Chapter 5
Drawing a cube
Math 4520, Spring 2015
R.
5.1
Math 452, Spring 2008
2002
Connelly
CLASSICALperspective
GEOMETI;lIES
One and two-point
In Chapter6.5 Drawing
we saw ahow
cube to calculate the center of vision and the viewing distance for a
square in one
perspective.
6.1 or
Onetwo-point
and two point
perspectiveSuppose that we wish to extend our construction to
a cube. In particular,
what
conditions
mustthethe
projection
of the
a cube
satisfy?
In Chapter 5 we saw how to calculate
center
of vision and
viewing
distance for
square in one or
point perspective.
Suppose
weone-point
wish to extend
our construction
For theaprojection
of two
a cube,
we say that
it that
is in
perspective
if only one of
to
a
cube.
In
particular,
what
conditions
must
the
projection
of
a
cube
satisfy?
the three vanishing points, corresponding to three non-parallel sides, is finite. A one-point
For the projection of a cube, we say that it is in one point perspectiveif only one of
perspectivethe
drawing
(or projection)
of a cube
is easy
to handle.
three vanishing
points, corresponding
to three
non-parallel
sides,isThe
finite.front
A oneface
point of the cube
projects onto
a perfect
square
as in Figure
5.1.is This
the extra
condition.
perspective
drawing
(or projection)
of a cube
easy toishandle.
The front
face of the cube
projects onto a perfect square as in Figure 6.1.1. This is the extra condition.
Horizon
One point perspective
Figure
6.1.1
Figure 5.1
Two point per.spective(where two of the three vanishing points correspond to finite
points) is somewhat more complicated. This the same situation as in Chapter 5, except
that the square is the bottom face of the cube that we are projecting. As in Chapter 5, we
project orthogonally onto the base plane, which in Chapter 5 was called the object plane.
Two point
(where
two
the 6.1.2.
threeIfvanishing
points
finite points)
Thisperspective
bird's-eye view
is shown
in of
Figure
the cube has
sides ofcorrespond
length s, thentothe
is somewhat
more complicated.
same
Chapter
5, except
that the
projection
orthogonal to the This
picturethe
plane
of thesituation
two sides ofas
theinbase
of the cube,
as in
Figure
6.1.2,
have
lengths
x
and
y,
say.
So
we
have
square is the bottom face of the cube that we are projecting. As in Chapter 5, we project
orthogonally onto the base plane, whichX2 in
Chapter 5 was called the object plane. This
+ y2 =-S2.
bird’s-eye view is shown in Figure 5.2. If the cube has sides of length s, then the projection
If we draw the line in the base object plane parallel to the picture plane through the front
orthogonal corner
to the
picture plane of the two sides of the base of the cube, as in Figure 5.2,
of the cube, then we have the line L as in Figure 6.2. We can find the line segments
have lengths
x andx y,
of length
andsay.
y in So
line we
L. have
We next project this construction into the picture plane as in Figure 6.1.3. The projection of the lengths x and y in L and s 2in the 2vertical2 side of the cube are all proportional
x +y =s .
,
2
CHAPTER 5. DRAWING A CUBE
MATH 4520, SPRING 2015
If we draw the line in the base object plane parallel to the picture plane through the front
corner
of the cube, then we have CLASSICAL
the line L as
in Figure 6.2. We can find the line segments
GEOMETRIES
2
of length x and y in line L.
L
Picture plane edge on
CLASSICAL GEOMETRIES
2
..Station
Figure
point
6.1.2
Figure 5.2
L
Picture plane edge on
Center ofinto
visionthe picture plane as in Figure 5.3. The projection
We next project this construction
of the lengths x and y in L and s in the vertical side of the cube are all proportional to their
..Station
point
true lengths. So a simple Euclidean construction
(constructing
a right triangle with x and y
Figure
6.1.2
as legs) allows us to find the projection of the vertical line segment in the picture plane.
