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ShrutheeshIR avatar ShrutheeshIR commented on June 12, 2024 1

Hi @jiawei-mo
Thanks. It works!

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jiawei-mo avatar jiawei-mo commented on June 12, 2024

Hi,

Thank you for being interested in this project. The Jacobians are located in ScaleOptimizer.cpp. The code is modified from DSO with SSE instructions, so the readability is not good (apologies for that).

The Jacobians in the paper is much clear. At a high level, it is the image gradient and then the projection by scale. I think your understanding is correct.

To calculate the Jacobians, you simply find the image gradient [du dv], then calculate ∂Π / ∂s; then you get the Jacobian of the pixel intensity w.r.t. scale as [du dv] * ∂Π / ∂s.

If your images are stereo-rectified, you can simplify the ∂Π / ∂s: [x', y', z'] = [x, y, z] the point position; tx is the baseline, ty=tz=0.

Please let me know for further questions.

Best,
Jiawei

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ShrutheeshIR avatar ShrutheeshIR commented on June 12, 2024

Fantastic! Thanks for the quick reply.
By [du dv] * ∂Π / ∂s, do you mean a simple dot product ? (since ∂Π / ∂s has two components, one of which is 0 since ty is zero which means the value is merely du * 'x' component of ∂Π / ∂s), or are both 'du' and the 'dv' components multiplied by the 'x' term of the ∂Π / ∂s ?

To be more clear, say, ∂Π / ∂s = [p, q]. Then is the Jacobian of the pixel intensity
Js = p * du + q * dv ---------(1)
or, is
Js = p * du + p * dv + q * du + q * dv -------- (2)

This is the tiny part where I am unclear.

Thanks!

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jiawei-mo avatar jiawei-mo commented on June 12, 2024

It is a dot product, eq. (1). For the stereo-rectified case, 'dv' is multiplied by zero.

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