCBHF, more the prism HBAOIF; these 4 Solids being added together make the whole Solid ACEPRIFV. For the whole is, 4½ the product is 40½ equal to the prism GHDEUF. HD, 3; HB, 1; therefore HBCD is equal to 3; which being multiplied by one third of the altitude HF, that is, 3; the pyramid HBCDF will be equal to 9; BA, 3; AO, 1; therefore ABHO will be equal to 3, which being multiplied by half HF the altitude, that is, by 4½ the product is 13½, equal to the prism OABHFI. Now these 4 composed Numbers being added together make 144 equal to the whole frustum pyramid PACEUFIR. Then as 14: 11:: 144: 113 1/7 the irregular Elliptic solid, whose Transverse PE, and conjugate PA diameters at the Elliptic Base are 6 and 4, and the diameters of the Circular Base is RV 3.  

Here note,  
Although Archimedes gives the proportion of the square of the diameter of any Circle, to the area of that Circle; as 14, is to 11. And also; that the diameter of any Circle is to the circumference of that Circle, as 7, is to 22; Yet you are not to underderstand that these proportions are in their just values; But that they are the least numbers that will best agree to that approach.  
If those numbers be thought not to agree near enough to truth, instead of those, these may be used; for, 14 and 11, take 452 and 355; for 7 and 22, take 113 and 355.   

Further note.  
In the 24th. proposition there is somewhat said of Cylindrick hoofs. Those cylinders aught to be upright and not inclining cylinders.   

Ye●… further note.  
By reason of much business, I could not attend the Press as I ought to have done; and by that means there are considerable faults, which I desire may be corrected before the Book is read, otherwise there may rise a misapprehension of what there is intended.   

Lastly note,  
Almost at the beginning of the first proposition you may find it printed, Thus, CN: NZ:: EH: HZ, 4. 6. the meaning is thus, as CN is to NZ; so is EH, to HZ, by the fourth prop. of the sixth of Euclid; The like in the rest.    
R. A.   


PROPOSITION I. To find the solidity of a Frustum Pyramid, whose bases are parallel and a like.  
LEt ABCNHEDF be a frustum pyramid, whose solidity is required; Let the Lines OF, BD, CE, and NH be continued till they meet at Z; Let GO, be the height of the frustum pyramid, OZ the height of the Continuation; Let FDEH be set at the Angle N, as NR QK. Then CN: NZ:: EH: HZ, 4. 6.  
And CN: EH:: NZ: HZ, 16. 5. CN— EH: EH:: NZ— HZ: HZ, 17. 5. That is, CK: KN:: NH: HZ, because the planes HFDE, and NABC are parallel, they cut the Line ZG, in G and O in the same proportion as they cut NZ, in N and Z, by 17. 11. Thus, NH: HZ:: GO: OZ, Therefore CK: KN:: GO: OZ, 11. 5. OZ being found, which added to GO makes the whole Altitude GZ; by which find the whole pyramid NABCZ, which done, find the pyramid HFD EZ, by the 7. 12. then N ABCZ less HFDEZ, the Remainder is the frustum HFDECB AN.  
A 2d way may be thus; First to prove that AK is a mean proportion between AC and F E; then C N: NK:: C A: AK, 1. 6. AN: NR:: NM: NQ, 1. 6. CN: NK:: AN: NR, 7. 5. CA: AK:: NM: NQ, 11. 5.  
Further, if GO the Altitude of the frustum FDEHCBAN, be multiplied by AC more AK more KR, that Product will be the triple of the said frustum; for the parallelepipedon made of GZ the altitude of the pyramid in the Base NABC is triple of the pyramid NABCZ, by the same reason the parallelepipedon made of GO in AC, together with the parallelepipedon made by ZO in AC is triple of the said pyramid by 7. 12. if the parallelepipedon made of OZ in FE, that is, in RK, that is, the triple of the pyramid HFDEZ be taken from the parallelepipedon NABCZ, there remains that which is made of GO in AC, together with those that are made of OZ in AQ and the same OZ in MC, the triple of the frustum ABC NHEDF, by 5. 5. for, CK: KN:: GO: OZ, CK: KN:: ARE: RN, Therefore GO: OZ:: ARE: RN, and CK: KN:: CM: MN, 4. 6. ARE: RN:: AQ: QN, therefore GO: OZ:: CM: MN, 11. 5. GO: OZ:: AQ: QN, therefore GO in MN = OZ in CM, GO in QN = OZ in AQ, add GO in AC to both, that is, GO in MN, more GO in QN, more GO in AC, is equal to OZ in CM, more OZ in AQ, more GO in AC; but OZ in CM, more OZ in AQ, more GO in AC are triple of the said frustum; therefore GO in MN, more GO in QN, more GO in AC are equal to the triple of the said frustum.  

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 multiplied 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 pyramid the Bases Squares and Parallel; Let HPOF be projected in the Base ACFZ, as ZURT; VR and TR be continued to D and B; then will TD be equal to RAMIRES, and RC a Square; most manifest it is, that the parallelepipedon ZURTPOFH, more the prism VABR OPEN, more the pyramid RBCDO, more the prism RDETOF is equal to the said frustum pyramid.  
For all the parts of any magnitude being taken, together are equal to the whole. Axiom 19 1.  
The prism V ABROP is equal to the prism 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 pyramid 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 added to 1560, is equal to 1945●…, this last composed number multiplied 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 ascribed within the frustum pyramid.    

PROPOSITION II. To find the solidity of a frustum pyramid, whose Bases are parallel, but not alike; that is, when their corresponding sides are not proportional.  
LET ADGOUZTS be the pyramid proposed, here AGNOSTUS is a square, SZ a parallelogram; Let STZV be projected in the base, as, QPKL; continue the sides, as QL, to B and I; PK, to C and H; LK, to M and F; PQ, to E and R; the Figure being completed, the whole frustum will be composed of these parts, namely 4 pyramids, as CDEPT, KFGHZ, LIOMV, RQ BAS, more 4 prismes, as BCPQTS, QR MLUS, LKHIUZ, KFEPTZ, more the parallelepipedon PQLKZUST; a further demonstration is needless.  
 

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

Thus.  The 4 pyramids are 6000. The 2 lesser prismes are 6120. The 2 greater prismes are 22500. The parallelepipedon is 30600. The whole frustum pyramid is 65220.   

A Second way thus.  
Let FC be the greater base, RQ the lesser, GO the height; make it as RH: HQ:: FAVORINA: AB; draw BE parallel to FAVORINA, then will 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 prism, BCUT QZ, more the pyramid TUDEZ: 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 multiplied by ½ the altitude, that is 32, makes 13312, equal to the prism BCUTQZ; TV, 26; VD, 34; TD, 544; being multiplied by ●… of GO that is 21●…, makes 11605⅓ equal to the pyramid TUDEZ; these three composed numbers being added together, makes 149418●… equal to the frustum pyramid ACD FRHQZ.  
The frustum ABEFHRZQ 124501●…. The prism BCUTQZ 13312. The pyramid TUDEZ 11605●…. The frustum ACDFRHQZ 149418●….   

