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Shift lens vs move body?
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Shift lens vs move body?
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This is a question that sounds basic, but I have yet to find an answer. When using a tilt-shift lens, if you shift the lens up 5 mm, or if you move the camera body up 5 mm, does the camera capture the same image?
tilt-shift
New contributor
add a comment |
This is a question that sounds basic, but I have yet to find an answer. When using a tilt-shift lens, if you shift the lens up 5 mm, or if you move the camera body up 5 mm, does the camera capture the same image?
tilt-shift
New contributor
add a comment |
This is a question that sounds basic, but I have yet to find an answer. When using a tilt-shift lens, if you shift the lens up 5 mm, or if you move the camera body up 5 mm, does the camera capture the same image?
tilt-shift
New contributor
This is a question that sounds basic, but I have yet to find an answer. When using a tilt-shift lens, if you shift the lens up 5 mm, or if you move the camera body up 5 mm, does the camera capture the same image?
tilt-shift
tilt-shift
New contributor
New contributor
New contributor
asked 11 hours ago
Dirk101Dirk101
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4 Answers
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No.
When moving camera body, the change significance is in relation to subject. For example, when taking photo of a person, 5 mm is quite unnoticeable. When shifting lens, the shift significance is in relation to imaging area size. In case of full frame, a 5 mm shift up moves the whole image by 21 percent of the image height.
In principle, shifting a lens allows you to select which part of its image circle will land on sensor. Therefore shift lenses have wider image circles than would be minimally needed for the sensor size from a non-shifting lens. So you can compare it to cropping from an image of a lens with a shorter focal length; the advantage of shift lens is that you can use the sensor area and frame on spot.
add a comment |
The other answers by xenoid and Imre are perfectly correct, but for visual reference, I've created a graphic to display the difference. The blue cone is the camera in the original position, the red demonstrates raising the camera, and the green is a camera in the original position with only the lens raised to the same position.
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
add a comment |
No.
If you shift the camera up 5mm, you shift the captured image by 5 "absoluet" millimeters. For instance, on a building, that would make the picture include or not the thickness of roof tiles.
If you shift the lens by 5mm on a 15mm-high sensor, you shift the capture image by one third of its relative size. If you shoot a building horizontally (to keep parallel verticals) the bottom half of your picture is the ground in front of the building. By shifting lens, you get one third of ground, and two thirds of building so you have more chances to include the full building.
See the diagram here
add a comment |
In the following diagram, the crossed lines represent the light rays. The green square represents the sensor. The shapes represent the physical object and image you wish to capture. In the normal position, the circle is captured on the sensor. A 40mm lens was used to take the photo.
Suppose you wish to capture the square. You could shift the lens up (or shift the sensor down). The sensor captures a different portion of the imaging circle, which contains the square. To take the photo, the 40mm lens was shifted upwards 1-2 cm.
If the imaging circle is too small, you will see its edge when you shift the lens. A 35mm lens was used to take this photo.
Suppose you want to move the camera up, with the sensor and lens in their normal alignment. The distance you'd have to move the camera is about the same as the distance between the circle and the square, which is much greater than the distance the sensor had to be shifted. This is why you can get the framing you want with a tiny shift lens, but might have to lift the camera in a crane. The photo was taken after raising the camera about a foot. The 40mm lens was used.
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
No.
When moving camera body, the change significance is in relation to subject. For example, when taking photo of a person, 5 mm is quite unnoticeable. When shifting lens, the shift significance is in relation to imaging area size. In case of full frame, a 5 mm shift up moves the whole image by 21 percent of the image height.
In principle, shifting a lens allows you to select which part of its image circle will land on sensor. Therefore shift lenses have wider image circles than would be minimally needed for the sensor size from a non-shifting lens. So you can compare it to cropping from an image of a lens with a shorter focal length; the advantage of shift lens is that you can use the sensor area and frame on spot.
add a comment |
No.
When moving camera body, the change significance is in relation to subject. For example, when taking photo of a person, 5 mm is quite unnoticeable. When shifting lens, the shift significance is in relation to imaging area size. In case of full frame, a 5 mm shift up moves the whole image by 21 percent of the image height.
In principle, shifting a lens allows you to select which part of its image circle will land on sensor. Therefore shift lenses have wider image circles than would be minimally needed for the sensor size from a non-shifting lens. So you can compare it to cropping from an image of a lens with a shorter focal length; the advantage of shift lens is that you can use the sensor area and frame on spot.
add a comment |
No.
When moving camera body, the change significance is in relation to subject. For example, when taking photo of a person, 5 mm is quite unnoticeable. When shifting lens, the shift significance is in relation to imaging area size. In case of full frame, a 5 mm shift up moves the whole image by 21 percent of the image height.
