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RM2AKR716–Descriptive geometry . ion when the plane shallhave been revolved about either of its tracesas an axis to coincide with a coordinate plane. Principle. This is identical with the prin-ciple of Art. 38, page 25, since either trace ofthe plane is an axis lying in a coordinate plane,one projection of which is the line itself, andthe other projection of which is in the groundline. Method. See Art. 38, page 25. Construction. Fig. 79. ffiVandFiVarethe traces of the given plane and 5 and 5% theprojections of the point. If the plane withpoint h thereon be revolved into H about HNas an axis, the point w
RM2AKFMX0–Descriptive geometry . Fig. 196. F-trace of the line (Prob. 1, § 37). The actual trace, f, lies inVQ. The projection A is now determined, since two points,sv and f, are known. Similarly, the projection Ah may be found if Av is given.. Fig. 19 A second example is given in Fig. 197. The lettering andexplanation are the same as for Fig. 196. Special Case I. Suppose that the given projection is par-allel (a) to the corresponding trace of the plane; or (b) to theground line. In either event the line should be recognized asone of the principal lines of the plane (§ 99), parallel to H orV as the case
RM2AKH6N2–Descriptive geometry . r adopted, whenapplied to an actual object, always gives a view of its rightside; whereas a view of the left side may be decidedly prefer- a+ -> ef+- + a° Fig. 84. able. For this purjiose, P is viewed from its left side, that is,from left to right, and then projected so as to bring this sideinto the drawing surface. This method is shown for a point inthe first quadrant in Fig. 84. That this method is directlyopposite to that previously used is shown also by the line ab,Fig. 85. Since the H- and F-prejections of a and b show that z 1 / 3 4 t 1 / iIi bh- — i / i i bv- b
RM2AKBB44–Descriptive geometry . and base, thedistances l-2, 2-3, etc. being equal to 1*2, 2P3D, etc. Re-placing the development of this base in line with its pro-jections, as shown, the points 12, 13, 14, 15 may be obtainedby projection. The points 19 and 20 are obtained from thepositions of 19* and 20p. The development, Z, of the tangent line is obtained bylocating in the development the triangle 4-14-i. Case III. The base of the cylinder does not lie in II, V or P. Construction (Fig. 309). The given cylinder lies in the thirdquadrant, and is intersected by the plane Q in the curve 11-12-13 • • • 18-1
![Descriptive geometry . Fig. 143. Construction (Fig. 143). Let it be required to pass the planethrough the line A. Assume any point con J; through c drawthe line D parallel to B. Pass the required plane Q throughthe lines A and D (Prob. 6, § 106).. Fig. 144. X, § 107] PARALLEL LINES AND PLANES 85 Special Case I. Suppose the second line is parallel to IIor V; then no auxiliary line is needed. Thus, in Fig. 144, letus find the plane which contains the line A and is parallel toB. The .ff-trace of any plane which is parallel to B must beparallel to Bh (§ 104). Hence, find the traces of A. DrawHQ th Stock Photo](https://c8.alamy.com/comp/2AKG9HY/descriptive-geometry-fig-143-construction-fig-143-let-it-be-required-to-pass-the-planethrough-the-line-a-assume-any-point-con-j-through-c-drawthe-line-d-parallel-to-b-pass-the-required-plane-q-throughthe-lines-a-and-d-prob-6-106-fig-144-x-107-parallel-lines-and-planes-85-special-case-i-suppose-the-second-line-is-parallel-to-iior-v-then-no-auxiliary-line-is-needed-thus-in-fig-144-letus-find-the-plane-which-contains-the-line-a-and-is-parallel-tob-the-ff-trace-of-any-plane-which-is-parallel-to-b-must-beparallel-to-bh-104-hence-find-the-traces-of-a-drawhq-th-2AKG9HY.jpg "Descriptive geometry . Fig. 143. Construction (Fig. 143). Let it be required to pass the planethrough the line A. Assume any point con J; through c drawthe line D parallel to B. Pass the required plane Q throughthe lines A and D (Prob. 6, § 106).. Fig. 144. X, § 107] PARALLEL LINES AND PLANES 85 Special Case I. Suppose the second line is parallel to IIor V; then no auxiliary line is needed. Thus, in Fig. 144, letus find the plane which contains the line A and is parallel toB. The .ff-trace of any plane which is parallel to B must beparallel to Bh (§ 104). Hence, find the traces of A. DrawHQ th Stock Photo")
RM2AKG9HY–Descriptive geometry . Fig. 143. Construction (Fig. 143). Let it be required to pass the planethrough the line A. Assume any point con J; through c drawthe line D parallel to B. Pass the required plane Q throughthe lines A and D (Prob. 6, § 106).. Fig. 144. X, § 107] PARALLEL LINES AND PLANES 85 Special Case I. Suppose the second line is parallel to IIor V; then no auxiliary line is needed. Thus, in Fig. 144, letus find the plane which contains the line A and is parallel toB. The .ff-trace of any plane which is parallel to B must beparallel to Bh (§ 104). Hence, find the traces of A. DrawHQ th
![Descriptive geometry . Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givendirection ahbh. Find th Stock Photo](https://c8.alamy.com/comp/2AKGMYX/descriptive-geometry-fig-101-repeated-construction-first-method-fig-100-reverse-offirst-method-prob-3-78-let-the-angle-with-h-andthe-direction-of-the-if-projection-be-given-let-the-point-a-bethe-fixed-end-of-the-line-viii-83-true-length-of-a-line-59-place-the-line-in-the-position-ac-avcv-ahch-parallel-to-vmaking-avcv-equal-to-the-true-length-of-the-line-at-the-givenangle-with-h-and-sloping-upward-or-downward-from-a-asgiven-take-an-axis-perpendicular-to-h-through-the-fixed-end-aand-revolve-the-line-until-the-projection-takes-the-givendirection-ahbh-find-th-2AKGMYX.jpg "Descriptive geometry . Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givendirection ahbh. Find th Stock Photo")
RM2AKGMYX–Descriptive geometry . Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givendirection ahbh. Find th
RM2AKFCGW–Descriptive geometry . the revolved position of a line lying in aplane, when the plane is revolved into H or V about the corre-sponding trace. Since a straight line is determined when two of its pointsare known, this problem can always be solved by revolvingtwo points of the line by the method already given. However,a simpler method can usually be found, as in the followingexamples. (1) Let the line A, Fig. 221, which is to be revolved aboutVQ into V, intersect VQ in the point t. Revolve any point ofA, as c(ch, cv) to cr. If now we attempt to revolve the point t,this point, being on the axis o
