Descriptive geometry
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A plane, which is parall l to the plane of projection, is called a le vel plane (fig. 3.5).
Fig. 3.5. Level planes:
a – a frontal level plane; b – a ho rizontal le el plane; c – a profile level plane
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Any line or figure contained in t he level p ane parallel to a projection plane, projects to t e last plane in true shape.
3.4. Mutual positions of two planes
According to mutual p osition of planes there are parallel plan es and intersecting planes.
In general case planes usually i ntersect. In the par icular ca se when the line of intersection is infi nity, the planes become para lel. The parallel pl anes coin cide when the dista ce between them i shortened to zero.
The planes are conside ed to be parallel if two intersecting lin es of one plane are parallel to two int rsecting lines of an other plane. It me ns in a drawing tw o parallel planes have two parallel projections of two inter secting lines with the same name laying in each projection. This is a parallelis m property of the planes. This property is used to solve problems requiring the construction of a parallel plane (fig. 3.7 ).
Fig. 3.7 Parallel planes α and β
A c ommon element of two inter secting planes is a line of int ersection which belongs to both planes. As it was menti oned above, plane s can be f several types according to their mutual location with the planes of projection. So there are several cases of plane intersection.
In t he first case when both planes are the planes of particular position he line of intersection is a p rojecting line whic h is projected as a point in a drawing. In other words, both plane s of parti ular position are projected in the line s, so their point of intersection is the line of intersection (which is projected in the point accord ing to the property of projecting line) (fig. 3.8).
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Fig. 3.8 Two inters ecting frontal projecting planes: a – a 3-dimantional model; b – a drawing of them
In t he second case, when only o ne plane is the plane of par ticular position and another is the plane of general positi on. In thi case, pla ne of particular position has one proje ction wh ch is a li ne (accord ing to th e properti es of lines of particular position); at the same ime the plane of g eneral position has all projections with distorted and non-true size images of the origin al object. So one projection of the line of intersection coincides with proj ction of the plane of particular position which is a line and anot her projection line o f intersection can b e constructed (fig. 3.9, a).
The third cas e is that w hen both planes are the plan es of the general position. It means that they do not have any proj ection as a line. All their projections are distorted and they are pro jected in non-true size.
If three planes interse ct, they have only one com mon poin t, it mean s they do not have any line of intersection, they have only the point of intersection (fig. 3.10). So, according to this i nformation, for solv ing the t sk by the third cas e it is nec essary to introduce an auxiliary plane (third plane) (intermediary) and draw the lines of intersec tion of this plane with tw o given p anes (fig. 3.9, b). The intersection of the drawn lines shows the common point of the planes. In or der to find another common point it is necessary to use another
auxiliar plane.
Fig. 3.10. Intersec tion of three planes
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Fig. 3.9. Plane intersection:
a – intersection of a fronta projecting plane (α) ith a plane of general position (A BC);
b– intersection of two planes of general position (ABC, D EF)
3.5.Relative positions of a line with a plane
The common element of intersecting lines with a lane is th e point ( n general case). According t o the clas ification of the line s and cla sification of the planes there are 3 cas es.
The first case that is when both the line a nd the plane are ele ments of particular position. In this c se a plane of particular position has the proje ction in a line. The
point of intersection can b |
found w ithout ad ditional |
ction only with the help of |
connecti n lines (fig. 3.11, a ). |
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The second case that is |
when only one ele ment – a line or a plane is an element |
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of partic ular position. In ot |
er words another lement is |
an eleme nt of general posi- |
tion. (fig. 3.11, b) In this c se the point of int ersection has to be found by auxiliary line which must be long to th e given plane in accordance with the following theorems: belongin g of the p oint to the plane, belonging of the straight line to the plan .
The third cas e that is w hen both a line and a plane have the |
general p osition. In |
this case a line lying on the plane an d intersec ing the given line |
may be o btained as |
the line |
of interse ction of an auxiliary plane passed thr ough the line with the given |
plane (fi |
. 3.12). |
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Fig. 3.1 1. Relative positions o f a line with a plane:
a – a projecting line (m) with a pro jecting plane (ABCD); b – a projecting line (m) with a plane of general position
Fig. 3.12. Intersection o a line of general posi ion with a plane of ge neral positi on: a – a 3-dimantional model; b – a drawing of them
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3.6. Line and plane pe rpendicularity
There are two basic theorems for solving tasks in case of perp endicularity:
–if a line is p erpendic lar to tw o intersec ing lines which bel ong to some plane, this line is perpend icular to this plane;
–theorem of a straight angle projection (it has been described above).
So rojection of a perpendicular can be d awn as th e true size only fro m frontal line of the level an d horizontal line o f the level, in other words fro m two intersecting lines of t he level, hich can be specified in any plane.
