PinholeCameraModel::project3dToPixel is correct?

asked 2016-12-26 08:00:38 -0600

baozhang
39 ●11 ●12 ●14

updated 2018-04-15 14:36:06 -0600

jayess
6155 ●26 ●85 ●92

I'm a beginner about camera geometry.

I would like to understand PinholeCameraModel::project3dToPixel function.

PinholeCameraModel::project3dToPixel

I derived uv_rect like below but the formula are a little different from API.

API tells,

uv_rect.x = (fx()*xyz.x + Tx()) / xyz.z + cx();
uv_rect.y = (fy()*xyz.y + Ty()) / xyz.z + cy();

I derived,

uv_rect.x = (fx()*(xyz.x + Tx())) / xyz.z + cx();
uv_rect.y = (fy()*(xyz.y + Ty())) / xyz.z + cy();

Is The unit of xyz.x and Tx() meter? If so, my answer is not wrong.

Could you tell me the deriving process of these formula?

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answered 2017-03-21 20:07:50 -0600

tfoote

58457 ●128 ●550 ●526 http://www.ros.org/

It's possible there's a bug /transcription error. Can you create a unit test and show it's invalid and see if your solution works correctly? If so I'm sure the maintainer would be interested.

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answered 2018-04-15 08:32:47 -0600

Hakaishin
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The function inverts all points at the midpoint so x and y are inverted. This was at least for me very surprising and I sounds wrong to me. So for anybody having a problem with this function try to invert both axis and see if that helps.

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answered 2017-03-22 04:13:43 -0600

Vincent Rabaud

1111 ●25 ●20

updated 2018-04-15 14:36:53 -0600

jayess
6155 ●26 ●85 ●92

How do you get to that?

[U V W]^T = P * [X Y Z 1]^T   (X=xyz.x, Y=xyz.y ...)

with P something like:

[[ fx 0 0 Tx]
 [ 0 fy 0 Ty]
 [ 0 0  1 0]]

That gives: u = U/W + cx which is what we have in the code no?

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PinholeCameraModel::project3dToPixel is correct?

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