Circular Roofs may be covered upon two different principles

First Method

Assume the vertical section, or axis, to be divided into a number of equal parts, and the roof, or figure, cut by planes through the points of division parallel to the base; and then consider the portions of the figure as so many frustrums of a cone; the surface of each frustrum can then be determined as by fig. 26, page 26.

Second Method

Divide the circumference of the base into a number of equal parts, and assume sections to be made perpendicular through these points of division; then estimate the surface of each of these di-visions on the surface of the figure.

Fig. 26. To Cover a Dome by the First Method.

Let ABC, fig. 26, be the section of a dome. Draw the axis DB; produce to J; divide the curve of one-half the figure into equal parts, as EFG and H, the width of these divisions being the width required by that of the metal with which the dome is to be covered; produce AE, EF, FG, GH, and HB, severally until they intersect the axis BD; then [for example] from the point I, with the radii IG and IF, describe the curves GM, FN; then set off that portion of the circumference of the base FL required for a pattern to cover the course FG.

In the same manner, the covering for the other portion can be found.

Fig. 27. To Cover a Dome by the Second Method.

Let ABC, fig. 27, be the section of a dome; then the length of a course of covering is obtained as follows: The length of the course BF is equal to the curve AB, and EG the breadth of it; join ED, and the lines 1, 2, 3, and 4, intersected thereby, will bo the half breadth (for the vertical BD) of the course at the corresponding lines on BF, through which points a line can be drawn which will give the form of the course required.

Fig. 28. To ascertain the Outlines of a Course of Covering to a Dome, without reference to a Section of the Dome.

Let AD be the breadth of the course. Bisect it at B by the perpendicular CE; make BE equal to the length of the arc from the base of the dome to the top of it (which may be found either by measurement or calculation); divide the semi-circle ACD into any number of equal parts, and draw the lines parallel to BD; divide BE into the same number of equal parts, and draw lines parallel to AD; mark ordinates on each side of BE; as 1, 2, 3, and 4 equal to the lines of BCD, and a curve drawn through their terminations 1, 2, 3, and 4 on both sides will give the outline of the course

Erect the perpendicular line GE; draw the line AB at right angle to GE; make EF equal to the Slant height, and draw the line CD parallel to AB; make AB equal in length to one side of the base; make CD equal in length to one side of the top or smallest end, draw the lines AG and BG, cutting the points AC and BD, G as a centre with the radii GC and GA. Describe the arcs KM and JI; set off on the arc JI, JA, BH, and HI equal in length to AB, and draw the lines JG, HG, and IG, also, the lines JA, BH, HI, and KC, DL, LM.

Edges to be allowed.