The Solution of the Pyramid Problem Part 12

PR = 277516

GD = 2245135

SP = 263932

The triangle DXH represents a vertical section of the pentagonal pyramid; the edge HX is equal to HN, and the apothem DX is equal to DE.

Let DH be a hinge attaching the plane DXH to the base, now lift the plane DXH until the point X is vertical above the centre C. Then the points A, E, B, O, N of the five slant slides, when closed up, will all meet at the point X over the centre C.

We have now built a pyramid out of the pentangle, whose slope is 2 to 1, alt.i.tude CX being to CD as 2 to 1.

Apothem DX = DE Alt.i.tude CX = HM or MN Alt.i.tude CX + CH = CM radius.

Apothem DX + CD = CM radius.

Edge HX = HN or PF

Note also that

MP -- = CH 2

OP = HR

Let us now consider the _Pentangle as the symbol of the Great Pyramid Cheops_.

The line MP = the base of Cheops.

The line CH = half base of Cheops.

The line HM = apothem of Cheops.

The line HN = slant edge of Cheops.

Thus: Apothem of Cheops = side of decagon.

Apothem of Cheops = alt.i.tude of pentagonal pyramid.

Slant edge of Cheops = slant edge of pentagonal pyramid.

Now since apothem of Cheops = MH and half base of Cheops = HC

then do apothem and half base represent, when taken together, extreme and mean ratio, and alt.i.tude is a mean proportional between them: it having already been stated, which also is proved by the figures in the table, that MC : MH :: MH : HC and apoth: alt :: alt : half base.

Thus is the four pointed star _Cheops_ evolved from the five pointed star _Pentalpha_. This is shown clearly by Fig. 68, thus:--

Fig. 68.

Within a circle describe a pentangle, around the interior pentagon of the star describe a circle, around the circle describe a square; then will the square represent the base of Cheops.

Draw two diameters of the outer circle pa.s.sing through the centre square at right angles to each other, and each diameter parallel to sides of the square; then will the parts of these diameters between the square and the outer circle represent the four apothems of the four slant sides of the pyramid. Connect the angles of the square with the circ.u.mference of the outer circle by lines at the four points indicated by the diameters, and the star of the pyramid is formed, which, when closed as a solid, will be a correct model of Cheops.

Calling apothem of Cheops, MH = 34 and half base, HC = 21 as per Figure 6. Then-- MH + MC = 55

and 55 : 34 :: 34 : 21018, being only in error a few inches in the pyramid itself, if carried into actual measures.

The ratio, therefore, of apothem to half-base, 34 to 21, which I ascribe to Cheops, is as near as stone and mortar can be got to ill.u.s.trate the above proportions.

Correctly stated arithmetically let MH = 2.

Then HC = v5 - 1 MC = v5 + 1

and alt.i.tude of Cheops = v(MH MC)

Let us now compare the construction of the two stars:--

Fig. 69.

TO CONSTRUCT THE STAR PENTALPHA FIG. 69.

Describe a circle.

Draw diameter MCE.

Divide MC in mean and extreme ratio at H.

Lay off half MH from C, to D.

Draw chord ADB, at right angles to diameter ECM.

Draw chord BHN, through H.

Draw chord AHO, through H.

Connect NE.

Connect EO.

Fig. 70.

TO CONSTRUCT THE STAR CHEOPS, FIG. 70.

Describe a circle.

Draw diameter MCE.

Divide MC in mean and extreme ratio, at H.

Describe an inner circle with radius CH, and around it describe the square a, b, c, d.

Draw diameter ACB, at right angles to diameter ECM.

Draw Aa, aE, Eb, bB, Bd, dM, Mc, and cA.

The question now arises, does this pyramid Cheops set forth by the relations of its alt.i.tude to perimeter of base the ratio of diameter to circ.u.mference; or, does it set forth mean proportional, and extreme and mean ratio, by the proportions of its apothem, alt.i.tude, and half-base?

The answer is--from the practical impossibility of such extreme accuracy in such a ma.s.s of masonry, that it points alike to all, and may as fairly be considered the exponent of the one as of the others. Piazzi Smyth makes Cheops 76165 feet base, and 48491 feet alt.i.tude, which is very nearly what he calls a [Pi] pyramid, for which I reckon the alt.i.tude would be about 48487 feet with the same base: and for a pyramid of extreme and mean ratio the alt.i.tude would be 48434 feet.

The whole difference, therefore, is only about six inches in a height of nearly five hundred feet. This difference, evidently beyond the power of man to discover, now that the pyramid is a ruin, would even in its perfect state have been inappreciable.

It appears most probable that the star Pentalpha led to the star Cheops, and that the star Cheops (_Fig_. 70) was the plan used by the ancient architect, and the ratio of 34 to 21, hypotenuse to base, the template used by the ancient builders.

Suppose some king said to his architect, "Make me a plan of a pyramid, of which the base shall be 420 cubits square, and alt.i.tude shall be to the perimeter of the base as the radius of a circle to the circ.u.mference."--Then might the architect prepare an elaborate plan in which the relative dimensions would be about--

R. B. CUBITS {Base 420 Base angle 51 51' 143? {Alt.i.tude 267380304 &c.

{Apothem 339988573 &c.

The king then orders another pyramid, of the same base, of which alt.i.tude is to be a mean proportional between apothem and half-base--and apothem and half-base taken as one line are to be in mean and extreme ratio.

The architect's plan of this pyramid will be the simple figure ill.u.s.trated by me (_Fig_. 70), and the dimensions about--