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UNIVERSITY OF MICHIGAN
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1
STEREOMETRICAL
Propoſitions: :
VARIOUSLY APPLICABLE ;
But
PAR TICULARLY
IN TENDED
For
GAGEING.
$
Ву
ROBERT ANDERSON.
LONDON,
Printed by William Godbid for
Joſhuah Conniers at the Sign of the
Black Raven in Duck-lane, 1668.
History of acconse
bookshop
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ܕܟ
To the
R E A D E R
years ſince
11 - Ĉ-44
Bout ſeven
I refolved theſe fol-
lowing propoſitions,
with others of the
like nature ; but did
not intend ever to
have publiſhed them : But being over-
perſwaded by ſome of my Friends,
I have here made ſo many of them
publick, as I thought would be con-
venient to what here is intended.
A3
3
Here
To the Reader.
Here I ought to acknowledge that
great Reſpect which I owe to my
Worthy Friend Mr. John Collins, one
of the Royal Society; which pro-
pounded to me about 6 years ſince,
that propoſition of the ſecond Sedi-
ons of the Spbere 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 Mathe-
maticks are very inuch oblieged 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
Mathematicks, and Mathematical
Men.
For
Totbe Reader.
Foraſmuch as throughout this little
Tract, here is inuch uſe made of Pa-
rallelepipedons, Priſmes and Pyramides:
I think fit, a little to infilt thereon ;
chiefly for the young Learners fakes,
for which this is only intended.
Y
In the Diagram,
Let there be a Solid, as POHG
V FIR; the plane P OHG parallel
equal and alike to the plane RIF V;
Further, POIR is parallel equal
and like to the plane GHFV : Yet
further, the plane OHFI is parallel
equal and like the plane PÅ VG:
ſuch a Solid is called a Parallelepipedon.
Let there be a Solid as GHD EV
F, the plane GEV is parallel equal
and like che plane HDF; Further,
the planes HGFV and DEV F are
alike and the line FV cominon to
boch : ſuch a Solid is called a Prifmen-
Ler there be a Solid as HBCDF,
the planes B CF, CDF, DHF and
HBF
A 4
To ibe Redder.
1
1
HB F all meet at the point F, ſuch a
Solid is called a Pyramide. :
Further,
Pyramides, Priſmes and Parallelepi-
pedons upon the lame Bafe and Alci-
tude, are as, one, one and a half, and
three, that is, if the Pyramide be 49
the priſme will be 6, and the parallele-
pipedon 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 multiplyed by 9
the Alcitide, the product is 432; the
parallelepipedon POHGVFIR, if
the Area 48. be multiplyed by half 9
that is 4 the product will be 216,
equal to a priſone of the Baſe POHG
and altitude HF.
If the Area POHG, 48, be mul-
tiplyed by one third of 9, that is by 3,
the product will be 144 equal to a py-
ramide whoſe Baſe is POHG and
altitude HF
TO
1
To tbe Reader
To find the ſolidity of the prifme HD EV F.
Let GE, voli GHS ; HF9;
multiply 10 by, and the produ& will
be 8o the Ares GHDE, this 80 be-
ing multiplyed by half the altitude
HF, 9; that is, 4 i, the product will
be 360, the priſme GHD EVF.
To find the ſolidity of the pyramide HBCDF,
Let HD, 10; HB, 4; multiply
10 by 4 the product will be 40, the
Area HBCD, this Area multiplyed
by 3, that is one third of the Altitude
HF, the product will be 120, the py-
ramide HBCD F.
ܘܰܚܬܳ
!
To
To the Reader.
RE
1
P
G
E E
of
H
D
A
B
С
To find the ſolidity of a Solid, tbat hath
one of its Baſes an Ellipfis and the
other Baſe a Circle, theſe two Baſes
are ſuppoſed to be parallel.
In ſuch a Solid, the ſides being conti-
nued will never meet at a point as in
circular and Eliptick cones, and there-
fore the general Rule will not reſolve
theſe kind of Solids : therefore icon-
ſider it thus ;
Let
To tbe Reader.
Let P E be equal to the Tranſverſe
diameter in an Ellipſis; P A its Con-
jugate diameter. Let R V and RI be
the diameters of a Circle. Firſt, Let
there be made a Rectangled figure of
the Tranſverſe and Conjugate diame-
ters of the Ellipſis, as PACE. Let
there be a ſquare made of the diame-
ter RV, as RIFV. Let thealcitude
of the Solid be HF, find the ſolidity
of the Solid PACEVFIR, thus, Let
it be cut into as many parallelepipedons,
priſmes, & pyramides as are neceſſary:
then find the folidity of thoſe paralle-
lepipedons , priſmes and pyramides, as
before is taught, thoſe parallelepipe-
dons, priſmes & pyramides being added
together will be equal to the whole
Solid, thus, the parallepipedon POHG
VFIR, more the priſme GEDHFV,
more the pyramide DCBHF, more the
priſme HBAOIF;
theſe
4
Solids be-
ing added together make the whole
Solid ACEPRIFV. For the whole
is
To tbe Reader.
is equal to all its parts taken together.
Axiom. 19. 1. of Euclid.
The Areas of Circles, are in pro-
portion one to another, as the Squares
of their. Diameters 2. 12, of Euclid.
The Areas of Elipſis are in propor-
tion one to another, as the Rectangled
figures of their Tranſverſe and Con-
jugate Diameters. Archimedes the 7.
Prop. of Conoides and Spheroides.
The Area of a Circle, is to the Area
of an Ellipſis; as the Square of the Di-
meter of that Circle, is to the Rectan-
gled figure of the Tranſverſe and Con-
jugate Diameters of that Ellipſis, the
6th of Archimedes Conoides and Sphe-
roides.
As 14, is to ; ſo is the ſquare of
the Diameter of any Circle to the
Area of that Circle. prop. the 2. of
Archimedes Circul. Dimenſ.
The Solid PACEVFIR may be
fuppoſed to have the nature of infinice
Rectangled figures, parallel one to
another,
To tbe Reader.
another, and parallel to the parallel
planes A CEP and IFVR, and from
che parallelogram ACEP to the ſquare
RIFV, every parallelogram coming
nearer and nearer a ſquare, even until
they come to the ſquare RIF V it
ſelf
, ſo in Elliptick Solids adſcribed in
ſuch plane Solids; the Areas of Ellip-
ſes comes nearer and nearer to the
Area of the Circle whoſe Diameter is
RV. Therefore, as 14, is to it, ſo
are ſuch plane ſolids to ſuch elliptick
Solids, adſcribed in thoſe plane ſolids.
An Example in Numbers.
Let R V or RI be equal to 3; PE, 6;
PA, 4; HF, 9; Therefore GE, 3;
OA, 1; Therefore POHG, 9; equal
RIFV, which being multiplyed by 9,
the altitude HF; makes 81,equal to the
parallelepipedon POHGVFIR, GE, 3;
being mulciplyed by GH, 3; the pro-
duct is GHDE, 9; which being mul-
tiplyed by half the altitude HF, that
ز
is,
To the Keader.
is, 4 : the produ& is 40; equal to the
priſme GHDEV F. HD, 3; H B, 1;
therefore HBCD is equal to 3;
which being multiplyed by one third
of the altitude HF, that is, 3 ; the
pyramide HBCDF will be equal to 9;
BA, 3; A0, 1; therefore ABHO
will be equal co 3, which bring mul--
tiplyed by half hf the alcicude, that
is, by 4 the product is 13s, cqual to
the priſme O ABHFI. Now theſe
compoſed Numbers being added toge-
ther make 144 equal to the whole
fruftum pyramide PACEV FIR. Then
as 14:11:: 144: 113; the irregular
Elliptick ſolid, whoſe Tranſverſe PE,
and conjugate Pa diameters at the
Eliptick Baſe are 6 and 4, and the
diameters of the Circular Baſe is
4
RV 3.
Here note,
.
Although Archimedes gives the pro-
portion of the ſquare of the diameter
of any Circle, to the area of that
Circle;
To the Reader.
Circle ; ; as 14, is to IT.
And alſo ;
that the diameter of any Circle is to
the circumference of that Circle, as 7,
is to 22; Yer you are not to under- .
derſtand that theſe proportions are in
their juft values; But that they are the
leaſt numbers that will beſt agree to
that approach
If thoſe numbers bè thought not to
agree near enough to truth, inſtead of
thoſe, there may be uſed; for, 14 and
452
and 355; for 7 and 22,
Il, take
take 113
and 355.
Further note.
In the 24th. propoficion there is
ſomewhat ſaid of Cylindrick hoofs.
Thoſe cylinders ought to be upright
and not inclining cylinders.
Yet further note:
By reaſon of much buſineſs, I could
pot 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,
other-
ned
Tätbe Reader
3
7
otherwiſe there may riſe a miſappre-
henfion of what there is intended.
Laſtly note,
Almoſt at the beginning of the firſt
propofition you may find it printed,
Thus, CN:NZ::EH:HZ, 4:6.
the meaning is thus, as ÇN is to N Z;
fo is EH, to Z, by the fourth prop.
of the ſixth of Enclid; The like in the
reft.
R.A.
(un)
+
#
1
PROPOSITION I
1
To find the ſolidity of a Fruftum: Pyramide,
whoſe baſes are parallel and a like.
Et A BCNHEDF be a
ſiun fyramide, whoſe foli-
dity iş required; Let the Lines
AFBD, CE, and NH be
continued till they meet at 2;
Let Go, be the height of the frufium pra-
mide, OZ the heigật of the Continuation ;
Let FDEH befệt at the Angle N, as NR
QK.: Then CN NZ::EH HZ, 4.6. "
And-CN:E :::NL:HZ, 1;6.3. CN-EH:
EH:NZ-HZ:HZ, 17.5. Thar is, CK:
KN;:14 H:HZ, becauſe the planes HFDE,
and NAD Ç are; parallel, they cut the Line
Ż G, in G and 0, in the fame proportion as
they cut NZ, in Nand Z, by 17. 11. Thus,
NÁ:HZ::GO:OZ Therefore CK:KN::
60:02, 11.5. O Z being found, which
B
added
1
(2)
added to GO makes the whole Altitude GZ;
by which find the whole pyramide NABCZ,
which done,
find the pyra.
H
E
1
mide H FD
E Z, by the
7. 12. then N
ABCZ leſs
HFDEZ,the
Remainder is
the frustum
HFDECB
AN.
2
may be thus;
Firſt toprove
that A K is a
c meanpropor-
tion between
AC and F
E; then C
GON: NKC
A: AK, 1.6.
AN:NR::
NM:NR.
I. 6. CN:
NK:: AN:
NR,7. 5.
CA: AK::
K
11
26
R
q
G
4
TIL
B
NM:NQ, 11.50
Further,
i
(3)
Further, if GO the Altitude of the fruſtım
ÉDEHĆ BAN, be multiplied by AC
more AK more KR, that Product will be the
triple of the ſaid frustum; for the parallelepi-
pedon made of G Z the altitude of the pyra-
mide in the Baſe N ABC is triple of the pyra-
mide N ABCZ, by the ſame reaſon the pa-
rallclepipeden made of G O in A C, together
with the parallelepipedon made by Zo in
AC is triple of the ſaid pyramide by 7. 12.
if the parallelepipedon made of OZ in FE,
that is, in R K, that is, the triple of the pyra-
mide H FDEZ be taken from the parallele-
pipedon N ABCZ, there remains that which
is made of GO in À C, together with thoſe
that are made of OZ in A Q and the fame
OZ in MC, the triple of the fruſtum ABC
NHED F, by 5. 5. for, GK:KN::GO:
O2, CK KN::AR:R N, Therefore GO:
OZ::AR:R N, and CK:KN::CM:MN,
4.6. AR:RN::AQ:QN, therefore GO:
OZ::CM:MN, 11.5. GO:O2 :: AQ:
QN, therefore GO in MN=OZ in CM,
GO in QN-OZin All, add GO in AC
to both, that is, GO in MN, mure GO in
QN, more GO in A C, is equal to OZ in
CM, more OZ in AR, more GO in AC;
but OZ in CM, more OZ in AQ, morc
GO in AC are triple of the ſaid fruftum ;
therefore GO in MN, more GO in QN,
B 2
more
1
(4)
:
more GO in AC are equal to the triple of
the faid fruftum.
An Example in Numbers, and firſt of
the firſt.
}
CK, 34:K N, 26 :: GO, 68:0Z, 52 ;
then 68 more 52=120, GZ: then AC 3600;
in G Z, 120 ; is equal to 432000, a third
part is 144000, ABCNZ, FE, 676 in O Z,
52 is equal to 35152.
A third part is 117171, HFDEZ,144000
leſs 11717. is equal to 132282 į the folidity
of the frujium A CEF.
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
82. 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 ACEZHPO F be a fruftum pyramide
the Baſes Squares and Parallel ; Let HPOF
be projected in the Baſe ACF Z, as Z VRT;
VR and TR be continued to Dand B; then
will T Dbe equal to R A, and R C a Squarez
nuft manifeſt it is, that the parallelepipedon
ZVRTPOFH, more the priſme V ABR
OP, more the pyramide R BCDO, more the
prilne RDETO F is equal to the faid fru-
jinm Pjramida.
For
(5)
For all the parts of any magnitude being
taken together are equal to the whole. Axiom
19. I.
H
I
0
The priſme V
ABROP is e.
qual to the priſme
R DETFO, be-
cauſe of equal ba-
ſes and altitude,
P
but the priſm VA
BROPis halfthe
parallelepipedon,
made of the baſe Z
E
V ABR, and the
altitude R O, by
28. 11. therefore y
R
D
the parallelepipe-
don inade of the
baſe Z ABT that
is the Rectangle
made of HF and
· B
C
tude 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 e-
qual to BZ, a third part of RC is 385}, which
sadded
.
ZA, and the alti. A
B3
(6)
added to 1560, is equal to 1945 i, this laſt
compoſed number multiplyed by 68 equal to
R O makes 132282; the ſolidity of the ſaid
frujium; then as 14 to it, or more near the
truth, as 452 to 355, ſo is 132282 to 103894
19* the fruftum cone adſcribed within the fru.
Atun pyramide.
PROPOSITION II.
L
To find the ſolidity of a fruſtum pyramide, whoſe
Bajes are parallel
, but not alike; that is,
azken their correſponding ſides are
not proportional.
ET ADGOVZTS be the pyramide
propoſed, here AG is a ſquare, sZ a
parallelogram; Let ST Z V be projected in
the baſe, as, QPKL; continue the ſides, as
QL, to B and I; PK, to Cand H; LK, to
Mand F; P Roto E and R ; the Figure be-
ing completed, the whole fruſtum will be
compoſed of theſe parts, namely 4 pyramides,
as CDEPT, KFGHZ, LIOMV, RO
B AS, more 4 priſmes, as BCPQTS, QR
MLVS, LKHIVZ, KFEPTZ, more
the parallelepipedon PQLKZV ST; a fur-
ther demonſtration is needleſs.
An
(7)
S
GO
34
TEK
N
A
RI
M
'B
Ga
L
34
P
60
K
235
12
DF
F F
An Example.
CD equal to 25, D E equal to 12, the
like for the other Angles, CE equal to 300,
B 4
which
.
(8)
i
which being multiplyed by*5, that is of
TP, makes the pyramide CDEPT equal to
1500, which multiplyed by 4 (becauſe there
are 4 of them) is boco; C, 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; P K, 60; Ø F, 1500; which be-
ing multiplyed by 7 makes 11250, equal to
the priſme PEFKZT, this being doubled
(becauſe there are two oppoſites equal) is
22500; P.Q, 34; Q1,60; LP, 2040; this
laſt number being multiplyed by 15 the alti-
tude of the pyramide TP, produceth 30500;
the parallelepipedon QPKLVZTS; theſe
4 compoſed numbers being added together,
makes the whole fruſtum pyramide.
Thus.
The 4 pyramides are
-6000
The 2 lefſer priſm's are
6120.
The 2 greater priſmes are
22500.
The parallelepipedon is
-306vo.
The whole fruſtum pyramide is 65220.
A Second way thus.
Let F C be the greater baſe, R Q the leſſer,
GO the height; make it as RH:HQ::FA:
AB; draw B E parallel to FA, then will
the
(9)
the plane F B be like the plane R Q; Let the
Figure be completed, then will the fruſtum
ACDFRHOZ be equal to the fruſtum
ABEFRH QZ more the priſme, BCVT
QZ, more the pyramide TV DEZ : herc
note that T B is made equal to QZ.
1
R
2.06, 64
H
D
34
Go
G
V
126
A 60
B16C
The Example in Numbers.
The fruſtum A BEFRHQZ, is equal to
1245015, found by the firſt Propoſition, BC,
16; CV, 26; BV, 416; being multiplyed
by the altitude, that is 32, makes 13312,
equal
(10)
equal to the priſme BCVTQZ; TV, 263,
VD, 34; TD,544; being multiplyed by
of G O that is 21 }; makes 11605 | equal to
the pyramicie TV DEZ; theſe three com-
poſed numbers being added together, makes
149418; equal to the fruftum pyramide ACD
FRH QZ.