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 easy it is to apprehend, that the frustum pyramid PACEUFIR, is equal to the parallelepipedon POHGUFIR, more the prism OA BHFI, more the pyramid HBCDF, more the prism HDEGUF, 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 POHGUFIR, 31200: OA, 12; AB, 30; AH, 360; being multiplied by 20, that is half the height PR., it makes 7200 equal to OABHFI. BC, 28; HB, 12; BD, 336; being multiplied by ●… of the altitude that is 13⅓, it produceth 4480, equal to the pyramid HBCDF; GH, 26; HD, 28; HE, 728; being multiplied by 20, the product is 14560 equalto the prism HDE GUF; these 4 composed numbers being added together, makes the solidity of the said frustum.  
The parallelepipedon-POHGFVRI = 31200 The prism— OABHFI = 07200 The pyramid— HBCDF = 04480 The prism— HDEGUF = 14560 The frustum pyramid— PACEUFIR = 57440  
Then,  
As 14 is to 11, so is the solidity of any frustum pyramid thus found, to any Elliptic frustum, ascribed in that frustum pyramid.    

PROPOSITION. III. To find the solidity of a Frustum Prism.  
LEt CAGDEF be a prism, who solidity is required, let BC be mad●… equal to FE, and BH parallel to CD, and th●…  
Figure being completed, the prisme●…CDGFE ●…CDGFE may be composed of the prisme●…CDHFE ●…CDHFE, more the pyramid BAGHF: further demonstration is needless.  

An Example in Numbers.  
BC, 52; BH, 64; BD, 3328; this last ●…umber being multiplied by 40, that is half ●…he height, makes 133120 equal to the prisme●…BCDEF ●…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 pyramid GABHF: this prisme●…nd ●…nd pyramid being added together, makes 187733●…, equal to the frustum prism AGD CEF.   

The second Case thus.  
Let ZBCDEH be a frustum prism; Let AC be equal to HE and AO parallel to ZB, the Figure being completed, the prism AC DOHE less the pyramid ABZOH is equal to the prism BCDZHE; or thus, the prism ZBCDFE more the solidity ZBHF, that is, half the pyramid 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 prism OACDHE; AC, 32; BZ, 64; AZ, 2048 being multiplied by 26●…, that is, one third of the height, it makes 54613⅓ equal to the pyramid ABZOH this pyramid taken from the prism leaves 160426●…, equal to the frustum prism proposed.  
Then as 14 is to 11, so are such prismes to Elliptic solids ascribed in those prismes.   

Hence follows a Fourth way for finding the solidity of these irregular frustums.  
Suppose ACDFGNIH be the frustum proposed, because FAVORINA is equal to DC, and AC equal to DF, and GN equal to IH, and NI to GH; and because GN is less than FAVORINA, 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 DFUZ, less NI HGUZ, 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, disjoined, thus.  
The pyramid BCDEZ, is to the pyramid RQIHZ, as the cube of BE to the cube of RQ; 8, 12. The prism FABEUZ, is to the prism GNQRUZ, as the square of FAVORINA, 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 pyramids; is because FE, GRACCUS, and US are equal, and the ratio riseth but from VG●… GN and VF, and FAVORINA; whereas in th●… pyramid the ratio riseth from three, that is from ZR, RQ, and RH; and from ZE●… ED, and EBB.   

In Numbers thus.  
Let US be 9, ZR, 3; ZE, 6; RQ, ●… EBB, 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●… pyramid 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 prism G●… QRZV; that is, 54 to the prism 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.  
What ever is said of these prismes and pyramids, the same is to be understood in Cones and Ecliptic 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 fixed; 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 pyramid are the Cube of the continuation, or the side RZ; and the pyramid RQIH, the like in the prism, 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 pyramid RQIHDC ●…E, the first number is the solidity of one ●…nch, the second number of two inches, the ●…hird of three inches; for the prism; 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 prism 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 Elliptic frustum.    

Proposition. IV. To find the solidity of a frustum pyramid, whose Bases are not parallel; and the inclination is from side to side.  
 
LEt R HE TCBA●… be a frustum pyramid, the base R●… ET, no●… parallel to the base ●… ABC; le●… RHD●… be paralle to AB●… F, the●… may th●… whole frustum pyramid RH ETAB●… F be composed o●… the frustum pyramid RHDZABCF more, the prism RHDZTE, by a composition of the last Propos. PD being the altitude of the frustum pyramid RHDZABCF: EX the height of the prism RHDZTE may be found, thus▪ BD: DP:: DE: EX; a further demonstration is needless.  

II. To find the solidity of a triangular frustum pyramid, whose bases are not parallel.  
 
Let CABDHF be a frustum pyramid, CAB not parallel to HDF, let QDE be parallel to CAB; the triangular frustum Q DEBAC is half a quadrangular frustum pyramid, found by the 2 Proposion, DZ the height of the pyramid QHFED; Therefore the frustum pyramid QDEBAC more the pyramid QHFED will be equal to the frustum pyramid FHDABC.   

III. To find the solidity of the frustum pyramid, ACFEDZ.  
 
Here the base AFC is not parallel to the base ZDE but to the plane ZOE; the frustum pyramid ZOE FCA may be found by the 2 Proposition, the part ZOE D will be a pyramid, its altitude may be found, thus, As OF: OG:: ODD: DR, the frustum pyramid ZOEFCA more the pyramid ZOED will be equal to the solid ACFEOZD.   

IV. To find the solidity of a frustum pyramid, whose bases are not parallel, and whose inclination is from Angle to Angle.  
Let CABDFXKQ be a frustum pyramid, the base CABD not parallel to the base FXKQ, but parallel to the plane FEH G; then the frustum pyramid C ABDHEFG found by the 2 Propos. more the pyramid EQXG F, found by the 2 case of this Propos. more the frustum pyramid GEHZXQ, found by the 2 Propos. more the pyramid XKZQ, found by the 3 case of this Propos. will be equal to the whole frustum pyramid CABDFXKQ.  
   

PROPOSITION V.  
LEt FHAOE be a Semicircle, HO parallel to FE; FB and ED parallel to CA, the Figure being completed, the Cone QCI is equal to the Cylinder GE less the portion of the Sphere FHOE.  
For the square of CO less the square of RO is equal to the square of RC, the square of RN (equal to the square of CO) less the square of RO is equal to the square of ON more the Rectangle twice under RONVERE; therefore the square of RC (that is the square of RI) is equal to the square of ON more, the double Rectangle RONVERE, by 4: 2. every one of these being considered one by one; but being collected (viz.) all the squares of RI is equal to all the squares of RN, less all the squares of RO; or if there be taken the quadruple of them, all the squares of QI is equal to all the squares of GN less all the squares of HO: by 2: 12, all the circles; therefore the Cone QCI is equal to the Cylinder GE less the portion of the Sphere HFEO which was proposed.  

