S T E

448

STE

the seat of the line in the plane w, and its inclination tothe said plane

Figure 9.—Draw g q perpendicular to e f, meeting e fin Q ; in e P, make E T equal to E Q, and join t a ; draw T Eperpendicular to a t ; make t e equal to Q G, and join e a ;then will t a be the seat of the line in the plane w, and tseits inclination to the said plane.

Demonstration. —For when e S is made to coincide with e f,as in the last Problem, the plane fcg will be perpendiculaito the plane of the section w ; but the line f c being now inthe plane w, and Q G being perpendicular to F C, Q G will alsobe perpendicular to the plane w ; but the point G, that is e,is a point in the line, whose seat is a b, and the point a isanother point in the line, whose seat is A b; therefore, E and aare two points in the line, whose seat is A b ; then join a b,which will be the part of the line in space on the other sideof the plane W, and a t its seat.

Scholium . —This amounts to the same as when the seatand distance of the eye are given with respect to the originalplane, and the position of a point in the said plane; to deter-mine the seat and inclination of the visual ray.

Peoblem XIII.— Two straight lines, a b and c d, tendingto an inaccessible point, being given; through a given point, e,to draw a third straight line, to tend to the same inaccessiblepoint.

Figure 10.—From any point, A, in ab, draw a straightline, a e, from a to the given point e, cutting c d in c ; andthrough any other convenient point, b, draw b f parallelto A E, cutting CD in D ; find D f, a fourth proportionalto A c, C e, b d, and join e f ; then will the lines A b, c d,and e f, tend to the same point of concourse.

Scholium . —It sometimes happens, that a number of linesradiate from the same point, and that from a given point itis required to radiate other straight lines, so as to meet theradiations given in a given straight line : some, or many, ofthese radiations, according to their number, will be inacces-sible ; and though they may be all found by this Problem,yet if *he several operations are combined in one, muchtrouble will be saved.

Figure 11.—Thus, let the radiations be A b, a c, ad,a e, a f, tending to the straight line b c ; and let a be agiven point, from which it is required to radiate other straightlines to meet or tend to the same points in b c, with thosedrawn from a. Join a a, cutting b c in g ; and throughany convenient point, c, in b c, draw f f parallel to A a;make c p equal to g a, and CQ in b c, produced to o, equalto g a; join P q; let a d, a e, meet c f in d and e ; draw d m,Eit, fo, parallel to p Q, cutting b o at m, n, o; make c d equalto c m, c e equal to C n, cf equal to Co; draw da, e a, fa,which are the lines required.

Example 1.— Given the meridian, a b, in the plane of thehorizon, x, the latitude of the place, the intersection, in, tf/the plane, w, with the horizontal plane, x, and the inclinationof the plane, w, to that of the horizon x ; to construct a dialin the plane w.

Figure 12.—In a b, take any point, A, for the foot of thestyle ; then A b will be the seat of the style, or of the linetending to the pole ; the latitude of the place is its inclina-tion. Find a , the representation of a, that is, the section ofthe style of the dial in the plane w, by the eleventh Pro-blem ; produce A b to meet i N in D ; draw D a in the dial-plane w ; then, by the twelfth Problem, find a t, the seat ofthe style in the plane w, and the angle of elevation toe:and by the tenth Problem, find the intersections, do, t a,z a, &c., of planes passing through the style, making anglesrespectively of 15°, 30°, 45°, 60°, &c., with the vertical planepassing through the meridian, A B ; that is, with the plane

whose intersection is Tin found by the last Problem ;

are a,s °

the inaccessible lines ‘ d a s

_ _, and tlie dial is constru^,^ j S

required : a t is the seat of the style, whose intei . a ndt a; toe its inclination; a D is the 12 o’clock 1 ’

u a, t a, z a, &c., are the hour-lines. . _ . n rodU<*

the sub-style is thus . P r n d7 - - to » ’

