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perpendicularly to touch the arch in h, and from v to r, andt to u, and draw the line r u, which halve in the point z, anddraw the line z y, parallel to v r, or t u. Then strain a line,or lay a straight rule from A through gr, towards a;; as alsofrom z through y, towards x, and these two lines will cut oneanother at x, from whence to the points v and t, draw thelines x v and x t. Now set off g h perpendicular to x t fromx to w, and from t to s, and draw the line s w, and divide g binto any number of equal parts at pleasure, from the divisionsof which, draw perpendiculars to g b, to touch the arch a o bbetween the points b and h, and divide v y and y t, the halvesof the base v t, each into the same number of equal parts asg b, is divided into: as likewise the base x t, and from thepoints of division draw parallel lines to touch the lines u rand s w. This being done, take the lengths of the lines thatwere drawn from the points of division of g b, perpendicu-larly to touch the part b h of the arch aob, and set them offin the correspondent parallels from y v to zr, and from y t toz u ; as likewise from x t to w s. Then, if at the extent ofeach line, as you set it off in the parallels, you stick in nails,as in the foregoing examples, and bend a thin rule about them,you will describe the sought arches v z t and w t, whereofv z t is the true intersecting arch, and w t the curve line ofthe groin that is correspondent thereto.

“ After the very same manner the arches km z and k p aredrawn.”

In his explanations of the diagrams, he is tolerably intelli-gent ; but he has departed from truth and reason in the twofollowing problems:—

“ The arch of a round tower, or any other circular build-ing, being given, icherein a semicircular window is to stand,how to find a centre, so that the mason or bricklayer shalltwin their arches thereon without crippling them. (SeePlate III. Figure 5.)

“ First, draw the arch a f b, from the centre e, to representthe arch line of the wall, and set the width of the windowfrom d to C, which halve at h, and draw the line l m, whichhalve at n ; from whence describe the semicircle l o m. Thisbeing done, divide the semi-diameter l n into any number ofequal parts, from the division of which, draw parallel linesto o n, the arch of the quadrant, which parallels continue outto divide the arch f c into the same number of parts asmis ; and from the points of division in the arch F C drawperpendiculars to the parallels, each equal in length to thecorrespondent parallel of the quadrant I. o; and from thepoints of the divisions of the line H C (made by continuing| out of each of the aforesaid parallels) draw right lines to theextreme points of the aforesaid perpendiculars, as from gto h. This being done, if the line g h be laid off in theparallel o N, continued out from h to i, and the rest ofthese lines last drawn be laid off in the respective con-tinuations of the parallels, the extreme points of theselines being joined, will form the curve C i; which, whenset in its due position, will hang perpendicular over thearch C F, having its points coinciding with the ex-tremities of the perpendiculars drawn from the divisionsj of the arch c f.”

Figure 6.—“ The centre whereon the arch of a bow-windowI is turned being given, how to find another centre that shall| answer parallel to it, according to the upper edge of thei surface of the arch. —First, describe the arch b k c, accordingto the directions laid down in the last problem, and set thewidth of the flat surface of the arch from b to A, and from cto d ; and draw the lines ad,bc, and halve them at f and e,from whence draw a perpendicular of a length at pleasure toh. Then in any convenient place ( Figure 6 , No. 2 ) draw a; line at pleasure, as from A to G, and from a draw to a g the

perpendicular a f. Then take e i, in No 1, and set it froma to b, No. 2, and f i from a to c. This being done, take thesemi-diameter b e, or e c, No. 1, and set it from A to d, No. 2.Also, take a b, or c d, and set it from d to e, and draw theline e c, which set in the line e h, from f to g. Again, takethe width of the flat surface of the arch A b, or C d, and setit in the line e h, from k to 7, and divide the remainder from7 to g, into seven equal parts. Also, divide the arch b k, intoseven equal parts. Then take k 1, in the line e h, betweenyour compasses, and setting one foot in 1 , with the otherstrike the arch 1 at pleasure: then take k 2 , and strike the arch2: also take k3,k4,k 5, and k 6 , severally, and strike thearches 3, 4, 5, and 6 . When this is done, open your coffl'passes, and divide from a to g, keeping the points of them onthose arches, till you have gotten seven equal distances froma to g; at the points of which, if nails be stuck in, and a thinrule be bent round them from a to g , along the edge thereof thearch A g may be drawn. And in like manner may the archD g be drawn.”

This description is so far intelligible, that we perfectlyunderstand his geometrical process ; but it is so void of truth,that no geometrical reasoning can be applied, unless it wereto prove the contrary of his assertion: the arch which wouldbe required to stand perpendicularly over such a plan upona semicircular centre, would not be in the same plane, whichis the case with the one he has found, and asserted to heright.

In the construction of the ribs of niches for plastering, heis extremely obscure, and takes only the most common andeasy cases ; such as might occur to every one, even to thosewho are not much in the habit of thinking, as the reader wmobserve in the following quotations:—

Figure 7.—“ How to form a semicircular niche with rib s ’as is usual when it is to be plastered. —First, describe thesemicircular plate a c b, as also the semicircular front n"a d b, equal to it, and fix the plate acb level in the pla cewhere it is to continue, and upon it set the front rib A ® 5perpendicular on a b. This being done, describe the quad'rantal ribs bc,de,df,dg, and d ii, each equal to a d or B P>and place them about 8 ^ inches from one another, on the pl ateA c B, as at c, e, f, g, and n, so as to meet in one point, at p ’.

on the crown of the front rib adb; and thus is one halfAnd after the same manner may

the work finished,other be done.”

Figure 8 .—“ How to form an elliptical niche, with f*for plastering. —First, describe Figure 8 , No. I and 2, kbeing a semi-ellipsis, representing the plate whereon the r*stand, and being equal to A d b, or a e b. The pricked h 1 ’ .L n, l o, i, p, i. q, x, r, and L m, represent the base lines of thee d,/d, g d, h d, i d, and bd; so likewise do the lines S t,S ’s v, s w, s x, and s y; and the perpendiculars a t,bu,c v,d ’e x, and f y, do represent the rising of the ribs e V,f Pi 3 ’h D, i d, and b d, which is equal in length to CD; obser'' 1 ^that within those lines the different arch of each rib is todescribed, viz. the arch s a is a quadrant of a circle, ha^®

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t for its centre, and is equal to the arch of the rib e d ■-lines s u, s z, equal to zb,b u, are the semi-transvetseconjugate axes of a semi-ellipsis, whose arch, s b, is equa^ 3the arch of the rib f D, which may be struck either ' V1 -

trammel, or by the intersection of lines. Moreover, the J ^& z,sv, equal to v c, c z, are the semi-transverse and conju- ^axes of a semi-ellipsis, whose arch is equal to the arch orib g d ; and so of the rest. p p

“ Now having the ribs all ready, set the front rib, ^perpendicular on the plate a e b, as at a b, and fix tb® ^of the short ribs on the plate acb, as at e,f g, k, h ' v .^3correspond with the points n, o, p, q, r, and their P

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