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 square of LV be equal to the square of SU. For ARE RD:: TV: VL, there fore the squares of them, and ARE: RD:: OV: US, 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 US; Therefore, as the square of EV, to the square of EV less the square of TU. 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 SU.    

PROPOSITION VI.  
IN the Hemisphere JAP, Let FE be parallel to IP, SZ and GX parallel to AC; the Figure being completed; the Cylinder QG is equal to the Cylinder IE less the Cylinder RH; for as the square of AC, to the square of BC; so the Cylinder IE, to the Cylinder RH; again, as the square of AC, to the square of AC less the square of BC, (that is the square of BS) so is the Cylinder IE, to the Cylinder IE less the Cylinder RH, (that is the Cylinder QG) it followeth that the Cylinder RH is equal to the excavetus Cylinder IQSFGEPV, for the Cylinder QG is equal to the excavetus Cylinder IROFHTPE.  
But if the excavetus Cylinder ORQSGV TH., be added to both; the Cylinder RH will be equal to the excavetus or hollow cylinder QIFSGUPE; but the cone OCH is equal to FISGPE by the first part of the Fifth Proposition; Therefore IQSGUP is equal to ORCTH, that is, ●… of the difference betwixt the circumscribed IE and inscribed QG cylinders; substracted from the circumscribed cylinder IE is equal to the portion of the Sphere ISGP; or if ●… of the difference be added to the inscribed cylinder QG it will be equal to the portion of the Sphere IS GP.  
The like in a Spheroid by the second part of the last Proposition.  

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 multiplied by the length ZS, 20; the Product is 5531 3/7 the solidity of the portion ZISG PX.    

PROPOSITION VII.  
IF two equal and alike Triangles as GCA and GFA be contrary put, (viz.) the Vertex of the one to the Base of the other, and Lines drawn parallel to the Base as MXV parallel to CA; If MX be drawn into XV, that so the Triangle GCA and GFA being drawn one into another; they will generate a solid equal to one sixth of the paralellepipedon ABDELLA CEFGH.  
For by 4; 2. the Line MV being cut into 2 parts in X, the square MX, (that is IZ) more the square of XV, (that is XO) more the double Rectangle of MXV, that is, the Rectangles XQ and OR, are equal to the whole square US; for that reason the whole paralellepipedon ABDCEFGH is composed of these parts, the pyramid ABDCH more the pyramid EFGHA more the two solids BDKOIRHE and ACMXIQFG, but the paralellepipedon ABDCEFGH is equal to three pyramids of the same Base and Altitude; therefore one of these solids as ACXMQIFG will be equal to ⅙ of the whole.   

PROPOSITION VIII.  
The second part of this Theorem may b●… that the Rectangle EC, CF; is to the Rectangle of the Tropezium EABD and the Tr●…angle BDF, as the Line AC, to ½ AB mo●… ⅓ BC; the demonstration of this differs no●… from the former part, except in this, that a●… the Rectangles DFD, ZRZ, ZR Z, are of the paralellepipedon, as appears by the la●… Proposition.  
This Proposition is the 30 of the second o●… Caval. Geomet. Indivis. and is of good use i●… solid Geometry.   

PROPOSITION. IX.  
LEt ADBEK be a cone and be cu●… through the axis, whose Section will b●… the Triangle AKB, Let it be cut by another Plane as EZD, the diameter of the Section Z●… parallel to BK, the Base of the Section ED at Right Angle at C with AB; Let it b●… as ABA: AKB:: PZ: ZK, then will PZC be equal to the square of CE;  
For,  
AB: BK:: AC: CZ, therefore by 23. 6.  
BASILIUS: AK:: CB: ZK, therefore by 23. 6.   

PROPOSITION X.  
LEt AUBXK be a cone, and be cut through the axis, whose Section shall be the Triangle AKB, Let it be cut by another Plane as NIEC whose diameter is NC, (being continued if need be) shall cut both sides of the cone, the Plane NIEC being at Right Angle with the Plane AKB, the Planes FED and TIG parallel to the Plane AUBX, Let it be made, as CRN: GRT:: CN: NP, then will QRN be equal to the square of RI; for CN: NP:: CR: RQ, taking the Altitude RN it may be CN: NR:: CRN: QRN:: CRN: GRT, in these last four terms CRN is twice, therefore QRN is equal to GRT, that is the square of RI.  
By reason of the similitude of the Triangle CHD to CRG, and of the Triangle NRT to NHF, it may be, as  
NR: TR:: NH: FH by 23. 6.  
RC: RG:: HC: HD by 23. 6.  
NRC: TRG:: NHC: FHD, but T ●… G is equal to the square of RI, and FHD ●…s equal to the square of HE; therefore as ●…RC to the square of RI, so is NHC to ●…he square of HERALD  
This Section of a cone is named by Apollo●…ius an Ellipsis.  
But if the Semi-Ellipsis CEIN be move●… about the axis, CN, until it return to tha●… point from whence it began its motion, it wil●… Generate a solid, Archimedes' calls such a soli●… a Spheroide; but if it be turned about th●… lesser Diameter, than he calls that solid a Spheroide prolatus.   

PROPOSITION XI.  
LEt ABCBF be a cone, and be cut throug●… the axis, whose section will make th●… the Triangle AFC, Let it be cut b●… another Plane BHD, the Diamete●… HG, and the side of the Triangle C●… being continued may meet at K, HN at righ●… Angle with HG, the Plane OIE parallel t●… ABCD, the Plane BDH at right Angl●… with the Plane AFC; Let it be made, a●… KRH: ERO:: KH: HN, draw RZ parallel to HN, and join ZNK, then will ZR●… be equal to the square of RI; For, as KH▪ ●… NH:: KR: ZR, take the Altitude RH an●… apply it to the two last terms, and it will be a●… KH: NH:: KRH: ZRH:: KRH: ERO▪ Therefore ZRH is equal to ERO that is the square of RI because of the similitude of the Triangle AGH to HRO, and of KGC to KRE, it may be as,  
KG: CG:: KR: ER by 23, 6.  
GH: GA':: RH: RO by 23, 6.  
This manner of Section Apollonius calls Hyperbola; but if the semihyperbola BIHR be moved about the Axis HG, until it retu●… to that point from whence it began its mtion, it will Generate a solid; such a solid called by Archimedes an Amblygon conoid●… but generally it is called a Hyperbolical conide.   