Another method of findin^

GQ to meet PE in H; join ha, which produce p,.jn'draw x a, the sub-styler line. Thus, upon one con ’ n j je f 0 un<hciple, the sections of lines, planes, and solids, ma ^ ect i 0 nSThe sections of solids are found by means of the s ^ se C'planes ; and the construction of a dial is only fin -rivif ra eth ° Ations of planes, whose positions are given. I' _ an d tois, perhaps, of all others the easiest to consi econstruct. -f in

Example II.— Given the base, a x Q e, of apy ra ^ lines*plane x, and the whole seat, ab, of one of its an 9 inch'the intersection, i x, of the cutting plane W, an 0 j th enation of the planes w and x ; to find the secpyramid.

Figure 13

-Find the vanishing line, v l,

and the vanishing points, V and L, of the lines > t)e q ; produce a e to k, x q to i,an to J, an Jj ok0 } j, an®

meet the intersecting-line i x ; join k v and M .Imfd°insi stingl i,; then an qr will be the section of the py iaupon a x Q e. . t ]ie s<i ,1>e

Example III.— To find the section of a p r >things as before being given. T of

Figure 14.—Find the vanishing-points,

and

,nd J ^

an u

plane angle ; the three angles being 0 f th^ c j gt

the

developed, or

upon one plane. Here p f will be the intei sect'® . ne &ting plane with the plane of the base. n t j ie i nteI tr e-a b c d be the base of the pyramid paralle eaC b eX ^ jtion, e p ; find the eentre, e ; produce A D ^ E dr^ 0 .mity, to meet p f in F, and p G in P; thro ^ p G a t I > f ^parallel to A d, or b c, meeting p f at H, a " a t 1 > a ‘ e tduce b c both ways, to meet p f at K, a " uce a b t0 ea chex parallel to ab, meeting PGatNi P a nd N e rfep G at m, and dc to meet p G at o ; draw ^ heigh* 0 cU tperpendicular to p G, and in length equa j e t v 9 D . 0 ;

pyramid ; join GQ at i. Q ; also O b an ^ an d J* 1 ]i i,by G Q in g, i Q in i, L Q in l, o e in o, » » p * . join / i,

transfer p f, p h, and p k, to p/ r «» a . fq at ‘ t he

d0 J, cutting wl n be

" to

as

— a.— -— - o i. . it v an u

lines A J and a k, and the representations, »■before ; draw x. r parallel to J n, and M n para eqn a r will be the section of the prism require • py r<i .

Plate 3. Figure I .— To find the section of a / y th re®mid. Suppose a trahedral, or solid angle, C ° I1S,S eaC h otl ,er ’plane angles, two of which are at right angles plane

let the base of the solid be disposed in ° n ® i flne be t

which form the right angle, and let the eutting^i 7 ^ or q? ’plane angle which subtends the right angle; jjd is t0 n »be the plane angle on which the base or * * e jj n0 - perp e .placed; y, or g p g, the other plane an S le > /? ' j, y poth e ^.dicular to the plane x; and gpf, °n z ’, n r unf°*

<f0>'

; parallel to p f, draw m a and oa, cand k l at b and c ; then the quadrilatei asection required. , baS e of tn

Figure 2.— To find the section when me ^ *

mid is oblique . . . n L l, ^

Let a b c d be the base ; find the P 011 ^ 5l and »as before; produce a b and DC to naee j 0 0 , ’tr^ 9 '

US UCiUIC, piuuuuo Ai- D uixva AS r

fer o f, o h, o k, o m, and o x, to °/’ 0 ’ g t ’o F in V 0 th e *’the diagonal b d, and. produce it till 1 nl eet ea v et? eri {fer o t to ot, and produce o G and b plane ’ ^ 0 jot 0from the point of section draw a line, > ( w ith f'., c ut *

dicular to o g, cutting o g produced > J , . „ f an®section ; also join gf, ih, and Ik, and &