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3b - Descriptive Geometry for Students of Engineering, Moyer 3rd 1K

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G E O ME T RY
D E SC RI P T I V E
FO R
STUD E NTS O F E NGI NEE RI NG
BY
J A ME S A MB RO SE
M
O YE R, S B
.
.
E B
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,
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,
AM
.
f M ec h a n i c a l E ngi n eer i ng i n C h ar ge of th e Il l ech a ni c a l L a bo r a to r i es i n
th e Un i ver s i ty of IlI w h i ga n; f o r mer l y I n str u c tor i n D es cr i p ti ve G eo m etry i n
Ha r var d U n i vers i ty : E n gi n eer w i th th e G en er a l E lectr i c Co mp a n y
a n d wi th W es ti ngh o u s e C h u r ch K er r a n d Co mp a n y
M em ber of th e A mer i c a n So ci ety of M echan i c a l E n gi n eer s ; Jl I i tgl i ed d es V er ei n es d eu ts ch er
I n gen i eu r e; M em ber of F r a n k l i n I n s ti tu te; B o s to n A ss oc i a ti o n f o r th e A d van c e
men t of Sci enc e; A mer i ca n S o c i ety of C i vi l E n gi n eer s ; S o ci ety f o r th e
P r omo ti o n of E n gi ne er i n g E d u c a ti o n etc
A ss i s ta n t P r of ess o r
o
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,
T H I RD
E D I TI ON
T H I RD T H O U S A N D
NE W YO RK
J O HN
LO ND O N :
W I LE Y
H A P MA N
C
1 9 09
SO NS
H ALL L M T D
,
I
I
E
copy ri gh t,
1 904 and 1905
BY
E
J AM
B OSE MO YER
S AM R
P RE FA CE
.
TH I S b ook i s the result Of teachi n g desc ri ptive geometry to
stude n ts o f e n gi n eeri n g My aim is to presen t the su bj ect SO as
to make it most easily applicab le to the requireme n ts of rece n t
e n gi n ee ri n g practice The methods Of prese n tatio n i n thi s b ook
therefore are no t tra di tio n al E xperie n ce has shown that most
stude n ts i n o u r b est tech ni cal schools have difficulty i n applyi n g
their k n owledge o f thi s subj ect to subs eque n t work i n structural
Two thi n gs have bee n attempted i n thi s
an d machi n e desig n
b ook to overcome thi s failure Of o u r stude n ts : ( 1 ) The n otatio n
is essen tially the same as that used i n mechani cal drawi ng Fo r
a lo n g time practical drafti n g and desc ri ptive geometry have had
2
to o little i n commo n
( ) The exercises have b ee n carefull y
graded to e n courage a studen t to do thi nk i ng fo r hims elf ; an d
to stimulate h i s i n terest man y co n crete exercises showi ng usually
practical applicatio ns have b ee n i ns erted S uch exercises I thi nk
should be i n troduced from the b egi n ni ng so that the studen t may
see the practical applicatio n o f his pro blems as he goes alo ng
The data for the exercises are stated by the system Of coor
Reaso n s for choosin g this
di n ates used i n a n alytic geometry
system are O b vious Fo r a class b egi nn i n g thi s subj ect there is
a great adva n tage i n statin g the exercises wi th ab solute defi ni te
If a defi ni te pro blem i s n o t gi ven man y studen ts i n order
n ess
to Show a sat i sfactory solutio n will waste much time selecti n g
data ; an d others will presen t drawi ngs that fo r their complicatio n
are mostly u ni n telligi b le
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“
2 36 38 4
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Illustratio n s are of more u se than much wordy desc riptio n
Fo r this reaso n an u n usually large n umb er of perspective an d
orthographic drawi n gs have b ee n i n serted The il lustratio n s i n
perspective are very helpful Whe n ever it is possi ble however
stude n ts should b e en couraged to make models o f bardbo ard an d
“
pe n cils that they may b uild W hat they are drawi n g
This b ook is n o t i n te n ded for self ins tructio n Like lan guages
this su b j ect can be learn ed successfully o n ly from a teacher an d
"
The stude n t must take the
n o t alo n e from b ooks an d lectures
time to work out ma n y exercises S pace h as b ee n left o n the right
han d pages for lecture n otes an d Sketches The studen t may well
put the solutio n s for man y of the exercises o n these pages
A good deal O f space is take n to explai n P ro b lems 6 7 an d 8
These are co n sidered fun dame ntal ; an d the teacher Shoul d be sure
they are mastered before the studen t goes further With these
pro blems well i n mi n d there should be no difficulty w ith those
that follow It has b ee n my O bj ect to make the explan atio n s
Of the pro blems throughout the b ook co n siste n tly b riefer as the
su bj ect matter i s developed
I am u n der great O bligatio n to P rofessor Ira N Hollis an d
P rofessor Le w is J J oh n so n fo r much assistan ce an d e n couragemen t
i n prepa ri n g this b ook
I o w e S pecial ack n owledgme n t however
to P rofessor He n ry S J aco by who led i n teachi n g thi s subj ect
with its practical app l icatio ns He has carefully read much Of
this book an d I have received man y suggestio n s from him
Fo r assista n ce i n ma n y ways I wish to tha nk my b rother Mr
J Clare n ce Moyer M E O f P hi l adelphi a Mr C B Lewis o f
C i n ci n n ati an d Mr B rya n t White O f C ambridge
J A MO Y E R
mb
CA M R D G
D
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B
I
E
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ece
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er,
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SE CO ND
T HE
P RE FA CE T O
E D I TI O N
.
gratifyi ng results with the first editio n showed that the
methods O f this b ook were appreciated b eyo n d my expectatio n s
I n the seco n d editio n I have added a n um b er O f n e w exercises
Man y O f these appear throughout the text
I n prepa ri n g the seco n d editio n the help O f Mr A E N orto n
Fo r valua ble suggestio n s an d
P h B has b ee n i n valuab le to me
criticisms I am much i n deb ted to C omma n der B arto n U S N aval
A cademy ; P rof A dams Mas s I n st of Tech n ology ; P rof Ke n n edy
H arvard Un iv P rof O gde n C orn ell Univ P rof Ra n dall B rown
U n iv P rof S pan gler U n iv O f P en n sylva n ia ; P rof Tilde n U n iv
Of Michigan ; P rof Tracy Y ale Un iv ; an d Mr W V Moses of
the Ge n eral E lectric C ompan y
The A merican B ridge C ompan y an d the B osto n B ridge “o rk s
have ki n dly supplied drawi n gs from whi ch the data for some O f the
exercises have b ee n take n
I am much gratified that i n P rof Ferri s s b ook o n descriptive
geometry which h as just b ee n pu blished an e ff ort is sho w n to
meet i n a d egree pra c tical requiremen ts S i n ce the first edi tio n
Of this b ook appeared I have received man y letters regard i ng the
relative importa n ce to b e given this sub j ect from a practical View
poi n t i n a course i n e n gi n eeri n g These i n qui ri es i n terest me much
an d i n replyi n g I have gladly give n the results Of my expe ri e n ce
J A MO Y E R
J
CA M R D
y 1 9 05
T
HE
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B
I
GE
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an u ar
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P RE FA CE T O T H E T H I RD
E D ITI O N
.
IN D U STRI AL educ atio n is bec omi ng every day more important
The te n de n cy i n educ atio n is toward
i n all systems O f teachi ng
the ec o n omic applic atio n s The advan tages o f teachi ng with the
help o f practic al pro blems an d exercises is more appreciated tha n
ever with c orrespo n di ngly more satisfactory results These n ew
requireme n ts are measured i n a degree by the success Of this
b ook
I n this editio n s me cha nges mostl y suggested by teache rs
o
have bee n made i n the text and an i n dex h as bee n added to make
the b oo k more c o n ve n ie n t f o r refere n ce Fo r very valuable cri ti
c i sm s I am especially i n de b ted to P rof Dr Li nse l O f B erli n Ge r
ma n y and P rof J aco by Of Ithac a Much Of the work o f revisio n
h as falle n to my c olleague Mr A E N orto n of C am b ridge whose
services I c ann o t to o highly appreciate
J A MO Y E R
LY NN De cem be 1 906
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CO NT E NT S
I NT R O D U CT I O N
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HA PT E R I
L E M E N TARY PRI N C L E S AND
IP
Pl an es o f P o j ecti o n
P ro j ecti o n s o f Po i n ts an d L i n
The
0
NOTA T I O N
r
es
.
H A PT E R II
PRO MS R A T N G TO H PO N T L N
P bl m
ti l E x m pl
d P
C
B LE
ro
e
s an
EL
r ac
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I
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Co n
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o l u te
S u rfaces
,
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AND PL AN
PL AN S TANG E N T
To
E
To
S OL I D S
.
Su rfaces
H A P T E R IV
I N T RS E T O N S AND D
LO M N T
C
E
I n terse cti o ns o f Su
De
C
r f ac e s
v l p m ent o f S
e o
u rf a c e s
E VE
P
PT E R
V
I
w i th
E
HA P T E R III
PRO B LE M S R LAT I NG
of
I
es
C
Class i fi cati o n
E
P l an
E
S
OF
S OL I D S
es
.
I ntersecti o ns o f Su rf aces
CHA
MI SCE LLAN E O U S PRA CT I CA L
E xE R CI SE s
.
C
SH
HA PTE R
AD S AND
E
VI
SH
AD O W S
HA P T E R VII
C
VVA R P E D SU R F
C
HA PT E R
P RS
E
ipl s f P sp
P p tiv D i t ti n
P p tiv Sk t h f
P acti al E xe ci s
Th e
P
ri n c
er
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e rs
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s or
e rs
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se
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es
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v
e c ti
e
PE C
T
V
A S
CE
III
I VE
D r aw i n g
.
ro
m
W ki
or
ng
D raw i n gs
D E SC RI P T I V E
G E O ME T RY
I N TRO D U CTI ON
treats of th e m eth o ds of maki n g draw
ings to represen t o b j ects w ith mathematical accuracy There are
two commo n methods for such represe n tatio n B y o n e method
all ed perspective d rawi n g the chief purpose is to produce a picture
which will b e plai n to a perso n u n familiar wi th the methods used
for tech n ical drawi ngs B y the other method however the chief
aim i s to Show an O b j ect with the true dime n sio n s that are n eeded
The drawi ngs are
i n the co n structio n O f b uildi n gs an d machi n es
the n made b y a method which does n o t give a pictorial effect ;
m
f
b
O
ther
ha
d
shows
views
the
O
j
ect
fro
which by
n
O
n
t
h
e
o
t
bu
very S imple processes true dime n sio n s Of all parts can b e quickl y
O b tai n ed This latter method is called o rth ograph i c proj e ctio n
I t i s the method with w hich the stude n t must b ecome most
familiar an d with which this treatise must most c o n c ern hi m
for some time Pers pective drawi n g will b e discussed later The
method Of orth ographic proj ectio n represe n ts the outli n es Of the
O b j ect as they might b e traced o n trans pare n t plan es placed arou n d
the Ob j ect as shown i n Figs 1 ( fro n tispiece ) an d 2 a where three
views o f a hexago n al pyramid are S hown pictorially o n horizo n tal
D raw i n gs represe n ti n g these vi ews b y the
an d vertical plan es
orthographic method are made i n the same way as i n mechani cal
drawi n g The O b j ect i s thus represe n ted as though the eye were
infi nitely distan t ; that is the vani shi ng Of the l i nes Of the O bj ec t
i n the dista n ce is n o t represe n ted
I
D escri ptiv e G eometry
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C HA PT E R I
E
L E ME NT A R Y P R I NCI P L E S
A ND N
O T AT I O N
The ho ri zo n tal and two vertical plan es upo n which the
three vrew s Of the pyramid are show n i n Fi g 2 a are called the
n
s
f
r
e
n
n
l
e
oj
ctio
These
pla
es
are
always
take
at
right
a
o
n
p
p
a ngles to each other an d are desig n ated accordi ng to their positio n
The li nes Of i n ter
as h ori zo n t al fro n t verti c al an d si de v e rtic al
sectio n o i the horizo n tal w i th the fro nt vertical and side vertical
plan es are called respectively the X an d Y axes The i n tersectio n
Of the fro n t vertical and side vertical plan es is called the Z axis
These axes are show n plai nl y i n the figure an d the poi n t where
they i n tersect is called the o ri gi n and is usually marked O
2
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r
n
the
pla
es Of proj ectio n are shown i n a pic torial
3
drawi ng where they are placed arou n d a pyramid which is the
O bj ect to b e represen ted The pla n es are arranged as we must
imagi n e them placed to Sh ow the top fro n t an d side views Of the
pyramid accordi ng to th e co n ve n tio n al methods used i n prac
tical drafti ng I n this figure the views Of the p y ra mi d Shown
o n the pla n es Of proj ectio n are its outli n es made b y rays O f li ght
reflected from poi n ts o n the pyrami d perpe n dicular to a pla n e
Of proj ectio n
The poi n ts where the rays pierce these pla n es
are called the proj ectio n s of poi n ts o n the surface Of the pyra
mid Thus i n the figure two corn ers O f the pyramid are marked
a an d b
From these poi n ts dotted li n es are draw n represe n tin g
rays of ligh t reflected from them perp en dic u l ar to th e pl an es
The i ntersectio ns Of these dotted li nes from
o f pro j ectio n
h
at an d a
a an d b with the pla n es are marked respectively a
h
f
O f th ese th e first three are called the pro j ec
an d b
n
b
b
a d
tio n s o f th e poi n t a ; an d the last three the pro j ectio n s o f th e
n ts are fou n d i n the same
f
oi
t
The
proj
e
tio
s
other
poi
n
o
n
c
b
p
.
I n Fi g 2 a
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8
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that the complete proj ectio n s O f the outli n es o f the pyra
mid can be made o n each Of the plan es Of proj ectio n The co m
l
n
f
e
t
e
o
n
proj
ectio
s
the
outli
es Of an o bj ect are called its h o ri
p
z o n tal fro n t v e rti c al
an d si de v e rti c al proj ec tio n s
Fo r o b j ects
with a ngular outlin es o nl y the proj ectio ns o f the corn ers are
us ually n eeded The views Of the o bj ect are the n o f course
drawn b y j oi ni n g the proj ectio ns Of the corn ers The dotted
li n es i n the fi gure represe n ti ng the rays from the poi n ts o n the
pyramid to the proj ectio ns are called proj ectin g l in es
I n Fi g 2 a we see the plan es O f proj ectio n the hexago n al pyra
mi d an d the proj ectio ns o f the outli n es Of the pyramid o n the
pla nes as i n a picture A l l is shown as o ne complete view such
as we see whe n several sides O f an O b j ect we are looki n g at are
see n from a si ngle viewpoi n t If however the eye is moved SO
that we see o nly o ne proj ectio n at a time a set o f three difi eren t
views would b e Ob tai n ed called o rth o gr aph i c proj ectio n s B y the
methods o f orthographic proj ectio n these views wo u ld be arrang ed
I n this latter fi gure a horizo n tal li n e is first drawn
as i n Fig 2 b
to represe n t the X axis usually n ear the middle O f the space to
The n at right angles to this li n e the
be take n for the drawi ng
To represe n t b y t hi s method
Y an d Z axes are drawn as show n
the views shown pictorially i n Fig 2 a the hori z o n tal proj ectio n
is drawn behi nd the X axis an d o n the left han d Side Of the vertical
The fro n t vertical proj ectio n is drawn b elow the X axis
Y axi s
The side ver
an d vertically b elow the horizo n tal proj ectio n
tical proj ectio n is also draw n b elow the X axis b u t o n the other
side Of the Z axis As the views are thus show n the hori z o n tal
proj ectio n shows the outli n es O f the pyramid when the eye is i m
mediately over the horizo n tal plan e The fro n t an d side vertical
proj ectio ns represe n t the outli n es as they appear whe n the eye
is moved i n fro n t Of the fro n t an d side vertical plan es This is a
very co n ve nie n t way to arran ge the proj ectio ns O f an Ob j ect an d
it h as the advan tage that it is more easily u n ders tood by mecha nics
Of ordin ary ab ility than some other arra ngeme n ts o f the same
proj ectio ns that will b e discussed later
I n Fi g 3a the same pyramid that is Show n i n Fi g 2 a is rep
w ay ,
SO
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rese n ted
in a
differe n t positio n with respect to the plan es of pro
i
n
n
i
2
e
c
o
I
F
a
the
pyramid
is
placed
b
e
l
o
w
h
l
t
e
h
r
i
z
n
o
t
o
a
t
g
j
plan e an d b ehi n d the fro n t plan e ; i n Fi g 3a it is placed above
the horizo n tal plan e an d b ehi n d the fro n t plan e Fo r this arra n ge
me n t the drawi n g made by orthographic proj ectio n is show n i n
This method however is scarcely ever used as it crowds
Fi g 3b
the proj ectio n s too much for ordi n ary purposes O b serve that i n
the orthographic proj ectio n the fro n t vertical proj ectio n is ab ove
the X ax
is an d th e h o ri z o n tal proj ectio n is b ehi n d as b efore It
thus happe ns that for this method Of arra n gemen t Of views b oth
the hori z o n tal an d the vertical proj ectio n s are shown i n ortho
graphic proj ectio n O n the same side O f the X axis I n Fi g 4 a
however a little b etter arra ngeme n t of views is S hown H ere the
pyramid is a bo ve the hori z o n tal plan e an d i n fro n t o f the fro n t
plan e I n orthographic proj ectio n the same views are show n i n
This arran geme n t o f the proj ectio n s was o n ce much used
Fi g 4 b
It does n o t crowd the views as i n Fi g 3b
i n practical drafti ng
I n Fi g 5 a the pyram id is bel o w the horizo n tal pla n e an d i n f ro n t
O rthographic proj ectio n s O f these Views are
O f the fro n t plan e
shown i n Fi g 5 b This last arra n geme n t O f the proj ectio n s i s also
likely to produce crowdi ng and h as n o Sig n ifica n ce i n practical
drafti n g
It should b e me ntion ed here that the arra ngemen t o f Views Shown
by orthographic projectio n i n Fi g 2 b is adopted almost exclusively
f o r e ngi n eeri ng drawi ngs i n the U n ited S tates This arra ngeme n t
is almost universally applied i n modern machi n e d rawi n g B this
method the views are placed i n the most n atural positio n s to S uggest
a me n tal picture Of the Ob ject represe n ted b y the drawi ngs ; the to p
proj ectio n is at the top Of the drawi n g an d the fro n t View j ust i n
fro n t of the top view The right han d side view is at the right
han d side of the Sheet The views are thus arran ged where they
can b est suggest a me n tal picture to a workma n
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n g a drawi n g like Fi g 2 b b y
The
actual
method
of
maki
4
orthographic proj ectio n will n o w b e take n up It can be most easi
l y explai n ed b y S howi n g how the pri n cipal poi n ts i n the drawi n g
.
.
,
.
w ere
lai d o ut A corn er o f the pyramid marked a has b ee n referred
h
af
were poi n ted o u t
to b efore ; an d i ts proj ectio ns a
an d a
They are als o marked in the orthograph ic proj ectio ns i n Fi g 2 b
These proj ectio ns are located by their distan ces fro m the X Y
h
Thus the dista n ce from the hori z o n tal proj ectio n a
an d Z axes
to the vertical Y axis measured ho ri z o n tally sho ws the dista n ce
i n Space from the poin t a to the Side pla n e
The distan ce from
h
a
to the X axis meas u red verti call y shows the dista n ce Of the
p oin t a b ehi n d the fro n t vertical pla n e The fro n t ver tical projec tio n
af is b elow the X axis an d the dista n ce measured vertically from
this proj ectio n to the axis shows the distan ce that the poi n t a
is b elow the horizo n tal plan e The Side vertical proj ectio n a is
fou nd by layi ng o ff a di stan ce measured hori z o n tally equal to
the dista n ce that the poin t a is b eh in d the fro n t vertical pla n e
O f course the tw o pro j ectio n s a f an d a are at equal distan c es
b elow the X axi s
.
s
,
.
,
.
.
w
.
,
,
,
,
,
,
,
,
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,
,
,
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,
,
,
,
,
,
,
s
.
,
,
,
,
.
8
,
.
r
t
r
d
s
e
r
t
i
f
e
c
a
c
c
a
l
e
m
o
e
The
methods
i
iv
o
tr
y
are
useful
o
p
p
5
g
to e ngi n eers and a rchitects i n ma n y structural an d mechanical
Operatio ns I n desig ni ng and co n structi n g b u ildin gs and machi n es
it is Ofte n n ecessary to fi n d the true size an d Shape Of parts Shown
Wh e n the n ecessary dimensio n li n es are no t shown
o n drawi n gs
and whe n some Of the actual le ngths are fo resh o rten ed i n di rect
geometric al methods Of measureme n t must be used The study
Of this su b j ect is useful however fo r more than its i n dustrial utility
The stude n t b ecomes accustomed to co n sideri n g very complicated
geometri cal comb i n atio ns an d to follow accu ratel y t h e correspo nd
“
e n ce b etwee n the drawin gs an d the O b j ects represe n ted
De
“
scriptive Geometry a well k n own e n gi n eer h as said trai ns o n e
”
I ndeed it exercises i n the most precise
to see arou n d corn ers
mann e r the power to vi sualize which is represe n tin g to ourselves
clearl y an d easily ideal Ob j ects as if they were really b efore us
The importan ce o f this sub j ect i n trai n in g stude n ts fo r work i n
des i gn i ng b ridges b uil di ngs an d machi n ery can hardly b e over
u
b
t
u
f
hi
s
s
e
h
t
c
d
e
s
n
t
timated
N
everthel
ss
s
d
y
o
j
o
t
t
e
if
o
e
es
l
i
s
m
s
s
e
a
d
h
t
e
n
k
hi
o
i
t
t
t
n
o
e
t
u
d
s
teach th e
g
.
,
.
,
.
,
_
.
,
,
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'
,
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-
,
,
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,
,
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,
,
,
.
,
,
.
The methods Of d escri p tive geome try are a bsolu tely gen eral i n
their app licatio n so tha t i n the w or k that follows if the solutio n
o f a p ro b lem is give n for an y o n e o f the arra ngeme n ts o f views
that have b ee n explai n ed it is applica ble as well for all the others
T h e pl an es o f proj ectio n also ar e i n de fi n it e i n ext en t ; that is
they ex tend withou t limit i n every directio n P roj e ctio n s o f l i n es
m ay be p r o d u c e d as f ar as the y are n e eded wh en th e solutio n o f
a p ro bl em r e q uir es it
,
,
,
.
,
,
,
.
,
.
other than the plan es Of proj ec tio n must be O fte n
repres en ted A ny plan e that can b e Show n withi n the limi ts o f
the drawing wi ll i n tersect o n e o r more o f the plan es Of proj ectio n
These li n es o f i n tersectio n called the tra ce s o f
i n s traight l i n es
the plane are made use Of to represe n t plan es i n orthographic pro
n
n
n
rsectin g the three pla n es
A
b
draw
gs
O
lique
pla
e
i
te
i
n
i
n
ec
t
o
j
n tersectio n o f the pla n e
The
i
o f proj ect i o n i s sho n i n Fi g 6a
w
with th e horizo n tal pla n e Of proj ectio n i s called its h orizo n tal
I ts i ntersectio ns with the fro n t an d side vertical pla n es
trace
are called respectivel y i ts f ron t vertic al an d si de vertic a l tr a c e s
The same pla n e is Shown i n Fig 6b by its traces as it is represe n ted
Fo r b revity however
b y the me thod O f orthographic proj ectio n
these traces will be called simply h o ri z o n tal fro n t an d si de tr a ce s
The Simpler n ame for the last two traces can lead to n o co n fusio n
Fo r
an d short n ames are prefera ble to a bb reviatio n s o r s y m bols
th e same reas o n the pla n es Of proj ectio n wil l b e called hereaf ter
simply h o rizo n tal fro n t and si de pl an e s ; and w e shall use the
correspo n di ng simpler n ames fo r the proj ectio n s o f poi n ts and li nes
6 P l an es
.
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N O TA T I ON
.
n
t
t
n
i
n
s
a
n
The
roj
io
Of
oi
Space
are
desig
ated
a
small
e
c
b
p
y
p
7
letter w i th h f o r s placed above it to i n dicate w hich proj ectio n is
meant ; thus the h ori z o n t al fro n t and si de proj ectio n s o f apo i n t
h
f
a
a
are
mark
d
respectively
an d a
I n th e drawi ng these
e
a
proj ec tio ns are located by the di stan ces o f the poi n t a from the
plan es o f proj ectio n measured parallel to t h e X Y an d Z ax es ;
.
,
,
,
,
s
,
,
,
.
,
,
,
the geometry o f space the di stances alo ng these axes are
represented respecti vely by the co Ordi nateS as y and z
D i st an ces al o n g th e X axi s ( represen ted by th e a: co o rdi n ates)
Lo n g
are m e asu red to th e l ef t o r ri gh t f r om th e si de pl an e
u sage h as e stabli sh ed th at th ese di stan ce s sh al l b e c o n si dered
n egativ e w h en m e asu r e d to th e l e f t o f th e si d e pl an e an d po si
tiv e w h en m easu red to th e rig h t
D i stan ces al o n g th e Y axi s ( represen t ed by th e y co o rdi n at es)
T h ese are
are m e asu red b ehi n d o r i n f ro n t o f th e f r o n t pl an e
n egati ve w h en b e hi n d an d po sitiv e w h en i n fro n t o f th e f r o n t
and
in
,
,
.
,
.
.
.
l
a
n
e
p
D i stan ces
.
lo n g th e Z axi s ( represen te d by th e z c oordi n ates)
are m easu re d b el ow o r ab ov e th e h ori z o n t al pl an e ; n eg ativ e
w h en bel o w po sitiv e w h en ab ov e
—
—
a
2
If a poin t is represe n ted as:
3
we mean that the
= — 3 an d z = — 4 — that
poi n t a h as fo r its c rdi n ates a:
y
;
the poin t a is 2 u ni ts Of le ngth to the left o f the side plan e 3 u nits
b ehin d the fro n t pla n e an d 4 u nits b elow the horiz o n tal plane
a
.
,
,
,
,
,
,
.
The proj ection s o f a l i n e are den o ted by the pro j ectio ns O f
tw o o r more po i n ts i n the li n e ; thus a li n e co n tai ni ng the tw o
poi n ts a and b i s called the li n e a b and is represen ted by the co br
di n ates o f these tw o po i n ts
The poin t where a l i ne i n tersects the horizo n tal plan e Of pro
b
s
h
i
is
marked
the
letter
the
i
tersectio
s
a
o
n
a
n
n
ec
t
i
d
n
f
o
y
;
j
lin e with the fro n t and side pl an es by i i and s i T O locate thes e
poin ts more plai nl y draw a small circl e arou n d th eir pro j ectio ns
a s show n i n Fi gs 8 and 9 o n pag es 2 1 an d 2 9
8
.
.
.
.
,
.
f
s
t
r
a
e
The
c
O
9
pl an es are repres e n ted i n a w ay sugges ted by
the method Of co Ordi nates fo r poi nts A plan e is shown in Fi gs
T h e horizo n tal and fro n t traces i n tersect o n the X
6a an d 6b
axi s to the left o f the origin and make with thi s axis the a ngles
mark ed a and fi B y symb ols such a pla n e P is represen ted as
T his n otatio n fo r planes is rememb ered b es t by
a
o b servi n g that the fi rst numb er ( the x co Ordi n ate) represe n ts the
.
.
.
.
.
°
,
,
,
i n tersectio n o f the traces o n the X axis ; and that the seco n d an d
thi rd refer to the n umb er o f degrees the horizo ntal an d fro nt
traces make with the X axis These a ngles are measured i n anti
cl ockwise direc tio n as i n trigo n ometry I n Fig 6b the a n gles a
I n such a n otatio n
an d fl are respectivel y ab out 2 0 an d
for the traces it is n o t n ecessary to desig n ate the side trace as
an y two traces deter mi n e the pla n e an d data co n cer n i n g a third
is usually superfluous I n the drawin gs the traces of a plan e are
marked with the same letter used i n nami ng the plan e with H
Fo r example i n Fi g 6b the hori
F o r S w ri tte n b efore i t
z o n tal
fro n t an d side traces of the pla n e P are marked
respectively H P F P a
nd S P
I n all drawin gs the horizo n tal an d fro n t traces must in tersect
o n the X ax i s
an d testi n g f or this i n tersectio n is a check o n the
accuracy O f the co n structio n s
10
Li n es an d traces O f pla n es give n or required are r epr e
sen ted i n orthographic proj ectio n s b y full li n es whe n visible by
dotted lin es whe n in v isible P roj ecti n g lin es are i n dicated by
short dashes Traces O f auxiliary pla n es are usuall y represen ted
by o n e lo n g an d two S hor t dashes
.
.
.
°
,
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,
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,
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.
.
C HA PT E R II
P R O B L E MS R E LA T I N G
T O T HE
P O I NT LI NE
,
,
A ND
P LAN E
The relatio n b etween the actual positio n of a li n e i n spac e
an d its pro j ectio n s o n a d rawi n g is easily co n ceived for a li n e tha t
is parallel to a pla n e O f proj ectio n D i fficulty is ho w ever usually
experien ced i n co n c eivi n g this relatio n for a li n e that is O b lique to
all th e pla n es Of proj ectio n A pictorial drawi n g of such a lin e is
Shown i n Fi g 7 a The proj ectio ns Of the li n e are drawn by j oi n i n g
Fo r the li n e show n
th e proj ectio n s O f an y tw o poi n ts i n the lin e
i n the figure the poi n ts most easily determi n ed i n the drawi n g are
These i n ter
i ts i n tersectio n s with the horizo n tal and fro n t plan es
h
h
i
sec tio n s are marked a and b The former a has its horizo n tal
proj ec tio n coi n cide n t w ith the poi n t itself and i ts fro n t proj ectio n
is of cour e i n the X axis at a f S imilarly the la tter i n tersec tio n
h
bf has its horizo n tal an d fro n t proj ectio n s at b ( i n the X axis )
The horizo n tal and fro n t proj ectio ns Of the li n e are
an d at bf
"
”
the n draw n by j oi n i n g a with b an d a f w ith bf
I I
.