In principle, shifting a lens allows you to select which part of its image circle will land on sensor. Therefore shift lenses have wider image circles than would be minimally needed for the sensor size from a non-shifting lens. So you can compare it to cropping from an image of a lens with a shorter focal length; the advantage of shift lens is that you can use the sensor area and frame on spot.
No.
When moving camera body, the change significance is in relation to subject. For example, when taking photo of a person, 5 mm is quite unnoticeable. When shifting lens, the shift significance is in relation to imaging area size. In case of full frame, a 5 mm shift up moves the whole image by 21 percent of the image height.
In principle, shifting a lens allows you to select which part of its image circle will land on sensor. Therefore shift lenses have wider image circles than would be minimally needed for the sensor size from a non-shifting lens. So you can compare it to cropping from an image of a lens with a shorter focal length; the advantage of shift lens is that you can use the sensor area and frame on spot.
answered 10 hours ago
ImreImre
27.6k9 gold badges96 silver badges169 bronze badges
27.6k9 gold badges96 silver badges169 bronze badges
add a comment |
add a comment |
The other answers by xenoid and Imre are perfectly correct, but for visual reference, I've created a graphic to display the difference. The blue cone is the camera in the original position, the red demonstrates raising the camera, and the green is a camera in the original position with only the lens raised to the same position.
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
add a comment |
The other answers by xenoid and Imre are perfectly correct, but for visual reference, I've created a graphic to display the difference. The blue cone is the camera in the original position, the red demonstrates raising the camera, and the green is a camera in the original position with only the lens raised to the same position.
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
add a comment |
The other answers by xenoid and Imre are perfectly correct, but for visual reference, I've created a graphic to display the difference. The blue cone is the camera in the original position, the red demonstrates raising the camera, and the green is a camera in the original position with only the lens raised to the same position.
The other answers by xenoid and Imre are perfectly correct, but for visual reference, I've created a graphic to display the difference. The blue cone is the camera in the original position, the red demonstrates raising the camera, and the green is a camera in the original position with only the lens raised to the same position.
answered 10 hours ago
LightBenderLightBender
7094 silver badges15 bronze badges
7094 silver badges15 bronze badges
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
add a comment |
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
@xiota The imaging circles are not represented at all in this diagram, we are just assuming that the circle is large enough to cover the entire area. This is a map of the extreme edges of the area visible to the sensor using only those rays that pass through the center of the lens. I admit it has been many years since I've worked with a rail camera, but I don't recall a tremendous difference in field of view based on tilt. Either way, this diagram is accurate when all planes are parallel.
– LightBender
9 hours ago
1
1
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
@xiota I think the confusion is that you have drawn your diagrams in relation to the full imaging circle of the lens while I have drawn mine to demonstrate only what the sensor sees. As I stated, my diagram does not show the imaging circle and no tilt is present. The two diagrams merely have a different point of reference. Here I have overlaid the two to demonstrate that both diagrams show exactly the same thing.
– LightBender
16 mins ago
That diagram is helpful.
– xiota
15 mins ago
That diagram is helpful.
– xiota
15 mins ago
add a comment |
No.
If you shift the camera up 5mm, you shift the captured image by 5 "absoluet" millimeters. For instance, on a building, that would make the picture include or not the thickness of roof tiles.
If you shift the lens by 5mm on a 15mm-high sensor, you shift the capture image by one third of its relative size. If you shoot a building horizontally (to keep parallel verticals) the bottom half of your picture is the ground in front of the building. By shifting lens, you get one third of ground, and two thirds of building so you have more chances to include the full building.
See the diagram here
add a comment |
No.
If you shift the camera up 5mm, you shift the captured image by 5 "absoluet" millimeters. For instance, on a building, that would make the picture include or not the thickness of roof tiles.
If you shift the lens by 5mm on a 15mm-high sensor, you shift the capture image by one third of its relative size. If you shoot a building horizontally (to keep parallel verticals) the bottom half of your picture is the ground in front of the building. By shifting lens, you get one third of ground, and two thirds of building so you have more chances to include the full building.
See the diagram here
add a comment |
No.
If you shift the camera up 5mm, you shift the captured image by 5 "absoluet" millimeters. For instance, on a building, that would make the picture include or not the thickness of roof tiles.
If you shift the lens by 5mm on a 15mm-high sensor, you shift the capture image by one third of its relative size. If you shoot a building horizontally (to keep parallel verticals) the bottom half of your picture is the ground in front of the building. By shifting lens, you get one third of ground, and two thirds of building so you have more chances to include the full building.
See the diagram here
No.
