The Reference Plane

The observer **O**, standing on the reference plane **π _{1}**, looks ahead of himself and sees:

– the horizon

**i’**;

– the line

**t**, that the reference plane and the picture plane have in common.

.These two parallel lines represent the reference horizontal plane

**π**.

_{1}The observer is standing on the horizontal reference plane

**π**:

_{1}**h**measures the height of the eye

**O**above this plane.

The reference plane

**π**intersects the picture plane in line

_{1}**t**, which is called trace of

**π**.

_{1}Imagine a plane

**π**

_{1}^{0}, belonging to point

**O**and parallel to plane

**π**, This plane intersects the picture plane in the

_{1}horizon

**i’**, which is called the

*Vanishing line*of

**π**.

_{1}Since the planes

**π**and

_{1}**π**

_{1}^{0}are parallel, the lines

**t**and

**i’**are parallel, too.

Since the picture plane is vertical, the horizon

**i’**passes through the principal point

**O**. The line

^{0}**i’**is the image of

the line at infinity

**i**that the plane

**π**has in common with all the planes parallel to it.

_{1}This is why we say that the vanishing line is the image of the laying of a plane and all the planes parallel to it.

If you want to explore this topic you can read:

*Perspective and infinity*.

**Practice**

Measure the height of the eye (**h**). Draw two parallel lines: **i’** and **t**. Note that, if the picture plane is vertical, **i’** belongs

to **O ^{0}**.

**Note**

The two lines

**i’**and

**t**

*represent*the plane

**π**, according to geometry, but they do not describe it effectively. So it is

_{1}always best to add a landscape allusion (e.g. the ground, the sky etc.).

Perspective of a line perpendicular to the picture plane

- Consider a straight line
**r**, belonging to the reference plane**π**, and perpendicular to the picture plane._{1} - Set up a straight line
**r**, parallel to^{0}**r**, and passing through the centre of projection**O**(we call that condition of a line or a plane simply*projecting*; therefore, a*projecting line*is any line that passes through**O**). - The line
**r**meets the picture plane in**T**, called_{r}*trace*of the straight line. Since**r**belongs to the reference plane,**T**belongs to the trace of the reference plane._{r} - The line
**r**meets the picture plane in^{0}**I’**, called vanishing point of the straight line. Since_{r}**r**belongs to the reference plane,**I’**belongs to the vanishing line of the reference plane._{r}

Perspective of points that belong to the straight line **r**

- Consider a plane
**π**^{0}, passing through the centre of projection**O**and parallel to the picture plane**π**: we

call this plane*front plane*. - Now we can fix these notable positions:
**T**– the point**T**that the line**r**has in common with the picture plane;**P**– any points like**P**, which are located beyond the picture plane;**I**– the point**I**, which is at infinity;**Q**– any points, like**Q**, which are located between the picture plane and the front plane;**A**– the point**A**that the line**r**has in common with the front plane**π**^{0}. - The above mentioned points have the following perspectives on the line
**r’**:**T**has a perspective**T’**coincident with**T**that is the trace of**r**, as we know;**P**has a perspective**P’**;**I**, point at infinity, has a perspective**I’**that is the vanishing point of the straight line**r**, but it is also the vanishing point of any line parallel to**r**;**Q**has a perspective**Q’**;

A has a perspective**A’**that is the point at infinity of the line**r’**: in fact, the straight line projecting the point**A**belongs to the front plane and it is therefore parallel to the straight line**r’**; this means that they meet at infinity.

**Notes**

You might consider the correspondence between the straight line **r** and its perspective **r’** as a good demonstration of the power of perspective, which allows to imagine and to view infinitely distant entities.**Practice**

The lines **r’** and **s’**, incident in **I’ _{r}**, are the perspectives of two parallel lines

**r**and

**s**; The lines

**u’**and

**v’**, parallel, are

perspectives of two lines incident in a point

**A**that belongs to the front plane.

**Notes**

If a circle has no points on the front plane, its perspective is an

*ellipse*.

If a circle is tangent to the front plane, its perspective is a

*parabola*, since one of its points has its perspective at infinity.

If a circle passes through the front plane, its perspective is a

*hyperbola*, since two of its points have their perspectives at infinity.

