QL-6: QL-Conversion CT -> DT-coordinates
Let L1.L2.L3.L4 be the Reference Quadrilateral with defining lines L1, L2, L3, L4.
Let La.Lb.Lc be the Diagonal Triangle of the Quadrilateral with bounding lines La, Lb, Lc.
The construction of the QL-Diagonal Triangle can be found at QL-Tr1.
Let L1.L2.L3 be the random Reference Component Triangle and let L4 be the 4th line.
The QL-Diagonal Triangle La.Lb.Lc is the AntiCevian Triangle of L4 wrt L1.L2.L3, where L4 is the perspectrix.
Conversion of a line
Since we deal with lines in a Quadrilateral we first will convert the equation of a line.
Let R be some line to be converted from CT- to DT-coordinates.
Let Rc (xc : yc : zc) be the presentation of R in barycentric coordinates wrt the Component Triangle.
Let Rd (xd : yd : zd) be the presentation of L in barycentric coordinates wrt the Diagonal Triangle.
Now Rc = xc.cfc1.L1 + yc.cfc2.L2 + zc.cfc3.L3 wrt the Reference Component Triangle
and Rd = xd.cfd1.La + yd.cfd2.Lb + zd.cfd3.Lb wrt the Diagonal Triangle,
where:
- (xc : yc : zc) are the barycentric coordinates of R wrt the Component Triangle,
- (xd : yd : zd) are the barycentric coordinates of R wrt the Diagonal Triangle,
- cfc1, cfc2, cfc3 are the Compliance Factors of the Component Triangle,
- cfd1, cfd2, cfd3 are the Compliance Factors of the Diagonal Triangle.
Explanation of Compliance Factors can be found at (See Ref-26b page 40).
Since the Component Triangle is the Reference Triangle, the Compliance Factors of the Component Triangle are all equal 1.
The Compliance Factors of the Diagonal Triangle are:
- cfd1 = Det [LI, Lb, Lc] / Det [La, Lb, Lc],
- cfd2 = Det [La, LI, Lc] / Det [La, Lb, Lc],
- cfd3 = Det [La, Lb, LI] / Det [La, Lb, Lc],
where LI = the line at infinity (1 : 1 : 1) and “Det” is the abbreviation for “Determinant”.
Calculation gives 2 presentations of the coordinates of R wrt the Component Triangle:
- Rc = (xc : yc : zc),
- Rd = ( l ( l m yd - l n yd + m n yd - l m zd + l n zd + m n zd) :
-m (-l m xd - l n xd + m n xd + l m zd - l n zd - m n zd) :-n (-l m xd - l n xd + m n xd - l m yd + l n yd - m n yd) )
Since Rc and Rd present the same point we can now calculate the coordinates of R wrt the Diagonal Triangle:
- (xd : yd : zd) = ((l m - l n - m n) (l m - l n + m n) (-m n xc + l n yc + l m zc) :
(l m - l n - m n) (l m + l n - m n) ( m n xc - l n yc + l m zc) :(l m + l n - m n) (l m - l n + m n) (-m n xc - l n yc + l m zc))
However we have to bear in mind that the variables in these coordinates are expressions in (a,b,c) and (l,m,n), which are variables wrt the Component Triangle. So also a conversion of these variables from CT to DT should be done.
Therefore the CT > DT-conversion of line R (x : y : z) consists of 3 consecutive steps:
1. Transform Line (x : y : z) -->
((l m - l n - m n) (l m - l n + m n) (-m n x + l n y + l m z) :
(l m - l n - m n) (l m + l n - m n) ( m n x - l n y + l m z) :
(l m + l n - m n) (l m - l n + m n) (-m n x - l n y + l m z))
2. Replace a2 -> l2 (m + n)2 (a2 (l + m) (l + n) + (m - n) (b2 ( l + m) - c2 ( l + n)))
b2 -> m2 (l + n)2 (a2 (l + m) (l - n) + (m + n) (b2 ( l + m) + c2 (-l + n)))
c2 -> n2 (l + m)2 (a2 (l - m) (l + n) + (m + n) (b2 (-l + m) + c2 ( l + n)))
3. Replace l -> m n (l + m) ( l + n)
m -> l n (l + m) (m + n)
n -> l m (l + n) (m + n)
Conversion of a point
Converting the coordinates of a point in the Quadrilateral-environment is converting the intersection of 2 lines.
This actually only changes the Transform of (x : y : z) and leaves the replacements of variables unchanged:
1. Transform Point (x : y : z) -->
((-l m - l n + m n) (m y + n z) :
(-l m + l n - m n) ( l x + n z) :
( l m - l n - m n) ( l x + m y))
2. Replace a2 -> l2 (m + n)2 (a2 (l + m) (l + n) + (m - n) (b2 ( l + m) - c2 ( l + n)))
b2 -> m2 (l + n)2 (a2 (l + m) (l - n) + (m + n) (b2 ( l + m) + c2 (-l + n)))
c2 -> n2 (l + m)2 (a2 (l - m) (l + n) + (m + n) (b2 (-l + m) + c2 ( l + n)))
3. Replace l -> m n (l + m) (l + n)
m -> l n (l + m) (m + n)
n -> l m (l + n) (m + n)