PROPOSITION XII. To find the solidity of the portion of the Sphere BNRE.  
LEt ABNE be a Sphere and BH a C●… linder, it is manifest that the paralellepipedon whose Base is the square of GE, a Altitude GN is equal to the paralellepipea whose Base is the square of GN and Altitu AGNOSTUS; it is also manifest that the Rectang AGN is equal to the square GE, the li for any other, as, AZN equal to ZR Therefore by the 8 Proposition, as AGNOSTUS t●… AGNOSTUS more ⅙ GN, so are all the squares Z to all the squares of ZR; Therefore, as A to ½ AGNOSTUS more ⅙ GN, so the Cylinder E ●…o the portion of the Sphere ERNB. The ●…ike in a Spheroide.  
 

In Numbers thus.  
Let AGNOSTUS be 150; GN, 54; GE 90; the paralellepipedon made of the square of BE and Altitude GN, in 1749600; then, as AG, 150; to ●… AGNOSTUS, 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.    

PROPOSITION XIII. To find the solidity of a Parabolical Conoide  
LEt BCOE be a semiparabola; by th●… 9 Proposition it is, as EBB, is to ER that is to ZQ; so is the square of BC that i●… the square of RI, to the square of RO, tha●… is, as the parallelogram BF, is to the Triang FEB; so is the parallelepipedon made of th●… square of BC and Altitude BE, to all th●… squares of RO, but the Triangle FBE is of the parallelogram BF; Therefore the parabolical conoide will be ½ of the cylinder of th●… same Base and Altitude.  
 

Thus in Numbers.  
Let AB, 10; BC, 10; Therefore AC, 20; Let BE, 30; the square of AC, 20; is 400, which being multiplied by ½ EBB, 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.    

PROPOSITION. XIV. To find the solidity of a Hyperbolical Conoide.  
LEt ZRHX be a semihyperbola, and OF parallel to ZK, ARE the Transverse Diameter, RX the intercepted Axis, the point H in the Hyperbolical Line by the 11 Proposiion it is, as AXR to AFR, so is the square of XZ, to the square of FH, that is, as the parallelepipedon made of the Rectangle RXZ ●…nd the Altitude AXE, is to the prism made of the Rectangle ARX and the Altitude ●…D, more the pyramid made of the square ●…f ZD and Altitude DR; so is the parallelepipedon made of the square of ZK and Altitude XR, to all the squares in the conoide●…HRK ●…HRK.  
But as the parallelepipedon made of the Rectangle RXZ and Altitude AXE, is to the prism made of the Rectangle ARX and Altitude RD, more the pyramid made of the square DZ and Altitude DR; so is the Line AXE to the Line BR (that is ½ ARE) more ⅓ of the Line RX, by the 8 Proposition; Therefore by the 11. 5. it will be as the Line AXE, is to the Line BR, more ⅓ of the Line RX, so is the parallelepipedon made of the square of ZK and Altitude XR, to all the squares in the conoide ZHRK; then as 14: 11, so are all the squares in the conoide ZHRK, to the conoide ZHRK.  

In Numbers thus.  
Let AXE be 150; RX, 54; ZX, 90; ARE, 96; BR, 48; the parallelepipedon whose Base is the Rectangle of RXZ and the Altitude AXE is 729000.  
The prism whose Base is the Rectangle ARX and Altitude RD is 233280.  
The pyramid 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:: AXE, 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 multiplied 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 Hyperbolic conoide ZHRK.    

PROPOSITION XV. To find the solidity of a whole Sphere, or a portion thereof.  
LEt ADEF be a Plane and the Semicircle DQIQE be in that Plane, Let the Planes ABCD and CDE be at Right Angle with the Plane ADEF; Let DC, AB and OF each of them be equal to DE; Let the Planes KHXR be parallel to the Base of the prism; namely, parallel to AB CD. The Rectangle DRE is equal to the square of RQ, by 353, that is, the Rectangle comprehended of IR and RX, for RX is equal to RE, and IR is equal to RD, therefore all the squares in ¼ of the Sphere E QIQD are equal to all the Rectangle in the prism CDFE, the like in any part; Therefore the prism RXHKFE, less the pyramid OIKHF shall be equal to all the squares in the portion of the Sphere QER. Then as 14: 11, so the prism XRIOFE to the solidity of the portion of the Sphere QER.  

In Numbers thus.  
Let AB, OF, 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 multiplied by RX, 54; is XK, 11016; which being multiplied by ½ RE 27; the Product will be 297232, equal to the prism KHXR OF; FK, KH, HO, and IK, every of them equal to 54; Therefore the pyramid KHO IF will be 52488; then KHXREF, 297 232; less KHOIF, 52488 there will remain JOXREF, 244944; equal to all the squares in the portion QER; that is, all the squares in the portion of ¼ of a Sphere,    

PROPOSITION XVI. To find the solidity of a whole Spheroide, or a portion thereof.  
LEt FADE be a Plane, the Ellipsis EQ D be in that Plane; Let AB and DC each of them be equal to DE, the Planes K RXH parallel to the Base of the prism AB CD. Find the Line FE by the 10th. Prop. by the said 10th. Prop. the Rectangle IRE will be equal to the square of RQ; but the Rectangle IRE is equal to the Rectangle IRX, because DC is equal to DE, therefore RX is equal to RE, therefore the Rectangle IRX is equal to the square of RQ; therefore the prism XHKREF less the pyramid KHOIF will be equal to all the squares in the portion QER.  

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, multiplied by R X, 9; XK will be 108; which being multiplied by ½ of ER, that is 4 ½ the Product will be 486, equal to KHXREF; for the finding of KI, work thus, OF, 36; to AD, 12; so is FK, 9; to IK, 3; then IK, 3, multiplied by KH 9; the Product is 27, that is, HI; this last number being multiplied by ⅓ of KF, that is by 3; the pyramid KHIOF will be equal to 81, which being substracted from the prism KHXREF there will remain 405 the prism JOXREF equal to all the squares in the portion of the spheroide QER.    

PROPOSITION XVI. To find the solidity of a hyperbolical Conoide.  
LEt MAEL be a Plane, and the hyperbola be in that Plane; to that Plane let the Planes BAM, CELIA and DEL be perpendicular; Let DE, CF and BASILIUS be each of them equal too AM; Let the Planes KIXH be parallel to the Plane ABDE the Base of the prism ABDELM; find the Lines ML and MG by the 11 Proposition, the Rectangle IKM, is equal to the square of KZ, but the Rectangle IKM is equal to the Rectangle IKHX, because AB is equal too AM, therefore HK is equal to KM; Therefore the Rectangle IKHX, is equal to the square KZ; Therefore the prism ABDELM, is equal to all the squares in the conoide AMZ.  