Center of vision
x
y
x
Two
y
~
point
perspective
y
Figure
x
Two
y
point
x
6.1.3
~
perspective
Figure
6.1.3
to their true lengths. So a simple Euclidean construction ( constructing a right triangle
Figure 5.3
with x and y as legs) allows us to find the projection of the vertical line segment in the
to their true lengths. So a simple Euclidean construction ( constructing a right triangle
picture plane.
with x and y as legs) allows us to find the projection of the vertical line segment in the
The moral
of this story is that for two point perspective, once the base of the cube is
picture
plane.
The
moral
this
storythisisstory
that
for two-point
perspective,
the
base
of the
cube is
drawn,
thenofthe
the
rest
determined,
asonce
wellof as
station
Theheight
moral ofand
is thatofforthe
twocube
point is
perspective,
once the
base
thethe
cubeunique
is
drawn,
then
thedrawn,
height
and
theand
rest
of
determined,
the
unique not
station
then
the height
thenot
restthe
of thecube
cube
isis
determined,
as well
thewell
uniqueas
station
point
in 3-space.
If this
height
is
drawn
correctly,
then
theasas
drawing
will
probably
point in 3-space. If this height is not drawn correctly, then the drawing will probably not
pointlook
in 3-space.
If
this
height
is
not
drawn
correctly,
then
the
drawing
will
probably
like a look
cube.
It will look like a box that is too high, too low, too short, or too long, not
like a cube. It will look like a box that is too high, too low, too short, or too long,
may
not
be look
able
tell
which
dimension
is the
largest,
you
can
telltoo
thatlong,
the etc.
etc.It
You
may
not beto
able
to
which
dimension
the
largest,
but
youbut
can
that
theor
look etc.
like You
a cube.
will
like
a tell
box
that
is toois high,
too
low,
tootell
short,
sides
are not equal.
'.'.
sides
are
not
equal.
You may not be able to tell which dimension is the largest, but you can tell that the sides
are not equal.
6.2 Three point perspective
6.2 Three point
perspective
If one draws
a cube where none of its
5.2
sides are parallel ( or perpendicular) to the picture
plane, we say the cube is in three point per.!pective. In other words, the vanishing point of
Three point perspective
If one draws a cube where none of its sides are parallel ( or perpendicular) to the picture
plane, we say the cube is in three point per.!pective. In other words, the vanishing point of
If one draws a cube where none of its sides are parallel (or perpendicular) to the picture
plane, we say the cube is in three point perspective. In other words, the vanishing point of
each of the three sets of parallel sides project onto a finite point in the picture plane. Of
5.2. THREE POINT PERSPECTIVE
3
DRAWING
A CUBE
3
course it could happen that these three points are not on the actual canvas, but nevertheless
each of the three sets of parallel sides project onto a finite point in the picture plane. Of
they arecourse
on the
Euclidean
plane
through
theare
canvas
not at
infinity.
Since the plane of
it could
happen that
these
three points
not on and
the actual
canvas,
but nevertheless
any facethey
of the
cube
not perpendicular
picture
plane,
our previous
constructions
for
are on
the is
Euclidean
plane through to
thethe
canvas
and not
at infinity.
Since the plane
of
any
face
of
the
cube
is
not
perpendicular
to
the
pictu~
plane,
our
previous
constructions
finding the center of vision and the viewing distance do not apply. However, there are other
for finding the center of vision and the viewing distance do not apply. However, there are
constructions
that are relatively simple. Consider the tetrahedron determined by the station
other constructions that are relatively simple. Consider the tetrahedron determined by
point and
three
vanishing
points
in thepoints
picture
plane
as plane
in Figure
5.4. 6.2.1.
the the
station
point
and the three
vanishing
in the
picture
as in Figure
Figure
6.2.1
Figure 5.4
Note that the three edgesof the tetrahedron incident to the station point are all perpendicular to each other, since they are parallel to the edgesof the cube. The viewing distance
is the length of the altitude of this tetrahedron, coming from the station point. The center
visionthe
is the
baseedges
of the of
altitude
in the picture plane.
