Stephen Barrett
Leading Member
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The Problem
I am having a recurring problem with discrepancies of approximately 5%. Most often, I notice the problem when I take a shot with a closeup lens and want to know the size of something in the picture. The steps are fairly straightforward:
The first problem occurs in step 2 with knowing the size of the frame on the sensor.
According to Wiki, dpreview and other sources, the size of the 1/2.3" sensor in the SX30, SX40 and SX50 is:
6.17 mm x 4.55 mm [A]
But the dimensions are not in a 4:3 ratio, so the largest possible frame size that will fit on the sensor is:
6.07 mm x 4.55 mm
It is possible that the frame that you see in your picture is smaller than that because the in-camera conversion from RAW to jpeg crops away a bit of the edges. The amount may vary with focal length and with focus distance. My thinking is that the amount of cropping is not significant at the higher focal lengths where distortion should be relatively small.
Steen Bay, replying (Nov 27/12) to one of my postings in this forum, suggested that:
"If we compare the actual FL and the equivalent FL (4.3-215mm vs. 24-1200mm) we'll get a 5.58x crop-factor, and if we divide the diagonal of a 24x36mm sensor (43.27mm) by 5.58 we'll get 7.754mm, meaning that the SX50 sensor should be app. 6.20x4.65mm":
6.20 mm x 4.65 mm [C]
This suggest that Canon may be using a 1/2.3" sensor that is about 2% larger than , which is certainly possible because Canon apparently makes its own sensors and I can find no information online where Canon specifies the sensor dimensions.
Yet another possibility is to calculate sensor size from the crop factor as Steen Bay did, but to use a different definition of crop factor:
"Crop factor can be defined in either of the following two ways:
Using the second definition, the area of the full-frame sensor is 36 mm x 24 mm = 864 mm^2
The area of the sensors used in the SX30, SX40 and SX50, with a crop factor of 5.58, would then be 864 mm^2 / 5.58^2 = 27.75 mm^2, leading to dimensions:
6.08 mm x 4.56 mm [D]
This is very close to so, if Canon is using this less-usual definition of crop factor, they could still be using the usual 1/2.3" sensor with the dimensions of [A] and the 4:3 image would approximately fit on it.
Testing with Closeup Lenses
Most of my test measurements have been at full optical zoom of my SX30 (150.5 mm) with the camera set to infinite zoom. One reason for using infinite zoom is that it allows you to know what the actual focal length of the camera lens is:
"It is a little known fact that telephoto lenses can have a radically different actual focal length depending upon focus distance (Google 'focus breathing' for more info). For example, a Nikon AF-S 28-300mm f/3.5-5.6G ED VR lens set to 300mm will have the expected actual focal length near to 300mm when focused far away -- but has an actual focal length less than half that when focused at 18 inches!" http://www.panohelp.com/lensfov.html
Step 3 for a closeup lens:
The magnification of a closeup lens is given by:
m = f / f_closeup (1)
with the camera set to infinite focus,
where f is the focal length of the camera (mm)
and f_closeup is the focal length of the closeup lens (mm)
The Raynox-250 has a nominal strength of 8 diopters or a nominal focal length of 1000 / 8 = 125 mm.
I performed thorough measurements on my Raynox-250, treating it as a thick lens, to obtain its focal length and location of the principal planes. The actual focal length was 125.0 mm +/- 0.5 mm, in agreement with its nominal value.
Here is a test shot of a reticle with 0.1 mm gradations taken with the Raynox-250 closeup lens on my SX30 with focal length set to 150.5 mm and infinite focus:
FOV width = 5.32 mm with SX30 @ 150.5 mm, infinite focus + Raynox 250
From Equation 1, the magnification is:
m = 150.5 mm / 125.0 mm = 1.204 (2)
The image size, which is the width of the frame on the sensor, is m times the object size (5.32 mm in the photo):
Frame width on sensor = m x 5.32 mm = 6.40 mm (3)
This is more than 5% larger than the the width B or D.