In Numbers thus.  
Let GM the Transverse Diameter be 18; MA, 12; A, 10; ML, 6; AB equal to AM multiplied by A, 16; the Product is 72 equal to ABCF, which being multiplied by ½ AM, that is, 6, the Product is 432 equal to the prism ABCFLM; FE, 4; multiplied by FC, 12; the Product is 48, equal to FCDE, which being multiplied by ⅓ of FL, that is, 4; the Product is 192, equal to the pyramid FCDEL; this pyramid being added to the prism before found, the whole prism ABDELM, will be 634, equal to all the squares in the conoide AZM.    

PROPOSITION XVIII. To find the solidity of a Conoide parabola.  
LEt AUHP be a Plane, the parabola PO RAMIRES be in that Plane, Let VC and AB each of them be equal to AP; Let the Planes QG, QF, and QE be parallel to the Base of the prism ABCV. By the 9 Proposition the parallelogram HQ is equal to the square of QO, but the Plane HQ is equal to the Rectangle QZGI, because IQ is equal to HP and QZ is equal to QP; Therefore the Plane QG is equal to the square of QO, the like in the rest; Therefore the prism AB CUHP is equal to all the squares in the conoide RPA.  
 

In Numbers thus.  
Let AP, be 10; AV, 8; Therefore AC will be 80; which being multiplied by ½ AP, 10; that is, by 5, the Product will be 400, equal to the prism ABCUHP; 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 itself.    

PROPOSITION XIX. To find the Area of a segment of a Circle or an Ellepsis.  
LEt AGE be a semicircle, the area HGF the segment required; Let there be given AC, 13; HR, 12; CR, 5; then, as 14▪ 11, so is the square of 13, that is, 169, to the area of ¼ of the Circle, 132 11/14, the area AGC. In the Triangle HRC, there i●… given HR, 12; CR, 5; and the Right Angle CRH; therefore there is given the Angle CRH; thus, 5: 12:: 100000: 240000, the Tangent of 67 Degrees 23 Minutes, again, as 90 Degrees, that is the Angle ACG, is to 67 Degrees 23 Minutes; so is the area ACG, 132 11/14; to the area of the Sector HC G, 98:: 88; the Triangle HCR, 30; being taken from the Secto●…, leaves the Segment HRG, 68:: 88; by the latter part of the 10 Proposition it will be as AC: BC:: IO: ZO, Therefore as AC: BC, so the segment of the Circle HRG, to the segment of the Ellipsis ZRG; Let BC be 10, then as 13: 10:: 68:: 88: 52:: 98 the segment of the Ellipsis.   

PROPOSITION XX. To find the solidity of a Circular or Elliptic Spindle.  
If from the Solid whose Base is KHG●… CBA and Altitude APK, there be take a Solid, whose Base is KGCA and Altitud●… AKP there will remain a Solid whose Ba●… and Altitude is equal to the segment GO●… or APK; equal to all the squares in the ¼ o●… the Spindle.  
By the 15 Proposition find all the squares in ¼ of a Sphere whose Diameter is AK.  
By the 19 Proposition find the segment A PR. which multiplied by KG makes a parallelepipedon, which being taken from ¼ of the forementioned Sphere, leaves all the squares in ¼ of the Spindle, this ¼ being multiplied by 4 gives all the squares in the Spindle; Then, as 14: 11, so are all the squares in the Spindle, to all the Circles, that is the Spindle itself.  
Further, as the whole so the parts are calculated; find the segment of a Sphere by the 15 Proposition, and the segment of the Circle KRP by the 19 Proposition, this segment KRP multiplied by KG, which taken from the corresponding part of the Sphere, leaves the corresponding part of the Spindle GOI.  
Let BOHP be supposed to be an Ellipsis, HB and PZO its Diameters; the Area K PA may be found by the last part of the 19th. Proposition: find ¼ of a Spheroide that hath KA for one of its Diameters, if the square of ZP be lessened by the square of ZR, there will remain a square whose Root shall be half the other Diameter; find all the squares in ¼ of such a Spheroide by the 16. Propos. then proceed as in the Circle, as well for the parts as the whole.   

PROPOSITION XXI. To find the solidity of the second Sections in: Sphere or Spheroide.  
LEt DABRCMNLFQ be ⅓ of ●… Sphere cut by two Planes, viz. the Plan●… RXDGZQ and MXAPZT being a●… right Angles, their Intersection XZ. MA Equal to MZ; RD equal to RZ; CN equal to CB the semidiameter of the Sphere MX and XR being known, the rest may be found thus.  
 

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. john Leek, in page, 116, of Mr. gibson's Syntaxis Mathematica, Which is this. If the excess of the three Angles above 180 Degrees be multiplied by half the Diameter of the Sphere, the Superficies of any Spherical Triangle is thereby produced.  
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 showeth the Areas of the two Figures, viz. GZID and OZPA, which added to the Area DIZOA gives the Superficie of the mixed 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 Pyramids, whose Bases lie in the superficies of the Sphere, and their vertices at the Centre, 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 multiplied by one third of the Semidiameter CA, CF or CZ produceth the Spherical pyramid ZFAC.  
This pyramid may be divided into three solids, viz. a pyramid whose Base is the segment ZFX and Altitude ZB, the pyramid whose Base is the Area ZAX and Altitude XE; and the solid AZXF the solid required. But the pyramids are given, because 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 pyramids ZXFC and ZXAC taken from the pyramid ZFAC leaves the solid FXAZ.   

10.  
Let ♄ ♃ ♂ ☉ ♀ ☿ ☽ ABDELLA ♄ 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.  
♀ C: XC:: ♂ H: OH, by the 10. Prop.  
♀ C: XC:: ♃ E: LE,  
♂ H: OH:: ♃ E: LE, 11. 5.  
♂ HOH: OH:: ♃ EEL: LE, 17. 5.  
♂ O: OH:: ♃ L: LE, 19 5.  
☉ F: ♃ E:: ☉ F: LE, 10th. Prop.  
☉ F- ♃ E: ♃ E:: ☉ I'LL: 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,  
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.    

PROPOSITION XXII.  
IF four Numbers be in proportion, that is, as the first is to the second; so is the third, to the fourth. It will always be, as the first, is to a Geometrical mean proportion between the first and second; so is the third, to a Geometrical mean proportion between the third and fourth.  
Let it be,  
A: B:: C: D, multiply the two first terms by A, and it will be, as AA: AB:: C: D, 17. 7. If the two last terms be multiplied by C, it will be AA: AB:: CC: CD, then, as A: √ AB:: C: √ CD, 22. 6.  

In Numbers thus.  
4: 9:: 16: 36.  
16: 36:: 16: 36.  
16: 36:: 256: 576.  
4: 6:: 16: 24.    

