How To Find Jacobian Given Camera Matrix And 3d Point

I've fabricated some progress.

Start, my camera equation was definitely wrong: I was missing a partitioning by range to object. This video is what caused me to see the light. Really:

$$ \begin{bmatrix} u \\ 5 \end{bmatrix} = K_{2\times3} \cdot \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}/z' $$

where

$$ \brainstorm{bmatrix} x' \\ y' \\ z' \cease{bmatrix} = R_{3\times3} \cdot \begin{bmatrix} x \\ y \\ z \finish{bmatrix} + \vec{T}_{3 \times 1} \tag{1}$$

The primed ($'$) variables represent coordinates of the thing you lot're trying to projection in to camera in the photographic camera'due south reference frame, and the unprimed variables stand for coordinates in the earth frame. I'g trying to find $H$ w.r.t. the world frame.

This makes

$$ u = \frac{\alpha_x x'}{z'} + u_0 $$ $$ v = \frac{\alpha_y y'}{z'} + v_0 $$

If you follow the Jacobian equation that I left above in the question (very carefully, taking derivatives w.r.t. $x$ and friends rather than $ten'$ and other friends) through a folio and a one-half of calculus and simplifying linear algebra, you somewhen end up with

$$ Jacobian_{2\times3} = \frac{KR}{z'} - \frac{K\vec{x'} \otimes R_3}{z'^2} $$

where $ \vec{ten'} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} $, $R_3$ is the 3rd row of the $R$ matrix, and $\otimes$ is an outer production.

Note that you can optionally include translation before rotation in equation (ane), and it makes no difference to the derivatives considering $x$, $y$, and $z$ don't appear in the translation term.

Second, there is a distinction to be made between system evolution and observation functions and the Jacobians. These tend to blur together because in the linear case multiplying by the Jacobian is the aforementioned as evaluating the part. Take the instance $y_1 = ax_1 + bx_2$, $y_2 = cx_1 + dx_2$. Then we tin can write the system as:

$$ \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \brainstorm{bmatrix} x_1 \\ x_2 \end{bmatrix} $$

But likewise

$$ \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\fractional x_2} \\ \frac{\fractional y_2}{\fractional x_1} & \frac{\partial y_2}{\partial x_2} \cease{bmatrix} = \brainstorm{bmatrix} a & b \\ c & d \end{bmatrix} $$

This is convenient if you demand to both notice $\vec{y}$ from $\vec{10}$ and exercise other things like project covariance matrices around, merely in a nonlinear system you do these things past separate methods:

Notice the first line uses $h(10)$, merely on subsequent lines we use the Jacobian $H$ evaluated at $x$. Likewise at the update step we utilize $f(ten)$ to evolve the system and $F$ evaluated at $x$ to update covariance.

Now to specifically answer my questions:

• You lot don't just get a single $H$; you need both $H(\vec{10})$ and $h(\vec{ten})$, evaluated at $\vec{10}$ because the nonlinearity makes the shape vary from place to place.

• No, don't use homogeneous coordinates. This is actually so nonlinear (thanks to that sectionalization past $z'$) that you lot're definitely going to demand to use the Jacobian anyway. Non sure whether a filter could always exist trusted to go along a state variable stationary. If at that place were no noise in that variable and the update equations were just and then, maybe? Irrelevant here now.

• That subtraction isn't..nope.

• I call back linearity isn't actually violated by that additive term, since that's just an offset. But linearity is violated here later on all.

Source: https://stats.stackexchange.com/questions/497283/how-to-derive-camera-jacobian

Posted by: gallawaysagell.blogspot.com

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