3
1
The fruftun ÆBEFHRZQ 124501
The prijme BCVTQZ
13312.
The pyramide TV DEZ 11605
The frustum A6DFRHQŹ 149418
A Third way thus.
Let PC the greater baſe; RF the leſſer ;
Let the upper bale be projected in the lower,
as RIF V be equal to POHG, moſt eaſie it
is to apprehend, that the fruſtum pyramide
PACEV FIR, is equal to the parallelepi<
pedon POHGV FIR, more the priſme OA
BHFI, more the pyramide HBCDF, more
the priſme H DEGVF, for the whole is
equal to all its parts.
In numbers thus.
Let A C, be 56; AP, 38; RI, 26; R V,
30; RP, 40; RI, 26; VR, 30; IV, 780
equal to PH; which being drawn into RP,
40; makes POHGVFIR, 31200: O A,
12; A B, 30; A H, 360 ; being multiplyed
by 20, that is half the height PR, it makes
7200
1
+
(11)
7200 equal to OABHFI. BC, 28;
HB, 12; BD, 336; being multiplyed by
of the altitude
that is 13 j, it R
produceth 4480,
equal to the pyra-
mide HBCDF;
JE
GH, 26; HD,
28; HE, 728;
being multiplyed
by 20,the product
is
14560 qiial.o
the priſme H DE
P
E
GVF; theſe 4
compoſed num-
bers being added
D
together, makes
the ſolidity of the A B C
ſaid fruſtum
The parallelepipedox-POHGF VRI=31200
The priſme
O ABHFI=07200
The pyramide
-H BCDF=04480
The priſme
HDEGVF=14560
G
H
The frustum pyramide--PACEVFIR=57440
1
Then,
As 14 isto 11, ſo is the folidity of any
fruſtum pyramide thụs found, to any Ellip-
tick frustum, adſcribed in that frustum pyra-
mide.
Propofition.
(12)
PROPOSITION. III.
1
To find the ſolidity of a Fruſtum Priſme.
LI
Et CAGDEF be a priſme, whoſe
ſolidity is required, let B C be made
equal to FE, and BH parallel to CD,and the
F
2
HF
80
E
1
1
G
D
H
6/4
4
A32. B 52
C
Figure
(13.)
7
Figure being completed, the frustum priſme
AČDGFĚ may be compoſed of the prime
BCDHFE, more the pyramide BAGHF:
a further demonſtration is needleſs.
An Example in Numbers.
BC, 52; BH, 64; BD, 3328; this laſt .
number being multiplyed by 40, that is half
the height, makes 133120 equal to the priſime
HBCĎEF; A B, 32; BH, 64; RG,
2048 : this laſt being drawn into 26, that
is, a third part of the altitude makes 546131
equal to the pyramide GABHF : this priſme
and pyramide being added together, makes
1877331, equal to the frustum priſme AGD
CEF.
The ſecond Caſe, thus.
Let ZBCDEH be a fruſtum priſme ; Let
AC be equal to H E and AO parallel to Z B,
the
Figure being completed, the priſme AC
DOH E leſs the pyramide A,BZOH is equal
to the priſme BCD ZHE; or thus, the
priſme Z BCDF E more the folidity ZBHF,
that is, half the pyramide ZB AOH, is equal
to the fruſtum propoſed.
Example in Numbers,
A C, 845, A0, 64; AD, 5376; this laſt
drawn into 40, half the height makes 215040;
equal to the priſme O ACDHE; A C, 32;
B2, 64; AZ, 2048 being multiplyed by
26", that is, one third of the height, it makes-
5
(14)
H H
F
¡
E
D
N
-
A B
$4613; equal to the pyramide A B ZOH,
Oh, this piramide taken from the prijine,
leaves 160426, equal to the fruftum priſme
propoſed.
Then as 14 is to Ir, ſo are ſuch priſmes to
Elliptick ſolids adſcribed in thoſe prijmes.
Hence follows a Fourth way for finding the
ſolidity of theſe irregular fruſtums.
Suppoſe AC DFGNIH be the fruſtum
propoſed, becauſe F A is equal to D C, and
AC equal to DF, and GN equal tol H, and
NI
(15)
N
1
1
1
IN
NI to GH; and
becauſe GN is
V
leſs than F A, and
HI than DC,
therefore if AN,
CI, DH, FG
be continned to-
wards Z and V, G
they will meet,
R
H
ſuppoſe at Z
and V, then AC
DFV Z, leſs NI
qiz
HGVZ, there F
D
remains A CDF
GNIH the fru-
ftum propoſed.
It alſo follows,
That ſuch priſmes
have ſuch propor- A
B!
tions one to anb-
ther, ás Squares and Cubes of their correl
ponding ſides, disjoyned, thus.
The pyramide BCDEZ, is to the pyramide
R QIHZ, as the cube of B E to the cube
of RQ; 8, 12. The priſme FABE V Z,
is to the priſme GNQR V Z, 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 pgra-
mides ; is becauſe FE, GR, and v Z are
equal,
(
fo!
I
(16)
equal, and the ratio rileth but from VG,
GN and VF, and FA; whereas in the
pyramide the ratio riſeth from three, that is,
from ZR, R Q, and RH; and from ZE,
ED, and EB.
In Nunzbers thus.
Let V Z be 9, ZR, 3; ZE, 6; RQ, 45
EB, 8; RH, 3 ;; ED, 7; Therefore GN
QRZV is equal to 54, and R QIHZ is
equal to 14 ; then as the cube of 3, that is
27, to the cube of 6, that is 216; fo is the
pyramide R QIHZ, that is 14, to the pyra-
mide. EBCDZ, that is 112... Then 112 leſs
14 is equal to 98, equal to RQIHDCBE.
Again, as the ſquare of 3, that is 9, to the
{quare of 6, that is, 36; fo is the priſme GN
QRZV; that is, 54 to the priſme FABEZ
V, that is,-216. Then 216 leſs, 54 is equal
to 162, equal to GNQREBAF. This laſt
fruftum, more the frustum R QIHDBCE
is equal to the fruftum GNIHDCAF
equal to 260.
Or thus:
As 27: 189:: (that is the cube of Z E leſs
the cube of Z R) 14:98, equal to the fru-
ſtum RQIHDCBE. Again, as 9: 27::
(that is the ſquare of Z E leſs the ſquare of
ZR) 54: 162, equal to the fruftum GNQ
REB AF.
What
(17)
What ever is ſaid of theſe priſmes and pura-
mides, the ſame is to be underſtood in Cones
and Ecliptick priſmes;
For they are in duplicate and triplicate
ratio of their bomologal ſides, further for as
much as that, the two firſt terms in each pro-
portion may be fixt; there may by the help
of a Table of Squares and Cubes, the folidity
of ſuch ſolids ealily be calculated gradually,
that is, inch by inch, or foot by foot, the two
Axed numbers in the pyramide are the Cube of
the continuation, or the ſide R2; and the
Pyramide R QIH, the like in the priſme, and
with the ſquare of the continuation R Z.
Thus .
C4127 32 14 :: 37: 19 o
27:14:: 98:
24:14:: 189:98, theſe three num-
bertshamely, 19., 56, and 98, are the
folidity of the frufium pyramide Á QIHDC
BE, (Ke firſt number is the ſolidity of one
inch, the ſecond number of two inches, tha
third of three inchess for the priſme; as 9:54::
7:42, 9:54:: 16:96,9:54:: 27: 162; thele 3
numbers, viz. 42, 96, and 162 are the ſolidity
of the fruſtum prijme GNQREBAF, taken
inch by inch; therefore 42 more 19 cqua!
to 61, 96 more 50 ; equal 1461;, 162 more
36 equal to 25%, are equal to the folidity of
the wbole frujium GNIHDCAF, taken inch
by inch; the like for any Elliptick fruftum.
C
Propofition.
uit
17
( 18 )
i'
}
1
$
3
Propoſition. IV.
To find the ſolidity of a fruſtum pyramide,
whoſe Baſes are not parallel ; and the initli-
nation is from ſide to ſide.
Et R
LE
T.,
HE
R
TCBAF
Z
be a fru-
ftum pyra.
mide, the
'baſe 'RH
ET, not
X
parallel to
the bafe É
ABC let
RHDZ
bepárallel
to ABC
F, then
.
wholefry.
Stum pyra.
mide RH
ETABC
E
H
D
4
ZI
!
kong
F be com:
A
B poſed of
the fru-
ftum
(19)
fium pyramide RHDZ ARCF more, the
priſine RHDZTE, by a compoſition of
the laſt Propoſ. PD being the altitude of the
fruftum pyramide RHDŽ ABCF: EX the
height of the priſme RHDZTE may be
found, thus. BD:DP::DE:EX; a fur-
ther demonſtration is needleſs.
II.
To find the folidity of a triangular fruſtum
pyramide, whojè baſes are not parallel.
H
T
9
F
E
C
A
3
C 2
Let
( 20 )
Let CABDHF be a fruſtam pyramide,
CAB not parallel to HDF, let QDE be
parallel to CAB; the triangular fruſtum
DEBAC is half a quadrangular fruſtum py-
ramide, found by the 12 Propofion, Ď Z the
height of the pyramide QHFED; There-
fore the fruſtum pyramide QDEB AC more
the pyramide QËFED will be equal to the
fruſtum pyramide FHD ABC.
III.
To find the folidity of the fruſtum pyramide,
ACFED Z.
Here the
baſe AFC is
not parallel to
the baſe ZDE
R but to
the
plane ZOE;
the frustum
py-
ramide ZOE
FCA may be
found by the
2. Propofition,
the part ZOE
G/F
D will be a pya
ramide, its al-
titude may be
found, thus,
As OF:OG::
B
OD: DR,
then
N
E
1
2
(21)
}
RE
then the frufium pyramide ZOEF CA more
the pyramide ZOED will be equal to the
ſolid ACFE OZD. IV.
To find the ſolidity of a fruſtum pyramide, whoſe
baſes are not parallel, and whoſe inclina-
tion is from Angle to Anglo
.
Let CABDFXKQ be a fruſtum pyramide,
the baſe CABD
K
not parallel to
the baſe FXKQ,
X
Z
but parallel to
P'
H
the plane FEH
G; then the fru.
štum pyramide C
A B DHEFG
found by the 2
Propoſ. more the
1yramide EQXG
F, found by the
2 caſe of this
Propos. more the
fruftum pyramide
GEHZXQ,found
by the 2 Propof.
morethepyramide
XKZQ, found by
the 2 caſe of this
Propos. will be e-
qual to the whole
fruſtum pyramide
CABDFXKQ.
A
F
E
V
1
M
(22)
PROPOSITION V.
LE
Et FHAO E ben Semicircle, HO pax
rallel 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.
B
D
1
G
H 1
G
H
2
R
R
h
11
12
GH
E
For the ſquare of CO leſs the ſquare of
RO is equal to the ſquare of R C, the ſquare
of RN (equal to the ſquare of CO) leſs the
square of R O is equal to the ſquare of ON
more the Rectangle twice under RON;
therefore the ſquare of R C (that is the ſquare
of R I) is equal to the ſquare of ON more,
the
.
(23)
1
the double Redangle RON, bý,4: 2. every
one of tlieſe being conſidered one by one; but
being collected (viz.) all the ſquares of RI
is equal to all the ſquares of R N, leſs all the
{quares of RO; or if there be taken the qua-
druple of them, all the ſquares of QI is equal
to all the ſquares of G N 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.
á
Let ROBC be of a Circle, and RSQC
of an Ellipfis complete the Diagram, then
will the ſquare of QC, that is I V leſs the
D
А
}
i
ܢ
1 S 1
be
B
1
64
ſquares
( 24)
ſquare of L V be equal to the ſquare
of SV. For AR:RD :: TV:V L, there
fore the ſquares of them, and AR:R Dv:
OV:VS, therefore the ſquares of them
therefore, as the ſquare of TV, to the ſquare
of V L, ſo the ſquare of O V, to the ſquare
of VS; Therefore, as the ſquare of EV, to
the ſquare of E V leſs the ſquare of TV. So
the ſquare of IV, to the ſquare of IV leſs
the ſquare of LV, but the ſquare of E V leſs
the ſquare of T V is equal to the ſquare of
OV; therefore the ſquare of IV leſs the
ſquare of L V equal to the ſquare of SV.
PROPOSITION VI.
N the Hemiſphere ÍAP, Let FE be pa:
rallel to IP, SZ and GX parallel to
AC; the Figure being completed; the Cy-
linder QG is equal to the Cylinder IE leſs
ilie Cylinder RH; for as the ſquare of AC,
to the ſquare of B C; ſo the Cylinder I E, to
the Cylinder RH; again, as the ſquare of
A C, to the ſquare of AC leſs the ſquare of
BC, (that is the ſquare of BS) ſo is the Cy-
linder IE, to the Cylinder I E leſs the Cylin-
der R H, (that is the Cylinder QG) it follow-
eth that the Cylinder R H is equal to the
exc. Vitus Cylinder I QSFGEPV, for the
cylin-
.
(25)
Cylinder QG is equal to the excavetus Cylinder
IROFHTPE.
· But if the excavetus Cylinder ORQSGV
TH, be added to both; the Cylinder R H will
be equal to the excavętus or hollow cylinden
QIFSGV PE; but the cone OCH is equal
to FISGPE by the firlt part of the Fifth
Propofition; Therefore I OSGVP is.es
qual to ORC'TH, that is, of the difference
betwixt'the circumſcribed ie and inſcribed
QG cylinders'; ſübſtracted from the circum-
feribed cylinder I E is equal to the portion of
the Sphere ISGP; or if şof 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.
:
:
11
A
D
F
B
H
E
'.
IM
19
SP
X
(26)
In Numbers thus.
Let IP bę 20; QV, 16; CB, 103
the ſquare of IP is 400, the ſquare
of Q V 256; the difference of the ſquares
iš 144,, a third part of that number is 48,
which ſubltracted from the greater ſquare
400, leaves 352: or iftwo thirds of the dif
ference, 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 the mean circle, which being mul-
tiplyed by the length 25, 20.; the Product is
5537} the ſolidity of the portion ZISG
PX
1
+
!
PROPOSITION VII.
TER
F 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 pa-
rallel to CA; IF M X 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
ABD CEFGH,
For
(27)
cream
G
HI
M
R.
M
E
V
x144
7
R
K
?
K.
K
4
"B
For
( 28 )
1
2
For by 4; 2. the Line MV being cut into
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
Ređangles XQ and OR, are equal to the
whole ſquare V Z; for that reaſon the whole
paralellepipedon ABDCEFGH is compoſed
of theſe parts, the pyramide ABDCH more
the pyramide EFGH A more the two folids
BDKOIRHE and ACMXIQFG,
but the paralellepipedon ABDCEFGH is
equal to three pyramides of the ſame Baſe and
Altitude; therefore one of theſe ſolids as
ACXMQIFG will be equal to of the
whole
PROPOSITION VIII.
I
F in the ſide of the paralellogram ACDE
there be taken a point, ſuppoſe at B, and
the Line BF drawn parallel to AE and CD,
the Diameter BD being drawn, and the Fi-
gure being completed; then will the Rect-
angle under E Band BD, be to the Rectangle
under the trapezium E ABD and the Tri-
angle BDC, aş A B, to the compoſed of;
AB more of BG; For the Rectangle EB,
BD; that is, all the Rectangles of AB, B C;
QR, R X; EF, FD; that is the paralelle-
pipedon
( 29 )
1
t
1
E
F
N
X
Za
R
IR
ROZ
K
Z
น
A
ti.
B
pipedon made of A B, BC, and the Altitude
BF: the Rectangle E ABD, .BDC; that
is, all the Redangles EF, FD; QR, R.Z;
that is a priſme of EFD and the Altitude
F B; more all the Rectangles of RZ, ZX;
bat a paralellepipedon is to a priſme of the fama
Baſe and. Altitude as I to, that may be as
AB, to; AB, but all the Rectangles of RZ,
ZX, are of the paralellepipedom: BCB and
the Altitude CD by the laſt Propoſition,
but
the paralellepipedon BCBFD; is to. BC, as
of the paralellepipedon BCBF E, to of B6
the priſme made of DFE and the Altitude
EA, more all the Rectangles of R Z, ZX;
that is of the paralellepipedon B CB and the
Altitude BF, is equal to the Rectangle EA
BD, BD C; that is all the Rectangles of
OZX;
ai
( 30 )
1
1
68: PR Oils
QZX; Therefore as AB, to A B more
:BC, ſo are all the Rectangles A B, BC;
QR, R X; EF, FD; to all the Rectangles
ef AB, BC; QZ, ZX,
The ſecond part of this Theorem may be
that the Rectangle. EC, CF; is to the Rect-
angle of the Tropezium E ABD and the Tri-
angle B D F, as the Line AC, to': A B mogę
BC; the demonftration of this differs riot
from the former part, except in this, that all
the Ređangles DFD, ZR Z, ZR Z, are
of the paralellepipedon, as appears by the laſt
Propoſition.
obThis Propofition' is the 30 of the ſecond of
Caval: GeometAsidivif. and is of good uſe in
Jolid Geometry etsii
SUU..
