RECTANGULAR INTERLOCKING TILE INSTALLATION
Rectangular [with the joint pattern laid perpendicularly or in parallel to the
edges of the covering] interlocking installation is a bit more complex than the
continuous joint method [continuous lines in both perpendicular directions formed by
tile-to-tile joints].
When we try to obtain a centered layout [symmetry axis] and absence of narrow
strips [tiles cut with a side shorter than half the length of the side of the full tile],
modulation is a bit more complex than with square geometry tiles, due to the
diplacement, to right or left, of a row with respect to the previous one.
Let us see an example: Assuming that we have a free length L = 3660 mm and
that we are going to install square tiles with a 300 x 300 x 7.5 mm work size W
following the interlocking method; according to the formula we use in the case of
continuous joint installation, we would have:
L 3660

 12.2 thereby E=12 full tiles
w 300
Te first cut tile would be:
A
L  WxE W 3660  3600 300



 180mm
2
2
2
2
From design to execution · Setting out
Rectangular interlocking tile installation
1
So, in the first row we would have eleven full tiles and two 180 mm cut tiles at
the edges. However, in the following row we would have a narrow strip of 30 mm, as
shown below, due to the displacement of the joint to left or right, being the
W
displacement
 15mm .
2
For interlocking installation we would use another formula that allows us to
guarantee that the cut tiles are not shorter than W/4. In this case, we will use the
following expressions for the first cut tile A and the last one Z:
L - WxE W

2
4
W
Z (ultima pieza cortada) = A 
2
A (1ª pieza cortada) =
These expressions are deduced from considering for the first piece half the
requirements observed in continuous joint installation:
L  WxE W L  WxE


2L  2(WxE) W L  WxE W
2
2
2
A




2
4
4
2
4
We will apply the expressions above to our example. Let us remind that:
L= 3660 mm
W= 300 mm
E= 12 full tiles
3660 - 300x12 300

 105mm
2
4
W
Z (ultima pieza) = A 
 105  150  255mm
2
A (1ª pieza) =
Thereby in a first row we would have 11 full tiles, starting with a cut tile of 105
mm and finishing with another cut tile of 255 mm.
The situation changes in the second row: we would start with a cut tile of 255
mm, continuing with the installation of 11 (E-1) full tiles and finish with the installation
of a cut tile of 105 mm of side.
From design to execution · Setting out
Rectangular interlocking tile installation
2
Using this method no piece will have a side shorter than W/4 (75 mm), we
observe the joint interlocking in the middle of the piece and obtain a symmetry with
respect to a fictitious axis if we consider two consecutive rows as a unit.
This calculation method can be applied to any square or rectangular tile, where
W will be the length of the tile side involved in the interlock. If we use the open joint
system, instead of using the W work size, we will use the coordination size C = W + J,
where J is the joint width.
In the other direction we will work with the formula we have presented in
continuous joint installation, since we intend to avoid the presence of narrow strips in
the direction perpendicular to the interlock, as well as succeed in obtaining the first and
last tile with a cut side no shorter than W/2.
Here is an actual example. Rectangular living room in a luxury house; ceramic
flooring with useful dimensions of 12100 x 6800 mm. We plan to install ceramic tiles of
W 595 x 295 x 11 mm, with 5 mm open joint and interlock centered with respect to the
bigger dimension of the tile, which shall be parallel to the longest side of the living
room. In this case:
L= 12100 mm
C= W + J = 595 + 5 = 600 mm
L 12100

 20,17
C
600
L  C x E C 12100  20x600 600
 

 200mm
2
4
2
4
C
Z = A +  200  300  500mm
2
A
thereby E = 20
From design to execution · Setting out
Rectangular interlocking tile installation
3
So, in the first row we would have a first tile of 200 mm, 19 full tiles (E-1) with
their 5 mm tile-to-tile joints, and a last tile of 500 mm. The total length shall be 200 +
19 x 600 + 500 = 12100 mm. It must be taken into account that the actual modules will
be a cut tile of 200 mm, nineteen 595 mm full tiles, twenty 5 mm joints and one 500
mm cut tile, which shall give a total length of 12105 mm.
This involves the study of the dimension of joint number 20, which must be
absorbed by the two cut tiles by assigning 2.5 mm to each one; hence the actual
dimensions would be:
A = 200 - J/2 = 200 – 2,5 mm = 197,5 mm
Z = 500 – J/2 = 500 – 2,5 mm = 497,5 mm
In the second row we will have a first tile of 497.5 mm, 19 full tiles (E-1) with
their tile-to-tile joints and a last tile of 197.5 mm. The total length occupied by the tiles
and their joints shall be: 497.5 + 19 x 595 + 20 x 5 + 197.5 = 12100 mm
In the other direction we have continuous joint installation with the following
data:
L = 6800 mm
C = W + J = 295 + 5 = 300 mm
L/C = 6800/300 = 22,67 mm
L  CxE C 6800  300x22 300
 


