Ellipsoid-of-Revolution to Tangential Plane - Map Projections: Cartographic Information Systems - page 280

Fig. 8.12. Ellipsoidal horizontal plane at the point P 0

In the frame that is located at the point P 0 , let us here derive the spherical coordinates of the
point P from the coordinates { α,β,r } :

X P − X 0 = r cos β cos α, Y P − Y 0 = r cos β sin α, Z P − Z 0 = r cos β,
tan α = Y P −

Y 0

X 0 = r 21 ( X P

− X 0 )+ r 22 ( Y P − Y 0 )+ r 23 ( Z P − Z 0 )

Z 0 ) ,

(8.99)

X P − r 11 ( X P − X 0 )+ r 12 ( Y P − Y 0 )+ r 13 ( Z P −

Z P − Z 0

tan β = ( X P −

Y 0 ) 2 . X 0 ) 2 +( Y P −

Finally, we transform the relative placement vector { X P − X 0 ,Y P − Y 0 ,Z P − Z 0 } E to the relative

X P − X 0 ,Y P − Y 0 ,Z P − Z 0 }

placement vector { E ∗ :

⎡ ⎤ ⎡ ⎤

X P − X 0

X P − X 0

⎣ ⎦ E ∗ ⎣ ⎦ E

Y P − Y 0

= R E ( Λ 0 ,Φ 0 , 0) Y P − Y 0 , (8.100)

Z P − Z 0 Z P − Z 0

tan α =

− sin Φ 0 ( X P − X 0 )+cos Λ 0 ( Y P − Y 0 )
sin Φ 0 cos Λ 0 ( X P

Z 0 ) ,

= (8.101)

− X 0 )+sin Φ 0 cos Λ 0 ( Y P − Y 0 ) − cos Φ 0 ( Z P −

tan β analogous .

The arctan leads to the orientation angle we need. But we have to pay attention to the quadrant
rule . The mapping α ∈ [0 , 2 π ] → tan α is not injective. Therefore, we must apply the quadrant
rule :

Y P − Y 0 positive ,X P − X 0 positive : 1st quadrant 0

≤ α<π/ 2 ,

Y P − Y 0 positive ,X P − X 0 negative : 2nd quadrant π/ 2

≤ α< 0 ,

(8.102)

Y P − Y 0 negative ,X P − X 0 negative : 3rd quadrant π ≤ α< 3 π/ 2 ,

Y P − Y 0 negative ,X P − X 0 positive : 4th quadrant 3 π/ 2

≤ α< 2 π.