· 09 ?
lu
12. 2:
3s will
!
:8 I
Tu ,,?ti, i suoi
Et ADBEK be a conei and be cut
Let through the axis, whoſe. Section will be
the Triangle A KB, Let it be cut by another
Plane as Ě ZU, the diameter of the Section
ZB parallel to BK, the Baſe of the Section
EDat Right Angle at C with A B ; Let it be
as' ABA: AKB::PZ:ZK, then will PZC
be equal to the ſquare of CE;
i
For,
(31)
b.
Fór,
AB:BK::AC:CZŁtherefore by 23.6.
SA
BA: AK ::CB:ZK, S
CABA:BKA:ACB:CZK::PZ:ZK, apply CZ to
the two laſt terms, aud it will be as 'ACB:
CZK::CZP:CZK, becauſe CZK is in
each proportion; Therefore AC B that is the
ſquare of CE is equal to CZP; the like in all
other planes parallel to the Baſe AEBD. This
manner of ſedion Apolloniks calls a parabola3bbe
""???
niin,
A.; 2;
..
1).
Ww.Liela
?:13
:«i ?i és escayni
"?!!..adito
Milli OA
vi list? iT bor:
VR
i
to
J
!!!
"Biļi!!
ů
و
1
KM
;
، رنو
1.
'
1
C
.
$
1
PRETO
H
iji i
.
2
1
B)
ti?
A
1
i
}
E:
if
( 3 )
.
if the femiparabola EIZRC be moved about
the axis ZC, until it return to that point from
whence it began its motion, it will Generate
a ſolide, Archimedes names ſuch a ſolid, a Rect-
angled conoide; s. but generally it is called by
the name of a parabolical conside
21
i
1
09
و ان
ننللننلهننيي
: 1
:!
سند
ܐ ܐ ܂ ܃ ܃ ܃
ELE
LE
is iis .
aidT PROPOSITION, X.
Et AV BXK be a tone, and be cut
through the axis, whoſe Section ſhall be
the Triangłe AKB , Let it be cut by another
Plane as NIEC whoſe diameter is NC, (be-
ing continued if need bet 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 AV BX,
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 R N it may be CN:NR::CRN:
QRN::CRN:GRT, in theſe laſt four
terms CR N-is-twice, therefore QR N is.
equal to GRT, that is, the ſquare of R I.
By reaſon of the ſimilitude of the Triangle
CHD to CRG, and of the Triangle NRT
to NHF, it may be, as
NR
(
33)
ENROTRANH: FH
TROURO HC HD by 23. 6.
}
YILL
DINRGY TŘG NHC:FHD, but T
R Gigual to the fetare bf R'I, and FHD
is eepuiak tö' tlie ſquare of HE; therefore as
NRC to the ſquare of RI, ſois NHC to
the ſquare of HE.
This Section of a cone is named by Apollo-
nius an.Ellipfis.
cin!!
4
:: ::
wilt wer: :
K
1050 1
cili!! UN
1 sigasi 1 to be
rumah Atvom
11
Gr
Dan
1
و مرا در ::
:ر !] و در رد ::
ji
A
I
Hilir
: 0 :
E
C
1
{
į
B
L
1
X
But
(34)
1
i
But if the Semi-Ellipſis CEIN be moved
about the exis, CN, until it return to that
point from whence it began its motion, it will
Generate a folid, Archiwides calls ſuch a ſolid
2. Spheroide ; but if it be turned about thc
leſfer Diameter, then he calls that folid a Sphe-
roide prelatus.
PROPOSITION XI.
3
Le
Et ABCBF be a cone, and be cut through
the axis, whoſe fection will make the
the Triangle' AFC, Let it be cut by
another Plane BHD the Diameter
HG, and the ſide of the Triangle CF
being continued may meet at K, H N at right
Angle with HG, the Plane QI E parallel to
ABCD, the Plane BDH at right Angle
with the Plane AFC; Let it be made, as
KRH:ERO::KH:HN, draw R Z paral-
lel to HN, and joyn Z NK, then will ZRH
be equal to the ſquare of RI; For, as KH:
NH::KR:ZR, take the Altitude RH and
apply it to the two laſt terms, and it will be as
KH:NH::KRH:ZRH:: KRH:ERO,
Therefore ZRH is equal to ER O that is
the ſquare of RI becauſe of the fimilitude
of the Triangle AGH to HRO, and of
KG C to K Ř E, it may be as,
KG:
(35)
CHIGA ORA:RD }by 23,6
::
KGH:CGA::KRHERO; COA is' equal
to the fquare of G B; and ERO is equal to
..
7!,
K K
F
J
M
H
::
f. ad
1
}
A
N
در ز.
A
BV
B
Anscs
D
the
(36)
the fquare of R. ITherefore as K GH, to
the fquäre of 'G BL fois KRH, to the ſquare
of RI.
w This manncr of Section Apolloniug calls a
Hyperbohaa but if the femihyperbola BIHRG
be moved about the Axis HG, until it returns
to that point from whence it began įts mo-
tiun, it will Generate a folid ; ſuch a ſolid is
called by Archimedes an Amblygon conoide 3
but generally it is called a Hyperbolical cono-
ide.
PROP:0S I TION XII.
To find the ſolidity of the portion of the
Sphere-BNR E.
LE
Et ABN E be a sphere and BH a Cg-
linder, it is manifelt that the paralellepi.
pedon whiple Bafe-is the ſquare of GE, and
Altitude GN is equal to the paralellepipedon
whoſe Baſe is the ſquare of GŃ and Altitude
AG; it fis alſo manifeft that the Rectangle
AGN is equal to the ſquare Ge, the like
for any other, as, A GN equal to ZRZ,
Therefore by the 8 Propoſition, as AG to
AG more GN, ſo are all the ſquares ZF,
to all the ſquares of ZR; Therefore, as AG,
to AG more; GN, ſo the Cylinder EK,
to
( 37
to the portion of the Sphere ERNB. The
like in a spheroide.
K
.
H
R
Z
Z
E
1
1
+
A
1
In Numbers thus.
Let AG be 150; GN, 54; GE 90;
the paralellepipedon made of the ſquare of BE
and Altitude GN, is 1749600; then, as AG,
150; to; AG, more GN; that is 75 more
9; that is 84; ſo is 1749600, to 999776, that
is all the ſquares in the portion of the Sphere
BNRE; then as 14:11:: 979776:7698243
the portion of the Sphere BNRE.
3
Propol.
.
D
(78)
PROPOSITION XIII.
L
To find the folidity of a Parabolical Conoide,
Et BCOE be a ſemiparahola; by the
9 Propoſition it is, as E B, is to ER
that is to ŽQ; fo is the ſquare of BC that is
the ſquare of Ř I, to the ſquare of RO, that
is, as the parallelogram BF, is to the Triangle
FEB; fo is the parallelepipedon made of the
ſquare of BC and Altitude BE, to all the
ſquares of RO, but the Triangle FBE is
of the parallelogram B F; Therefore the para-
bolical conoide will be of the cylinder of the
ſame Baſe and Altitude,
FzzzzzzzE
NI
q
RO
q
R
2 IR
4 R
9 IR
IVP
NR
A
B
oleto
C
Thus
(39.)
!
२.
Thus in Numbers.
Let AB, 10; BG, 10; Thereforc A C,
30; Let BE, 30; the ſquare of AC, 20;
is 400, whieh being multiplyed by EB, that
is by 15š the Product is 6000, equal to all
dhe ſquares in the conoide A EC; but if it be
made as 14:11::6000:4714 the folidity of
the conoide A EC.
PROPOSITION, XIV.
L.
To find the ſolidity of a Hyperbolical Conoide.
Et ZRHX be a femihyperbola, and EF
parallel to ZK, A R the Tranſverſe Di-
ameter, R X the intercepted Axis, the point
H in the Hyperbolical Line by the ir Propofi-
tion 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
and the Altitude AX, is to the priliy made
of the Rectangle AŘ X and the Altitude
RD, more the pyramide made of the ſquare
of ZD and Altitude DR; fo is the parallele-
pipedon made of the ſquare of Z K and Alti.
tude XR, to all the ſquares in the conside
ZHRK,
But
1
D 4
(40)
ތާ
But as the parallelepipedon made of the Recto
angle RXZ and Altitude AX, is to the
priſme made of the Rectangle ARX and Al-
titude RD, more the pyramide made of the
ſquare D Zsand Altitude eR; fo is the Line
A X to the Line BR (that is ; 4R), gore
of the Line RX, by the 8 Propoậtion;
Therefore by the 11. 5. it will be as the Line
AX, is to the Line B R, more of the Linç
RX, fo is the parallelepipedon made of the
ſquare of Z K and Altitude XR, to all the
squares in the conoide ZHR K; then as 14:11,
fu are all the ſquares in the conoide ZHRK,
to
the conoide ZHRK.
1
A
+B
R
c
E
11
F
Z
X
Ķ
Let
(41)
In Numbers thus.
V2:20 1,0
Let AX be 1503 RX, 545 ZX,90; ÅR, 963
BRO:48s the parallelepipeglana whatsappose is
the Rectangle of Rwandte Altitude à x
is 729000.
The prime whoſe: Boyle is thg Reggangte
ARXwand Altitude RD is 28328000
The pyramidei wholer Befe is the loupe, of
R X and Altitude R Divisi 87.489 gigai
The parallelepipedon, whoſe: Baſe is the Aquare
of ZX and Altitude XR is 437400.
Then 729000: 233280 more 87480, that is,
320760 :: AX, 150: BR, 48, more of
RX, 18; that is 66:: 437400: 192456 equal
to all the ſquares in the 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
14:17: 769824: 533433, the ſolidity of
the Hyperboliek conoide ZHR K.
5
PROP.
(4)
1
ROPOSITION XV.
i to frid'tkie folidity of a whole sphere, or a
portion thereof.
Út ADEF be a Plane and the Semi-
circle DQIQE be in that Plane, Let
the Planes ABCD and CDE be at Right
Angle with the Planc ADEF; Let DC,
AL and E F each of them be equal to DE;
E
1
K
PINO
K
H
K
X
RS
X
IR
o
X
q
A
D
B
Let
(43)
Let the Planes ĶHXR be parallel to tho
Baſe of the priſme; namely, parallel to AB
CD. The Rectangle DR E is equal to the
ſquare of R Q by 35 3, that is, the Rectan-
gle comprehended of IR and R X, for RX
is equal to RE, and IR is equal to R D,
therefore all the ſquares in ; of the Sphere E
QIQD are equal to all the Rectangle
in the
priſme CDFÉ, the like in any part There-
fore the priſme RXHKFE, leſs the pyra-
mide OIKH Fſball be equal to all the ſquares
in the portion of the Sphere QER. Then
as 14:11, ſo the priſme XRIOFE to the
ſolidity of the portion of the Sphere QER.
In Numbers thus:
Let AB, AF, FE, AD, DC, BC each
of theſe be equal to D Е, and DE be equal
to 204 : from E to the firſt R be 54, apd.
from that firſt R to Dbe 150; KR, 204 mul-
tiplyed by R X, 54; is XK, 11016; which
being multiplyed by RE 27; the Product
will be 297232, equal to the priſme KHXR
EF; FK, KH,HO, and I K, every of them
equal to 54; Therefore the pyramide KHO
IF will be 52488; then KHXREF, 297
232; lefs KHOIF, 52488 there will remain
IOXREF, 2449443 equal to all the ſquares
in the portion QER; that is, all the ſquares
in the portion of ; of a Sphere,
PROP.
(442
.: il
.
''!)
L
)
el
>
5
.,,
J-PROPOSITION XVI.
!!!
To find the Solidity of a whole Spheroide, or
portion thereof.
Et FADE be a Plane, the Elliplás 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 priſme AB
CD. Find the Line FE by the roth. Prop.
by the fajd 19th. Prop, the Rectangle IRE
Leben
K
q 원
​90
H
X
K
no
H 표
​X
K ҡ
H
А.
D
B;
+
will
(145)
I
1
&
1
1
3
will be equal to the ſquare of RQ; but the
Re&angle TR E is equal to the Rectangle
IRX, becauſe DC is equal to D:E, -there-
fore R X is equal to R E, therefore the Red-
angle BR X is equal to the ſquare of RQ;
therefore the prijme X HKR E F leſs the py-
ramide KHOFF will be equal to all the
ſquares in the portion QER.
...
In Number's Thus,i.digis, i
i costulpo:, ?
Let D E the longeft Diameter beizby the
Conjugate'or ſhorteſt 18,1 therefore Ke will
be 12, for it is 36:18 : 12. 7Let from E
to the firſt å be 9, then' E E, 12,1 dmtltiplyed
by Ri Xi99; X K will be 1083 which bring
multiplyed by u of ER, thád-ligl 4 lghe
Produ: will be 486, equal to KH XE EF
for the finding of Kilwórk thuşzri Abj16
to AD, 10.; fo'is AK,9; to bilB;
then
IK,3, meltiplyed by KH9the Product is27,
1 bat is, Hl; this laſt number being multiply-
ed by of K F, that is by. 3; the pyramide
KHIO F will be equal to 81, which being
ſubkracted from the priſme K'HXREF
there will remain 406 the priſme IQ XREF
equal to all the ſquares in the portion of the
Spheroide. QER.
.
5
A
PROP.
A
1
(46)
ܕ ܙ
is
P. ROPOSITION XV I.
To find the.ſolidity of a layperbolical Conoide.
.
LE
Et MAEL bea Plane, and, 'the hyper-
bola be in that Plane; to that Plane let the
Planes BAM, GEL and DEL be perpek-
dicular ; Let DE, CF and B A be each of
them equal to AM; Let the Planes KIXH
be parallel to the Plane ABDE the Baſe of
the priſme ABDELM;: find the Lives ML
and MG by the:11 Propoſition, the Recto
angle IKM, is cqual to the ſquare of-KZ,
but the Rectangle IKM is equal to the Red-
angle IKH X, becaufe A B is equal to A M,
therefore HK is equal to KM; Therefore
the Rectangle IKHX, is equal to the ſquare
KZ; Therefore the priſme ABDEL M, is
equal to all the ſquares in the conoide AMZ:
$
In Numbers thus.
Let GM the Tranſverſe Diameter be 18;
MA, 12; AE, 10; ML, 6; A B equal to
AM multiplyed by AE, 16; the Product is
72 equal to ABCF, which being multiplyed
by ; AM, that is, 6, the Product is 432
equal to the priſme ABCFLM; FE, 4;
multi-
0:47)
1
1
5
A
M
hing
'!."
1
3
risu
ils.
K
j
"!?
built
A
F
B
D
1
CG
multiplyed by FC, 12; the Product is 48,
equal to FCDE, which being multiplyed
by of FL, that is, 4; the Product is 192,
equal to the pyramide FCDEL; this
Pyra-
mide being added to the priſme before found,
the whole priſme ABDELM, will be 634,
equal to all the ſquares in the conoide AZ M.
PROPOSITION XVIII.
To find the folidity of a Conoide parabola.
LE
Et AVHP be a Rlánė, the parabola PO
R A be in that Planes Let VC and A B
each of them be equal to AP; Let the Planes
QG, QF, and QE bie parallel to the Baſe
of the priſme ABCV. By the 9 Propoſition
the parallelogram HD is equal to the fquare
of QO, but the Plane H Q is equal to the
Rectangle Q-ZGI, becauſe IQ is equal to
HP and QZ is equal to CPTherefore
the Plane QG is equal to the ſquare of QO,
the like in the reſt; Therefore the priſme A B
CVHP igequal to all the ſquares in the co-
noide Re
RA.
In
(49)
H
PA
INE
G
q
N
E
o
2
A
E
C
B.
+
In Numbers thus.
Let AP, be 10; AV, 8; Therefore AC
'will be 80; which being multiplyed by: AP,
10; thàt is, by 5, the Product will be 400,.
equal to the priſme A BEVHP; that is, all
the ſquares in of a parabolical conoide whole
Altitude is AP, 10; and ſemidiameter of the
Baſe is AR.
In theſe 4 laſt Propoſitions it ought to be
made, as 14:11; ſo are all the ſquares in the
portion to the portion it felf.
E
Prop.
(50)
PROPOSITION XIX.
To find the Area of a ſegment of a Circle or
an Ellepfis.
L.
Et 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 of the Circle, 132 the area
AGC. In the Triangle HRC, there is
given HR, 12-; CR, 5; and the Right
G
T
H
I
I
Z
T
Z
A B
C
Angle
(50)
Angle CRH; therefore there is given the
Angle CR H; thus, 5: 12 :: 100000: 240000,
the Tangent of 67 Degrees 23 Minutes, a-
gain, as 90 Degrees, that is the Angle A ĆG,
is to 67 Degrees 23 Minutes; ſo is the area
ACG, 132; to the area of the Sector HC
G, 98188 ; the Triangle HCR, 30; being
taken from the Sector, leaves the Segment
HRG, 68188; by the latter part of the 10
Propoſition it will be as A C:BC::10:ZO,
Therefore as AC:BC, ſo the ſegment of
the Circle HRG, to the ſegment of the El-
lipfis ZRG; Let B C be 10, then as 13 :
10::68188: 52L98 the ſegment of the Ellip-
fis.
