Construction Adhesive Tube Count
Adhesive tubes for a subfloor or panel job, where the bead size sets the coverage.
Example
You enter
- Total bead length (ft) 1200
- Cartridge volume (in³, ~50.6 for 28 oz) 50.6
- Bead diameter (in) 0.375
You get
- Bead area (in²) 0.1105
- Coverage per tube 38.2 ft
- Adhesive tubes 32
Checked against published worked examples from: Henkel Corporation (Loctite).
Details, formula, and sources
ScopeLength per tube = tube volume / (pi/4 x bead dia^2) / 12; tubes = ceil(total / length per tube). A 28 oz tube run as a 3/8 in bead covers 38 ft, so 1,200 LF of joist tops takes 32 tubes; a 1/2 in bead cuts coverage to 21.5 ft and takes 56 tubes - the bead diameter enters squared, so a step up dominates. The spec sets the bead.
Formula or tablebead_area_in2 = (PI/4) x bead_dia_in^2; lf_per_tube = tube_volume_in3 / bead_area_in2 / 12; tubes = ceil(total_lf / lf_per_tube).
Source and editionAdhesive bead-yield identity by name (tube volume divided by the bead cross-section gives the run per tube); first-principles geometry.
Free accessThe bead-yield geometry is public first-principles; the bead diameter follows the manufacturer's nozzle cut and the spec.
What governsEstimate. AHJ and licensed professional govern.
API field namestotal_lf, tube_volume_in3, bead_dia_in, bead_area_in2, lf_per_tube, tubes
- Bead diameter ~3/8 in for subfloor adhesive; the nozzle cut and spec govern (enters squared)manufacturer / spec
- Tube volume ~50.6 in^3 for a 28 fl-oz cartridgeadhesive manufacturer