Obtaining depth to an object of known dimensions from a monocular calibrated camera
I am using notation defined in this page: Camera Calibration
In summary:
- The standard frame is the frame that has the origin at the projection center and z axis pointed "out" from the lens.
- The image frame is the frame onto which the image is projected. This is what we get measurements from. It sits in front of the standard frame.
I know from the derivation on that page that we can convert the image points to coordinates in the standard frame modulo some scaling. That is
$[x_i, y_i]$ in image frame will be $z_s[x_s, y_s, 1]$ in the standard frame, where $z_s$ is unknown or arbitrary. So we can't get depth information.
However, assume that I observe four points in the image, and I know the distance between these four points in the standard frame. So the four points in the image frame are:
$p_i(1)$, $p_i(2)$, $p_i(3)$, and $p_i(4)$,
where each $p_i$ has just $x$ and $y$ coordinates in the image frame and is of the form $[x,y]$. I convert them to the standard frame
$z_1*p_s(1)$, $z_2*p_s(2)$, $z_3*p_s(3)$, $z_4*p_s(4)$,
where each $p_s$ has the $x$ and $y$ coordinates in the standard frame and is of the form $z[x,y,1]$, and $z$ is the unknown scale factor. The question is, does knowing the distance between these points (in the standard frame) help me solve for $z_1$, $z_2$, $z_3$, and $z_4$? Or is there something fundamental I'm missing here?
computer-vision
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
add a comment |
I am using notation defined in this page: Camera Calibration
In summary:
- The standard frame is the frame that has the origin at the projection center and z axis pointed "out" from the lens.
- The image frame is the frame onto which the image is projected. This is what we get measurements from. It sits in front of the standard frame.
I know from the derivation on that page that we can convert the image points to coordinates in the standard frame modulo some scaling. That is
$[x_i, y_i]$ in image frame will be $z_s[x_s, y_s, 1]$ in the standard frame, where $z_s$ is unknown or arbitrary. So we can't get depth information.
However, assume that I observe four points in the image, and I know the distance between these four points in the standard frame. So the four points in the image frame are:
$p_i(1)$, $p_i(2)$, $p_i(3)$, and $p_i(4)$,
where each $p_i$ has just $x$ and $y$ coordinates in the image frame and is of the form $[x,y]$. I convert them to the standard frame
$z_1*p_s(1)$, $z_2*p_s(2)$, $z_3*p_s(3)$, $z_4*p_s(4)$,
where each $p_s$ has the $x$ and $y$ coordinates in the standard frame and is of the form $z[x,y,1]$, and $z$ is the unknown scale factor. The question is, does knowing the distance between these points (in the standard frame) help me solve for $z_1$, $z_2$, $z_3$, and $z_4$? Or is there something fundamental I'm missing here?
computer-vision
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
add a comment |
I am using notation defined in this page: Camera Calibration
In summary:
- The standard frame is the frame that has the origin at the projection center and z axis pointed "out" from the lens.
- The image frame is the frame onto which the image is projected. This is what we get measurements from. It sits in front of the standard frame.
I know from the derivation on that page that we can convert the image points to coordinates in the standard frame modulo some scaling. That is
$[x_i, y_i]$ in image frame will be $z_s[x_s, y_s, 1]$ in the standard frame, where $z_s$ is unknown or arbitrary. So we can't get depth information.
However, assume that I observe four points in the image, and I know the distance between these four points in the standard frame. So the four points in the image frame are:
$p_i(1)$, $p_i(2)$, $p_i(3)$, and $p_i(4)$,
where each $p_i$ has just $x$ and $y$ coordinates in the image frame and is of the form $[x,y]$. I convert them to the standard frame
$z_1*p_s(1)$, $z_2*p_s(2)$, $z_3*p_s(3)$, $z_4*p_s(4)$,
where each $p_s$ has the $x$ and $y$ coordinates in the standard frame and is of the form $z[x,y,1]$, and $z$ is the unknown scale factor. The question is, does knowing the distance between these points (in the standard frame) help me solve for $z_1$, $z_2$, $z_3$, and $z_4$? Or is there something fundamental I'm missing here?
computer-vision
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
I am using notation defined in this page: Camera Calibration
In summary:
- The standard frame is the frame that has the origin at the projection center and z axis pointed "out" from the lens.
- The image frame is the frame onto which the image is projected. This is what we get measurements from. It sits in front of the standard frame.
I know from the derivation on that page that we can convert the image points to coordinates in the standard frame modulo some scaling. That is
$[x_i, y_i]$ in image frame will be $z_s[x_s, y_s, 1]$ in the standard frame, where $z_s$ is unknown or arbitrary. So we can't get depth information.
However, assume that I observe four points in the image, and I know the distance between these four points in the standard frame. So the four points in the image frame are:
$p_i(1)$, $p_i(2)$, $p_i(3)$, and $p_i(4)$,
where each $p_i$ has just $x$ and $y$ coordinates in the image frame and is of the form $[x,y]$. I convert them to the standard frame
$z_1*p_s(1)$, $z_2*p_s(2)$, $z_3*p_s(3)$, $z_4*p_s(4)$,
where each $p_s$ has the $x$ and $y$ coordinates in the standard frame and is of the form $z[x,y,1]$, and $z$ is the unknown scale factor. The question is, does knowing the distance between these points (in the standard frame) help me solve for $z_1$, $z_2$, $z_3$, and $z_4$? Or is there something fundamental I'm missing here?
computer-vision
computer-vision
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
asked Dec 28 '18 at 0:01
Mr. Fegur
517313
517313
add a comment |
add a comment |
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