E=I (Q φ) CosQ/(PA) 2

Because PA · cosQ=H, PA=H - medical cosQ

So PA2=H2 case - case cosQ (18-2)

I.e. E=I (Q φ)/ H2 · cos2Q

H - height of the light source/m;

Q - The angle between the perpendicular line of the light source to the ground and PA.

In the above example, if we calculate the illuminance of the lamp on point B on the ground (as shown by the dashed line in Figure 18-11) and find ∠ BPO in the right triangle BOP, then:

Tg (90 degree - ∠ BPO)=H/B=7/13=0.54, which can be calculated as ∠ BPO=62 degree , cos ∠ BPO=cos62 degree =0.47

Then, based on the light distribution curve of the lamp DDY400, find the light intensity I of point B (or from the equal light intensity curve). According to the curve, the light intensity at ∠ BPO=62 degree is 66cd. Because this light distribution curve is based on the light flux of the light source of 1000lm, and the light flux of the light source GGY400 we selected is 21500lm, so 66cd needs to be multiplied by 21.5 times. Therefore, the illumination of point B is

E=I (Q φ)/ H2 · cos3Q

=I/H2 · ∠ BPO

=66 × 21.5/72 × zero point four seven three

=6.40 (lx)