Transcript Slide 1

The Concept of Daylighting
Lighting buildings prior to the late 1800s relied
chiefly on daylight from windows.
Today in America, non-residential buildings
consume nearly 60% of electrical energy. Why
in America and not the rest of the world?
Electric lighting was developed in America
America soon learned it had plenty of energy
Modern building design evolved in America
Energy consumption played a major role in improving
the comfort of users of buildings.
– Until the 1970s. We realized a source of free energy.
The realization of need for conserving energy
brought a new awareness of daylight as free
energy – something that Europe and the rest of
the world knew all along. They did not have the
abundance of petroleum energy that came and
went in America.
Daylighting is a different concept:
– A process of designing from outside in
– A careful consideration of light without glare, and
energy without heat radiation
– Daylight is healthy
– Daylight is free
A concept developed in Europe considered
daylight to be primary, artificial to be secondary
Since most of the workday was during the light
of day, why not utilize daylight for the work
tasks at hand?
Utilize daylight to its maximum and supplement
that with artificial light only as needed.
The P.S.A.L.I Concept -- Permanent
Supplementary Artificial Lighting Inside.
Three Theorems of PSALI :
– 1 Large variation of light during workday
does not adversely affect visual performance.
Lightness Constance
 Variations in daylight occur slowly over a period of
 Contrast ratios remain constant
– 2 Daylight and artificial light are easily
– 3 Supplemental systems must be
coordinated so lighting levels do not change
Example Problem in Daylighting
As with the artificial light solution, consider
Room 102 of the Architecture Building:
– North exposure of windows
– 3 Units, each 5’ wide x 8’ high
– Reflection criteria; 80% ceiling, 50% walls
– Room is 32’ x 31’
– Glass is bronze tinted with transmittance of
– The average available exterior illumination
for Lubbock at 34 degrees north latitude is
10,000 lux on a cloudy, overcast day.
Lubbock is located at 34
degrees north latitude.
(bottom of chart)
Right side of chart is in
thousands of lux of
available light on a
cloudy day.
Intersect lux with
latitude and find
available light is in a
range of 6,000 to 13,000
Use 10,000 lux for the
example problem
The example problem will follow the guidelines
of the C.I.E. method of daylight design. A
European method known as “Commission
Internationale de l’Eclairage”
Originated in Europe, recognized worldwide.
The method assumes the worst condition for
daylight, which is a cloudy, overcast condition,
calculating light levels available. Any
improvement in the condition will mean
additional daylight.
Sequence of Problem:
1 Find the available minimum illumination at a
distance of 20’ from the window.
2 Find the distance from the windows where
the daylight illumination is twice the minimum
3 Find the distance from the windows where
the daylight illumination is four times minimum
4 Determine the percentage of the daylight
workday when the calculated values are
C.I.E. method
considers min
daylight to be
2’ from wall
opposite the
A Determine the Basic Daylight Factor
First, on the window wall, find the total
WIDTH of the wall, then the AGGREGATE width
of the windows. Calculate the percentage
width of windows,
– Width of windows divided by total width of
wall . . .
– The total window width equals 15’.
– the total window wall width is 32’ So,
percentage of windows is 15/32 of window
wall = .47, or 47%. Then on the DAYLIGHT
FACTOR chart, find the point between the
curved, dashed lines that would be 47.
Second, find the depth of the room in terms
of multiples of window heights:
The depth of the room measured from the
window wall is 31 feet. The CIE method
calculates the minimum daylight available a
distance of two feet from the back wall.
The depth of the room equals 31’ – 2’ = 29’.
But in terms of multiples of window height (8’)
= 29/8 = 3.625. In other words, the point
where daylight will be the minimum amount is
3.625 window heights from the window wall.
Enter the DAYLIGHT FACTOR chart to find
basic Daylight Factor.
Third, at the bottom of the DAYLIGHT
FACTOR chart, left to right, find 3.625. Then
follow vertically upward until you reach the
percentage of 47. Then follow horizontally left
to the border of the chart and read the figure,
approximately 1.1.
Remember that this number is a percent,
which represents the percentage of available
daylight on the exterior that will reach the
point 29’ into the room opposite the windows.
but you must consider the clarity of the window
The Basic daylight factor is found to be 1.1,
but since the number is a percentage,
mathematically the number is 0.011.
Since light passes through glass in a
building, corrections must be made in the
daylight factor to compensate for the amount of
light the glass diminishes, plus an allowance for
how much dirt might collect on the outside of
the glass to reduce the amount of light even
Four, find the chart for TRANSMISSIBILITY
of glass:
This is a Correction Factor because the glass is
tinted, which means it does not allow as much
light through as it would if it were clear.
So the correction factor for a transmissibility
of 80% equals 0.95
Apply this correction factor to the basic daylight
Corrected DF (t) = 1.1 x .95 = 1.045
Five, Correction Factor for dirt accumulation.
If the building is located in an area where the
surrounding elements may cause excessive dirt
particles to collect on the glass, in which case
would further diffuse the amount of daylight
that penetrates the glass.
Note that the Architecture Building is located in
a clean outer suburban area. (no industry or
farming) The glass is installed 90 degrees to
horizontal, so dirt does not rest on the surface.