RM2AKPHM7–Descriptive geometry . B tangent to theparallel, and, therefore, in the plane of theparallel. That portion of A^ lying withinthe circle, which is the horizontal projectionof the ellipsoid, will be the horizontal projec-tion of the meridian drawn through point e.Revolve this meridian line about fk as an axisuntil it coincides with the principal meridian(Art. 102, page 70), the vertical projection ofwhich will be shown by the ellipse. The re-volved position of the vertical projection ofpoint e will now be at e^^ and a line may bedrawn tangent to the meridian at this point,its projections being A
RM2AKGRK5–Descriptive geometry . lution. Since a*is the projection of the point a in space, the distance, aah, ofpoint a from the axis is equal to the distance of the point afrom H, which shows in the ^projection as the distance from avto GL. Hence to find the revolved position ar, make ahar per-pendicular to ahbh, and the distance ahar equal to ave. Simi-larly, bhbr is perpendicular to ahbh, and is equal to bvf. Thenarbr shows the true length of the line ab. In Fig. 104 the F-projection is taken as the axis, and theline is revolved into V. The distances avar and bvbr are thedistances of the points a an
![Descriptive geometry . Fig. 19 A second example is given in Fig. 197. The lettering andexplanation are the same as for Fig. 196. Special Case I. Suppose that the given projection is par-allel (a) to the corresponding trace of the plane; or (b) to theground line. In either event the line should be recognized asone of the principal lines of the plane (§ 99), parallel to H orV as the case may be. Thus, in Fig. 198, if Ah is given parallelto HQ, Av is parallel to GL; or if Av is given parallel to GL,then Ah is parallel to HQ. The line A in this case is a horizontal XIV, § 133] A LINE IX A PLANE 12 Stock Photo](https://c8.alamy.com/comp/2AKFMF9/descriptive-geometry-fig-19-a-second-example-is-given-in-fig-197-the-lettering-andexplanation-are-the-same-as-for-fig-196-special-case-i-suppose-that-the-given-projection-is-par-allel-a-to-the-corresponding-trace-of-the-plane-or-b-to-theground-line-in-either-event-the-line-should-be-recognized-asone-of-the-principal-lines-of-the-plane-99-parallel-to-h-orv-as-the-case-may-be-thus-in-fig-198-if-ah-is-given-parallelto-hq-av-is-parallel-to-gl-or-if-av-is-given-parallel-to-glthen-ah-is-parallel-to-hq-the-line-a-in-this-case-is-a-horizontal-xiv-133-a-line-ix-a-plane-12-2AKFMF9.jpg "Descriptive geometry . Fig. 19 A second example is given in Fig. 197. The lettering andexplanation are the same as for Fig. 196. Special Case I. Suppose that the given projection is par-allel (a) to the corresponding trace of the plane; or (b) to theground line. In either event the line should be recognized asone of the principal lines of the plane (§ 99), parallel to H orV as the case may be. Thus, in Fig. 198, if Ah is given parallelto HQ, Av is parallel to GL; or if Av is given parallel to GL,then Ah is parallel to HQ. The line A in this case is a horizontal XIV, § 133] A LINE IX A PLANE 12 Stock Photo")
RM2AKFMF9–Descriptive geometry . Fig. 19 A second example is given in Fig. 197. The lettering andexplanation are the same as for Fig. 196. Special Case I. Suppose that the given projection is par-allel (a) to the corresponding trace of the plane; or (b) to theground line. In either event the line should be recognized asone of the principal lines of the plane (§ 99), parallel to H orV as the case may be. Thus, in Fig. 198, if Ah is given parallelto HQ, Av is parallel to GL; or if Av is given parallel to GL,then Ah is parallel to HQ. The line A in this case is a horizontal XIV, § 133] A LINE IX A PLANE 12
RM2AKMF6E–Descriptive geometry . A is the generatrix of an hyper-boloid of revolution.. A is the generatrix of an hyper-boloid of revolution.
RM2AKMXX9–Descriptive geometry . Problems 1-4. Draw the traces of a plane which is tangent at point a of the surface. (Art. Ill, page 78.) Problems 5, 6. Draw the traces of a plane w^hich contains point a and is tangent to the surface at the given parallel. (Art. 112,page 79.) Problems 7, 8. Draw the traces of planes tangent to the sphere and containing line A. (Art. 114, page 80.) Unit of measure, | inch. Space required for each problem, 7 x lO inches. Angles bet^veen GL and traces ofplanes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type PLATE 26
RM2AKN45X–Descriptive geometry . 33 28 Problems 1-4. Determine the true size of the diedral angle between planes N and T. (Arts. 83-85Problems 5-8. Determine the true sizes of the diedral angles of the objects. (Arts. 83-85, pages pages, 57,57, 58.) Unit of measure, J inch. Space required for each problem, 2^ x 3 inches. Angles between GL and traces of p| A-fC 1Qplanes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. »-A It. 1 O l4 2 8 4 ^-^ ^^ J^^ 7 6 12/ 6 12/^ ^ 78 . 8 /16/ / ^y^ ^^ S 9 vs . 10 11 I^S , 12 HS ^^ c * 0 //S ^ 1 Determine the an
RM2AKPG9T–Descriptive geometry . 88 DESCRIPTIVE GEOMETRY development has been found. The accuracyof the development depends upon the numberof elements used, the greater number givinggreater accuracy. The development of the curve of intersectionis obtained, as in the pyramid, by laying offfrom the apex, on their corresponding ele-ments, the true lengths of the portions of theelements from the apex to the cut section,and joining the points thus found. 124. To determine the curve of intersectionbetween a plane and any cylinder. Principle. A series of auxiliary cuttingplanes parallel to the axis of the cyli
RM2AKHEJ1–Descriptive geometry . tive position of plan and eleva-tion is usually prescribed by the custom of the office, so thatit is known by the position on the sheet which view is tobe read as an //-projection, and which as a P-projection.In studying the theory, however, this is not the case ; objectsmay be placed in any position in any quadrant, and someindication as to which is the l/-p rejection and which the V*projection must always be given. Forthe present, we shall do this by the nota-tion, using the index letters A and • (§ 12). 25. Illustrative Examples. The stu-dent should visualize the soli