PROPOSITION X X.
To find the folidity of a Circular er Elliptick
Spindle.
LE
Et ABCOHP be a Circle, HB and
PVO be two Diameters at right Angle
at Z; Let AK and GC be parallel to H B,
the Lines P O parallel to PVO; Let GC
equal to AK be the axis, Vo che Dia-
meter of the Spindle. Suppoſe there be
a Solid whoſe Baſe ſhall be the ſegment. KHG
OB A and Altitude the ſegment APK; ſuch
E 2
al
(52)
H
K
G
F
P
PE
R
9
I
P
GE
R
N
1
Pad
R
E
1
A
C
D
B
a Solid is equal to all the ſquares in of a
Sphere whoſe Diameter is AK; for the Rect-
angle ORP is equal to the Rectangle ARK,
35,3, and by the ſame 35, 3, the Rectangles
ARK are equal to all the ſquares In å of a
Sphere whoſe Diameter is A K.
If from the Solid whoſe Baſe is KHGO
CB A and Altitude APK, there be taken
a Solid, whoſe Baſe is KGCA and Altitude
AKP there will remain a Solid whoſe Baſe
and Altitude is equal to the ſegment GOC
or A PK; equal to all the ſquares in the of
the Spindle.
By
('53)
in
By the 15 Propofition find all the ſquares
25 of a Sphere whoſe Diameter is AK.
By the 19 Propoſition find the ſegment A
PR which multiplyed by K G makes a paral-
lelepipedon, which being taken from of the
forementioned Sphere, leaves all the ſquares
in á of the Spindle, this being multiplyed 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 it felf.
Further, as the whole ſo the parts are cal-
culated; find the ſegment of a Sphere by the
15 Propoſition, and the ſegment of the Circle
KRP by the 19 Propolition, this ſegment
KR P multiplyed 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 of a spheroide that hath KA
for one of its Diameters, if the ſquare of
ZP be leſlened by the ſquare of ZR, there
will remain a ſquare whoſe Root ſhall be half
the other Diameter; find all the ſquares in
of ſuch a Spheroide by the 16. Propof. then
proceed as in the Circle, as well for the
parts
as the whole.
E 3
3
Proposa
1
(54)
PROPOSITION X X I.
To find the ſolidity of the ſecond sections in a
Sphere or Spheroide.
Et D A BRCMNLFQ be of a
Sphere cut by two Planes, viz. the Plane
RXD GZQ and MX APZT being at
right Angles, their Interſection X Z. MA
Equal to MZ; RD equal to RZ; CN e-
qual to C B the ſemidiameter of the Sphere :
MX and XR being known, the reſt may be
found thus.
N
D
M
1
x
A
H
?
T
1
F
B
C
R
(55)
I.
The Areas ZXD and ZX A may be found
by the 19th. Propofition.
2.
The Spherical Superficie of BDZQ is
equal to a Surface whoſe Saſe 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.
Ler RDGZQ and MAPZT be Qua-
drants of leſler Circles of the fame Sphere,
RD and MA their Semidiameters; CEZL
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 B Z equal to
the Arch BD; the right lined Angle DRZ
equal to the Spherical Angle DB Z, the Angle
Z M A equal to the Angle ANZ.
4
As 90 Degrees, is to the degrees and parts
of a Degree in the Angle DBZ; fo is the
Spherical Superficie BADGZQ, to the Sphe-
rical Superficie of the Triangle BADGZ.
1
E
In
(56)
5.
In the Triangle IZ BDI; 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 A OZID, there is given
the Arch AD and the Angles Z AD and
ZDA; Therefore there is given the Angle
AOZID, by the 8th. Caſe of Oblique angled
Spherical Triangles.
In the Triangle AOZID there are given
the three Angles, Therefore the Area may
be
found, thus. As 180 Degrees, is to the ex-
ceſs of the three Angles over and above 180
Degrees; fo 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.
Johor Leek,in page, 116,of Mr.Gibſons Syntaxis
Muthematica, Which is this. If the exceſs of
the three Angles above 180 Degrees be mut-
tiplyed' by half the Diameter of the Sphere,
the Superficies of any Spherical Triangle is
thereby produced.
By
*
(57)
1
By this Rule and by what Mr. Gibſon delive-
red 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 B DIZ and NAOZ: the
difference between the Areas of theſe, and
thoſe found by the 4. of this ſbeweth the Areas
of the two Figures,viz. G ZID and OZPA,
which added to the Area DIZO A gives the
Superficie of the mixc Lined Triangle APZ
GD.
R
A
F
X
C
B
D
1
9.
(58)
9.
In the gth. Chapter of the forementioned
Syntaxis Mathematica there is given ſome
Rules about ſome folids; 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 ſu-
perficies of the Sphere, and their vertices at
the Center, the folidity of one of theſe is found
by multiplying the third part of the Baſe by the
Altitude (which here is the ſemiaxis) the pro-
sluct is the folidity : theſe Fragments are thoſe
which are uſually called Solid Angles. Thus far
Mr. Gibſon.
In the Diagram, Let the Triangle AZF,
be equal to the Triangle ZP ADG in the
laſt Diagram, the Spherical Superficie of that
Triangle multiplyed by one third of the Semi-
diameter CA, CF or CZ produceth the
Spherical pyramide Z FAC.
This pyramide may be divided into three
folids, viz. a pyramide whoſe Baſe is the ſeg.
ment Z FX and Altitude Z B, the pyramide
whoſe Baſe is the Arca Z AX and Altitude
XE; and the folid AZXF the folid requi-
red. But the pyramides are given, becauſe
the
(59)
the Areas ZXF and ZX A may be found by
the I of this, and the Altitudes X B and XÉ
are given to limịt the Propoſition ; There-
fore the pyramides ZXFC and ZXAC
taken from the pyramide Z FAC leaves the
ſolid FX AZ
1
IO.
8
Let ħ 480PABDhHFCT be
of a spheroide; boOXVIABDh HFCT,
be of a Sphere. Unto the Planes ħ 48096
and + LOXC, Let there be Planes at right
Angles as 8 QDH, O QB F and PPCA
C.
OIDH, OIBF and XIAC. From the
points L parallel to 8C draw the Line 4L
E, becauſe the Line 4 LE is parallel to ! Cg
Therefore the Line 4 T is equal to LT, and
the Line TQ is equal to TI, Therefore the
Quadrants of the Circle T 48 and TLI
are equal. In theſe Planes, viz. QDH,
LO QBF and P&AC, Let there be Lines
drawn, viz. P comma, and Q N, parallel to
& H. P comma, and ON parallel to o F.
comma, and N parallel to 4 C. The points
SPQD, OPQD and PPO A in chole El-
lipſes, the points OVID, OVIB and XV
"A in thoſe Circles,
1
☺
(60)
h
L
ES
H
ZI
D
RZ
1
B
of
I
V
hot
A
C
:
,9C :
XC:::H: OH,by the 1o. Prop.
OC
XC::HE: LE,
TH : OH::YE : LE, 11.5.
H-OH:OH:14 E-LE:LE, 17. 5.
80 : OH::4L : LE, 19.5.
OF : 4 E::OF :
4 E:: 0F : LE, ioth. Prop.
OF-4E: 4E :: F-LE:LE, 17.5.
OR: #E::OZ : LE, 19.5.
R+J0:4E :: ©Z+OZ:LE, 12.5.
HE : LE::IC: XC,
2C XC:: 0R+80: Z+OZ, 11. 5.
9 C XC:: Area R094 ; Area ZOOL,
Again,
:
an
(61)
Again.
& H : #E ::(QNJOH:IN, 10. Prop.
H-QN:QN::O HIN(ZH)IN, 17.5.
ſo :
: OH:: OZ
ZH, 19.5.
:
LE
P, QN::V, : IN,
P, QN:QN::V,-IN:IN, 17.5.
P. : QN::V,: IN 19.5.
ŠO+P.:QN::(HE)OZ+VI:IN; (LE) 12.5.
30+P.: 4 E::OZ + VI: LE,
PC : XC:: 2 E
9C : XC:: 304P. : OZ#VI, that is,
9C:XC::Area OQP8: Area Ziyo, the Plane.
O P QB F 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 8C
ſo the ſegment ILZO to the ſegment Q4
08, the ſecond Section in the spheroide.
PROP.
(62)
PROPOSITION XXII.
IRE
F four Numbers be in proportion, that is,
as the firſt is to the ſecond ; fo is the
third, to the fourth. It will always be, as the
firſt, is to a Geometrical mean proportion be-
tween the firſt and ſecond; fo is the third, to
-a Geometrical mean proportion between the
third and fourth.
: Let it be,
A:B::C:D, multiply the two firſt recrmis
by A, and it will be, as A A:AB :: C:D,
17. 7. If the two lali tearms be multip yed by
C, it will be A A: AB::CC:CD, they, as
A:VAB::C: VCD, 22.6.
1
In Numbers thus.
:.4
4:9 4 16 : 36.
16 : 36 :: 16 : 36.
16 : 36 :: 256 : 576.
4 : 6 :: 16
: 24
.
PROP.
(63)
1
سه
1
PROPOSITIO N. XIII.
>
To find the Relation of one Hyperbola
to another.
1
5
1
LEE
Et FLBC be a ſemihyperbola, its Tranſ-
verſe Diameter FH, its intercepted
Axis FC. Let GMBC be another ſemihy-
perbola, its Tranſverſe Diameter KG, its in-
tercepted Axis GC.
Let it be made, as,
1.
KG
: GC
HF
: FC
KG4GC: GC HF+FC: FC, 18. 5.
KC
: GC
HC : FC
KC
: HC
GC : FC, 16.5.
KC
GC
: FC, 17.7.
2
::
..
: HC
.
2
: FE
: HC
: HE
: HC
KC
::
KF
..
KC
GC
:
FC
KCG : HCF ::
KCG : KFG ::
HCF :
HEF ::
KCG :
KFG ::
GF
GF
KF
GF
KFG
HCF
CBC
CBC
: FE, 19.5.
: HE, 11.5.
: FE, 11.5.
: HEF, 23. 6.
; HEF, 16.5.
; ELE, 11 Pro.
: FMF=ELE,
11.5. 11. Prop.
Therefore
(64)
Therefore FM is equal to the Line L E,
and for as much as GC is divided in F, in
the ſame Ratio as F C is in E, Therefore,
GG
: FC ::
GF
: FE, 17.7.
GC :
FC
: EC, 19.5.
GC
MD : LD,
FC :: area CBG : ATBA CBF 12.5.
FG :::
: FC
::
GC :
म
TK
G
H
H
А В
1
म
M.
L
ABD
From
N
(65)
2.
/
From the Vertex of the Hyperbola CBF,
Let there be a Semihyperbola as CAF; FH
its Tranſverſe Diameter.
Then as,
:ENE,
HCF: HEF :: CBC : ELE II. Prop,
HCF: HEF :: CAC
11. Prop.
CBC:CAC:: ELE :ENE,
CB : CA :: EL
22.6.
CB :CA :: area CBF:area CAF, 12.5.
11.5.
: EN
3.
:: CV
.:
2
3
Let CV Q be of an Ellipſis; its Tranſ-
verſe Diameter FV, its ſemiconjugate CQ,
Let CGQ be of another Ellipſis its Tranf-
verſe Diameter EG, its conjugate CQ.
Let it be made, as,
FC : EC
: EG,
CV ; CG,
FC : EC
17.7.
FC : EC
: CD,
17.7.
FC : CG
:CD,
FC :FC+CG :: EC : EC+CD, 18.5.
FC :FG
: ED,
CV : GV
:: CG
: DG,
FCV : FGV :: ECG :EDG;
23.6.
FCV: FGV :: CQC :GLG, 10. Prop.
ECG: EDG :: CQC : DHD=GLG,
10. Profi
F
Therefore
:: CG
:: EC
16.54
:: EC
18.5.
/
17. 7.
(66)
1
Therefore DH and GL are equal, Be-
cauſe che Line CV is divided in G in the
fame 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 coy: area CQG,
CG
}
4
1
From the vertex of the Quarter of the
Ellipſis CGQ;. Let there be of another
Ellipſis, as GGR, its Tranſverſe Diameter
EG the ſemiconjugate CR,
i
Then, as,
ECG:EDG :: CQC :DHD,
ÉCG:EDG :: CRC : DKD,
CO.C:DHD:: CRC :DKD,
CQ:CR :: DH : DK,
CQ :CR :: area CDQ: area CGR, 12. 5.
10. Pr.
10. Pr.
11.5.
22.6.
5.
Let CGB be half a parabola, its Diame-
ter CG, its Ordinate @ B. Let CVB be
half of another parabola, its Diameter CV,
its Ordinate C B.
Let
I
(67)
G
K
N
C
ILIR
E
F
,
2
3
Let it be,
CV: CG :: CV : CG,
CV
:CG,
CV:CG ::
CV: CG :: GV : GD,
CG: OD :: CBC: DND,
9. Prop.
CV:GV ::CBC : GTG=DND, 9. Prop:
1
1
1-2
Therefore,
(68)
Therefore GT is equal to DN; becauſe
the Line CV is divided in G in the famc
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 femiparabola, as CGZ,
Then, as,
GC :GD
: DND, 9. Pr.
GC
:DOD,
CZC : CBC :: DODY : DND, II.5.
ZC : BC
ZC : BC
: BC ,
:: 'area ZGC : area BGC, 12. 5.
:: CBC
:: CZC
: GD
9. Pr.
:: OD
:ND,
22. 6.
7.
Hence,
It follows that the areas of hyperbolas, Ellip-
ſes and parabolas may be increaſed or decreaſed
in any proportion aſſigned ; either according
to their Tranſverſe diameters or Ordinates :
or both together.
It
(69)
It follows alſo
That the areas of hyperbolas, ellipſes and pas
rabolas of the ſame baſes are as their Alti-
tudes.
And alſo,
That the areas of hyperbolas , ellipſes and
parabolas of the ſame Altitudes are as their
bafes.
Further it follows,
If there be two hyperbolas, namely, A and B;
if the Tranſverſe diameter of A, be to the
Tranfverſe 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 inter-
preted 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 Rectan-
gled 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 di-
ameter, and Ordinate of A, is to the Rectan-
gled Figure of the interpreted diameter and
Ordinate of B.
F3 Here
(70)
8.
Here note,
In the byperbolas CF B, CFA and E
CGB,
the Lines CB, EL; and, CA, EN; and
CB, FM, are called Ordinates.
In the Ellipſes,
CGQ, CGR and CV; the Lines DH,
and DK and GL are Ordinates; the Lines
CR and CR conjugate diameters:
In the parabolas.
CGB, CGZ and CVB; the Lines CB,
DN; and CZ, DO; and CB, GT are
called Ordinate's.
A
7
9.
In the hyperbola CFB; CF, EF, CB
and EL being given, to find the Tranſverſe
dameter FH.
Let FE=A.FG=B.CB=D.EL=E.Z=HF.
Therefore HE=Z+A. HC-Z + B.
Therefore,
DD;EE :: ZB+BB: ZAAA, 11. Prop.
ZADD+AADD=ZBEE+BBEE,
16.6.
ZBE E
LAADD.
112
(71)
1
1
In Words thus.
- The difference between the tqüäre of iß in
the ſquare of E, and the ſquare of A 'in the
{quare of D; being divided by the difference
between A in the ſquare of :D Hd B in the
ſquare of E; the Ouotient is the value of Z:
1
IO.
In the Ellipſis CRG; CQ, DH, and
CD being given, to find the Tranſverſe dia-
meter EG
Let CQ=A. DH=B. 'CD D, GC32 .
Therefore Z+D-ED. Z-D=DG.
ZZ: Zt Din Z_D:: AA:BB, 10. Prop.
LA
ZZBB_ZZAA — DDAA.
:6.
DDAA=ZZAA
ZZBB
:
!
In Words thys.
nili
· As the ſquare Root, of the difference be-
tween the ſquare of A and the fquare of B, is.
to A, ſo is D, to Z; the ſemitranſverſe diame-
ter. 22, 6,
F4
In
(72)
II.
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
C+2=CG. DAA : BB :: C+Z : Z, 9. Prop:
-tom
AAZ=BBC+-BBZ
166,
-BBZ
In Words thus.
As the difference betwixt the ſquare of A
and B, is to the ſquare of B; fo is C, to Z,
Or thus.
AA : BB
: Z, 9. Prop.
AA-- BB: BB :: C+Z-Z:Z, 17.5.
:: C+Z
12.
Yet further, it may be made manifeſt, that
by the points of B and F there may be infinite
hyperbolick lines pafs : the area of the leaſt,
ſhall not be ſo little as half the parallelogram,
whoſe haſe is B.C. and Altitude CE: nor the
area:of the greateſt, ſo great, as three fourthş
of the ſaid parallelogram,
.
PROPE
(23)
ËROPOSITION XXIV.
of all manner of cylindrick hoofs.
I.