,
b
,
,
.
‘
.
.
.
.
,
.
.
,
,
,
s
,
.
,
,
,
.
.
reversin g the process the i n tersectio ns o f the li n e w ith
the pla n es o f proj ectio n can be fOu n d whe n the proj ectio n s O f the
li n e are give n ; thus i n Fi g 7b the i n tersec tio n Of the li n e a b with
th e horizo n tal pla n e is the poi n t a Show n by its horizo n tal pro
h
n of the fro n t proj ectio n Of the
n
ehi
d
the
i
tersectio
n
n
t
i
a
b
ec
o
(
j
li n e with the X axi s ) an d its fro n t pro j ectio n a f (i n the X axis )
The in tersectio n Of the same lin e with the fro n t pla n e is the poi n t
"
f
n
n
s
I
a
b
d
show
Similarly
its
proj
ectio
by
n
b
Wh e n o n l y tw o pl an e s o f proj e ction ar e m en tio n ed ( as i s
the case for mo st pr o bl em s) it i s assu me d th e si de pl an e i s n o t
12
By
.
.
,
,
,
,
,
,
.
,
)
,
.
,
n ee de d
1
3
.
.
P ROBLE M
the i n ter secti o n s
o
To draw the pr o j ecti ons o f a li n e havi ng gi ven
the l i ne wi th the ho r i zo ntal and fron t planes
1
f
.
.
Method Draw th e horizo n tal pro j ectio n Of the li n e by j oi ni ng
the horizo n tal proj ectio ns Of the two in tersectio ns Draw the
fro n t proj ectio n Of the li n e by j oini n g the fro n t proj ectio ns o f the
tw o in tersectio ns
.
.
.
E XE
Fo r
n
expressed
1
otatio n
see A rts
i n i n ch es
.
R CIS E S
—
7 9
.
—Th e
c
oor di n ates
poi n ts
of
are
.
n
n
n
n
li
e
i
t
e
rsects
the
horizo
tal
pla
e at the poi n t a
( )
— 3 — 2 0 an d the fro n t pla n e at the poi n t b — 1 0
)
Draw the proj ectio n s Of the li n e
—
—
n
n
e
li
e
passes
t
rough
the
poi
t
b
A
h
2
2
n
a
d
)
(
i n tersects the horizo n tal an d fro n t plan es at o (0 0
Locate the proj ectio ns Of the lin e
a
.
A
,
,
,
,
.
,
,
,
,
.
;
Gi ven the proj ecti o n s of a l i ne to find the poi nt
4
wher e the l i ne i n tersects (a) the hori z on tal pl an e ( b) the fro n t pl an e
Method Fo r (a) produce if n ecessary the fro n t proj ectio n o f
I
.
PROBLE M
2
.
,
,
.
.
the li n e to the X axis This is the front proj ectio n o f the required
poi n t Of i n tersectio n TO fi nd the hori z ontal pro j ectio n draw a
pe rpen di cular to the X axi s through the fro n t proj ecti o n j us t fou nd
to meet the hori zo n tal pro j ectio n Of the give n li n e ; Fo r ( b) pro
duce the horizo n tal pro j ectio n o f the li n e to the X axis This is
the hOri z onta l proj ectio n o f the requi red poi nt o f i n tersectio n Th e
front pro j ectio n o f the poi n t is the n e asily fou n d
.
“
.
.
,
,
‘
.
.
.
E XE
2
.
R CI SE S
Give n
a li n e p assi ng through the po rn ts c 3
—
2
Fi n d the poi n ts where this li n e
1
an d d
i n tersects ( 1 ) the horizo n tal plan e (2 ) the fro n t plan e
n
3
the
side
pla
e
( )
w
n
t
n
n
s
o
o
Draw
the
proj
ectio
Of
li
es
a
e parallel to th e
( )
horizo n tal plan e an d O blique to the fro n t pla n e the other
parallel to the fro n t plan e and O bl ique to the hori zo n tal
,
,
,
,
,
.
3
.
—1 — 2
,
)
,
,
n
b
F
i
d
)
(
the in tersectio ns o f each li n e wi th the horizo n tal
an d fro n t pla n es o f proj ectio n
O n the li n e give n i n E x 2 locate a poi n t k 1 i n ch b elow
the horizo n tal pla n e an d a n other poi n t l i i i n ches
b ehi n d the fro n t pla n e
S ho w three proj ectio n s of each
poi n t located
Give n the li n e through the poi n ts e — 3 — 1 4 2 ) an d f
— 1 — 2
Locate its proj ectio n o n the side pla n e
A ssume any l i rie i n space an d draw its proj ectio n s
Fi n d
the n ew proj ectio ns o f this li n e ( a) when the X axis is
perpe n dicular to its former positio n ( coi n cidi ng with the
vertical Y axi s ) ( b) whe n the X axis is revolved 30
n
a
( ti clockwise ) from its first positio n
S uggesti o n The exercise i s easily solved b y fi n di n g h i
Whe n the plan es of proj ectio n are
an d f i o f the li n e
revolved the positio n s of these poi n ts with respect to the
axes will n dt b e cha n ged
.
.
,
,
,
,
,
.
.
5
.
,
,
,
,
6
.
.
,
.
°
,
-
.
.
.
,
.
li n e lyi ng i n a give n pla n e i n te rsects the horizo n tal plan e
o f proj ectio n at a poi n t i n the hori z o n tal trace ; that is i n the lin e
showi ng the i n tersectio n of the give n pla n e with the horizo n tal
plan e The same li n e i n tersects also the fro n t pla n e at a poin t i n
the fron t trace o f the given plan e This is illus trated b y a pictorial
d rawi n g i n Fi g 8 a where a li n e a b lyi ng i n a plan e P is shown
The li ne is here produced upward u n til i t i n tersects the hori z o n tal
plan e at h i and down ward to i n tersect the fro n t plan e at f i
The pro blem n o w takes this form (see Fi g 8b) z G ive n o nly
the fro n t proj ectio n af bf of the li n e a b determi n e the h o rizo n t al
h
h
ne P
h
n
a
i
n
n
roj
e
tio
that
li
e
b
shall
lie
the
pla
a
so
t
e
b
c
p
h
h
I nstead of fi ndi ng at o n ce the poi n ts a b it is first n ecessary
to loc ate b oth proj ectio n s of th e poin ts h i an d i i where the
give n li n e a b i n tersects the hori z o ntal and fro n t planes respect
The n the hori z o n tal projectio ns of the poi n ts hi an d i i
i vel y
provide wh e n j o i ne d th e required horizo n tal projectio n of the li n e
O bvi l y th h i tal p j ti f th p i t h i i i n th h i t l
m
h
i
i
i
h
h
t
p
j
i
p
t
d
t
t
P
t
f
f
f th
l
t
t
p
f fi
e p
ti v l y
S i m i l l y t h f t d h i t l p j ti
X xi
1
5 A
.
,
.
.
.
,
,
.
,
,
.
.
,
,
,
,
.
,
,
,
.
,
ous
e
r ac e o
a
s
.
.
, ,
e
an e
ar
or z o n
,
e
an
ro n
on o
ro e c
e
an
ro n
e
ro ec
or z on a
o n
on
o
ro e c
e
s
e
sa
ons o
e
o n
a re r s
or zo n a
s
ec
n
e
e
i n th e f r o n t trace
hi
to o
an d
i
of
i i n th e X
.
m any l e tte r s
P
an d
i n th e X
axi s
.
m ar k ed to
axi s a r e n o t
j
I n th e fi gu r es th e pro ec ti o n s o f
a o i d co n f u si n g th e dra w i n gs w i th
v
.
the horizon tal projectio n of th e li n e b etwee n h i an d f i
h
h
thus fou nd the required projectio ns a an d b are loc ated b y p ro
f an d bf
i
n
s
n
X
li
perpe
dicular
to
the
ax
through
i
n
e
a
s
ec
t
j
g
1 6 P R O BLE M 3
Gi ven o ne proj ecti o n o f a l i n e o r a po i n t i n a
On
,
,
'
.
.
.
,
e
r
o
ec
n
d
o
t
h
er
t
o
n
i
a
o
t
h
e
n
l
n
e
t
v
i
,
p j
g
fi
p
Metho d (a) Give n so n e proj ectio n
,
.
a li n e Determi n e the
po sitio ns o f the in tersectio ns o f the li n e i n the give n pla n e w ith
th e horizo n tal an d fro n t pla n es and draw the re q ui red proj ectio n
M ethod ( b) Given o ne pro j ectio n o f a poi n t Draw a pro
nd
n
oi
t
i
i
n
f
an y li n e through the give n proj ectio n o f the
F
o
c
t
o
e
j
p
the other proj ectio n o f this li n e as descri b ed ab ove A t the i n ter
sectio n o f the proj ectio n that has b ee n fou n d with a proj ecti n g
li n e from the given proj ectio n o f the poi n t is the required pro
of
.
.
.
,
.
.
.
.
,
,
j
ecti o n
.
E
7
.
X E R CI SE S
—2
—1
i s in a
an d b
The lin e through a
plan e P (0
Draw the fro n t an d side pro
o
n
s o f the li n e
i
ec
t
j
G iven o n e proj ectio n o f a li n e which is o b lique to the X
axis i n a plan e Q with traces parallel to th is axis Fin d
an other proj ectio n o f the lin e
Locate a poin t m i n the pla n e P give n i n E x 7 that
shall b e i n ch b ehin d the fro n t plan e I n the same
plan e locate also a po i n t n 1 1 in ches b elow the hori
z o n tal pla n e
Draw the three projectio n s o f a poin t 0 i n a plan e R
which is perpen dic ul ar to tw o plan es o f pro j ection
Draw the three proj ectio n s o f a po i n t at in a plan e S
which is parallel to the horizo n tal plan e
The s10 pe o f a roof o n a b uildin g is represen ted by i ts hori
maki ng an gles respectivel y of
z o n tal an d fro n t traces
There is a hole i n th e
1 50 an d 2 2 5 with the X axi s
roof for a chimn ey of which the horizo n tal projectio n is
shown i n th e drawin gs as a regular hexago n ( length o f
,
,
,
.
8
.
.
,
,
.
9
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,
,
,
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,
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»
,
,
,
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,
,
,
,
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,
,
.
,
0
°
.
"
,
ides is i n ch ) The horizo n tal pro j ectio n o f the cen ter
o f the hole is located 1 ? i n ches to the left o f the in ter
sectio n o f the traces and i n ch b ehin d the X axis Draw
th e fro n t proj ectio n of the hole
The
required
r
o
c
e
(
p j
tio n is fou n d readily by drawin g the proj ectio n s of the
diago n als of the hexago n ) S tate the usual n otatio n fo r
the plan e of the roof (A rt
A circular wat r pipe o f which the diameter is 1 i n ch
e
passes through an i n clin ed floor represen ted o n a draw
i n g b y h orizo n tal an d fro n t traces makin g a n gles respect
i vel y of 1 5 0 an d 2 1 0 with the X axis
T h e ce n ter lin e
o f the pipe is parallel to the horizo n tal pla n e
The fro n t
proj ectio n of the pipe is therefore a circle an d its cen ter
is shown 3 i n ches to the left o f the i n tersectio n of the
traces an d 5 i n ch b elow the X axis C omplete the draw
in g b y fi n di n g the horizo n tal proj ec tio n of the hole
through which the pipe passes ( D raw a n umb er of
diameters of the circle, produci n g them whe n n ecessary
The
an d fi n d the horizo n tal proj ectio n s of these li n es
poi n ts i n the outli n e of the hori z o n tal proj ectio n of the
hole will b e vertically over the correspo n din g poi n ts i n
the g i ven fro n t proj ectio n )
The hori z o n tal trace of a plan e i n tersects the X axis at
—
=
A poi n t
1 % an d makes with it an a n gle of
x
—
—
—
4 ) lies i n the pla n e
D etermin e the fro n t
1
1
0
trace o f the plan e ( D raw through 0 an y lin e that will
have its hori z o n tal i n tersectio n (h i ) i n the horizo n tal
trace The i i of this li n e determin es the fro n t trace )
s
.
.
.
.
.
13
.
,
°
,
°
.
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,
?
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.
,
.
.
14
.
5
,
.
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.
.
1
to
a
7
.
P R O B LE M 4
T o draw thro ugh
.
a
e
i
v
g n poi n t
a
li ne para llel
e
a
n
e
n
l
i
v
p
g
.
Method Draw o n e proj ectio n of an y l i n e an d b y the method
prob lem determi n e its oth er pro j ectio n so that the
o f the last
lin e will lie i n th e given plan e Through the two proj ectio n s of
th e gi ven poi n t draw the proj ectio n s of the required li n e parallel
i
n
n
the
correspo
di
g
proj
ectio
s
of
the
l
e which has b ee n drawn
n
n
to
,
.
,
.
,
'
the plan e The li n e through the poin t and the lin e in the plane
shown t hus by parallel proj ectio n s are parallel to each other ; and
the lin e passi n g through the poin t b ei ng parallel to a lin e lyi ng i n
the pla n e is parallel to the pla n e
P
h v p
ll l p j t i n
ll l l i
n
ny p l
in
.
,
,
,
.
,
a ra
n es
e
a
a ra
e
e
ro e c
E XE
15
—3
o
an e .
a
s o
R CISE S
Through a poin t a
draw a li n e parallel to
—
the pla n e R
3
Draw through the poi n t of i n tersectio n o f th e li n e b — 3
—
l%
c
0
{ ) with the horizo n tal pla n e
a lin e parallel to the pla n e T (0
Fin d the i n tersectio n with the side pla n e 8 i of a lin e with
b oth proj ectio n s paralle l to the X axis
—
3
Fin d the i n tersectio n o f the li n e d
e
Through thi s poin t draw
1 3) wi th the side pla n e
1
the side proj ectio n of a lin e wh ich is to b e parallel to the
pla n e U o f which the hor i zo n tal and fro n t traces are li nes
parallel to the X axis The former trace is 1 5 i n c h es
behi n d an d th e latter is 2 i n ches b elow the X axis
.
— 1
,
,
,
16
.
,
-
s
,
,
,
,
17
,
.
,
.
18
,
.
7
,
.
.
.
that the lin e a b i n Figs 8a and 8b which are
agai n shown o n the Opposite page is in the pla n e P O bse rve
also that the i n tersectio n s h i an d i i o f th i s li n e are respectivel y
No w if the pro
i n the horizo n tal an d fro n t traces of the pla n e
its
i
ter
o
n
s O f a seco n d li n e i n the same pla n e were give n
i
n
c
t
e
j
sectio ns h i an d i i wo uld b e poi n ts also i n the traces of th e pl an e
The n suppose the co n di tio n s of the pro blem are reversed and the
i
n a pla n e P are gi v e n an d we are to co n
two
li
n
es
s
f
o
o
n
e
c
t
i
o
r
p j
struct the traces H P an d F P The horizo n tal an d fro n t i n ter
sectio n s h i an d f i o f each lin e are usually very easily fou n d T he
tw o horizo n tal i n tersectio n s will determi n e th e horizo n tal trac e ;
and the tw o fro n t i n tersectio n s will determin e the fro n t trace
Le t us n o w appl y this method to a co n cre te pro blem
I n Fi g
9 the horizo n tal an d fro n t projectio n s o f a tria n gu lar p yramid are
shown and w e w i sh to sh o w th e pl an e o f a si de a b c by i ts traces
18
Rememb er
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,
,
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,
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,
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,
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,
‘
,
,
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,
,
,
'
.
,
,
,
.
.
.
,
.
.
The proj ectio n s o f two li n es a b an d a c in this side are marked
The horizo n tal an d fro n t i n tersectio n s of each lin e
i n the figure
with the pla n es of proj ectio n are marked respectively h i an d f i
The horizo n tal trace O f the plan e is O b ta in ed by j oin i n g the two
horizo n tal i n tersectio n s an d the fro n t trace by joi n in g the two
If we call the side a b c the plan e P the
fro nt i n tersec ti o ns
horizo n tal trace should b e marked H P an d the fro n t trace F P
The accuracy o f the drawin g should b e checked b y O b servin g
whether these trace s i n tersect o n the X axis
I n the example j ust explai n ed the li n es determin in g the pla n e
i n te r se ct ed each other ; b u t the solutio n is the same for paral l el r
li n e s A pla n e is determi n ed also b y an y three poin t s n o t i n the
same straight lin e o r b y a poi n t an d a l i n e The first case that
of the three poi n ts is most easily solved by j oi n i ng them by straight
”
“
I n th e seco n d case that of the poin t an d
li n es two an d two
li n e solve by drawin g through the poi n t an y lin e i n tersecti ng the
given lin e B oth bases are thpn resolved in to that of drawi n g a
pla n e through tw o i n te r se cti n g l i n es
.
.
,
'
,
.
.
.
,
’
.
.
,
,
,
.
,
,
.
.
r
n
h
r
h
n
w
n
t
n
e
e
l
The
pro
lems
of
d
i
l
oug
li
ar
ll
a
a
a
e
a
e
o
b
p
g p
9
to an oth e r o r th r ou gh a giv en poin t p a rall e l to tw o gi v en lin e s are
scarcely more tha n variatio n s of this same pro blem I n the firs t
case o n e lin e i n the required pla n e is give n b y its proj ectio ns
Through an y poin t i n the first li n e
an d the directio n Of a n other
draw a li n e parallel to the seco n d an d the problem is easily
solved I n the last case the directio n s of two li n es are given an d
the pla n e is determin ed b y drawi ng two li n es parallel to them
through the give n poi n t B oth cases are resolved agai n i n to
passi ng a pla n e through two i n tersecti n g li n es
1
.
,
,
.
,
.
.
,
,
.
-
.
.
p
I f tw o l i n es i n te rse c t, th e h o riz o n tal ,
I n tersecti ng l i nes m e et i n a o i n t
r o ec ti o n s o f th e
fr o n t , an d si d e r o e cti o n s i n te r se c t i n th e c o rres on d i n g
p j
p
o int
p
.
v
s ha
l
i
n
e
l
l
e
l
r
a
P
a
T
th a t i s , i f tw o l i n e s
tw o
.
si
de
p j
e
are
ro e c ti o n s are
th e i r h o ri z o n tal , f ro n t ,
p
p
p j
r o ec ti o n s
th e tw o h o ri z o n tal , th e tw o fro n t ,
a r al l el
ar al l el
an d si d e
p j
.
p
a ral l el
an d
;
th e
20
PR OBLE M
.
r a l l el
a
p
l i n es
5
T o pass
.
a
l
a
n
e
h
o
r
t
u gh tw o i n ter secti n
p
g
or
.
Method Fi n d the poi n ts of in tersectio n of the given li n es with
the hori z o n tal an d fro n t pla n es The lin e j oin i n g the i n tersectio n s
The
i n the hori z o n tal pla n e i s the hori z o n tal trace of the pla n e
lin e j oi n i n g the i n tersectio n s i n the fro n t plan e is the fro n t trace
The side trace is located by the horizo n tal an d fro n t t races
.
.
.
.
.
E
X E R CIS E S
the f o llowin g exerci ses m ark the hori zon tal fro n t an d
si de trace s o f a pl an e n o t other w ise desi gn ate d by the l etters
H P P P an d S P
—
—
n
1
a d b
2
1 9 The li n e through a
i
2
0) an d d
m ter sec ts the li n e thro u gh c
—
—
Pass a pla n e through
1
l%
at e
these two li n es
s
e
e
n
t
r
c
t
i
n
n
i
f
n
a
raw
the
proj
ectio
s
O
two
li
es
n
D
20
g at the
y
—
Determi n e the pla n e i n which
1
poi n t m
these two lin es lie
n
e
a
r
a
l
l
l
n
f
raw
the
traces
a
pla
e
through
the
li
es
o
D
21
p
—
—
—
—
—
l
I
m
i
a
2
b
1
i
i
a
d
g t) ;
2
)
i
In
,
,
,
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,
,
“
,
,
.
-
,
,
,
.
.
,
.
.
_
-
,
—1
22
)
,
d
-
,
P ass a plan e through the three poi n ts :
.
—
1,
b
23
— 1
i
— 1 —
0
n
d
a
t) ,
,
—
t
t
w
,
,
— i‘
r
a
—2
—
,
i
,
)
—1
.
,
,
,
,
,
n
d
n
i
d
the
pla
e
passi
g
through
the
li
e
n
n
F
—
1
2
n
o
i
t
the
n
d
a
§
e
f
p
h
n
Through
the
li
e
a
g
( )
pass a plan e parallel to the li n e i
—
,
1
-
—
—1
h
g
( O bserve
.
that the traces O f the two plan es are parallel )
—
n
pass
a
pla
e
parallel
1
n
0
Through the poi t
—
—
n
d
a
2
1
l
4
i
to the li n es It
)
g
.
25
.
,
—
i)
i
i (
l
n
Through
pass
a
pla
e
para
lel
to
i
b
j
( )
~
,
,
,
m
— 1
,
n
,
—2 — 2
,
,
,
Pass a plan e whi ch is parallel to the X axi s through the
—
—
Draw the trace of the plan e
1
2
poi n t p
o n the side pla n e
The corn ers
A skew b ridge is shown i n the figure below
of the portal Of the b ri dge are give n by the poi n ts a
‘
,
,
.
.
—4
-
,
2i
b
.
—2
i
—2
.
s a )
,
,
a
(
c
s
,
w
it
the figure The po rtal is the plan e
The li n es a b
b ou n ded b y the li n es a b c d an d b d
an d c d are called the en d po st s an d b d is the po rtal
stru t
D etermi n e the pla n e o f the portal
d
(0,
as
in
.
.
,
,
.
.
The
n
n
o
b
posts
the
portal
Of
a
skew
ridge
are
show
of
en d
—
—
2
r
i
n
e
gi
eer
s
drawi
gs
b
the
l
es
n
i
n
n
an
y
—
—
—
1
u
a
—
d
t
n
}
a
s
i
ll
i
s
H
>
i ;
a
The horizo n tal proj ectio n shows the plan an d the
n es
n
o
D
ro
t
proj
ectio
the
elevatio
raw
the
three
pla
n
n
f n
Of proj ection the traces o f the plan e i n which the li n es
n
n
b
a
n
i
lie
Measure
degrees
the
gle
etwee
r s an d t u
n
n
r
i
z
n
d
a
pla
e
Of
the
posts
the
ho
o
tal
pla
e
en d
n
th e
—
c
2
b
Th e poi n ts a
—
deter
i
e
a
surface
n
m
1
an d d
n
?
n
you
a
pla
e
which
will
i
clude
these
poi
ts
n
fin d
Can
?
Is it then a plan e surface
—
2
a
n
pyramid
is
give
the
poi
ts
b
n
y
A tria n gu lar
—
f
o
r
n
at
the
co
ers
— 2
c
n
d
a
b(
2)
its b ase and its vertex d which i s 1 % i n ches ab ove the
n d the traces
F
n
0
d
ase
equidista
t
from
b
a
i
a
n
an d
b
’
,
,
-
v
,
,
,
,
,
,
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.
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,
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,
.
1
7
u
,
,
,
,
.
plan es
respectively
Of
th e
of a
db
an d
bd
c,
mark i ng them P
an d
Q
.
practice the prob l ems o f des c ri pti ve geome try are
mai nl y those o f fi n di n g the i n tersectio ns o f plan es with other
plan es ( P ro blem
the in ters ectio ns o f l i n es with planes ( P rob
lem
o r the true le n gt h s o f lin es ( P ro b lem
These may be
spoken o f as the fun dame n tal prob lems o f descriptive geometry
E very stude n t mus t n o w learn these solutio ns so that he can use
them immediately u n der any co n ditions
I n Fi g 1 0a two pla n es P an d Q are shown by their tra ces
A
The plane
tria n gular pyramid i s also shown by its proj ectio n s
o f its side a b d is the pla n e marked P ; an d a n other side b d o
O b viously the i n tersecti o n o f these tw o sides
i s the plan e Q
f
n
d
b
a
n
i
pla
n
es
is
the
edge
o
the
pyram
d
d
co
struct
o
o
n
n
i
is
;
)
(
n ecessary whe n the poi n ts b an d d are give n b y their proj ecti o ns
The first o f the fun damen tal problems me n tio n ed ab ove arises
however when the poin ts b and d are n o t k n own the traces
an d the li n e in which the pla nes in
o f the plan es are give n
The pro blem to be solved is shown
tersect must b e foun d
more simply in Fi g l ob where the traces H P F P H Q an d F Q
shown without the pyrami d Explan atio n is u nn ecessary
are
to show that if a li n e i n the pla n e P i n tersects a li n e i n th e
plan e Q a po in t i n th e lin e o f i n terse ctio n o f th e tw o pl an es
The hori z o n tal trace H P is a l i n e i n the plan e
i s determ i n ed
P ; an d the hori z o n tal trace H Q is a lin e in the plane Q Where
H P and H Q i n tersect is a poin t then i n the requi red lin e o f i n ter
sectio n Of the two pla n es The horizo n tal proj ectio n o i this poi n t
"
f
n
n
n
the
fro
t
pro
j
ectio
is eviden tly in the X axis S im
n
d
is n a
f
n
F
a
d
F
n
where
the
fro
t
traces
P
n
i
tersect
is
o
r
l
th e
il a y
Q
fro n t proj ectio n of an other poi n t i n the li n e o f i n tersectio n The
k
hori z o n tal proj ectio n o f this poi n t is o i n the X axis The hori
z o n tal an d fro n t proj ectio n s Of the li n e o f i n te rsectio n are fou n d
h
h
the n by drawi n g n o an d rd o f If these proj ectio n s are dra wn
also i n Fi g l 0a we can check the accuracy o f o u r work by O bserv
i n g whether n o coi n cid es with d b as i t sh o ril d
2 1
.
In
,
.
,
.
.
.
.
.
,
,
.
'
,
,
,
,
.
.
,
,
,
,
.
,
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,
,
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,
,
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,
,
,
,
,
.
,
,
.
.
.
.
,
.
some pro blems where the li n e o f i n tersectio n Of tw o
plan es is to b e fou n d the traces o f the plan es do n o t in tersect
S uch a case is shown
o n the sheet o n which the drawi n g is made
i n Fi g 1 1 where the fro n t traces do n o t i n tersect i n the limi ts of
the dra w i n g Fo r the solutio n o f this pro blem an auxiliary plane A
parallel to the fro n t plan e o f proj ectio n is draw n to i n tersect the
pla n es P an d Q The hori z o n tal trace of this plan e is shown b y H A
Thi s auxiliary pla n e cuts auxiliary li n es from the plan es P an d Q
The horizo n tal proj ectio n s o f these li n es are Of course coin ciden t
with the trace H A an d the fro n t proj ectio n s are parallel to the
fro n t traces The i n tersectio n o f o f the fro n t proj ectio n s Of
these auxilia ry li n es determi n es o n e poi n t i n the li n e Of i n tersec
h
tio n an d th e h o ri z o n tal proj ectio n o is vertically over o f i n
the ho ri zo n tal pro j e cti o ns o f b oth lin es A s explai n ed in the pre
cedi n g paragraph the poi n t n is also a poi n t i n the lin e o f i n ter
h
h
sectio n o f the plan es P an d Q B y drawi n g the n n o an d n f o f
we O b tai n the tw o proj ectio n s of the required li n e o f i n tersectio n
I f i n this pro blem the hori z o n tal traces did n o t i n tersect o n the
paper an auxi liary plan e would b e n eeded parallel to the hori
z o n tal pla n e
2 2
.
In
,
.
.
,
.
,
,
.
.
.
,
,
,
.
,
,
‘
,
,
,
.
,
.
.
,
.
a case where b oth plan es are parallel to the X axis
their li n e o f i n tersectio n is likewi se parallel to the axis ; an d the
req u rred li n e is determi n ed by the i n tersectio n o f the side traces
o f the give n pla n es
I n Fig 1 2 the pla n es P an d Q are repre
se nted by their horizo n tal an d fro n t traces parallel to the X axi s
The side traces o f the plan es are shown at the right o n the draw
”
The ho ri z o n tal an d fro n t pro
i ng i n tersecti n g at the poi n t as
"
"
n
f
n
n
x
x
b
the
required
li
n
e
i
tersectio
are
show
ec
o
n
f
o
i
s
t
o
y
j
2
3
.
In
.
.
.
.
,
"
7
n
x
at
a d
2
4
.
.
P R OBLE M
6
.
To
nd
fi
th e li ne
o
f
i ntersecti on
o
f
two pl an es
.
M etho d The poin t where the ho ri z o n tal traces i ntersect i s th e
poi n t o f i n tersectio n o f the required li n e with the ho ri z o n tal plan e
The poin t where the fro n t traces i n tersect is the poi n t o f i ntersec
tio n Of the required lin e with the fro n t plan e A fter locatin g the
.
.
.
proj ectio n s o f each o f these poin ts j oin the like pro j ecti o ns
O b tai n the li n e o f i n tersectio n
tw o
to
,
.
E
31
i d the li e
Fn
.
°
an d
32
.
in tersectio
of
n
N
X E R CIS E S
—
n
—3
plan es M
o f tw o
,
1,
Draw the horizo n tal and fro n t proj ectio ns o f the li n e o f
i n tersectio n o f the plan e of the end posts i n E xercise 2 8
with ario th er plan e parallel to the X axis making an
an gle o f 30 with the ho ri zo n tal pla n e an d pas si n g through
a poi n t 1 i n ch b ehi nd the fro n t plan e an d l i n ch b elo w
the horiz on tal plan e ( Use the side plan e )
Fin d the proj ectio n s o f the i n tersectio n o f the plan e M
—
1
i n E xercise 3 1 with the pla n e P
b
O
(
serve that the li n e o f i n tersectio n is parallel to the fro nt
plan e an d that the fro n t proj ecti on i s parallel to th e
fro n t traces )
Fi n d the lin e O f in tersectio n O f the pla n e Q ( 0
with a vertical plan e paral lel to the fro n t plan e and 1
i n ch b ehin d i t (The fro n t proj ectio n of the lin e o f
i n tersectio n is parallel to the fro n t trace of Q T he
h o ri o z ntal projectio n is parallel to the X axis )
Fin d the l i n e O f i n tersectio n o f the pla n e M i n E xercise 3 1
with the plan e Q in E xercise 34 ( Draw an auxili ary
plan e parallel to the fro n t plan e cutti ng the planes M
It will cut from each of these pla n es a lin e
an d Q
parallel to the fro n t plan e The in tersectio n of the two
l i n es thus o b tai n ed is one poi n t in the required li n e o f
in tersectio n )
—
—
2
1
b
as t he li n e o f
Take the lin e a
in tersectio n o f tw o plan es R and S I n an y direc tio n
through the fro n t proj ectio n of a draw the fro n t traces
Determin e the horizo n tal traces
o f the two pla n es
an d elevatio n s o f the roofs of a stab le
I n Fi g 1 3 the pla n
,
,
°
.