RM2AKFY0B–Descriptive geometry . rojections are, however, usuallyconsidered a part of the problem, and are obtained bj project-ing from the profile view. 106 DESCRIPTIVE GEOMETRY [XII, § 121 121. The Projection of a Point or Line on a Plane. The projection of a point on a plane is the foot of the perpendiculardropped from the point to the plane. This definition is notconfined to the coordinate planes of projection, but applies toany plane in space. However, when a point is projected on tosome oblique plane represented by its traces on H and V, theprojection must in turn be represented by its projections
RM2AKPAX4–Descriptive geometry . lements, of the cylinder. This linewill be common to all auxiliary cutting planes,and its horizontal trace, b will be a pointcommon to all their horizontal traces. JTNisone such trace which cuts, or is tangent to,the base of the cylinder at t?^ and cuts thebase of the cone in d^ and e Since these arepoints in elements cut from the cylinder andcone by the auxiliary plane iV, the horizontaland vertical projections of the elements maybe drawn. Line U will be the element cutfrom the cylinder, while da and ea are tlieelements cut from the cone. The intersectionof these elem
RM2AKFP19–Descriptive geometry . Fig. 193. Vir,. mi. Example 4 (Fig. 194). Given a sphere and a circular cylin-der. In this example it may not be so easy to vizualize theintersection by considering the plan view. Nevertheless alittle Btudy will show thai there must be two closed figureshaving a common point a. Asa matter of fad flic int.! lectionwhen found in elevation will he shaped like an irregular figure 124 DESCRIPTIVE GEOMETRY [XIII, § 131 8, the curve crossing itself (not tangent), at a. It is calleda one-curve intersection. The solution is left to the student. 132. The Intersection of a Cylinder
RM2AKGH4A–Descriptive geometry . Fig. liiiJ Fig. 107. The true size of the section is found by revolving the sectionuntil parallel to V about an axis parallel to VQ. The truewidth at any point, as 28, equals the true distance 2-8 whichappears in the fl-projection. Example - (Fig. 107). A roue of revolution cut by a planeparallel to V. The curve is visible in elevation, since it iawholly on the front half of the cone. The true si/e of the .section appears in the ^projection. 64 DESCRIPTIVE GEOMETRY [IX, § 86 Example 3. (Eig. 108). A sphere cut by a plane perpen-dicular to V. Here the intersection, as see
RM2AKFJ1N–Descriptive geometry . Fig. 202. Fig. 203. Corollary II. To find the second projection of a line lyingin a plane when the general solution fails, partially or wholly. Let Q, Fig. 203, be the given plane, and let ah bh be given sothat it cannot be produced to intersect either HQ or GLwithin the limits of the figure. Then a and bv, or in generalany two points on the line, may be located by Corollary I. Corollary III. To find a line of maximum inclination ofa plane. Let Q, Fig. 204, be the given plane. A line of maximuminclination to 77 is perpendicular to HQ (§ 111). Hence assumeAh perpendicular
RM2AKGG3H–Descriptive geometry . Fig. 110. Fig. ill. Example 6 (Fig. 111). A torus, cut by two planes parallelto V. The X section, wholly on the hack of the surface, isinvisible. The FTsection, entirely on the visible part of thetorus, is wholly visible. The true, size of each section appears at once. 66 DESCRIPTIVE GEOMETRY [IX, § 86 Example 7 (Fig. 112). A hyperbolic spindle, cut by tv:oplanes parallel to V. Each plane cuts a section composed of. Fig. 112. two parts. The visibility of the curves should be evident froman inspection of the position of the cutting planes. 87. Developments. The developmen
RM2AKR45N–Descriptive geometry . Fig. 93.. 3Q <^ J^^^^%^ f. ^ ^ <^7 -^HS K5 ^^ ^^4/ 1 --^7^<^^^^^ 4Q a: <^ ^Vx Fig. 95.
RM2AKGEYY–Descriptive geometry . Fig. 115. lay off the distance 0-1; with 0 and 1 as centers, and radii 0-2and 1-2 respectively, strike arcs intersecting in point 2. Thisgives the development of the triangular face 0-1-2. With 0 and 2 as centers, radii 0-3 and 2-3 respectively,locate point 3, thus obtaining the development of the face0-2-3. In the same way obtain the faces 0-3^ and 0-4-1. The base, 1-2-3-4, of the pyramid is divided into trianglesby the diagonal 1-3, and is plotted to join one of the edgespreviously located, as 2-3. The result thus far obtained is thedevelopment of the complete pyramid.
RM2AKHBBH–Descriptive geometry . wholly in V. In Fig. 57, the actual shadow falls partly on //and partly on V. To find this shadow, Ave may find the com-plete shadow on each plane, regarding the other plane as trans-parent, and then take the actual portion of each shadow. Butthe two shadows, asbs and atbt, one in // and one in V, must in-tersect in a point, e, in GL, since this line is the intersection ofthe planes // and V. Hence we may find one complete shadow,as a b, note the point e in which ajb, crosses GL, and draw frome to the actual shadow at. 43. Shadows of Surfaces. Shadows of surfaces and sol
RM2AKR0A1–Descriptive geometry . ane (Art.9, page 8). 2. Determine the trace of thisline, thus determining one point in the re-quired trace of the plane. 3. Draw thetraces parallel to those of the given plane. Construction. Fig. 114. Let JV be thegiven plane and b the given point. Throughb pass line A parallel to V and in such a di-rection that it will lie in the required plane,A^ being, parallel to GrL and A^ parallel to VN(Art. 35, page 22). Determine d, the hori-zontal trace of line A, and through d^ drawHS^ the horizontal trace of the required plane,parallel to ^iV. VjS will be parallel to A 72. Ca
RM2AKP4KW–Descriptive geometry . surface of the hyperbolic paraboloidhaving A and B for its directrices, and JV forthe plane director. Tluough m draw mg per-pendicular to H, and determine its intersectionwith the surface, as follows: Determine two elements, cd and ef^ near theextremities of the directrices (Art. 150), thework not being shown in the figure. Dividethe jDortion of each directrix limited by theelements cd and ef into an equal number ofparts to obtain elements of the surface (Art.149, page 114). Pass an auxiliary plane Xthrough the perpendicular mg. This will in-tersect the elements at k, Z,
