3

In the Wikipedia article about Homography, the projective transformation between picture A to picture B is defined as:

$H_{ab} = R - \frac{t n^T}{d}$

However, the formula lacks the derivation. How is it derived?

Saurav Maheshkar
  • 135
  • 7
Tommaso Bendinelli
  • 185
  • 1
  • 5

1 Answers1

2

Given a point on a plane $p$, $n^T p$ for the directed unit normal vector for the plane gives you the projection of the given point onto the line of the normal vector, which gives you the distance $-d$ away from the origin on that plane (negative because the normal vector is in the opposite direction to the point vector). i.e.

$$ n^T p = -d $$

For a homography from $a$ to $b$, we want

$$ p_b = H_{ab}p_a = (R_{ab}p_a+t_{ab}) $$

for some rotation and translation. The translation term is annoying because it doesn't have $p_a$ on the right even though H should be a single multiplicative term for p. We have to split up $H$ into a single multiplicative factor for $p$ with two terms, one for the $R$ which doesn't have to change, and one for $t$ for which we need to split out so that multiplying by $p$ results in the $t$ translation.

We do this by looking back at our $n^T p = -d$ equation, dividing by $-d$ on both sides so LHS = 1. The t can now be multiplied by the LHS, which doesn't change the value but introduces the $p$ to be factored out $$ p_b = H_{ab}p_a = (R_{ab}+\frac{t_{ab}n^T}{-d})p \\ = R_{ab}p+ t_{ab}\frac{n^T p}{-d} \\ = R_{ab}P + t_{ab} $$ This recovers our expected relationship

Raggy
  • 128
  • 7