5 Spatial Relations on Lines

Transcription

1 5 Spaial Relaions on Lines There are number of useful problems ha can be solved wih he basic consrucion echniques developed hus far. We now look a cerain problems, which involve spaial relaionships beween lines, and beween lines and planes. 5. PRLLEL LINES N PLNES We know ha a line is parallel o a plane if i has no common poin wih he plane. This fac can be used o es if a given line and plane are parallel: Simply consruc an edge view (see onsrucion 4- on page 0) of he plane and projec he line ino he same view; if he line appears in poin view or parallel o he edge view, i canno mee he plane in a poin and is herefore parallel o he plane. The same fac can be used o consruc a plane parallel o a given line or a line parallel o a given plane. In boh cases, here exis infiniely many soluions o he problem. The consrucion below demonsraes his for he second case. onsrucion 5- Line parallel hrough a given poin o a given plane Suppose we are given and a poin O in wo adjacen views and, we are required o find a line hrough O parallel o plane. See Figure 5-. There are four seps:. onsruc an edge view of he plane in an auxiliary view,, and projec O ino.. raw a line hrough parallel o he edge view of he plane and selec wo arbirary poins, and N, on his line.

2 . raw projecion lines hrough and N perpendicular o folding line. 4. raw a convenien line hrough O p ha inersecs boh projecion lines in poins p and N p ; his line is parallel o plane. Projec p and N p ino q o obain a view of he line in q. O The consrucion is illusraed below in Figure 5-. O Two possible lines parallel o plane hrough poin O O N O N Line hrough O parallel o he edge view of plane N How do we locae poins and N? O 5- onsrucing a line hrough a given poin parallel o a given plane (Top Righ) Problem configuraion (oom) Soluion consrucion 46

3 When wo planes are parallel, any view showing one plane in edge view mus show he oher plane in a parallel edge view. This fac can be used o es wheher wo planes are in fac parallel. Recall ha onsrucion 4- (on page 0) serves o find he edge view. We show he consrucion below in Figure 5-. E E F F parallel edge views indicae parallel planes F E 5- Parallel edge views indicae parallel planes 5. PERPENIULR LINES N PLNES line is perpendicular o a plane, referred o as normal o he plane, if every line in he plane ha passes hrough he poin of inersecion of he given line and he plane makes a righ angle wih he given line. 47

4 N Q L P O Lines L, NO, PQ all lie in he plane Line is perpendicular o he plane 5- line normal o a plane We can apply he same principles as before o solving problems involving perpendicular lines and planes. For example, a line normal o a given plane can be found in an auxiliary showing he plane in edge view; in his view, he perpendicular appears in TL perpendicular o he edge view and can be projeced ino he given views. This mehod also yields he poin where he line inersecs he plane, which is called he piercing poin (see he haper 6 on inersecions). p N P, Q p is a plane Line is a normal o i Lines L, NO and PQ lie in he plane L O L,P 90º,Q L,O 90º,N 5-4 line perpendicular o plane in views showing he plane in edge view 48

5 We can also find he normal wihou consrucing an auxiliary view by he following wo-view mehod. The normal mus be perpendicular o every line in he plane. If he plane is given in wo auxiliary views, we find a line in he plane shown in TL in one of he views. The normal mus be perpendicular o ha line in he same view. However, his mehod does no auomaically yield he piercing poin. See Figure 5-5. direcion of he normal in view # direcion of he normal in view # direcion of he normal in view # TL 90º TL 90º 90º TL direcion of he normal in view # direcion of he normal in view # direcion of he normal in view # 5-5 Two-view mehod for finding he direcion of he normal o a plane Similarly, we can find a plane perpendicular o a given plane. How? 5. PERPENIULR PLNES eermining wheher wo planes are perpendicular generally requires finding heir inersecion and his is deal wih in anoher chaper and he consrucion is omied for now. 5.4 SHORTEST ISTNE ONSTRUTIONS We nex consider various shores disance consrucions, namely, from poin o plane, from poin o line and beween lines. 49