RM2AKB5GG–Descriptive geometry . ase is the initial point 1, and we have com-pleted the circuit. In an actual construction, the duplicate numbers on thebases should be placed before any elements are drawn. Thereis a noticeable symmetry in their placing, provided we startwith a point in a limiting plane, which enables this to be done.The elements are then drawn, and the points in the intersec-tion are marked with the same numbers. The points maythen be readily connected in order. C. Visibility of the Curve of Intersection. A pointin the intersection is visible only when each of the elements,one of each s
RM2AKPYJB–Descriptive geometry . n in which thegiven plane iVis parallel to GL, which neces-sitates the use of an auxiliary profile plane.Through the given point b pass the profileplane P. Determine b^ and PA^, and throughb^ draw PjS^ the profile trace of the requiredplane, parallel to PiV, whence Fas and HjS^ thetraces of the required plane, are determined. This condition may also be solved, Fig. 119,by passing a line A through the given pointb, parallel to ani/ line C of the given planeJST. Through the traces of line A the tracesof the required plane S are drawn parallel toffiVand VN^ respectively. Th
![Descriptive geometry . is perpendicular to a given straight line. The point maybe on the line, or at any distance from it. One plane, and one only, may be found which contains agiven straight line and is perpendicular to a given plane. Anexception occurs when the line is itself perpendicular to thegiven plane, in which case an infinite number of planes may befound. XI, § 115] PERPENDICULAR LINES AND PLANES 95 Problem 10. To find the plane which contains a given point andis perpendicular to a given line. Analysis. Through the given point draw an auxiliary linewhich shall be perpendicular to (bu Stock Photo](https://c8.alamy.com/comp/2AKG3ND/descriptive-geometry-is-perpendicular-to-a-given-straight-line-the-point-maybe-on-the-line-or-at-any-distance-from-it-one-plane-and-one-only-may-be-found-which-contains-agiven-straight-line-and-is-perpendicular-to-a-given-plane-anexception-occurs-when-the-line-is-itself-perpendicular-to-thegiven-plane-in-which-case-an-infinite-number-of-planes-may-befound-xi-115-perpendicular-lines-and-planes-95-problem-10-to-find-the-plane-which-contains-a-given-point-andis-perpendicular-to-a-given-line-analysis-through-the-given-point-draw-an-auxiliary-linewhich-shall-be-perpendicular-to-bu-2AKG3ND.jpg "Descriptive geometry . is perpendicular to a given straight line. The point maybe on the line, or at any distance from it. One plane, and one only, may be found which contains agiven straight line and is perpendicular to a given plane. Anexception occurs when the line is itself perpendicular to thegiven plane, in which case an infinite number of planes may befound. XI, § 115] PERPENDICULAR LINES AND PLANES 95 Problem 10. To find the plane which contains a given point andis perpendicular to a given line. Analysis. Through the given point draw an auxiliary linewhich shall be perpendicular to (bu Stock Photo")
RM2AKG3ND–Descriptive geometry . is perpendicular to a given straight line. The point maybe on the line, or at any distance from it. One plane, and one only, may be found which contains agiven straight line and is perpendicular to a given plane. Anexception occurs when the line is itself perpendicular to thegiven plane, in which case an infinite number of planes may befound. XI, § 115] PERPENDICULAR LINES AND PLANES 95 Problem 10. To find the plane which contains a given point andis perpendicular to a given line. Analysis. Through the given point draw an auxiliary linewhich shall be perpendicular to (bu
RM2AKB24P–Descriptive geometry . n the elements 0-1 and 0-2. It intersects the spherein a circle, of which 3A-4A is a diameter. The center of thiscircle projects at oh, the middle point of 3A-4 Let the planeW be revolved about an axis passing through o until it coin-cides with the plane Y, parallel to V. The elements 0-1 and0-2 will now appear in ^projection as 0-lr and 0-2r. Thecenter, 5, of the circle is found at 5r. Xote that, although wehave not the ^projection, 5, of point 5, it is not necessary,since point 5 and the center, e, of the sphere are at the samedistance above H. With 5r as center, radi
RM2AKP1NB–Descriptive geometry . >^ follows: J., passing through 2 Q and 3 Q B, passing through 3 Q and 4 Q ; (7, passing through 2 Q^ 3 Q^ and 4 Q (Arts. 27, 28, page 19). 46. Determine the prs. of a point on plane S which is 7 units from P^and 4 units from H(Arts. 36, 37, page 24). Through this pointdraw three lines on the plane as .a^ follows: A^ parallel to V B^ paral- <^/lei to H C, passing through 2 Q^ 3 /Q, and 4 Q (Arts. 27, 28, page 19). 47. Determine the prs. of the followingpoints on plane iV, but not lying in a profile plane, a, 6 units from I^in ^ 3 Q 5, 4 units from R in 3 Q ;
RM2AKCR4W–Descriptive geometry . 160 DESCRIPTIVE GEOMETRY [XVI, § 145 respectively. Draw crs2 and crs3. The angle between theselines equals the required angle, 6, between the given planes. Additional examples are given in Figs. 242 and 243. Eachof them falls under the general case, but they differ in themethod of finding the line of intersection of the planes. In Fig. 242 the line A passes through the point s„ inwhich HQ, HR, VQ aud VR intersect. A second point is. R is in quadrants I and 3,30° with H Fig. 242. Fig. 243. obtained by means of the auxiliary plane X (Prob. 12, SpecialCase 5, § 118). The se
RM2AKMFPE–Descriptive geometry . Problems 1, 2. Draw an element of the -w^arped surface through point a. (Art. 145, page llO.) Problems 3, 4. Draw an element of the warped surface through point a. (Art. 146, page 111.) Problems 5, 6. Draw an element of the warped surface parallel to line C of the plane director, N. (Art. 147, page 112.) Problems 7, 8. Dravsr an element of the hyperbolic paraboloid through point a, of a directrix. (Art. 150, page 116.) Unit of measure, | inch. Space required for each problem, 5x7 inches. Angles bet^veen GL and traces ofplanes, multiples of 15°. Measurements from GL, in l
RM2AKCEAC–Descriptive geometry . ent to a cone contains the vertex. Problem 31. To pass a plane tangent to a cone at a given pointin the surface. Analysis. The plane is determined by the element whichpasses through the given point, and a line tangent to the baseat the point where this element intersects the base (§§ 155,162, &). Construction. Case I. 77/<= base of the cone lien in H or V. Example 1 (Fig. 274). The base of this cone lies in H.Let a (oh. a), lying in the element E (Eh. E*) be the given pointin the surface. Since the required tangent plane contains theelement E, find the traces sx and t