Theto
following
Lemmapoint
tells us
thatall perpenNoteofthat
three
the tetrahedron
incident
the station
are
the
center
of
vision
is
just
the
altitude
center
of
the
three
vanishing
points
in
the
picture
dicular to each other, since they are parallel to the edges of the cube. The viewing distance
plane.
is the length of the altitude of this tetrahedron, coming from the station point. The center of
6.2.1:
Consider
a tetrahedron
T with one
vertex
0 having
its three
incident
edges
vision isLemma
the base
of the
altitude
in the picture
plane.
The
following
Lemma
tells
us that the
mutually perpendicular. Then the foot of the altitude of T starting at 0 is the altitude
center ofcenter
vision
is
just
the
altitude
center
of
the
three
vanishing
points
in
the
picture
plane.
of the face opposite 0.
CallConsider
the vertices
the tetrahedron
0, A,one
B, C,
whereO
OA,
OB, OC
mutually
LemmaProof:
5.2.1.
a oftetrahedron
T with
vertex
having
its are
three
incident edges
perpendicular. Let X be the foot of the altitude to the tetrahedron in the plane of ABC.
mutuallyThen
perpendicular.
Then the foot of the altitude of T starting at O is the altitude center
the plane of 0 X A is perpendicular to the plane of AB C. Then the plane of 0 X A is
of the face
opposite
O.
perpendicular to the plane of ABC, since it cOn\ajnsthe line OX, which is perpendicular
to the plane of ABC.
through
is also perpendicular
theC,
plane
OBC,OA,
sinceOB,
it contajns
the mutually
Proof. CallThe
theplane
vertices
ofOX
theA tetrahedron
O, A,toB,
where
OC are
line OA, which
OBC. to
Thus
plane OXA isin
perpendicular
perpendicular.
Let Xis perpendicular
be the foot to
of the
theplane
altitude
thethetetrahedron
the plane of ABC.
to the intersection of the planes OBC and ABC, which is the line through BC. But the
Then the
plane of OXA is perpendicular to the plane of ABC. Then the plane of OXA is
plane OX A and the plane ABC intersect in the line through AX .Thus the line AX and
perpendicular to the plane of ABC, since it contains the line OX, which is perpendicular to
the plane of ABC.
The plane through OXA is also perpendicular to the plane OBC, since it contains the
line OA, which is perpendicular to the plane OBC. Thus the plane OXA is perpendicular
to the intersection of the planes OBC and ABC, which is the line through BC. But the
plane OXA and the plane ABC intersect in the line through AX. Thus the line AX and
the line BC are perpendicular, since the line BC is perpendicular to the plane OXA and the
line AX intersects the line BC (being in the plane of ABC). Thus X lies on the altitude of
ABC from A. Similarly, X lies on the other two altitudes. Thus X is the altitude center of
ABC. This is what we were to prove. See Figure 5.5.
CLASSICAL GEOMETRIES
4c
the line RC are perpendicular, since the line RC is perpendicular to the plane OXA and
the line AX intersects the line RC (being in the plane of ARC). Thus X lies on the
altitude of ARC from A. Similarly, X lies on the other two altitudes. Thus X is the
CHAPTER
5. DRAWING
CUBE
MATH 4520, SPRING
altitude
center of ARC.
This is what weA
were
to prove. SeeFigure 6.2.2.
4
2015
Figure
5.5 three propositions from Euclidean
In the proof of the Lemma we have used
the following
geometry in three-space.
1.
If a line L is perpendicular to a plane P, then any other plane through p is
perpendicular to p .
In the proof2.ofT\\'o
theintersecting
Lemma lines
we have
used
following
propositions
in 3-space
arethe
perpendicular
, if athree
plane through
one is per-from Euclidean
pendicular
to
the
other
.
geometry in three-space.
3. If one plane is perpendicular to two others, then it is perpendicular to their intersection (if the intersection
is nonempty).
1. If a line L is perpendicular
to a plane
P , then any other plane through P is perpendicular
to P .