Correction factor from the chart is 0.90
Apply the dirt accumulation correction factor to
the DF corrected for glass transmissibility:
Corrected DF = 1.045 x 0.90 = 0.9405
Remember this number is a percentage, so
mathematically it is written, 0.009405.
The result is the Daylight Factor to be used in
the C.I.E. formula for minimum interior
So the minimum amount of daylight equals,
Corrected Daylight Factor x Available Exterior
From the formula, the available interior
illumination at a point 29’ from the windows
0.009405 x 10,000 = 94.05 lux.
Remember that 10,000 lux is the available
amount of outside daylight.
Six: Convert the available light from lux, to
footcandles, since we use English notation. A
lux is one lumen on one square meter, and a
footcandle is one lumen on one square foot. So
look at the relationship of a square foot and a
square meter.
A square meter is 3.28’ x 3.28’, which equals
10.76 square feet. So, if each area has only one
lumen, that makes a footcandle 10.76 times
brighter than a lux; or a lux equal .092
Since a footcandle is 10.76 times as bright
as a lux, the figure reverts to 94.05 / 10.76 =
8.74 footcandles is the amount of daylight
available at a point 29’ from the window wall.
Logic says that light from the windows is
brightest at the window, and diminishes in
intensity to the minimum amount at the back of
the room.
The C.I.E. method establishes three levels of
light within the room:
– The minimum amount, which is 2’ from the
back wall
– An amount at a point where the DF is twice that
of minimum
– An amount at a point where the DF is four times
Two charts are used to find the distances from
the windows where DF is twice minimum, and
where DF is four times minimum:
Consequently, the amount of available daylight
is the same multiple:
Seven; In the chart
19.18(a) find at the
bottom the point that
is 3.625 window
From that point
follow vertically
until it intersects
with zero degree
Then move to the
left horizontally to
find the distance
of 1.87 window
The distance is equal
to 1.87 x 8’ = 14.9
At a point 14.9 feet from the window wall,
the amount of daylight available is equal to
twice the minimum amount at 29’ from the
window wall.
Eight; Next, calculate from the chart
19.18(b), the distance from the window wall
where the available amount of light will equal
four times the minimum amount at 29’ from the
window wall.
In the chart find at
bottom the point of
3.625 window heights.
From that point follow
vertically until it
intersects with zero
degree obstructions.
Then move to the left
horizontally to find
the distance of 1.14
window heights.
The distance is equal
to 1.14 x 8’ = 9.12 feet
From the calculations, the distances from
window to:
2 x minimum light = 14.9’, which equals 8.74 x 2 = 17.48 FC
4 x minimum light = 9.12’, which equals 8.72 x 4 = 34.96 FC
Realize that none of the 3 figures calculated
are sufficient daylight to serve the room if 70
footcandles are desired.
However, each of the levels of light is a
percentage of that required, which, if
daylighting is primary, then less artificial light,
and therefore energy, is needed for the task.
– At minimum, daylight = 10/70 = 14%
– At 2x min, daylight = 17.48/70 = 25%
– At 4x min, daylight = 34.96/70 = 50%
– The average daylight in the room = 20.81 FC = 30%,
so a saving of 30% of energy can be realized if
daylight is utilized.
A room that is 40’ x 32’ has a windows 6’ tall
on one 40’ side. Total width of the windows is
32 feet. The total correction factor
(transmissibility and dirt accumulation) is
0.855. Available outside light = 10,000 lux.
Basic daylight factor
Corrected daylight factor
Minimum illumination from daylight
Distance where daylight is 2 x minimum
Distance where daylight is 4 x minimum
Room depth is 32’ – 2’ = 30’. Room depth in
terms of window heights = 30 / 6 = 5.0
Percentage of window wall is 32’ / 40’ = 80%.
1 Basic Daylight Factor from chart = 1.0
2 Corrected daylight factor = .855 x 1.0 =
.855, but since it is a percentage, then
mathematically written is 0.0085
3 Minimum illumination = 0.0085 x 10,000 =
85 lux, which = 85/10.76 = 7.9 footcandles.
4 Distance where daylight is twice minimum;
go to 5 window heights on the chart and read
at 0 obstruction, 2.60. So 2.6 x window height
= 2.6 x 6’ = 15.6 feet. Amount of light = 7.9 x
2 = 15.8 footcandles.
5 Distance where daylight is 4 x minimum;
go to 5 window heights on the chart and read
at 0 obstruction, 1.37. So 1.37 x window
height = 1.37 x 6’ = 8.22’. Amount of light =
7.9 x 4 = 31.6 footcandles.
THE FINAL consideration for the P.S.A.L.I.
technique of daylighting is to find how much of
the workday can be expected to produce the
calculated amount of daylight.
The chart labeled, Figure 19.17 maintains
external illumination as a function of latitude
for a given percentage of the normal working
day – assuming the workday period is
approximately centered within the daylight
hours. This concept approximates the
reasoning behind “daylight savings time.”
The chart at right
shows the percentage of the
workday, for which
levels of illumination
will be available.
The latitude of
Lubbock is shown on
a vertical dashed line.
Find 10,000 lux on
vertical column on
right, follow to left
to intersect with
latitude line.
illumination in
Lubbock will be
approx 88% of
the day.