RM2AKFT97–Descriptive geometry . el to K. This shows that K, and therefore also Jand L are nearly parallel to the face bcfe, so their points ofintersection are inaccessible. Therefore choose next anotherface, as abc. Line K is found to intersect at point 5, and J atpoint 16 in the face produced. Joining these points, the part5-17 is the actual intersection of JK with the face, and isinvisible in both views. Line L does not actually intersect theface, but will intersect the face produced. The plane Z throughL intersects the edges ac and be extended in the points 28 and27, and this line when produced will
RM2AKCNXF–Descriptive geometry . volved position to GXLU revolve abouto to bxv, which lies in V{Q produced below GXLX. Then pro-ceed as for the point a. As before, there is a cheek on theconstruction ; the distance from b to GL equals the distancefrom bx to GXLX. 166 DESCRIPTIVE GEOMETRY [XVII, § 147 Note. The student does not always see readily why the F-projectionsav and bv should be located by means of auxiliary lines in the plane Q,since the distances of these points from GL appear at once in the second-ary projection. Indeed, the projections av and bv can be located by trans-ferring from the secon
RM2AKG7XM–Descriptive geometry . Fig. 145. auxiliary line E parallel to C. Pass the plane Q through thetwo parallel lines D and E. Then if the required plane is tocontain ab and be parallel to C, plane Q is the required plane.If, however, the required plane is to contain C and be parallelto ab, the plane It, passed through C parallel to plane Q (§ 103),is the required plane. 86 DESCRIPTIVE GEOMETRY [X, § 107 Problem 8. To find the plane which contains a given point and isparallel to each of two given lines. Analysis. Through the given point draw two auxiliary lines,one parallel to one given line, the ot
RM2AKN3B6–Descriptive geometry . HIP RAFTER. Determine length—down cut-heel cut— side cut — top bevel. JACK RAFTER. Determine side cut. PURLIN. Determine down cut — side cut-angle between face and end. Determine the bevels, cuts, and lengths of roof members, as above. (Art. 87, page OO, inch. Space required for the problem, 7 x lO inches, PLATE 20. Determine angle of cut on top of purlin (A). Bevel on web of purlin (B). Angle bet-w^een plane of -web of hip rafter and purlin, orbend of gusset (C). Angle between top edges of gusset (D). (Art, 87, page 61.) Unit of measure, i inch. Space required for each
RM2AKCT5N–Descriptive geometry . Fig. 238. dicular to Ah. Project the given lines A and B on the second-ary vertical plane Vx thus assumed (§ 68). The line A willproject as a point, Axv. The required line no is perpendicularto both A and B. Since A is perpendicular to Vx, no, which isperpendicular to A, is parallel to Vx. Again, since no and 5are perpendicular, and no is parallel to Vx, the projection nfofand Bf are perpendicular (§ 110). Hence, from Af drawniOx perpendicular to By. Project from of on Bf to o* onBh, thence to ov on B. From oh draw ohnh perpendicular to Ah(see Prob. 25, Special Case I, §
RM2AKG278–Descriptive geometry . Fig. 1G0. Fig. 167. A second example, with planes differently situated, is givenin Fig. 167. The general case may fail because of (a) parallel lines;(6) intersections inaccessible ; (c) points s and t coincident. Special Case I. Suppose that one pair of traces is parallel(Fig. 168). Let Q and R be the given planes, with HQ and 11Rparallel. The intersection of VQand VR gives the F-trace, t, of therequired line of intersection A.Consider the planes Q, R, and //.Since the intersections of H withQ and R are parallel, H must beparallel to the line of intersection,A, of these
RM2AKH5G5–Descriptive geometry . Fig. 89. Fig. 89, note that the directions of the points a and & fromH, as shown by the relations of a and bv to OL, must be pre-served in the secondaryprojection. In Fig. 90 aprofile line is shown pro-jected on a profile planeby this method. (See§§ 64, 66.) 70. Simplification ofProblems by Means ofSecondary Projections. Inthe solution of a problem,there is no advantage inintroducing a secondaryplane of projection unlessthe new projection is insome way simpler thanthe original projections.A point always projects as a point, and cannot be madeany simpler. G L ^<P< b
RM2AKPG04–Descriptive geometry . lement, A^and intersects the given plane in the line Gr.Since lines 6r and A lie in plane X, they in-tersect in a, one point in the required curveof intersection between the cylinder and thegiven plane iV. Likewise plane Z intersectsthe cylinder in elements 0 and i>, and theplane iV^in line K, the intersections of whichwith lines O and D are c and t?, two otherpoints in the required curve. The true size of the cut section has beendetermined by revolving it into V about F7Vas an axis. 125. To develop the cylinder. Principle. When a cylinder is rolled upona plane to det
RM2AKPEWX–Descriptive geometry . ual to the rectified distances dcch etc. Through points c?, e, 5, etc., perpen-diculars to line dd have been drawn equal tothe true length of the elements. The portionsabove and below dd are iequal to the lengthsof the elements above and below the cutsection. 126. If the axis of the cylinder be parallelto a coordinate plane, the development may beobtained without the use of a right section. Fig. 164 represents an oblique cylinderwith its axis parallel to V and its basesparallel to H. The elements are representedin vertical projection in their true lengths,and in horizo
RM2AKP38B–Descriptive geometry . and the surface generated will be aright helicoid. This type is illustrated by thesquare-threaded screw (Fig. 146, page 69). 154. If the generatrix does not intersectthe axis, as in the preceding cases, a moregeneral type will be generated, as shown inFig. 193. In this case the generatrix is gov-erned by two helical directrices and a conedirector, the generatrix being tangent to thecylinder on which the inner helix is described. 155. Hyperboloid of Revolution of one Nappe.This is a surface of revolution which may begenerated by the revolution of an hyperbolaabout its con
