PERIMETER CALCULATION

The calculation methods described below permit fast and accurate calculation of the perimeter of the surfaces to be sealed.

Basic Geometric Shapes



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Joint Width

To determine joint width, you must know:

The following flowchart shows how to establish joint width using this information.

The gap between materials (new surfaces)

The gap between materials must allow for expansion. When it’s too narrow, warping occurs under the summer heat. Some defects can also be hidden by applying sealants (improperly cut clapboards, scratches, etc.). Finally, joints must cover at least 1/8 in. (3 mm.) over each of the surfaces to be joined. Joint width is therefore computed as follows:

gap between the materials (GM) + 1/8 in. (3 mm.)

Outdoors, this width must be compared with how much the materials move. The highest value must be selected for each material to be joined.

Width of the joint to be replaced (already sealed surfaces)

For proper adhesion to the surfaces, the new joint must be a little larger than the joint to be replaced (at least 1/8 in. or 3 mm.). The width of the joint can therefore be computed as follows:

width of the joint to be replaced (WJR) + 1/8 in. (3 mm.).

Outdoors, this width must be compared with how much the materials move. The highest value must be selected for each material to be joined.

Material moves (outdoors)

Outdoors, some materials have more significant thermal movements than others. The following chart shows the required joint widths for various 10-foot long outdoor materials. Those widths depend on the types of materials involved and on the elasticity of the selected sealant. Joint widths showed in gray being more difficult to achieve, we suggest you select a sealant with better elasticity. Attention! The values of this chart do not apply to materials less than 10-foot long such as door and window frames.

Chart of Required Joint Widths
for Various 10-Foot Long Outdoor Materials

The following examples show how to compute joint widths using the previous section’s chart.

Example 1
A customer wants to seal a kitchen counter. Using the following information, establish the joint width required to seal the surfaces.

• Environment: indoors
• Types of surfaces to be sealed: already sealed surfaces
• Width of the joint to be replaced: ¼ in. x ¼ in.

Joint width = WJR + 1/8 in. = ¼ in. + 1/8 in. = 3/8 in. Since we’re indoors, we don’t need to compare this value with the material moves value (MM). The required joint width is therefore 3/8 in. x 3/8 in.

Example 2
A customer wants to seal wooden baseboards. Using the following information, establish the joint width required to seal the surfaces.

• Environment: indoors
• Types of surfaces to be sealed: new surfaces
• Gap between materials: 1/8 in.

Joint width = GM + 1/8 in. = 1/8 in. + 1/8 in. = ¼ in. Since we’re indoors, we don’t need to compare this value with material moves (MM). The required joint width is therefore ¼ in. x ¼ in.

Example 3
A customer wants to seal three pine window frames and a brick wall. Using the following information, establish the joint width required to seal the surfaces.

• Environment: outdoors
• Types of surfaces to be sealed: already sealed surfaces
• Width of the joint to be replaced: 1/4 in. x 1/4 in.
• Elasticity of the selected sealant: 10%

Joint width = WJR + 1/8 in. = ¼ in. + 1/8 in. = 3/8 in. Since we are outdoors, we must compare this value with the material moves value (MM). The pine and the brick require a joint ½ in. x ¼ in. with 10% elasticity. However, the material moves value does not apply for the pine window frames because they are less than 10-foot long. Since the highest value must be used for each material to be joined, the required joint width is therefore 3/8 in. x 3/8 in.

Example 4
A customer wants to seal an aluminum wall and a brick surface. Using the following information, establish the joint width required to seal the surfaces.

• Environment: outdoors
• Types of surfaces to be sealed: new surfaces
• Gap between materials: 1/8 in.
• Elasticity of the selected sealant: 10%

Joint width = GM + 1/8 in. = 1/8 in. + 1/8 in. = ¼ in. Since we are outdoors, we must compare this value with the material moves value (MM). Aluminum and brick require a joint ½ in. x ¼ in. with an elasticity of 10%. Since the highest value must be used for each material to be joined, the required joint width is therefore ½ in. x ¼ in.

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Le recouvrement

To estimate the amounts of sealant required, you must know the coverage rate of the selected sealant. Coverage is defined as the length of joint that the sealant can cover. It depends on the required width joint and is expressed in linear feet or linear meters per 300 ml cartridge. 

For instance, the coverage of a sealant applied 6 mm. x 8 mm. wide is 7.9 linear meters. This means that each cartridge can cover a length of 7.9 meters.

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NUMBER OF CARTRIDGES REQUIRED

To obtain the number of cartridges required to do the work, the perimeter is divided by the selected sealant covering value.

Example

A customer wants to seal a pine garage door frame and a brick wall. Using the following information, find out the number of cartridges required to seal the surfaces.

40 feet ÷ 13.8 feet = 2.9 cartridges, thus 3 cartridges

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EXAMPLE OF HOW TO COMPUTE QUANTITIES

A customer wants to seal three pine window frames and a brick wall. Using the following information, compute the number of cartridges required to seal the surfaces.

1. • Frame dimensions: 4 feet × 4 feet
2. • Environment: outdoors
3. • Types of surfaces to be sealed: already sealed surfaces
4. • Width of the joint to be replaced: 1/4 in. x 1/4 in.
5. • Elasticity of the selected sealant: 35%

One frame P = 2 × (S1 + S2) = 2 × (4 + 4) = 16 feet

All three frames 16 feet × 3 = 48 feet

The required joint width = WJR + 1/8 in. = ¼ in. + 1/8 in. = 3/8 in. Since we are outdoors, we must compare this value with the material moves value (MM). The pine and the brick require a joint ¼ in. x ¼ in. with 35 % elasticity. However, the material moves value does not apply for the pine window frames because they are less than 10-foot long. Since the highest value must be used for each material to be joined, the required joint width is therefore 3/8 in. x 3/8 in.

Coverage in linear feet per 300 ml cartridge for a width of 3/8 in. x 3/8 in. = 13.8 feet.

48 feet ÷ 13.8 feet = 3.5 cartridges, thus 4 cartridges.