1.The frontal projection of the l ine m’’ which is perpendicular to a pla ne is perpendicul ar to the f ontal projection of the fronta l level line f’’ (m’’ f’’);
2.The horizontal projection of the line m’ which is perpendicular to a plane is
perpendi cular to the horizontal projection of the horizontal level li ne h’ (m’ |
h’). |
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All tasks of perpendicularity can be divided into thr ee big groups. |
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The first gro p sets constructio of a perpendicular line from a plan |
to space |
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(fig. 3.13) (draw a perpendic ular). |
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Fig. 3.13. Drawing a parallel pl ane to the given plane
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There is next algorithm for these tasks: |
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1. |
D raw proje ction of |
he frontal level line f (f’’, f’) and projection of the hori- |
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zontal le vel line h (h’’, h’) i n the given plane of |
general p osition α ( ABC) in the fol- |
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lowing w ay: f’ x, f’’ by the additional point 1; h’’ |
x, h’ by the additiona l point 2. |
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2. |
D raw proj |
ctions of the line m from a ny point |
(for exam ple A) of the given |
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plane α t o space by |
following actions: m’’ f’’; m’ |
h’. |
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3. |
Set required distance |
on proje ction of th e line m ccording to the giv en condi- |
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tions (set addition al point D |
with distance in 1 |
5 mm). |
s it is ob vious lin e m is the |
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line of a general p osition, so firstly i t has to |
e limited as a seg ment. Th n its true |
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length sh ould be found by any method (only method of the right triangle has been consider d above).
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4. The required plane β with distance 20 |
mm fro m the giv en plane α can be |
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drawn by |
two intersecting l nes (k and l). Thes e two lines have to |
be parallel to two |
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given int ersecting lines (sides of the given triang le). l’’ |
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A’’B’’, n’’ |
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A’’C’’, l’ |
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A’B’, |
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n’ |
A’C’. β (l |
∩ |
n) |
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α (( ABC). |
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The |
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from a |
point in space to a |
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second g roup sets drawing of a per pendicula |
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plane (o |
it a perp endicular). The cas e, which is used more often, is drawin g an inter- |
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section point of this perpendicular with the plan e (fig. 3.1 4). |
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Fig. 3 .14. Drawing a perpendicular to a plane
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Graphical algorithm for these tasks is the following:
1. D raw proje ction of he frontal line f (f’’, f’) and projectio n of the horizontal
line h (h’’, h’) |
in he given plane of the gener al positio α ( AB C) in the following |
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f’ |
x, f’’ |
by the additional point 1; h’’ x, h’ by the additional poin . |
way: |
2. |
D raw a perpendicular line m t hrough the given po int (K). m ’’ f’’; ’ h’. |
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3.Set an aux liary projecting pla ne β through the l ne m (the se steps have been consider d above).
4.D raw an ad ditional line 3–4.
5. Specify the point of intersection O by ntersectio n of the additional line with the perpendicular.
6. If it is necessary to determine the real distance between the given point and the given plane it c an be found by the method of the righ triangle.
By he third roup of tasks of th e perpen icularity it is nece ssary to draw some auxiliary plane (geometrical locus) which is perpendicular to the line of gen eral position (fig. 3.15). Ge ometrica locus is set only b y two intersecting lines (a fr ontal line and a horizontal li ne (lines of level)).
Fig. 3 15. Drawin g a perpen dicular plane to a line f general position
There are four steps for solving the task. |
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1. |
D raw an auxiliary pla ne β which is perpendicular t o the give n line l by two inter- |
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secting lines – a fro ntal line |
m and a h orizontal line n. m’ |
e |
x, m’’ l’’; n’’ |
, n’ l’. |
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2. |
Specify point O as |
a point o f intersection of th |
given line l with |
auxiliary |
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plane β by the way which is given abo ve.
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3. |
Connect projections of the points O and K with e ach other. |
4. |
D etermine the true s ze of the received egment of general position K. |
4. M ETHODS OF DRAWING TRANSF ORMAT ION
Different methods of orthogonal projectio ns transformation are used to make the solution of metric and posi ional problems sim pler. After such transformations new projections help to solve the problem by minimum graphic means.
4 .1. Method for Replaceme t Projection plane s
The Method for replacement of planes of projection is made up of the |
substitu- |
tion of a plane wit h a new one. The n ew plane should be perpendi cular to the remain- |
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ing one. The position in sp ace of the geometr c figure emains unchanged |
in space. |
The plane should |
e positio ned in suc h a way that the geometric figure has a particu- |
lar position to it |
which is convenie nt for solving the problem. This ma nipulation |
makes it possible by changing general position of lines and planes into a particular one (fig. 4.1).
Fig. 4.1. 3-dimentio al model of a plane replacement
Four principal problems solve d by repl cement o f projection planes
1. Transform tion of a line of general position into a line of the level. Such transform ation helps to determine the true size of the line-segment and its inclination angles to the projection planes (fig. 4.2).
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Fig. 4.2. Transformation of a line of general position i nto a frontal level line
In order to so lve the problem it is necessary to dr w a new plane V1 which is parallel to the segment. In this case the new coordinate axis pass es paralle lly to the horizonta l project on of the given line. It is necessary to dra w connection lines through the horizontal projections A’ and B’ w hich are perpendicular to the new axis.
Then z-c oordinate |
of the p oints (distance from the x-axis to the frontal projection of |
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the points) are laid |
off on th em. The new proje ction A′1 ′1 is equa l to the t ue length |
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of the segment, an d the an |
le φH is equal to the inclin ation angle contained by the |
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segment and the h orizontal |
lane of p rojection H. |
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2. Transform |
tion of a line of the level into a projecting line, i. e. set it perpen- |
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dicular to the projection pla |
e and pro ject the line as a point in this plane (fi g. 4.3). |
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Fig. 4.3. Transformation of a horizontal level line into a frontal pro jecting line
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