6.3 Constructing
the center
of vision
in 3-space
2. Two intersecting
3-space ofare
if a6.2.1
plane
through
If we "open lines
up" theintetrahedron
the perpendicular,
construction in Lemma
and lay
the right one is perpensides
flat in the picture plane next to the triangle of vanishing points, then we get
dicularangled
to the
other.
the following diagram.
3. If one plane
perpendicular
tobetwo
others,indicated
then itin is
perpendicular
intersection
Lemma is6.3.1:
Let dl, d2, d3,
the lengths
Figure
6.6 determined to
by their
the
of ,rision Xis and
the vanishing points A, B, C. Then the viewing distance ( the
(if the center
intersection
nonempty).
length ofOABC
5.3
from 0) is
d=~.
Constructing
the center of vision in 3-space
Proof: We use Figure 6.3.1 In the tetrahedron OABC after folding it back into 3-space,
oxy
is a right triangle with OY as hypotel1bse.
Then OY has length ,fiI;iI;
from
If we “open up”
the5.tetrahedron
construction
in Lemma 5.2.1 and lay the right angled
Chapter
So the length ofof
OXthe
is Vd1d2
-4.
sides flat in theThus
picture
plane next
to the
vanishing
then we
the expression
of Lemma
6.2.1 triangle
gives a way of
of calculating
thepoints,
viewing distance
d3 ifget Figure 5.6.
A CUBE
one knows the three vanishing points, DRAWING
ABC. Of
course, the center of vision is5 the altitude
center of ABC.
Figure 5.6
It is interesting to note that, when the cube is on the opposite side of the picture plane
from the station point, the triangle ABC is always acute, not matter how the cube is
projected onto the picture plane. Also the center ofvision X is always inside a circle with
diameter AB, if A and B are vanishing points as above.
Another interesting point is that it is not necessaryto use the fact that all three lengths
of the sides of the cube are equal. All that was used was that the three sides are mutually
1perpendicular
2
3 .In fact, the three vanishing points can be calculated from a picture (in
three point perspective) of any "box" with arbitrary three values for length, width, and
height. Knowing the three vanishing points allows one to compute the center of vision
and the viewing distance. This is different from two point perspective, where the center
of vision cannot be determined unless, say, we know that the base is a square. We only
know that the center of vision in two point perspective lies on the horizon line.
2 points of three point perspecOn the other hand, if one only knows the three vanishing
tive, then one can construct a cube starting1at 2
any point.3 The key is to construct the
vanishing points for the 45° lines of the base and sides by bisecting the O angles in the
construction of Figure 6.3.1. This is shown in Figure 6.3.2.
Once one has the 45° vanishing points determined, then it is an easy matter to draw a
Lemma 5.3.1. Let d , d , d , be the lengths indicated in Figure 6.6 determined by the center
of vision X and the vanishing points A, B, C. Then the viewing distance (the length of
OABC from O) is
q
d=
d d −d .
5.3. CONSTRUCTING THE CENTER OF VISION IN 3-SPACE
5
Proof. We use Figure 5.5. In the tetrahedron OABC after folding√
it back into 3-space, OXY
d1 d2 from Chapter 5. So
is a right triangle with
OY
as
hypotenuse.
Then
OY
has
length
p
2
the length of OX is d1 d2 − d3 .
Thus the expression of Lemma 5.3.1 gives a way of calculating the viewing distance d3 if
one knows the three vanishing points, ABC. Of course, the center of vision is the altitude
center of ABC as in Figure 5.6.
It is interesting to note that, when the cube is on the opposite side of the picture plane
from the station point, the triangle ABC is always acute, not matter how the cube is projected
onto the picture plane. Also the center of vision X is always inside a circle with diameter
AB, if A and B are vanishing points as above.
Another interesting point is that it is not necessary to use the fact that all three lengths
of the sides of the cube are equal. All that was used was that the three sides are mutually
perpendicular. In fact, the three vanishing points can be calculated from a picture (in three
point perspective) of any “box” with arbitrary three values for length, width, and height.