2
2
2
2
 100  150  250mm
A
thereby E = 22
Therefore, for the width of the living room we would have a first cut tile of 250
mm, twenty-one 295 mm full tiles, twenty-two 5 mm joints and a last cut tile of 250
mm, with a total length of 6805 mm for the flooring. Also here, cut tiles must absorb the
width of the joint, hence we will cut at 247.5 mm; i.e., A = Z = 247.5 mm.
At present, we find ceramic coverings with big rectangular formats (in one
dimension) installed with different types of interlocking. We will see below that we can
carry out the setting out, in interlocking installation, by previously selecting the
displacement of the interlock one fraction with respect to the continuous joint, i.e. ½ (in
the case already seen), 1/3, 1/4, etc, with respect to the side of the tile where that
interlock takes place.
Let us see a first example where the joint between tiles is displaced 1/3 to the
right with respect to the previous row, as shown in the figure below.
From design to execution · Setting out
Rectangular interlocking tile installation
4
L  WxE W

2
6
2W
Z (ultima pieza) = A +
3
A (1ª pieza) 
This last piece can be also obtained by subtracting the length occupied by the
number of full tiles [Wx(E-1)] and the length of the first piece (A) from the total length
available; i.e. Z = L-A-W(E-1).
Let us see an example of installation without joints. The interlock is displaced
1/3 with respect to the bigger side of the tile, with work size W (600 x 300 x 7.5) mm,
for an available space of 3660 mm:
L - WxE W 3660  600x6 600




2
6
2
6
L = 3660 mm
A(1ª pieza) =
L 3660

 6,1mm
W 600
 30  100  130mm
Z (ultima pieza) = A +
2W
2x600
 130 
 130  400  530mm
3
3
thereby E = 6
Thereby we will have a first cut piece of 130 mm, 5 full units installed along
their longer side and a last cut piece of 530 mm, with a total length occupied of 130 + 5
x 600 + 530 = 3660 mm.
In the second row, the situation changes, starting with a piece of 530 mm and
finishing with another cut piece of 130 mm, always with 5 full tiles between them.
From design to execution · Setting out
Rectangular interlocking tile installation
5
If we use the open joint installation system, we should remind that we must use
the coordination size and subtract half the width of the joint from the cut pieces.
Using these formulas we succeed in avoiding narrow strips shorter than the
expression of the second part of the formula for the first cut piece; i.e.:
L  WxE W

2
2
►
Rectangular continuous joint installation (1/1) A 
►
Rectangular interlocking joint installation (1/2) A  L  WxE  W
2
4
►
Rectangular interlocking joint installation (1/3) A 
L  WxE W

2
6
According to the expressions above, we can express the rectangular interlocking
joint installation, with a displacement representing a fraction 1/n with respect to the side
where the interlock takes place, using a setting out formula for the first cut piece:
 1  L  WxE W
A  

 n
2
2n
(n  1)W
 1
Z   A 
 n
n
Let us apply this formula to a specific case. We pretend to install tiles of W
1200x600x11 mm in an available length of 6100 mm, at interlock displaced 1/6 with
respect to the longer side of the tile. In this case, we will have:
L= 6100 mm
W= 1200 mm
L 6100

 5,083
W 1200
 1  L  WxE W 6100  6000 1200
A  



 50  100  150mm
 6
2
12
2
12
5x1200
Z  150 
 1150mm
6
Thereby E = 5
From design to execution · Setting out
Rectangular interlocking tile installation
6
Thereby the layout in the first row will start with a cut piece of 150 mm, 4 full
tiles (E-1) and a last cut piece of 1150 mm, with a total length of 150 + 4 x 1200 + 1150
= 6100 mm occupied by the tiles installed without joint.
In the second row the situation changes, starting with a cut tile of 1150 mm,
following with 4 full tiles and finishing with a last cut tile of 150 mm. We can see here
that the shortest tiles are those of 150 mm in opposition to the 100 mm (W/12)
guaranteed by the calculation formula. The figure below illustrates this example.
From design to execution · Setting out
Rectangular interlocking tile installation
7