Point-Method Required Candela for a Target
The luminous intensity a fixture must aim toward a point to hit a target illuminance.
Example
You enter
- Target illuminance at the point 10
- Unit fc
- Mounting height above work plane (ft) 10
You get
- Required intensity toward point 1,000 cd
Checked against published worked examples from: San Diego Gas & Electric Energy Innovation Center (James Donson, PE, LC).
Details, formula, and sources
ScopeI = E x mounting-height^2 / cos(angle)^3 (E in footcandles, lux / 10.764). Hitting 10 fc from 10 ft up straight below needs 1,000 cd; the same 10 fc at a point 30 deg to the side needs about 1,540 cd (the cos^3 penalty off-nadir). Answers 'what candlepower do I need' instead of the illuminance from one fixture. Direct component only; the photometric file and the IES target govern.
Formula or tableE_fc = target (lux / 10.764); required intensity I = E_fc x mount_height^2 / cos(angle)^3. The inverse of E_fc = I x cos(angle)^3 / mount_height^2.
Source and editionPoint-by-point (inverse-square and cosine) method solved for the luminous intensity, IES Lighting Handbook / IESNA point method, by name.
Free accessThe result is the intensity toward the point for the direct component from one source, ignoring interreflection; lux = fc x 10.764. A design relies on the manufacturer's photometric file and the IES target level.
What governsEstimate. AHJ and licensed electrician govern. Verify against the NEC edition adopted in your jurisdiction.
API field namestarget_illuminance, illuminance_unit, mount_height_ft, required_cd
- Inverse-square + cosine the required intensity rises with mount_height^2 and with 1/cos^3(angle) off-nadirIES point method
- Single source the intensity for the direct component from one luminaire, ignoring interreflected lightIES Lighting Handbook