Perspective of any plane perpendicular to the picture plane

As we know), any plane is described by the trace **t** (intersection with the picture plane) and by the vanishing line **i’** (imagine of the line at infinity of plane **i**, namely its laying in the space).

Remember that the trace **t** and the vanishing line** i’** are always parallel.

Consider, to start, a vertical plane **α** perpendicular to the picture plane.

Since the picture plane is vertical and perpendicular to the reference plane, the trace **t _{α}** of this plane is perpendicular to trace

**t**of the reference plane.

_{π1}Since the plane is perpendicular to the picture plane, its vanishing line

**i’**passes through the principal point

_{π1}**O**.

^{0}The trace

**t**and the vanishing line

_{α}**i’**are parallel.

_{α}Similar considerations allow us to set up the perspective of any plane sloping to the reference plane. If these

planes are perpendicular to the reference plane, the angle formed by the trace of the sloping plane and the trace of

the reference plane is the sloping angle of the plane.

To set up the perspective of a vertical plane **α**, perpendicular to the picture plane, you draw, first of all, the trace**t _{α}** of the plane, and then the vanishing line

**i’**, passing through

_{α}**O**. To set up the perspective of a sloping plane,

^{0}perpendicular to the picture plane, you draw the trace

**t**remembering that the angle (

**t**^

**t**) is the sloping angle of

_{π1}the plane.

Then, draw the vanishing line

**i’**, parallel to the trace.

_{α}**Notes**

The two lines **i’ _{α}** and

**t**

_{α}*represent*the plane

**α**, according to geometry, but they do not describe it effectively. So it is

always best to add some architectural element, like a wall, a roof, a ramp etc.

Perspective of straight lines in notable positions

The perspectives of straight lines that are parallel to the picture plane are parallel to each other.

The straight lines sloping at 45° to the picture plane have the vanishing point on the distance circle.

The straight lines parallel to the picture plane can have two possible positions:

– they can belong to the picture plane and in this case all their points belong to the picture plane;

– they are out of the picture plane and in this case they have no point in common with the picture plane.

If the line belongs to the picture plane, it is possible to draw and measure it without performing any projecting operation.

If the line is out of the picture plane, it meets the picture plane at infinity and therefore:

– the trace of the straight line is at infinity and we say that it is the direction of the line;

– the vanishing point of the perspective of this line is at infinity, too, and it is the same direction;

therefore, the perspectives of straight lines that are parallel to the picture plane are parallel to each other.

Now, consider a horizontal line like d, which forms an angle of 45° with the picture plane:

– set the point **T _{d}**, that is the trace of the line; construct the straight line

**d**, which passes through the centre of projection

^{0}**O**and that is parallel to

**d**;

– since the lines

**d**and

**d**are parallel, they form the same angle of 45° with the picture plane;

^{0}– the line

**d**meets the picture plane in

^{0}**I’**, which is the vanishing point of the perspective

_{d}**d’**of

**d**;

– as we know, the line (

**O-O**) is perpendicular to the picture plane and the triangle (

^{0}**O-O**–

^{0}**I’d**) is therefore isosceles;

– thus: the distance (

**O-O**) is equal to the distance (

^{0}**O**–

^{0}**I’d**).

Consider, furthermore, a straight line s that belongs to a vertical plane α which is perpendicular to the picture plane.

As you know if a line belongs to a plane, it has the trace on the trace of the plane and the vanishing point on the vanishing line of the plane.

Therefore, the vanishing point **I’ _{s}** of the straight line

**s**is the point that the distance circle and the vanishing line

**i’**have in common.

_{α}This means that the straight lines sloping at 45° to the picture plane have the vanishing point on the distance circle.

As we will see shortly, these points can be used to measure distances on lines that are perpendicular to the

picture plane and, for this reason, they are also called distance points.

About the history of the distance points see: Kirsti Andersen,

*The Geometry of an Art – The History*

of the Mathematical Theory of Perspective from Alberti to

Monge, Springer 2007.

of the Mathematical Theory of Perspective from Alberti to

Monge

**In practice**

Draw two vertical planes on two horizontal parallel lines forming an angle of 45° with the picture plane.

