Determination of Field Boundary Line

■B|-VB.2 -4AC <y<-B, + ,/B,2 -4A, C,

2 A

2 A

z = Xx + УУ + H

And

0.009308 I 2 2 ,, 2 0.009308 / 2 2 ,, 2

xє[------------ 2— Vx, + У + Ht, +—2— + У + Ht ].

Put the facula on Plane bt into some area elements dsj ( (j=l,2,3,...is the elements’ number).

Assume that dsj is relatively small enough, and its coordinates is expressed as Sj (x]bi, y]bi, z]bi), so that:

(6)

If the shape of receiver aperture is rectangle, it can be assumed that the height and width of receiver aperture are Lr and Wr respectively. When element ds1ai is in the range of receiver aperture, its coordinates need to satisfy the following condition.

-1W < x}. <1W

2 r a 2 r

L. , L.

-ysinaR < ya, <уsinaR

(7)

When all elements ds]ai are in the range of receiver aperture, Heliostat i is in the range of heliostat

field, otherwise Heliostat i is out of in the range of heliostat field. According by this, calculating and distinguishing all locations in a large area around the tower, field boundary line can be educed.