LE
Et 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' H B
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 fuperficie
of the ſaid cylinder. Let it be cut by the Plane
O ADQM, the line AOD in the ſuper-
ficie of the cylinder, and the line DQM in
the Plane CFPM; the Ordinate A M the
common Section of the Planes ABCM and
AODQM. Further, Let it be cut by the
Plane AIELM, the line AI E 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
interfe&tion of GR and ME, that is from L,
to the interſection of the lines H B and AE,
that is to I, let there be a line drawn, as LI;
and
(74)
-
and likewiſe the line QO; then the lines
LI and QO will be equal and parallel to the
line R B. Becauſe the Planes, AOD QM
and AIELM, cut the Plane GFPM at
right Angles, and the, points B, O,1,H, are
in the ſuperficie of the cylinder and in the line
ВН.
Betwixt the Planes ACM apd AODQM
there is a ſolid made, as ABCMQDO A;
alſo betwixt the ſaid baſe ABCM and the Plane
AIEL Mthere is another folid made as ABC
MLEIA : ſuch folids are called cylindrick
boofs, and they take their particular names
from ſuch cylinders as they are part of; viz.
of the baſe ABCM be ball, or a quarter of a
citele, it may be called a circular cylindrick
hoofs. If half or a quarter of an ellipſis, then
an elliptick
cylindrick hoof. If the baſē be half
or a whole parabola, then a parabolick cylin-
prick: hoof.; If half; or a whole hyperbola, then
a hyperbolick.cylindrick" hoof; the like for any
other.
TO
(75)
e
G
H
?
T
F .
:
S.
;
D
►
N
M
R
5.24
B
K
A
Το
(76)
2.
1
To prove that the Section AODQM is
of the ſame kind and degree as the baſe AB
CM.
The Angle BKO is equal to the Angle
RMQ; becauſe BK and R M are parallel,
equal and in the Plane A B C M, the Lines
OK and QM are parallel, equal and in the
Plane AOD QM, alſo the Lines B O and
R Qarq parallel, equal and in the Plane BH
GR; Therefore the Triangles BKO and
RMQ are equal and a like.
Becauſe it is, as, CM:CR::MAM:R
BR. 9th. Prop. as MC:MR::MD:MQ.
4.
6. but A M is common to both and
equal to BR; Therefore by the 5th. of the
23. Propof. as, DM:DQ:: MAM:QOQ,
Therefore the Section AODQM is a para-
bola by the gth. Propoſition.
3.
To find the relation of one hoof to another.
The Triangles MC D and MCE are upon
the ſame baſe MC, Therefore they are as
their Altitudes CD and CE. 1.6. Again,
the Triangles K BO and KBI are upon the
fame baſe KB, Therefore they are as their
Altitudes BO and BI, 1.6.
1
Further,
(77)
Further, the Triangles KBO and MCD
are alike, and alſo the Triangles KBI
and MCE are alike, therefore their fides
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. II.5.
KBO + MCD:KBI MCE::CD:CE. 12.5.
Hence it follows,
That the ſolidities of hoofs upon the ſame
baſe are as their Altitudes, that is, the hoof
ABCMQDO A, is to the hoof A BCML
EIA ; as CD, is to CE.;
Further it follows.
Becauſe it is BO:BI::CD:CE, There-
"fore the ſuperficies of hoofs, upon the ſame
baſe are as their Altitudes, that is, the ſuper-
ficie CBAOD, is to the ſuperficie CBAIE;
as CD, is to CE.
Te
(78)
1
4.
2
To find the folidity of cylindrick hoofs.
- The Triangles "MCD' and KBO are
alike, - Therefore as, as KB:BO::MC:
CD, 4.6. and KB:BO::MC:CD, by the
converſe of 17.7. then as, KB, to a Geome-
trical mean proportion between K B and half
BO, ſo is MC, to a Geometrical mean pro-
portion between MC and half CD, by the
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 KB
and half BO, and À ZX M will be a ſemi-
parabola. by the 5th. of the 23 Propſition;
that is, as CM:XM::B K:Z K. Further,
the area of the Triangle MC D is equal to
the Product of MC in half CD, that is, a
Square whoſe ſide is a Geometrical mean pro-
portion between M C and half CD, that is,
equal to the Square of MX. Alſo the area of
the Triangle KBO is equal to the Product of
K B. in half BO, that is, equal to a ſquare
whoſe fide is a Geometrical mean .proportion
between K B and half BO; that is equal to
the Square of KZ.
Hence
(79)
1
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 femidiameter is MX; but all the ſquares.
in a parabolick Spindle is to a parallelepipedon
of the ſame Baſèand Altitude; as 8, is to '15.
Bonao, Caval. Exerc. quer, page 282,
1
6.
It further follows.
As all the ſquares in the whole folid AZ
X.M are equal to the whole hoof MQDO
AC, 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.
3
7.
If the boof ABCMQDO A 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 K AZ: then take away
the
(80)
the priſme KBRM QO; Laſtly, take
away
the parabolick Semicylinder RBCNOR,
there will remain the little hoof QOND.
8.
If ACM be a ſemihyperbola its Tranſverſe
diameter C4, the hyperbolick cylinder may
be MABCFHTP: this cylinder being cut
by Planes according to the Firſt of this Pro-
poſition, the Sections AODOM and AIF
LM will be ſemihyperbolas, DS and EW
their Tranſverſe diameters by the First of the
230. Propoſition. If between M4 and half
'S there be a Geometrical mcan proportion
found, ſuppoſe it MD, and if between MC
and half DC there be taken a Geometrical
mean proportion, ſuppoſe it MX: And alſo
if a Geometrical mean proportion be taken
between K B and half BO, ſuppole it KZ;
the Plane A ZXM will be the area of a hyper-
bola its Tranſverſe diameter Xh; by the 22d.
Propoſition and by the Firſt of the 23d. Pro-
poſition all the Squares in one Fourth of the
hyperbolick Spindle A ZXM taking A M for
its axis, will be equal to the hyperbolick boof
ABCMQDOA, by the 4th. and 5th.
of this Propoſition. The relations of the ſoli-
dities and ſuperficies, by the Third of this.
If
(81)
.
9.
IF ABC M be a quarter of a Circle, the
circular cylinder will be A B CMTHFP;
this cylinder being cut by Planes, according
to the Firſt of this, the Sections A ODOM
and AIÉLM will be quarters of Ellipſes
by the Third of the 23d. Propoſition. If a
Geometrical mean proportion be taken be-
tween MC and half CD, ſuppoſe it MX,
and alſo a Geometrical mean proportion be
taken between K B and half B O, ſuppoſe it
KZ.' The area AZXM will be one quarter
of an Ellipfis by the 22d. Propoſition, and by
the Third of the 23d. Propoſition.
All the ſquares in one fourth of the ſemi-
Spheroide , taking A M for its semiaxis, and
MX for its jemidiameter , will be equal to
the circular cylindrick hoof. AODOMC
BA, by the Fourth and Fifth of this Propo-
fition. 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.
G
If
}
|
*
(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.
IO.
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 AODOM
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 B O, ſuppoſe them to be M X 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 femiſpheroide taking A M for its
ſemiaxis and MX for its ſemidiameter, will
be equal to the Elliptick cylindrick hoof AO
DOMCBA by the Fourth and Fifth of
this Propofition. The parts KABOBC
R QON and QOND are found as in the
Sixth and Seventh of this.
A
(63)
A Table of Squares and Cubes.
Roots.
Squares,
Cubes.
Roots.
Squares.
Cubes,
I
2
26
27
16
64
216
64
1
8
3
9
4
5 25
125
36
3 49
343
SI2
9
81
729
10, I001 IOOO
LI I21 1331
12 1441 1728
13 169 2197
14 196 2744
IS 225 3375
16 2564095
171289 49,13
18 3241 5832
19 361 6859
201 400 8000
21 441 9261
22 48410648
23 529 12167
24 57613834
256251 15625
675 17576
27 729) 19683
28
784121952
29 841 24389
301 900 27.000
311 9611 29791
32 1024 32768
331089 35937
34.11539304
351225 42 875
36L290 46658
371 369 50653
38114441 54872
391521 59319
401600 64000
411681 689-1
421764 74088
431849 79907
441936 85184
45120251 91125
46211697336
4712209103823
14812304110592
A
(84)
A Table of Squares and Cubes.
Roots.
Squares.
Cabes.
Roots.
Squares
Cubes.
49,2401117649
5012500 125000
512601132651
522704140608
532809148877
542916157464
553025166375
56:3136|175616
573249185193
583364195112
59348120.5379
636ool2 Looo
61137.2 722 6981
623844238328
633969250047
6414096262144
654225274625
66 43562 87496
67 1489300763
68 462 4314432
69 4701328509
704900343000
21 5041357911
7215784373248
735329309017
7415 4761405224
75 56251421875
7657761438976
77,5929456533
7816084 474552
79 6241493039
80 6400512000
81 65611531441
826724551368
836889571787
847056592704
857225614125
8673961636056
877569658503
887744/681472
89179211704969
9081007129000
91 82 811753571
9084641778688
938649 804357
948836830584
9590251857375
9692161884736
I
А
1
(85)
A Table of Squares and Cubes.
Roots.
Squares.
Cubes.
Roots.
Squares
Cubes.
971 9409 912673
98 9604941192
99 9801 970299
100 10000 1000000
101102011030301
102 104041061208
103 106091092737
104108161124864
105110251157625
106112 361191016
10711 449|122 5043
108 116641259712
109118811295029
11012100 1331000
I1112 321 1 367631
1121 2 544 1 404928
13127691 442897
114129961481544
IIS 132251520875
11613456 1 560896
11 7136851601613
II813924/1643032
11914161 1685159
120 14400 172 8000
12114641177156I
I 22 148841815848
123151291 860867
1 24153761906624
1 25156251953125
1 26158762000376
127 161292048383
1 2815384120971.52
129136412146689
1301690021 97000
131171612248091
132/1742412299968
133|176892352637
1341 795 6 2406104
135 1 8225 2460375
1361 84962515456
137 187692571353
13819044 262 8072
139193212685619
1401960012744000
14119881|2 8032 2 1
142/201642 863288
143204492924207
144207362985984
1
II
G3
A
(86)
A Table of Squares and Cubes
145-10253048625
14631316112136
1472 16093176523
148_19043241792
1491222013307949
1 502 2500 3 375000
15122 801 3442951
1522 310435 11 808
153234093581577
1541237163653264
15524025 3723875
156243363796416
157243493869893
15 81249643944312
159252814019679
160256004096000
161125921 4173241
161 26244425152
8
163265694330747
164126896 441 0944
1652722514492125
166275564574296
167 278894657463
168/282244741632
169285614826809
170 2 8900 4913000
17129241500021 I
17229584 5088448
173299295177717
174302765268024
175306255369375
176309765451776
177313295 545233
1783168415639752
1793204115735339
1801324005832000
181327615929741
182/331245028568
183133489612 8487
184338566229504
185 3422516331625
1 86 3459616434856
187/349696539203
188343446644672
18913572 16751269
1903610068590001
191 364816967871
192 3686417077888
A
2
(87)
A Table of Squares and Cubes.
Roots.
Squares.
Cubes.
Roots.
squares
Cubes.
222
193372491 71 89057
19437630 7301384
1951380251 7417875
196138416,7529536
197|38009 7645373
19839204 7762392
19939601 7880599
20040000_8000000
20140401 8120601
20240804 8242408
203412098365427
20441616 8489664
20542 0251 8615135
20642436 8741816
207|42849 8869743
208|43264 8998912
209143681 9129329
210/44100 92 61000
2 1 1 44521 9393931
2 1 2 449441 952 812 8
21345 369 9663597
214457969800344
215 46225 9938375
216646656110077696
217 4708910218313
2184752410360232
219 4796110503459
220 4840010648080
2214884110793861
492 8410941048
223/4972911089567
224 5017611239424
225150625 11390625
2265107611543176
227 51529'11697983
22151994!11852352
229152441|12008989
23015290012167000
2315336112326391
232 5382412487168
23315428912649337
23415475612812904
23515522512977875
33615-69613144256
23715616913312053
238 5664413481272
2295712113651919
240157600'1 3824000
G
4
1
(88)
PROPOSITION XXV.
LE
Et AYDG be a Sphere, O Y its axis,
C the Center of the sphere, the Lines
AD and BI be at right Angles, and at right
Angles with the axis G Y.
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(89)
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Let this Sphere be cut by a Plane through
the axis GY, and the Baſe AD; the section
makes the Circle AYD G.
Let the Sphere be cut by another Plane, viz.
as by the Plane CBHSI; HC its diameter,
BI'its diameter of the Baſe. In the inclining
Solid whoſe diameter of the Bafe is Bl, and
diameter of the Section CH and the altitude
CZ or HN, that is the inclining Solid BH
SIC is equal to the Zone A HFD, leſs the
Cone HWFQC, or the inclining Solid BH
SIC 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 R S. 47. I.
Therefore the Area of the Circle who femidi-
diameter is O s, is equal to the Areas of the
two Circles whoſe femidiameters are OR and
R S, that is, the Area of the ſemicircle OK
SE, leſs the Area of the ſemicircle ORL
is equal to the armille 4 RKSE, that is,
the Area of a ſemicircle whoſe femidiameter 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
A
(90)
AHFD and the Cone HWFQC, Further,
this inclining Solid BHSIC is equal to a
hemiſpheroide whoſe axis is H N, the altitude
of the ſection BHSIC, and the ſemidiame-
ter of the Baſe of the hemiſpheroide is CA
By the ioth. of the 6th. of Exclid, and by the
3d. of the 23. Propofition.
2.
Let AKD be a Hemiſphere; the Plane A
KD and the Plane L GFS cutting one ano-
ther at right Angles their common ſection the
ļine 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
!
K
H
N
G
700
B
27
F
R
A
C
„D
and
(91)
and M let there be lines drawn as GM and
LF their common fe&tion the point Z. From
the point L, to the line M F let there be a
perpendicular line as LE. Let YP be equal to
58; then will the Spheroide whoſe axis is EL
and its conjugate diameter Y P be equal to
the Excavatus part MZLGZF of the Zone
MLGF; or the Zone MLGF leſs the cones
GZ L and FZM will be equal to the sphe-
roide EYLP. By the 47. of the 1. of Euclid,
and by the joth. 6th. of Euclid, and by the
3d. of the 23d. Propof. Here note, Z is the
vertex of the two cones. Here alſo note, that
L 4 is equal to 4 F. Further note, the lines
IO, ñ 4 and BO are ſuppoſed to be at right
Angles with the line LF.
G
F
R
L
F F
✓
A B в
C
DE
3
(92)
.
I
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 R L c-
qual to ZI; then will the Hemiſphere AGE
leſs the cone whole diameter of che 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.
I. and roth, and 6th, and by the third of the
23d. Prop.
4.
Let AGE be a Hemiſphere; the Planes
AGE and GQ, cutting one another at
right Angles; their common ſection the line
GB. Let G! be equal to ! B being conti-
nued till it meets with the Circle GY A be-
ing continued; Let RO be equal to ở L.
Then will the Hemiſphere AGE, leſs the cone
whoſe diameter of the Baſe is B D and alti-
tude CG, that is the cone B GD, be equal
to the fruſtum Spheroide whoſe axis is B P
and conjugate femidiameter is OR. By the
47. I. and 10.6. and by the third of the 23d.
Propofition.
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
(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 spbereides, will always be ſpheres
er Spheroides, or parts thereof their demon-
ſtrations by the 47.1. 10. 6. and by the third
of the 23d. Prop.
2
:
5.
The Excavatus parts of a frustum cone, will
have relation to the hyperbola, ellipſis and pa-
rabola; 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 le&tions, viz.
the hyperbola, ellipfis and parabola, their de-
monſtrations 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 para-
bola, their demonſtrations from the 47. I. and
10.6. and from the third and fifth of the 23.
Propoſition.
(94)
8.
The folidity of every fruftum cone, is equal
to a cylinder whoſe Baſë is the leſſer Baſe of
the frustum, and its altitude the altitude of
the fruftam, 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 hyperholick conoide will be a line intercep-
ted, betwixt the continuation of the other
ſide of the frustum cone, and the intercepted
Diameter, being continued.
9.
The folidity of every fruftum parabolick
L'onoide, is equal to a cylinder whofe Baſe is the
lefſer 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 fruftum, and its
altitude the altitude of the fruftum.
10.
The folidity of every fruſtum hyperbolick
conoidi, is equal to a cylinder whoſe Baſe is the
Jeſler Baſe of the fruſtum, and altitude the
altitude of the ſaid fruftum ; more a hyperbo-
lick conoide whoſe Baſe is equal to the diffe-
rence betwixt the Baſes of the ſaid frustum z
and
(95)
and its altitude the altitude of the fruftum.
This latter hyperbolick conoide, is like to
that hyperbolick conoide, of which the fruftum
is a part.
II.
A sphere being cut by two parallel planes,
both of them equidiſtant & parallel to a plane
paſſing through the Center of the sphere; in-
cludes a part of the sphere, which for diſtin-
&ion fake may be called a middle Zone.
The folidity of every middle Zone of any
sphere, is equal to a cylinder of the fame Baje
and altitude as the Zone; more a Sphere whole
diameter is equal to the altitude of the Zone.
