Illuminance in Complex Situation

UNIVERSITY OF HONG KONG

M.Sc. (Eng) in Building Services

MEBS 6002 Lighting Engineering

Illuminance in complex situation

Whilst the floodlighting diagram is useful for situations where there are few floodlights and the illuminances at only a few points on a single surface are required, it is not suitable for more detail calculations. For this, an illuminance table has to be used and a computer program will be required.

To table the calculation in a systematic way, a coordinate system in 3-dimensions (x, y, z) is used.

In the above diagram,

L denotes the point where the floodlight is positioned,

A is the point where the floodlight is aimed at,

P is the point at where we wish to calculate the illuminance,

C is the point where the camera is located.

Let , , be the unit vectors parallel to x, y, and z axes,

then, vector

vector

and

To find the vertical angle and horizontal angle, we have to relate x, y, z coordinates to a new coordinate system.

1) Shift the x, y, z coordinates to x’, y’, z’ so that L becomes the centre of origin, i.e. (0,0,0) so that x’ = x – xL, and y’ = y – yL

2) Rotate the coordinate until u, z’ plane passes through A and L, i.e. coordinate of A becomes (uA, 0, zA)

Here ζ is the angle of rotation, and

Procedures for finding new u,v,w coordinates:

Consider triangle XPR

PR = y’p sinx, XR = y’p cosx

Consider triangle XLQ

LQ = x’p cosx, XQ = x’p sinx

-vp = XQ – XR

wp = z’p

Alternative diagram for consideration :-

Now consider the floodlight on the u, v, w coordinate system

Let B be the vertical angle (or called V)

Let b be the horizontal angle (or called H)

Let , , be the unit vectors parallel to the u, v, w axes, then

Vector = uT + wT (vT = 0)

= uP + wP

Vector = uA + wA (vA = 0)

Vector = uP + vP + wP

Length LT =

For horizontal angle

and,

or,

where

and

For camera at xC, yC, zC, the angle of incidence between LP and CP is given by

To find the horizontal illuminance on a surface facing upwards, we take the normal as , then

If cosθC, or cosθH is negative, then the facet concerned does not receive light.

A spreadsheet can be written to simplify the calculations.

K.F.Chan (Mr.) Jan 2008 Page 7 of 7