Analyse af solafskærmninger mht. termiske og ... - Viden om vinduer

Z 2
2
min
sin F fra d Z 0.0646
The part covered by the scan gave a (partial) hemispherical optical property of Rm, in this
case 0.220. The corrected value is therefore Rh,
R m 0.220 R h R m Z R h 0.2846
Example of a Gaussian profile
Suppose that the profile of the spatial distribution is well fitted by a Gaussian function
(Input-værdierne er fundet i excelregnearket rinsax.exl)
G R a R exp a 2
R
2.864910 5
28
a 0.00024 1802
2
where R is the value of the optical property at = 0, (corrected for the detector resolution
) and a is the parameter that determines the width of the profile. The parameter a as
defined above is used here since it is more convenient to determine the fit in the Excel
sheet using in degrees. The factor (180/ ) 2 converts it to the value with in radians, as
required here.
For the parameter values chosen here, the total hemispherical optical property would be
given by
z 2
0
2
sin G R a d z 0.3545
For a partial rectangular scan with and set as above, the missing part of the
For hemispherical a partial rectangular scan with
hemispherical optical property optical is property is
and set as above, the missing part of the
Echo of scan ranges set at the top of the sheet: 34deg 25deg
zz 2
2
min
The ratio is
sin G R a fra d zz 0.2521
zz
z
0.7111
In fact, in this case we don't even need R, since it cancels in the ratio, i.e. we can use the
relative values to fit the data to the Gaussian. We can simply take the measured (partial)