33
.
.
,
,
,
.
34
.
,
.
.
.
35
.
,
.
,
,
,
,
.
.
.
36
.
,
,
.
,
,
.
.
.
P la n i s
th e f ro n t
an o th er
o r si d e
n am e
pro j ecti ons
.
f o r th e h o ri z o n tal
pr oj ecti o n
,
an d elevati on
fo r
adjoin i n g shed are shown C omplete the draw
i ngs by showin g the proj ectio n s of the li n e where the
two roofs i n tersect The dime n sio n s an d slopes n eces
sary for drawi n g the traces of the roof plan es are give n
Mark the stab le an d shed roofs respectively R an d S
Take scale
an d an
.
.
.
.
th r o u gh th e
.
T h e f r o n t t r a ce s
.
ti o n
c o i n c i de n t
ti o n s
w i th th e l i n e s
of
ro
.
m i n e s th e
th e i n te rse c ti o n s
sh o w
T h e l in e
’
mpe
i n th e
d ra
5
.
X
of
th e
HS
xi
a
th e
pl
ro o f
th e
,
s,
an es ,
r o o f sl o
an d
r o o fs
p
O P — th
e
F S,
are ,
i n th e fr o n t pro j e c
th ro u gh th e i n te rsec
a ral l el
to th e l in e s i n th e
pl
w i th th e h o ri z o n tal
sh o w i n g
l ine
an d
pe s
d raw n
ar e
FR
pl
an es ,
R
an d
S
,
th e i n te rse c ti o n s
an e o f
d e te r
of
th e
.
e
ec
d raw i n g
4
1
F
i
) is
( g
.
add e d
k
m erel y to m a
e
th e
o th e r
a n er
s
2
an
e rs
e
k
j
i n te rse cti o n Of th e se tw o
Of
O f th e l i n e
pl
p tiv
Th p
w in g p l i
r o o fs
an d
th e f ro n t trac e s w i th th e
pl an w h i ch
p j ecti o n
Of
sho w i n g
Th e h o ri z o n tal tr ace s , H R
.
p
,
i n th e d raw i n g
th e n ,
p
—I n th e fi u re th e h o ri z o n tal l an e o f pro e cti o n i s ta en
g
b o tto m o f th e sh ed r o o f thu s l aci n g th e X axi s as i t i s sh o w n
Su ggesti on
.
O f the fun dame n tal prob lems the fir st that of fi n din g the
,
,
in tersectio n of a pla n e with a n other pla n e should b e no w well i n
We shall n o w take up the seco n d of these most importa n t
min d
prob lems that of fi ndi ng the i n tersec tio n of a give n lin e with a
—
h
r
a
n
e
a
n
e
give n pla n e It should b e plai n that if
ot
called
pl
usuall y an auxiliary plan e —i s passed th rou gh th e gi v en lin e th e
o
iv
l
i
t
n
rs ec t s th e giv en pl an e i s i n th e l in e
i
n
e
h
n
e
e
h
r
t
e
e
e
i
n
t
w
g
p
l an e with th e giv en pl an e To
o f i n te r sectio n o f th e a uxi li a ry p
illustrate take two sheets of card board and a pe n cil Place o n e
sheet o n a tab le an d hold the pe n cil i n an y way to in tersect this
sheet N ow hold the other sheet soli th at o n e of its edges touches
the first sheet while at the same time the pe n cil lies i n its pla n e
O b serve that th e pen c il n o w to u ch es th e sh ee t o n th e t abl e i n th e
This same prin ciple is
li n e o f i n ters ecti oir o f th e tw o sh ee t s
show n also in the pictorial drawi n g i n Fi g 1 5 H ere we wa n t to
show the in tersectio n of the li n e 0 d with the plan e Q The lin e
h
h
f
n
o
is shown b y its projectio n s c d a d df an d the plan e by its
traces H Q an d F Q The auxiliary plan e A marked by its
traces H A an d F A is passed through the lin e c d The plan es
,
.
,
‘
,
.
,
,
.
.
,
'
~
.
.
,
'
.
.
.
.
,
.
,
,
,
,
,
,
.
Q
and
A in tersect i n
ml 111
an d
the li n e represen ted
the pro j ectio ns
by
Rem ember th at th e po i ri t w h ere th e li n e
.
in
h
n
i n ter
cd
This re q uir ed poin t is
also o f course i n the lin e 0 d We k now then that the poi n t we
are seekin g is in both O f the lines m n and c d and therefore at
the poin t i where they in tersect
The same pro blem is shown in orthographic proj ectio n in Fi g
The lin e 0 d i s shown by its proj ectio n s an d the plan e Q by
16
ts traces The a u xil iary pla n e A is passed through the li n e 0 d
The solutio n is ge neral so tha t any other plan e in cludin g the l i n e
Fewer lin es are n eeded
0 d could be take n for the auxili ary pla n e
for the drawing ho wever if the auxiliary plan e is drawn perpen
I n Fig 1 5 it is per
di cul ar to o n e o f the pla n es o f proj ectio n
n t plan e
l
to
the
fro
so that the horizo n tal trace
H
A
a
r
n
di
c
u
e
p
is perpe n dicular to the X axis The plan es Q and A in tersec t
i n the li n e m n ; an d the poin t where the lin e c d in tersects the
—
n
n
a
n
is
at
the
i
n
t
ersectio
of
m
d
at the poin t i shown
d c
pla n e Q
h
I n this partie
b y its horizo n tal an d fro n t proj ec tio ns i and 17
h
the projectio n i is determi ned by the in tersectio n o f
ul ar case
the horizo n tal proj ectio n s o f m n and e d The fro n t proj ectio n
h
17 is fou n d b y drawi n g a proj ecting lin e thr o ugh i
to i n tersec t
the fro n t proj ectio ns o f m n and c d which are coin cident
th e pl an e
sects
m u st be i n th e l in e
,
"
Q
,
,
mn
.
.
,
,
,
,
.
,
.
.
,
,
.
.
,
.
,
,
.
.
,
,
,
.
,
,
,
,
_
.
,
.
,
,
,
2
6
.
PR OB L E M
7
e
n
l
a
n
v
e
i
g
p
Method Pass
sects a
.
To
nd
fi
.
the poi nt i n whi ch
a
v
e
i
g n li ne i nter
.
plan e through the given li n e tusrfii l ly a
plan e perpe n dicular to the fro n t plan e) The l in e o f in tersecti o n
o f this auxiliary pla n e with the give n plan e i n tersec ts the g
i ven
l i n e at the poin t that is requi red
.
'
an y
.
,
.
E XE
R CIS E S
—
—
n
i
d
the
poi
t
where
the
li
e
n
Fn
k
15
j
1
— 1 in tersects the pla n e U 0
)
(
Determi ne where the lin e j lc give n i n the precedi ng exer
c ise in tersects the pla n e W ( 0
,
,
,
,
Draw the plan e P
—
a
Fi n d
i)
,
where the lin e
i n tersects it
—
2,
b ( 0,
.
—1
§
Locate
—
—1
the poin t where the lin e c
d
i
—1
passes through the plan e Q o f which b oth the
horizo n tal an d fro n t traces are parallel to the X axis
an d are respectively 5
5 in ch b ehi n d an d i i i n ches b elow
this axis
The top of th e desk shown i n the accompa n yi n g figur e is
—
—
12
10
b
located by the poin ts a
8
—
—
2
4
c (0
an d d
6
A light placed
—
4
1
1 0 + 2 ) h as its rays reflected so that the
at l
most i n te n se light is i n cli n ed 60 from the verti cal S how
the curve O f i n tersectio n o f the rays o f maximum i n ten
si ty with the pla n e Of the desk
,
,
,
.
.
42
.
,
,
,
"
-
,
,
,
,
-
,
,
°
.
,
‘
.
43
.
The plan e M
the poin t where a ray
represe n ts a mirror Fi nd
light passi ng through the poin ts
.
of
r
-
—1
2,
)
an d
—
s
5) is reflected from
the surface of the mirror
A steam pipe i n a b uilding passes through a slopin g floor
The axis of the p i pe i s located b y the poi n ts v
—
w
1}
The plan e of the floor ma y
an d
Fin d
b e represe n ted b y the plan e F
the in tersectio n o f the axis o f the pipe with the floor
.
44
-
.
.
1
.
h
r
e
r
r
u
i
n
t
t
n
e
a
l
i
l
n
a
a
e
e
e
h
t
7
p
pp
g
I n Fi g 1 7 the horizo n tal and
i n th eir proj ectio n s o n th at pl an e
fro n t proj ectio n s o f a hip roof are shown Lin es that are parallel
either to the horizo n tal plan e o r to the fro n t plan e such as the
lin es o d or c e i n the figure are here drawn i n their true len gth
A lin e that is o blique however to b oth pla n es of proj ectio n as
fo r example the lin e b i i s n o t sho wn in its true le n gth i n the
drawin g ; b u t i n this case a thi rd view may b e made to show its
true dime n sio n s This is illustrated i n the right han d drawi ng i n the
fi gure
The roof is here proj ected o n a vertical plan e V drawn
through the lin e b f an d the n this vertical plan e is revolved i n to
the plan e of the drawin g This proj ectio n is co n structed b y draw
A t g o n the proj ectin g
i ng the b ase li n e V V parallel to V V
h
lin e b b produced the li n e g b is laid o ff perpe n dicular to V V
Its le ngth is equal to the altitude Of the poin t b o r to the
dista n ce bf gf i n the fro n t view The li n e b f shows the n the
true le n gth O f b f The completed drawin g shows also the true
le ngth of the lin e a c as it is parallel to the vertical plan e V
2
.
Lin e s th at ar e paral l el to
a
.
.
.
,
.
,
,
,
,
,
-
.
,
.
,
,
.
’
’
’
,
,
.
’
’
’
’
’
.
,
,
’
’
.
.
,
.
,
practice these co n structio n s are simplified by passi ng the
h h
auxiliary vertical plan e throu gh b i The lin e V V is the trace of
this plan e i n the plan e of the b ase Of the roof If this plan e i s
rev olved together with the lin e b i about V V to coi n cide with
the plan e o f the b ase the lin e b f wil l then lie i n a pla n e where
true dimen sio n s can b e measured
The revolutio n o f o n e lin e about an other lin e as an axis is ao
complish ed usually by the revolutio n of two poi n ts i n th e lin e A
poin t revolved thus ab out a lin e as an axi s describ es a circle with
2
8
.
In
.
.
,
,
,
.
.
radius equal to the actual perpe n dicular distan ce from the poin t
to the axis I n this case the po i n ts b an d f must b e revolved ab out
the axis V V The axis is here i n the pla n e o f the b ase so that the
perpen dicular dista n ces o f the poin ts from it can b e measured in
Fo r the poi n t b this dis ta n ce is gf bf which is laid
th e fro n t view
h
perpe n dicul ar to V V The revolved
O ff o n the lin e through b
”
positio n of b is at b
The poin t f i s i n the axis so that i n r evo l u
” h
tio n it is statio n ary The lin e b f i s then the positio n Of b f when
revolved i n to a plan e of the drawin g an d shows its true le ngth
a
.
,
.
,
.
.
,
.
,
.
.
If it is required to l ay off a gi ven di stan ce o n a lin e n o t shown
i n its true le ngth i n a drawi n g we must first revolve the li n e ab out
an axis to b ri n g it i n to a pla n e where its true le n gth i s shown an d
the n measure the dista n ce Fo r example if we wish to measure a
dista n ce b 33 alo n g the li n e b i ( Fig 1 7 ) from the poin t b we lay
” h
locating the
o ff the give n le n gth o n the revolved positio n b f
poi n t x an d then reversi ng the preceding process we revolve at
h
h
b ack i n to the li n e b i
The horizo n tal proj ectio n o f b x i s the n b sc
3 0 Usually whe n the true le n gths of lin es are to b e fou n d i t is
most co nve n ie n t to revolve them i n to either the horizo n tal or
fro n t pla n es rather tha n in to an auxiliary pla n e which is parallel
to o n e Of the plan es o f proj ectio n O f course an y lin e which lies
i n o n e of the pla n es o f proj ectio n i s sho w n there i n its true le ngth
This method o f solutio n is illustrated i n Fig 1 8a where a li n e
m n is show n as i t is located in space an d also by its proj ectio n s
"
h
in
n
an d mi ni
The poi n t i n in this lin e lies i n the horizo n tal
h
pla n e so that this poi n t is marked o nl y by its proj ectio n m
Th i s
figure shows the lin e m n also whe n it is revolved in to the plan es
Of proj ectio n Let us co n sider first how i t is revolved in to the
hori z o n tal pla n e The poin t 777 b ei ng already i n this plan e i n r evo
l u ti o n remai n s statio n ary
O nly th e poi n t n therefore must b e
revolved to s h ow the revolved positio n o f the lin e This poi n t
h
moves i n the arc of a circle ab ou t its horizo n tal proj ectio n m as a
ce n ter an d with a radius equal to the distan ce the poin t n is b elow
the horizo n tal plan e Its revolved positio n is at n
In a
h
drawi ng however this revolutio n is shown by constructing at n
2
9
.
,
.
,
,
,
.
,
,
,
.
.
,
,
.
'
.
.
,
,
,
.
,
.
.
.
.
,
.
'
.
,
’
.
,
,
"
n
an d layi g
n
l i ne perpen dic ul ar to th e axi s m
o ff o n the per
cul ar the dis ta n ce the poi n t n is b elow the horizo n tal pla n e
e
n
di
p
B y the d i mensio n lin es in the figure this dista n ce is marked s
h
The lin e m n is thus ob tai n ed It i s a lin e i n the horizo n tal
plan e and is therefore shown i n its true length I n the same
way the true length Of the lin e w as foun d b y revolving i nto the
fro n t plan e The po i n ts m an d n revolve n o w respectively to
Dimen sio n lin es t an d u show the
m an d n i n the fro n t plan e
distan ces that i n this case are e qual The true length is shown
If the lin e m n is revolved in to the side plan e the same
by m n
r esult is ob tai n ed
Figure 1 8a shows all th i s very pla i nl y as i n a picture
This
method however is the same fo r prob lems in orthographic pro
i
c
n
I
n
n
n
i
s
n
n
i
b
this
way
the
same
li
e
m
represe
ted
1
e
t
o
i
F
8
g
j
b y its horizo n tal an d fro n t proj ectio ns an d its true le n gth is shown
h
The studen t shoul d meas
n o w very accura tely by m n an d m n
u re these le n gths to satisfy himself that they are equ al
A case occurs sometimes where the horizo n tal an d fro n t plan es
have n o t b ee n advan tageously located so that lin es pass through
The n the pro jec
an d co n tin ue b eyo n d the pla n es of proj ectio n
tio n s o f parts o f the li n e fall o n Opposite sides of the X axis Whe n
such a l i n e is revolved i n to the pla n e which i t p asses through
distan c es must b e laid o ff o n opposite sides o f the axis o f rev o
h
a
,
,
.
.
,
'
.
,
,
.
.
’
’
.
,
’
.
,
’
.
,
.
.
,
,
.
.
,
,
’
’
’
.
.
,
.
.
,
l u ti o n
.
a lin e a c is shown i n tersecti ng and p assin g thr ough
If this lin e is revolved in to the horizo n tal
th e horizo n tal pla n e at b
h h
plan e ab out a c as an axis the poi n t b remai ns station ary an d
The distan ce that ai is b elow
a an d c revolve i n o ppo si te di r ecti o n s
h h
the X axis is me asured o n o ne side o f a c an d the dista n ce that
The n a c is the true
a is ab ove is measur ed o n the other side
le ngth o f the lin e If however the same li n e is revolved in to the
"
fro nt plan e this diffi culty is n o t met as a c shows n o w the true
le ngth
I n Fig 1 9
.
.
,
,
.
,
'
‘
’
.
.
,
,
’
,
,
.
31
.
PRO B LE M
ro ecti on s; or
p
j
to
8
.
To
nd the
fi
nd
fi
the tr ue l ength
o
f
a
l i ne gi ven by i ts
di stance betw een tw o poi nts
.
Method P ass an auxiliary plan e through the li n e perpendi cu
lar to a plan e i n which lin es are shown i n their true le ngths o n the
drawin g ( usually a pla n e o f proj ectio n ) With the lin e of i n ter
section of the two plan es as an axis revolve the lin e i n to the seco n d
plan e where it is shown in its true length
.
.
,
.
,
E
Fi n d
b
Fi n d
X E R CIS E S
the distan ce b etwee n the poin ts
—l
-
,
a
—5
an d
,
3,
the t rue dista n ce b etwee n the poi n ts
an d
l
,
—6 — 1 —2
)
,
,
0
— 2
,
d
Locate , i n
a pla n e parallel to the fro n t pla n e two poi n ts
that are two i n ches apart an d are n o t equal dista n ces
b elow the horizo n tal pla n e
—
5 0
an d f (
l
A li n e passes through the poi n ts e
D raw the side proj ectio n of this li n e
Fi n d
the len gth o f this li n e i n cluded b etwee n its i n tersectio ns
with the horizo n tal an d fro n t plan es ; an d w ith the fro n t
an d side pla n es
Two statio n s 3 1 an d 3 2 are to b e co nn ected by a telegraph
—
3
3
an d 3 2
li n e The locatio n s are 1
Fi n d the le n gth O f the shortest li n e that
will co n n ect the two statio n s
Fi n d the le n gth Of the shortest b elt to co n n ect two pulleys
B oth are two i n ches i n diameter an d are i n the same
—
1
plan e Their ce n ters are at b
an d
,
.
,
,
,
.
.
,
.
.
50
.
.
,
.
c
—
i
,
i n ches lo n g The proj ectio n s of i ts en ds are
lin e is
—
—
n
a
5
1
a
d
1
n
b
located b y the poi ts
3
Determi n e the pro jectio n that is n o t give n
3
O n e proj ectio n
A li n e lies i n the pla n e M
—
n
1%
an d
o f the l rn e i s give n b y the poi ts g
—
O n this li n e lay O ff 1 } i n ches from its
1
h
i n tersectio n with the fro n t pla n e
A
1
43
.
-
-
,
,
,
.
-
,
7
,
.
,
plan es P an d H are
shown in Fig 2 0 A lin e a c is perpe ndicular to the plan e P
h
From the poin t a the pro j ecti ng li n e a a is draw n perpe n dicular
h
to the pla ne H
The plan e determin ed by the lines a c an d a a
is perpen dicular to the plan es P an d H It is therefore perpen
di cu l ar to the trace H P which is the i n tersectio n o f the plan es
h "
Then a c which is a lin e lyin g i n th is plan e is perpe n dicular to
H P C o ns ider n o w the plan e H as a plan e of proj ection ; the n
h
h
is the proj ec tio n O f the li n e a c while H P is the trace o f the
a c
plan e P We have show n then that when a li n e a c i s perpen dic ular
to a plan e P its horizo n tal proj ectio n is perpe n dicular to the hori
The same relatio n holds o f course fo r the other
z o n tal trace
proj ecti o ns o f the li n e an d the correspondin g traces o f the plan e
I f th en a straigh t li n e i s to be dr aw n perpen di cu l ar to a
r aw i ts proj e ctio n s pe rpen di cu l ar
t
d
r
i
o
n
l
y
n
ece
ss
y
o
i
a
s
t
l
e
a
n
p
An d to dr aw a pl an e
to th e co rrespo n din g tr aces o f th e pl an e
raw th e t race s perpen di cul ar to th e
e
d
l
l
i
n
n
u
a
r
t
o
a
r
e
d
i
c
e
p p
ro j e ctio n s o f th e l in e
p
32
.
Li n es P erpen di cul ar to
a
P l an e
Tw o
.
.
.
.
.
.
,
,
.
,
,
.
,
,
,
.
,
,
.
,
.
,
,
,
.
,
.
33
.
PR OBLE M
e
a
n
l
n
v
e
i
p
g
Method
9
.
To
nd
fi
the di stan ce
from
a
t
o
v
e
n
n
to
i
i
g
p
a
.
through th eg ve n poi n t a l i n e perpendi cu lar to
the plan e Fin d the poin t where this lin e i n tersects the give n plan e
“
The required distan ce is the true length o f the l i n e j oin in g t hi s
last poin t with the give n poi n t
.
D raw
.
.
.
E
—2
X E R CISE S
is perpendicular to a l i n e drawn
plan e Q
through the poin t j 2
Draw the proj ectio ns
Of the lin e ; an d fi n d the distan ce from j to Q
I n E x 4 2 fi n d the dista n ce from th e poi n t t to the pla n e
Of the top O f the desk
Fi n d the dista n ce from the poin t w in the axis of the steam
pipe give n i n E x 4 4 to the pla n e O f the floor
A
A hillside is represe n ted by the pla n e S (0
A
,
,
.
54
.
.
.
.
.
,
ole
is
pla
The
n ted o n it perpe n dicular to its plan e
p
—
n
t
li
t
b ottom of the pole passes through the poi
—1
1i )
Measuring by the same scale as for th e
c o ordin ates of t locate a poi n t that is 1 i n ch ab ove th e
grou n d What is the length O f the part o f the pole i n
the groun d ?
A poin t i
is ab ove the roof represen ted
A shaft for tra n smitti n g
by the plan e R ( 0
“
power to an other b uildi ng passes through this poin t
perpe ndicular to the roof Locate poin ts o n the shaft
1 i n ch ab ove the roof an d 2 i n ches b elow
Draw the proj ectio n s Of a cub e an d fi n d the distan ce from
The
any cor n er to an o b lique pla n e passin g through it
plan e should n o t b e parallel to an edge
Fi n d the dista n ce b etwee n two parallel pla n es o f which
all the traces are Ob lique to the X axis
.
,
-
.
,
,
.
,
.
.
.
.
,
.
PROBLE M 1 0 To proj ect a gi ven li ne u pon a gi ven plan e
Method S elect any tw o poi n ts i n the li n e an d through each
draw the proj ectio n s of a lin e perpen dicular to the pla n e T h e li n e
j oin ing the poin ts where these perpe n dicular lin es in tersect the given
plan e is the pro j ectio n o f the lin e upon tha t plane
34
.
.
.
,
.
.
.
E
—1 ) upon
1
1
n
the li n e m
—
1
the plan e P
The maj or an d min or axes of an elliptical cam wheel are
determi n ed respectively b y the lin es d
14
60 P roj ect
.
X E R CISE S
-
-
,
,
-
,
e
— 1 t,
(
—
1 t,
—l
i
—
,
i
,
Draw the proj ectio
this cam wheel o n the pla n e
—
5
by proj ecting a n umb er O f diameters
T
o f the ellipse
The vertices Of a triangle shown i n the figure are at
th e po in ts : a
b ( 1F
c
li
i
n of
-
’
,
.
_
_
_
’
1
r
'
)
Proj ec t this triangle upon
0
( ,
th e pl ane
R
n
n
i
s
b
n
n
If
n
li
e
give
its
proj
ectio
s
a
d
a
pl
e
is
e
a
n
o
to be
y
35
draw n perpe n dicular to it it follows from the explan atio n i n A rt
32 that the traces o f the required pla n e must b e draw n perpen di cu
lar to the c orrespon di ng proj ectio n s O f the give n li n e I n Figs
2 1 a an d 2 1 b a lin e a b is shown by its tw o proj ectio n s an d it is re
quired to draw through a give n poi n t 0 a plan e perpe n dicular
Through the poin t 0 draw a lin e m 0 parallel to
to the lin e a b
A
th e horizo n tal pla n e which shall lie i n the required pla n e
l in e which is parallel to th e horizo n tal pla n e is commo n ly called
“horizo n tal ” an d is a lin e with its fro n t proj ectio n parallel
a
“
A n d if this
horizo n tal shall lie i n the required
to the X axis
z o n tal proj ectio n must b e parallel to the horizo n tal
la
e
its
hori
n
p
I n other words i n this case the
trace O f the pla n e to b e draw n
co nditio n s to b e satisfied are : ( 1 ) that the fro n t proj ectio n W o f
”
horizo n tal must b e drawn parallel to the X axis ; an d (2 )
o f the
h
h
that the horizo n tal projectio n m o must b e drawn perpe n dicul ar
,
.
,
.
.
.
,
.
.
,
.
,
.
,
,
—
b
the horizo n tal proj ectio n o f the g i ve n lin e a
the same as sayin g
i t mus t b e parallel to the horizo n tal trace Of the required pl an e
H avi n g thus determin ed a lin e m o lyi n g in the required pla n e
the fro n t trace F P is drawn through the in tersectio n ( f i ) o f
the lin e m o with the fro n t plan e an d perpe n dicular to a) bf The
horizo n tal trace H P is drawn through the in tersectio n of F P
h
h
with the X axi s an d perpe n dicular to a b A pla n e P perpen
di c u l ar to the lin e a b an d passi n g through a give n poin t 0 is thus
determi n ed
A lin e parallel to the fro n t p an e an d i n tersec ti ng the horizon ta l
l
pl an e (d e termi n i ng the horizo n tal trace first) might O f c ourse b e
“
”
used i n the place o f the hori z on tal to Ob tai n the same result
to
.
,
,
,
,
.
,
,
.
,
,
,
‘
i
.
,
,
.
,
36
.
PR OBLE M
di cu l ar to
a
11
T o pass
.
e
n
v
e
n
l
i
i
g
a
h
t
l
a
n
e
r ou gh
p
agi ven
poi nt perpen
.
Method D raw through the give n poin t a lin e which will lie
“
”
horizo ntal lin e is usually most e asil y
i n the requ i red plan e ( a
used ) Through the i n tersectio n Of this li n e with o n e of the plan es
o f pro j ectio n draw o n e trace Of the required pla n e perpe n dicular
to the correspo n di n g proj ectio n o f the give n li n e The other
trace is drawn perpe n dicular to the correspo ndi ng proj ectio n o f
the li n e an d through the i n tersectio n o f the first trace with the X
axis
.
.
.
,
.
E
X ER CIS E S
Draw throu gh the poi n t e — 1 — 1
which shall be perpe ndicular to the li n e
,
—1 ) h (
,
“
Z
—
i
a plan e
,
g
—3
,
U,
—l
,
}
,
,
The poi ts b
0
5
are i n the
eaves of a roof The poi n t g
is i n the
ridge D raw through the poi n t w (0
the
s
r
o
e
c
t
o
n
f
n
i
a
arrow
that
will
show
the
directio
O
n of
j
p
a force due to win d pressure o n the roof Make the tip
o f the arrow tough the surface Of the roof
P
ressures
(
o f fluids are perpe n dicular to the pla n es o n which they
are exerted )
D raw a pla n e parallel to the pla n e P ( 0
at a
distan ce of two in ches from it o n th e right han d side
n
,
an d 0
—
,
.
.
,
.
.
.
,
-
.
a complicated drawing it is ofte n n ecessary to fin d the
true size o f the a n gle b etwee n li n es i n a pla n e surface which is n o t
parallel to any pla n e showing true dimen sio n s The usual process
is to imagi n e the plan e Of the surface exte n ded to i n tersect the plan es
thus determini n g the horizo n tal an d fro n t traces
o f projectio n
A b ou t o n e o f these traces
o f the pla n e i n which the surface lies
as an axis the pla n e is the n revolved till it coin c ides with the
hori z o n tal o r fro n t plan es
The simplest case of fi n din g the true size o f a plan e surface is
illustrated at the left han d side of the roof shown i n Fig 2 2 The
side a b c is n o t shown i n its true size n either are its angles If
however the side is revolved ab out its in tersectio n with the plan e
h h
o f the b ase of the roof b c
i n to the plan e of the b ase its true size
will b e see n The n the lin es a b an d a c will be see n i n their true
le ngths and with th e tru e si z e o f th e an gl e betw een th e m I n this
h
h
revolutio n ab out the axis a b the path traced by the poi n t a
is show n i n its fron t proj ectio n by the dotted li n es of the arc o f a
with its ce n ter at bf d The poi n t a shows its
c ircle through of
h
h
revolved positio n i n the pla n e o f the b ase an d a b c shows true
dime n sio n s for any part Of the outlin e of the surface
37
In
.
.
,
.
,
.
-
.
.
,
.
,
,
,
,
,
.
.
,
,
'
.
,
’
,
.
I n the preceding case the problem was simplified b ecaus e
the surface to b e determ i n ed was perpe n dicular to the fro n t plan e
o f proj ectio n so that the radius for the revolutio n of the cor n er a
A t the right
w as show n i n its true le n gth i n its fro n t proj ectio n
ha n d side o f Fi g 2 2 where the side o f the roof d e f is represen ted
H ere whe n the revolutio n
a more diffi cult pro b lem is prese n ted
h h
is made ab out the lin e e f ( the in tersectio n o f the side wi th the
plan e o f the b as e) as an axis the poi n t d will move i n the arc Of a
c ircle wi th a radius equal to the perpe n dicular dis tan ce from d to
the axis O bviously the lin e d g is this radius The p erspe ctive
vie w in Fi g 2 3 shows this radius and b y the arrow o n d d the
*
n
t
e
direc tio n of revolutio n The r u l e gth is fou nd e asily from its
d b y th e h y p th n
f th e i gh t t i n gl e
T h e t e l n gth i m
I n th i t i n gl d dh i l ai d o ff eq u al to
d dh gh i n th e h o i o nt l p o j ti o n
38
'
.