I 2.
A Spheroide being cut by two parallel planes,
both of them equidiſtant and parallel to a
plane paſſing through the Center of the Sphe.
roide and cutting the Axis of the ſaid ſpheroide
at right Angles, includes a part of the Sphe-
roide, which part for diſtinction fake may be
called the middle Zone of a Spheroide.
The folidity of the middle Zone of
any
Spheroide, is equal to a cylinder of the fame
Baſe, and altitude as the Zone; more a Sphe-
roide whoſe Axis is equal to the altitude of
the Zone.
The
(96)
The ellipſis which generates this laſt sphe-
oroide, is like to that ellipſis which generated
that spberoide of which this middle Zone is á
part.
13.
The folidities of hyperbolick conoides upon
the fame Baſei, are as their altitudes. Here
the Tranſverſe Diameter is increaſed or de-
creaſed in the ſame proportion as the inter-
..cepted Diameter. By the firſt of the 23d.
Propofition.
And alſo
The folidities of hyperbolick conoides under
the ſame altitudes, are as their Baſes. Here
the conjugate Diameter is increaſed or de-
creaſed in the ſame proportion as the ordi-
dinate. By the ſecond of the 23d. Propofi-
tion.
The folidițies of like hyperbolick Conoides,
are în a triplicate nie of their correſponding
terms, that is, their Tranſverſe Diameters,
or their intercepted Diameters, their con-
jagate Diameters, or the Ordinates.
1
14.
ju: The ſolidities of bumniſpheroides upon the
fame Baf, are as their Altitudes. By the 3d.
of the 23d, Prop.
The ſolidities of bëmipheroides under the
ſame Altitude, are as their Bujes, by the 4th.
of the 23d. Prop.
The
(97)
$
The folidities of like Spheroides, are in a
triplicate ratio of their correſponding terms.
15.
The folidities of parabolick conoides upon
the fame Baſe, are as their Altitudes. By the
5th of the 23d. Prop.
The folidity of parabolick conoides under
the ſame Altitude, are as their Baſeś. By the
6th of the 23d. Prop.
The ſolidities of like parabolick conoides,
are in the triplicate ratio of their correſpon-
ding terms.
An Example thus.
Suppoſe there be two parabolick conoides,
A and B, the ſolidity of A, will be to the ſa-
lidity of B; as the Cube of the axis of A, is to
the Cube of the axis of B.
Further, as the parabolick cönoide 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 conoicles 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 Gube of the Diameter of B. The like in
the reſt.
If there be two parabolick conoidis unlike,
ſuppoſe A and D. Then,
À will have that proportion to D, as is
H
com-
(98)
!
1
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 conside
of B, then their Baſes and altitudes are reci-
procal, and if their Baſes and altitudes are re-
ciprocal; thoſe conoides are equal.
Thus.
As the Baſe of A, is to the Baſe of B; fois
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 hy-
perbolick conoides; for there may be infinite
hyperbolick conoides,yet having the fame Baſe &
altitude;' the leaſt not ſo little as a Cone, nor
the greateſt ſo great as a parabolick conoide of
that lame Baſe and altitude. In this condition
the femidiameter 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
paſleth 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 fix Spheres to be added toge-
ther,
(99)
1
ther,' the axis of the Spheroide will be equal
to one of their Diameters, but the area of
that Circle, that paſſech through the Center
of that ſpheroide, and cuttech the axis at right
Angles, ſhall be equal to the areas of fix Cir-
cles whoſe Diameter is equal to the Diameter
of one of the Spheres. The areas of the Circles
are added, or fabſtracted, by the 47th. of
the i. of Euclid.
Spheres and spheroides of the ſame axis,
may be added to, or ſubſtracted from each
other, their fum, and difference will be jpheres
or ſphèroides.
17.
Hyperbolick conoides of the ſame axis may be
added too, óf ſubſtracted from each other,
their ſum and difference will be hyperbolick co-
noides. Here you are to take the ſum of the
Baſes, for the baſe of the ſum, and the diffe-
rence of the baſes, for the baſe of the diffe-
rence. Further, We are to take both the
parameters for the parameter of the fum, 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 fun and difference will be parabolick co-
noides, uſing the former rules for their baſis
aud parameters,
Here
2
12
(100)
.
Here note, the line P Z in the Diagram
for the 9th. Propoſ. the line PN in the Dia-
gram for the ioth. Propof. the line N H in
the Diagram for the 11th. Propof. are called
parameters.
.
19.
If the axis, Tranſverſe and conjugate Dia-
meters 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 Bole and altitude equal to the
Baſe and altitude of the luyperbolick conoide,
its parameter the double of the Diameter of
the sphere.
If there be a hyperbolick concide, 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 safe
the ſame as the hyperbolick conoide; its pura-
meter equal to the parmeters of the other co-
noides, 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 Bajë equal to the Baſe
of
1
(101)
of the Cone, its parameter equal to the Dia-
meter of the Sphere:
1
20.
Parabolick Conoides, may always be added
toand ſometimes ſubſtracted from hyperborck
conoides of the fame axis ; that fum 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 fum, and the diffe-
rence of the parameters for the parameter of
the difference.
Here note, when the parameter of the hy.
perbolick 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 Propof.
9, 10 and 11. of this Book.
}
/
APPLI-
(107)
Meaning and continue
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​ကြံတို့ကလေး
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1900024
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The Application.
He firſt Propoſition is to find the foļi-
dity of pyramides and Coxes, or fru-
ftum pyramides 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 applyed to
tủe meaſuring of irregular ſolids, and may be
uſeful for the exact meaſuring of all ſorts of
Stone and Timber: alſo for the exact meafu-
ring of all ſorts of elliptick, parabolick and
hyperbolick irregular ſolids, or Veſſels that are
made in that form : for ſuch ſolids may
be
cut into parallèlepipedons, priſmes and pyra.
mides, and then reduced to their own na-
ture, by the proportion of the parallelogran
adſcribed about thoſe Figures to the Figures
themſelves, Thus.
The proportion of parallelograms of the
fame baſe and altitude with the areas of para-
bolas are as 4 to 3, Therefore,
AS
(103)
As 4, is to 3; fo is any ſuch ſolid to any
ſuch parabolick irregular folid.
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 ſhewed in the third caſe of the third Pro-
poſition; 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 fruftum pyramides
when their Baſes are not parallel.
In the fifth Propoſ. there is the relation of
the Sphere and spheroide to the cylinders of
their hafes 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 ta-
ken generally for the Figure repreſenting a
Cask ; ſo, the meaſuring of one, the other is
meaſured.
In the twelfth Propoſ. there is the meaſu-
ring of a portion of a sphere, which may be
applyed to the meaſuring of the inverted
Crown of Brewers Coppers, or ſeveral other
uſes.
In the thirteenth Propoſ
. there is the mea-
ſuring of parabolick conoides, which may be
taken for a Brewers Copper, the inverted
crown.
H4
Some-
(104)
Sometimes it may be a portion of a Sphere,
or Spheroide, but ſometimes the portion of a
paetbolick conóide ; other times the portion of
a hyperbolick conoide, they ought to be taken
as diſcretion ſeems convenient.
** In the fourteenth Propoſ. there is the mea-
ſuring of a hyperbolick conoide, which may be
token for a Brewers Copper.
* In Propof. 15, 16, 17, and 18. there is the
meaſuring of a Sphere, Spheroide , parabolick
conoide and hyperbolick conoide, as well the
whole as their parts.
If the parabolick or hyperbolick conoides be
taken for Brewers 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 maſu-
ring of Circular and Elliptick Splindles.
å
The middle Zone may be taken for a Cask.
In the twenty firſt Propoſ
. there is the mea-
ſuring 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
meeíuring of right cylindrick hoofs, viz. Cir-
cnlar, Elliptick, parabolick and hyperbolick ;
and
I
t
(105)
and may be uſed for the meaſuring of Brewers
leaning Veſſels.
: If a Brewers Copper be taken to be of that
Figure that parabolick or byperbolick conoides
àre, and they ſtand leaning, the meaſuring of
them is almoſt the ſame as though they did
not lean.
Here I ought to have ſhewed the making,
and uſe of an Inſtrument for taking the lean-
ing of ſuch Veſſels; But my buſineſs calls mė
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 thac incomparable
Inſtrument maker Mr. Henrj Sútton.
Here note, the Table of Squares and Cubes
is very ready’and uſeful in finding the porti-
ons of a ſphere, Spheroide, parabolick and hyper-
bolick conoides.
7
To find two fuch numbers, that their Product
being added to the ſum of their Squares, the :
ſum ſhall be a Square, and it's Root commen-
Jurable.
Le
Et one of the numbers be A, and the
other Z. The product may be A Z; 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, 2+2
that
1
(106)
that is the ſquare Z Z — 47 +4, therefore
ZZ+ZA+AA=ZZ --42 + 4, that is,
ZA+ AA=-47 + 4; Let A be a unit,
then 5Z=3, that is, 2= 7, 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 Girąd obſerves from
this ſeventh Propoſ. of the sth. of Diophantus,
That if there be a Triangle made of 3 ſuch
numbers, the Angle oppolite to the greateſt
ſide will be 120 degrees. It may be further
obſerved, that if there be a right angled Tri-
angles 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 ope ſhall
be equal to the area of the other.
Thus,
1749 17149
5,25 - 319
58 The ſum of their Squares.
24 40 The difference of their Squares.
170 42 Their double Rectangles.
98 189 The ſum of their fides.
18401 840) Their Areas.
Here note, the double Rectangles are
taken for their Altitudes.
74
Here
(107)
Here note,
In Progreſſions from a Unit, the ſum, of
the ſum and difference of the greateſt number
in that progreffion, and any one number be-
twixt the greateſt and Unity, is equal to the
ſum, of the ſum and difference of that fame
greateſt number; and any other number be-
twixt 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 hypo-
teruſe and baſe of the one, is equal to the kypar
tenuſe and Bafe of the other.
A general Theorem for the finding of two
ſuch numbers ; Take the ſquare of any num-
ber, from which take a Unit ; take the double
of that number, to which adde a Unit; that
ſum and difference will be the two numbers
required. Thus,
The number taken is
32 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
3
!
ta
(108)
This Propoſition was publickly propoſed in
Paris in the year 1633, which Renatos' des
Cartes reſolved, and Francis Schooten publiſh-
ed in Sect. 12. Miſcel
, incumbred with a ſquare
adfected equation, with furds.
1
F 1 N I S.
>
1
Eryata.
Page 17, line 3, for Ecliptick read Eliptick. p.25, for Excave=
W r. Excavatus. p.29, 1.12, for-r. In the ſame line, for BC
BPE; t. B CBFD:"P:39, 1. 22, for Z B, r. 2 C. p. 32. 1. 19
for NR, r. NP. p.39, 1. 13, put ; after Line. p. 46, 1.21,
for 634, r. 624. p. 53, 1. 3, for APR, 1. APK. p. 48,1. 25,
for Z B, r. XB. p. 60, 1. ult. for RO94, r. RO 4. P.
61,5, 6, for V, 1. VI. p. 64, l. 4, for GG, r. GC. p:66;
1,18, for CDQ, 1,CGQ: P.70, P: 711. 5, pat , after D.
P: 74,1.4, for GFPM; r. CFPM. p.80, 1.21, put : before
all. p. 82, 1. 25, after K ABO put,
1
អណ្តូ obimo ဒီလို့ဝလို့ဝရှိလို့ရှိ
Imprimatur.
Tbo. Tomkins R. Rmo in Chrifto Patri ac Domino Dno.
GILBERTO Divinâ Providantiâ Archiep.
Cant. à Sac. Dom.
Ex Ads Limb.
Aug.25.1668.
******* : *:t** cto conto che tocou* *
i
GAGING
PROMOTED.
AN
APPENDIX
TO
Stereometrical Propofitions.
By
ROBERT ANDERSON,
LONDON
Printed by 7. W. for Jofbua Coniers,
at the Raven in Ducklane, 1669.
+
Thurtouz.un steises
scindi
ola Mcha, bocaschap
)
go a quella
s1022
Gaging Promoted
AN
APPENDIX
Τ Ο .
Scercometrical Propofitions
}
-
1. Note.
Ś an Abftra&t from the andoubred Axioma
of Geometry, it is generally obſerved, that
in a Rank of numbers, having equal diffe-
rence,the ſecond differences of the ſquares
of thoſe numbers are equal, che third differences
of the cubes of thoſe numbers are equal and ſo is
order in the higher powers. Thus,
&
A
(2)
In Squares
I
I
lalala
4
1.100 100 con
5125
I
if I
214 39
39
2
51 25
24
4
16
747
81
21314.
1_213 14
Obſerve in the firſt of theſe Examples, in the firſt
collum are the numbers of a progreſſition, having
equal difference, to wit, a unice. In the ſecond co-
lum, che ſquares
of thoſe numbers. In the third co-
lum, che firſt differences. In the fourth colum, the
ſecond differences, to wit, 2, 2, 2. In the ſecond
Example, in the firſt colum are a Rank of numbers,
having equal difference, to wit, 2. In the fecond
colum, their ſquares. In the third colum, the firſt
difference. In the fourth colum, the ſecond diffe-
rence.
II. Note.
Hence it follows, that by the help of fuch diffe-
rences the table of ſquares may be calculated : thus,
in the firſt Example, the ſum of 1 and 3, is 4 ; che
ſquare of 2. The ſum of 2, 3 and 4, is 9; the ſquare
of 3. The ſum of 2, 5 and 9, is 10 ; the ſquare of 4.
The ſum of 2, 7 and 16, is 25; the ſquare of 5. The
fum of 2, 11 and 36, is 49; the ſquare of 7. &c.
111.
.
(3)
III. Note.
9:39
Like plain numbers are in the ſame proportion
one to another, that a ſquare number is in, to a
ſquare number : "Euclide the 26 Propoſition of the
Eighth Book. Therefore the ſecond difference in
fuch a Rank of plane numbers are equal. Further,
what planes and folids are either equal or propor-
tionable co ſuch Ranks may be gradually calculated;
as in the laſt.
IV. N 01.C.
1
بهتي
doo
3. *27
48.
.I
7
2 8
1 2
IG 6
3 27
i 8
37
71
5125
21 3' 4:5
26
72
98
5 I 25 120
218
7343 186
386
9729
-6
24
4 64
48
in
I 2.3
A is
In che firſt Example, in the first colum are the
numbers in a Rank having equal difference, co wit,
a unite. In the ſecond colum, the cubes of thoſe
numbers. In the third colum, the firſt differences
of thoſe cubes. In the fourth colum, the ſecond
differences. In the fifth colum, ehe third differen-
ces, to wit, 6, 8. The like in the ſecond Example:
V. Note.
Hence it follows, charrlic rable of qu'es math.
Az
117.1 dia,
(4)
"
made, thus : In the firſt Example 1 and 7, is 8; the,
cube of 2. The ſum of 8 and 19, is 27; the cube of
3. The ſum of 18, 19, and 27, is 64; che cube of 4.
The ſum of 6, 18, 37 and 64, is 125;che cube of s.
The ſum of 6, 24, 61 and 125, is 2 16; che cube of
6. The ſum of 6, 30, 91 and 216, is 343; the cube
of 7. The like in the ſecond Example.
VI. Note.
Like folid, numbers are in the ſame proportion
one to another, that a cube number is in, co a cube
number.' Euclide che XXVII Prop. of the Eighth
Book. Therefore the third differences in fuch a
Rank of ſolid numbers are equal : further, ſuch
planes and ſolids as are either equal or proporcio-
nable to ſuch Ranks, may be gradually calculated,
as in the laſt.
VII. Note.
O
If a Rank of Squares, whoſe Roots have equal
differences, be multiplied by any number, the ſe-
cond differences of ſuch a Rank of prouds are e-
qual. Let the number multiplying be 10.
In the firit colum are the num-
bers bearing equal difference. In
che ſecond colum are the ſquares
of thoſe numbers. In the third
celum the products. In the fourch
colum the firſt differences. In the
fiſch colum che ſecond differences
and they equal.
1
이 ​in
I
40
em con
3 91 9u
4 16 160
5 25/250
919
11:1
20
70
90
21 2
6)
VIII. Note:
.
If unto luch a Rank of Produ&ts, as in the laft,
there be added a Rank of Cubes, whoſe Roots are
equal to the Roots of che Squares, the third diffe
rences of ſuch a Rank will be equal.
00 OC
© III
26
37
8 48
40
32
6g 6
3 90 27117
416064224
1077
6
44
151
5.2501125375
.2 13
3'4 5 E
olalalm' &
!
38
37
In the firſt colam are the numbers baving equal
difference. In the ſecond colum the Products of
their ſquares by a given number. In the third co-
lum che cube of the numbers in the firſt colum. To
the fourth celum the ſum of the products and cubesi
In che fifth colum their firſt difference. In the fixe
colum che ſecond differences. In the ſeventh cor
lum the third differences which are equal.
