the vector of sensor in SVF calculation

radiance-general
1

Dear Radiance experts,

According to the paper "Compagnon, R. (2004). Solar and daylight
availability in the urban fabric. Energy and Buildings, 36(4), 321-328", we
can use Radiance to calculate Sky View Factor by "using a uniform sky model
of arbitrary luminance L and by computing the illuminance Ei without taking
into account the contribution of interreflections to ensure the integration
is performed over the unobstructed part of the sky vault only".

The formula is: SVF = Ei / (PI * L).

We use the following to define the uniform sky with a horizontal diffuse
irradiation of 100 w/m^2:

···

#########################################################
!gensky 6 20 +15.0 -a 1.0 -o 103.0 -m 105 -u -B 100.0 -g 0.0

skyfunc glow sky_glow 0 0 4 1.0 1.0 1.0 0

skyfunc glow ground_glow 0 0 4 1.0 1.0 1.0 0

sky_glow source sky 0 0 4 0 0 1 180

ground_glow source ground 0 0 4 0 0 -1 180
#########################################################

We use the following specification for rtrace to calculate the irradiation
of a sensor point:
aa=0.1, ab=1, ad=4096, ar=512, as=1024

We then divide the irradiation value of the sensor point as obtained by 100
(the unobstructed horizontal diffuse irradiation as specified for the "-B"
option of gensky) and this will give us the SVF of that sensor point.

According to the way of calculating SVF as introduced by Compagnon (2004),
my question is:
Should the vector of the sensor point being measured always point upward to
the zenith (in other words, the xyz vector for the sensor is 0 0 1) no
matter how the normal of that sensor point might oriented to?

For example, if we try to calculate the SVF of the centroid of a tilted
polygon (a non-horizontal and non-vertical plane), should we specify the
vector of the sensor as 0 0 1 or as the xyz vector values of the actual
normal of that polygon?

Thanks!

Ji

2

Hi Ji!

Should the vector of the sensor point being measured always point upward
to the zenith (in other words, the xyz vector for the sensor is 0 0 1)
no matter how the normal of that sensor point might oriented to?

For example, if we try to calculate the SVF of the centroid of a tilted
polygon (a non-horizontal and non-vertical plane), should we specify the
vector of the sensor as 0 0 1 or as the xyz vector values of the actual
normal of that polygon?

sky is the hemisphere above the horizon, and the sky view factor is the fraction of it visible from a given point. As the horizon does not change with the tilt angle of the surface your point is located on, the normal should still point up. If you would rotate the normal, the result would not be the fraction of the sky visible, but include the ground. And as the ground is nothing but a given obstruction of the sky if you do not consider indirect / reflected light, it should not be counted.

I am not sure why the sky is not defined to produce 1 W/m2 in the first step so that you could read out the SVF without any further division, but this is the easiest part of the calculation.

I hope my understanding of the sky view factor is correct..?

Cheers, Lars.

···

From my understanding: You want to calculate the sky view factor. The

3

Thank you for your prompt reply, Lars!

Actually, one of our colleagues who leads me to post this question also has
the same conclusion that for SVF calculation using Radiance, the vector of
the sensor should always be pointing to the zenith regardless of the
orientation of the normal of the surface.

···

On Wed, Sep 1, 2010 at 2:56 PM, Lars O. Grobe <[email protected]> wrote:

Hi Ji!

Should the vector of the sensor point being measured always point upward

to the zenith (in other words, the xyz vector for the sensor is 0 0 1)
no matter how the normal of that sensor point might oriented to?

For example, if we try to calculate the SVF of the centroid of a tilted
polygon (a non-horizontal and non-vertical plane), should we specify the
vector of the sensor as 0 0 1 or as the xyz vector values of the actual
normal of that polygon?

From my understanding: You want to calculate the sky view factor. The sky
is the hemisphere above the horizon, and the sky view factor is the fraction
of it visible from a given point. As the horizon does not change with the
tilt angle of the surface your point is located on, the normal should still
point up. If you would rotate the normal, the result would not be the
fraction of the sky visible, but include the ground. And as the ground is
nothing but a given obstruction of the sky if you do not consider indirect /
reflected light, it should not be counted.

I am not sure why the sky is not defined to produce 1 W/m2 in the first
step so that you could read out the SVF without any further division, but
this is the easiest part of the calculation.

I hope my understanding of the sky view factor is correct..?

Cheers, Lars.

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--
ZHANG Ji 张冀 (PhD) :: Research Fellow :: Centre for Sustainable Asian Cities
:: School of Design and Environment :: National University of Singapore :: 4
Architecture Drive, Singapore, 117566 :: Contact: 65-6516 5046 :: Email:
[email protected]