diminish as the length of the line increases; that is, the pressure on any point would be inversely proportional to the length of the line. Therefore, if we suppose the thrust of the dome to be resisted by a force acting all round the base, and tending every where towards the centre, the thrust upon each point of the line equal to A B, will be to that on each point of the base inversely, as that line is to the circumference of the base: that is inversely as the semi-diameter of a circle is to its circumference. But the radius being 1, the circumference is 6.283. Consequently, to find the horizontal thrust on every point of the base in the above example, we may say 6.283:1::500 tons : 79 tons, nearly the thrust required.

If the dome be supported on a circular wall of masonry, whose height is A G, the above thrust must be multiplied by the arm A G of the lever, at the end of which it acts, and the product will express the force with which the dome endeavours to overset the wall. This, of course, must be resisted, as in the case of common vaulting, by a force expressed by the area of the section A D of the wall, multiplied by the half breadth D H (supposing the wall and dome constructed of materials having the same

specific gravity, and the pier rectangular), and as all the terms are given, except DG, this may also be found.

If the dome is intended to be hooped with iron at the base as usual, and we would give such dimensions to the ring that it shall resist the strain, we have only to find from the tables that are published, how much a bar of iron of given dimensions (whose section is one square inch for example), will support, without breaking, when a load is diffused uniformly over it; then the horizontal thrust divided by this tabular number, will give the number of square inches in a section of the intended ring.

THE END.

G. Woodfall, Printer,

Angel Court, Skinner Street, London.