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.
(1) Imagine a straight line r, perpendicular to the picture plane, which belongs to the horizontal plane π1. On this plane, π1, set up two straight lines m, sloping at 45° to the picture plane. These two lines meet the line r and the trace tπ1 of the reference plane π1. Therefore the two above mentioned lines m identify on the line r and on the line tπ1 two segments: (AB) and (A”B”), respectively, which are equals.
– 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 π1°, which passes through the centre of projection O and is parallel to the plane π1. This plane meets the picture plane in the vanishing line i’π1, which is parallel to the trace tπ1, as we know.
On the above mentioned plane, construct the projecting straight lines r° and m°, which meet the i’π1 vanishing line in the vanishing points I’r º O° and I’m.
– 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.
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’π1 and the trace tr of the reference plane π1. Note that:
– the distance between i’π1 and tπ1 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’π1 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 π1.
(4) – On the trace of the reference plane π1, 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 p1 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
– (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.