A Treatise on Mensuration, Both in Theory and Practice

If the diameter AB be 50, the height AD 120, and the versed fine AF 10; what is the folidity of the hoof?

Or fuppofing a cylindric veffel ABCD, containing a fluid, to be placed in fuch a pofition that the furface of the fluid, difpofing itfelf parallel to the horizon, may cut the bafe in GE, leaving 40 inches of the diameter dry, and the fide of the cylinder in D, 120 inches diftant from the bafe; to find how many ale gallons are in it; the diameter of the base being 50 inches.

Here b dv = 25

=√252 — 152

And, to find

120, d = 50, and v = 10.

= =

Then

10 15 c, and ✔dd-cc
= √40 X 10 = 20 = S..
the bafe by the table of fegments,

== 2; this being found in the column of

verfed fines, oppofite to it is the area 1118238: hence 50X 50 X 1118238 = 279 5595b is the feg279-5595 ment or bafe.

Then

353 - bc

་

b = 12× (×8000—15×279°5595)

= 12 × (5333+ — 4193*3925) = 12 X 1139′9408 13679 2896 the folidity in inches; which, divided by 282, the inches in a gallon, give 48.50939 ale gallons for the content.

EXAMPLE II.

Suppofe the cylinder fo placed, that the surface of the liquor may bifect the bafe, and rife up the fide to the fame distance of 120 inches from the base: to find the content.

Here, by note 1, we have ddb = 50 X 50 X 20 = 50000 folid inches 177 3049645 gallons, for

the content in this cafe.

EXAMPLE III.

Suppofe, now, the fame veffel fo placed, as that the furface of the liquor may leave only 10 inches of the diameter dry, ftill rifing to the fame diftance of 120 inches along the fide; to find the content.

Here the part of the cylinder's bafe left dry, is equal to the bafe in the firft example, viz. 279'5595 which, therefore, taken from

50 X 50 X 78539816 = 1963 4954, the
whole circle, leaves

1683.9359b,

the base of the ungula in this example.

Now v = 40, c = 15, and s=20.
Whence 3-be

b = (3 × 8000 + 25259*0385)

I 20

= 30592*3718 × 3 = 91777°1154 folid inches = 325 45076 gallons, the content in this cafe.

PROBLEM XXVI.

To find the Solidity of the Elliptic Hoofs of the Fruftum of a Cone, made by a Plane cutting Diagonally the Oppofite Extremities of the Ends.

* From the fquare of the diameter of the base of the hoof, subtract the product of the diameter of the

other

Let AEBF be the base of a cone, or of any other pyramid, right, or oblique ; AVE a fection through the vertex by a plane perpendicular to the bafe; and EVF, ECF two other fections perpendicular to AVE, the former through the vertex, and the latter through the fide, at c, between v and B. On AB let fall the perpen

diculars VH, CI; and on DC N

the perpendiculars VK, BL;

and draw Gc parallel to AB, and meeting AV and VH in G and м.

Then

other end of the fruftum, and a mean proportional between the diameters; divide the difference by the difference of the diameters; multiply the quotient by the height of the hoof, and the product by the diameter of its bafe; fo fhall the laft product multiplied by 2618 give the content of the hoof.-That is, putting

DAB the diameter of the greater end,
dGc the lefs diameter, and

bc the perpendicular height.

Then

D2-dDd

D-d

And

PDd-d

D-d

Xo 2618 Db the greater hoof ABC,

Xo.2618 db the lefs hoof AGC.

A plane being conceived to be drawn through c and A.-The proof is in corollary 2.

E X

Then it is evident that EFBV is a pyramid whofe bafe is EFB (A), altitude VH (a); and therefore its content is equal to Aa; and that EFCV is a pyramid whofe bafe is EFC (B), height VK (b), and therefore its content equal to вb; and moreover that the difference of thefe pyramids, or ¦ sa 6 is the content of the hoof EPEC.

But, by the fimilar triangles ABV, GVC, it is

AB - CG CI or HV VM :: AB: HV, or a =

GC X CI

AB - GC

alfo AB GC: CI :: AB: HV :: GC: VM = and DC: BD (by the fimilar triangles 1CD, DBL) CI: BL :: (becaufe of the fimilar triangles BCI and CVM, VKC and CLE)

VM (= CCX CI) : VK (b) =

ABGC

GC X CIX DB

DC X (AB-GC)

Wherefore the hoof EFBC will be

Corol. 1. if the bafe be circular, or the pyramid a cone, and the angle CDB be lefs than the angle va ; or, which is the fame, if co and VA, produced, interest in w; the fection ECF

EXAMPLE I.

If a conical veffel, whofe bottom diameter is 30 inches, be inclined to the horizon till the liquor in it juft cover the bottom, and its surface, difpofing itself into the pofition Ac, cut the fide BC of the veffel in c, at the distance of 18 inches from the bottom AB; how many wine gallons of liquor are in it, fuppofing the diameter GC of the veffel at c to be 19.2 inches?

Here D = 30, d = 19°2, and b = 18.
Whence D2-da/pd

D--d

X.2618Db=

302—19°24/30× 19°2

30-19.29

X 30 X 18 X 2618 439 x 30 x 18 x 2618

2196 × 2618 X 10 =

Which being divided by 231,

will be a fegment of an ellipse whose transverse axe is cN, and conjugate NO X GC, No being drawn parallel to AB, and meeting VB produced in o. And then the above general

theorem will become

EBF

part 3,

GC X DB

x elliptic fegment ECF), by prob. 6 fect. 3

X (AB X cir. feg. EBF
ABGC

GC X DB X NO X GC

DC X CN

x cir. feg. whofe diameter is CN and height CD) =, fince fimilar

fegmenta

EXAMPLE II.

Let there be taken here the fame dimenfions as in the last example, fuppofing the veffel to be narrowest at the bottom, to find how many ale gallons are in