,
,
.
.
,
,
.
,
.
.
’
,
.
.
ru
e
r z
easu r e
s
a
r
ec
o
.
s
r a
th e d i stan ce f ro m df to th e l in e o f th e b ase o f th e
u se O
e
e
ro o f
r
r a
s
bf
of cf
f
f
.
two pro j ectio n s The true len gth d gle o f th i s radius whe n laid o ff o n
h
h
d g produced shows the revolved positio n Of the cor n er d at d
The true size O f the gi v eri surface is show n by the tria ngle d eh f"
fro m which true an gles can b e measur ed
The perspe cti ve v i ew o f this roof represen ted in Fig 2 3 shows
h
h
more plai n ly the right tria n gle called d d g i n th e precedin g co n
structi o n togethe rwith the ar e describ ed by the poi nt d i n rev o lv
h h
i n g to d a b out e f as an axis
f
-
.
,
,
’
,
,
,
.
’
,
.
/
.
,
’
.
tri an gl e d dh g ( Fi g 2 3) sh o w s al sO th e tr u e an gl e th e si d e d
w i th th e l an e o f th e b ase
T h i s angl e 1s m ar ed a i n th e fi gu re
The
mak es
r i gh t
.
p
k
.
e
f
.
Fig: 2 4 sh o w s th e sirnpl est
sta temen t o f the problem to fin d
N on e
th e tru e an gl e betw een tw o in t erse cti n g l in e s m n an d o p
Of the angles b etwee n the proj ectio n s Of these lin es show the true
a ngle b etwee n the lin es ; an d they must b e revolved i n to a plan e
where true dime n sio n s are shown B efore however the lin es
c an b e revolved in to o n e o f the pla n es of proj ectio n the traces o f the
pla n e i n which they lie must b e determi ned The i n tersectio ns
o f the li n es with the horizo n tal an d fro n t pla n es are marked respect
The lin es j oi n i ng these po in ts that are correspo n d
i vel y h i an d f i
The li n es mus t the n b e
i ng are the traces o f the required pla n e
revolved ab out o n e o f the traces in to the correspo nding plan e o f
proj ectio n ; that is we must revolve ab ou t the horizo n tal trac e
i n t o the hori z o n tal plan e ab out the fro n t trace i n to the fro n t p l an e
etc I n the figure the li n es are show n revolved i n to the hori z o n tal
plan e ab out the horizo n tal trace H P as an axis The true le ngth
o f the two
o f the radius for revolvi n g the poi n t of i n tersectio n i
lin es is fou n d by co n structin g the right tria ngle shown at the righ t
The hypothe nuse of this triangle is equal to the
O f the drawin g
true distan ce from i to H P The revolved positio n of i is shown
at i by layi n g o ff a d i stan ce equal to th e le ngth of the hypothe n us e
at right a ngles to H P The e nds of the li n es m n an d o p are
39
5
.
.
,
.
,
,
.
.
.
,
,
,
.
.
,
,
,
,
.
.
’
(
.
In
th at
the
p
r a c ti ce , u su a l l
o nl y o n e
ri gh t
v
hy
p
,
thi s
r i gh t
tr i a n gl e
l e g m u s t b e t ra n s f e r r e d
tri an gl e
w i th d i i d e rs
y
on
as s h o w n
th e
o th en u se c o ul
X
xi
a
,
.
Fo r
th e d i s ta n c e
s , a t ri
h
t
g
d b e m ea su red i n
e
x
a
be
p
p
pl
a ced o n
m l e i n s tea d
m a rk e d
an gl es
one o
c an
tO th e
e rati o n
.
so
o f c o n s tr u c ti n g
v
m i gh t h a e b ee n l a i d O ff
d i stan c e m ark e d b an d th e
a
‘
,
t h e d ra w i n g
H P; and i n revolving the lin es these poin ts remain s tationary
The lin es show n by heavy dashes i n the figure represe n t the lin es
m n and o p whe n they are revolved i nto the horizo ntal plane
The angle [9 is the true size o f the acute a ngle b etween the tw o lin es
O b serve that for the co nstructio n o nl y o ne trace is n eeded the o ther
might be omitted
in
,
.
.
.
,
.
40
.
PR OBLE M
i ntersecti ng li nes
12
To
.
nd
fi
the tru e
si z e o
f
the
angle
between two
.
Method Determi n e o n e o f the traces o f the plan e i n which
With this trace as an axis revolve the tw o
th e give n li n es lie
li n es in to the plan e o f pro j ectio n i n which the axis lies The angle
b e twee n the li n es i n their revolved positio n is the tru e siz e o f the
re q u i red angle
.
.
,
.
.
E XE
66
.
De termin e the plan e
co brd i n ates :
c
—l
i
—
,
1
of
—2
a
R CISE S
the lines
a
b
and
b0
gi ve n
b
,
by the
an d
With the horizo n tal trace O f this
axis revolve these lin es in to the horizo ntal
,
plan e as an
plan e
—
—
2
b
1
The li n e a
1) i ntersect s
—
—
2
d
the li n e 0
l % ) at 6
1
Fin d the true size o f the ob tus e a ngle betwee n
these two lin es
A b oat is towed alo n g a ca n al by tw o mul es o ne walki n g o n
each b ank I n proj ec tio n drawi ng the b oat is shown at
The to w paths are n o t at the same
b —2
level an d the po si ti o n s o f the mules are sho wn at m 1
—1 —1 —
—
Fi n d th e true
i ) and m2
1
size of the angle b etwee n the to w ropes
If i n the precedi ng exercise the forces exerted by the tw o
mules are the same show the course the boat will take
S uggesti o n If the forces in the tw o to w ropes are
equal
the c ourse of the b oat is shown by the bis ec to r
Of the a ngle b etwee n them
,
.
-
,
,
,
,
.
68
.
,
,
.
-
,
,
c
,
,
,
-
.
,
.
-
.
,
.
The vertices
of
—2
a triangle are at the poin ts
—1
i
—3
a
,
Draw the
projectio n s o f the b isector o f the a ngle b etween the sides
a b and b c o n the horizo n tal fro n t an d side pla n es
Th e mast o f a derrick i s shown b y the li n e c — 2
—2
2
d
Gu y ropes attached to the to p
)
g
i)
o f the mast at c are faste n ed to the grou n d at a
b
,
an d c
,
,
,
.
,
-
-
,
,
.
—2 — 2 ) and b — 4 — i
( a) Fin d th e a n gle b etween the guy ropes a c and b c
F
n
b
i
d
the
a
n
gle
etwee
n
the
n
m
b
c
a
d the mast
b
)
(
m
,
,
,
-
i
c
d
.
.
C heck this las t res ul t
plan e
revolvi ng
b
0
i n to the fro n t
.
Su ggesti o n
drawi ng
th e
by
.
Ob serve th at the
h i fo r
c
is
d
at c
h
h
d in
.
The a ngle which a line m ak es wi th i ts pro j ection o n an y
ngle b etw ee n the li n e an d the plan e
t
la
e
measures
the
rue
a
n
p
e from a poi n t i n the
B y drawi ng a lin e perpe n dicular to the plan
—
give n li n e a right a ngled triangle is formed i n which the perpen di c u
lar is o ne leg the proj ectio n of the li n e is an other leg an d the li n e
itself is the hypothen use The required a ngle is that b etwee n
the hypothe n use and the leg lyi n g i n the plan e ; b u t the other acute
a ngle o f this triangle the compleme n t Of the angle b etwee n the lin e
an d the pla n e is more easily fou n d b y the method of Ar t 4 0
41
.
.
,
,
,
.
,
.
,
42
.
P R OBLE M
e
l
a
n
e
v
n
i
p
g
Metho d
13
.
To
fi
nd
the
angle
betw een
a
.
e
v
e
i
n
l
i
n
g
and a
.
poi n t i n the give n lin e draw a lin e perpen
The
Fi n d the angle b etwee n these tw o li n es
di cul ar to the pla n e
complemen t o f this angle is th e required a ngle b etwee n the lin e
an d the plan e
.
From
any
.
.
.
E
72
.
Fin d
X E R CIS E S
the angl e the li e
makes with the plan e P
n
a
— 1 —l
,
— 1
-
1,
) b
,
—5
—2
—
n
i
n
c
i
d
the
a
gle
the
l
e
d
n
F
4
1
makes with the plan e M — 3
( O b serve
that the an gle most con ve n ie n tly foun d i n this case is
the supplemen t o f the an gle b etwee n the perpe n dic ular
an d the give n li n e as show n i n the right tria n gle men
—1
,
—
,
,
,
,
ti o n ed i n Ar t
.
The slope Of the b ank o f a can al makes an angle of 30
with the water level so that the plan e o f the b ank is
as shown i n the figure
A
r eprese n t ed b y P ( 0
mule o n the b ank draws a b oa t i n the can al The traces
o n the mule are attached to a rope at a po i n t m
—3
The other en d Of the rope is faste n ed to a
—
4
l
As sumi n g there is n o s ag
b oat at b
t the poi n ts m an d b i n the figure are i n a
i n the rope ( th a
straight li n e ) fi n d the true an gle b etwee n the rope an d
the b ank o f the ca n al
°
-
,
,
.
.
°
,
-
,
,
,
.
75 I n
.
co n structi ng shadows o n
we take rays o f light
that are reflected down ward from the left han d side an d
are show n b y lin es whose hori z o n tal an d fro n t proj ectio n s
-
,
make respectively angles o f 4 5 an d 3 1 5 with the X
axis
( a) Fi n d the a ngle b etwee n these rays an d a roof R
°
°
.
(0
,
Fi n d
the an gle b etween these rays and the hori
z o n tal an d fro n t pla n es of proj ectio n
— 1
A ray of light passes through the poin t p
)
an d is reflected from a poin t r
i n a pla n e
mirror repres e n ted b y M (0
Fin d the a n gle
*
Of refl ec ti o n an d the distan ce o f the poin t p from the
mirror
Fi n d the an gle b etwee n a li n e with b oth proj ectio n s parallel
to the X axis an d the pla n e Q (0
S u ggesti on S i n ce the pla n e Of the give n lin e an d th e
perpe n dicular must i n clude a lin e parallel to the X axis
its traces are parallel to the axis
( b)
.
,
,
.
,
.
‘
,
.
n
L
M
a
n
L
two
pla
es
N
d
N 0 are shown i n
43
a pictorial drawi n g an d we wish to determi n e the a n gle b etwee n
them If from an y poin t p we draw the lin es p g an d p r perpen
d i c u l ar respectively to the pla n es L M N an d L N 0 these perpen
d i c u l ar li nes will determin e a plan e whi ch is perpe n dicular to b oth
The
o f the give n pla n es an d i n tersects them i n the lin es g e an d e r
a ngle q e r is therefore the a n gle b etwee n the two give n plan es
This angle is n o t usually determin ed directly The method o r
di n ar il y used is to fi n d the true an gle g p r b etwee n the i n tersec t
This angle is the suppleme n t o f
i ng lin es p g an d p r ( A rt
the an gle g e r O f course either o f the an gles g e r ( gt) o r q p r
n
n
n
1
is
the
a
gle
b
etwee
the
two
pla
es
80
(
I n Fi g 2 5
.
,
.
,
,
.
,
,
.
.
.
.
.
°
.
T he
a
th e l i n e s
th e m ,
od
is
p
ra c t i c al l y
er
v
f ro m th e tw o gi e n
e n d i c ul a r to th e i r l i n e
cut
p p
T he
mi rro r
b o ve m eth
a n gl e o f re fl e c ti o n
b efo re refl ec ti o n
,
is
w h i ch
pl
th e
an e s
Of
by
as
an au
i n te rse c ti o n
equa l
is
sam e
to th e
c all ed
th e
fi n d i n g th
xi l i
a ry
pl
a n gl e
e
an e
p
a ssed
be tw ee n
th ro u gh
.
a n gl e
b etw ee n
an gl e o f
th e
i n ci d en ce
.
ray an d
th e
'
44
.
PROBLE M
14
To
.
the
fi
nd
an gl e
o bl i qu e
between two
s
l
a
n
e
p
.
Method From any poin t in space draw a lin e perpe n dicular to
The true a ngle b etwee n these tw o li n es
each o f the give n pla n es
is th e angl e b e twee n the plan es
.
.
.
E XE
Fin d
the angle b etwee n the pla es
n
T
S
—3
by
the poi ts
an d
,
(0
,
The corn ers
_
R CIS E S
i
a hip roof are give n
Of
-
b
'
:
(
"
F
_
r
i
’
1) y
_
i
C
—
n
i
an d
21
_
'
a
i
—2
—2
§
d
The ridge is the li n e e
—1
n
n
1
F
i
d
the
a
gle
b
etwee
the
sid
n
es
(
)
g
Of the roof c d e g an d a b g e ( Use the side traces o f the
pla n es ) ( 2 ) Fi n d the an gle betwee n the sides o f the
roof a b g e and g b c
—
n
Fi n d the angl e b etwee n th e pla e R
1
an d
a plan e parallel to the X axis with its horizo n tal an d fro nt
traces respectively 1 i n ch b ehin d an d 1 } in ches b elow
the X axis
The i n tersectio n o f o ne o f th e perpen
Su ggesti on
di cu l ars with either th e horizo n tal o r the fro n t plan e
is foun d easil y by usin g the side pro j ectio n s
Make a drawi ng similar to the figure o n p 7 6 taki ng th e
"
h
followi ng data : Draw i n n with a slope o f 30 to the
X axis Take fi = 60 a = 4 b = 2 c = 1 % d = 2 e = l i
2
h
n
1
C
omplete
the
dr
w
i
g
determ
i
g
a
b
i
n
1
n
5
y
f 1 g
a ccurately the l i n es o t an d 0 q
n
n
o
b
n
t
d
the
a
gle
etwee
the
roof
pla
e
the
i
n
a
n
F
d
1
o
n
e
)
p
(
i n which the li n e r s lies
s
n
i
The
li
n
e
r
represe
ts
a
rafter
n
the
ro
f
i
d
o
F
n
2
t
h
e
)
(
an gle o f the b evels to b e cut at its e n ds so that it may
b e j oi n ed to the ridge at r an d to th e valley rafter o g
at s
The corn ers an d ridge of the i rregular hip roof shown In th e
fi gure at the top o f p 78 are loc ated by the follow in g
—2
,
,
,
.
.
80
.
,
1
.
.
.
,
.
°
°
,
.
,
,
,
,
,
,
.
.
.
,
,
.
-
.
,
poin ts :
d ( 0,
-
a
b
0
0
,
,
(
0,
— l
4,
%
0
e
,
-
5,
— 3
,
min e the poin t e by fi n ding the lin e Of in tersectio n
sides meeti ng in the ridge (g e)
D eter
of
th e
.
n
b
n
the
a
gle
etwee
the
sides
of
the
roof
meet
i ng
( )
i n the ridge
n
b
t
i
n
d
the
a
gle
of
evel
for
the
o
side
Of
the
hip
F
b
)
p
(
rafter e c ; tha t is fi nd the an gle b etween the plan es
o f the sides d e c an d b c e g
n
m
n
n
i
n
The
li
e
represe
ts
a
rafter
the
side
c
b0 eg
)
(
Fi n d the a n gle of b evel for its en d at m to j oi n the hip
rafter e c The fro n t proj ection of m n must be supplied
a
Fi n d
.
,
.
.
.
.
shows a square b utt j oin ted h O pper
The edges o f the outside to p an d b ottom squares are re
The outside mpi ng
spec ti v el y -5 i n ches an d 2 in ches
edge is 3 in ches long Find the a ngle o f b evel for th e
j oints b etween the sides o f the hopper
83 Th e
fi gure
b elow
-
.
.
.
.
gen eral method used i n the precedin g exercises
an d
fi n d the an gle b etwee n the pla n e P (0
the horizon tal plan e o f proj ectio n
The horizo n tal trac e o f a plan e makes an an gle of 4 5 with
the X axis an d the plan e itself makes an an gle o f 60
with the fron t plan e Draw the fro n t trace of the pla n e
Take the data given fo r the skew b ridge i n E x 2 7 and
By th e
.
,
.
°
°
,
.
.
.
fi nd
n
n
The
a
n
gle
etwee
the
pla
e
Of
the
portal
b
)
(
a
a horizon
an d
tal plan e
n
n
The
a
n
gle
etwee
n
the
pla
e
of
the
portal
a
d
a
vertical
b
b
)
(
plan e through o n e O f the en d posts
n
n
d
n
The
a
gles
etwee
the
p
o
rtal
strut
the
posts
0
b
b
a
n
d
d
e
)
(
.
.
a
b
an d
c
d
.
The a ngle b etwee n the portal strut b d an d the li n e Of
i n tersectio n o f a vertical plan e through o n e of the end
posts with a plan e through the portal strut perpe n dicular
to the pla n e o f the portal
S u ggesti on A simple way to pass a plan e through
a give n lin e perpe n dicular to a give n pla n e is to draw
through any poi n t in the give n l i n e a lin e that will b e per
i
n
cu l ar to the give n pla n e
The
required
pla
e
is
e
d
n
p
determin ed by these tw o i n tersecti n g li n es
e
i
n
n
i
Measur
degrees
the
a
gle
fou
n
d
n
d
n
the
a
gles
a
d
n
)
(
( )
e
marked a an d [i i n the figure B y trigo n ometry the tan
ge n t O f the angle foun d i n (d) is equal to si n a tan fl
With this formula check the an gle fou nd
d
( )
.
.
.
.
.
.
.
very s i mple method for fi n ding the angle b etwee n tw o
It is shown there h o w
sides o f an O b j ect i s explai n ed i n A r t 38
the a ngle a b etwee n the side O f the roof d c i i n Fi g 2 2 and the
plan e of the b ase is fou n d immediately from the proj ectio n s The
an gle b etwee n the side d e f an d the fron t plan e can n ot however
The
be fou n d so easily without fi n di n g th e traces o f the pla n es
method to b e explain ed i n this article applies particularly to
fi n di n g the an gles b etwee n a plan e give n b y its traces an d the
horizo n tal an d fron t plan es O b serve that the prob lem is the same
as that solved i n E x 84 The problem ho w
ever appears so Ofte n
for solutio n that a Special case should b e made o f it
If an ob lique plan e is pepen di cu l ar to either the horizo n tal o r
the fro n t pla n e Of proj ectio n o n e o f its traces w il l b e perpe n dicular
to the X axis an d the other trace will show by its s 10 pe the a ngle
the plan e makes with o n e of the plan es o f projection I n Fig
2 6a a plan e P is show n perpe n dicular to the horizo n tal pla n e an d
45
.
A
.
.
.
.
,
,
.
.
.
,
.
,
.
fi
,
,
.
.
'
,
,
,
th e s 10pe
horizon tal trace shows the angle b etwee n the plan e
P and the fron t plan e The same plan e i s shown by orthogr aphic
projection in Fig 2 6b and the an gle mark ed 0 measures accur atel y
the same angle
The prob lem is a little more complicated when the plan e is
oblique to b oth plan es o f proj ectio n as shown in Fig 2 7a
In
this case o f c ourse the a ngle b etwee n o ne o f the traces and the X
axis does no t show the an gle b etwee n the plan e P an d a pla n e
ction
If ho wever an auxi liary plan e marked by its
o f proj e
trac es H A and F A is placed so that it in tersects the plan e P
an d is perpe n di cul ar to th e horizo n tal pla n e as shown i n Fi g
2 7 a th e a ngle 0 b etwee n the lin e of in tersection o f the plan es P
an d A an d the trace H A is the angle b etwee n the plan e P an d
the hori z onta l plane The method o f findin g this angle from th e
The given plan e
orthographi c pr o j ec ti o ns is shown in Fi g 2 7 b
i s shown by its tr aces H P an d F P ; and the auxiliary plan e
by its traces H A ( perpe n dicular to H P) an d F A ( perpe n dicular
The lin e of in tersectio n o f these tw o plan es is
to the X ax i s )
"
h
shown by its pro j ectio n s i j and 17 j f This l i n e whe n revolved
h
"
n
n
a
l
n
into the horizo t pla e ab out its horizo tal proj ection i j as
h
n
n
a
d
shows
the
a
gle
true
size
0
i
n
i
an axi s takes the positio n i
t
s
j
A di me nsio n l in e marked s is shown to make the co n struction
plainer It sho ul d b e evi den t that the auxiliary plan e A can
b e drawn through any poin t i n the trace o f the give n plan e
S imilarl y we can fin d the angle b etween the same plan e P
an d the fro n t pla n e b y drawin g the fro n t trace F B O f an au xfl i ar
y
plan e B perpe n dicular to the fro n t trace Of P an d makin g th e
horizon tal trace H B perpe n dicular to the X axis The required
angle b etwee n the plan e P an d the fro n t plan e o f proj ection is
shown by the true size o f the angle b etwee n the fron t plan e and
the lin e o f in tersectio n O f the auxil iary plan e B with the plan e P
of
th e
.
,
.
.
.
.
,
,
,
,
.
,
,
,
,
.
,
.
.
.
,
,
.
.
,
,
’
,
.
.
,
,
,
,
,
,
,
,
,
,
.
,
,
,
,
.
The reverse o f this last operation is m ade use o f in draw in g
th e traces o f pla n es whe n o nl y o n e o f them is give n o n the draw i n g
If the horizon tal trace o f a plan e is given an d the angle the plan e
makes with the horizo n tal plan e is kn own the fro n t trace c an
46
.
.
,
constructed A ssume n o w that i n Fig 2 7b the horizon tal
trace of the pla n e P an d the angle 0 the pla n e makes with the
horizo n tal plan e are given Through any poin t in H P draw th e
h
horizon tal trace H A o f an auxiliary plan e an d the line i j
makin g with H A an angle equal to the angle the plan e makes with
the horizo n tal plan e This l ast li n e is the revolved positio n in the
horizon tal plan e of a l i n e i j of which we k n ow the horizo n tal
h h
proj ectio n i
an d the dista n ce ( 3 ) o f the poin t j b elow the hori
The fro n t proj ection o f j is located by layin g O ff o n a
z o n tal pla n e
h
lin e perpe n dicular to the X axis through j the distan ce marked s
The lin e i j determi ned n o w by two proj ecti on s is a li n e i n th e
The requir ed
required pla n e an d j f is a poin t i n its fro n t trace
fro n t trace is drawn the n through the proj ectio n j f and the in ter
sectio n o f the give n horizo n tal trace with the X axis
The meth o d is ex ac tly sim ilar when the fro n t trace and the angle
the plan e makes with the fro n t plane are give n and the horizo n t al
trace is to b e co nstru cted
be
.
.
,
,
.
’
”
,
,
,
.
°
,
,
.
.
.
,
.
,
.
47
PR OBLE M
.
ei ther
15
the hori z on ta l
.
or
To
nd
fi
the
an gl e
fron t pla n es of
made by
j
on
ec
t
ro
i
p
a
h
v
e
n
l
a
n
e
w
i
t
i
g
p
.
Metho d To measur e the in cli n atio n to the horizo n tal plane
Through any poin t i n the horizo n tal trace of the grven plan e pass
auxiliary plan e perpe n dicular to this trace
an
( It will b e
perpe n dicular to the horizo n tal plan e ) The true angle b etwee n
the horiz o n tal plan e an d th e li n e of i n tersectio n O f the auxiliary
plan e with the give n plan e is the required an gle
n
n
a
n
:
me
ure
the
i
cli
atio
to
the
fro
t
pla
e
P
ass
auxil
n
n
n
a
s
TO
n
n
b
n
pla
e
perpe
dicular
to
the
fro
t
tra
e
The
true
a
gle
etwee
c
n
n
i ary
the fro n t plan e an d the lin e Of i n tersectio n Of the a u xiliary pla n e
n
n
i
ith
the
give
pla
e
s
the
required
a
gle
n
w
.
.
.
.
.
.
E
87
.
X E R CISE S
Fin d the a n gles b etwee the pla e P ( 0 1 2 0
horizo n tal an d fro n t plan es Of proj ectio n
Take the data give n fo r the skew b ridge in
n
n
an d
°
,
th e
.
88
.
Ex 2 7
.
an d
the angle b etween the plan e o f the portal
horizon tal an d fron t plan es o f proj ection
find
an d
th e
.
48
PROBLE M
.
16
Gi ven
.
one
trace
the
an d
mak es wi th the correspondi n g plan e of pr oj ecti on
Method
.
Revers e
the method
E
of
,
angl e a
to
find
l
a
e
n
p
e
v
i
g n
the o ther trace
.
the precedin g problem
.
XE R CIS E S
The fro n t trace Of a plan e Q makes an an gle o f 315 with
the X axi s The plan e itself makes an angle of 4 5
wi th the fro n t plan e D raw the horizo n tal trace
The horizo n tal trac e o f a plan e R makes an angle of 30
with the X axis The plan e itself makes an an gle O f 4 5
with the horizo n tal plan e Draw F R
—
The pla n o f a dry dock is shown b y the poin ts a 4 — 2
°
,
,
°
.
.
.
°
,
,
°
.
.
.
-
b
—2
,
— 4
,
c
(
,
—1
,
an d
d
—3
,
T he
,
plan es o f the sides through a b an d c d slope toward the
middle Of the dock an d make an gles of 4 5 with the
hori z on tal D raw the traces o f the plan es of the two
sides me n tion ed
°
,
.
.
n
n
Whe
n
n
o
e
of
the
traces
Of
a
pla
e are give n an d we k n ow
49
onl y the angles the plan e makes with the plan es o f projection
we have a more complicated problem than the last I n Fi g 2 8a
a plan e P is shown i n a pictorial drawing A sphere is also repre
sen ted w ith its ce n ter o n the X axis at e an d is ta n ge n t to the plan e
P at i An auxiliary plan e A is represe n ted as drawn perpen di cu
lar to the fron t plan e an d passin g through the po int i an d th e
ce n ter o f the sphere e This auxiliary plan e cuts a great circle
from the Sphere an d a li n e a b from the plan e P This last lin e is
It is evide n t that the angle or b etween th e
tangen t to the circle at i
lin e a b and the trace F A shows the an gle between the plan e P and
the fro n t plan e If the n the lin e a b is revol ved ab out the trace
F A into the fron t plan e its revolved position wil l b e shown b y a b
Lik ewise if the circular sectio n cut from the sphere bythe auxiliary
plan e is also revolved in to the fro n t plan e it wi ll be a c i rcle
.
,
.
,
,
.
.
,
,
,
.
,
,
.
,
.
.
.
,
,
’
,
.
,
coi n ciding however w ith the outli n e o f the sphere By this rev o
l u ti o n in to the fro n t plan e the lin e a b an d the circular sectio n O f
the sphere are shown in their true relative positio n and the poin t o f
tange n cy i is shown at i The angle b etwee n a b an d F A
marked a is the true size Of a
S imilarly an auxiliary plan e B is represen ted i n the figure
perpe ndicular to the horizo n tal plan e an d passing through the poin ts
e an d i
This plan e cuts from the give n plan e P the lin e 0 d an d
a circular sectio n fr om the sphere The an gle fl b etwee n the lin e
c d an d the trace H B shows the a n gle b etwee n the pla n e P an d the
horizo n tal pla n e B y revolvi ng c d an d the c ircular sectio n in to
the horizo n tal plan e the true size of Bcan be show n
This method of a n alysis is most useful whe n the co n dition s
—
are reversed whe n o nl y the angles a an d fi are give n and the traces
H P an d F P are to be determi n ed The actual process for the so
Through a poin t e o n the
l u ti o n of this case is shown i n Fi g 2 8b
X axis as a ce n ter draw a circle of any co n ve n ie n t radius This
represe n ts i n orthographic proj ectio n the revolved position in either
the horizo n tal o r the fro n t plan es of an y sectio n s of the sphere
cut b y auxiliary plan es passin g through the ce n ter I n order t o
show the revo l ved po si ti on of the l i n e o f i ntersecti on with the plan e P
o f an auxiliary pla n e perpe n dicular to the fro n t pla n e draw the lin e
a b ta n ge n t to the circle at any co n ve n ie n t poi n t an d draw a radius
e a so that the a ngle b a e is equal to the give n a ngle or b etwee n
the plan e P an d the fro n t plan e The lin e e b is drawn perpen di cu
lar to a e The triangle a b e that is thus formed correspo n ds to the
revolved positio n O f the trian gle a b e shown very much fo resh o rt
e n ed i n Fig 2 8a If n o w the le ngth e b ( Fig 2 8b) is measured
an d laid O ff perpe n dicular to the X axis b ehi n d the poi n t e the
h
poi n t marked b is Ob tained A n d this is a poi n t i n the hori
C on ti nui n g with the same ge n eral method draw
z o n tal trace H P
c d ta nge n t to the circle at a co n ve n ie n t poi n t an d by drawin g the
radius 0 e lay o ff the angle 5 The i n tersectio n Of c d with e d
perpen dicular to e c determines the poi n t d which is the revolved
positio n of a poin t located o n the trace F P Layin g O ff then e d
f
n
d
equal to e d the poi t
is Ob tain ed If n o w a e happen ed to
,
,
~
.
,
,
’
.
,
,
’
,
’
,
.
,
,
,
,
.
,
.
.
.
,
.
,
.
.
.
.
.
,
,
,
’
,
’
,
,
’
.
’
.
’
.
.
.
,
.
.
,
’
,
’
.
’
,
’
,
,
.
’
.
"
taken in such a position that it is perpen dicul ar to F P i n i ts
true locatio n this trace could b e drawn immediately through d}
"
similarly H P c oul d be drawn through b an d 0
an d a ; an d
S i n ce however a e was drawn in an y con ven ien t direction i t
c ould n o t b e taken n ecessarily perpe n dicular to F P bu t O b vi
o u sl y if an arc with a radius a e is draw n as shown i n the figure
F P must b e draw n ta n ge n t to this are an d through the poi n t d
Fo r the same reaso n H P will b e draw n ta nge n t to the arc with
h
a radius c e and thr ough the poi n t b
be
,
.
,
,
,
,
,
,
,
,
,
’
.
50
.
PR OBLE M
17
.