Knowing the three vanishing points allows one to compute the center of vision and the viewing
distance. This is different from two-point perspective, where the center of vision cannot be
determined unless, say, we know that the base is a square. We only know that the center of
vision in two-point perspective lies on the horizon line.
On the other hand, if one only knows the three vanishing points of three point perspective,
then one can construct a cube starting at any point. The key is to construct the vanishing
6
points
for the 45◦ lines of the baseCLASSICAL
and sides by
bisecting the O angles in the construction of
GEOMETRIES
Figure 5.6. This is shown in Figure 5.7.
Figure 6.3.2
Figure 5.7
Once one has the 45◦ vanishing points determined, then it is an easy matter to draw a
typical cube, as well (as a whole grid) as in Figure 5.8.
Figure 6.3.2
6
CHAPTER 5. DRAWING A CUBE
MATH 4520, SPRING 2015
Figure 6.3.3
Figure 5.8
5.4
Exercises:
1. Why is it true that the vanishing points of three point perspective are so often off the
paper?
2. Explain what a picture looks like when viewed from a distance greater that the proper
viewing distance for two and three point perspective. Is it merely lengths or even
relative lengths that appear distorted? Look at the pictures from Descargues’ book
Exercises: points are on the paper.
where the vanishing
DRAWING
I.
A CUBE
7
'\Thy is it true that the vanishing points of three point perspective are so often off
the paper?
3. Figure 5.9 is the2.projection
of the edges of a cube in three point perspective. Assume
Explain what a picture looks like when viewed from a distance greater that the
viewing distance for two and three point perspective. Is it merely lengths or
that the cube is onproper
the
other
side of the picture plane from the station point. Which
even relative lengths that appear distorted? Look at the pictures from Descargues'
book
where
the
vanishing
are on the
is the point nearest
to the picturepoints
plane,
A paper.
or B? Explain your answer mathematically.
3. Figure 6.E.l is the projection of the edges of a cube in three point perspective.
Assume
that
the
cube
is
on
the
other
side
of
the picture plane from the station
Does your answer “look”
right?
point. Which is the point nearestto the picture plane, A or B? Explain your answer
mathematically. Does your answer "look" right?
4. Suppose that one has a picture of a rectangle in two point perspective as in Figure
Figure 5.9
6.E.2 ( Assume that the plane of the rectangle is perpendicular to the picture plane. )
At what points for the center of vision will it appear that x .$:y?
4. Suppose that one has a picture of a rectangle in two-point perspective as in Figure 5.10
(Assume that the plane of the rectangle is perpendicular to the picture plane.) At what
points for the center of vision will it appear that x ≤ y?
4. Suppose that one has a picture of a rectangle in two point perspective as in Figure
5.4. EXERCISES:
6.E.2 ( Assume that the plane of the rectangle is perpendicular to the picture plane. )
At what points for the center of vision will it appear that x .$:y?
7
Figure 5.10
5. Suppose that Figure 5.10 is a picture of a square, but in three point perspective. That
is, the plane of the square is not perpendicular to the picture plane. What is the locus
of the possible centers of vision?
GEOMETRIES
8
6. Locate the
center of vision andCLASSICAL
the viewing
distance for the Escher drawing of a lattice.
Are the blocks
cubes?
can
make
a transparancy
usepoint
theperspective.
sketches in Geometer’s
5. Suppose
thatYou
Figure
6.E.2
is a picture
of a square, butand
in three
Sketchpad. That is, the plane of the square is not perpendicular to the picture plane. What
7.
8.
is the locus of the possible centers of vision?
6. Locate the center of vision and the viewing distance for the Escl1erdrav.'ing of a
Explain the “mistakes”
thecubes?
handout of odd drawings.
lattice. Are thein
blocks
7.. Explain the "mistakes" in the handout of odd drawings.
8. drawing
In technicalthe
drawing
thecone
term cone
of visionis
is sometimes
used.used.
