In the Center Line **method**, sum **total length** of the **center lines** of walls, long and short has to be found out. **Find **the total length of the **center lines **of **walls** of the same type.

**Long and short** have the same **type of foundation and footing **and they find the **quantities **by multiplying the** total center length** by the **respective **breadth and **height**.

In this method, the **length **will remain the same for the **excavation **in a foundation for **concrete **in the **foundation**. For all footings and for the **superstructure**. (With slight **difference **when there are** cross walls** or the **number of junctions**).

## Center Line Method Of Estimation

In center line **method **is quick but requires **special **attention and **consideration **at the junction, **meeting **points of cross **walls**, etc.

### One Room Estimation

**Centerline length of one wall = Inner dimension of room + Wall thickness**

**= Inner dimension of room + Â½ x Wall thickness of one side + Â½ x wall thickness of other.**

**Centre line length of horizontal wall**

= 4 + 0.45 + 0.45

**= 4.90 m**

**Centre line length of vertical wall**

= 3 + 0.45 + 0.45

= 3.90 m

**Total Length of excavation for a foundation**

= 2(4.90 + 3.90)

= 17.6 m

**Quantity of excavation for a foundation**

= Total length of centre line of walls x breath of trench x Depth of trench

= 17.6 x 0.9 x 1.2

**= 19.01 m3**

**Read More: Building Estimate | Building Estimate Format In Excel | Estimation and Costing of Building**

For **rectangular**, **circular**, or **polygonal buildings **having no inter or cross walls, this **method **is quite simple but for **buildings **having a cross or **partition **walls, for every **junction** of partition or cross **walls** with **main walls,** a half breath of the **respective **item or **footing **is to be **deducted **from the **total center length.**

**Net Centre line length â€“ (Â½ x breath of wall x number of the junction)**

**The following points in relation to the centerâ€“line method should be noted.**

The term **junction **is used to indicate the **meeting points **of cross **walls** or partition walls with** main walls.**

For **instance**, if there is **one cross wall** there will be **two junctions. **At the corner of a **building**, no junction is **formed**.

For **rectangular buildings **with No **cross wall,** there will not be any junction and hence **no deduction** from the **center line length **will be **required**. It is fact that a** certain portion** taken twice is counterbalanced by the **same** **amount **of portion** left out **at the **corner**.

The **breadth** **indicates **the width of an **item **whose center line is being **worked**, for **instance**, it will be the **width **of a trench of **excavation** and foundation concrete. But for **brick masonry** up to the **plinth** or for the superstructure will be **equal **to the **corresponding **thickness of the **wall**.

**Read More: Building Estimation Excel Sheet**

### Two-Room Estimation By Center Line Method

For a** two-room** building with an inner** cross wall, **two junctions will be formed **shown **in fig.

**Total center line length**

= (2 x AB + 2 X BC + EF)

= (2 X 9.50 + 2 X 5.5 + 5.5)

If this **center line length** is multiplied by the **width **and **depth**, the quantity at junctions **P and Q will come** twice, the **quantity **of each being **equal toÂ **( Â½ x width x Depth )

Hence the **quantity **of these **two junctions **must be **deducted **to get the** net centerline length**.

**Net center line length **

**=( Total Centre line length â€“ ( width/ 2) x No. of the junction)**

= 35. 50 â€“ (Â½ x 0.90) x 2

**= 34. 60 m**

## Example of Building Estimation By Center Line Method

**House Estimation by center line method,**

Item No. | Item Description (Work Details) | No | Length (L) | Breadth (B) | Height (H) | Quantity |

1. | Earthwork in excavation for the foundation | 1 | 44.4 | 0.9 | 1.10 | 43.96 |

Total Centre Line Length = 48.9 m | ||||||

No. of Junction = 10 | ||||||

L = 48.9 – (1/2) x 0.90 x 0.90 x 10 | ||||||

L = 44.4 m | ||||||

OR | ||||||

1. | Brickbat cement concrete (1:4:8)for the foundation | 1 | 44.4 | 0.9 | 0.2 | 7.99 m^{2} |