To
nd
fi
the traces
o
f
a
l
a
e
n
w
h
h
e
e
n
t
p
angl es
front plan es are gi ven
Method Imagine a sphere placed tange n t to the plan e Of which
the traces are to b e foun d an d tw o auxiliary pla n es are passed
through t h e ce n ter Of the sphere an d through the poi n t where the
—
n
o n e perpe n dicular to the fro n t pla n e
sphere touches the pla e
the other perpen dicu lar to the horizo n tal plan e
These plan es
will each cut a c i rcle from the sphere an d a tange n t li n e from the
give n plan e The a ngles these tange n ts make with the fron t an d
horizo n tal plan es resp ec tively are the a ngles the give n plan e make s
with the fro n t an d horizo n tal plan es
the pl an e ma kes w i th both the hori z on tal
an d
.
.
‘
,
,
.
.
,
,
.
E XE
R CISE S
Draw the traces o f a plan e P which makes angles o f 4 5
an d 60 respectively with the horizo n tal an d fro n t pla n es
Draw the traces of a plan e which is i n clin ed 1 2 0 to the
hori z o n tal plan e an d 7 5 to the fro n t plan e
The side of a b ridge pier i n a river makes an angle o f 60
with the plan e o f the water an d 50 with a vertical plan e
at right an gles to the course o f the river Represe n t
the plan e o f the side of the pier by its traces o n ass umed
plan es o f projectio n
°
,
,
°
.
°
°
.
°
°
,
.
.
1
5
to
a
.
PRO BLE M
ne
n
l
i
v
e
i
g
Method
.
18
.
To
nd
fi
the tru e di stance
from
a
gi ven poi nt
.
D etermi n e
the
l
e
a
n
p
Of the poin t
an d
the li n e
by
drawi ng through the poin t a l in e parallel to or in tersecti ng the
given lin e Fi n d the i n tersectio n s of b oth lin es with either the
horizo n tal or the fro n t pla n e With a trace of this plan e as an
axis revolve these lin es i n to the correspon ding plan e o f proj ection
and measure the dista n ce required
.
.
,
,
.
E
XE R CIS E S
the distan ce from the poi n t i — 2 — l
L} ) to the
—
—
—
—
—
—
lin e 7
2
li
t
i) k
i
t)
The telepho n e wi res ru n n in g from a village to a house o n
the side o f a mou n tain follow the shortest lin e b etwee n
—
n
2
n
h
A camp
the poi ts v
3 0
)a d
at c
is to b e co nn ected to the telephon e
system o f the v illage by erecti ng a li n e j oi n in g the o ne
Draw the proj ectio n s of the shortest
b etwee n v an d h
li n e th at can b e put up ; an d find the le ngth of the lin e
thus located
I n the tria ngle give n i n E x 62 fi n d the dista n ce from
the poin t e to the li n e a b
Fo r supporti ng a cran e a cab le is to b e attached at o ne
—
—
—
1
2 ) o n its mast an d at the
1
en d to a poi n t c
other en d to a steel b eam a
b
— 1 o n an adj oi n i ng b uildi ng
S how the pro jec
)
tio ns o f the shortest cab le that c an b e u sed
Fi n d
-
,
,
°
~
,
,
,
.
,
-
,
,
.
'
.
.
.
,
,
,
.
.
2
5
n ot
PROBLE M
.
i n the
19
me plan e
sa
Method
To
.
fi
nd
the
shortest
di stance betw een two l i nes
.
Tw o li n es a b an d c d n o t i n the same plan e are sho wn
P ass a plan e M N through a b parallel to c d ( Ar t
i n Fig 2 9
Proj ect 0 d upo n this plan e A t the poi n t e where the pro
n c d in tersects a b draw a li n e perpe n dicular to the pla ne
i
c
t
o
e
j
i n tersecti ng c d at e The true leng th o f the perpen dicular e e
i s th e r e q ui red distan c e
.
.
,
.
.
,
’
.
’
’
.
,
'
.
.
E XE
99
.
Fin d
R CIS E S
the shortest distan ce b etween
Ex
25
.
th e
lin es to l
m n in
an d
.
Fi n d
the distan ce b etwee n a lin e from a to b an d the m ast
o f the derrick i n E x 7 1
Through each o f tw o lin es pass a pla n e parallel to the other
l in e Fin d the shor test d i stan ce b etwee n these tw o
planes
S olve Ex 65 by the method o f this article
.
.
.
.
.
.
o
f
PR O B LE M
Gi ven the proj ecti on s of the cen ter of a ci rcl e
draw i ts pr oj ecti o n s s o that i t sha l l l i e i n a
20
53
k n o wn di ameter , to
.
.
l
a
n
e
v
e
n
i
p
g
ethod
.
M
The cen ter o f a c ircle is shown by its proj ectio n s c
The horizo n tal
an d d i n Fi g 30 as lyi n g i n the give n pla n e P
an d fro n t proj ectio ns o f the circle are to b e drawn so that the
circle also shall lie in the plan e The ce n ter o f the circle is revolved
ab out F P as an axis in to the fro n t plan e an d the true siz e o f the
c ircle is draw n When th i s circle is revolved b ack i n to the plan e P
b oth proj ectio n s will b e ellipses b ecause the proj ectio n s o f a circle o n
la
The
n es of proj ectio n that are O b lique to its pla n e are ellipses
p
le ngth o f the maj or axis of the ellipse in the fro n t projectio n is the
same as the diameter o f the circle an d is o f course parallel to F P
As all le ngths which are perpe n dicular to the axis o f revolutio n
ar e foreshorte n ed i n the fro n t proj ectio n the mi n or axis is the fore
The fro n t proj ectio n s
h o rten ed diameter perpe n dicular to F P
o f the min or axis is the n o n a li n e o f i n defi n ite le n gth drawn through
Mark the poin t where this proj ectio n
c perpe n dicular to F P
crosses the fro n t trace i f an d revolve the lin e i 0 i n to the fro n t plan e
ab out i f cf as an axis The revolved positro n 1s shown by i f c
F o m c lay o ff the le ngth
o n which true le n gths c an b e measured
r
I n coun ter revolutio n
c e equal to the radius ( r ) of the circle
the poi n t 0 revolves b ack to c} and e to ef Then er of is the semi
mi n or axis o f the ell iptical fron t proj ectio n of the give n circle
The horizon tal proj ection o f the circle is the n easily foun d by
h
.
.
.
.
'
,
.
,
,
.
,
,
.
,
.
,
,
.
,
’
.
,
’
.
’
’
-
.
’
’
.
.
h
a
s
n
r
i
g
li
es
that
lie
the
pla
e
P
suc
diameters
through
i
n
n
(
n
a
w
d
n
n
o
e
t
n
d
r
n
a
c
i
ce
ter
ta
ge
ts
to
the
fro
t
projectio
n
n
n
d
c
a
)
n
p j
g
th e
poin ts o n them from the fro n t proj ection o f the circle The hori
z o n tal proj ectio n o f the circle could b e fou n d also b y repeati n g the
method used for ob tain ing the fro n t proj ectio n
,
,
,
.
.
n
n
i
n
n
n
This
ro
lem
has
n
importa
t
applicatio
determi
i
g
a
b
54
p
th e plan e Of guide pulleys to direct b el ts ru nn ing b etwee n pulleys
?
which are o n shafts at right angles to each other I n Fig 3 1 two
pulleys are shown with ce n ters at e an d b The directio n of motio n
The b elt must b e led O ff the pulley at b i n
i s shown b y arrows
To accomplish this
i ts plan e an d led o n the pull ey at ai n i ts pla n e
Usually th e guide pulley
a guide pulle y is n eeded to direct the b elt
may b e placed at any co n ve n ien t poin t b etween the two pulleys
S elect a poin t where the directio n Of the b elt is to
at a an d at b
be cha n ged as the poi n t d i n the figure an d draw the l i n es f d an d
d e ta nge n t to the p ulleys
The plan e of the guide pulley mus t
This plan e
be the n in the plan e o f these tw o ta n ge n t li n es
is shown by the traces H P and F P The proj ection s o f the
guide pul ley are fou n d b y revolvi n g the lin es f d an d d e i n to th e
horizo n tal plan e of proj ectio n ab out H P as an axis The revolved
positio n o f d is show n at d The actual size O f the guide pull e i s
shown tan ge n t to the revolved positio n s O f f d an d d e A fter r ev o l v
i ng b ack i n to the plan e P the cen ter ( c ) O f the guide pulley is shown
h
f
n
n
c
c
its
proj
ectio
s
a
d
The proj ectio n s O f the gu ide pu ll ey
by
are fou n d b y the method explai n ed i n the last article The shaft
Of the guide pulley b ei n g perpe n dicular to its plan e is shown by
proj ectio n s perpe n dicular to the co rrespo nn i ng traces
I n Fi g 31 o n o n e side the b elt is led O ff the pulle y a i n l i n e with
the rim o f the p ul ley b so that o n this side there is n o cha nge i n
the direction o f the b elt I n practice however very often the give n
pulleys are n o t placed so advan t ageousl y an d tw o guide pul leys
are n eeded
.
.
.
.
.
.
,
,
.
.
“
,
,
"
.
.
.
.
.
’
y
.
’
,
.
.
.
,
,
.
,
.
.
,
,
E
1 03
.
X E R CIS E S
Draw the proj ectio n s o f a circle diameter 1 5 i n ches l yi ng
—
i n the plan e Q
3
Two pulleys revolvi ng o n li n es of shafti ng at right angles
to each other are to b e co nn ected by a b elt Determ in e
the plan es of i n termediate pulleys to properly d i rec t
the b elt drawi ng also the proj ectio n s of all the pulley s
that are n eeded
,
,
,
1 04
.
.
,
.
55
a
PR OBLE M
.
s
t
l
a
n
e
i
b
y
p
21
sl o pe,
Gi ven
.
a
f
su r ace
to determi n e the
by i ts
secti o n o
f
co ntou r ed
the
l
a
n
,
p
f
su r a ce cu t
a nd
by the
Method Draw the horizo n tal proj ection s o f a n umb er o f
horizon tal lin es lyi n g i n the give n plan e an d havin g the same
i ndices as the co n tour lin es The poin ts where these li n es mee t
th e co n tour li n es with the same i n dices are poi n ts i n the required
The compl ete sectio n is Ob tain ed by draw i ng a fair curve
s ectio n
thr ough the poin ts th u s O b tai n ed
E X E R CIS E S
1 05 The figure b elow represe n ts a hill b y its co n tour li n es
D raw plan es to show the emb a n kme n ts o f a rai lroad
”
“
passi ng through it S how the i n tersectio n s o f
cut
these plan es with the surface of the hill
.
,
.
.
.
.
.
.
.
p
j
v
v
p
Co n to u r l i n es a re u se d to o i n o i n ts a t th e sam e el e a ti o n ab o e a l an e
Num b e rs o n th ese l i nes , c al l ed
o f w h i c h th e el e a ti o n i s a ss u m e d to b e z ero
i n dices re r ese n t th e el e ati o ns
,
p
v
.
v
.
CHA P T E R
P
R O B L E MS R E LAT I N G
TO
E very s u rface may
II I
P LA NE S
T
ANGE NT
SO LI D S
TO
gen erated by the motio n of a li n e
an d the diff ere n t positio ns assum ed by this li n e are called the
el e m en ts o f the surface
A pla n e s urface o r a pl an e is ge n erated by a straight li n e movi ng
alo n g a n other s traigh t lin e an d remain in g always parallel to its fi rs t
*
ositio
n
p
A si n gl e curved surf ace IS ge n erated b y a straight lin e movi n g
so that any tw o o f its co ns ecutive positio n s are i n the same pla n e
A w arped o r tw i sted su rf ace is ge n erated b y a straight lin e
movin g so tha t n o tw o o f i ts co nsecutive position s are i n the same
la
n
e
p
Plan es sin gle c u rved sur faces and warped surfaces are ge n erated
by the motio n of a straight lin e an d therefore all have eleme n ts
that are straight l ines E very warped surface however is curved
an d it is therefore possib le to co n ceive i t also as b ei ng ge n erated by
a curve which as i t moves c o n tin ual l y cha n ges its form accordi n g
to a de fi n ite l aw
A d o u bl e c urved su rf ace is ge n erated b y a curve movi n g alo n g
an other curve I t has no elemen ts that are straight lin es
56
.
be
,
.
.
.
.
,
,
,
.
,
,
,
,
,
.
.
.
n
n
n
o
e
is
a
si
gle
curved
surface
ge
erated b y a straigh t
57
lin e moving along a curve an d also passi ng through a poin t n o t i n
the plan e o f the curve This poin t is the v ertex Of the con e
A cy lin der is a si ngle curved surface ge n erated b y a straight li n e
movi ng alon g a cur ve with all its positio n s parall el It may b e
regarded as a Special case o f a co n e with th e vertex at in fi n ity
.
A
c
.
.
.
.
A kn o w l edge o f
T h e m eth o ds o f
di scu sse d i n th e
re
p
p p ti
ti g p l
g h pt
th e
re sen
p recedi n
ro
er
pl
a n es , a n d
n
c
es o f
a
e rs
.
an e s
p
w a s assu m e d i n th e
o in ts an d
l i n es i n th em ,
b egi nn in g
h ave b een
.
P roblems relating to cylin ders are solved therefore
the same
methods that are applied for similar prob lems relati n g to co n es
A pla n e cuttin g all the straight li n e eleme n ts of a c on e or o f
a cylin der i n tersects it i n a curve called the base
If all the eleme n ts of a co n e make the same a n gle with a straight
lin e passin g through the vertex it is a righ t c on e ; otherwise it is
If all the eleme n ts of a cylin der are pe rpen dicular
an O bli qu e c o n e
to the b ase it is a righ t cyl in de r ; otherwise it i s an o bli qu e cyl i n
by
.
,
,
.
,
,
.
,
der
,
)
.
o n vol u te is a Single curved surface ge n erated by a straight
li n e m o vrng alo ng a curve of doub le curvature SO that it is always
tange n t to the curve The co n secutive positio n s o f the straigh t
li n e ge n erating the surface i n tersect two an d two n o three i n ter
sec ti n g i n a commo n poi n t
There are as many ki n ds o f co nvolutes as there are curves o f
doub le curvature S ome are importan t for their practical appl i
catio n s The methods o f co n struct in g an d represe n ti ng them will
b e discussed i n Ar t 7 6
58
A
.
c
,
.
,
.
.
.
.
.
u rf a c e o f revo l utio n is o n e that is ge n erated b y the rev
o l u ti o n of a straight li n e or a pla n e curve ab out a straight li n e i n
the same pla n e as an axis
—
There are o nly two Si ngle curved surfaces of revolutio n the
right co n e an d the right cyli n der whe n they have circular b ases
The prin cipal doub le curved surfaces O f revolutio n are the S phere
—
a
n
the ellipsoid the torus the parab oloid
d the hyperb oloid sur
faces which are explain ed later
59
.
A
s
,
.
,
,
.
,
,
,
,
.
poi n t i n an y plan e or curved surface is determi n ed b y
If the n a su rfa ce is give n by the proj ectio n s of
tw o proj ectio n s
its outlin es an d a poi n t o n the surface is give n by o nly o n e proj ectio n
an other projection c an b e located after determi n ing two proj ectio n s
o f an el emen t of the surface passi ng thr o u gh the po i n t
I n Fi g 32 a con e is Show n b y the hori z o n tal an d fro n t proj ectio n s
Of i ts outlin es A poi n t a o n the surface Of the co n e is give n b y its
60
.
A
.
,
,
,
,
,
.
.
~
.
horizont al pro j ection The horizo n tal proj ectio n o f an element
h
and the vertex
o f the co n e is draw n through a
As the poi n t a
may be o n the top side o f the co n e n earest the horizo n tal plan e o r
o n the lower side there are two eleme n ts b v and c v with hori z o n tal
h
proj ectio n s passi ng through a
The fro n t proj ectio n s of these
two eleme n ts bf vi an d cf vf are drawn through the vertex and
through the fro n t proj ectio n s of the i n tersectio n s of the two ele
me n ts wi th the b ase, The required fro n t proj ection s O f the poi n t
a are the n o n the eleme n ts bf vi an d o
f vi at a l f an d a zf
Fo r the case Of the cyl i nder the same method o f solutio n is
applicab le
.
.
,
,
,
,
.
,
,
,
.
.
61
.
PROBLE M
22
Gi
.
ven
on e
j
o
e
t
i
o
r
c
n
p
o
f
poi nt on
a
the
su r
face o f a con e to find the o ther pr oj ecti on
Method Through the give n proj ectio n O f the poin t an d th e
vertex of the c o ne draw eleme n ts Of the surface Fi n d the required
projectio n s o f the poin t o n the other proj ectio n s o f these eleme n ts
,
.
.
‘
.
.
62
.
PR OBLE M
face of a
cyl i nd er ,
23
Gi ven
.
find
to
the
on e
o ther
j
r
o
e
o
c
t
n
i
j
p
o
e
c
o
n
r
t
i
p
o
f
o
o
n
t
n
i
p
a
the
su r
.
Method Through the give n proj ection o f the poin t d raw
eleme n ts o f the surface parallel to an y eleme n t shown A s i n th e
case o f the co n e precedin g the required proj ection s o f the poin t
are fou n d o n the other proj ectio n s o f these eleme n ts
.
.
,
.
Whe n a po in t o n a double curved surface such as a Sphere
ellipsoid torus etc is give n by o nly o ne pro j ectio n a differe nt
method is used for fi n din g the other projectio n o f the poin t o n the
surface tha n that employed for the co n e an d the cyli n der I n Fig
33 the horizo n tal an d fro n t proj ectio n s O f an ellipsoid are shown
A poi n t a o n the surface o f the ellipsoid is give n b y its fro n t pro
i
n through a an d the ce n ter o f the
ne
i
A
pla
P
is
draw
n
a
e
c
t
o
j
ellipsoid perpe n dicular to the fro n t plan e Its fro n t trace i
marked F P If this plan e is the n revolved SO that it is parallel
to the horizo n tal plan e the fro n t trace is shown by F P and the
fron t projectio n o f a is at M The plan e F P cuts from the ell i p
so i d a secti on o f which th e h o ri z o n tal proj ectio n o f the surface is an
63
.
,
,
,
.
,
,
.
.
.
,
,
,
.
,
s
.
.
’
,
,
’
.
~
,
’
’
exact represen tatio n ; an d a ( the revolved positio n o f a) h as i ts
h
I n revolvi n g the pla n e P b ack to its
horizon tal proj ectio n at a
o rrgl n al positio n the poi n t a moves to a i n a circular arc l yi n g m
a vertical plan e The fro n t proj ection of this ar e is M of ; an d its
h
h
horizon tal proj ectio n is O f course a a parallel to the X axis
h
The required hori z o n tal proj ectio n a is then determ i n ed b y
f
n
n
a
n
drawi g proj ecti g li es from
O b serve that the solution gives
also an other projecti o n a l h s h ow n at the top o f the fi gure
.
,
,
.
,
,
,
,
.
,
.
,
,
.
PROBLE M 2 4 Gi ven o n e proj ecti on of a poi nt o n a dou bl e
cu rved su r face to find the o ther proj ecti o n
Method Through the give n projectio n of the poi n t an d the
cen ter o f th e double curved surface draw the trace of a pla n e
perpe ndicular to the pla n e of proj ectio n i n which the proj ectio n o f
the poi n t is give n Revolve this plan e SO that it b ecomes parallel
to a seco n d plan e O f prO jec ti o n
D etermi n e the projectio n s of
B y revolvi n g b ack to the
the poi n t i n its revolved positio n
origi n al positio n the required proj ectio n o f the poi n t is located *
64
.
.
,
.
.
.
.
.
.
,
.
gen eral a pla ne is tangen t to a surface at a give n poi n t
when it passes through tange n ts to tw o li n es of the surface meeting
i n the give n poi n t
If the n through a given poin t any two i n ter
se cti n g li n es of the surface are drawn an d a ta n ge n t to each l i n e is
drawn at the poi n t the required pl an e is determi n ed by the ta n gen ts
Fo r drawin g a pla n e ta n ge n t to a si n gle curved surface the
ge n eral method is somewha t simplified I n the case of the co n e
we may ob serve that if a plan e is tan ge n t to the surface at a give n
poin t it is tange n t to the surface thoughou t the le ngth of an
65
In
.
,
,
,
.
,
,
,
,
.
,
.
,
,
An o th er m e th o d i s to
th e d o u b l e c u rved su f ac e
r
j
j ec ti o n
ec ti o n O f
of
b,
h
th e
c,
h
th e
of a s
b ase
i ts f ro n t
o in t
i n th e
Of a
h
b
an d
,
I f th e
p
p j
p j
d r aw
th at i s gi
p
w i th
and
,
v
fi
b as
.
p
ro ec ti o n
is in
th e
el l i
p
so i
p
an arc
.
p
xi
a
s on
xi
th e
a
s of
th ro u gh th e pro
th e pro
gi en
as s i n g
v
.
o int 1s
d
es
p
.
.
ro ec ti o n o f a
o u tl i n e o f
i ts
g
ri gh t c yl i n d e r
T h e h o ri z o n tal
h
nd a
r
n
a
e
a
a
a d O f th e
oin t a
,
su rf a c e as a c e n te r , a n d a rad i u s
th e
one o f
w i th i ts
by
p ro j ec ti o n
th e b a se a e
I n Fi g 34 th e o i n t a i s
ra w th ro u gh (Lf th e f r o n t
re
D
u
g
en
re c e d i n
a ri gh t c yl i n d e r
gi
v
by
en
T h e f ro n t
to
ro ec ti o n s o f
i ts h o ri z o n tal
w i th th e f ro n t
e q u al
p j
p j
ro
r
p j
ec ti o n o f
ro ec ti o n
th e
xi
a
s Of
p j
ro
ar e
th en dl f
an d
th e
"
th e h o ri z o n tal d i stan ce f r o m
ec ti o n s
dh ,
as
d
.
eh
to
element passing through the give n poin t and is therefore tangent
T he
to the co n e at the poi n t where this elemen t meets the b ase
con structio n o f tange n ts at the gi ven poi n t makes it n eces sary
to represe n t sectio n s o f the con e which is usually a lab orious
process A tange n t to the base at its i n tersection with the eleme n t
is more easil y co n structed an d is used i n stead o f the ta nge n t to a
sectio n at the g ive n poin t
I n Fig 35 a co n e i s Shown with a poin t a marked o n its surface
The plan e ta n ge n t to the co n e at this poi n t is determi n ed by th e
elemen t b v through a an d by a ta ngen t b c to the curve o f the b ase
at its i n tersectio n with b v T w o i n tersectin g lin es b v and b c
are thus represe n ted They determi n e the plan e P which is tan
gen t to the co n e at the poi n t a
As a cylin der may b e regarded as a special case o f a co n e wi th
th e vertex at i n fi n ity the problems relati n g to the cyli n der will
be solved b y the s ame methods as fo r similar pro blems relati ng
to the c o n e
,
.
,
.
,
.
.
,
.
,
.
,
.
,
.
,
.
PROBLE M
66
.
der thro ugh
a
To pass a plan e tangent to
s u rface
o
n
t
o
h
e
i
n
t
p
25
gi ven
.
a con e or a cyl i n
.
Method Through the give n poi n t draw an eleme n t O f the
surface A t the i n tersectio n Of this eleme n t with the b ase draw a
line tange n t to the b ase The required plan e is determin ed b y the
elemen t and this tange n t l i n e
.
.
,
.
.
E XE
1 06
1 07
.
.
R CISE S
Pass a plan e tange n t to
an
Oblique cyli nder through
a
poin t b o n its surface :
The b ase o f an i nverted right c i rcular c on e is i n the ho ri
z o n tal pla n e an d the a ngle b etwee n the eleme n ts of the
surf ac e and the axis is
P ass a plan e tangen t to
the co n e through a poin t c o n its s urface
,
.
plan e may b e drawn tange n t to a co ne a l so through a
poin t which is ou tside the surface B oth the vertex o f the con e an d
the given poin t must b e o f course in the tangen t plan e A lin e
67
.
A
.
.
j oini ng the vertex
the b ase o f
ta nge n t plan e
the give n poin t and another line tangent
the con e an d i n tersecting the fi rst lin e determin e the
to
an d
,
,
.
a poin t 0 outside o f it are shown A
lin e 0 v is drawn through the poin t 0 an d the vertex v Through the
intersectio n o f o v with the plan e B o f the b ase the lin e i t is drawn
ta nge n t t o the b ase The tangen t plan e P is determin ed by these
li n es o v an d i t
Whe n for the same co n dition s a cyli nder is used instead of a
con e the solutio n is the same except that the lin e drawn through
th e vertex for the c o n e i s replaced by o n e through the give n
poin t parallel to an eleme n t
I n Fig 36
.
a
co n e an d
.
.
,
,
,
.
.
,
,
.
68
.
PROB L E M
der throu gh
a
26
T o pass
.
v
e
n
n
t
i
o
i
g
p
a
o u tsid e
n
e
e
o
l
a
t
a
n
n
t
t
p
g
the su r face
a cone o r a eyli n
.
Method Through the give n poi n t draw the proj ection Of a lin e
which i n th e case o f the c o ne passes through th e vertex an d
i n the case o f the cyli n der is parallel to an eleme n t
Produce this
lin e to in tersect the plan e o f the b ase and draw a tange n t to the
The required plan e is de
b ase through the poin t o f in tersectio n
term i n ed by this ta nge n t li n e and the lin e already drawn through
the give n poin t
.
,
,
,
,
,
.
.
.
E
X E R CI SE S
P ass a plan e tange n t to
i n the X axi s
P ass a pl an e tangen t
an
oblique co n e and through a poin t
.
a cylinder which h as o n e b ase
i n the horizo n tal plan e an d a n other i n the fron t plan e
an d throu gh a poi n t i n the fro n t pla n e outside the b ase
'
to
,
.
69
to
a
.
PROB L E M
v
e
e
i
n
l
n
i
g
27
.
To pass
a
l
a
e
n
t
a
n
e
t
o
n
t
p
g
nd
a co n e a
a
r
a
ll el
p
.
Method Through the vertex Of th e co n e draw a lin e parallel
to the given lin e A t the poin t where this lin e in tersects the plan e
Of the b ase draw a lin e tangen t to the b ase These tw o lin es de
termin e the required plan e
.
.
,
.
.
E
X E RCIS E S
Pass a plan e tangen t to
an
in verted Obli que co ne and paral
lel to an Oblique li n e
P ass a plan e tange n t to a right c ircular con e
to a lin e i n the side plan e
P ass a plan e tangen t to a right c ircul ar con e
to an Ob li q ue lin e
.
an d
parallel
an d
parallel
.
.
70.
ll
e
l
r
a
a
p
PR O B L E M
to
28
T o pa ss
.
plane tangent to
a
a
cyl i nder
and
e
n
v
l
i
n
e
i
g
a
.
Method Through any poin t i n the give n lin e draw a li n e parallel
The pla n e determ i n ed by the give n
to an eleme n t o f the cyli n der
line an d the l i n e j ust drawn will be paral l el to the required plan e
A pla n e ta nge n t to the cyli nder is the n determi n ed b y a lin e tan
e
n t to the b ase an d p arallel to the pla n e already fou n d and the
g
eleme n t o f the c y linder which the t ange n t l i n e in tersects
.
.
.
,
.
E XE
R CIS E S
plan e tangen t to an Obli q ue c yl inder and parallel
to the X axis
P ass a plan e tangen t to a righ t circular c y lin der wi th i ts
b ase i n an oblique plan e an d parallel to a li n e i n the
same Oblique plan e
P ass
a
.
,
.
1
7
.
PRO B L E M
l uti on thro ugh
a
29
To pass
.
e
n
v
o
i
n
t
i
g
p
on
a
e
t
a
e
l
a
n
n
t
t
o
n
p
g
i ts
f
su r ace
f
f
a su r a ce o
revo
.
Metho d Draw through the given poi n t a lin e tangen t to the
i ntersectio n o f the surface with a plan e passi n g through the po in t
Whe n this lin e is revolved ab out the axis it ge n erates
and the axis
a right con e tange n t to the surface i n a circum fere n ce which c o n
tain s the give n poi n t A plan e tan ge n t to the con e at the given
poin t is the re quired plan e tange nt to the surfac e
.
.
.
.
.
R CISE S
Th ro ugh any point a o n the surface o f an ellipsoid wi th a
vertical axis pass a tange n t plan e
A circle 1 in ch i n diame ter with its plan e perpe n dicular
to the horizo n tal plan e r evo l ves a bo u t a ver ti ca l a xi s
which is three in ch es from the cen ter o f the circle A
surfac e o f revolutio n called a to ru s is thus ge n erated
Thro ugh a poin t b o n t his surface draw a tangen t plan e
E XE
1 15
.
,
,
.
,
1 16
.
,
.
.
.
2
7
PRO B LE M
.
thro ugh
a
30
.
To
p
ass
a
n
l
a
n
e
t
a
e
n
t
t
o
g
p
a
pher e
s
and
v
l
i
i
e
n
ne
g
.
Method P ass an auxil iary plane perpendicular to the give n
line through the ce n ter o f the sphere (This plan e cuts a great circle
from the sphere an d wi ll cu t a li n e ta nge n t to this circle from th e
requi red tan ge nt plan e ) Revolve the auxil iary plan e with i ts
in tersec tio ns with the given l i n e and the sphere in to a plan e o f pro
n o f the give n li n e with
n
n
Through
the
poi
t
Of
i
tersectio
n
c
i
o
e
t
j
the auxiliary plan e draw a lin e ta ngen t to the circular i n tersectio n
A plan e passed through
o f the auxiliary pla n e with the Sphere
th e ta n ge n t li n e ( whe n revolved to i ts o riginal o r true position )
and th e giv en li ne is the o n e r eq uired
.
.
.
.
.
.
R CIS E S
Draw a pl ane tange n t to a Sphere and passi ng thro ugh
an y ob lique lin e
P ass a plan e through a lin e parallel to the X axis an d
tange n t to a sphere wi th i ts cen ter in the horizo n tal
plan e
E XE
.
