Chap. VIII.

BRITAIN.

421

ten, to enable it to support the same weight on the crown as the catenarian, and the haunchof the segment aec is likewise subject to the same rule, and requires but seven, by reasonthe haunch is not so much elevated, which gives the weight a greater gravitation.

“ In like manner may the difference of weight be discovered in any two arches, which willalways be in proportion as the chord lines of their respective segments are to one another,or in proportion as the sides of the inscribed pediments are to one another, which impliesthe same thing. To shew the application of the steelyard to the nature and properties ofthe arch, let a b c represent the arch proposed: now if we suppose that the half archb c was in one stone, and its weight 28^ tons, and that it was by some power to be liftedupon its pier w , until it come to its centre of gravity, which may be supposed to be at d,then it is very plain that the pier must bear all its weight, and if from thence it was to belet down to a horizontal position as at x , and a support put under the end at o, thesupport owould bear 14] tons,and the pier w would bear the same weight; and was it fromthence to be lifted up to its proper place at B and there supported, then will the pier w bear20] tons, and the support at B bear 8] tons, for the end B being elevated, the support o isreleased of 6 tons weight,which consequently must beadded to the pier, for as thishalf arch is supposed to be inone stone, it may be consideredas a lever, and be subject to thesame laws of gravitation as hathbefore been demonstrated, andmay be explained by thefigure.

“ This method seems to me togive some insight whereby thereal weight of the abutmentsmay be discovered ; for if thesupport o was taken away, andthe other half arch, A B, wasset up against it, in its proper aposition, that must certainlythrow the whole weight of bothon their respective piers; butas A B is set at the same angleas B C, they would thrust each other down, if there was not an abutment placed at A andC to prevent it. Now, if we consider that before A B was set up against B C, thesupport o supported 8] tons, the thrust of A B must be equal to that weight, if therewas no friction at c ; and if so, it appears very plain to me, that a weight of 8] tons,placed at A and C, will be a sufficient abutment for the arch, ABC; for as such aweight is a counterbalance to the thrust, there will be the friction of A B C on thepiers, for a further security; which, considering the great weight of A B C, may perhapsbe equal to 2 tons more, but this must be referred to experiment.

“ And if the arch ABC was to be considered in parts as the half A B, the weightor pressure would still be the same; for drawing parallel lines, from the joints to thechord lines, plainly sheweth how the weight is diminished, as appears by the figure. Bywhat I have endeavoured to demonstrate, it appears very plain, that the centre which isto support the arch, ABC, will be released of 12 ton weight, in the thickness of thewhole breadth, which is supposed to be 40 feet, and the number being multiplied by 12,the product is 480, which being deducted from 2320, which is supposed to be theweight of the whole, the remainder is 1840, which will be nearly the weight, whenthat of the key-stone is deducted out, that will lie on the centre, before it can be relievedby the keystone. How much of the spandrills’ weight will press on the arch is very diffi-cult to discover, because it hath a communication with the upper pier or breast-work,which takes off a great deal of the pressure ; but if that communication be destroyed,and it be considered as one stone, and no friction on the arch, it could then be onlysaid to lean on the arch, and what weight that leaning is may in some measure be dis-covered by the same rule as the former; that is, to draw a line as d, in such a manner thatif the spandrill was lifted up on its lower point, until that line become perpendicular, itwould then be the centre of gravity, and if the line d was continued to cut the arch P,and from thence let fall a perpendicular to the chord line, which if divided according to theweight of the whole spandrill, it will give the weight that will press on* the arch if the lined be so drawn as the quantity of matter on both sides are equal, which ought to be regardedin the half arch, B C, as well as in the present case.

« This method will not exactly discover the weight or pressure of any arch, but it will be

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