The pylonic series=n2=the square of the 1st t ordinate, or the last ordinate of the obelisk.

The ordinate of obelisk x axis will, during the descent, generate a curvilinear obeliscal or parabolic area; while the t ordinate will generate a curvilinear area, similar to the outline, or section of the massive curved cornice projecting from an Egyptian propylon. Hence the curve traced by the tordinate may be called the pylonic curve.

If the last ordinate of the obelisk the first ordinate of the pylonic area, the common axis will = n2, and the area of the series of rectangled parallelograms along the sectional axes will

=

n2.

The circumscribing rectangled parallelogram of the obelisk or pylonic area will

n3.

In Fig. 34. the series of rectangled parallelograms have been constructed to the 9th ordinate, the end of the common axis, but they may be continued along the produced axis. Thus the areas of the rectangled parallelograms, how numerous soever they may be, will all be equal.

The pylonic curve will be continually approaching to the axis, and to each other if a similar curve were constructed on the other side of the axis, while the sides of two obeliscal areas will be continually receding from each other and from the axis, but still continually approaching to parallelism with each other, with the axis, and with the pylonic curve.

As the sectional axes 1, 3, 5, 7, &c., are described in equal times, the series of equal rectangled parallelograms or equal areas, along the sectional axes, would be described in equal times by the m t ordinates of the sections. But during the descent the time t ordinate continually varies, so the area described will be curvilinear.

Generally, when the last ordinate n of the obelisk is made the first ordinate t of the pylonic area, each rectangled parallelogram will = n squares of unity, a line of squares of unity of the length of the ordinate n=nx 1=n.

2

=

Sum of the series of rectangled parallelograms = n2 = ordinate axis, or a line of square units of the length of the axis.

=

Circumscribing rectangled parallelogram = axis x ordi

nate = n2 x n=n3.

An area of square units the length of the axis and breadth of the ordinate; or an area of square units = n_times the

2

ordinate.

=

Circumscribing square an area of square units n times the circumscribing rectangled parallelogram = n times the axis x ordinaten x n2 x n = n2 × n2 = n1 = axis

Hence we may say,

2

=n=ordinate axis

2

= = n2=ordinate = axis

area of a rectangled parallelogram series of rectangled parallelograms circumscribing rectangled parallelogram=n3=ordinate axis circumscribing square

3

4

2

=n1ordinate axis

Having found the sum of the series of squares of 1, 4, 9, 16, 25, 36, or of 14, 24, 34, 44, 5a, 64, when placed along an axis, fig. 22.

Let each square in this series be made a square stratum of the depth of unity, and placed in the order 36, 25, 16, 9, 4, 1, such a series of square strata will form a solid like a teocalli, fig. 23.; the height, 6, will the square root of the side of the base or of the lowest terrace, 36, and

2

=

Content=1(n+1.n .n+1−±n+1.n.n+) cubes of unity.

The content of a pyramid having its base the side of a cube, and the height or axis = the length of the side of the cube, will the content of the cube.

=

A cube has 6 square sides all equal. radiate from the centre, and the axes

Suppose 6 axes to to be rectangular

to each other, or perpendicular to the sides of the supposed cube.

Then let 6 pyramids having axes of equal length be

2

2

generated by square ordinates x axis, or distance from that central point or common apex; these 6 pyramids will have equal square bases and equal heights, so they will be equal to each other, and these 6 bases will form the 6 sides of a cube having a content = the content of the 6 pyramids.

Let the cube be divided into 2 equal rectangled parallelopipeds by a plane parallel to one of the sides of the cube; then each rectangled parallelopiped will the content of 3 pyramids, one of which pyramids will be entire.

=

As each pyramid = the content of the cube, this pyramid will the content of the rectangled parallelopiped = area of base of rectangled parallelopiped multiplied by the height.

So the content of pyramid having the same base and height as the rectangled parallelopiped will = the content of the circumscribing rectangled parallelopiped.

Or a pyramid having the same base and twice the height will the circumscribing cube.

=

The horn of Jupiter Ammon, like the ammonite, represents the spiral obelisk, and is typical of infinity.

IS GENERATED THE PYRAMID AND HYPERBOLIC SOLID, THE ORDINATES OF WHICH VARY INVERSELY AS EACH OTHER, THAT OF THE PYRAMID VARIES AS D2, THAT OF THE HYPERBOLIC SOLID VARIES 1 1 &C. THE HYPER22 32'

1

AS

SERIES 12, 22, 32, &C., AND 1,

D2

--

BE REPRESENTED BY THE ORDINATE OF PYRAMID, OR BY THE SOLID OBEBISK. GRAVITY REPRESENTED SYMBOLICALLY IN HIEROGLYPHICS BY THE HYPERBOLIC SOLID. ——— THE OBELISK REPRESENTS THE PLANETARY DISTANCES, VELOCITIES, PERIODIC TIMES, AREAS DESCRIBED IN EQUAL TIMES, TIMES OF DESCRIBING EQUAL AREAS AND EQUAL DISTANCES IN DIFFERENT ORBITS HAVING THE COMMON CENTRE IN THE APEX OF THE OBELISK. THE ATTRIBUTES OF OSIRIS SYMBOLISE ETERNITY.

Hyperbolic Areas and Solids.

LET fig. 37. be a series of 6 rectangled parallelograms, all of equal areas and rising from the side or base of the 1st, which is a square, and the side of the square to =6, then the area will 36; the height of the second rectangled parallelogram =2 × 6, and breadth =16, then 12 × 3=36; 3rd rectangled parallelogram =18x2; 4th, 24 x 1.5; 5th, 30 × 1·2; 6th,

36 x 1.

=

=

=

=

x

Or 1st, 6×6; 2nd, 2×6×16; 3rd, 3×6×16; 4×6×16; 5th, 5×6×16; 6th, 6×6×16. So that the axis of each rectangled parallelogram ∞ D, and

4th,

The area of each rectangled parallelogram = 6.

Hence it follows that the ordinates will be bounded on one side by the asymptote, and on the other by the hyperbolic curve; or the series of rectangled parallelograms will be an hyperbolic series. The 1st ordinate will 6, the whole axis or asymptote =62, and the area of the series of rectangled parallelograms =62×6=63, the circumscribing rectangled parallelogram.

=

Next take the areas between every two of these ordinates in succession, and let n = 6.

These different areas so cut off will form another series of rectangled parallelograms, which will be as 1, 4, 1, 1, 1, ↓ of n2. So that if

Or the area of the series of rectangled parallelograms = 88.2. Let the equal sides of such a series of rectangled parallelograms be placed in the same straight line or axis, fig. 36. Then