From 98383c6c1a2616704561b8d60a63ea168dfa589d Mon Sep 17 00:00:00 2001 From: Matt Sonic Date: Sat, 11 Sep 2021 07:34:50 -0700 Subject: [PATCH] Update cameraModel.tex Fix typo --- chapters/cameraModel.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/chapters/cameraModel.tex b/chapters/cameraModel.tex index e33b53d..055356b 100644 --- a/chapters/cameraModel.tex +++ b/chapters/cameraModel.tex @@ -53,7 +53,7 @@ \subsection{Pinhole Camera Geometry} X' = f\frac{X}{Z}, \quad Y' = f\frac{Y}{Z}. \end{equation} -Readers may ask why we can arbitrarily move the imaging plane to the front? In fact, this is just a mathematical approach to handle the camera projection, and most of the images captured by the camera are not upside-down. The camera's software will flip the picture for you, so what we actually get is the symmetric plane's image. Although the pin-hole image should be inverted from the physical principle since we have pre-processed the picture, it is not wrong to take the symmetric one. Therefore, without causing ambiguity, we often emit the minus symbol in the pin-hole model. +Readers may ask why we can arbitrarily move the imaging plane to the front? In fact, this is just a mathematical approach to handle the camera projection, and most of the images captured by the camera are not upside-down. The camera's software will flip the picture for you, so what we actually get is the symmetric plane's image. Although the pin-hole image should be inverted from the physical principle since we have pre-processed the picture, it is not wrong to take the symmetric one. Therefore, without causing ambiguity, we often omit the minus symbol in the pin-hole model. The formula~\eqref{eq:P2Pprime} describes the spatial relationship between the point $P$ and its image, where the units of all points are meters. For example, a focal length maybe 0.2 meters, and $X'$ be 0.14 meters. However, in the camera, we end up with pixels, where we need to sample and quantize the pixels on the imaging plane. To describe how the sensor converts the perceived light into image pixels, we set a pixel plane $o-u-v$ fixed on the physical imaging plane. Finally, we get \textit{pixel coordinates} of $P'$ in the pixel plane: $[u,v]^T$.