– Set, as you like, the points** t _{d}** and

**t**, which are the traces of the two horizontal lines

_{e}**d**and

**e**;

– since

**d**and

**e**are horizontal, the vanishing point of their perspective,

**I’**≡

_{d}**I’**belongs to the horizon (vanishing

_{e}line of the horizontal planes) and to the distance circle;

– draw

**d’**and

**e’**;

– now draw, as you like it, the sides

**f’**and

**g’**of the two planes, which are parallel to the picture plane and to each other, since the planes are vertical, like the picture plane is.

If you want to draw the perspective s’ of a line that belongs to a vertical plane **α**, perpendicular to the picture plane:

– set the vanishing point **I’ _{s}** on the vanishing line

**I’α**and on the distance circle;

– draw the perspective

**s’**.

Points in given positions on a line perpendicular to the picture plane

Imagine a straight line **r**, perpendicular to the picture plane, which belongs to the horizontal plane **π _{1}**. On this plane,

**π**, set up two straight lines

_{1}**m**, sloping at 45° to the picture plane.

These two lines meet the line

**r**and the trace

**t**of the reference plane

_{π1}**π**. Therefore the two above mentioned lines

_{1}**m**identify on the line

**r**and on the line

**t**two segments: (

_{π1}**AB**) and (

**A”B”**), respectively, which are equals.

Indeed:

– the triangle (

**A T**) is isosceles so that (

_{r}A”**A T**) = (

_{r}**T**);

_{r}A”– the triangle (

**B T**) is isosceles so that (

_{r}B”**B T**) = (

_{r}**T**);

_{r}B”– therefore (

**AB**) = (

**A”B”**). It means that, if we draw the perspectives of the two straight lines

**m**(

**AA”**) and

**m**(

**BB”**) we shall obtain on

**r’**the perspectives

**A’**and

**B’**of the points

**A**and

**B**.

Consider now the projecting plane

**π**, which passes through the centre of projection

_{1}^{0}**O**and is parallel to the

plane

**π**. This plane meets the picture plane in the vanishing line

_{1}**i’**, which is parallel to the trace

_{π1}**t**, as we

_{π1}know. On the above mentioned plane, construct the projecting straight lines

**r**and

^{0}**m**, which meet the

^{0}**i’**vanishing

_{π1}line in the vanishing points

**I’**≡

_{r}**O**and

^{0}**I’**.

_{m}Observe that:

– the triangle (

**O I’**) is similar to the triangle (

_{r}I’_{m}**AT**) and it is therefore an isosceles triangle, having

_{r}A”equal the sides (

**I’**) and (

_{m}I’_{r}**I’**).

_{r}OWe can resume the above observations in this statement:

*the distance of the vanishing point*;

**I’**of the straight line_{m}**m**from the vanishing point**I’**of the line_{r}**r**is equal to the distance of the vanishing point**I’**from the projection centre_{r}**O**– if the line

**r**is perpendicular to the picture plane, the vanishing point

**I’**belongs to the distance circle.

_{m}**In practice – first exemple**

We want to set up two points **A **and **B** on the straight line **r** perpendicular to the picture plane, knowing that **A** is distant two meters from the picture plane and that **B** is distant four meters from** A**.

1. Draw the vanishing line** i’ _{π1}** and the trace

**t**of the reference plane

_{r}**π**. Note that:

_{1}– the distance between

**i’**and

_{π1}**t**represents the elevation of the eye

_{π1}**O**on the reference plane;

– usually the reference plane is the ground and the elevation of the eye is the height of a standing

man;

– the vanishing line

**i’**is the horizon and the image of any person standing on the ground, having the same height of the observer, has the head on the horizon;

_{π1}– this first operation determines the reduction ratio; in fact, when you draw on a paper and not on a wall (like a painter of a trompe-l’oeil), you have to reduce the dimensions of the space so that the represented objects can be included on

the paper; e.g., if the eye of the observer is 170 cm above the reference plane, you could choose a reduction ratio of 1/20 and, consequently, set the distance between

**t**and

_{π1}**I’**at 8,5 cm (170/20).

_{π1}2. Draw the distance circle having in mind that the radius of this circle represents the distance of the eye from the picture plane; this distance is reduced in function of the above mentioned reduction ratio, so if your draw a circle that has a radius of 20 cm e.g., the principal distance is equal to four meters.