Testing the Camera Alone
With my SX30 set to 150.5 mm and infinite focus, I measured a field of view of 2,530 mm at a distance of E = 59,863 mm from the camera's entrance pupil. http://www.panohelp.com/thinlensformula.html
(For those camera settings, I measured the entrance pupil to be 300 mm behind the camera's mounting screw.
http://www.panohelp.com/lensfov.html ) The half view angle for width is given by:
alpha = arctan ( 0.5 x 2,430 / 59,863) = 1.21 degree (4)
So the full view angle for width is 2.42 degrees.
(Frame width on Sensor) / F = 2 tan (alpha) = 2,530 / 59,863 = 0.0423 (5)
where F = (1 + m/P) f
= f at large distance and infinite focus
So
Frame width on sensor = 0.0423 f (6)
If f = 150.5 mm, Frame width on sensor = 6.37 mm
which, again, is about 5% larger than the width B or D.
Focal Lengths Shorter than Nominal Values?
Here is a possible explanation of why I am obtaining values that appear to be about 5% too high:
"Lenses are rarely exactly as marked when it comes to focal length (or aperture, though that's another issue...), especially telephoto lenses."
"As you can see, it doesn't much matter who made the lens or what the maximum focal length was, in every case the measured maximum focal length is less than the marked maximum focal length, and the average difference is about 5%. The difference is always on the short side. I didn't find any examples where the actual focal length of a telephoto was longer than marked!"
-Bob Atkins, discussing the focal lengths of SLR lenses
Bob Atkins describes the methods that he uses to measure the focal lengths, his "hard way" and his "easy way". His "easy way" is to set the focus to infinity and use:
f = Image size / [2 tan( object angle / 2)].
This is very similar to method E that I used above in Equation 5, except that, with an SLR lens, Atkins was able to make a direct measurement of image size, whereas we are unable to remove our PowerShot lenses to get a direct measurement of the image size. Nevertheless, if SLR telephoto lenses have actual focal lengths that are on average 5% smaller than specified values, it seems likely that this practice may also apply to PowerShot lenses. For example, the maximum equivalent focal length of the SX50 might be only 95% x 1200 mm = 1140 mm.
If f were to be reduced by 5% in my calculations, the magnification value in Equation 2 and the frame width in Equation 2 would also be reduced by 5% and the discrepancy for closeup lenses would vanish. Likewise, for the camera alone, the frame width calculated in equation 6 would also be reduced by 5% and that discrepancy would also vanish.
Conclusion
Use of the nominal value f = 150.5 mm at infinite focus on my SX30 leads to frame sizes on the sensor that are about 5% too large to fit on the typical 1/2.3" sensor.
I am having a recurring problem with discrepancies of approximately 5%. Most often, I notice the problem when I take a shot with a closeup lens and want to know the size of something in the picture. The steps are fairly straightforward:
- Measure the object of interest (a bug, say) on your computer screen and divide by the width (or height) of the full frame on your computer screen.
- Multiply the ratio obtained in step 1 by the width (or height) of the frame on the sensor to get the size of the bug's image on the sensor.
- Divide the image size from step 2 by the magnification to get the size of the bug.
The first problem occurs in step 2 with knowing the size of the frame on the sensor.
According to Wiki, dpreview and other sources, the size of the 1/2.3" sensor in the SX30, SX40 and SX50 is:
6.17 mm x 4.55 mm [A]
But the dimensions are not in a 4:3 ratio, so the largest possible frame size that will fit on the sensor is:
6.07 mm x 4.55 mm
It is possible that the frame that you see in your picture is smaller than that because the in-camera conversion from RAW to jpeg crops away a bit of the edges. The amount may vary with focal length and with focus distance. My thinking is that the amount of cropping is not significant at the higher focal lengths where distortion should be relatively small.
Steen Bay, replying (Nov 27/12) to one of my postings in this forum, suggested that:
"If we compare the actual FL and the equivalent FL (4.3-215mm vs. 24-1200mm) we'll get a 5.58x crop-factor, and if we divide the diagonal of a 24x36mm sensor (43.27mm) by 5.58 we'll get 7.754mm, meaning that the SX50 sensor should be app. 6.20x4.65mm":
6.20 mm x 4.65 mm [C]
This suggest that Canon may be using a 1/2.3" sensor that is about 2% larger than , which is certainly possible because Canon apparently makes its own sensors and I can find no information online where Canon specifies the sensor dimensions.
