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43mφ3211212312313143m176044020011002830.9m from wall/1 lamplx5020010521052480960144019200.9 m from wall4m3m3milluminance angleilluminance anglelx20001000500200100Rule of the inverse square of the distanceRight side of the graphThe direct horizontal luminance dataindicates the range within which thehorizontal illuminance expressed by thecurve can be obtained, using distance fromthe fixture as the X-axis and horizontaldistance from a point directly underneaththe fixture as the Y-axis.The graph indicates that for a light sourceheight of 2 m, an illuminance of at least 200 lx can be obtained in an area with a radius of 0.5 m.100cd1mLight sourceIlluminated area: Am2Illuminance: 100 lxIlluminated area: 4Am2Illuminance: 25 lxIlluminated area: 9Am2Illuminance: 11.1 lx2m3mThe illuminance of an illuminated surface is inverselyproportional to the square of the distance from the lightsource. Axial luminous intensity （cd）÷Square of distance （m）= Illuminance （lx）Illuminance at point A: 3,600÷22= 900 （lx）Illuminance at point B: 900÷2 = 450 （lx）（1/2 illuminance at point A）Illuminance at point C: 3,600÷32= 400 （lx）Illuminance at point D: 400÷2 = 200 （lx）（1/2 illuminance at point C）Left side of the graphIndicates the relationship between the spread ofthe fixture’s light and its illuminance （lx）. Theangle and light spread shown in the graphindicates 1/2 illuminance, indicating the 1/2illuminance angle （ø） and the center illuminancefor each height level. The graph indicates thatfor a light source height of 2 m, illuminancedirectly under the fixture is about 440 lx, with a1/2 illuminance of φ960 mm.This graph indicates the illuminance distribution for a fixture installed0.9 m from a wall, as shown in the diagram to the right.Calculating illuminance from product dataDirect horizontal illuminanceWall illuminance distribution （included for wall washer models）