.
regular prism wi th sixteen sides
Th e studen t must n o w f orm a me n tal pic ture o f a righ t
is Shown
h
k
r
n
n
f
b
n
ria
gle
made
o
thi
card
oard
esti
g
o
t
h
pro
ectio
e
n
3
n i
t
j
This proj ectio n shall be al so the horizon tal leg o f
in the figure
the trian gle The vertical leg is Shown in the fron t pro j e ction o f
the prism by the distan ce from 17 to the bas e We c an imagi n e
then suc h a paper triangle wrapp ed aro und th e pri sm wi th th e ver
73
.
Convo l utes
.
I n Fig 37
.
a
.
,
.
.
.
tical leg through i held station ary and the long leg of th e triangle
always touches the b ase of the prism The fro n t proj ectio n
o f the hypothe n use o f this paper tria n gle is the n a b roke n li n e
co nn ectin g the front proj ectio n s Of the poin ts i h g
c b a
E ach portio n of this b roke n li n e i s equ al l y in cl in ed to the edges
The horizo n tal proj ectio n Of the h ypothe n use is
o f the prism
the broke n lin e j o in ing the horizo n tal proj ectio n s of the same poi n ts
The surface o f the paper trian gle i s n o w u n wrapped tak i ng it o ff
A n d if i t is tur n ed o n each
o n e S ide of the prism at a time
edge of the prism i n successi on as o n a hinge till each un wrapped
portio n co i ncides with the plan e of the n ext Side the portio n o f
the hypothe n use that has b ee n released b ecomes an exte n sio n an d
a tange n t Of the portio n o f the b roke n li n e o n that side of the prism
Thus whe n the portio n b etwee n a and b is u nwr apped it b ecomes
tange n t to the S ide in which b c lies ; an d whe n also the portio n
b etwee n b an d c is u n rolled the portio n a b c o f the h ypothe n use i s
tan ge n t to the side co n tain i n g 0 d
Let us n o w co n sider the properties Of this b roke n li n e from
The portion a b i n tersects b c b c in tersects c d etc ; tha t
a to i
”
“
is the y in tersec t tw o an d two ; b u t a b does n o t i n tersect c d
no r does b c i n tersec t d e etc
.
,
,
,
,
,
.
.
.
,
.
,
,
.
,
,
.
,
.
,
.
,
.
,
li n e o n the
prism If the n umb er of S ides of the prism is in creased in defi n itely
the b roken lin e o n its surface b ecomes a curve called a h eli x It is
a curve ge n erated by a poin t movi n g o n the surface Of a cylin der
Of revolution SO as to cut all the eleme n ts at a co n stan t a ngle The
movi n g poin t has u n iform motio n ar o u n d an d at the same time
n struc ti n g
ll
l
to
the
axis
The
method
for
co
o f the cyli n der
e
r
a
a
p
a helix i s Shown i n Fig 38 The axis of the hel i x is vertical SO
that the horizo n tal proj ection is a circle with its cen ter at o in
the axis N ow if the ge n erati n g poin t moves the d istan ce from
f
i
to
i n maki n g o n e complete revolutio n a b out th e axis
v
m
passin g th rough the poi n ts m p s t u an d v the vertical dista nc e
f is called th e i tch Of th e helix
etwee
n
v
a
d
W
n
b
p
74
.
C o n sider also an other property of the
b roke n
,
.
.
.
,
.
,
.
.
,
.
,
,
,
,
,
,
,
.
is formed o f the tangen ts to th e
broke n li n e in Fi g 37 whe n the n umb er O f S ides O f the prism
has b een in creased in defi n itely an d it b ecomes a cylin der
The elemen ts o f this surface are the li n es m b n c o d p e q f
etc S uch elemen ts when tange n t to a cyli n de r form a co n
volute su rf ace
P ortion s o f such surfaces are represe n ted i n
Figs 39 an d 4 0
They are surfaces ge n erated by a str aigh t
lin e m ovi n g alon g a c u rve o f dou bl e cu rv atu re so th at th e li n e
I n this surface agai n an y two
i s alw ay s t an gen t to th e cu rv e
b u t n o three c on secutive straight l i n e eleme n t lie i n the same
plan e
The co nvolute surface i n Fig 4 0 can b e regarded as formed
by the co n secutive positio n s of the hypot he n use O f a paper tria n gle
as it is u n wou n d from the surface of a cylin der o f which the b ase
h
h
h "
is the circle a d n e
The poin t a at the en d of the u n wi n din g
h
h h h
tria ngle will always lie i n the curve a c i f i n the hori z o n tal plan e
This curve is the in v o lute of the circular b ase of the cyli n der ;
The
b u t it is also the hori z o n tal trace o f the co n volute surfac e
cyli n der itself i s n o part of the s u rface O b serve the striki ng
resemblan ce b etwee n the way this surface wi n ds arou n d a cyl i n der
S hell
an d the c o n volutio n s o f a sea —
Whe n a helical co n volute is to be represe n ted o n a drawi n g
the curve of the hel i x to which the surface is ta nge n t should
—
draw i ng first the top and fro n t views
b e accurately co n structed
Of the cylin der o n which it lies If the axis is vertical with the
b ase i n the horizo n tal pla n e it is show n ab ove that the horizo n
tal trace of the co n volute is the in volute Of the c i rcular b ase
75
A
.
v
cu r e d
su rf ace
.
.
,
,
,
,
,
,
.
.
.
.
,
.
,
,
,
,
.
.
.
.
.
.
.
,
.
,
,
.
If a poi n t m o n the con volute surface i n Fi g 4 0 is given
h
o nly by the horizon tal proj ectio n m the fro n t proj ectio n can be
foun d by co n structio n The horizo n tal proj ectio n o f an elemen t
76
.
.
,
.
If
a
v
cu r
on
tan gen t
in
e c al l e d an
th e
th re ad
i n vo l u te
c i rc u l a r
at a ,
of
ro l l s u
v
on a
o l u te O f
b ase
th en th e
th e
p
c i rcl e
.
an d
th e
fix d c u ve an y p i n t
fi st I n Fi g 4 0 su pp
k pt
e
en d a t a
e
r
r
.
tau t
o
,
.
as
i t is
w i l l d es c ri
be
o
mi t d esc ri b es
o se a
u n w o un d
th e
c ur
v
a s ec o n d
th re a d to b e w o u n d
f ro m th e
h h
e a c
i
h
e nd O f
th e
ed
th e
"
al l
f
,
c
pass i ng through the poin t m can b e drawn tangen t to the c i rcul ar
"
"
an d it will in tersect the in volute at i
The front
b ase at n
proj ection s o f n an d i are respectively in the helix at rd an d at 17
h
f
B y drawin g the proj ectio n i n
} i a
n d a projecti n g lin e from m
the
fro n t proj ectio n o f m is determi n ed
The solution can b e reversed : If the fron t proj ection Of a poin t
o n the c o n volute surface is give n the horizo ntal proj ectio n c an b e
fou n d
.
,
.
,
,
.
,
.
77
.
PR OBLE M
31
.
Gi ven
o ne
j
e
o
t
o
n
o
r
c
i
p
f
a
n
t
a
v
o
o lu te
o
n
co
n
i
p
Method Draw a proj ec tio n o f an eleme n t through the give n
proj ec tio n Of the poin t D etermin e the other proj ection of this
eleme n t an d locate o n it the required projec tio n o f the poi n t
.
.
.
E
X E R CIS E S
Draw the pro j ectio n s o f ten equidis tan t elemen ts o f a
helical co nvolute surface with a vertical axis The
diameter an d pitch of the helix are respectively 2 i n ches
an d 3 in c hes
Fi n d the i n tersectio n s o f the con volute surface give n i n
E x 1 1 9 with tw o plan es perpen dic ul ar to the ax
is of
the hel i x Ob serve the Shape o f the c urves cut by these
plan es
Draw the pro j ections o f six elemen ts o f a con volute sur
f ace O f which the axi s is parallel to the X axis
The
diameter and pitch o f the helix are respectivel y2 % i n ches
A ssume o n e pro j ectio n of a poi n t o n the
an d 4 i n ches
surface an d locate the other proj ection (Use the Side
plan e )
D raw tw o helices o n e right han ded an d the other left
han ded o n a cylin d er 1} in ches I n di ameter an d 2 i n ches
high the former to b e Of 1 in ch pitch an d the latter
1
Of in ch pitch
Represe n t a square thread ed screw Of the followin g
.
.
.
.
.
.
.
.
.
-
,
7
,
,
2
.
-
dimen sio n s : O u tside diameter o f the thread is 2 } in ch es
Diameter at the b ottom o f the thread is 1 5 i n ches
Pitch is 1 inch
Thick n ess O f the thread is 4 i n ch
S how two complete turn s of the thread
A Spiral sprin g is O f the form of a square screw thre ad
The cross sectio n is 4 i n ch square and the outside diam
eter an d pitch are respectively 3 i n ches an d 2 in ches
Draw two complete turn s of the sprin g
”
A Spiral Sprin g is made Of rou n d wire as S hown i n the fi gure
The diameter o f the wire is 5 in ch The outside
b elow
diameter O f the spri n g is 2 } in ches an d the pitch is 2
i n ches Draw the plan an d elevation o f tw o tur ns O f
the Sprin g
Su ggesti on
If a Spiral Spring is made Of roun d wire
w e co n ceive its surfa ce to b e ge n erated by a sphere m o v
i ng alo ng a helix which is the ce n ter li n e o f the wire
The proj ec tion s o f the helix are fi rst drawn and the n
th e projectio n s o f the sphere i n a n um b er o f di fferent
positio n s
7
.
.
.
.
.
.
-
,
.
.
«
.
.
.
.
.
,
.
.
8
7
.
PROBLE M
face through
a
32
gi ven
To pass a plane tangen t to
su rface
o
i
h
e
n
t
o
n
t
p
.
a co nvo lu te sur
.
Method (The same as for the con e o r the cylin der ) Through
A t the in tersectio n
th e give n poi n t draw an eleme n t of the surface
The requi red plane
o f this eleme n t with the b ase draw a ta nge n t
is determin ed b y the eleme n t an d the tan ge n t lin e
.
.
.
.
.
79
and
.
PROBLE M 33
throu gh
Metho d
a
e
i
v
g n
.
To pa ss a pla n e tan gen t to
o u tsi de the su r face
t
o
n
i
p
f
a co n vo l u te su r ace
.
Through the give n poi n t pass a p lan e w hi ch is per
n tersectio n of thi s pla n e
The
i
i
cu l ar to the axis of the Surface
n
d
e
p
with the surface is an i n volute to which a tan ge n t is then drawn
from the give n poi n t Draw an eleme n t o f the surface at the poin t
.
.
,
.
ta n ge n c y T his element
req uired plan e
of
.
an d
the tan gen t lin e determi n e
.
E XE
12 6
.
Draw a plan e tange n t to the helical c onvolute c onstructed
i n E x 1 1 9 at any poi n t o n the surface
Draw a plan e tangen t to the con volute surfac e c onstruc ted
in E x 1 2 1 and through any po in t o u tside th e surface
.
127
.
R CISE S
.
,
.
.
C HA P T E R
I N T E R SECT I O NS
A ND
DE
IV
V E L O PME NT S
O F SO LI D S
The i n tersectio n o f an y su r f ace with a gi v en pl an e is fou n d
by passi n g a series o f au xi li ary pl an e s I n such a way that they cu t
from the sur face strai gh t lin es c i rcl es o r oth er cu rv es th at can
b e qu i ck l y dr awn ; an d from the give n plan e strai gh t l i n e s
The
in tersectio n s o f these li n es cut from the surface with the l in es cut
from the plan e determin e poin ts o n the required i n tersectio n
Whe n straight li n e elemen ts can b e cut from the surface the aux
il i ary plan es are usually passed perpe n dicular to o n e o f the pla n es
Wh en circles c an b e cut from the surface the aux
o f proj ectio n
i l i ary plan es should b e drawn par a
llel to the plan e o f pro j ection o n
which the proj ection s o f the circles w il l appear i n their true form
The tan gen t to a cu rve o f i ntersectio n at a give n poin t o n a su r
face lies i n the plan e which cuts from the surface this lin e o f i n
tersecti o n
It is als o a li n e i n the tange n t plan e to the surface at
the give n poin t
80
.
,
,
,
.
,
,
,
.
.
,
.
.
.
The curve o f i n tersection o f an o bli qu e c on e with a plan e P
The required curve o f in tersectio n is shown
is Shown i n Fig 4 1
i n the figure by its elliptical proj ectio n s o n which the po m ts s an d t
are marked The curve is Ob tai n ed b y drawi n g a series o f au xili ary
h
r
n
h
e
n
h
h
l
r
u
t
h
n
n
n
u
a
r
t
o
t
t
t
o
g
v
o
r
l
f
o
a
e
s
e
e
r
t
e
f
t
e
d
e
e
di
c
x
e
o
c
a
p
p p
a
n
e
n
l
the
horizo
tal
traces
wi
l
perpe
n
dicular
the
X
axis
t
l
e
o
b
p
(
)
E ach auxiliary plan e cuts from the c on e el em en t s an d from the
given plan e a strai gh t li n e I n the fi gure o ne o f these auxiliary
plan es is marked b y its traces H A an d F A This auxiliary plan e
cuts from the c on e the elemen ts a v an d b v an d from the plan e
a straight lin e m n The in tersection s o f the proj ection s o f this last
lin e with the proj ection s o f a v an d b v give respectivel y the poin ts
3 an d t o n the re q uired li n e o f i n tersec ti o n
81
.
.
.
.
.
i
,
.
,
.
.
.
The curve o f in tersection o f an obli que c ylinder with a given
plan e is foun d by the same ge n eral method usin g auxiliary planes
that will cut eleme n ts from the surface o f the cyli n der and straigh t
lin es from the give n plan e
,
.
A pl an e w h i ch i s tan gen t to a gi ven c o n e o r a cyli n der,
If , then , the surface o f a co n e ,
c o n tai n s an el em en t o f i ts su rf ace
82
.
.
a cylin der is rolled o n a tangen t plan e u n til each o f its eleme n ts
has come in to this plan e the par t Of the plan e passed over and
i n c luded b etween the extreme elemen ts is a pl an e su rf ace equ al to
th e gi v en su rf ace
The surface thus p as sed over is called a dev elop
men t I n o rder to determin e the positio n s o f the diff ere n t ele
me n ts o f the surface as they come in to the plan e o f developme n t
it is n ecessary to lo cat e so m e c u rv e o n th e su rf ace w h i ch wi ll
dev elo p i n to a str aigh t l i n e Or a sim pl e curv e upo n which the actual
distances betw een th e eleme n ts can b e laid o ff
The sectio n cu t
I n Fig 4 2 an Ob lique cyli n der is Shown
f rom thi s c y li n der b y a hori z o n tal pla n e happe n s to b e a circle
Through the poin t a a plan e is passed cutting the surface i n the
straight li n e through the poi n ts a c and e The curve of in ter
sectio n whe n revolved through 90 in to the plan e o f the drawi n g
is shown as the circle 1 2 3 4 5 T h e ci rcumferen ce o f th i s
c i rc l e i s th e a ctu al l en gth o f th e dev elo pm en t o f th e cyli n de r
If we wish to develop the part Of the cylin der ab ove the i n ter
sectio n we draw the b ase li n e a a making its le n gth equal to the
c ircumfere n ce of the circular i n tersectio n A poi n t b o n the curve
of the dev elopme n t is immediately located over the poin t a at a
distan ce laid o ff o n an eleme n t equal to that from a to b i n the
to p b ase
O ther poin ts are take n in the same way To locate
an other poi n t o n the curve of the devel o pm en t th e distan ce 1 2
is laid O ff o n the b ase li n e equal to the arc 1 2 an d the le ngth
c d is the width at this poi n t
o f the surface we are developi n g
B y co n ti n uin g the process a series o f poin ts is Ob tai n ed an d the
surface in cluded b etwee n the curve drawn through these p oin ts
an d the b ase li n e is the developme n t o f the surface o f the cyli n der
ab ove the in tersectio n with the plan e passed through a
or
,
,
.
.
,
,
.
.
.
.
.
,
.
,
,
°
,
,
,
.
,
.
,
,
,
.
,
'
,
,
.
.
'
,
,
,
,
.
,
.
The method for the developmen t o f a con e b ei ng similar it
ne eds n o expl an ti o n
The problems relating to developme nts Of surfaces are co n
O f course theoretic
stan tl y appl i ed by w o rk e rs i n sh ee t m etal
ally i t makes n o diff ere n ce which elemen t o f the surface is cut
“
for the developmen t ; b u t practically it is eco n omical to ou t
”
so as to make the shortest seam unless however there are good
reaso n s for doi ng otherwise
,
.
.
,
,
,
,
,
,
,
.
83
e
l
a
n
p
.
PROBLE M
34
.
To
nd
fi
‘
the i n tersecti on
o
f
an y cone
wi th
an y
.
Method P ass through the co ne a series o f aux i liary plan es
take n either perpe n dicular to its axis or through the vertex an d
perpe n dicular to o ne of the plan es o f proj e ctio n Fo r the curve
o f i n tersectio n j oi n the poi n ts where the l i nes cut from the c o n e
in ters ect th e l in es cu t from the plan e
.
,
.
.
’
.
.
E XE
R CIS E S
Draw the c urve o f i n tersectio n o f a right c one axis ver
tical with an Ob lique plan e Draw the devel opmen t
o f the portio n Of the co n e b etwee n the li n e o f i n ter
sectio n an d the vertex
Taking the data of E x 4 2 determin e the curve o f i n ter
sectio n o f the co n e of rays of maximum in te ns ity with
the plan e Of the desk
Make the draw i ngs for the patterns o f the b ath tub
shown i n the figure o n page 1 34
—
s
o
e
S u gg ti n Th is exercise requires merely the devel
n es with vertices at n and
O pm e n t of porti on s o f tw o c o
As sume sui table
o
an d the patter n s o f pla n e surfaces
dimen sio n s for those n o t give n i n this exercise and i n
those that follow
Draw a pattern for the sheet metal for a regularly flaring
n
n
n
n
a
l
e
o
roof
co
ectio
lik
the
o
n
i
c
e Shown i n the
c
)
(
,
.
,
.
.
,
.
-
.
.
,
,
.
figure
xis o f the cyli ndri cal p1pe is vert i cal and
th e pla n e o f the roof is represen ted by R( 0
A con ical tower is to b e placed o n the righ t han d s i de
S how the
82
o f the irr egul ar hi p roof shown i n E x
true size of the hole to b e cut i n the roo f
—
s
Fi n d the true size o f the curve o f in ter
Sugge ti on
sectio n b y r evo l vi ng the cu rve ab out o ne o f the traces
C f A rt 5 3
o f its pla n e in to a plan e of proj ectio n
Dr aw the pattern fo r a b oot to j oin two pieces o f stove
pipe o n e o f which is c ircular and the other oval
The circul ar pipe is 6 in ches i n diameter an d the oval
pipe is represen ted by parallel sides 2 } in ches apart
long wi th se micircles at the en ds o f
an d 5 4 in ches
I n other words the over all di men
th e parallel Sides
Mak e the
s io ns o f the oval are 8 in ch es by 2 k in ch es
“ oo ” 1 0 in ches long
b t
Th e
.
a
,
-
.
.
.
.
.
,
.
.
.
,
,
1
,
-
,
.
.
.
.
84
any
.
PRO B L E M
l
n
e
a
p
35
.
To
nd
fi
the i n tersecti o n
o
f
wi th
an y cyli nder
.
Method Pass through the cylin der a series o f auxili ary plan es
taken either perpen dicul ar to its axis o r parallel to its axis and
r the cu rve o f
n
F
n
f
n
f
o
erpe
dicular
n
the
pla
es
proj
ectio
o
t
o
o
e
o
p
in tersectio n joi n the poi n ts where the l in es cut from the cyli n d er
in tersec t the l i n es cut from the give n plan e
,
.
,
.
.
EXE
R CIS E S
Fin d
the true size and shape o f the hole cut fo r a circular
chi mn ey i n a ti n coveri n g fo r the right ha n d S ide o f the
hi p roof give n i n E x 7 9
Draw the pattern for the i n cli n ed en d o f a b ath tu b with a
semicircular cross —sectio n
The
i
n
cli
n
ed
n
is
as
d
e
(
sumed to be a plan e surface
Fi n d the true size of the Ope n i n g to b e cut in the wal l
Shown i n Fig 5 1 for a pipe 3 feet in diameter maki n g
an gles o f 30 an d 2 0 respecti vel y wi th the hori z o n tal
an d with the vertical pla n e Of the b ack o f the wall
I n the figure a flue from a b oiler house is Shown pass
in g through the roof o f a smal l Shed
Fi n d the true
-
.
.
-
.
.
.
°
,
°
‘
,
.
-
.
i e o f the hole that w as cut in the roof for the flue
Make a pattern to Show a Sheet o f ti n to cover this roof
D evelop th e i n
w ith the hole cut o u t f o r the flue
cl i n ed portio n o f the flue ab ove the roof
(S elect suit
ab le data fo r the co Ordinates o f the poin ts that are
marked in the fi gure )
s z
.
.
.
.
85
o
.
PRO B LE M
f revo l u ti on
wi th
nd
fi
36
To
any
a
n
l
e
p
.
the
o
f
i nter secti on
f
o
f
an y su r a ce
.
M ethod Pass through the surface o f revolutio n a series o f auxil
i ary plan es perpe n dicular to its axi s
These plan es cut circles from
.
.
curve of in tersectio n
.
E
1 38
.
X E R CIS E S
Fi n d
the i n tersectio n o f the torus given i n E x 1 1 6 with
an O b lique pla n e
Draw the proj ecti o n s of the hexago n al n u t S hown i n the
figure represen ti n g accurately the lin es o f in tersectio n
.
.
139
.
,
1 40
.
.
“
Draw accurately the curves o f the stub end
i n g rod
of a
co nn ec t
-
.
14 1
.
The b lades Of a ve n tilatin g fan are plan e surfaces attached
S how the in tersectio n s
to a Spherical h u b
.
.
The in tersecti o n o f any two curved su rf aces is fou nd by
p as sing auxiliary plan es to cut from each surface l in es tha t can
The in tersectio n s Of these li n es give
be easily co n structed
poin ts o n the required curve o f in tersectio n
The auxil iary
plan es should b e selected so that the li n es cut f rom the surfaces
are straight lin es circles parallel to a plan e o f proj ectio n o r some
other curve o f which the proj ectio ns can be easily drawn
86
.
.
.
,
,
.
shaded drawi n g of tw o i n tersecti ng co n es is shown
The same co n es are shown i n orthographic pro j ectio n
I n Fi g 4 3
The curves o f i n terse ctio n that are shown were o h
i n Fi g 44
tai n ed b y the use o f auxiliary pla n es passed SO as to cut elemen ts
from each co n e ; i n other words the auxiliary plan es were passed
through the vertices Of b oth c o n es The elemen ts cut from each
co n e b y an auxiliary plan e are drawn through its i n tersectio n s with
the b ase an d through the vertex
S in ce it is a
I n Fi g 4 4 an auxilia ry plan e A i s marked
plan e passin g through the vertices o f b oth c on es its horizo n tal
trace H A is draw n through the i n tersection o f the lin e v w with
"
the p lan e of the b ases marked x
O b serve that the horizo n tal
rojectio
n s O f the co n es are represe n ted o n the pla n e of the b ases
p
i
i
d
x
that
the
li
e
through
d
is used as the axi s S howin g
n
w
n
o
an d
the intersectio n Of the hori z o n tal an d fro n t plan es The auxiliary
plan e cuts from the co n e with vertex v the eleme n ts c v an d d v;
the eleme n ts a w an d b w
an d from the co n e with vertex w
The i n tersectio n s of the horizo n tal proj ection s of these elemen ts
”
two an d two determin e the horizo n tal proj ectio n s Of four poin ts
The curves in the fron t proj e ction
o n the curves o f i n tersectio n
are Ob tain ed b y drawin g the proj ection s Of the eleme n ts Of o n e
con e an d proj ecti ng o n them the poin ts in the horizon tal pro
n s o f two
The
pro
j
ectio
i
o n s o f the curves of i n tersectio n
c
t
e
j
oi
ts
i
n the curves are m arked m an d n
n
p
87
.
A
.
.
.
.
,
'
.
.
.
,
,
.
,
,
,
.
,
,
,
.
,
.
,
,
.
,
.
.
PROBLE M 37 To find the curve of i ntersecti on o f two cones
Method P ass through the con es auxiliary plan es drawn through
88
.
.
.
line j o mmg their vertices E lemen ts cut from each con e by
these pla n es determin e by their in tersection s the requir ed curv e
th e
.
.
E X E RCI SES
Fi n d
the curve o f in tersection of tw o right con es with axes
that do n o t in tersect The axis of o n e co n e is horizo n
tal the o ther verti cal Draw the developme n t of the
co n e with the horizon tal axis showing the curve o f
i n tersection
Draw the curve o f in tersection o f two oblique con es with
i n tersecting axes S how the developmen t o f o n e o f the
c on es
.
,
.
‘
,
.
.
.
the meth od most comm onl y used for fi n din g
the in tersectio n of two cylin ders is shown The solutio n is a
si mplified met h od o f passing plan es perpen dicular to the fro n t
plane— o ne set parallel to the axis o f the smaller cylin der an d
an other set parallel to the axis of the larger cyli n der The figure
is lettered so that n o other explanatio n is n eeded
89
.
I n Fi g 4 5
.
~
.
.
.
90
.
P R O B LE M 38
cyl i nders
.
To
nd
fi
the
cu r ve
o
f
i n ter secti o n
o
f
two
.
Method P ass through the cylin ders tw o sets of auxiliary
plan es o ne set parallel to each axis The eleme n ts cut from each
cyli nder b y these planes determi n e by their in tersection s the re
quired curve
.
.
,
.
E X E RCI SES
D raw
accurately the li n es o f i n tersectio n appearing o n th e
surfac e o f the flanged pipe fitti n g shown i n the figure
o n page 1 44
The fitti ng is made u p of two cyli n ders
with their axes i n tersecti n g each other symm etrically
Fi n d the i n tersectio n of tw o ob lique cylin ders with their
b ases i n the hori z o n tal plan e
A vertical steam drum is to b e put o n a horizo n tal cyl i n
,
.
.
.
-
Make a pattern to show the size
h ole to b e cut i n the b oiler plate
(C f Fig
dri cal b oiler
.
.
147
.
.
of
th e
.
Fin d
the intersection o f tw o cyli n ders forming a b ran ch
“
Y
fo r a b lowpipe as show n i n the figure o n page 14 6
D evelop b oth cylin ders to show the curve of in tersection
.
.
n
f
n
n
a
very
simple
case
o
the
i
tersectio
of
a
co
e
9
The horizo n tal proj ectio n of the c yli n der
an d a cyli n der i s shown
shows immediately the hori z on tal proj ectio n of the curve of in ter
section an d th e front proj ection o f the curve is fou n d by proj ection
1
I n Fi g 4 6
.
.
.
s
.
.
Horizon t al
lines
drawn through the fro n t pro j ections of the
po in ts a b and c to show a simple method for fi n di ng the tr ue
le ngths o f elemen ts of the c on e an d the true distan ces from the
vertex o f poi n ts o n an eleme n t These true le ngths are needed for
a d evelopme n t o f the surface o f the co n e
are
,
,
,
.
.
.
This is a special c ase
n ext pr o bl em
.
The ge n eral method
is
ed
stat
i n the
f
a cyli nder
.
92
.
P RO B LE M 39
and a con e
.
To
find
the
cu rve o
f
i nter secti on
o
.
Method P ass through the vertex o f the c o n e auxiliary p lan es
parallel to the axis o f the cylin der The eleme n ts cu t from each
surf ace by these plan es determin e b y their i n tersectio n s the requ i r ed
curve
When th e two surfaces have circular b ases i t is most ad van
tageo us to use aux i liary pla n es which cut circles from each s u rface
.
.
.
,
.
E
X E RCI SES
c ircular tower has a co n ical roof through which a hori
Fi n d the size of a coverin g for the
z o n tal pipe passes
roof an d show i n the developme n t the hole cut for the
pipe
Fin d the i n tersectio n o f an o b lique c o n e with a righ t
cylin der The axes do n o t in tersect and o ne axis is
parallel and the other perpe ndicul ar to the front plan e
Make the n ecessary pa tter ns fo r an arch stone o f a coni c al
arch i n a c ircular w all
A
.
.
.
,
.
.
horizo n tal steam pipe 1 1 i n ches i n diameter is in tersected
b y a co n ical n o z z le an d two smaller vertical pipes as
shown i n the figure The s10 pe o f the curved s u rface o f
the n oz z le is 7 5
n
D
a
f
etermi
e
the
developed
t
rue
size
o
the
hole
c
u t in
)
(
the hori z o n tal pipe by the n o z zle
n
Develop
the
c
o
ical surface of the n ozzle
b
( )
A
7
;
.
3
.
.
.
1 52
.
Fin d
the i ntersectio n of the co n e an d the cylin der formin g
the steam exhaust head shown i n the figure o n page 1 5 0
-
93
.
P RO B L E M
40
.
To
.
nd
fi
the l i n e
o
f
i n ter secti o n
o
f
an y
tw o
o n ( spher e el li pso id to ru s
e
o
l
u
t
i
r
v
f
f
Method If the axes of the two s u rfaces of revolution in tersec t
su r a ces o
,
,
,
.
.
poi n t of in tersection of the axes is taken as the cen ter fo r a
series o f auxiliary spheres These plan es cut c ircles from each
surface The in tersection s o f these circles with each other are
poin ts o n the required curve
If the axes do n o t in tersect pass through the surfaces auxiliary
n
n
i
n
la
es
perpe
dicular
t
o
the
ax
s
of
o
e
surface
a
n
cut
i
g
circles
d
n
t
p
th e
.
.
.
