Extracted text
Editorial transcription, lightly normalized from the scan. Line-break hyphenation, spacing, capitalization, and unmistakable OCR errors have been repaired; italics follow the printed text and the claims remain those of The Flaming Sword.
EDITOR FLAMING SWORD:—It is held by astronomers that the earth's equatorial diameter is about 26 miles longer than its polar diameter. This would place the equator about thirteen miles higher all around, than it would be if the earth were a perfect sphere. In other words, there would be a rise in the earth from the poles to the equator, of about 10 feet to the mile; the Mississippi river would be running up hill at this ratio. Now how could this be? And how does this ratio of the rise tally with the 8 inches to the mile curvature? 10 feet to the mile would be so much in excess of the 8-inch ratio, that it would produce an enormous difference in the direction and ratio of the meridian arc from pole to equator. Please explain.—SUBSCRIBER.
There is a great deal of foolish speculation concerning the rise between the pole and the equator. It is certain that the earth cannot have two ratios of curvature in the same place and at the same time. If the earth is larger in diameter at the equator than through the poles, it is so because the static planes of materialization of the substances comprising the shell, constitute a spheroid instead of a perfect sphere; the waters of the sea are in their static plane or equilibrium from pole to equator, and there is no up hill, though the equator be farther from the center of the earth than the poles.
The Mississippi river flows down from Lake Itasca to the Gulf; that is, its source has a greater altitude above the sea level than its mouth; consequently, the water gravitates from the source to the Gulf. If the earth is bulged at the equator, no matter whether the earth is convex or concave, it is flatter at the poles than at the equator.
The actual ratio of curvature of a sphere 7,935 miles in diameter, is 7.92 inches to the mile, increasing its angles and distance from a given tangent, as to the square of the distance. If the normal curvation of a perfect sphere 7,935 miles in diameter is 7.92 inches to the mile, it would follow that if the earth is flattened 13 miles at each pole, it curves a little less than 7.92 inches in the vicinity of the poles, and a little more than that ratio near the equator. The curvature would still be "about 8 inches to the mile."
The difficulty with the above inquirer is, that the ratio of curvature from a given tangent, and the ratio of divergence of two lines, are confused in the mind. The curvature on both a perfect sphere 7,935 miles in diameter, and a sphere of the same size, flattened 13 miles at the poles, would be about 3,967 miles in one fourth the earth's circumference. If we place a tangent at the pole, it is obvious that the sphere curves away from that tangent to the equator; the amount of curvature is the semi-diameter.
The curvature of a sphere from an external tangent at the pole would amount to the semi-diameter of the sphere, or 3,967 miles, in the distance that lies between the pole and equator along the meridian. The ratio of divergence of the arc of a spheroid from the arc of the perfect sphere, amounts to only 13 miles in the 6,229 miles of the meridian arc, or about 11 feet for each mile, without increase as to the square of the distance. This latter ratio is but the ratio of divergence of two curved lines not parallel. The two arcs would curve so nearly in the same direction that there would be but very little difference in the actual ratio of curvation.
On a spheroid 10 feet in diameter, with difference of equatorial and polar diameters in the proper proportion, to represent the spheroid of the earth under consideration, would amount to only .39 of an inch, swelling only .195 of an inch on each side at the equator, which would be scarcely appreciable to the eye. The curvation of such a spheroid is practically the same as that of a perfect sphere 10 feet in diameter, amounting to 5 feet, or the semi-diameter, in one fourth of its circumference.
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The source scans remain the record. This page preserves the correspondence and editorial reply without presenting the Koreshan System's conclusions as independently established results. The claims remain attributed to The Flaming Sword.
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The Flaming Sword · Vol. 13, No. 43 · 15 Sep 1899 · PDF p. 12