PROPOSITION. XIII. To find the Relation of one Hyperbola to another.  
LEt FLBC be a semihyperbola, its Transverse Diameter FH, its intercepted Axis FC. Let GMBC be another semihyperbola, its Transverse Diameter KG, its intercepted Axis GC.  
Let it be made, as,  

1.  
KG: GC:: HF: FC, KGGC: GC:: HFFC: FC, 18. 5. KC: GC:: HC: FC KC: HC:: GC: FC, 16. 5. KC: HC:: GC: FC, 17. 7.   2 2  
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: EEL, 11 Pro.  
KCG: KFG:: CBC: FMF = EEL, 11. 5. 11. Prop.  
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: EEL, 11. Prop.  
HCF: HEF:: CAC: EVEN, 11. Prop.  
CBC: CAC:: EEL: EVEN, 11. 5.  
CB: CA:: EL: EN 22. 6.  
CB: CA:: area CBF: area CAF, 12. 5.   

3.  
Let CUQ be ¼ of an Ellipsis, its Transverse Diameter FV, it's semiconjugate CQ, Let CGQ be ¼ of another Ellipsis its Transverse Diameter EGLANTINE, its conjugate CQ.  
Let it be made, as,  
FC: EC:: CV: CG, FC: EC:: CV: CG, 17. 7.   2 2  
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.  
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 EGLANTINE 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 CUB be half of another parabola, its Diameter CV, its Ordinate CB.  
Let it be,  
CV: CG:: CV: CG, CV: CG:: CV/ 2: CG/ 2,  
CV: CG:: GV: GD,  
CG: GD:: CBC: DND, 9 Prop.  
CV: GV:: CBC: GTG = DND, 9 Prop.  
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 CUB: 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:: ODD: 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.  
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 CUQ; 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, OF, 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 EGLANTINE.  
Let CQ = A. DH = B. CD = D, GC = Z. Therefore ZD = ED. ZD = DG.  
ZZ: ZD in ZD:: 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 hyperbolic 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 fourth's of the said parallelogram.    

PROPOSITION XXIV. Of all manner of cylindric hoofs.  

1.  
LEt MACFTP be half a parabolic 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 HUKB 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. AYE 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 GRACCUS and ME, that is from L, to the intersection of the lines HB and A, that is to I, let there be a line drawn, as LI; 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 cylindric hoof; If half or a quarter of an ellipsis, than an elliptic cylindric hoof. If the base be half, or a whole parabola, than a parabolic cylindric hoof. If half, or a whole hyperbola, than a hyperbolic cylindric hoof; the like for any other.   

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 BOY 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 BOY and BY, 1. 6.  
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:: BOY: BY. 1. 6.  
But BOY: BY:: 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 BOY: BY:: 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 cylindric hoofs.  
The Triangles MCD and KBO a●… alike, Therefore as, as KB: BOY:: MC CD, 4. 6. and KB: BOY/ 2:: MC: CD/ 2, by the converse of 17. 7. then as, KB, to a Geometrical mean proportion between KB and ha●… BOY, 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 BOY, 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 BOY, that is, equal to a square whose side is a Geometrical mean proportion between KB and half BOY; 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 parabolic Spindle whose semiaxis is AM and semidiameter is MX; but all the squares in a parabolic 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 the prism KBRMQO; Lastly, take away the parabolic Semicylinder RBCNOQ there will remain the little hoof QOND.   

8.  
If ACM be a semihyperbola its Transver●… diameter Crum, the hyperbolic 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 EWE their Transverse diameters by the First of the 23d. Proposition. If between Mrum and ha●…rumS 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 BOY, 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 hyperbolic Spindle AZXM taking AM for its axis, will be equal to the hyperbolic 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 BOY, 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 cylindric 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 ascribed in that cylinder.  
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, than 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 Elliptic 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 BOY, 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 Elliptic cylindric 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.    

A Table of Squares and Cubes.  
Roots. Squares. Cubes. 1 1 1 2 4 8 3 9 27 4 16 64 5 25 125 6 36 216 7 49 343 8 64 512 9 81 729 10 100 1000 11 121 1331 12 144 1728 13 169 2197 14 196 2744 15 225 3375 16 256 409●… 17 289 4913 18 324 5832 19 361 6859 20 400 8000 21 441 9261 22 484 10648 23 529 12167 24 576 13824 25 625 15625 26 67●… 17576 27 729 19683 28 784 21952 29 841 24389 30 900 27000 31 961 29791 32 1024 32768 33 1089 35937 34 1156 39304 35 1225 42875 36 1296 46656 37 1369 50653 38 1444 54872 39 1521 59319 40 1600 64000 41 1681 68921 42 1764 74088 43 1849 79507 44 1936 85184 45 2025 91125 46 2116 97336 47 3209 103823 48 2304 110593 49 2401 117649 50 2500 125000 51 2601 132651 52 2704 140608 53 2809 148877 54 2916 157464 55 3025 166375 56 3136 175616 57 3249 185193 58 3364 195112 59 3481 205379 60 3600 216000 61 3721 226981 62 3844 238328 63 3969 250047 64 4096 262144 65 4225 274625 66 4356 287496 67 4489 300763 68 4624 314432 69 4761 328509 70 4900 343000 71 5041 357911 72 5184 373248 73 5329 389017 74 5476 405224 75 5625 421875 76 5776 438976 77 5929 456533 78 6084 474552 79 6241 493039 80 6400 512000 81 6561 531441 82 6724 551368 83 6889 571787 84 7056 592704 85 7225 614125 86 7396 636056 87 7569 658503 88 7744 68147●… 89 7921 704969 90 8100 729000 91 8281 753571 92 8464 778688 93 8649 804357 94 8836 830584 95 9025 857375 96 9216 884736 97 9409 912673 98 9604 941192 99 9801 970299 100 10000 1000000 101 10201 1030301 102 10404 1061208 103 10609 1092727 104 10816 1124864 105 11025 1157625 106 11236 1191016 107 11449 1225043 108 11664 1259712 109 11881 1295029 110 12100 1331000 111 12321 1367631 112 12544 1404928 113 12769 1442897 114 12996 1481544 115 13225 1520875 116 13456 1560896 117 13685 1601613 118 13924 1643032 119 14161 1685159 120 14400 1728000 121 14641 1771561 122 14884 1815848 123 15129 1860867 124 15376 1906624 125 15625 1953125 126 15876 2000376 127 16129 2048383 128 16384 2097152 129 16641 2146689 130 16900 2197000 131 17161 2248091 132 17424 2299968 133 17689 2352637 134 17956 2406104 135 18225 2460375 136 18496 2515456 137 18769 2571353 138 19044 2628072 139 19321 2685619 140 19600 2744000 141 19881 2803221 142 20164 2863288 143 20449 2924207 144 20736 2985984 145 21025 3048625 146 21316 3112136 147 21609 3176523 148 21904 3241792 149 22201 3307949 150 22500 3375000 151 22801 3442951 152 23104 3511808 153 23409 3581577 154 23716 3652264 155 24025 3723875 156 24336 3796416 157 24649 3869893 158 24964 3944312 159 25281 4019679 160 25600 4096000 161 25921 4173241 161 26244 4251528 163 26569 4330747 164 26896 4410944 165 27225 4492125 166 27556 4574296 167 27889 4657463 168 28224 4741632 169 28561 4826809 170 28900 4913000 171 29241 5000211 172 29584 5088448 173 29929 5177717 174 30276 5268024 175 30625 5369375 176 30976 5451776 177 31329 5545233 178 31684 5639752 179 32041 5735339 180 32400 5832000 181 32761 5929741 182 33124 6028568 183 33489 6128487 184 33856 6229504 185 34225 6331625 186 34596 6434856 187 34969 6539203 188 35344 6644672 189 35721 6751269 190 36100 6859000 191 36481 6967871 192 36864 7077888 193 37249 7189057 194 37636 7301384 195 38025 7417875 196 38416 7529536 197 38809 7645373 198 39204 7762392 199 39601 7880599 200 40000 8000000 201 40401 8120601 202 40804 8242408 203 41209 8365427 204 41616 8489664 205 42025 8615125 206 42436 8741816 207 42849 8869743 208 43264 8998912 209 43681 9129329 210 44100 9261000 211 44521 9393931 212 44944 9528128 213 45369 9663597 214 45796 9800344 215 46225 9938375 216 46656 10077696 217 47089 10218313 218 47524 10360232 219 47961 10503459 220 48400 10648080 221 48841 10793861 222 49284 10941048 223 49729 11089567 224 50176 11239424 225 50625 11390625 226 51076 11543176 227 51529 11697083 228 51984 11852352 229 52441 12008989 230 52900 12167000 231 53361 12326391 232 53824 12487168 233 54289 12649337 234 54756 12812904 235 55225 12977875 236 55696 13144256 237 56169 13312053 238 56644 13481272 239 57121 13651919 240 57600 13824000   

