Given known intrinsic and extrinsic parameters, for an image point with known depth, calculate the point at distance X
As mentioned, we assume that we have a fully calibrated camera, which all intrinsic and extrinsic parameters known, as well as known distances od all the points.
Assuming for a point (100,100,500), how can we calculate the position of the point if it were to have a distance of 1000?
Converting from image point to 3d world coordinate point is something I have already implemented, I am interested on how to find the image / 3d world coordinate point of point if its depth wasn't 500, but 1000, the collinear of the initial point for a specific depth.
opencv math camera
asked Nov 10 '18 at 9:21
Konstantine
549
|
show 2 more comments
As mentioned, we assume that we have a fully calibrated camera, which all intrinsic and extrinsic parameters known, as well as known distances od all the points.
Assuming for a point (100,100,500), how can we calculate the position of the point if it were to have a distance of 1000?
Converting from image point to 3d world coordinate point is something I have already implemented, I am interested on how to find the image / 3d world coordinate point of point if its depth wasn't 500, but 1000, the collinear of the initial point for a specific depth.
opencv math camera
asked Nov 10 '18 at 9:21
Konstantine
549
have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
1
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39
|
show 2 more comments
As mentioned, we assume that we have a fully calibrated camera, which all intrinsic and extrinsic parameters known, as well as known distances od all the points.
Assuming for a point (100,100,500), how can we calculate the position of the point if it were to have a distance of 1000?
Converting from image point to 3d world coordinate point is something I have already implemented, I am interested on how to find the image / 3d world coordinate point of point if its depth wasn't 500, but 1000, the collinear of the initial point for a specific depth.
opencv math camera
asked Nov 10 '18 at 9:21
Konstantine
549
As mentioned, we assume that we have a fully calibrated camera, which all intrinsic and extrinsic parameters known, as well as known distances od all the points.
Assuming for a point (100,100,500), how can we calculate the position of the point if it were to have a distance of 1000?
Converting from image point to 3d world coordinate point is something I have already implemented, I am interested on how to find the image / 3d world coordinate point of point if its depth wasn't 500, but 1000, the collinear of the initial point for a specific depth.
opencv math camera
opencv math camera
asked Nov 10 '18 at 9:21
Konstantine
549
asked Nov 10 '18 at 9:21
Konstantine
549
edited Nov 10 '18 at 10:37
asked Nov 10 '18 at 9:21
Konstantine
549
asked Nov 10 '18 at 9:21
Konstantine
549
asked Nov 10 '18 at 9:21
Konstantine
549
549
have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
1
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39
|
show 2 more comments
have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
1
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39
have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
1
1
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39
|
show 2 more comments
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have a look at this: stackoverflow.com/questions/31265245/… to compute the global coordinates you'll have to transform with camera ectrinsics, too.
– Micka
Nov 10 '18 at 9:29
I improved my question a bit. Converting from image points to 3d world coordinate point is not an issue. What I need is to find those coordinates for the collinear of the said in a different depth, in either image or 3d worlds coordinates.
– Konstantine
Nov 10 '18 at 10:38
1
simplest code would be to just replace the depth=500 to depth=1000 and run the "reconstruction" again. However you could instead construct the ray from camera projection center through the 3D point and parametrize this line for a unit and evaluate that line-ewuation for multiple depth values.
– Micka
Nov 10 '18 at 12:30
e.g. direction=(point3d - camera_center)/depth and then: newPoint1000 = camera_center + 1000*direction
– Micka
Nov 10 '18 at 12:33
Is the camera center actually the translation matrix?
– Konstantine
Nov 10 '18 at 12:39