6 5.. Shores perpendicular disance o a plane We can use he principles underlying Figure 5-5 o deermine he shores beween a poin and plane. For any given poin, we consruc a line perpendicular o he given plane. In edge view, his perpendicular line will be seen in rue lengh and a righ angles o he plane. The consrucion is given nex. O E Lines and EF lie in he plane P Shores line (OP) from poin O o plane F Observer's line of sigh plane is seen as an edge and rue lengh of OP appears 5-6 Shores disance o a plane from a poin onsrucion 5- Shores disance from a poin o a plane Suppose is he given plane and is he given poin.. onsruc he edge view of plane (see Figure 4.6) by consrucing an auxiliary view #.. In his view, consruc a perpendicular from o he edge view of he plane. Le P be he foo of he perpendicular. P is he required disance.. To deermine where P is locaed in he views # and #, see Figure 5-7. (Righ) Problem configuraion 50

7 Seps & onsruc edge view and hen perpendicular from o edge view Sep Locaing P in views # and # True lengh of he shores line from o he plane True lengh of he shores line from o he plane P P Edge view of plane Edge view of plane P P lies on he perpendicular from o he rue lengh line in view # P P is locaed by using he ransfer disance from view # 5-7 Shores disance o a plane from a poin 5.. Shores disance beween a poin and a line 5-8 The Problem Given adjacen views of poin and line, find he rue lengh of he shores disance from o he line We consider wo mehods: he line mehod and he plane mehod. 5

8 Line mehod We solved his problem in a previous chaper by consrucing he line in poin view and measuring he disance beween he PV of he line and he view of he poin. See onsrucion -7 (on page 08). We may call his he line mehod as we deal direcly wih he given line. The mehod is illusraed again below in Figure s line is in rue lengh, he consruced perpendicular from o produces poin True lengh of he shores disance TL, Poin view of line Projec back from view # o ge Projec back from view # o ge 5-9 The line mehod for deermining he shores disance beween a poin and a line Plane mehod second mehod is based on a plane consruced from he poin and he given line. We consruc a view showing he plane in rue shape (see onsrucion 4- on page 4) and he perpendicular from he poin o he given line in his view gives he shores disance. This mehod is illusraed in Figure

9 4 Edge view of True shape of Projec back from view #4 o ge is he shores disance from o Projec back from view # o ge defines a plane Projec back from view # o ge 5-0 The plane mehod for deermining he shores disance beween a poin and a line 5.. Shores disance beween wo skew lines Skew lines are non-inersecing non-parallel lines. Skew lines are parallel o he same plan, bu here is no plane in which he wo lines are coplanar. For any wo skew lines, here exiss exacly one line perpendicular o boh lines; his line is called he common perpendicular. The shores disance beween wo skew lines is he disance beween heir inersecion poins wih heir common perpendicular. This disance can be deermined by finding a view in which one of he lines is shown in poin view. 90 Line is he shores disance beween skew lines and as i is perpendicular o boh lines Two skew lines and heir common perpendicular 5

10 We highligh below, wo consrucions o deermine his shores disance, one is a line mehod and he oher, a plane mehod. In boh cases, we are given wo skew lines in wo adjacen views, and, where he lines are described by he segmens, and, he consrucion below finds he shores disance beween he lines. 5- Two skew lines onsrucion 5- Shores disance beween wo skew lines (line mehod) The consrucion is shown Figure 5-., 90 4 True lengh of shores line is seen in view #4 is in rue lengh in view # 90 ommon perpendicular beween skew lines and in view # ommon perpendicular beween skew lines and in view # 5- Line mehod for consrucing he shores disance beween wo skew lines There are five basic seps: 54

