Plastering & Insulation Formulas
• 03 Sep 2023
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# Plastering & Insulation Formulas

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Article Summary

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.

• 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.

TermExplained
`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: PartExplanationFormulaUMO
Wall lengthHow many `total metres` measured`ceil(len)`m
CorniceHow 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``len*2.4/10.4`rolls
Total wall areaHow many `total metres` measured multiplied by the ceiling height `2.4``len*2.4`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: PartExplanationFormulaUMO
Plasterboard 6m x 1.2mHow 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(len*2.4/7.2)`sheets
CorniceHow 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``len*2.4/10.4`rolls
Total wall areaHow many `total metres` measured multiplied by the ceiling height `2.4``len*2.4`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: PartExplanationFormulaUMO
Plasterboard 6m x 1.2mHow many sheets are required for every `6m` multiplied by `2 sheets` required for wall height of 2.4m`ceil(len/6*2)`sheets
CorniceHow 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(len*2.4/10.4)`rolls
Total wall areaHow many `total metres` measured multiplied by the ceiling height `2.4``len*2.4`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: PartExplanationFormulaUMO
Total wall areaCalculating `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`-count*5.28`m2

## Window Removal

Example Parts for the removal of any length window with a height of 1.2m using a Length Measurement: PartExplanationFormulaUMO
Total wall areaCalculating 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`-len*1.2`m2

For Formula assistance, please reach out to Support. Go to: Help > Send us a Message.

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