2. | Brick masonry up to plinth in C.M.1:6 | |||||

First Step: L= 48.0-1/2X0.5X10=46.40m | 1 | 46.4 | 0.5 | 0.3 | 6.96 | |

Second Step:L= 48.9-1/2X0.4X10= 46.9m | 1 | 46.9 | 0.4 | 0.3 | 5.63 | |

Third Step:L= 48.9-1/2X0.3X10= 47.4m | 1 | 47.4 | 0.3 | 0.85 | 12.08 | |

Steps: | ||||||

First Step: | 1 | 1.1 | 0.9 | 0.15 | 0.15 | |

Second Step: | 1 | 1.1 | 0.6 | 0.15 | 0.10 | |

Third Step: | 1 | 1.1 | 0.3 | 0.15 | 0.05 | |

For steps L =D1= 1.1m | 24.97 m^{2} | |||||

3. | Brick masonry above plinth up to slab level in C.M. 1:6 | |||||

L= 48.9=1/2X0.2X10= 47.9m | 1 | 47.9 | 0.2 | 3.0 | 28.74 m^{2} | |

Deduction For Doors & Windows | ||||||

D1 | 3 | 1.10 | 0.2 | 2.1 | 1.39 | |

D2 | 2 | 0.90 | 0.2 | 2.1 | 0.76 | |

G1 | 1 | 1.2 | 0.2 | 2.1 | 0.50 | |

W1 | 4 | 1.8 | 0.2 | 1.4 | 2.02 | |

W2 | 1 | 1.2 | 0.2 | 1.4 | 0.34 | |

W3 | 1 | 1.5 | 0.2 | 1.4 | 0.42 | |

V | 2 | 0.6 | 0.2 | 0.6 | 0.14 | |

Deduction for door/ windows | (-)5.57 m^{2} | |||||

Deduction For Linter Above Door & Windows with 15 cm Bearing at each End | ||||||

D1 | 3 | 1.4 | 0.2 | 0.15 | 0.126 | |

D2 | 2 | 1.2 | 0.2 | 0.15 | 0.072 | |

G1 | 1 | 1.5 | 0.2 | 0.15 | 0.045 | |

W1 | 4 | 2.1 | 0.2 | 0.15 | 0.25 | |

W2 | 1 | 1.5 | 0.2 | 0.15 | 0.045 | |

W3 | 1 | 1.8 | 0.20 | 0.15 | 0.054 | |

V | 2 | 0.9 | 0.2 | 0.15 | .054 | |

Deduction for lintel | (-)0.646 m^{2} | |||||

(Same as that obtained in the long wall-short wall method) | Net quantity | 22.52 m^{2} | ||||

OR | ||||||

3. | Smooth plaster inside the rooms and ceilings in C.M. 1:3 | 238.39 m^{2} | ||||

Plaster For Wall | ||||||

Drawing Room | 4 | 4.0 | 3.0 | 48.0 | ||

Bed Room | 2 | 4.0 | 3.0 | 24.0 | ||

2 | 3.0 | 3.0 | 18.0 | |||

Kitchen | 2 | 4.0 | 3.0 | 24.0 | ||

2 | 3.0 | 3.0 | 18.0 | |||

Bath | 2 | 3.0 | 3.0 | 18.0 | ||

2 | 1.7 | 3.0 | 10.2 | |||

W. C. | 2 | 1.3 | 3.0 | 7.8 | ||

2 | 1.1 | 3.0 | 6.6 | |||

In Front Of W. C. | 2 | 1.5 | 3.0 | 9.0 | ||

2 | 1.1 | 3.0 | 6.6 | |||

Ceiling Plaster: | ||||||

Drawing Room | 1 | 4.0 | 4.0 | 16.0 | ||

Bedroom | 1 | 4.0 | 3.0 | 12.0 | ||

Kitchen | 1 | 3.0 | 4.0 | 12.0 | ||

Bath | 1 | 3.0 | 1.7 | 5.1 | ||

W. C. | 1 | 1.3 | 1.1 | 1.43 | ||

In front of W. C. | 1 | 1.5 | 1.1 | 1.65 | ||

238.39 m^{2} | ||||||

Deduction For Plaster: | ||||||

D1 | 5/2 | 1.1 | 2.1 | 5.78 | ||

D2 | 4/2 | 0.9 | 2.1 | 3.78 | ||

G1 | 2/2 | 1.2 | 2.1 | 2.52 | ||

W1 | 4/2 | 1.8 | 1.4 | 5.04 | ||

W2 | 1/2 | 1.2 | 1.4 | 0.84 | ||

W3 | 1/2 | 1.5 | 1.4 | 1.05 | ||

(-)9.01 m^{2} | ||||||

Net quantity | 219.37 m^{2} |

### Important Notes:

- The area of
**ventilation**is less than**0.5 sq m,**therefore**no deduction**is made. - For Doors,
**Half**the number of faces is**deducted**. - For calculation of
**internal plaster**, inner doors have two**faces**and the external door has**one inner face** - For
**calculation**of internal plaster,**windows**have one internal face. - for windows,
**half the number**of faces is**deducted**.

**Reference & Credit:** **ESTIMATING COSTING AND VALUATION** by Dr. *R P Rethaliya*

**FAQs:**

### what do you mean by the center-line method?

In this **method**, sum total length of the center lines of walls, long and short has to be found out. Find the total length of the **center lines **of walls of the same type.

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