,
from
that surface an d cuttin g some other curve from the
The i n tersectio n s of the correspo n din g proj ectio n s o f the
surface
curves give the poin ts to b e foun d
T he
Fig 47 shows the in tersectio n o f a sphere with an ellipsoid
poin t m is take n as the cen ter for the auxiliary spheres ( at the i n
The arcs of the auxiliary circles shown de
terse c ti o n of the axes )
B y co n ti n ui n g this
termin e b y their i n ter sectio n the poi n t xf
co n structio n a series o f poin ts is ob tai n ed which whe n conn ecte d
f
t
n
f
a
b
a
smooth
curve
gives
the
li
e
o
i
n
tersectio
n
The hori
by
z o n tal proj ectio n is ob tai n ed easily b y proj ectio n
,
.
.
.
.
.
.
.
.
E
1 53
.
X E R CI SES
Draw the curve o f i n tersectio n o f a sphere and an ob lique
co n e
Draw the lin e of i n tersectio n o f the ellipsoid and th e toru s
shown i n the first figure o n p age 1 52
.
1 54
.
.
U se th e
Suggesti o n
th e
el li
p se
oh H
1 55
i n th e
h
E
o
r
.
i
p
x
m
p
ro
a
fi gu re
,
a te
meth o d
mak i ng C D
t
o f ol rc u l ar arc s
A B
F
'
E Oh
f or
Oh F
co n str u cti n g
an d
G 0h
.
is shown i n the figure bel ow The cente r
"
line of o n e portio n is the arc o f a circle i n tersectin g
the axis of the smaller portion whi ch is cylin drical
Make the diameters respectively 1 } i n ches an d 1 i n ch
Draw the curves o f i n tersectio n o n the outside
o f the fitt i n g
A pi pe fitting
.
.
7
.
.
T h e sewer
show n i n the figure b elow by its sectio n is i n ter
se cted by two elliptical sewers with ma j or an d m in or
The dimen sion marked r
axes o f 4 feet an d 2 % feet
for the large sewer i s 5 feet S how i n a pl an drawi n g
the li n es o f i n tersectio n to be made i n the mason ry where
the sewers meet The b ottoms o f the three sewers
The small s ewers are o n o p
are i n the same pla n e
o si te sides o f the large sewer
p
.
.
.
.
,
.
CHA PT E R V
MI S CE LLANE O U S P R A C TI CAL
E XE
RCI S E S
'
A shaft o f a min e follows approximately th e l i n e b etwee n
—
an d b
n
a
3
1
the poi ts
—
n
A tu nn el is to be made from a poi t e
1§
o n the side of a mou n tai n to i n tersect the shaft
Fin d
the shortest le n gth o f the tu nn el an d the a n gle be
twee n the ce n ter lin e o f the shaft and the ce n ter li n e
of the tu n n el
—
nn
n
n
4
a
d
a
I
b eam are show n as the y
a
cha
el
8
I n Fi g
i n tersect i n a roof Take the followi n g co ordi n ates
fo r the poi n ts mark ed i n the figure : a
-
,
,
.
,
.
.
.
b
e
a
,
_
s)
e
,
4)
(w
O) h
(
—l
i)
+ 5%
C
—
2
e
,
,
,
47
_
d
i (w e
,
4
4
,
n
u)
,
Draw the proj ections
a secti o n drawi n g
o f a bent pl a te con n ecti o n an d in
Show the true a n gle b etwee n its sides
A cylin drical pipe 2 feet i n diameter passes through the
roofs shown i n Fi g 1 4 a t a poi n t i n the i n tersectio n
The axis o f the pipe is perpen dicular
o f the roofs
to the l in e o f i n tersectio n an d is i n cl i n ed 4 5 from th e
vertical Fin d the true size of the hole cut o u t o f
the roofs for the pipe
O b tain by developme n t the true si z e and shape o f the
coverin g n eeded for a sy m metrical dome with eigh t
sides
I n Fi g 4 9 a portio n of a locomotive b oiler is shown
Make the to p elemen t o f the slope sh eet at an in cli n a
tion o f 30 to the hori z o n t al and show the true srz e
o f a steel plate to b e used i n maki n g the slope sheet
The san d —b o x o n a locomotive sta n ds partly o n the slope
—21
5,
1
3
4
3,
(9
-
7
5
-
,
2,
.
,
,
.
.
°
.
.
.
.
.
°
,
.
sheet and partl y o n the cyl in drical portio n Fin d
the shape of the b ottom o f the sa n d b o x to fit the
b oiler draw n for E x 1 61
Make the pattern for the co ni cal portio n o f the eave
t ro ugh outlet shown in the figure
.
-
.
163
.
.
.
164
.
A
b loc k
wood with a s quare cross sec tion has b ee n
turn ed i n a lathe to the shape shown i n the figure
S how the lin es of in tersection b etween the part with
the square sectio n and the c o n ical portio n
of
-
.
.
Make the patterns for the
—
oil can
an d
grocer s scoop
’
shown o n page 1 60
—
The stack of a b oiler is supported b y gusset plates as
show n i n Fi g 50 Make the pattern s for the gusset
plates
Fin d the i n tersectio n o i a sphere with a cyli n der whose
axis does n o t pass through the ce n ter of the sphere
Fin d the in tersectio n of the cyli n drical ceilin g o f a corridor
with the hemispherical ceil i n g of a vaul t
.
.
.
.
.
.
DO ME
4
O
D EVE L P ME NT O F T H E SLO P E SHE ET .
FI G
.
49
.
A semicircul ar wire loop 1 } in ches i n diameter rotates on
a vertical axis supported at the poin ts a — 1
—1
— 1f
an d b
}) at the en ds o f its vertical
diameter A b all in ch i n diameter is attached to the
en d o f a hori z o n tal supportin g arm
an d revolves ab ou t
the poi n t c
The distan ce from th e
cen ter o f the b all to the ce n ter of revolution is 1 inch
Will the ball meet the wire loop ?
,
7
,
,
’
,
.
,
,
.
A metal shade for an electric lamp is made up o f a hemi
sphere and half o f a circular cylin der The axis o f the
n
cyli der passes through the ce n ter o f th e Sphere D raw
n
the i tersection of the two surfac es an d a pattern to b e
s
u ed i n cutti ng o u t the metal to m
ake the shade
Fin d the i n tersectio n of a hexago n al prism with
an o b liqu e
n
D
pla e
raw the developme n t of the par t o f the prism
ab ove the li ne o f in tersectio n
.
.
.
.
.
1 72
.
Make a pattern for o n e of the section s
el b ow shown i n the figure
of
the stovepip e
.
I n Fi g; 5 1 a b uttress i n a slopi ng wall is
d i me n sio n s c an d 9 are 1 0 feet an d 8 feet
shown T h e
respectively
As sume other suitab le dime n sio n s and make all the
pat tern s a sto n e cutter wil l n eed for makin g the top
sto n e of the buttress
Make the patter n s for the furn ace setter s o ffset b oot
shown i n the figure The sectio n of the top portio n is
oval an d of the lower portio n circular
.
.
.
,
-
.
’
-
.
.
1 75
.
the size o f the hole tha t must be cu t i n the roof of
th e shed i n E x 1 3 7 to allow a tigh t b el t to ru n betwee n
the pulleys q a n d r
Make the pattern s for the ash chute head show n i n F1g 52
Fi n d
.
.
1 76
-
.
.
.
1 77
D evelo p
a portio n of the vertical pipe i n the accompan y
i n g fi gure to show the true si z e of the hole to b e cut s o
that the smaller pipe i n clin ed 30 to the vertical may
B oth pipes have circular sectio n s
b e j oi n ed to it
.
°
,
,
.
1 78
.
.
Make the pattern s for the tran sitio n conn ectio n b etween
a square duct an d a circular pipe
.
CHA PT E R VI
S HAD O WS
SH A D E S A ND
p erson w h o un derstan ds descriptive geometry can usuall y
get a correct co nception o f an ob j ect from the orthographic pro
f
n
i
n
o
its
edges
or
other
li
es
of
its
co
n
tour
A
proper
s
t
o
ec
j
c o n ceptio n o f the form an d relatio n s of the d iff ere n t parts of some
however more careful study o f the proj ection s
o b j ects requires
Fo r this reaso n drawi ngs
than is desirab le fo r practical purp oses
are sometimes made to show an e ffect simil ar to that produced b y
the shad ows from illumin ation This eff ect is a great assistan ce
also i n makin g draw i n gs plain to person s who are un familiar with
th e methods of ort hographic proj ectio n
The subj ect o f sh ades an d sh adow s treats o f the application
ry to produce the e ffect o f
f
descriptive
geomet
o f the methods
q
il lumi n atio n which gives a more real appearan ce to the proj ectio ns
o f an o b j ect
If an Opaque b ody is placed n ear a source of light par t of th e
surface wi ll b e bright and the remain der will b e dark A portio n
o f the li ght from the lumi n ous b ody wil l b e in ter cepted an d a
portio n of the space b ehin d the b od y will b e i n dark n ess Thi s
dark space is call ed the sh adow o f the b ody and the lin e o n its
surface separatin g the b righ t side fro m the dark is the sh ade
94
.
A
.
,
,
.
.
.
,
.
,
.
.
,
lin e
.
co n e is shown near a vertical plan e P P aral
lel rays of light are represe n ted pass in g through the poin ts a b c d
The i n te rsectio n s o f these rays wi th the p l a ne determi n e
an d v
the sh ado w o f the co n e o n the plan e The sh ade lin e is foun d by
j oini ng poin ts o n th e s urface where ravs are ta nge n t
95
.
I n Fi g 5 3
.
a
.
,
,
,
,
.
.
‘
.
U nless it i s o therwise specified the rays o f light are
represe n ted by parallel lin es B y the usual co nve n tio n
6
9
be
.
,
.
to
in
practical drafti n g the rays are take n as comin g over the left
shoulder so that the horizo n tal an d fro n t proj ectio n s o f a ray
make respectively an gles o f 4 5 an d 3 1 5 with the X axis
,
°
°
.
The method for co n structi n g the shadow of a co n e o n the
hori zo n tal plan e through its b ase is shown i n Fi g 5 4 The pro f
n s of a ray through the vertex v are first draw n
i
c
o
n
e
t
d
h
a
e
t
j
i n tersecti o n o f this ray with the plan e of the b ase is fou n d at the
1
2
poi n t marked by the proj ectio ns v and v Li n e s are drawn i n
1
the hori z o n tal proj ectio n tan ge n t to the b ase and through v These
are the limiti n g l i n es of the shadow an d determi n e at the poi n ts
o f ta n ge n cy the shade li n es e v an d d v
“
Fi g 55 shows the shadow of a circular cap o n a cyli n drical
colum n S carcely an y expla n atio n is n eeded Rays are drawn
”
through poi n ts o n the cap and their in tersectio n s with the sur
face o i the colum n are easily foun d The horizo n tal projecti o n s
of the rays determi n e the poi n ts of i n tersectio n The xerti c al
shade li n es at ti and 37 are determi n ed by ta ngen t rays
97
.
.
,
.
,
,
.
.
.
.
.
.
.
.
.
The shadow of an ellipsoid o n a hori z o n tal pla n e P is show n
The l i n e o f shade is made use o f to determi n e the shado w
i n Fi g 5 6
“
”
O bviously it is the li n e of co n tact with the ellipsoid of a cyli nder
of rays ta nge n t to the surface This li n e of co n tact is of course
I n the figure two proj ectio n s of an ellipso i d are shown
an ellipse
If a vertical auxiliary plan e A is draw n parallel to th e dire ctio n
o f the rays an d through the ce n ter 0 o f th e ellipsoid i ; will c u t an
elli p se from the ellipsoid an d two eleme n ts ta nge n t to the ellipse
”
“
from the cyli n der of rays These poi n ts of tan ge ncy w ill b e the
highest an d lowes t poi n ts in the ellipse of shade I f the pla ne 4 i s
revolved ab out a vertical ax i s through 0 so that it is parallel to
the fro n t pla n e the sectio n cut from the ellipsoid by the auxiliary
pla n e will b e made to coin cide with the fro n t proj ectio n A ray
drawn through the ce n ter 0 i n tersects the horizo n tal pla n e P at th e
2
A fter revolutio n wi th
poin t marked b y the proj ectio n s o an d 0
h
the auxiliary plan e th is ray is shown b y the proj ectio n s o 0 1 an d
D r aw the fro n t proj ectio n s o f rays parallel to
ta nge n t
0 70 2
98
.
.
,
,
.
.
.
,
,
.
.
’
,
,
‘
,
,
,
.
'
.
.
,
.
'
.
to the revolved sectio n o f the ellipsoid at a l f an d bl f Whe n the
plan e A is revolved b ack to its former positio n these tange n t
poin ts are at a an d b I n the fron t proj ection then ai bf is the
maj or axis of the ellipse of shade I n the horizo n tal proj ectio n o f
h
h
the ellipsoid a b is the min or axis of the ellipse of shade an d
h
h
h h
perpe n dicular to a b is the maj or axis
e i
The shadow o f the ellipsoid o n the horizo n tal plan e P is the
shadow ca st by a sec ti on of the ellipsoid i n cluded b y the ellipse of
shade T 0 fi n d this shadow determi n e the i n tersectio n s with the
plan e P o f rays through the poin ts 0 a b e an d f These poi n ts
determi n e the maj or an d mi n or axes a b an d e f of the elliptical
shadow
E X E R CI SES
Fin d the shadow of an ellipsoid o n a vertical plan e
Fi n d the shadow of a sphere (a ) o n a pla n e b elow an d
parallel to th e hori z o n tal plan e ; ( b) o n an o blique plan e
D raw the shadow of a dormer wi n dow o n a roof
D raw the shadow o f a chi m n ey o n a roof
The stair ramp show n i n the figure i s parallel to the slope
The i n cli n atio n o f the stairway i s 1 ver
o f the steps
tical to 2 hori z o n tal C o ns truct the shadow o f the
I m p o n the ste ps
.
’
.
,
,
.
,
,
.
,
,
.
,
,
,
,
’
,
.
’
’
,
’
,
.
.
,
.
.
.
.
.
'
.
Draw the cap an d column i n Fig 5 5 with the axis
ho ri z o n tal i n stead of vertical S how the shade an d
shadow
Make a simple drawing o f a sectio n of an e n gin e cyli n der
through the cen ter li n e Represe nt the whole p isto n
i n the cylin der b y showin g the shadow o n the i n s i de
o f the cyli n der o f the po rtio n that proj ec ts b e y on d the
se ctio n
.
.
.
.
.
CHA PTE R VI I
W A R PE D
W arped
S U R FACE S
are distinguished from plan e surfaces
and surfaces of si n gle curvature ( co n es cyli n ders an d co n volutes ) b y
a differen t positio n of the ele me n ts with respect to each other I n
all the surfaces that have been discussed a plan e could always be
passed through at least tw o o f the co n secutive eleme n ts This
property was very serviceable for the solutio n of pro blems I n
warped surfac es however n o two co n secutive eleme n ts lie i n the
same plan e ; t hat is li n e s draw n th rough an y tw o co n s ec utive
h
f
h
i
n
n
n
n
t
s
e
e
a
h
r
n
ositio
tr
a
t
li
e
r
ti
t
m
ith
r
s
o
e
e
a
e
e
e
r
a
g
g
g
p
p
allel n o r i n te rse ctin g E xamples o f warped surfaces are show n i n
Figs 5 7a 5 7 b an d 58
A Warpe d su rf ace therefore may be ge n erated by a s trai gh t
lin e movi ng so th at it al w ays to u ch es tw o giv en lin es an d rem ai n s
*
n
l
a
n
e
n
i
e
n
l
g
The
fixed
li
es
which
the
movi
g
li
e
r
e
l
t
o
a
v
n
a
a
l
p
p
touches are called t h e di re ctri ces an d the plan e is called the pl an e
di re cto r
99
.
su rf ac es
,
,
.
,
'
.
.
,
,
,
.
,
,
.
.
,
,
.
.
.
The h yperb oli c parab o l o i d is a warped surface with a
plan e director an d two straight li n e directrices which are n o t i n
the same plan e It takes its n ame from the fact that curved
sectio n s made by plan es cuttin g the surface are either hyperb olas
«
i
s
A
para
olas
cf
F
the
directrices
approach
paral
b
r
o
(
g
This
l el is m the surface approaches a pla n e as a l i mi ti n g surface
surface is o f some practical importan ce as it is used i n maso n ry
co n structio n an d the cow—catcher o f a locomotive ( Fi g 60) is
usually o f this form
1 00
.
.
.
.
.
,
,
.
.
Th
of
e re
is
a
ge n e rati o n a n
v
d p p
of
w ar
e rti es ; b u t
ro
t h a t w i l l b e d i s c u sse d
y
ari e t
r
t
e
a
g
.
p
ed
th i s
e
s u r f a c e s , al l
x pl
an ati o n
is
d i ff e ri n g i n th e i r m o de
s uffi c i en t f o r th e s u rf aces
a hyperb olic parab oloid any plan e parallel to a plan e director
will o u t each directrix i n a poin t an d the l in e j oi n i ng these two
poi n ts will b e an eleme n t o f the surface The eleme n ts may be
regarded as lyi n g i n a series o f pl an es which are paral lel to the
plan e director an d dividing the two directrices proportio n ally
If the n an y two straight li n es n o t i n the same plan e are divi ded
i n to proportio n al parts the straight li n es j o i n in g the correspo n d
i n g poi n ts o f divisio n are elemen ts o f the surface o f the hyperb olic
9
parab oloid
In
,
,
.
.
,
,
,
.
The characteristic properties o f this surface can b e mos t
co n ve n ien tly i n vestigated by refere n ce to a pictorial drawi n g
showi n g the directrices an d the plan e director I n Fig 61 we
shall take for simplicity of represe n tatio n the horizo n tal plan e
marked H for the pla n e di recto r The li n es a c and b d shall be
the di r ectri ces From poi n ts o n these li n es the elemen ts o f the
surface a b and c d are drawn parallel to the plan e director H
Fo r the n ecessary co ns tructio n draw through the directri x a c
a vertical plan e V 1 parallel to b d an d cutti n g the plan e director
Then draw an other vertical plan e V2 parallel
i n the lin e x1 1171
to V I through b d cutti ng the plan e H i n the li n e x2
The lin e
c f is draw n perpe n dicular to xl x1 an d d g is draw n perpe n dicular
to x2 x2 ; therefore i g will b e parallel an d equal to c d Through
o n a c draw i j parallel to the pla n e director H an d
an y poi n t i
cutti n g b d i n j ; the n i f is an other eleme n t o f the surface The
li n e i h is drawn perpe n dicular to x2 x2 an d i t perpe n dicular to
rallel an d equal to i i A lso
x1 ; the n l k will b e p a
101
.
.
.
,
.
.
,
,
,
.
,
,
,
,
,
.
,
,
.
,
,
,
,
.
,
.
}
4
235—
'
c
Therefore
a
t
b h
,
1
l
an d I t
f
foll ows that
l ie
f 9 cut
an d
a
b
In
same poi n t r N ow draw a plan e parallel to V I cutti n g a b i n
poin t n the n its i n tersectio n wi th H w i ll b e parallel to xl x1
,
.
,
,
W l ll
cut
lh
In 0 ,
jg
In
m,
an d
n
o
-
5771
li
—
c
also
0
9
is
parallel
th e
an y
an d
an d
e qual to
ti
an d
is parallel
mp
which proves
three
elemen ts a b
portion ally so tha t
,
i
j
n
i
s
qp
an d c
,
d,
equal
an d
to
i
Therefore
a
.
a straight lin e i n tersectin g the
,
an d
dividi ng these elemen ts
r
o
p
,
a n
n
b
i q cp
=
d
qj p
'
If the elemen ts a b an d c d are take n as directrices an d the vertical
plan e V1 as a plan e director the surface which is ob tain ed will be
ide n tical with that havin g a c and b d as directrices an d the plan e
The elemen ts a b i j etc are called
H as its plan e d i rector
eleme n ts o f the fi rst gen e rati on an d those o f the other set as a c
which are parallel to the vertical plan e VI are call ed
n p etc
eleme n ts of the sec on d gen eratio n
We have show n the n that
every eleme n t of eith er ge n eratio n i n tersects all those of the other
Th is surface is do u bl y ru l ed; that is it h as two s ets o f
straight li n e eleme n ts an d through an y poin t o n the surface two
straight lin e eleme n ts can b e always drawn E ach elemen t o f
o n e ge n eratio n i n tersects n o eleme n t of its o w n se t b u t meets all
the eleme n ts of the other set The lin es o n the surface i n Fig 5 8
show t h is doub le ruli n g very plai nl y
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a hyperb olic parab oloid are
represe n ted i n Figs 62 an d 63 Two straight li n es a b an d c d
o f defi n i te le n gth an d n o t i n the same pla n e are divided i n to th e
same n umb er of equal parts The lin es j oin in g correspon din g poi n ts
A n eleme n t marked my
o f divisio n are eleme n ts o f the surface
is shown i n each figure O b serve that the di ffere n ce b etween
the two figures is merely that the same poi n ts of division o n a b
I n Fi g 62 the method of j oin i n g
an d c d are j oi n ed up di ffere ntly
the poi n ts of divisio n is the same as shown i n the pictorial drawi ng
102
E lemen ts
.
of
the surface
of
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i n Fi g 61
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r
r
e
r
c
f
w
a
d
s
e
rec
i
If
the
e
s
o
a
f
a
c
i
n
ge
eral
are
di
t
u
n
no t
3
p
limited i n l en gth it is n ecessary to have give n the directrices o f
“
b oth systems o f rulin g o r the directrices an d plan e director o f
'
19
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,
sy stem This las t case is il lustrated i n Fi g 64 The lin es q r
an d s t are the direc trices an d th e pla n e director is give n b y its
To draw an elemen t of the surface then
tr aces H P an d F P
Through this poin t
s ome poi n t as e o n the directrix s t is assumed
draw two lin es d l bl an d cl d l parallel to a b an d c d which are
The in tersection o f the plan e
an y two lin es i n the pla n e director
o f the tw o lin es through e with the directrix q r i s a poi n t i n the
required eleme n t of the surface This i n tersection can b e fou n d
very simply with out fi n di ng the traces o f the plan e of the lin es
First pass through the li n e q r a plan e perpen
a l bl an d c l d l
di cu l ar to the fro n t pla n e an d fi n d the in tersection s o f the li n es
The i n tersectio n o f a l bl with
a ; b1 an d c l d l with this pla n e
th i s plan e has of course its fro n t proj ectio n i n the fro n t trace at
f
n e determ i n es the horizo n tal proj ectio n
a
proj
ecti
n
g
li
n
a
d
g
S i milarly cl d l i n tersects th e pla n e of q r at f an d the plan e o f
The
th e lin es a l bl an d c l d l i n tersects the directrix q r at i
lin e c i is in a plan e parallel t o the plan e director an d touches the
two directrices It is therefore an eleme n t of the surface through
B y this process any n umb er o f eleme n ts c an
an assumed poin t e
b e draw n
o ne
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If o ne proj ectio n o f a poin t o n any warped surface is
k n own an other proj ectio n may b e located b y drawi ng through
the give n proj ectio n a li n e perpe n dicular to the plan e of projectio n
Through this per
i n which the give n proj ectio n is represe n ted
i
l
r li n e pass a pla n e which will i n tersect the eleme n ts o f
n
c
a
e
d
u
p
the surface i n poi n ts which whe n j oin ed will give a lin e o n the
surface The in tersectio n of this li n e with the other proj ectio n
of the perpe n dicular li n e is the required proj ectio n o f the poin t
1 04
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The h ype rb o l oi d o f revolution is a warped surface ge n e
rated by a straight lin e revolvi n g ab out an axis n o t lyin g i n the
same plan e wi th it Usually a vertical lin e is take n for the axis
P roj ectio n s o f this surface showi n g a n umb er o f eleme n ts are
represe n ted i n Fi g 65 The method for co nstructi ng this surface
is shown i n Fi g 66 The axis a b is vertical A n elemen t of th e
105
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.
surface as m n is called a gen eratrix b ecause the surface can be
gen erated by revolvi n g this lin e a b out the axis E ach poin t i n
T he
m n as it revolves ab out the axis descri b es a hori z o n tal circle
true radius o f each circle is seen i n the horizo n tal proj ec tio n
Thus the poin t r o n the ge n eratrix n earest the axis descri bes the
circle of which the diameter is shown i n the fro n t proj ectio n of
the surface as u f rf vf This smallest circle described by a poin t
i n the ge n eratri x i called the ci r cl e o f the go rge
The diameter
of the circle describ ed by a poi n t p is gl f (M which is equal to
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s
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,
h
h
2 >< b p
The hyperb oloid o f revolutio n is a dou bl y ru led surface as
it h as tw o sets o f straight li n e eleme n ts O n e eleme n t o f each set
A n eleme n t o f
c an b e draw n through an y poi n t o n its surface
o n e ge n eratio n c an b e produced to i n tersect all those of the other
gen eratio n for the s imple reason that they lie upo n the same sur
face If the n an y three eleme n ts of either ge n eratio n are taken
as directrices an y eleme n t o f the other ge n eratio n may b e take n as
a ge n eratrix which when movin g will produce the same surfa ce
The method for drawi n g a plan e tan ge n t to a hyperb oloid o f
revolutio n is show n i n Fig 67 The plan e P shown i n the figure
i f o n ly o n e proj ectio n O f
is ta n ge n t to the surface at the poi n t e
h
draw an eleme n t of th e surface d 6 through
0 is give n as s ay c
f
n
n
c
b y proj ectin g to the fro n t proj ectio n of d e
d
fi
d
O ne
c a
o f the other se t of eleme n ts passi n g throug h the poi n t is show n
b y f g The two eleme n ts determi n e th e pla n e t ange n t to the
surface at the poi n t e
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u rf ace is a surface ge nerated by a
straight lin e moving so that it is always touchi ng tw o h el i cal d i rec
trices lyi ng upo n co n ce n tric cyli n ders I n Figs 68 an d 69 tw o cy l i n
ders of revolutio n are shown O n each is a port i o n o f a helix
The straight lin e m n ge n erates the surface o f the helicoid I n bot h
fi gures The differe n ce is merely i n the slope of th e ge nerating
lin e Whe n b oth o f the directi n g helices of a helicoid h ave the
same pi tch i t is called a helicoid of u n i for m pi tch
If the pitches
ar e n o t the same the surface is called a h elicoid o f va ryi ng pi tch
106
.
A h eli co i d
or
screw s
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The tw o examples of helicoids of un iform pitch in Figs 68 and
69 di ff er o nly in the relative positio n o f the movin g o r ge n erati n g
lin e I n b oth cases this lin e moves un iformly aroun d the vertical
axis while at the same time all the poi n ts o n it move u n iformly
I n either figure the poin t b
in a directio n parallel to the axis
i n the ge n eratin g li n e m n will describ e the helix a b c d lyi ng o n the
inn er cylin der an d the po in ts m an d n will describ e helices o f the
same pitch lyi ng o n the outer cyli n der These helices traced by
m an d n are marked respectively c i g an d h i j The ge n eratin g
li ne m n will always lie i n a plan e tan ge n t to the i nn er cyli n der
an d will in tersect the vertical eleme n t at the poin t o f ta nge n c y
at a co n stan t angle
I n the helicoid o f Fi g 68 t h e ge n erati ng lin e m n is always
perpen dicular to the vertical eleme n ts o f the i nn er cylin der an d is
called a ri ght heli coi d The helicoid of Fi g 69 has the gen erati ng
lin e m n i n cli n ed to the vertical eleme n ts of the i nn er cylin der
an d is called an o bl i qu e hel i co i d
Whe n the elemen ts o f a helicoid are perpen di cu l ar to the axis
the surface i the same as that o f a square threaded screw (Fig
an d whe n th e a n gle is less it i s that of a V —
thr eaded screw ( Fig
Fo r the V thr ead of the U n ited S tates stan dard screw th e
an gle b etwee n the elemen ts and the axis is
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s
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a satisfactory represe n tation of a helicoid it is not
Usu
n ecessary to draw a large n um b er of eleme n ts of the surface
ally i t is shown b etter b y drawi n g a few elemen ts for a small portion
I n Fi g 7 2 an ob lique helicoid is sho w n resemb li ng
o f the surface
the o n e i n Fi g 69 The small po rti o n o f the outlin e that is repre
sen ted is easily recog n ized as like that i n the groove of an auger
or a twist drill
If o n th i s surface it is required to locate say the fro n t pro
draw
n o f a poi n t whe n the hori z o n tal proj ectio n is give n
i
o
ec
t
j
the hori z o n tal proj ectio n of the eleme n t o f the surface at the poi n t
The requi red fro n t proj ectio n of the poi n t is the n foun d by
projectin g to th e fro n t proj ectio n of the eleme n t If o nly the
fro n t proj ection o f a poin t is k n own the horizo n tal proj ectio n
1 07
.
Fo r
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determi n ed by drawin g a lin e through the fro n t proj ectio n
parallel to the horizo n tal plan e A n y plan e passed through this
li n e will cut the eleme n ts i n poi n ts determin i ng a curve The
h ori z o n tal proj ectio n of this curve will i n tersect the correspo n d
i n g proj ectio n of the li n e at a poi n t which is the required pro
is
.
.
j ecti o n
.