PROPOSITION XXV.  
LEt AYDG be a Sphere, GY its axis, C the Centre of the Sphere, the Lines AD and BY be at right Angles, and at right Angles with the axis GY.  
Let this Sphere be cut by a Plane through the axis GY, and the Base AD; the Section makes the Circle AYDG.  
Let the Sphere be cut by another Plane, viz. as by the Plane CBHSI; HC its diameter, BY its diameter of the Base. In the inclining Solid whose diameter of the Base is BY, and diameter of the Section CH and the altitude CZ or HN, that is the inclining Solid BHSIC is equal to the Zone AHFD, less the Cone HWFQC, or the inclining Solid BHSIC is equal to the Excavatus part of the Sphere ACHFCD. Let the plane KSE be parallel to the plane AID; RS the common section of the planes KSE and CHI: OE, OS and OK are equal; the square of OS is equal to the squares of RO and RS. 47. 1. Therefore the Area of the Circle who semididiameter is OS, is equal to the Areas of the two Circles whose semidiameters are OR and RS, that is, the Area of the semicircle OKSE, less the Area of the semicircle OR ☉ ♃ is equal to the armille ♃ ☉ RKSE, that is, the Area of a semicircle whose semidiameter is RS. Wherever the plane KSE be drawn parallel between the two parallel planes AID and HWF it will have the same qualification by the 47. 1.  
Whence it is manifest the inclining Solid BHSIC is the difference betwixt the Zone AHFD. and the Cone HWFQC. Further this inclining Solid BHSIC is equal to ●… hemispheroide whose axis is HN, the altitude of the section BHSIC, and the semidiameter of the Base of the hemispheroide is CA▪ By the 10th. of the 6th. of Euclid, and by the 3d. of the 23. Proposition.  

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 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 BOY 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 GOE 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 A 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 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 hyperbolic 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 parabolic 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 hyperbolic 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 hyperbolic 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 parabolic 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 parabolic 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 hyperbolic 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 hyperbolic conoide whose Base is equal to the difference betwixt the Bases of the said frustum; and its altitude the altitude of the frustum.  
This latter hyperbolic conoide, is like to that hyperbolic 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 Centre 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 Centre 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.  
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 hyperbolic 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 hyperbolic 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 hyperbolic 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.  
The solidities of like spheroides, are in a triplicate ratio of their corresponding terms.   

15.  
The solidities of parabolic conoides upon the same Base, are as their Altitudes. By the 5th. of the 23d. Prop.  
The solidity of parabolic conoides under the same Altitude, are as their Bases. By the 6th. of the 23d. Prop.  
The solidities of like parabolic conoides, are in the triplicate ratio of their corresponding terms.   

An Example thus.  
Suppose there be two parabolic 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 parabolic conoide of A, is to the parabolic 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 parabolic conoides unlike, suppose A and D. Then,  
A will have that proportion to D, as is composed of the Base and altitude of A, to the Base and altitude of D.  
In parabolic conoides, as A and B.  
If the conoide of A be equal to the conoide of B, than 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 hyperbolic conoides; for there may be infinite hyperbolic conoides, yet having the same Base & altitude; the least not so little as a Cone, no●… the greatest so great as a parabolic 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 Centre 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, the axis of the spheroide will be equal to one of their Diameters; but the area of that Circle, that passeth through the Centre 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.  
Hyperbolic conoides of the same axis may be added too, or substracted from each other, their sum and difference will be hyperbolic 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.  
Parabolic conoides of the same axis, may be added too, or substracted from each other, the sum and difference will be parabolic conoides, using the former rules for their bases and parameters,  
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 hyperbolic conoide be equal one to another; and the axis equal to the axis of a Sphere: this hyperbolic conoide and Sphere being added together, they make a parabolic conoide; its Base and altitude equal to the Base and altitude of the hyperbolic conoide, its parameter the double of the Diameter of the Sphere.  
If there be a hyperbolic 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 parabolic conoide; its Base the same as the hyperbolic 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 parabolic conoide; its Base equal to the Base of the Cone, its parameter equal to the Diameter of the Sphere:   

20.  
Parabolic Conoides, may always be added to, and sometimes substracted from hyperbolic conoides of the same axis; that sum will be a perbolick conoide, and that difference when a difference may be, will be a hyperbolic 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 hyperbolic conoide is greater than the parameter of the parabolic conoide then a difference may be: but if it be lesser, than 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.    

