242

DE S

J;D. E: S

then, the line g h be drawn perpendicular to e f terminating,in G and H, at the traces of the two given planes, and meet-ing e/ in the point i ; this line will be the base of thetriangle intended to be constructed. In fact, let us supposethat the plane of this triangle turns on its base, gh, as ona hinge, to adapt itself to the horizontal plane; in thismotion, its apex, which was in the first instance placed onthe intersection of the two planes, continues in the verticalplane drawn through such intersection, because the verticalplane is perpendicular to GH; and when the plane of thetriangle is laid down, this apex will be found on one ofthe points of the line E f. It therefore remains only todiscover the heights of the triangle, or the extent of theperpendicular dropped from the point i, on the intersectionof the two planes.

“ But this perpendicular is comprised in the vertical planedrawn from E-to^j if, therefore, we conceive this plane torevolve about the vertical line f r, in order to apply itselfto the vertical plane of projection ; and if we carry f e from fto e, fl from f to i, the line ev will be the extent of theportion of the intersection comprised between the two planesof projection ; and if from the point i, the perpendicular ikbe dropped upon this line, it will give the height of therequired triangle.

“ Hence, by carrying i h from i to k, and completingthe triangle gk h, the angle in k will be equal to the angleformed by the two planes.

“ Seventh Question. — Figure 10. Two right lines inter-secting each other in space, being given by their horizontalprojections A b, a c, and by their vertical projections ah, a c;to construct the angle formed between them.

“ Before we enter on the solution of this question, we mayremark, that as the two given right lines are supposed tointersect each other, the point, A, of the coincidence of theirhorizontal projections, and the point, a, of the coincidenceof their vertical projections, will be the projections of thepoint in which they cross each other, and will, consequently,be in the same right line, a G A, perpendicular to L M. "Werethe two points a and a not in the same perpendicular toL m,the given right lines would not intersect each other, and ofcourse would not be in the same plane.

“ Solution. Conceive the two given right lines to beproduced so as to meet the horizontal plane, each in a point,and then construct these two points of coincidence. Toperform this, produce the lines a b and a c, till they cut L min the points d and e, which will be the vertical projectionsof these two points of coincidence. From the points d and e,draw upon the horizontal plane, and perpendicularly to l m,two indefinite right lines, d e and e e, which, as theymust pass through one of these points, will determinetheir positions by their intersections, i> and e, with therespective horizontal projections a b and a c, producedif necessary.

“ This done, draw the right line d e, which, with the twoparts of the given lines comprised between their intersecting-point and the points D and E, will form a triangle, of whichthe angle opposite to d e will be the angle required: wehave, therefore, only to construct this triangle. To do so,having dropped from the point A, the indefinite perpendicularA F, upon d e, conceive the plane of the triangle to turn asupon a hinge on its base de, till it lie flat on the horizontalplane ; the apex of this triangle, during its movement, willnot depart from the vertical plane described by a f, andwill at length apply itself in some degree upon the prolonga-tion of f a in a point, H, of which it remains only to find thedistance from the base D E.

“ Now the horizontal projection of this distance is the right

~‘- - --- - -; _ tre mi tlc "

line A f, and the vertical height of one of its e dingabove that of the other is equal to a G; hence, acc p-a#the property of Figure 3, if upon l M, a f be ineasu 1 ^^ |lU seG to f and if the hypothenuse a f be drawn, such hyp 0 car riedwill be the distance required. Finally, if a J D g p an“from f to h, and if from the point h the two l>n es ^ \h e be drawn, the triangle will be complete, and nbe the angle sought for. ,- ne

“ Eighth Question. —The projections of a righ *the traces of a plane, being given ; to construct

formed by such line and plane. ^ e giv' e °

“ Solution. Suppose a line perpendicular to r ;gtdplane to be drawn from a certain point in the gif 6 "line, the angle formed by such perpendicular with ^ an gl«>right line, would be the complement of the req u,rthe construction of which will resolve the question- t"'°“ Now, if upon the two projections of the rig j n ter 9eC "points be taken, in the same perpendicular with th ^ jj-jov'i'

tion of the two planes of projection ; and if h ne9 . g tra 1

ice 9

ctivey.^d

of the apex of th ' i

its sides, so that Sou « ,u for, and ab iu»» ...

Conceive the vertical °^ er side may be re P reseflte -;ln,: having drawn th P ane -° f P ro J ection to passtJ) e point a anv t Vert,cal indefinite line A f’. slI re, f

the vertical ’nrtlc ?■ ,nt> as ^ may be taken at P e . '

if f™, S S' r.K rf ,h t ° ! "»*

wake with flip h • ’ iine ds be drawn* ^

made by tL }**, an angle, d V A, eqf

the coincidence oJ. e WJtb the horizon, the po 111J 'f from the no' S,de with the horizontal p hnie '

‘he point d, the line do be drawn, so as ^ , d e

■ ^naltoth^A,

I ;

from these two points, perpendicular to the resp e p l * ^ (a lof the given plane, they will describe the hoi ^ aeS ti<^vertical projections of the second right line,will therefore be reduced to the construction ox ^ rl )l b 6formed by two right lines which cut each other, 8of the same nature with the former. t0 imaS'”

“It is usual, in projecting a chart of a country, h 11the remarkable points to be connected by means ot * c ha r _forming triangles, which are to be transferred or j e r 9on a smaller scale, but placed in the same relatij ^ nlA othose they represent. The operations necessary 1 ^ a ngj e ^on the earth consist chiefly of the measuremen g be* 1 ’®and of these triangles ; and in order to the a ”® t0 be ’ n edescribed correctly on the chart, they ought eac > pl^horizontal plane, parallel to that of the chart. ^ r«P^j,

of the angle be oblique to the horizon, it must tirb 1 ^

sented, but its horizontal projection must be ‘ ^le i 19 ^may always be found, if after measuring the ^ r izo°those angles which its two sides form with th e ^also measured. Hence we derive the -o

known under the appellation of the reduction ojthe horizon. r jirlit I" 1 [

“■Ninth Question .—The angle formed by tw ^ hot ,l0 °^ 0 ttand the angles formed by such lines with pi-oj eCplanes, being given ; to construct the horizonof the first of these angles. . .a pi'oj eC of

“ Solution.—Figure 11. Let Abe the horizon ^ 0 f o a

with the horizontal line an angle, d C A, efl'V n th e f° (

and ll rlefin ite ft «■the m den 0 ee l

ond side ca cof the ai_^ b et' v

by the second side with the horizon ;as from a centre, with the radius A C,a circle, c e f, be described, the secon- - arChorizontal plane only in the points of ‘jj gta iiceremains, therefore, only to ascertain thethis point and some other, as b. . e of s<> K

“Now this latter distance is in the p* dra' v, ’i oi’observed. If, therefore, the ri{

■ight

that the angle d d b may be equal to

of tl>ein the Phan® ^line d nb ser' e

the

angle'