DE S

241

DE S

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Plane, iVe trace °f the same plane upon the horizontal“If •

for, a ?j lns tead of a horizontal right line in the plane soughtC( 1 I > 8 tru .!• P ara llel to the vertical plane were conceived, thefitted !° n Showing, by parity of reasoning, would be“]} a :

r >glit ]j avv from the point G, parallel to lit, the indefinitePfotJu j e G d ; from the point g, draw g h parallel to c b,lm in the point n, whence draw h da p. ( 1 . < j U ar to l m : this latter will cut g d in d, whence^ a ,“ el b e drawn to a b, one of the traces of the plane

h,

ba<: e or he obtained ; and if, having produced thislv e th ,flee ^ r ‘ M in e, b p be drawn parallel to bc, we shall'• 'j'fo lrace on the vertical plane.

a a * Question. — Figure 6 . A plane whose two traces

1 * n g a' ^ Bc > an< I a point whose two projections are D, d,9 6 f a {jj Ven ’ to construct, 1 , the projections of the right*> the n r n -" Pp r P en< licularly from the point upon the plane;^•th ti °J«ction cf the point of coincidence of the right line“ ane -

. 'tits n t0n ' The perpendiculars d g, dg, falling from thef h‘e upon the respective traces of the plane, will

1 j n ** e projections of the right line required : forVli Plan ' le . Pedicular, a vertical plane be conceived,

Poi

j 1 * h v o ri^i ' Vl .h cut both the horizontal and the given planes,l(; hnes, both of them perpendicular to the commonof bein»’ A B ’ l * ie two planes : now, the first of thesei " lp r> ® t *' e Projection of the vertical plane, is also that, 'o n ‘p'ifudicular which is included ; therefore the pro-* this

^ be nl 11113 Perpendicular must pass through the point d,

- Tl Perpendicular to a b.

|lf «j-, c ,io n Sat “e demonstration will serve for the verticaltli to th

t j® Plane ■ P°' nt of coincidence of the perpendicular with

ti °f tl'ii* ev ’d en t that it must be found at the intersec-P ;uie , with the vertical plane drawn along theiii n ’ suc h intersection being projected indefinitely

i-^ersecti "y °btaining the vertical projection,ye, of thisli^'fed , ’ s b a ll find it to contain that of the pointft it m 88 t ' lls P°’ nt must be projected upon the rightO'* d 9 T *" oun d at the intersection, < 7 , of the lines\'f e. ‘ remains, therefore, only to discover the rightO^'l Pla' V> t *' e intersection of the given plane with theI'l'o ' 1 ^'e i)" 6 ’- w b‘cl> are perpendicular to each other, willii^ Je cti 0n ^ 0I izon tal plane in the point E, whose verticaln l. Ax ’ "dl be found by dropping e e perpendicularlyfoil,/ .... ,t; "'ill meet the vertical plane of projection

'ti

Hi

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ni e

1 ’ ' V tiAU V VI AIVUA Itiliv VI J'l 'fJOVA'Vi.

j lose horizontal projection is the intersection ofariq ' VHl p r , ■ 'T lt b D G, produced, if necessary, and whoseti. . dip .. Action must be at once upon the vertical line f f

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•i.jbtter^.®. CB » °f course, it will be at the point,/, of

ip ^ projection,^, of the foot of the perpendicular

; f 0 ’ r ' e construction of its horizontal projection willtl,~ xr uy dropping the indefinite perpendicular g G

be

tl,

' e h • ’ a r' 1 . Pl ,ll, o

'via ' 1 ' 1 rpql •'t will be obtained, which will containbglit, b p - fon "1 e< * : an< * as tbe line D f must also contain it,

UndQ Ue

at the point, G, of intersection of these two

a es tion.—Figure 7. A right line whose two

bein~ A B ’ a al 'd a puiuf whose two projectionsg'ven ; to construct the traces of a plane

yti f ’ u eing‘ $ 'otn t | 0

\t \\* ut io n perpendicularly to the riglit line.

t Wo ‘ t e have seen from the preceding question,‘'°Ms „(• '. lc<is must be perpendicular to the respective^ 0ll| ta e „ ,! e right lines: it remains to he discoveredc 1 ot them ought to pass through. For this3 q

purpose, let an horizontal line from the given point be con-ceived in the plane required, produced so as to meet thevertical plane of projection, and we shall find its verticalprojection, by drawing the indefinite horizontal d g throughthe point d, and its horizontal projection by drawing a per-pendicular to A B, through the point D, produced till it cutL m in h, which will be the horizontal projection of the pointof coincidence of the horizontal with the vertical plane ofprojection. This point of coincidence, which must be foundin the vertical line h g and the horizontal line d G, and con-sequently at the point, G, of the intersection of these twolines, will be one of the points of the trace on the verticalplane ; we shall then find this trace by drawing the line F C,from the point g, perpendicular to a b; and if from thepoint C, where the first trace meets LU, C E be drawn per-pendicular to a b, we shall have the second trace required.

“ The same process would discover the point of coincidenceof the plane with the right line.

“ Were it necessary to drop a perpendicular from thegiven point upon the right line, we should construct, as hasjust been described, the coincidence of the right line withthe plane drawn by the given point, and which would beperpendicular to it; and we should obtain, from each of thetwo projections of the required perpendicular, two pointsthrough which it must pass.

“ Fifth Question.—Figure 8 . Two planes being given inposition, by means of their traces a b and A b for one, andC d and c d for the other ; to construct the projections ofthe right line upon which they intersect each other.

“ Solution. All the points of the trace a b being found inthe first of the two given planes, and all those of the tracec d being found in the second, the point, e, of intersectionof these two traces is evidently in the two planes, and isconsequently one of the points of the required right line. Inlike manner, the point, f, of intersection of the two tracesupon the vertical plane, is also another point of this rightline. The intersection of the two planes is therefore so !

placed as to meet the horizontal plane in e, and the verticalplane in F.

“ If, therefore, the point f be projected on the horizontalplane, which may he performed by dropping the perpen-dicular f f on 1. m, and if the line /e be drawn, it will be thehorizontal projection of the intersection of the two planes.

So, if the point E be projected on the vertical plane, bydropping the perpendicular E e on lm, and if the right linee f be drawn, it will be the vertical projection of the sameintersection.

“ Sixth Question. — Figure 9. Two planes being given, bymeans of the traces A B and A b for the first, and c D and c dfor the second; to construct the angles formed by them.

“ Solution. Having constructed, as in the preceding ques-tion, the horizontal projection, e/ of the intersection of thetwo planes ; by conceiving a third plane perpendicular tothem, and consequently perpendicular to their common inter-section ; this third plane will cut the two given planes intwo right lines, containing between them an angle equalto the one required.

“ The horizontal trace of this third plane will be perpen-dicular to the projection, e/ of the intersection of the twogiven planes, forming with the other right lines a triangle,of which the angle opposed to the horizontal side will be theone required. It remains, therefore, only to constructthis triangle.

“ It is quite indifferent through what point of the inter-section of the two first planes the third passes ; and we areat liberty to mark its trace upon the horizontal plane, atpleasure; provided that it be perpendicular to E / Suppose.