11 . onsruc a firs auxiliary view showing one of he segmens, say, in rue lengh (TL) and a second auxiliary view 4 showing he same segmen in poin view (PV).. Projec he oher segmen ino boh auxiliary views.. In view 4, he common perpendicular appears in TL as a line hrough he poin view of,, and perpendicular o. Find he inersecion of his perpendicular and,. is he required shores disance. 4. Projec ino view. In view, he common perpendicular is he line hrough parallel o folding line 4 meeing he line a 5. Projec and ino and o obain all views of he common perpendicular. onsrucion 5-4 Shores disance beween wo skew lines (plane mehod) gain, as before, we are given wo skew lines in wo adjacen views, and, where he lines are described by he segmens, and. The consrucion below finds he shores disance beween he lines by consrucing planes firs. We consider four seps, alhough he saed problem is solvable in wo.. onsruc a plane parallel o segmen conaining he segmen in boh views. We do his by drawing a line parallel o hrough a seleced assumed poin W on. Projec W ono he oher view and, in he oher view, draw he line hrough his projeced poin W parallel o. See sep for a good choice of or. is a plane parallel o. is parallel o in view # and passes hrough W W HL is a plane is parallel o in view # and mees a W is parallel o folding line W. onsruc an auxiliary view showing he plane in edge view. We do his by drawing a horizonal line in one of he given views, say, and projec i ino he oher view. is now shown in rue lengh. folding line is drawn perpendicular o and he projecion of he plane in ha view will be seen in edge view. In his view is parallel o. The disance beween he wo lines is he shores disance beween he wo skew lines. 55

12 Shores disance beween skew lines and, is parallel o in view # and passes hrough W, W Plane seen in edge view in view # HL. To find he common perpendicular we need one furher sep. reae anoher auxiliary view 4 in which is shown in rue lengh. The apparen poin of inersecion in view 4 shows he common perpendicular in poin view. Incidenally, is also shown in rue lengh. This consrucion proves ha for any wo skew lines, here acually is a view ha shows hem boh lines in TL. 4. This poin is projeced back ino views, and o give he acual locaion of he common perpendicular, RS. The complee consrucion is illusraed in 5-4. R,S 4, is parallel o in view # and passes hrough W Shores disance RS beween skew lines and R S R, Plane seen in edge view in view # S W HL is a plane is parallel o in view # and mees a W is parallel o folding line S W R 5-4 Plane mehod for consrucing he shores disance beween wo skew lines 56

13 5..4 Shores horizonal disance beween wo skew lines The shores horizonal disance beween wo skew lines is he lengh of shores horizonal line beween hem. See Figure 5-5 in which represens he shores horizonal disance in rue lengh. Tha is, mus appear as a horizonal line in an (verical) elevaion view Horizonal projecion plane parallel 5-5 Shores horizonal disance beween wo skew lines Shores horizonal disance beween he wo skew lines For his, we need o creae an elevaion view in which he required line appears in poin view and projec back ino he original views. y he plane mehod i is possible o consruc a view in which he wo skew lines appear parallel as he planes hey lie in are seen in edge view. We have he following consrucion given in Figure 5-6. onsrucion 5-5 Shores horizonal disance beween wo skew lines s before, we are given wo skew lines in wo adjacen views, and, where he lines are described by he segmens, and. We employ he plane mehod. There are five seps.. In he elevaion view, consruc a line L parallel o passing hrough an assumed poin on. onsruc L o be parallel o he folding line, ha is, horizonal.. Projec L and back ino he op view and again draw L parallel o in his view. L is now shown in rue lengh. (Righ) Two skew lines 57

14 L is parallel o in view #. Now creae an auxiliary view showing he plane L in edge view. This plane conains, and herefore, i is seen collinear wih he edge of his plane. lso he skew lines and appear parallel. L L HL TL L is in rue lengh 4. View is an elevaion. Knowing he direcion of a horizonal line in his direcion consruc a view 4 perpendicular o his direcion. In his view he shores horizonal line beween and will appear in poin view, ha is, as a poin of inersecion beween and in view Projec and back o views, and., he shores horizonal disance beween he skew lines, is seen in rue lengh in view and. L is parallel o in view # (Lef) fer sep, is parallel o he edge view of he horizonal plane is also he rue lengh of he shores horizonal line 4 is parallel o he edge view of plane L in view #,L L TL L is parallel o in view # L is in rue lengh View # is an elevaion L HL L is parallel o in view # 5-6 Shores horizonal disance beween wo skew lines 58

