### Note:

For an easier reading, I suggest that you open two ‘windows’ side by side: on your screen, one for the figures and the other for the text.

### Theory

(1) Imagine a straight line **r**, perpendicular to the picture plane, which belongs to the horizontal plane π**1**. On this plane, π**1**, set up two straight lines **m**, sloping at 45° to the picture plane. These two lines meet the line **r** and the trace **tπ****1** of the reference plane π**1**. Therefore the two above mentioned lines **m** identify on the line **r** and on the line **tπ****1** two segments: (**AB**) and (**A”B”**), respectively, which are equals.

Indeed:

– the triangle (**ATrA”**) is isosceles so that (**ATr**) = (**TrA”**);

– the triangle (**BTrB”**) is isosceles so that (**BTr**) = (**TrB”**);

– 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**.

(2) Consider now the projecting plane π**1°**, which passes through the centre of projection **O** and is parallel to the plane π**1**. This plane meets the picture plane in the vanishing line **i’π****1**, which is parallel to the trace **tπ****1**, as we know.

On the above mentioned plane, construct the projecting straight lines **r°** and **m°**, which meet the **i’π****1** vanishing line in the vanishing points **I’r** º **O° **and **I’m**.

Observe that:

– the triangle (**OI’rI’m**) is similar to the triangle (**ATrA”**) and it is therefore an isosceles triangle, having equal the sides (**I’m I’r**) and (**I’r O**).

We can resume the above observations in this statement:

– the distance of the vanishing point **I’m** of the straight line **m** from the vanishing point **I’r** of the line **r** is equal to the distance of the vanishing point **I’r** from the projection centre **O**;

– if the line **r** is perpendicular to the picture plane, the vanishing point **I’m** belongs to the distance circle.

### Practice

**First example. **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’π****1** and the trace **tr** of the reference plane **π****1**. Note that:

– the distance between **i’π****1** and **tπ****1** represents the elevation of the eye **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’π****1** 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;

– 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 **tπ1** and **I’π1** at 8,5 cm (170/20).

(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.

(3) – Draw the perspective **r’** of a straight line **r** (**Tr I’r**) perpendicular to the picture plane, which belongs to the reference plane **π****1**.

(4) – On the trace of the reference plane **π****1**, mark the point **A”**, which is distant two meters from **Tr** and mark the point **B”** which is distant four meters from **Tr**.

(5) – Draw the perspectives **m’**(**A” I’m**) and **m’**(**B” I’m**) of two straight lines **m**, which belong to the reference plane **p****1** and pass through the point **A”** and **B”**. These two lines meet the line **r’** in point **A’ **and **B’**, which we would so find.

Note that the vanishing point **I’m** of the line **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**(**Tr** **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’α**, passing through **I’r** and the trace **tα**, passing through **Tr**. Note that **i’α** and **tα** are parallel, as we know.

(7) – Mark four points (**A”**,**B”**,**C”**,**D”**) on **tα**.

(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’m** of **m’** must belong to the distance circle and to the vanishing line **i’α**. These lines **m’** meet **r’** in **A’**, **B’**, **C’**, **D’**.

### Theory – First elementary observations on the projection properties (1)

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**.

### 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.

### Theory – First elementary observations on the projection properties (2)

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 **r** oriented). This could be a good example of how the perspective 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. We will resume this circle example again in the upcoming lessons.