RM2AKG7FY–Descriptive geometry . xiliary lines, D parallelto A, and E parallel to B. Pass the required plane Q throughthe lines D and E (Prob. 6, § 106). The special cases of this problem are too similar to those ofProblem 7 to require a detailed discussion. Problem 9. To find the plane which contains a given point and isparallel to a given plane. Analysis. Through the given point, pass a line parallel tothe given plane. Through this line, pass the required planeparallel to the given plane. Construction (Fig. 147). Let Q be the given plane and a thegiven point. A line may be drawn through a and parallel
RM2AKCW1X–Descriptive geometry . is also in Q, and since L is not parallel to A becauseB is not, L and A will intersect in a point, n. Let the per-pendicular ek come to rest when point k coincides with point n.The line will then be in the position on, intersecting both Aand B and perpendicular to each. Hence on is the commonperpendicular required. The actual distance between thelines A mid Ji may ho found by measuring the length ofeither ek or on. 156 DESCRIPTIVE GEOMETRY [XVI, § 144 Construction (Fig. 237). Let A and B be the given lines.Through A pass the plane Q, parallel to B (Prob. 7, § 107).To do
RM2AKR5BR–Descriptive geometry . r auxiliary plane parallel to Vor H^ and the required line will be determined. Construction. Fig. 88 represents twoplanes, T and Jf, with their horizontal tracesintersecting, bat their vertical traces inter-secting beyond the limits of the drawing. By Case 1, point d is one point in the lineof intersection between the planes. To obtaina second point an auxiliary plane -X has beenpassed parallel to V. Then HX is parallel toGrL^ and C^ the horizontal projection of theline of intersection between planes X and M,lies in HX, while (7^ is parallel to VM (Case1). Likewise B^, t
RM2AKR4JM–Descriptive geometry . ssing through 1Qand 3Q at an angle 0 with V. As in the pre-vious example the profile auxiliary plane isrequired and PN^ the profile trace of iV, isdetermined as before. PS^ the profile traceof S, is next drawn through 1Q and 3 Q andmaking an angle 6 with VP. Then c?^, theintersection of PiVand PS^ is the profile projec-tion of one point in the required line of inter-section between planes N and S^ and d^ andd^ are the required projections of this point.The horizontal and vertical traces of this lineof intersection are at 5, for it is here that VNand VS intersect, and als
![Descriptive geometry . Fig. 299 (repeated). GL, and the projections Eh, Ev are determined by the factthat the line E passes through the point c (reverse of Fig. 221,§ 138). Pass the required tangent plane Q through the linesE and A (Prob. 6, § 106). The tangent line F does not intersect HX within reach. In-stead of attempting to find the projections of F, let us deter-mine the tangent plane by means of the normal. Through oTdraw Nr perpendicular to FT. The line N lies in the plane X XX, § 171] SURFACES OF REVOLUTION 215 (see the Analysis); hence the intersection of Nr and HX is the//-trace of Stock Photo](https://c8.alamy.com/comp/2AKBJM1/descriptive-geometry-fig-299-repeated-gl-and-the-projections-eh-ev-are-determined-by-the-factthat-the-line-e-passes-through-the-point-c-reverse-of-fig-221-138-pass-the-required-tangent-plane-q-through-the-linese-and-a-prob-6-106-the-tangent-line-f-does-not-intersect-hx-within-reach-in-stead-of-attempting-to-find-the-projections-of-f-let-us-deter-mine-the-tangent-plane-by-means-of-the-normal-through-otdraw-nr-perpendicular-to-ft-the-line-n-lies-in-the-plane-x-xx-171-surfaces-of-revolution-215-see-the-analysis-hence-the-intersection-of-nr-and-hx-is-the-trace-of-2AKBJM1.jpg "Descriptive geometry . Fig. 299 (repeated). GL, and the projections Eh, Ev are determined by the factthat the line E passes through the point c (reverse of Fig. 221,§ 138). Pass the required tangent plane Q through the linesE and A (Prob. 6, § 106). The tangent line F does not intersect HX within reach. In-stead of attempting to find the projections of F, let us deter-mine the tangent plane by means of the normal. Through oTdraw Nr perpendicular to FT. The line N lies in the plane X XX, § 171] SURFACES OF REVOLUTION 215 (see the Analysis); hence the intersection of Nr and HX is the//-trace of Stock Photo")
RM2AKBJM1–Descriptive geometry . Fig. 299 (repeated). GL, and the projections Eh, Ev are determined by the factthat the line E passes through the point c (reverse of Fig. 221,§ 138). Pass the required tangent plane Q through the linesE and A (Prob. 6, § 106). The tangent line F does not intersect HX within reach. In-stead of attempting to find the projections of F, let us deter-mine the tangent plane by means of the normal. Through oTdraw Nr perpendicular to FT. The line N lies in the plane X XX, § 171] SURFACES OF REVOLUTION 215 (see the Analysis); hence the intersection of Nr and HX is the//-trace of
RM2AKRANX–Descriptive geometry . e A and B^ the traces ofwhich cannot be found within the limits of THE PLANE OF LINES 21 the drawing. Line C is an assumed line join-ing point e of line A and point/ of line B^ thetraces of which are easily located at g and h.Line i> is a second similar line. HT^ connect-ing the horizontal traces of lines C and Z>, isthe required horizontal trace of the plane oflines A and B. Likewise YT, connecting thevertical traces of lines C and i), is the requiredvertical trace of the plane of lines A and B.32. Case 3. Method. If one of the given lines is parallel to the groun
RM2AKH35T–Descriptive geometry . Fig. m. Fn;. 94. here the projection on the secondary Fplane gives informationabout the size, shape, and position of the prism which is notdirectly obtainable from the original ^-projection. 72. The Two V-Projections Compared. The relation betweentwo ^-projections of an object is the same, whichever be con-sidered the original, and which the secondary, projection (§ 67).Hence, in Fig. 94, instead of constructing the Pi-projectionfrom the //- and F-prejections, the ^-projection might be con-structed if the H- and p^-projections are known. That is. ifan object is known to
RM2AKCEGY–Descriptive geometry . Fig. 272.. 184 DESCRIPTIVE GEOMETRY [XIX, § 162 162. Tangent Planes to Cones and Cylinders. At every pointin the surface of a cone or cylinder, a plane tangent to the sur-face can be drawn. Planes tangent to cones and cylinders can also be passed tofulfill certain other conditions. Thus, the tangent planes maybe required to contain a given point outside the surface, orto be parallel to a given straight line. In determining these tangent planes, we shall make use ofthe following propositions. (a) Through every point in the surface of a cone or cylinder,a rectilinear eleme