15 5..5 Shores grade disance beween wo skew lines The grade is an alernaive way of measuring he slope of a line favored by civil engineers, paricularly ion reference o he incline of a road. Percen grade is defined as he number of verical rise for each hundred unis of horizonal disance. For any wo skew lines, for a given grade (slope), here is a unique line ha provides his shores disance. To deermine his we employ a consrucion similar o 4. using he plane mehod o obain a view in which he skew lines appear parallel o each oher from which we can deermine he view ha shows he required disance (line) in poin view. onsrucion 5-6 Shores grade disance beween wo skew lines. Suppose ha we wan o find he shores disance wih a slope of 5 from o. We consider wo possibiliies an upward slope from o, or a downward slope. The firs hree seps are he same as above. The remainder of he consrucion is given below and illusraed in Figures 5-7 and 5-8., 4 is also he rue lengh of he shores upward 5 grade line O 4 5,L N 4 L TL L is parallel o in view # L is in rue lengh View # is an elevaion L HL L is parallel o in view # 5-7 Shores grade (upward slope) disance beween wo skew lines 59

16 The las wo seps are: 4. We consruc a line PQ in view on he folding line wih he required slope. This line deermines he direcion of he shores disance. onsruc a view 4 perpendicular o his direcion. In his view he shores line wih he specified grade beween and will appear in poin view, ha is, as a poin of inersecion beween and in view Projec and back o views, and., he shores disance a he given grade beween he skew lines, is seen in rue lengh in view and., 4 5 L is parallel o in view # is also he rue lengh of he shores downward 5 grade line,l L TL L is in rue lengh View # is an elevaion L HL L is parallel o in view # 5-8 Shores grade (downward slope) disance beween wo skew lines 60

17 5..6 Visibiliy of skew lines revisied s skew lines do no inersec, i becomes imporan o consider he visibiliy of lines, paricularly when he lines represen solid objecs. We have considered visibiliy before in onsrucion - on page 7. We show he consrucion again wihou explanaion in Figure 5-0. Observers line of sigh in which line is above line 5-9 Visibiliy of lines Q, loser o fronal projecion plane P loser o horizonal projecion plane P,Q 5-0 Visibiliy es for lines 6

18 5.5 WORKE EPLES 5.5. poin on a line equidisan o wo given poins ou are given wo poins and and a line l. ou are asked o find he poin on l ha is equidisan o and. l (Righ) The problem The se of all poins ha are equidisan beween wo poins is a plane, which passes hrough he midpoin of he line formed by he wo poins and is perpendicular o he line. lernaively, consider a rue isosceles riangle wih as base and he desired poin on l as he verex of he riangle. f l onsequenly, we consruc a view showing in rue lengh. raw a line perpendicular o a is mid-poin. This line will mee l a he desired poin. ack projec o he op and fron views. See Figure 5-. l midpoin TL l Projec back from view # o ge f l Projec back from op view o ge 5- Finding a poin on a line equidisan o wo poins 6

19 5.5. line hrough a given poin and inersecing wo skew lines Such a problem occurs in engineering siuaions, for insance, when connecing wo skew braces by a siffening brace anchored a a specific poin. See figure o he righ. (Righ) The problem To solve he problem, consider a view in which one of he skew lines is shown in poin view. We simply connec he poin view wih he specified poin by a line and inersec he oher skew line a a definie poin. ack projec o he original views o complee he line. f The consrucion is shown in Figure 5-. In view #, we exend he poin view of hrough o mee a. This line gives he race of he required line. Projecing back ino view #, he line hrough inersecing gives he desired line. This line can now be projeced back ino he op and fron views. line hrough beween and, poin view of f line hrough beween and race of he line hrough beween and 5- onsrucing a line hrough a specified poin beween wo skew lines 6

20 5.5. line a a cerain grade beween wo skew lines ypical pracical problem migh be he following. Lines and specify cenerlines of wo exising sewers as shown in he figure below. f Sewer configuraion The sewer pipes are o be conneced by a branch pipe having a downward grade of :7 from he higher o he lower pipe. Given ha poin is 0' Norh of poin, he problem is o deermine he rue lengh and bearing of he branch pipe and show his pipe in all views. Line (in plan) measures 0'. The consrucion is shown in Figure 5-. The scale is defined wihin he drawing. We follow he seps in onsrucion 5-6 (on page 59). 64

21 , TL = 0'-8" :7 grade 0', bearing = S 5.5 E f The rue lengh is seen view #. (This can be appropriaely scaled by lengh o obain he acual disance of he new pipe) The bearing is seen in op view. 5- line a a cerain grade beween wo skew lines 65