IX. Note.
Let a conſtant number be added to a Rank of
Products, ſo that one of the numbers, multiplying
be
AAN
(6)
3
1
be a conſtant number, and the other of the num-
bers be the ſquares of numbers having equal diffe-
rence, and this Rank of ſums be added to a Rank
of cubes, whoſe roots are the ſame with the roots
of the ſquares; ſuch a compounded Rank will have
Cheir chird difference equal. Thus,
18 19
11 191
28,40
37
8 56
69
3
8
90 27125
107
4 8160 64232
6
44
151
5 8250 125 383
I 2' 3 4 15
415 1 61 78
In the firſt colum are the numbers having equal
difference. In the ſecond colum is the conſtant
number to be added. In the chird colum are the
Rank of products, that is, the ſquares of the num-
bers in the firſt colum multiplied by a given num-
ber. In the fourth colum are the cubes of the num-
'bers in the firſt colum. In the fifth calum are the
ſum of the numbers in the ſecond, third and fourth
colums. In the fixth colum are the firſt differences
of their ſums. In the ſeventh colum are the ſecond
differences. In the eighth colum the third differen,
çes, and they equal.
X. Note.
Ina Rank of numbers, having equal difference,
and equal in number; if the third part of the cubes
of each of theſe numbers, be ſubſtracted from the
products of the ſquares of each of theſe numbers,
in half the greateſt number of that Rank, the re-
mainders
(7)
9
09 108
I
>
be lo lalu lo
7
54
00
parent
108 54
mainders will bea Rank of numbers, equal to all the
ſquares in the ſeveral porcions of one fourth of a
ſphere, whoſe diameter is equal to the greateſt
number in that Rank, and the third differences of
this Rank of portions are equal; but the firſt and
fecond differences will increaſe and decreaſe, diffe-
rently one from another. Thus,
00 oolool
99
3
99 162
261
54
6
72_432 360 108
369
54
91 243 972 729
54
12 75617281152 423
423 54
15 1125 27001575
54
369
18.1944 3888 1944
361
54
2130875.2922205 162
96
2446086912 2304
T
I
3 4 5 1.67
In the firft colum are the numbers having equat
difference. In the ſecond colum are, che pyramides
adſcribed within the cubes of the numbers in the
firſt colum. In the third columare the products of
the ſquares of the numbers in the firft colum, by
half the greateſt number in the firſt colun. In the
fourch colum are the differences of the numbers in
che ſecond and.chird coluns, thac is, all the ſquares
in ſeveral portions of one fourth of a ſphere, whidſe
diameter is 24. In the fifth colum are tire firſt dif-
ferences, In the fixt colum are their ſecond diffe-
rences. In the ſeventh colum are the third differen-
ces, and they equal.
XI.
ci
2
(8)
1
XI. Note,
The Application or Uſe of the
Preceeding Notes.
The application or Uſe may be, to calculate Pya
Famides and cones, either che whole or their pares;
as alſo to calculate the parabolick and hyperbolick
conoides, either the whole, or their fruftims; yet
allo, to calculate the ſphere or ſpheroide, either
the whole or their portions or Zones, and that
gradually, that is, to find the ſolidity upon every
inch or foot.
To find the ſolidity of a parabolick Conoide upon
every two inches,
To do which, conſider the Diagram of the 18
Prop. of my Stereometrical Prop. Ler PA be 16;
AR 12; therefore A Vor PH will be 9.; for it
ought to be as PA, is to AR; fois A R, CO A V.
Let the axis AP be divided into eight equal parts;
viz. 2,4,6, 8, 10, 12, 14, 16. Let there be planes
drawn parallel to the baſe, through every one of
there diviſions, though in the Diagram there is noc
ſo many. From P.to the firſt ſuppole to be 2, its
{quare 4; the half thereof 2, which multiplied by 9
equal to P H, the Product will be 18; that is, the
Priſm QZGIH P.; equal co all the ſquares in the
portion of the conoid QOP. Let from P, to the
kecond Qbe 4, its ſquare 16, the half is 8; which
mu'ti-
(9)
3
muleiplied by 9, the Produ& is 72; equal to the
Prilo QZFIHP, equal to all the ſquares in the
portion of the conoid ROP. Lee from P to the
chird Qube 6, its ſquare 36, the half of it is 18,
which multiplied by 9, che product is 162;, the
Priſm QZETHP; equal to all the ſquares in the
third fortion of the conoid. QOP.
Having obtained two portions, the reft may be
obtained thus : having obtained the ſecond diffe-
rence, which is 36, we may proceed co find the reſt
by the Seventh Note; chus, add 36 to 54, and it
makes 90:which added to 72, the ſum is 162, equa!
to all the ſquares in chac porcion, and ſo in order
36, 90 and 162; the ſum is 288. 36,126 and 288,
the ſum is 450. 36, 162 and 450, the fum is 648,
&c.
In the firft colum are che pares
of the altıcude of the conoid. In
che ſecond colum are all the
36
ſquares in ſeveral porcions of one
fourth of a conoid. In the third
colum che firſt differences. In the
36
fourth colum che ſecond differen-
1264
36
ces. Theſe porcions in the ſecond
36
colum may be reduced to circular
portions, thus, as 14 is to 11; ſo
are all the ſquares in tlieſe parti-
ons to the portions chemſelves,
o 001-
18
,
2
361
54
4 ? 36
90
126
36
62
1o 455
198
6 I2
283
141 882234
270
1152
1
2
3 14
I. Scholium:
The uſe of this gradual calculation may be chus:
Suppoſe a Brewers Copper be in form of a para-
bolick
1
(10)
bolick conoid; the quantity of ţiquor therein con
tained may be found, thus, having calculated a
cable upon every inch, or two inches, or as is
thought convenient"; then having a ſtraight' Ru-
1er divided equally into inches, putting the Ruler
into the liquor to the bottom of the Copper, fee
how many inches of the Ruler is wet, with the
number of wer inches enter the firſt colum of your
cable, and in the next colum are the number of cu-
Dick inches which that portion contains; the num-
beř of cubick inches thus found, being divided by
the number of cabičk inchies in a Gallon, the quo-
tient'fhews the number of Gallons in that portion
ofthe Copper.
:
II. Scholium:
3
To compoſe ſeveral works into one,
t
A's 14, is to 11; ſo are all the ſquares in one
fourth of the conoid, to one fourth of the conoid
įt' ſelf. becauſe this one fourth 'ought to be di-
vided by the number of cubick inches in a Gal-
lon, ſáppoſe it 288, to thew the namber of Gal-
lons in each portion, we may multiply 14 by 288,
that is 40323 Tlienas 4032 is co 1.1 ; fo are all che
fquares in one fourch of the conoid, to the Gala
lons in that one fourth. Further, becauſe this one
fourth ought to be multiplied by 4; to reduce it to
a whole conoid; therefore, divide the conſtant
diviſor, that is, 4032, by 4, and it will be 1008.
Then, as 1008, is to 11 ; lo are choſe ſeveral pór-
tions in the ſecond colum of the laſt table, to the
num-
II
number of Gallons in thoſe ſeveral portions of a
parabolick conoid. By ſuch compoſitions may the
Pra&icioner compoſe conſtant diviſors or divi-
dends, which will much breviate the work; this is
onely for an Example.
ܪܕ
i
Every parabolick conoid hath its ſecond diffe-
rences equal. To find the fecand differences,
work chus, Square one of the equal ſegments of
the axis, and multiply that Square by the Parame-
ter, that product will be the ſecond difference. In
chis Example, the equal fegment of the axis is 2
the ſquare of ic 4; which mulciplied by the Para-
meter 9, the product is 36, the ſecond difference.
Half of the ſecond difference, is always the firſt of
the firſt difference. Half 36 the ſecond difference,
is 18, the firſt of the firſt difference, &c. Here
note, this 36 is the ſecond difference of one fourth
of all the ſquares in a parabolick conoid ; if 36 be
multiplied by 4, it makes the ſecond difference 144;
whoſe half is 72, the firſt of the firſt differences.
Or, the firſt differences are found by taking half
che difference of the ſquares of any cwo ſegments,
which mulciplied by the Parameter, chereby the
firſt differences are obtained. Thus, to find the
firſt difference anſwerable to che Segments 6 and
8; the Square of 8, is 64 ; the Square of 6, is 36
che difference of thoſe Squares is 28, whoſe hali
is 14 ; which multiplied by the Parameter 9, che,
product is 126; the firſt difference anſwerable be-
twixt 6 and 8,
XII
(12)
XII, Note.
To find the ſolidity of as hyperbolick conoid gräs
dually, to wit, upon eyerj three incbes.
For the performance of which, take notice of
the XVI Prop. of Stereom. Prop. in chat Diagram,
Lec A M equal to A B be g. Let ML equal to AF,
be 6. Let AF be 15 : therefore FE will be g.
Ler the reſt of the conſtruction be as in that Pro-
poſition. Let from M to the firſt K be 3, whoſe
Tquare is 9, whoſe half is 4i, the area KHM; which
being multiplied by ML, 6; the produd will be 27,
the priſm KHNOLM. Becauſe F E, F C and FL
are equal, that is, each of them 9 : Therefore, the
firſt pyramid ONXILM will be 9. Then this
priſm and pyramid being added, will make 36, che
whole priſm KHXILM, equal to all the ſquares
in the portion KZM. Ler from M to the ſecond
K be 6, whoſe ſquare is 36, its half 18, che area
KHM; 'which being multiplied by ML, 6; the
product will be 108, equal co the priſm KHNO
LM. The cube of 6, is 216; a chird parc is 73,
the pyramid ONXIL, this priſm and pyramid be-
ing added together, the ſum will be 1 So; the priſm
KHXILM : equal to all the ſquares in che por-
tion KZM. Lec from M to A, be 9; its ſquare
81, che half 40, equal to the area ABM, which
being mulciplied by ML, 6; the product will be
243 : the cube of 9, is 729; a third part thereof
is 243; equal to the pyramid FCDEL: this priſm
and pyramid being added cogecher, is 486, che
whola
( 13 )
1
whole priſm ABDELM, equal to all the ſquares
of the portion AZ M.
Theſe three porcions being obtained, they may
be continued by the VIII Note, thus:
*52
151800
1116324
378
For if the third dif-
00
36
ferences which are e-
3_46 108
qual, and in this Ex-
144
54
180
ample is 54, be added
306
54 to che firſt of the fe-
486 216
cond differences, be.
522
54
121008 270 ing 108, it makes 162,
792
54 and by fuch additions,
the ſecond differences
812216
in the fourth colum
1494 54 are made. Furcher, by
214410 432
1926
adding theſe ſecond
246336
differences to the firſt
I 2
3 14 15
of the firſt differences
which is 36, it makes
144, &c. So the numbers in the third colum are
made. Yec further, by adding theſe firſt differen-
ces to the firſt number in the ſecond colum, che
Rank of portions of ſuch a conoid is made.
Then,
By making uſe of the directions in the firſt and
ſecond scholiums, the number of Gallons are ob-
tained. The parabolick and hyperbolick conoides
may well be made uſe of for Brewers Coppers; the
parabolick, when the crown is ſomewhat blunt ;
but the hyperbolick conoid when the crown is
more ſharp.
)
XIII.
.
(14)
$
XIII. Note.
To calculate a sphere gradually, to wit, upor
every three Inches.
1
Conſider the XV Prop. of Stereon, Prop. Let
E D equal to EF, be 24. The reſt of the conſtru-
dion as in that Prop. Let from E, co che firft R be
3, whoſe ſquare is 9 , whoſe half is 41, the area
RXE, which being mulciplied by 24, the produ&
will be 108; the priſm KHXREF. The cabe of
3, is 27; a third part thereof is 9, the pyramid
KHOIF; this pyramid taken from the former
priſm; leaves the priſm R XQ I FE, 99 : equal to
all the ſquares in the portion R QE. From E, to
che ſecond R, 6; its ſquare 36, the half 18, which
multiplied by 24, makes 432; che priſm RXHK
FE. The cube of 6, is 216, a third part of it is 72;
the pyramid KHOIF : this pyramid being taken
from that priſm, there reſt 360; the priſm RXO
IFE, equal to all the fquares in the portion RQE.
Let from E to the third R be 9; its ſquare 81, the
half thereof 407, the area RXE, this area being
multiplied by 24, the product will be 972, the
priſm KHYREF : the cube of 9, is 729, a third
part of it is 243, the pyramid KHOIF : chis py-
ramid being ſubftracted from chae priſm, the re-
mainder is 729; the priſm R XOIFE, equal co all
che ſquares in the third portion R QE. Having
obtained theſe three portions, the reſt may be
found by their chird difference, according to the
X. Note.
- The
(15)
la hilo
1
3) 09 108
10854
1
54
18
369
108
1
2
4 15 16
00
99
99 162
261
54
이 ​72_432 360
369
2 343 972 729 54
423
54
12 57017281152
423 54
15112527001575 54
54
1944 38881944
261
54
21 30875294 2305 162
99
2446086912 2304
3
7
The numbers in the ſeventh colum are the third
differences, and they equal; the numbers in che
Gxe colum are che fecond differences, and are com
poſed by ſubſtrading the numbers in the ſeventh
from the firſt and laſt numbers in the Gxt colum.;
the numbers in the fifcb colum are the firſt differ
rences, and are compoſed by adding thoſe num-
bers in the fixt colum to the firſt and laſt of thoſe
in che fifth colum; the numbers in the fourth colum
are all the ſquares in ſeveral portions of one
fourth of a ſphere whoſe diameter is 24, thoſe por-
cions are made by adding the numbers in the fifchi
colum to the numbers in the fourth, thus, 261, and
99, is 360. 369, and 350, is 729. 423, and 720,
1$ 1152, &c.
Then making uſe of the firſt and ſecond ſchio
lium the number of gallons are obtained. Or if
it be made, as 44, IS 1.0.31, fo is 54, 60 a fourch
B
numbers
(16)
.
S
number, with that fourth number proceede to
make tables of the fecond and firſt differences,
and then the table of porcions ic felfe. Every
ſphere bath its third differences equall. To find
che chird difference, doe thus. Cube one of the
equal ſegments of the axis and mulciply thac cube
by 2, and that product will be the third difference,
thus, the cube of three is 27, which mulciplyed
by 2, the product is 54 ; the third difference of
all the ſquares in one fourth of a ſphere. Here nore,
ebat it is to be underſtood, that che axis of the
ſphere is equally divided into an equal number
of ſegments; ſo then, if the number of fegments
in the lemiaxis, leſs by one ; be multiplyed by
the third difference, it gives the firſt of the ſecond
differences. Thus, the number of ſegments in the
femiaxis is 4, then 4
which being
mulciplyed by 54, the product is 162: the firſt
of the ſecond differences,
To find the third difference in one fourth of
all the ſquares in a ſpheroid, do thus: The axis
being divided as above in the ſphere; cabe the
difference betwixt two Segments, which being
multiplyed by 2, makes a product ; then, as the
ſquare of the ſemiaxis, is to che ſquare of the
ocher femidiameter ; ſo is chac fermer product co
a fourth number, which will be the third diffe-
rence. For the ſecond differences, uſe the Rules
given for the ſphere.
XIV. Note.
To calculate a pyramid or cone gradually. To
find the chird difference in a pyramid work thus,
che
lels 1, is 3;
1
(17)
en
1
the Alcitude of the pyramid being equally divided
cube che difference of the two legmencs, which
being doubled, makes a number, then, as the
ſquare of the Alcicude of che pyramid, is co chic
area of the baſe of that pyramid ; ſo is that for-
mer number, to the third difference of chac
pyramid.
To find the ſecond differences in a pyramid:
As the difference of two of the ſegments of che
Alcicude, is to the following ſegmenc ; 'ſo is the
chird difference, to the ſecond difference anſwet-
able to chat ſegment.
To find the firſt differences in a pyramid.
Cube cwo of the ſegments, and cake a third part
of their difference. Then, as the ſquare of che
Altitude of the pyramid; is to the area of the
baſe of that pyramid, fo is that former difference;
to the firſt difference anſwerable to thoſe two
ſegments.
Let there be a pyramid whoſe Altitude is 1ő
and one ſide of che baſe is 40, and the ocher ſide
5, cherefore che area of the baſe is 200. Let the
Altitude be divided into five equall parts, and co
calculate accordingly. To find the third differ-
ence, che cube of 2, is 8; whoſe double is 16.
Then, as 100 the ſquare of the Alcitude, is có
200 che area of the baſe; fo is 16. to che third
difference 32. To find any of the ſecond differ-
ences ac demand, co find the ſecond difference än-
Iwerable co 8. As 2, the difference betwixe che
ſegments 6, and 8, is co 8; fo is che firft difference
32,00 128 the ſecond difference anſwerable co 8.
The ſecond differences are in proportion one co
another, as their anſwering ſegments; ás 2, is to
B 2
32
4
(18)
7
3.2; fo is 8, to 128. To find any of the firſt dif-
ferences, cube the two Segments, to wit, 2 and
4, and the cubes will be 8 and 64, then take 3
from 64 and the Remainder is 56, a third part
· is 18;. chen, as the ſquare of the Altitude too, is
to che area of the baſe 200; fo is 182, CO 37,
the firſt difference anſwering to 2 and 4. Then
by a continuall adding of the third difference to
che fecond differences chey are made, and by ad-
ding the firſt of the ſecond differences to the
firſt of the firſt differences and ſo in order the
firſt differences are made. Laſtly by adding the
firft differences the Segments of the pyramid,
are made aceording to the Ill. Note; or thus.
