ClueLuma

Required lighting power reduction calculator

A space-based allowance sums area times the applicable density for each space. Reduction is the positive excess of installed power; the supplied limits are scenario criteria, not a code-version claim.

Units
dP=max⁡(Pactual−A1×LPD1−A2×LPD2, 0)\mathit{dP} = \max\left(P_{\text{actual}} - A_{1} \times \mathrm{LPD}_{1} - A_{2} \times \mathrm{LPD}_{2},\ 0\right)

Calculator

Enter the values you know and leave one box empty. Calculate finds the empty one, the result or any input.

W
ft²
W/ft²
ft²
W/ft²
W

Worked example

Given

  • PactualP_{\text{actual}} = 8,000 W
  • A1A_{1} = 5,000 ft²
  • LPD1\mathrm{LPD}_{1} = 1 W/ft²
  • A2A_{2} = 3,000 ft²
  • LPD2\mathrm{LPD}_{2} = 0.5 W/ft²
dP=max⁡(8000−5000×1−3000×0.5, 0)=1,500\mathit{dP} = \max\left(8000 - 5000 \times 1 - 3000 \times 0.5,\ 0\right) = 1{,}500

Symbols and units

SymbolMeaningUnit
dP\mathit{dP}minimum installed lighting power to removeW
PactualP_{\text{actual}}installed lighting input within scopeW
A1A_{1}first eligible space areaft²
LPD1\mathrm{LPD}_{1}stipulated first-space allowanceW/ft²
A2A_{2}second eligible space areaft²
LPD2\mathrm{LPD}_{2}stipulated second-space allowanceW/ft²

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Results are estimates for study and screening. Check the inputs, units and assumptions before you use them for design, billing or contract decisions.