22 5.5.4 How far from he neares face? onsider he views shown on he righ of a ransiion piece shape in which a cable is passed hrough he opening of he piece from poin. Wha is he shores disance from o he neares face? '-0" The firs sep of he consrucion is given in Figures 5-4. The neares face is p. We consruc he perpendicular from o p. The perpendicular from does no mee he face bu mees he plane represened by he face a a poin off he face. We exend he projecion line from view so ha i mees he edge of face p. This is he neares disance. 5'-0" 45 '-6" 4'-0" '-6" We sill have o deermine is rue lengh (see Figures 5-5 and 5-6). is no on face p need o find poin on face a neares in he same plane p f is he foo of he perpendicular from o he plane of face p edge view of face p 5-4 Neares disance o a face (sep ) 66

23 p Z is projeced back from view # Z f, Z Z Z is projeced back from view # rue lengh of Z is perpendicular from o in view # 5-5 Neares disance o a face (sep ) In Figure 5-5 we deermine he poin view of he perpendicular o face p shown in view. In his view he line is shown in rue lengh. The neares poin from o he face is he hypoenuse of he righ-angled riangle Z shown in view as he perpendicular from o he line. The las sep shown in Figure 5-6 deermines he rue lengh of he line Z, which is given in view. Z = '-" rue shores disance beween and Z is given in view # p Z f, Z Z 5-6 Neares disance o a face (sep ) 67

24 5.5.5 learance beween a spherical ank and pipeline In his problem, we consider he possible inerference beween an exising pipeline of a given diameer (") and a spherical ank also of given diameer (0'). Poins and are locaed as shown in Figure 5-7, which has been drawn o scale (" = 8'). We are required deermine he clearance, if any, beween he pipe and he ank and show all consruced views. f 5-7 learance beween a spherical ank and a pipeline (indicaed by i cenerline) problem definiion I is imporan o noe ha he pipeline and he cener of he spherical ank form a plane, which means we can consruc he edge view of his plane and hen is rue shape fro m which we can deermine, if any, he clearance beween he ank and pipeline afer accouning for is diameer. For his we consruc he perpendicular from he cener of he ank o he cenerline of he pipeline. The consrucion is shown in Figure 5-8. Noice ha he spherical ank appears as a circle in every view. The plane is shown only for purposes of illusraion. The rue shape of he pipeline is shown in view #. 68

25 Edge-view of f learance = '-5" True shape of 5-8 eermining he clearance beween a spherical ank and a pipelin dding a new pipeline We are given wo exising sewer lines and ha inersec a a manhole, wih and locaed wih respec o as follows: is 5' Norh 0' Eas of and 0' above ; is 0' Norh 60' Wes of and 5' above. new line is o be locaed in he plane of a a poin 0' due wes of. See configuraion o he righ. We are required o deermine he rue lenghs of each sewer line and he angle of he plane. s lies in he plane of, we can readily deermine is locaion in fron view. The remaining consrucion o deermine rue lenghs is sraighforward. See Figure

26 49'-8" 47'-" 65'-0" 40-8' rue lenghs of he pipes edge-view of 5-9 dding a new pipeline Shores disance beween non-inersecing diagonals of adjacen faces of a cube onsider he op and fron view of a cube wih non-inersecing diagonals and of adjacen faces as shown on he righ. s is parallel o he folding line in fron view, we can easily consruc he poin view of line in an auxiliary view aken from he op view. The perpendicular EF o in his view gives he shores disance. The consrucion is given in Figure 5-0. f,, 70

27 bearing N 45 E E F why is F locaed here?, a f, F,,F 5.6 E E shores disance beween diagonals 5-0 Shores disance beween non-inersecing diagonals of adjacen faces of a cube ompleing he view of a plane parallel o anoher Imagine we are given an incomplee view of wo planes as shown in he figure below. E F F 7

28 We are required o complee he view of plane EF parallel o wihou consrucing any addiional views. Since parallel lines are seen as parallel in adjacen views we can use his propery o complee he consrucion. The consrucion is given in Figure 5-. Here is parallel o EF and parallel F. E F F E 5- ompleing he view of a plane parallel o a given plane 7