Site
| 37
41 42
}
101)
lalalala inlo
32
32
64
32
96
32
128
6 144
1
1341, 1975
325
3
10, 666
Il 2
1
3
4's
The numbers in the fifch colum are the third
differences, the firſt number in the fourth colum
being found by the Rule before given, all the
numbers in that fourch colum may be made by
adding the third difference, thus, to 32 adde 32,
the ſumme is 64. adde 32, to 64; the ſumme is
96. adde 32, to 96; the ſumme is 128. The firſt
number in the chird colum being found by the Rule
above, thens, added to 32; the ſumme is 371.
adde
1
(19)
adde 64, to 37; ; the fumme is 101;. adde 96, to
1015, the ſum is 1975. adde 128, c9 1971 ; the
fum is 3251. further, adde the firſt of the third
colum, to the firſt of che ſecond colum ; thus, adde
St, co o; che ſum will be si, adde 575, co sj
the ſum is 422, adde roi. Co 42; che ſum is 144.
adde 197', co 144; the ſum is 341;, adde, 32575
10 341}; che fum is 666.
If it be to calculate a cone whoſe diameters of
the baſe are 40 and 5. Let ie ic be made, as 14,
is to 11; ſo is 32, to che third difference of the
ſame cone. Then proceede with the chird dif-
ference to make che ſecond and firſt; and laſtly,
the table it felf.
XV. Note
The calculation of fruſtum pyramides whoſe
baſes are unlike, To the performance of which
conſider the third caſe of che ſecond propoſicion
of Stereen. Prop.
Every ſuch ſolide liąth its third differences
equall, but the ſecond and firſt differences will be
complicated according to the IX. Note.
To find the third difference proper to the
pyramid BCDHF, Let the conſtruction and
numbers be the ſame as in that diagram, and let
it be co calculate it upon every two inches, chus.
The cube of 2, is 8; the double chereof is 16,
Then, as the ſquare of the Alcitude 40, thac is
1600, is to clie arca of the bare BCDH, 336;
To is 16, to 3-2.. by the R.:le delivered in the
14 Note,
B 3
( 20 )
14 Note, the firſt of the ſecond differences is
Sites and the firſt of the firſt differences is not
The ſolide HDEGVF hath its ſecond differ-
ences equall by the Vil. Note.
To find its firſt and ſecond differences. The
ſquare of 2, is 4. which multiplyed by FV, 26;
che product will be 104. chen, as 40 the Altitude,
is to HD, 28; ſo is 104, CO 7260. che ſecond
difference. Therefore the firſt of the firſt differ-
ences will be 36-40
To find the ſecond differences of the ſolide
ABHOIF the ſquare of 2 is 4, which multi-
plyed by IF, 30; the product is 120, then, as
40, the Altitude; is to O A, 12 : fo is 120, to
36, the ſecond difference. Therefore the firſt of
the firſt differences are 18. For the complication
of theſe differences.
I
2
TO
3
3, 3ion in the pyramid BCDHF
360 72,00 in the priſm HGEDFV
18 36
in the priſm ABHOIF
54,112,61 375 their fumme.
Rejecting the denominators they may be writ-
țen Thus,
5496 | 11216 | 336
Becauſe che denominators are Rejected, there-
fore the two laſt figures toward the Right hand
are decimals.
(21)
3342 08 172712
1
Н
161496,
1216
336
4.
1552
6, 518472
184264
336
1888
19652
8 74624
336
12 2 2 4
2. 8376
336
10! 92 3000
12!143936
22896
1616 24832
247064
181885464
26o632
13904
14240
336
14576
14912
336
15 248
220936' 2560
336
141377768 338;
336
13568
336
336
20216oooo 2745 36
336
222448776288776
427521 28 33352
336
озо70392 318264
283403904 333512
зо/3753ooo 34909615584336
336
32)4118o16 365016
344499288 38127716256336
36489715, 397864 1659:336
16928 336
57440oo 432056
17264336
1 32 32
15920
414792
38| 5 3 1944
2
3
4
5.
ВА-
-
:
(22)
The conſtru&ion of the table may be chus;
the numbers in the firſt colum are the third dif-
ferences. The firſt number in the fourth colum is
the complicated ſecond difference, and the other
number in that fourth colum are made chus, to
the firſt 11216,adde 336; the ſum is 11552. Then.
co t hat 11552, adde 336; che qum is 11888, &c.
The firſt number in the third colum is compli-
cated from the firſt complicated difference and
a parallelipepidon whoſe baſe is the plane RIFV,
and the Altitude the firſt Segment of the Altitude
of the fruſtum, thus, the plane RIFV, is 780;
which being doubled is 1560 ; then, 156000 more
5496 is 161496 ; the firſt of the firſt differences,
then 161496 more 11216, is 172712. Further,
172712 more 11552, is 184264.' Yer further
184264 more 11888, is 196152, &c.
The numbers in the ſecond colum are made
chus, the firſt number in the ſecond colum, is the
ſame as the firſt number in the third colum, then,
161496 more 172712, is 334208, and 334308
more 184264, is 518472, &c.
Then makeing uſe of the firſt and ſecond ſcho-
lium, the quantity of Liquor chat ſuch peffels
fontain may eaſily be obtained.
XVI. Note.
To calculate Elliptick ſolides whoſe baſes are
unlike. The calcalation of ſuch ſolides are the ſame
as in the 15.note for if the first, ſecond and third
complicated differences be found , then make-
ing uſe of this propotion as 14, is to 11; ſo is
336; to the chird difference.
re
And
I
C
(23)
And
As 14, is to it, ſo is 11216, to the firſt of
the ſecond differences.
Further
As 14, is to il; fo is 161496, to the first of
the firſt differences, then proceede to make the
table it ſelf, as in the 15 note. Or make uſe of
the ſecund ſcholium of the 11 note and you will
have che quancicy in Gallons,
Or
Such Elliptick ſolides may be calculated by the
12 noce : for every ſuch Ellipcick ſolide is equall
to a fruftum hyperbolick conoide whole circular
baſes of the conoide, are equall to the Elliptick
baſes of the Elliptick ſolide ; and the Alicude
of one fruſtum is equall to the Altitude of the
other.
XVII. Note.
Every hyperbolick conoid hach its third diffe-
rences equal.
To find the third, ſecond and firſt differences
in an hyperbolick conoid, and conſequently to
calculate that conoid gradually. In the foremen-
cioned diagram of the 17. prop. Scereom. Prop.
Let GM, the Tranſverſe diameter be 12. ML,
the parameter 6. MA, che axis of the conoid
24,
10
(24)
I
To calculate the folidity of this conoid
vpon every three Inches.
To find the third difference of this conoid.
)
+
Take the difference of two of the Segments,to
wit, 3 ; whoſe cube is 27: whoſe double is $4.
Then, as GM. 12; is to ML,6:fo is 54,00 27. The
third difference of all the ſquares in one fourth
of that conoid proper to chae pyramid FCDEL.
By the Rule in the laſt note the firſt of the
fecond differences is 27.
For the first of the firſt differences, worke
thus; take the firſt Segment which is 3; whoſe
cube is 27, a third part is 9, then, as G M, 13;
is to ML, 6: ſo is 9,00 4. chc firſt of the firſt
differences proper to che pyramid FCDEL.
The ſecond and firſt differences of all the
ſquares in one fourth of this conoids is complica-
ted from the ſecond and firſt differences of the
pyramid FCDEL, and tlie ſecond and firſt diffe-
rences of the priſm ABCFLM. Every ſuch
priſm hath it fecond difference equall.
To find the ſecond and firſt difference of the
priſm ABCFLM.
1
Square che difference of cwo of the Segments
of the axis, co wic, 3; that is 9, which being
multiplyed by the parameter ML, 6; che product
is 54, che ſecond difference. The firſt of the firſt
d fferences of every ſuch prilm is half of the
fecond
(25)
ſecond difference ; therefore the firſt of the firſt
differences is 27:
a:
To complicate theſe differences.
I
2.
3 differences
41 27 27 in the priſm FCDEL.
27 154 in the priſm ABCFLM,
31;181 | 27 in all che ſquares of one fourth
of that hyperbolick conoid.
O
112
27
27
720
311
31
81!
6
144 108
2-20
93641
135
355) 27
12:
162
5171
15 1237 189
18 1.944
706 27
216
922
12
27
21.2866
24 1032
2 3 14 15
27
1165, 243
5
In the firſt colum are the third differences. In
the fourth colum che ſecond differences. In the
third colum che firſt differences. In the ſecond
colum the portions of all the ſquares of one
foarch of an hyperbolick conoid, upon every cliree
inches,
(26)
>
inches, whoſe Tranſverſe diameter is 12, ånd
parameter is 6, and axis is 24.
The conſtruction of this table is the ſame as
the former ; thus, 8 1 more 27; is 108. more 27;
is 135. more 27, is 102. &c.
31.1. more 81 ; is 112.. more 108 ; is 2201. &c.
o more 31; is 31. more 112; is 144.more
220, is 364. more 355; is 78O.
Here remember that the Tranſverſe diameter
is found, by the g of the 23 Propoſition of
Stereom. Prop. Alſo the parameter found by the
converſe of the firſt part of the 11 Prop. of
Stereom. Prop. The parameter of the parabolike
conoid is found, by che converſe of the 9 Prop.
Of Stereom. Prop.
XVIII. Note.
Cautions concerning Reduction,
I
If it be to calculate pyramids whether Regular
or Irregular, whole or fruftums ; che chird, ſecond
and firſt differences are to be found as above:
then Reduce thoſe differences into Gallons and
parts of a Gallon, or Barrells, or parts of a
barrels;
Thus
Suppoſe 288 cubick inches make one Gallon,
and 36 Gallons make one Barrell.
Then,
If the mcaſure be taken in inches, divide the
third, ſecond and firſt differences by 288, and fo
there will be three quociencs in Gallons or parts
of
(27)
1
1
of a Gallon, then with thoſe three quotients
proceede to make the table of ſolid Segments,
and that cable will be in Gallons or parts of a
Gallon, If it be to calculate a table in Barrells
multiply 288 by 36 and che produ& will be 10368
the number of cubick inches in one Barrell. Then
divide the third, ſecond and firſt differences. by
16368, there will be three quotients in Barrells
or parts of a Barrell: Then with theſe chree
quocients proceede to make the table of ſolides
Segments.
That cable being ſo made will be in Barrells
or parts of a Barrell,
2.
To calculate Cones and Elliptick ſolids, whether
the whole or their fruftums.
Haveing found their third ſecond and firſt
differences, as above, and it be to calculate them
in cubick inches, Let it be made as 14, is to 11;
ſo is the third difference, to a fourch,
And,
As 14, is to it; ſo is the ſecond difference,
to a fourth,
Furcher,
As 14, is to 11; fo is the firſt difference,
to a fourch with theſe three number thus found,
proceede to make che table of ſolid Segments,
and chat table will be in cubick inches.
To calculate theſe ſolids in Gallons.
Multiply 14 by 288 the product will be 4032.
Then,
As 40329 to 11; ſo is the third difference, te
a fourch,
And
1
1
+
(28)
And,
: 'As 4032, to 11; ſo is the ſecond differences
to à fourth,
Furcher,
As 4032, to 11; ſo is the firſt difference, to
a fourth.
With theſe three numbers thus found, proceed
to make che table of folid Segments. So chat
rable will be in Gallons.
To calculate thele folids in Barrells.
Suppoſe 288 cubick inches makes one Gallon,
and 36 gallons makes one Barrell, then multiply
288, 36 and 14 one inco another and they make
145152.
Then,
As 145152, is to 11; fo is the third difference,
to a fourth.
And,
As 145152, is to 11, ſo is the ſecond difference,
to a fourch,
Further,
As 145:52, is to 11; ſo is the firſt difference,
to a foruth.
With theſe chree numbers thus found make the
table of ſolid Segmencs: that table will be in
Barrells.
III.
Having found the third, ſecond and firſt diffe-
rences of all the ſquares of one fourth of a ſphere,
Spheroid and hyperbolick Conoid, as in the iz and
13 notes and the ſecond and firſt differences of
all the ſquares of one fourth of a parabolick
conoid as in the 11 noce: they may be Reduced
to Circular differences.
Thus
(29)
Thus,
As 14, is to 11 ; ſo is the third difference, to
a fourth.
And,
As 14, is to 11, ſo is the ſecond difference, to
a fourch.
Further,
As 14, is to 11; ſo is the firſt difference, co
a fourch.
With theſe numbers thus found make a cable,
of ſolid Segments of cubical inches of one fourch
of any of theſe for:ds.
Thele folid Segments ought to be multiplyed by
four, to reduce them co folid Segments of a whole
ſphere, ſpheroid, hyperbolick and paraboliek conoidë
but to ſhun chać work divide 14, by four, and then
find the new differences, but becauſe 14 cannot be
juft divided by four, therefore div.de 14, by cwo,
and multiply 11, by two, and then work; Thus,
1
Then,
As 7, to 22; ſo is that third difference, to a
fourch.
And,
As 7 to 22; fo is that ſecond difference, to
fourch.
Further,
As 7 to 22; ſo is chat firſt difference, tot
fourch.
With theſe numbers thus found, proceed to
make cables as is taught in thoſe Noces: cables fd
made, will be tables of ſolid Segments of choſe
Lolids, in cubick inches.
To calculate theſe folids in Gallons.
Multiply
(30)
Multiply 288 by 14, wiroſe produ& is 4032;
one fourth chercof is 1008
;
Then
As 1008, is to ¡I; fo is the third difference,
to a fourth,
And
As 1008, is to II ; ſo is the ſecond difference;
to a fourth,
3
Further,
As 1008, is to II; fo is the firſt difference,
to a fourth,
Tables being made, with numbers thus found
according to the former directions in the ſphere,
ſpheroid , hyperbolick and parabolick conoids
will be tables of ſolid Segments of a whole ſphere,
Spheroid, hyperbolick and parabolick conoid, in
Gallons or parts thereof.
To calculate theſe ſolids in Barrells.
Multiply 4032 by 36, the product will be
145152, one fourth chereof will be 36288;
Then,
As 36288, is to st; so is the third difference,
to a fourth.
And,
As 36288, isto11, ſo is the ſecond difference,
co a fourch.
Furcher,
As 36288, is to sI; fo is the firſt difference,
to a fourth.
Tables being made, with numbers chus found,
according co the former directions, will be cables
of ſolid Segments in Barcells. &c.
Then,
1
(31)
Then,
Uling a Rod or Ruler equally divided into
inches as in ſcholium the first, the number of
Gallons or Barrells may ſpeedily be obtained.
As for che juſt magnitude of the Gallon, it is
not my buſineſſe to diſpute ; that being deter-
mined by cuſtom or Authority : I took 288
onely for Example fake.
XIX. Note.
.
In a Rank of numbers having
equal differences.
Let the firſt term in the Rank be Z, its ſquarë
ZZ. the ſecond term 2Z, its ſquare 4ZZ, there-
fore the firſt of the firſt differences is 3 ZZ, che
third term in that Rank 3Z, its ſquare 9Zz, then
OZZ, Leſs 4ZZ, the ſecond of che firſt differences
SZZ, cherefore ZZ Leſs 3ZZ che ſecond
difference will be 2ZZ.
Furcher,
The fourth cerm in chac Rank is 4Z, its ſquare
is 16ZZ, then 16ZZ Leſs 9ZZ tle chird of the
firft differences 7ZZ; again, 7ZZ Lcfc sZZ
the ſecond difference is 2ZZ.
Hence it follows,
That the ſecond difference is equallco the ſquare
of the firſt term doubled.
Or alſo,
The ſecond difference is equall to the ſquare
$
$
of
1
(32)
of the difference of two of the terms, (in order
taken) doubled.
By the Tame method we find that the third
differences in a Rank of cubes are equall, and the
third difference is equal to the firſt term multi-
plyed by 6.
$
Or,
The third difference is equal to the cube of the
difference of two of the terms, taken in order,
multiplged by ,
The index and equal difference, of every power
agrees; co wir, the index of the ſquare is 2, and
the ſecond differences are equal. The index of the
cube is 3, and the third differences are equal. The
index of the ſquare ſquared is 4, and the fourth
differences are equal. &c.
The equal difference of every power, is com-
plicated from the index of that power,and the equal
difference of the next Lefler power.
Let the Rank be in naturall order. Thus;
!, 2, 3, 4. &c.
?
The indices of che powers, Thus.
Il | 2 | 3 l 4
5
ZT ZZZZZZZZZT ZZZZZ
A unity the equal difference in chac naturall
Rank, whoſe ſquare is I, which multiplyed by 2
che index of the ſquare the product is 2, the equal
difference in the ſquares. 3, che index of the cube
multiplyed by a the equal difference in the ſquares,
c!le
,1
A 546330
UNIVERSITY OF MICHIGAN
3 9015 06525 5823
1