RM2AKRA5T–Descriptive geometry . ace of lineA; a^ is in GL (Art. 24, page 16). Con-tinue A^ to meet GL in 5 the horizontal pro-jection of the vertical trace; 6^, the verticaltrace, is in FiV. Connect a and h^ to obtain^^ the required vertical projection of line A. If A had been the given projection, A^would have been similarly determined. If the given projection, B^, Fig. 65, be par-allel to (7X, B will be parallel to HS for, ifa line lying in an inclined plane has one pro-jection parallel to the ground line, the other pro-jection is parallel to the trace of the plane; and PROJECTION OF LINES IN PLANE
![Descriptive geometry . Fig. 100 (repeated).. Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givend Stock Photo](https://c8.alamy.com/comp/2AKGNGK/descriptive-geometry-fig-100-repeated-fig-101-repeated-construction-first-method-fig-100-reverse-offirst-method-prob-3-78-let-the-angle-with-h-andthe-direction-of-the-if-projection-be-given-let-the-point-a-bethe-fixed-end-of-the-line-viii-83-true-length-of-a-line-59-place-the-line-in-the-position-ac-avcv-ahch-parallel-to-vmaking-avcv-equal-to-the-true-length-of-the-line-at-the-givenangle-with-h-and-sloping-upward-or-downward-from-a-asgiven-take-an-axis-perpendicular-to-h-through-the-fixed-end-aand-revolve-the-line-until-the-projection-takes-the-givend-2AKGNGK.jpg "Descriptive geometry . Fig. 100 (repeated).. Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givend Stock Photo")
RM2AKGNGK–Descriptive geometry . Fig. 100 (repeated).. Fig. 101 (repeated). Construction. First Method (Fig. 100). [Reverse ofFirst Method, Prob. 3, § 78.] Let the angle with H andthe direction of the if-projection be given. Let the point a bethe fixed end of the line. VIII, § 83] TRUE LENGTH OF A LINE 59 Place the line in the position ac (avcv, ahch), parallel to V,making avcv equal to the true length of the line, at the givenangle with H, and sloping upward or downward from a asgiven. Take an axis perpendicular to H through the fixed end, a,and revolve the line until the //-projection takes the givend
RM2AKBT2J–Descriptive geometry . Fig. 284. Problem 35. To pass a plane tangent to a cylinder through agiven point without the surface. (Two results.) Analysis. Through the given point pass a line parallel tothe elements of the cylinder. Since the recpiired tangentplane contains some element of the surface, this line must liein the required tangent plane. Find the point in which thisline pierces the plane of the base of the cylinder. From thispoint draw a line tangent to the base. Pass the required tan-gent plane through the tangent line and the line first drawn.In general, two tangent lines, and hence t
RM2AKN3R8–Descriptive geometry . 33 28 Problems 1-4. Determine the true size of the diedral angle between planes N and T. (Arts. 83-85Problems 5-8. Determine the true sizes of the diedral angles of the objects. (Arts. 83-85, pages pages, 57,57, 58.) Unit of measure, J inch. Space required for each problem, 2^ x 3 inches. Angles between GL and traces of p| A-fC 1Qplanes, multiples of 15°. Measurements from GL, in light type, and from right-hand division line, in heavy type. »-A It. 1 O l4 2 8 4 ^-^ ^^ J^^ 7 6 12/ 6 12/^ ^ 78 . 8 /16/ / ^y^ ^^ S 9 vs . 10 11 I^S , 12 HS ^^ c * 0 //S ^ 1 Determine the an
![Descriptive geometry . Fig. 131. drawn, however, it would be difficult to recognize the parallel-ism. Thus, in Fig. 132, both the lines A and B are parallel tothe plane B, but this fact can hardly be seen by inspection.. Fig. 132. Fig. 133. The fact that a line is parallel to a plane can be recognized atonce only when the line is parallel to one of the systems ofprincipal lines of the plane. In Fig. 133, it is evident thatboth lines C and D are parallel to the plane S. X, § 106] PARALLEL LINES AND PLANES 79 105. A Plane Parallel to a Line. Conversely, a plane is par-allel to a straight line if Stock Photo](https://c8.alamy.com/comp/2AKGBBP/descriptive-geometry-fig-131-drawn-however-it-would-be-difficult-to-recognize-the-parallel-ism-thus-in-fig-132-both-the-lines-a-and-b-are-parallel-tothe-plane-b-but-this-fact-can-hardly-be-seen-by-inspection-fig-132-fig-133-the-fact-that-a-line-is-parallel-to-a-plane-can-be-recognized-atonce-only-when-the-line-is-parallel-to-one-of-the-systems-ofprincipal-lines-of-the-plane-in-fig-133-it-is-evident-thatboth-lines-c-and-d-are-parallel-to-the-plane-s-x-106-parallel-lines-and-planes-79-105-a-plane-parallel-to-a-line-conversely-a-plane-is-par-allel-to-a-straight-line-if-2AKGBBP.jpg "Descriptive geometry . Fig. 131. drawn, however, it would be difficult to recognize the parallel-ism. Thus, in Fig. 132, both the lines A and B are parallel tothe plane B, but this fact can hardly be seen by inspection.. Fig. 132. Fig. 133. The fact that a line is parallel to a plane can be recognized atonce only when the line is parallel to one of the systems ofprincipal lines of the plane. In Fig. 133, it is evident thatboth lines C and D are parallel to the plane S. X, § 106] PARALLEL LINES AND PLANES 79 105. A Plane Parallel to a Line. Conversely, a plane is par-allel to a straight line if Stock Photo")
RM2AKGBBP–Descriptive geometry . Fig. 131. drawn, however, it would be difficult to recognize the parallel-ism. Thus, in Fig. 132, both the lines A and B are parallel tothe plane B, but this fact can hardly be seen by inspection.. Fig. 132. Fig. 133. The fact that a line is parallel to a plane can be recognized atonce only when the line is parallel to one of the systems ofprincipal lines of the plane. In Fig. 133, it is evident thatboth lines C and D are parallel to the plane S. X, § 106] PARALLEL LINES AND PLANES 79 105. A Plane Parallel to a Line. Conversely, a plane is par-allel to a straight line if
RM2AKPX63–Descriptive geometry . he complement of therequired angle. ANGLE BETWEEN LINE AND PLANE 55 Construction. Fig. 125. Given line Aand plane W. Through any point o, of line .4,pass line B perpendicular to plane iV (Art.62, page 44). Determine one of the traces ofthe plane of lines A and B, as VS. Revolvelines A and B into J^ about VS as an axis, asat A and B (Art. 43, page 31). Thenangle d^ce is the true size of the angle be-tween lines A and B, and its complement,ec% is the required angle between line A andplane iV(Art. 44, page 31). 0 B Fig. 126 illustrates a condition in which thegiven plane iV