If you shift the camera up 5mm, you shift the captured image by 5 "absoluet" millimeters. For instance, on a building, that would make the picture include or not the thickness of roof tiles.
If you shift the lens by 5mm on a 15mm-high sensor, you shift the capture image by one third of its relative size. If you shoot a building horizontally (to keep parallel verticals) the bottom half of your picture is the ground in front of the building. By shifting lens, you get one third of ground, and two thirds of building so you have more chances to include the full building.
See the diagram here
answered 11 hours ago
xenoidxenoid
6,1131 gold badge10 silver badges25 bronze badges
6,1131 gold badge10 silver badges25 bronze badges
add a comment |
add a comment |
In the following diagram, the crossed lines represent the light rays. The green square represents the sensor. The shapes represent the physical object and image you wish to capture. In the normal position, the circle is captured on the sensor. A 40mm lens was used to take the photo.
Suppose you wish to capture the square. You could shift the lens up (or shift the sensor down). The sensor captures a different portion of the imaging circle, which contains the square. To take the photo, the 40mm lens was shifted upwards 1-2 cm.
If the imaging circle is too small, you will see its edge when you shift the lens. A 35mm lens was used to take this photo.
Suppose you want to move the camera up, with the sensor and lens in their normal alignment. The distance you'd have to move the camera is about the same as the distance between the circle and the square, which is much greater than the distance the sensor had to be shifted. This is why you can get the framing you want with a tiny shift lens, but might have to lift the camera in a crane. The photo was taken after raising the camera about a foot. The 40mm lens was used.
add a comment |
In the following diagram, the crossed lines represent the light rays. The green square represents the sensor. The shapes represent the physical object and image you wish to capture. In the normal position, the circle is captured on the sensor. A 40mm lens was used to take the photo.
Suppose you wish to capture the square. You could shift the lens up (or shift the sensor down). The sensor captures a different portion of the imaging circle, which contains the square. To take the photo, the 40mm lens was shifted upwards 1-2 cm.
If the imaging circle is too small, you will see its edge when you shift the lens. A 35mm lens was used to take this photo.
Suppose you want to move the camera up, with the sensor and lens in their normal alignment. The distance you'd have to move the camera is about the same as the distance between the circle and the square, which is much greater than the distance the sensor had to be shifted. This is why you can get the framing you want with a tiny shift lens, but might have to lift the camera in a crane. The photo was taken after raising the camera about a foot. The 40mm lens was used.
add a comment |
In the following diagram, the crossed lines represent the light rays. The green square represents the sensor. The shapes represent the physical object and image you wish to capture. In the normal position, the circle is captured on the sensor. A 40mm lens was used to take the photo.
Suppose you wish to capture the square. You could shift the lens up (or shift the sensor down). The sensor captures a different portion of the imaging circle, which contains the square. To take the photo, the 40mm lens was shifted upwards 1-2 cm.
If the imaging circle is too small, you will see its edge when you shift the lens. A 35mm lens was used to take this photo.
Suppose you want to move the camera up, with the sensor and lens in their normal alignment. The distance you'd have to move the camera is about the same as the distance between the circle and the square, which is much greater than the distance the sensor had to be shifted. This is why you can get the framing you want with a tiny shift lens, but might have to lift the camera in a crane. The photo was taken after raising the camera about a foot. The 40mm lens was used.
In the following diagram, the crossed lines represent the light rays. The green square represents the sensor. The shapes represent the physical object and image you wish to capture. In the normal position, the circle is captured on the sensor. A 40mm lens was used to take the photo.
Suppose you wish to capture the square. You could shift the lens up (or shift the sensor down). The sensor captures a different portion of the imaging circle, which contains the square. To take the photo, the 40mm lens was shifted upwards 1-2 cm.
If the imaging circle is too small, you will see its edge when you shift the lens. A 35mm lens was used to take this photo.
Suppose you want to move the camera up, with the sensor and lens in their normal alignment. The distance you'd have to move the camera is about the same as the distance between the circle and the square, which is much greater than the distance the sensor had to be shifted. This is why you can get the framing you want with a tiny shift lens, but might have to lift the camera in a crane. The photo was taken after raising the camera about a foot. The 40mm lens was used.
edited 2 hours ago
answered 10 hours ago
xiotaxiota
16.6k4 gold badges22 silver badges79 bronze badges
16.6k4 gold badges22 silver badges79 bronze badges
add a comment |
add a comment |
Dirk101 is a new contributor. Be nice, and check out our Code of Conduct.
Dirk101 is a new contributor. Be nice, and check out our Code of Conduct.
Dirk101 is a new contributor. Be nice, and check out our Code of Conduct.
Dirk101 is a new contributor. Be nice, and check out our Code of Conduct.
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