Plastering & Insulation Formulas

Updated on 03 Sep 2023
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Before reading this article, ensure you have read Measurement - Parts & Labour and Using Formula within Groundplan articles to aid in understanding the following Formula examples.

Please note:

Doubled-sided and single-sided walls would need to be separate Measurements as the Parts and Labour are different.
If there is more than one ceiling height on the Project, each height will need to be a separate Measurement with the correct Formula, Parts and Labour applied.

📝 The Formula values provided in this article are for example purposes only.

Term	Explained
`count`	total counts
`len`	total metres measured
`area`	total square metres measured
`count, len or area / x`	total count/length/area divided by an `amount`
*`count, len or area x`**	total count/length/area multiplied by an `amount`
`ceil(x)`	rounding up a decimal value to the next whole number

Plasterboard

There are options to produce quantities or lengths of materials depending on what is required and how it is charged.

Below are three examples:
1. A per metre rate
2. The size of sheets required
3. Requiring 2 sheets to cover 6m length when wall height is 2.4m

1. Using a per metre rate

Example Parts using a Length Measurement and assign the cost per metre:

Part	Explanation	Formula	UMO
Wall length	How many `total metres` measured	`ceil(len)`	m
Cornice	How many `total metres` measured	`ceil(len)`	m
Insulation (covering 0.58m x 18m)	How many rolls are required when the ceiling is `2.4m high` divided by the area a roll covers `0.58m x 18m = 10.4m2`	*`len2.4/10.4`**	rolls
Total wall area	How many `total metres` measured multiplied by the ceiling height `2.4`	*`len2.4`**	m2

2. Using the size of sheets required

Example Parts for a 6m x 1.2m plasterboard sheet with a wall height of 2.4m Length Measurement:

Part	Explanation	Formula	UMO
Plasterboard 6m x 1.2m	How many sheets are required when `total metres` are multiplied by the ceiling is `2.4m high` divided by the coverage of a sheet `6m x 1.2m = 7.2m2`	*`ceil(len2.4/7.2)`**	sheets
Cornice	How many `total metres` measured	`ceil(len)`	m
Insulation (covering 0.58m x 18m)	How many rolls are required when the ceiling is `2.4m high` divided by the area a roll covers `0.58m x 18m = 10.4m2`	*`len2.4/10.4`**	rolls
Total wall area	How many `total metres` measured multiplied by the ceiling height `2.4`	*`len2.4`**	m2

3. Using 2 sheets required to cover 6m length when wall height is 2.4m

Example Parts for the length of a 6m plasterboard sheet and requiring 2 sheets for a wall height of 2.4m Length Measurement:

Part	Explanation	Formula	UMO
Plasterboard 6m x 1.2m	How many sheets are required for every `6m` multiplied by `2 sheets` required for wall height of 2.4m	*`ceil(len/62)`**	sheets
Cornice	How many `total metres` measured	`ceil(len)`	m
Insulation (covering 0.58m x 18m)	How many rolls are required when the ceiling is `2.4m high` divided by the area a roll covers `0.58m x 18m = 10.4m2`	*`ceil(len2.4/10.4)`**	rolls
Total wall area	How many `total metres` measured multiplied by the ceiling height `2.4`	*`len2.4`**	m2

Removals

There are different ways to calculate window and door removals using Parts with negative formula applied. All involve Grouping of Parts. Alternatively, if usage permits, use the Area Cut-Out Tool to remove windows from elevations Plans as well.

Sliding Door Removal

Example Parts for the removal of a 2400mm x 2200mm sliding door using a Count Measurement:

Part	Explanation	Formula	UMO
Total wall area	Calculating `negative m2` of the total area covered by a sliding door `2400mm x 2200mm = 5.28m2` to be removed from the Total wall area Grouped Part	*`-count5.28`**	m2

Window Removal

Example Parts for the removal of any length window with a height of 1.2m using a Length Measurement:

Part	Explanation	Formula	UMO
Total wall area	Calculating a `negative m2` of the total area covered by a window with a height of `1.2m` to be removed from the Total wall area Grouped Part	*`-len1.2`**	m2