RM2AKRJ4P–Descriptive geometry . V has been POINTS revolved to coincide with //. The distancefrom the point in space to H is equal to thedistance from its vertical projection to theground line, and the distance from the pointin space to T^is equal to the distance from itshorizontal projection to the ground line. Apoint on either coordinate plane is its ownprojection on that plane, and its other projec-tion is in the ground line, ;;s points c, c?, and g,Figs. 12 and 13. The points represented in Figs. 12 and 13are described as follows: Point «, in 5$, 5 units from Tand 8 units from H.Point 6, in 2Q, -^ u
RM2AKG1JP–Descriptive geometry . ityof the intersection, planes parallel to H or V are often intro-duced as auxiliaries when the general case fails. Special Case IV. The traces are not parallel, but one orboth pairs fail to intersect within the limits of the drawing.Let Q and R, Fig. 171, be the given planes. The intersectionof the P-traces of these planes gives the point t, as in thegeneral case. Since HQ and HR do not intersect withinreach, pass the auxiliary plane, X, parallel to H. Planes Xand Q intersect in the line M (Case III); planes X and Rintersect in the line X; the lines M and N intersect i
![Descriptive geometry . Fig. 214. XIV, § 137] THE REVOLUTION OF PLANES 137 with V. We find that HS intersects the axis at d, and theground line at g. Hence, d is the fixed point on HQ, while VQmust be tangent to the arc gj, drawn with center o and radius og.. Fig. 215. We now have one point each on HQ and VQ, and an arc towhich each is tangent. Therefore HQ and VQ are determined. Check. Since HQ and VQ are independently determined,they must intersect the ground line at the same point, n. If « and /3 are given as acute angles, as is usual, a solutionis possible only when the sum of a and /3 is n Stock Photo](https://c8.alamy.com/comp/2AKFD5H/descriptive-geometry-fig-214-xiv-137-the-revolution-of-planes-137-with-v-we-find-that-hs-intersects-the-axis-at-d-and-theground-line-at-g-hence-d-is-the-fixed-point-on-hq-while-vqmust-be-tangent-to-the-arc-gj-drawn-with-center-o-and-radius-og-fig-215-we-now-have-one-point-each-on-hq-and-vq-and-an-arc-towhich-each-is-tangent-therefore-hq-and-vq-are-determined-check-since-hq-and-vq-are-independently-determinedthey-must-intersect-the-ground-line-at-the-same-point-n-if-and-3-are-given-as-acute-angles-as-is-usual-a-solutionis-possible-only-when-the-sum-of-a-and-3-is-n-2AKFD5H.jpg "Descriptive geometry . Fig. 214. XIV, § 137] THE REVOLUTION OF PLANES 137 with V. We find that HS intersects the axis at d, and theground line at g. Hence, d is the fixed point on HQ, while VQmust be tangent to the arc gj, drawn with center o and radius og.. Fig. 215. We now have one point each on HQ and VQ, and an arc towhich each is tangent. Therefore HQ and VQ are determined. Check. Since HQ and VQ are independently determined,they must intersect the ground line at the same point, n. If « and /3 are given as acute angles, as is usual, a solutionis possible only when the sum of a and /3 is n Stock Photo")
RM2AKFD5H–Descriptive geometry . Fig. 214. XIV, § 137] THE REVOLUTION OF PLANES 137 with V. We find that HS intersects the axis at d, and theground line at g. Hence, d is the fixed point on HQ, while VQmust be tangent to the arc gj, drawn with center o and radius og.. Fig. 215. We now have one point each on HQ and VQ, and an arc towhich each is tangent. Therefore HQ and VQ are determined. Check. Since HQ and VQ are independently determined,they must intersect the ground line at the same point, n. If « and /3 are given as acute angles, as is usual, a solutionis possible only when the sum of a and /3 is n
![Descriptive geometry . ppes of the cone which arehere shown; but the upper nappes of the cones will intersectin the second branch of the same hyperbola. 195. Parallel Elements. If two elements, one in each sur-face, are parallel, the intersection of the surfaces may, thoughnot necessarily, have one or more infinite branches. XXV, § 195] CYLINDERS AND CONES 275 A. Two Cylinders. In the case of two cylinders, if thereis one pair of parallel elements, all the elements of both sur-faces are parallel, and the intersection can consist only ofstraight lines. B. A Cylinder and a Cone. In the case of a Stock Photo](https://c8.alamy.com/comp/2AKB3JW/descriptive-geometry-ppes-of-the-cone-which-arehere-shown-but-the-upper-nappes-of-the-cones-will-intersectin-the-second-branch-of-the-same-hyperbola-195-parallel-elements-if-two-elements-one-in-each-sur-face-are-parallel-the-intersection-of-the-surfaces-may-thoughnot-necessarily-have-one-or-more-infinite-branches-xxv-195-cylinders-and-cones-275-a-two-cylinders-in-the-case-of-two-cylinders-if-thereis-one-pair-of-parallel-elements-all-the-elements-of-both-sur-faces-are-parallel-and-the-intersection-can-consist-only-ofstraight-lines-b-a-cylinder-and-a-cone-in-the-case-of-a-2AKB3JW.jpg "Descriptive geometry . ppes of the cone which arehere shown; but the upper nappes of the cones will intersectin the second branch of the same hyperbola. 195. Parallel Elements. If two elements, one in each sur-face, are parallel, the intersection of the surfaces may, thoughnot necessarily, have one or more infinite branches. XXV, § 195] CYLINDERS AND CONES 275 A. Two Cylinders. In the case of two cylinders, if thereis one pair of parallel elements, all the elements of both sur-faces are parallel, and the intersection can consist only ofstraight lines. B. A Cylinder and a Cone. In the case of a Stock Photo")
RM2AKB3JW–Descriptive geometry . ppes of the cone which arehere shown; but the upper nappes of the cones will intersectin the second branch of the same hyperbola. 195. Parallel Elements. If two elements, one in each sur-face, are parallel, the intersection of the surfaces may, thoughnot necessarily, have one or more infinite branches. XXV, § 195] CYLINDERS AND CONES 275 A. Two Cylinders. In the case of two cylinders, if thereis one pair of parallel elements, all the elements of both sur-faces are parallel, and the intersection can consist only ofstraight lines. B. A Cylinder and a Cone. In the case of a