PRO
331
PRO
t^g 8 ^ c .d e • This may be done as in Problem I., Method Y.**<> ax artlc le Ellipsis (Yol. I., page 365), or by finding theas ,n Method I, (page 365), of the same article.
V an XII .—Any two conjugate diameters, b d and e C,
l P Sls , being given ; to find the representation of the
Jp. '
^ a ’"- —Find the representations, bd and e c, of the
b' M,, eter B D ant * E c 5 and b d and e c will also be conjugate. lp sis S /Tr**, en i l) y Problem I., Method L, of the article
B D and e c ;
s s ’^hen, by Problem I., Method L, - ...-
U ’ c h Wli °^' P a S e 365), describe the ellipsis bcdeb,^V>bl ' 3e t * le re P r(, sentation of bcdeb, as required.*j! a ce, i<p ^ XIII.— To find the projection of any point in
^Ce
Plft;
? a plane given in position; and to determine theJ trie noint fnrnm. the nlnriP
- lfle P°i n t from the plane.
ne :
Make choice of any three points, a, b, c, on the
j! th e D ? asure the distance of the point in space from each„ to t» ( , ac j| lnt;s A, b, c, and the distance of the points A, b, c,le/f'tk* 1 ' ’ t ^ len t ^ ie P r °jection may be found on paper,
“'Us. 1
t 6 be ^ ^ R be a scale, representing feet and inches ; letfoot 2 ^ Sure< f 1 foot 4| inches, bc 1 foot 1^ inch, and caj Be Se dim lnc ? les » describe the triangle, a b c, according toJ! 8 pace j- ens ' ons from the scale; let the distance of the pointt ' st| bi Ce J?° 1 m A be 1 foot 4 inches; with the centre, A, andjt^ce of ti * ^ 00t ^ ' nc bes, describe an arc at d ; let the dis-hes ! th ' e P°’ nt ’ n space from B be measured 1 foot 3^ eri > with the centre, b, and distance of 1 foot 3■ ^ B n. Scnhe an arc cutting the former at D, and join A D
et th e the same distance, from b, describe an arc at>°t ; ,,“'stance of the point in space from c be measuredrib e an 60 ’ t ^ le centre c, and distance of 1 foot,
arc > cutting the former at E, and join b e and c E;ill f r P en( f'cular to A B, and up perpendicular to b c ;
7*111
’ w ul bp ft -• ’ r
0 /J* j , tBe projection of the point in space.
</ b(l t} ifi * _/> ,7. __ _ • ♦ _ T
fl t j f , ' A ^ te altitude of the point in space. —Let d p cutp H perpendicular to v p; with the centre f,
^<W draw
Ml . 1 st an,
* b e t /' c ® F D, describe an arc, cutting p n at n ; and p hof the P oint required.
if '• Qn g l e . ^ — Supposing a tetraedral, or three-sided
\fi r °j e ctio ° ^ P^ ace d with one of its sides upon the plane
I tiJ ln at; n ° n ’ an d one of the other two sides, and the two
fide.ifiyns J. ... ... , , .... .
1 tj > b,
, e fidjoining that side, with the intersection of thaty en : to find the projection of the given side, and
\
p C l7 Ue
'‘jfitirjft ctl °n of the remaining side, upon the plane oft -Uiig ■
li'! U(ljoi r . ’ - 11 ) ’ s the sam e as if one of the sides, and thef be p ro - n ® inclinations, of a tetraedral, were given ; to•UaH* 0 fiod'n, Ct '° n t ^ le g‘ ven s 'de on one °f the ot bers, and“e. he side of the solid on which the projection isHifree.