The Application.  
THe first Proposition is to find the solidity of pyramids and Cones, or frustum pyramids and Cones, and may be applicable to the measuring of all solids or Vessels in that form, whether whole or in part, or gradually, that is, foot by foot, or inch by inch.  
The second Proposition may be applied to the measuring of irregular solids, and may be useful for the exact measuring of all sorts of Stone and Timber: also for the exact measuring of all sorts of elliptic, parabolic and hyperbolic irregular solids, or Vessels that are made in that form: for such solids may be cut into parallelepipedons, prismes and pyramids, and then reduced to their own nature, by the proportion of the parallelogram ascribed about those Figures to the Figures themselves, Thus.  
The proportion of parallellograms of the same Base and altitude with the areas of parabolas are as 4 to 3, Therefore,  
As 4, is to 3; so is any such solid to any such parabolic irregular solid.  
By the help of a Table of Squares and Cubes any such solids may be calculated foot by foot, or inch by inch, without any great trouble, as is showed in the third case of the third Proposition; This second and third Propos. are the general use in such kind of solids.  
In the fourth Propos. with its several cases, there is the measuring of frustum pyramids when their Bases are not parallel.  
In the fifth Propos. there is the relation of the Sphere and Spheroide to the cylinders of their bases and altitudes, as well of the parts as the whole.  
In the sixth Propos. there is the measuring of the middle Zone of a Sphere and Spheroide; the middle Zone of a Spheroide, hath been taken generally for the Figure representing a Cask; so, the measuring of one, the other is measured.  
In the twelfth Propos. there is the measuring of a portion of a Sphere, which may be applied to the measuring of the inverted Crown of Brewer's Coppers, or several other uses.  
In the thirteenth Propos. there is the measuring of parabolic conoides, which may be taken for a Brewer's Copper, the inverted crown.  
Sometimes it may be a portion of a Sphere, or Spheroide, but sometimes the portion of a paetbolick conoide; other times the portion of a hyperbolic conoide, they ought to be taken as discretion seems convenient.  
In the fourteenth Propos. there is the measuring of a hyperbolic conoide, which may be token for a Brewer's Copper.  
In Propos. 15, 16, 17, and 18. there is the measuring of a sphere, spheroide, parabolic conoide and hyperbolic conoide, as well the whole as their parts.  
If the parabolic or hyperbolic conoides be taken for Prewers' Coppers, with the help of the Tables of Squares and Cubes, they may easily be calculated foot by foot, or inch by inch according to the third Prop.  
In the twentyeth Prop. there is the masuring of Circular and Elliptic splindles.  
The middle Zone may be taken for a Cask.  
In the twenty first Propos. there is the measuring of the second Section in a sphere and spheroide.  
The use may be to measure the middle Zone of a spheroide, cut by a plane parallel to the axis; that is, when the superfice of the liquor cuts the heads of the Cask.  
In the twenty fourth Propos. there is the meesuring of right cylindric hoofs, viz. Circular, Elliptic, parabolic and hyperbolic; and may be used for the measuring of Brewers leaning Vessels.  
If a Brewer's Copper be taken to be of that Figure that parabolic or hyperbolic conoides are, and they stand leaning, the measuring of them is almost the same as though they did not lean.  
Here I ought to have showed the making, and use of an Instrument for taking the leaning of such Vessels; But my business calls me off; However, they may be had of Mr. john Marks Instrument maker, living at the Sign of the Ball near Somerset House in the Strand, who was formerly Servant to that incomparable Instrument maker Mr. Henry Sutton.  
Here note, the Table of Squares and Cubes is very ready and useful in finding the portions of a sphere, spheroide, parabolic and hyperbolic conoides.  

To find two such numbers, that their Product being added to the sum of their squares, the sum shall be a square, and its Root commensurable.  
LEt one of the numbers be A, and the other Z. The product may be AZ; the sum of their squares ZZ + AA; the sum of their squares and product may be ZZ + ZA + AA, equal to a square whose side is, Z + 2 that is the square ZZ— 4Z + 4, therefore ZZ + ZA + AA = ZZ— 4Z + 4, that is, ZA + AA =— 4Z + 4; Let A be a unit, then 5Z = 3, that is, Z = ●…, that is, Z is 3, and A, 5. These two numbers make good the question, for 3 in 5, is 15; the square of 3, 9, and the square of 5, is 25; their sum is 49, whose Root is 7. Albert Girad observes from this seventh Propos. of the 5th. of Diophantus, That if there be a Triangle made of 3 such numbers, the Angle opposite to the greatest side will be 120 degrees. It may be further observed, that if there be 2 right angled Triangles made of these 3 numbers, the sum of the hypotenuse and base of the one, will be equal to the sum of the hypotenuse and base of the other; and also the area of the one shall be equal to the area of the other.  
Thus,  
7 49 7 49  5 25 3 9   74  58 The sum of their Squares.  24  40 The difference of their Squares.  70  42 Their double Rectangles.  98  89 The sum of their sides.  840  840 Their Areas.  
Here note, the double Rectangles are taken for their Altitudes.  
Here note,  
In Progressions from a Unit, the sum, of the sum and difference of the greatest number in that progression, and any one number betwixt the greatest and Unity, is equal to the sum, of the sum and difference of that same greatest number; and any other number betwixt Unity and that greatest.  
Further note,  
This seventh Proposition is a Lemma to the eighth, to find three Triangles of equal areas; therefore the areas are equal, and the hypotenuse and base of the one, is equal to the hypotenuse and Base of the other.  
A general Theorem for the finding of two such numbers; Take the square of any number, from which take a Unit; take the double of that number, to which add a Unit; that sum and difference will be the two numbers required. Thus,  
The number taken is 3, its square 9, less a unit is 8; the double of 3, is 6 more, a unit is 7; these two numbers makes good the question.  
For 56 + 64 + 49 = 169, whose root is 13.  
Two right angled Triangles being made of these three numbers, viz. 7, 8, 13. according to the former method, will have that same qualification.  
This Proposition was publicly proposed i●… PARIS in the year 1633, which Renatos de Cartes resolved, and Francis Schooten published in Sect. 12. Miscel. encumbered with a squar●… adfected equation, with surds.    
FINIS.   

Errata.  
Page 17, line 3, for Ecliptic read Elliptic. p. 25, for Excavetu●… r. Excavatus. p. 29, l. 12, for ½, r. ⅙. In the same line, for B●… BFE; r. BCBFD p. 30, l. 22, for ZB, r. ZC. p. 32. l. 19▪ for NR, r. NP. p. 39, l. 13, put; after Line. p. 46, l. 21 for 634, r. 624. p. 53, l. 3, for APR, r. APK. p. 48, l. 25▪ for ZB, r. XB. p. 60, l. ult. for R ☉ ♀ ♃, r. R ☉ ♂ ♃. p. 61, l. 6, for V, r. VI p. 64, l. 4, for GG, r. GC. p. 66▪ l. 18, for CDQ, r. CGQ. p. 70, p. 71 l. 5, put, after D▪ p. 74, l. 4, for GFPM; r. CFPM. p. 80, l. 21, put: before all. p. 82, l. 25, after KABO put,   

Imprimatur.  
Tho. Tomkins R. Rmo in Christo Patri ae Domino Dno. GILBERTO Divinâ Providantiâ Archiep. Cant. à Sac. Dom.  
Ex Ad. Lamb. Aug. 25. 1668.