- Draw the perspective
**r’**of a straight line**r**(**T**) perpendicular to the picture plane, which belongs to_{r}I’_{r}

the reference plane**π**._{1} - On the trace of the reference plane
**t**, mark the point_{π1}**A”**, which is distant two meters from**T**and mark the point_{r}**B”**which is distant four meters from**Tr**. - Draw the perspectives
**m’**(**A” I’**) and_{m}**m’**(**B” I’**) of two straight lines_{m}**m**, which belong to the reference plane**π**and pass through the point_{1}**A”**and**B”**. These two lines meet the line**r’**in points**A’**and**B’**, which we would so find. Note that the vanishing point**I’**of the line_{m}**m’**must belong to the distance circle.

**Second example**

We want to construct the perspective of four point (**A,B,C,D**) which separate three equal intervals and belong to the straight line **r**(**T _{r} I’_{r}**) perpendicular to the picture plane.

6. Choose any plane belonging to the straight line

**r**, e.g. the vertical plane

**α**(imagine it being a wall). Draw the vanishing line

**i’**, passing through

_{α}**I’**and the trace

_{r}**t**, passing through

_{α}**T**. Note that

_{r}**i’**and

_{α}**t**are parallel, as we know.

_{α}7. Mark four points (

**A”,B”,C”,D”**) on

**t**.

_{π1}8. Draw the perspectives

**m’**of the lines

**m**, which belong to the plane

**α**and pass through the points (

**A”,B”, C”, D”**), remembering that the vanishing point

**I’**of

_{m}**m’**must belong to the distance circle and to the vanishing line

**i’**. These lines

_{π1}**m’**meet

**r’**in

**A’, B’, C’, D’**.

First elementary observations on the projection properties

As anybody can easily observe, the perspective modifies the length of the segments.

If you draw the perspectives of three equal segment (**AB**), (**BC**) and (**CD**), e.g., you obtain

three segments (**A’B’**), (**B’C’**) and (**C’D’**) of unequal lengths.

We say that perspective changes the metric properties of the shapes.

Nevertheless there exists a metric property that is constant, also in perspective. This constant is called *cross-ratio* and can be calculated as follows:

– consider four collinear points (**A, B, C, D**); – now set-up the segments (**AC**), (**AB**), (**AD**) and (**BD**); – the cross-ratio (**ABCD**) is:

(**AC/BC**)**/**(**AD/BD**).

For instance, suppose the intervals among these points to be the unit, in that case the value of the cross-ratio is:

(**AC**) = 2; (**BC**) = 1; (**AD**) = 3; (**BD**) = 2;

therefore:

(**AC/BC**)**/**(**AB/BD**) = (**2/1**)**/**(**3/2**) = **4/3**.

Now, draw the perspectives (A’, B’, C’, D’) of any collinear and equidistant points (**A, B, C, D**) and calculate the cross-ratio (**A’B’C’D’**): you will find, without fail, the value **4/3**.

Now, imagine a circle **c**, having centre in **C** and a diameter (**AB**). Imagine extending the line (**ACB**) to the point at infinity **I**.

Now calculate the cross-ratio (**ABCI**) considering the positive/negative direction of the line **r**:

(**AC**) = 1; (**BC**) = – 1; (**AI**) = ∞; **(BI**) = ∞, – since the ∞/∞ ratio is an indeterminate form we cannot find a valid result (without the help of analysis), but we can draw the perspective of this shape.

Well, draw the perspective **c’** of a circle and let’s say that **d’** is the perspective of the diameter **d** perpendicular to the picture plane.

Now consider the points **A’, C’, B’** and **I’** where: **A’**, **B’** are the perspectives of the extremes **A** and **B** of the diameter; **C’** is the perspective of the centre **C** and **I’** is the perspective of the point at infinity **I** of the diameter.

Calculate the cross-ratio (**A’B’C’I’**) and you will find the value **-1** (being the straight line **d** oriented).

This could be a good example of how the per spective translates the infinity in finite terms.

This special cross-ratio is called *harmonic ratio*.

Note that any cross ratio involving the vanishing point **I’** and the perspectives **A’, C’, B’ **of three equidistant points **A, B,** **C,** belonging to the same straight line, is a harmonic ratio.

An application to the antique perspective

There is a common question whether perspective was known in ancient times and, particularly, in the Roman world. Observe the fresco found in Augustus’s home in Rome, and calculate the cross-ratio of the intervals among the columns: you will find a value very close to **4/3**. Therefore, this perspective is correct!