er ^et a b c be the given side, and A b the edgett't J °int, Q 6 . 0ts or meets the plane of projection in i n ; from^ m b c, draw c d perpendicular to IN, cutting INa ,° th® • I> ’ ,^ le projection of bc; that is, make dal°juctio n lnc ^ nat ' on °f the given plane towards the planetb to Or), ^ake al equal to A c ; draw l d perpendicu-! A * Join,i fi n j e ^ uir ed\ h °jecn ndt >‘
I C|,t tincf^ e an y Point) <5; and draw gf perpendicular to
!^ Slt tt. 0 ^ B atF -
D b ; and dba will be the projection of
1 eu,
te intersection of the other side, with the plane of
s I , — * ) draw F I perpendicular to B D,
. 1 the centre f, and distance, f g, describe an
cutting
th j, ” llle centre f, and distance, f g, describe an-tit,,." °thp 5 j. • Bat K ; join fk ; make the angle, f k i, equalop . mclination ; join b i ; and b i will be the inter-other piA
Xv.—,
*0h
i“h °f the ’ J ul “ “ x »
XV 61 S ' de ’ as required.
C a. nt V ''—Given the representation, a, of a point,* ane i whose original is given ; to find the repre-
sentation of a line from the original point, perpendicular tothe original plane.
Plate IV. Figures 1, 2.—Draw G k perpendicular to theintersecting line, and make G K L equal to the inclination ofthe plane to that of projection, if not already made; draw a Lperpendicular to G K, and L M perpendicular to kl; makel m equal to the line ; draw a b parallel to G k, and m b par-allel tola; then a b will be the projection x-equired.
Problem XVI.— Given the representation, a, of the inter-section of a line with the original plane, and the seat in, andinclination of the line, to the original plane; to find therepresentation of the line.
Figures 3, 4.—Draw gk perpendicular to in; make theangle g k l equal to the inclination of the plane to that ofprojection, and produce k l to m, if necessary; draw a aperpendicular to IN, and c b parallel to A a ; make K M equalto C d ; draw M p perpendicular to K L : make M P equal tothe tangent of the seat, ac, made radius ; draw c b parallelto A a, and F b perpendicular to A a ; then join a b, and a bwill be the representation of the line required.
Problem XVII.— Given the representation , ab c d, of oneside of a parallelopiped, with the inclination of the plane, andthe other dimension perpendicular to the original of thatplane; to find the representation of the whole solid, supposingthe planes of the original solid at right angles to each other.
Figure 5.—Find the representation, afi of a line perpen-dicular to the original of the plane represented, by ProblemXV., and complete the parallelograms afgd and ab efwhich will form the representation of the whole solid.
Problem XVIII.— To find the inclination of two adjoiningplanes of the regular solids.
Plate V. Figures 1,2, 3, 5.—Draw the base, ab c, a b c d,or a b c d e, of the pyramid which forms one of the solidangles ; draw the triangle bgc, the development of one ofthe sides of the pyramid ; draw G F perpendicular to BC;bisect the angle abc, and let the bisecting line meet G f inf, their point of concourse ; draw F H perpendicular to b f ;from b, with the radius b g, describe an arc, cutting f h atH ; draw F I perpendicular to b h, cutting b h at I; in fbmake f i. equal to f i; let f h and b c meet in m ; also letfh and b a meet in k; join m l and k L ; then will klube the angle of the planes, as required.
As the sides of the tetraedron are equilateral triangles, andas three of the triangles form each solid angle, both the baseof the pyramid, abc, and the developed side, bgc, will beequilateral triangles in Figure 1.
In the construction for Figure 2, as the sides of the hexa-edron are squares, and as each solid angle is formed by threeof the right angles, the base of the pyramid will be anequilateral triangle, and the vertical angle of the developedside will be a right angle.
In the construction for the octaedron, Figure 3, as eachside is an equilateral triangle, and as four of the angles of thetriangle form the sides of the solid angle, the base of thepyramid will be a square, and the developed side will be anequilateral triangle.
In the construction for the dodecaedron, Figure 4, No. 2,as each solid angle is formed by three planes, and as each ofthe planes is a pentagon, the base of the pyramid will be anequilateral triangle, and the vertical angle of the developedside will be equal to the angle formed by the two sides of apentagon.
In the construction for the icosaedron, Figure 5, as eachsolid angle is composed of five sides, and as all the sides ofthe solid are equilateral triangles, the base of the pyramid isa pentagon, and the developed side an equilateral triangle.
In the construction for the dodecaedron, Figure 4, the