If the surface of the helicoid i n the last figure is cut b y a vertical
n
n
n
n
la
e
h
the
a
is
d
perpe
dicular
the
fro
t
pla
e
the
n
h
r
o
u
a
t
o
t
g
y
x
sectio n cut out will be like the shaded drawi n g at the ri ght h a n d
side of the figure Half o f the curve shown i n the sectio n i s marked
3 1 x1 y1
This figure is easily co n structed by drawi n g eleme n ts
The poin ts i n which the eleme n ts are i n tersected
o f the surface
Thus
by the cutti n g pla n e determi n e the curve of the sectio n
an eleme n t i j i n tersects the cutti n g pla n e at x
This poi n t is
located On the curve i n the sectio n b y proj ecti n g hori z o n tally
from xf and layin g o ff from the axis the len gth 0 1 x1 equ al to
h
k
as
The other poi n ts n eeded to determi n e the curve are fou n d
o
i n the same way
S uch a sectio n cut from the surface by a plane
through the axis is called a meri di a n secti o n
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A plan e is drawn tan gen t to the surface of a helicoid at a
n t b y drawi n g an eleme n t through the give n poi n t an d
ive
poi
n
g
a ta n ge n t to the helix lyi n g i n the surface an d passi n g through the
l
n es determi n e the required t a n ge n t
w
i
ive
po
t
These
t
i
n
n
o
g
plan e As the surface is warped the pla n e w ill n o t be tan ge n t
throughout the le ngth of an eleme n t an d will n o t co n tai n there
fore a tan ge n t to the b ase at its i n tersectio n with an eleme n t
p assi n g through the poi n t
1 08
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The i n tersectio n of the helicoid with an y give n plan e is
fou n d by passin g hori z o n tal auxiliary pla n es through it These
”
“
plan es will cut from the surface S pirals equ al to the base an d
from the give n plan e horizo n tal li n es The in tersectio n s O f these
li n es will give poi n ts o n the required li n e of i n tersectio n The
”
“
proj ectio n s of the differe n t S pirals are no t n eeded if a curve of
the b ase is made o n a tra n spare n t shee t
1 09
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E X E R CI SES
Locate
a poi n t o n the surface o f a hyperb olic parabo l o id
give n b y two limited lin es Draw a tangen t plan e
through thi s poin t
Three o b lique li n es are give n as b elo n gin g to o n e set o f
eleme n ts o f a hyperb olic parab oloid Fin d three ele
me n ts o f the other set
S how several elemen ts of the surface of a hyperboli c
parab oloid give n by tw o directrices and a plan e director
Draw the proj ectio n s o f a hyperb oloid o f revolutio n an d
assume a poi n t o n the surface Draw a plan e tan ge n t
to the surface at thi s poi n t
A ssume the proj ectio n s of an o blique helicoid
Locate
a poi n t o n the surface an d draw a tan ge n t plan e at
thi s poin t
Take the proj ectio n s o f the h elicoid co nstructed for th e
preceding exercise an d draw a meri dian sectio n an d
three tra nsverse sectio ns
Draw the proj ectio n s of ( a) a sq uare thre ad ed screw ( b)
—
a V thre aded sc re w
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CHA PT ER V II I
PE
R S P E CT I VE
the art of represen ti ng ob j ects as they
appear to the eye The pri n ciples are n o t diffi cult for a stude n t
who u n derstan ds the methods of orthographic proj ectio n
The sig n ifican ce o f perspective drawi n g can be explain ed b est
by Showi ng i n what way it di ffers from orthographic proj ectio n
I n perspective the eye o f the o b server is at a fi n ite dista n ce from
the o bj ect I n orthographic proj ectio n the views r eprese n t the
o bj ect as see n whe n the eye is i n fi n itely distan t B y the per
Specti ve method the ir the li n es drawn from poi n ts o n the o b j ect
to the eye co n verge an d i n tersect at the poi n t o f sight
I Io
.
P E RSP E CT I V E
Is
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A perspective drawi ng represen ts an o bj ect as s ee n through
a vertical plan e which is assumed to b e tra nspare n t I n Fig 7 3
the eye at S the o bj ect ( a square pyramid v a b c) an d the perspe c
tive proj ectio n v a? bp 0 are plai n ly shown The light solid li n es
i n the figure represe n t the rays from the corn ers o f the o b j ect to the
eye The straight li n es j oi n i n g the i n tersectio n s o f these rays with
the vertical plan e form the outli n e o f the o b j ect If this perspec
tive picture is shaded an d co lored it will exactly represe n t the
o bj ect as see n from the viewpoi n t
The vertical plan e o n which the perspective drawi n g is made
is called the pi ctu re pl an e and the positio n of the eye is called
the poin t o f si gh t
The simplest co n structio n o f a perspective drawi ng is o b tai n ed
by the use of two proj ectio ns o f the poi n t o f si ght together with
the two correspo n di ng proj ectio ns o f the obj ect T he method is
i l lustrated i n Fi g 74 The light solid li n es show the horizo n t al an d
fro n t projec tion s o f a rectan gular c ard held n early hori o n tally T w o
1 1 1
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19
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corn ers are marked a an d b The proj ectio ns of the poi n t of sight
"
f
n
are S a d S Dotted li n es are drawn j oi ni ng the corn ers o f the
"
card i n the horizo n tal proj ectio n with S an d j oi ni ng the corn ers
i n the fro n t proj ectio n with Sf
These are respectively the hori
z o n tal an d fro n t proj ectio ns of rays from the corn ers to the poi n t
From the poi n ts where the hori z o n ta l proj ectio ns of
o f sight
these rays i n tersect th e X axis li n es are drawn perpe n dicular to
the axis to i n tersect the correspo n di n g fro n t proj ectio n s o f the
rays The corn ers of the perspective proj ectio n are thus deter
mi n ed an d by j oi n i ng these corners the outli n e of the card show n
Fi g 75 shows a perspective
by the heavy solid li n es is o b tai n ed
“
drawi ng of a scree n with three blades made by this method
The poi n ts a b and c are marked tomake the co n structio n plai n er
Usually whe n represe n ti ng an o b j ect i n perspective the horizo n tal
plan e o n which the o b j ect rests is shown I n the figure the li n e
marked GL is the fro n t trace o f the horizo n tal pla n e throug h th e
If a house is to be represe n ted i n a
b ottom edges o f the scree n
drawi n g the level o f the grou n d is represe n ted by a hori z o n tal
plan e The fro n t trace is the n very properly called the grou n d
l in e
This n ame is give n however to the fro n t trace of any hori
z o n tal pla n e o n which an o b j ect is imagi n ed to b e placed
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A picture may be made very simply by the met h od de
scrib ed if there are n o t ma n y details to b e shown I n practical
drawi ng however this method is n o t ofte n used The difficulty
is that for a picture o f suitab le proportio n s the horizo ntal proj ect i o n
h
must be usually lo cated so far from the X
o f the poi n t of S ight S
axis that it is b eyo n d the limits o f th e drawi ng board Al so f o r
an o b j ect with ma n y details the n umb er o f co nstructio n li n es
It is the n ofte n difficul t
b ecomes so great as to b e very co n fusi n g
to decide which poi n ts are to be j oi n ed Fo r these reas o ns the n
an ab ridged method o f perspective drawi ng is commo n ly used
h
k
f
f
n
a
n
b
i
n
b
a
a
a
li
n
e
is
show
its
proj
ec
t
o
s
a
b
d
b
i
n
F
76
I
y
g
This lin e is parallel to the horizo ntal plan e an d makes an a n gle
of 4 5 with the fro n t plan e B y the method tha t h as bee n j ust
explained its perspective proj ectio n is fou n d at a' bp The other
1 12
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°
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,
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gen eral method for co ns tructi n g perspective drawi ngs may be
illustrated with this figure The lin e a b i s produced i n defi n itely
h
as shown i n b oth proj ectio n s A proj ectin g li n e from S to the
en d of the li n e through a b must b e represe n ted parallel to it an d
makin g an a ngle of 4 5 with th e X axis This proj ecti n g li n e
meets the X axis at u ; an d th e perspective of a poi n t o n a b at an
i n fi n ite distan ce is at M The perspective of an y poi n t o n the li n e
produced through a an d b is therefore b etwee n af an d M I n
i
h =
=
f
r
n
v
t
t
u
H orizo n tal l i n es
this co ns tructio O b se e that S
S M
maki ng the same angle with the fro n t plan e b u t slopi n g i n the
opposite directio n would co n verge toward a similar poi n t located
at an equal distan ce to the left of S It follows the n that all
h ori z o n tal li n e s at 4 5 to the pictur e p
l an e ( fro n t plan e ) c o n verge
i n p e r sp ective to w a r d poin ts o n e it h er si de o f Sf an d at a distan ce
from th i s p roj ectio n e qu al to th at o f th e poi n t o f si ght from th e
i
n
h
n
n
ce
t
e
n
i
ture
la
e
locatio
of
the
poi
t
M
depe
ds
o
ly
S
n
c
n
p
p
an y hori z o n tal 4 5
o n the di recti o n of a b an d n o t o n its po s i ti o n
li n e i n the drawi n g will co n verge toward M Th i s po rn t is called
the m easuri n g poi n t
S ome dimen sio n lin es are shown i n Fig 76 to Show how this
poin t can b e used for layi n g o ff distan ces Fo r example the
perspective proj ectio n of the poi n t b could be fou n d by layi ng
h
to the X axis o n the left—han d side o f bf
o ff the d ista n ce from b
an d b? would b e fou n d at the i n tersectio n of this li n e with the
proj ecti n g li n e j oi n i n g bf an d S7 I n the same way the perspective
h
of a poi n t e is fou n d by layi ng o ff the distan ce from c to the X
axis o n the left—han d side o f cf an d locati n g c at the i n tersectio n
of this li n e with cf Sf If the measuri n g poi n t ll had b ee n located
o n the left ha n d side of Sf the n i n these same cases the dista n ces
me n tio n ed would b e laid o ff o n the right ha n d side of the fro n t
proj ectio ns
This method o f co ns tructi ng perspective drawi n gs with the
help of measuri n g poi n ts is shown also i n Fig 7 7 A perspective
drawi ng o f a cub e with circles i n scrib ed i n its sides is illustrated
The measuri n g poi n t M is located b y maki n g Sf M equal to the
h
distan ce from the poi n t of sight to the picture plan e (S to the
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°
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°
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X axis ) O bserve that there are a n umb er of parallel lin es i n a
cub e The co nstructio n can b e simplified if we n otice that all
“
parallel li n es must co n verge or van ish at the same poi n t E very
se t of parallel li n es has the n a v an i sh i n g
oi
n
t
The
li
es
which
n
p
are perpe n dicular to the picture plan e have of course their van ish
i ng poi n t at S 7 ; an d hori z o n tal 4 5 li n es van ish at M
Fo r an y
drawi n g the van ishin g poi n ts the measuri n g poi n ts an d th e fro n t
proj ectio n o f S must b e i n a li n e parallel to the X axis
The j udicious u se of measuri n g an d van ishi ng—poi n ts saves
much lab or i n maki n g perspective drawi ngs I n Fi g 77 the
va n ishin g poi n ts of the horizo n tal edges o f the cu b e are at V 1
an d V 2
E ach i s located by drawi n g an edge fo r which the dirc o
tio n has b ee n determi n ed to i n tersect the li n e through Sf parallel
to the X axis
.
.
”
.
-
.
,
,
0
-
.
-
-
,
,
.
.
.
-
.
,
.
The perspective drawi ng o f the cub e illus trates also a
very satisfactory I ii eth o d for drawi ng the perspective proj ectio n s
of circles P oi n ts i n the perspective drawi n g of the circles are
fou n d by locati n g diago n als o f the circumscri bi n g squares formed
O ther li n es are also draw n parallel to
by the edges o f the cu b e
the sides o f the same square through the i n tersectio n s of the diago
h als
with the circle The i n tersectio ns of these straight li n es
locate usually en ough poi n ts to determin e the perspective of
the circle The ( o nstru cti o n for tw o poi n ts a and b o f the circle
are illustrated i n the figure
1 1
3
.
.
.
.
.
.
The pri n ciples o f perspective are
n o t di fficult to apply i n a mecha n ical way b y those who have n o
artistic trai n i n g ; bu t distorted results are o b tai n ed from ab solutely
correct applicatio n s whe n ab surd co n ditio n s are assumed If for
example a large house is represe n ted with the poi n t of sight ab out
twe n t y feet from the fro n t of the house o b viously a poor result is
o b tai n ed N evertheless fo r such a case the pri n ciples can b e applied
as well as to an y other
A perspective draw i n g should show the ob j ect as it appears to
the eye If is importan t therefore that the b est viewpoi n t is
1 14
P e rspe ctiv e D i sto rtio n —
.
.
.
,
,
.
‘
.
.
,
,
,
ob tai n ed ; and care i n selecti ng the viewpoin t is as esse n tial as a
kn owledge o f the rul es
If a house ab out forty feet high is to be
sketched the poin t o f sight should b e taken ab out eighty feet from
the picture plan e A good rule to follow is to make this distan ce
a b out twice the greatest di m ens ro n When large o bj ects are to
b e represe n ted the most satisfactory results are o b tai n ed usually
when the poi n t o f sight is take n n early i n fro n t o f the o bj ect It is
preferable the n tha t the plan drawi ng should b e shown i n cli n ed
to the picture plan e
.
,
.
.
,
.
,
,
)
.
1 1 5.
P erspecti ve Sk et ch es
f r om
“Wo rk in g ” D raw in s
g
—
A
very pro fi tabl eappli catio n of the methods o f perspective draw i n g
”
“
“
”
—
i s fou n d i n maki n g free ha n d sketches from worki n g or s h O p
drawi ngs show n i n orthographic proj ectio n The worki n g
drawi ng represe n ts certai n i n formatio n ab ou t an o bj ect by a
co ll ecti o n of views
S everal views are n ecessary to represe n t the
obj ect completely I n perspective drawi ng the same i n formatio n
i s sho w n i n a si n gle sketch
When b egi n ni ng a perspective draw
i ng which is to b e made from an y proj ectio n drawi n gs it is n eces
sary to acquire a thorough k n owledge of the form an d details
T h e work ot herwise ca n n ot b e do n e i n tellige n tly
o f the o b j ect
The perspective sketches for most o bj ects should
an d rapidly
b e comm e n ced by drawi n g i n perspective the edges of either
—
circumscri bed o r i n scrib ed solids usually square prisms Mos t
machi n es an d architectural forms are easily treated i n this
way The pri n cipal edges of the o b j ect should b e the n grouped
i nto three syste ms correspo n di n g to three n o n parallel ed ges o f
the prism O n e of these will b e represe n ted b y vertical li n es an d
two o thers must b e show n with their proper co n vergen ce A fter
the pri n cipal edges of the o bj ect have b ee n drawn the other
li n es are very e as ily represe n ted Whe n deali n g with complicated
forms it is ab solutely n ecessary to follow some defi n ite system to
o b tai n results showi n g reaso n able accuracy
E ve n with simple drawi ngs some care shoul d b e exercised i n
selecti n g the poi n t of sigh t It should b e take n so that th e
details which are co nsidered most import an t will appear i n the
'
.
.
.
.
,
.
,
,
.
.
,
.
-
,
,
.
.
,
.
.
.
perspective sketch as plai nly as possible A drawi n g is o f little
value i n which the importan t parts are crowded so that they are
n o t clearly show n
.
.
P
1 93
RA C T ICAL E X E RCI SES
“
.
Make a perspective drawi n g o f a scree n with four blades
usi n g for the co ns tructio n measuri ng an d vanishi ng
poi n ts
Make a perspective drawi n g of the end co nn ectio n o f a
truss as shown i n the figure
,
.
1 94
.
.
Make a perspective draw i ng
of
a s i mple wooden b ridge
.
Draw a cottage i n perspective
S how i n perspective the flight o f steps an d the ramp i n th e
figure i n E x 1 83
Make a perspective drawi ng of a locomotive b oiler
Make a perspective drawi n g of the mil l b uildi n g shown i n
the figure S how also a tall chim n ey b ehi n d the b ui ld
.
.
.
.
.
I ng
.
I ND E X
.
No te — I n
r
O
th e
In
x
de
al l
figu
art rc l e s , e xe rc rse s , o r
fi gu re s
re s
re fe r
to
r
s,
,
,
r
r
s
,
-
r
74 , 76
,
,
,
,
,
126
,
B th t b P b l m 1 30 1 34
B l ti g P b l m 1 36
B t Pl t C
1 5 6 1 57
ti
B v l f R ft 7 4 7 6
B t P bl m 66 7 0
B il h
Fl
1 34
B i l P bl m 1 4 2 1 4 4 1 58 1 5 9
B tf
S t v p i p 1 32
B i dg Pi P b l m 90
B i dg P b l m 32 7 8 80 84 90
B tt
P b l m 1 62 1 63
e
n
ro
u
ro
en
e
e s
oa
s,
e
o nn e c
a
or
ro
er-
o u se
ue,
o
er
ro
s,
r
e
r
e
u
e
o
ro
re ss
e
e
e
,
,
e,
s,
ro
,
,
e-
ro
er
,
,
o
or a
on ,
ers ,
s,
e
,
s,
e
a e
oo
,
,
,
,
,
,
,
Cam -w h eel , 56
P o b l em 1 58
Ch i m n ey P o b l em 1 34
Chi mn ey Sh ad o w o n a Roo f
Ce il i ng
r
,
r
,
,
Ci rc l e
,
1 70
th e G o rge , 1 80
of
Ci rcl es in
O bl i qu e
an
Pl an e Pro j ecti o n s o f
,
Co n e , 1 00
Co n e
to th e
.
rs
-
no ne
,
A ngle
A ngl e
A ngl e o f R efl ecti o n 7 2
A ngl e fo R afte s o f R o o fs
A h c h u te H ead 1 62 1 65
A uxil i a y Pl an es 4 0 4 6 5 2
a
age s ;
be tw een a L i n e an d a Pl an e 68
be tw een a Pl ane an d a Pl an e o f P o j ecti o n
betw ee n I n te e cti ng L i n e 64 66
An gl e
s
p
of
Co n i c al
R y
a
A
rc h ,
1 30, 1 32 , 1 34 , 1 36
14 6
v T
N o le
Co n i c a l E a
Co ni cal
s,
e
zz
,
ro u
l
gh O u t
148
et ,
1 58
,
8 0—
84
n um
be
rs
of
l
lR
l To we
Co ni c a Roo f Co nn ecti o n , 1 30 1 34
Co ni ca
o o f P r o b em , 1 4 6
Co ni c a
.
l
Ro o f
r on a
1 32
,
Co nn ecti ng- ro d , 1 38
L in e s 9 8
Co n to u re d P l an Ou t b y P l an es
Co nvo lu te S u rface 1 1 8
Co n to u r
Co n
,
98
,
,
v
o l u te s ,
1 02 , 1 1 4
Co ordi n ate s , System E
xp l i ned
R p f
a
Le ngth o f G u y o es
Cu be P ro b l em 56 1 9 4 1 9 7
Cu ve o f I n terse cti o n 1 2 6
Cyli n der 1 00
Cran e ,
,
,
12
,
92
o r,
,
r
,
,
D e rri c
k P bl
k P bl
v l pm
v l pm
v l pm
v l pm
D es
De
De
De
De
ro
em s ,
em s ,
ro
68 , 9 4
4 4 , 54 , 1 30
e o
e nt o f a
Co n e , 1 4 2
e o
ent o f a
Cy li n de r , 1 28
e o
ent o f a
e o
en t o f a
D i re cti o n
R
of
‘
S l o pe Sh ee t 1 59
Steam D o m e 1 59
,
,
1 67 , 1 68
ay s ,
W p
S u rface 1 7 2
betw ee n P arall el Pl an e 56
betw een T w o L i ne s 9 2
f ro m a Po i nt t o a L in e 90
f ro m a P o i n t to a Pl an e 54
D i re ctri ce s
D i stan ce
of a
ar
ed
,
s,
D i stan ce
,
,
D i sta n ce
,
D i s tan ce
,
D i stan ce s
”
tru e
) Measu re d
,
4 8 , 50
D o m e w i th E igh t Si des , 1 56
D o tte d i n e s , 1 4
L
D o rm e r
Wi
P o b l em s
n do w
r
,
1 70
b
ved S u rf ace s 1 00
D o u b l y R u l e d S u rf ace s 1 7 6 1 80
D ry D o c k P ro b l em 86
D o u l e Cu r
,
,
,
,
v T
Ea
e
l
O u t et , 1 58
r o u gh
E l e m e n ts
S u rf ace
of a
v ti n D w i ng
E l li p i d S h d w
E l li p t i l C m w h
E le
a
ra
o
so
a
,
ca
a
ee l ,
-
“
E n d Co n n e cti o n
y
x
H
e ad
r
e
P o j ect i o n s
r
of a
T
ru s s ,
r
r
P o b l m 1 38
”
i i I n te rse cti o n
Fan
1 68—1 7 0
,
56
1 98
P o b lem i n Sh ado w s
P o b l em 1 4 8 1 50
E ngi n e C li n de r ,
E h au st
”
1 00, 1 26
38
,
o f,
o
,
s,
,
,
12
Fi rst Gen e rati o n , 1 7 6
,
Mi l l bu i ldi ng
1 98
-
,
Mi n e Pro bl em 1 5 6
Mirro r Pro b l ems 4 4
,
72
,
Negative D irecti o n
N o tati o n 1 0 1 2
Nut P o b l em 1 38
,
,
12
,
r
,
O b li qu e Co n e , 1 02
O b l i qu e Cy lin d e r , 1 02
el i c o i d ,
O b i qu e
H
l
O b li qu e
( Li n e
s
)
P l ane
to
s of
Pro j ecti o n
O ff se t B o o t f o r a Fu r n ace , 1 62
O i l -c an P r o b l em , 1 58 , 1 60
O n e Pro e cti o n o f a L i n e i n a Plane gi
O ri gi n , 2
j
p
v pp
O rth o gra h i c P r o j e cti o n , 1 , 4
“
—
O al S to e i e , Co nn ecti ng
v
B
oot
”
,
16
ven
,
t o Fi n d th e
O th
fo r , 1 32
Pa al l l Lin e a Pl an e th o ugh 28
P a a l l e l L i n P o j e ti o n f 2 6 2 8
Pa all el (L i n ) to P lan o f P o j ecti o n 1 6
Pa al l el Pl an D i tan e betw een 56
Pa al l l P l an e D aw n at a G iven D istance A p art 62 94
Patte n fo an A ch S to n e 1 46
Perp n di cu l a (L in e ) to a Plan e 54
Pe pe ctive D aw i ng 1 1 88
Pe pe ctive f a Cu be 1 9 4 1 9 7
Per pective o f a Ci le 1 9 4 1 9 7
Pe p ective D i t rti o n 1 9 4
Pe p ective Sketch e fro m Wo k i ng D rawi ngs 196
Pi ctu re Pl an 1 88
Pi p e fi tti ng P o blem s 1 4 2 1 44 1 52
Pi pe P bl em s 2 4 4 6 5 4 1 30 1 32 1 34 146 1 48 1 56
Pi tch o f a H elix 1 1 6 1 2 0
Pi t h o f S cre w Su rface 1 80
Pl an D aw i ng 38
Plan e D i e to o f a Warp ed Su rf ace 172
Pl ane o f P o j ecti o n 2
Pl an e 1 0
Pl an e S u f a e 1 00
Pl an T angent to a Co n e o r a Cyli nder 1 08 1 1 2
Pl an e T angen t to a Co nvo lu te Su rf a 1 22
Pl an T angen t to a H yp e b o lo i d o f R vo lu ti n 1 80 1 8 3
Pl an T ang n t to a Sph ere 1 1 2—1 1 4
Plan e th ro ugh a Po int Perp endi cu l a to a Li ne 60
r
r
s,
e
es,
r
es
r
es ,
s
r
r
r
,
s
r
,
,
o
,
,
,
rc
s
rs
,
,
r
rs
,
,
r
e
,
,
c
s
r
rs
s o
es
s
e
r
c
r
r
,
,
s o
,
,
s
rs
r
,
e,
r
-
ro
,
,
,
,
,
,
,
,
,
,
,
,
,
,
s,
c
r
,
r c
s
r
r
,
,
s,
r
c
,
e
,
,
ce ,
r
e
e
e
e
o
,
,
,
r
,
:
,
er ,
16
P late Co nnecti o n fo r Steel Rafters 1 56 1 57
P o i nt o f S i gh t 1 88
Po rtal o f a B idge 32
Po i ti ve D i ecti o n 1 2
P o j ect a L i n e u p o n a Plane 56
Pro j e ti ng L i ne 4 1 4 2 2
Pro j ecti o n s o f I nte se cti ng L i nes 28
P o j e cti o n o f a L i n e 1 2 1 6
P o j ecti o n o f P a al l el L i n es 2 6 2 8
P o j ecti o n o f a P o i n t 2 1 0
P o j ecti o n o f P o i nt n a Co n e 1 04
P o j ecti o n s o f P o in t o n a Co nvo l u te S u f a e 1 2 0
Pro j e ti n o f P o i n t o a Cyl i n de 1 04
P ro j ecti n o f P o ints o n a D u b l e cu ved S u rface
Pro j e ti n o f P i n t o n H l i co i d o r S c ew S u rf ac
Pu l l y P o b l em 9 6 9 8 1 36 1 62
Pu l ley P o j ecti on o f 9 6—9 8
,
,
,
r
s
,
r
,
r
,
c
s,
,
,
r
r
s
r
s
r
s
r
s
,
,
,
r
,
,
,
,
s o
r
,
s
c
c
o
s
o
s
o
s
o
-
e
s
,
,
e rs
a
ro a
1 04
,
r
es ,
1 82
,
s
r
r
cu
or,
s
oo s,
-
,
r
s
R ft o f R f A ngle f 7 4
R i l d t P o blem 9 8
R igh t C n 1 02
R igh t Cy li n d 1 02
R igh t H l i i d 1 82 1 85 1 87
R o o f P bl m 2 2 38 39 56
a
c
r,
o
s,
s,
rr
s
r
e
r
,
76
,
e,
o
er ,
co
e
San d b o x
of a
-
Sco o
s,
e
ro
p P bl
ro
,
em ,
re
7 4 , 1 30, 1 32 , 1 34 , 1 4 6, 1 5 6 , 1 62 , 1 70
1 56 , 1 58
,
1 58 , 1 60
B l ade
Sc re e n w i th Fo u r
qu a
,
,
,
L o co m o tive
-
Sc rew , S
,
,
,
s
Perspe cti ve
.
,
1 90, 1 9 3 , 1 9 8
—
th re ad e d , 1 2 0, 1 82 , 1 8 5
Scre w , V th re ad e d , 1 8 2 , 1 8 5
-
Sec o n d G e n e r ati o n , 1 7 6
Se w e r
P o bl
r
Sh a de f o r
an d
m , 1 54
E l e ctri c
an
S h ade L i n e
Sh ad es
e
L amp
S h ado w s
of a
Cap
Sh ad o w
of a
Co ne
S h ado w
of
Shado w
of a
an
1 66
,
on a
E lli
Co
so i d ,
S ph ere
,
r
e
s,
mn , 1 68—1 70
P l an e
on a
p
l
u
,
,
,
1 36, 1 44 , 1 45
Sh o rte st Se am , 1 30
”
I nterse cti o n , 1 2
si
l
1 68 , 1 69
1 70
-
Si de
,
1 70
1 68 —
P o bl m s 7 0
—
6
98
m
56
9
P
b
l
e
o
Sh aft
m etal P attern s 1 30
Sh eet—
r
1 60
1 66
,
Sh ad o w
Sh ado w
,
Verti cal P ro j ecti o n
,
2, 4
-cu rv e d Su rf ace
e
n
i
S g
, 1 00
1 56,
1 5 8- 1 65
k
B
P o b l m 3 2 7 8 80 84
S l o pe S h eet o f a L o c m tiv 1 5 6 1 5 9
S p h ere Sh ad o w o f 1 70
S p h e ri cal Hu b fo a Fan 1 38
S p i al S p i ng P o b l em 1 2 2
Squ a e th ad e d S ew 1 2 0 1 8 2 1 85
S
ew
ri dge
s,
e
r
o
o
r
r
,
s,
cr
re
-
R
e,
,
r
r
Stai r
,
,
,
r
,
,
m
a
p
,
,
,
1 70, 1 9 8
,
P o b l em 1 4 2
Ste am I i p P b l em 4 6 54
1 70
St p P o b l m i n S h ado w
p i pe E l bo w 1 62
Sto v —
S t ve p i p e P o b l m 1 32
Stu b E n d f a Co n n cti ng o d 1 38
S u fa f R vo l uti o n 1 02
Ste am D rum
r
ro
e
e
s,
,
e
r
s,
s,
s
e
o
,
r
-
e
,
o
e
ce o
r
,
-
e
r
,
,
T l p h P b l m ( S h t t Li ) 9 2
Th P i t D t mi
Pl
28
T
1 38
T w P bl m 1 4 6
10 1 2
T
f Pl
T
C i t
t f m T
Angl
f Pl
T i ti C
1 64
ti
T i gl P b l m 56 68
T A gl b tw I t
ti g L i
64 66
T w P i t 50
T Di t
b tw
T L gth f L i 50
T S i f Pl S f
62
T
l P b l m 1 56
1 00
Tw i t d S f
one
e e
ro
o n s
re e
ne
o r es
e
an e ,
ne a
e er
,
o ru s ,
e
ro
er
o
,
r ac e s o
ane s ,
ra ce s o
an e s
r an
e
ru e
n
ro
en
ru e
ru e
o
ze o
a
ro
e
u n ne
n e r se c
e
een
a
ne ,
ane
o
,
g
s
,
r
,
r
V e rte
x
,
1 38
Co n e , 1 00
of a
Sc
V th re ad ed
-
re w
,
1 8 2 , 1 85
W pdS f
1 00 1 7 2
W t p ip P bl m 2 4
W i L p P bl m 1 60
W d B i dg P bl m 1 9 8
ar
u r ace s ,
e
a e rre
oo
Y
”
e
oo
ro
en
fo r
r
a
,
e
ro
e
,
,
e
ro
e
Blo w pip
-
e,
n es ,
o n s,
u r ace ,
V an i h i n P o i n t 1 9 4
Va l t P o b l em 1 58
Ve n ti l ati ng Fan P o b l em
u
n
u r ace s ,
s e
r ue
es ,
,
ee n
e
e
ro
on,
s,
e
s a nce
ru e
o rs r u c
.
o n n ec
on
r an s
,
,
144 , 145
,
SH O RT
-
C A T A LO G U E
T I T LE
O F TH E
PU B L I C A T I O N S
WI LEY
J O HN
SO N S
N E W Y O RK
L ON D ON
C H A P MA N
:
HA LL , LI MI TE D
A RRAN G E D U N D E R SUBJ E C T S
v i
t p i
D escri pti
d
sol
l
ppli ti
A ll b k
b
e c r c u ar s s en t o n a
a t ne
r c es o n
ly
ca
s are
oo
.
on
k d w i th
th
l th l
B k
oo
.
o un
d
in
m ar
s
e
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