This isThis
the cone
In technical
term
of vision
sometimes
is the
from the station along the line of sight to the center of vision. SeeFigure 6.E.3.
cone from
the station along the line of sight to the center of vision. See Figure 5.11.
Cone
of vision
Figure 6.E.3
Figure 5.11
a.
b.
If the cone of vision is restricted to 30° (as is often done), how many vanishing
points for a cube can there be inside the cone?
If all three vanishing points for a cube are in the cone of vision, what is the
smallest that the angle can be? What is the position of the vanishing points
at this minimum angle?
◦
(a) If the cone of vision is restricted to 30 (as is often done), how many vanishing
( Challenging)
points9.for
a cube can there be inside the cone?
a.
In the usual orthogonal Cartesian coordinate system let a, b, c be the distances
(b) If all threealong
vanishing
for respectively,
a cube are
in the
cone
of A,vision,
the x-axis,points
y-axis, z-axis,
for three
separate
points
B, G, what is the
respectively. What is the equation of the plane through A, B, G?
smallest b.that
the angle can be? What is the position of the vanishing points at
For the same points as in part a, what is the distance d of the plane from the
this minimum
originangle?
?
Let dl, d2, d3, be the distances between the three pairs of points of Part a.
Find 8Jl expressionfor a, b, c in t~rms of these dist8Jlces.
followingd.exercise
provides
a formula
for pairs
the of
viewing
in two-point
perspecGiven the
distances between
the three
vanishingdistance
points in three
point
◦
perspective,
:find
an
expression
for
the
viewing
distance.
when the distance between the side vanishing points and one of the 45 vanishing
c.
9. The
tive
points is known.
(a) Suppose we have a right angled triangle with legs of length x and y, while the
hypotenuse has length a + b, where a and b are determined as in Figure 5.12 by
8
CHAPTER 5. DRAWING A CUBE
MATH 4520, SPRING 2015
b
y
c
45
a
d
c
x
Figure 5.12
the inscribed square with side length c. Show that
conclude that xy = ab .
a
c
=
a+b
y
and
b
c
=
a+b
x
and
(b) The altitude of triangle in Figure 5.12 has length d. By calculating the area of the
right triangle two ways, show that d(a + b) = xy.
(c) Use the calculations of Part (a) and Part (b) to show that
d=
ab(a + b)
.
a2 + b 2
(d) Show that the formula of Part (c) can be used to calculate the viewing distance
for two-point perspective as in Figure 5.13.
b
45 vanishing point
a
Figure 5.13
10. (Challenging) The following exercise provides a formula for the viewing distance in
three-point perspective in terms of the distances between the three vanishing points.
This calculation is easy with a simple calculator.
(a) In the usual orthogonal Cartesian coordinate system let a, b, c be the distances
along the x-axis, y-axis, z-axis, respectively, for three separate points A, B, C,
respectively. What is the equation of the plane through A, B, C?
(b) For the same points as in part a, what is the distance d of the plane from the
origin?
5.4. EXERCISES:
DRAWING
9
A CUBE
9
Figure 6.E.4
Figure 5.14
(c) Let d1 , d2 , d3 , be the distances between the three pairs of points of Part (a). Find
an expression for a, b, c in terms of these distances.
(d) Given the distances between the three pairs of vanishing points in three point
perspective, find an expression for the viewing distance. (Hint: The answer is the
following, where d is the viewing distance.)
s
1
d=
2
+ d2 −d22 +d2 + d2 +d22 −d2
−d2 +d2 +d2
1
2
3
1
2
3
1
2
3
11. Use the formula in Problem 10d to calculate the viewing distance in the following
Escher picture Figure 5.16, OR find your own example of a drawing or picture in 3point perspective. Then find the correct viewing distance for Figure 5.16
...
Figure 5.15
10
CHAPTER 5. DRAWING A CUBE
MATH 4520, SPRING 2015
Figure 5.16: This is Figure 5.16 rescaled by a factor of 4.