Yet another possibility is to calculate sensor size from the crop factor as Steen Bay did, but to use a different definition of crop factor:
"Crop factor can be defined in either of the following two ways:
- Crop factor = the ratio of the diagonal dimensions of the respective formats
- Crop factor = the square root of the ratio of the areas of the respective formats."
Using the second definition, the area of the full-frame sensor is 36 mm x 24 mm = 864 mm^2
The area of the sensors used in the SX30, SX40 and SX50, with a crop factor of 5.58, would then be 864 mm^2 / 5.58^2 = 27.75 mm^2, leading to dimensions:
6.08 mm x 4.56 mm [D]
This is very close to so, if Canon is using this less-usual definition of crop factor, they could still be using the usual 1/2.3" sensor with the dimensions of [A] and the 4:3 image would approximately fit on it.
Testing with Closeup Lenses
Most of my test measurements have been at full optical zoom of my SX30 (150.5 mm) with the camera set to infinite zoom. One reason for using infinite zoom is that it allows you to know what the actual focal length of the camera lens is:
"It is a little known fact that telephoto lenses can have a radically different actual focal length depending upon focus distance (Google 'focus breathing' for more info). For example, a Nikon AF-S 28-300mm f/3.5-5.6G ED VR lens set to 300mm will have the expected actual focal length near to 300mm when focused far away -- but has an actual focal length less than half that when focused at 18 inches!" http://www.panohelp.com/lensfov.html
Step 3 for a closeup lens:
The magnification of a closeup lens is given by:
m = f / f_closeup (1)
with the camera set to infinite focus,
where f is the focal length of the camera (mm)
and f_closeup is the focal length of the closeup lens (mm)
The Raynox-250 has a nominal strength of 8 diopters or a nominal focal length of 1000 / 8 = 125 mm.
I performed thorough measurements on my Raynox-250, treating it as a thick lens, to obtain its focal length and location of the principal planes. The actual focal length was 125.0 mm +/- 0.5 mm, in agreement with its nominal value.
Here is a test shot of a reticle with 0.1 mm gradations taken with the Raynox-250 closeup lens on my SX30 with focal length set to 150.5 mm and infinite focus:
FOV width = 5.32 mm with SX30 @ 150.5 mm, infinite focus + Raynox 250
From Equation 1, the magnification is:
m = 150.5 mm / 125.0 mm = 1.204 (2)
The image size, which is the width of the frame on the sensor, is m times the object size (5.32 mm in the photo):
Frame width on sensor = m x 5.32 mm = 6.40 mm (3)
This is more than 5% larger than the the width B or D.
Testing the Camera Alone
With my SX30 set to 150.5 mm and infinite focus, I measured a field of view of 2,530 mm at a distance of E = 59,863 mm from the camera's entrance pupil. http://www.panohelp.com/thinlensformula.html
(For those camera settings, I measured the entrance pupil to be 300 mm behind the camera's mounting screw.
http://www.panohelp.com/lensfov.html ) The half view angle for width is given by:
alpha = arctan ( 0.5 x 2,430 / 59,863) = 1.21 degree (4)
So the full view angle for width is 2.42 degrees.
(Frame width on Sensor) / F = 2 tan (alpha) = 2,530 / 59,863 = 0.0423 (5)
where F = (1 + m/P) f
= f at large distance and infinite focus
Loading…
www.panohelp.com
So
Frame width on sensor = 0.0423 f (6)
If f = 150.5 mm, Frame width on sensor = 6.37 mm
which, again, is about 5% larger than the width B or D.
Focal Lengths Shorter than Nominal Values?
Here is a possible explanation of why I am obtaining values that appear to be about 5% too high:
"Lenses are rarely exactly as marked when it comes to focal length (or aperture, though that's another issue...), especially telephoto lenses."
"As you can see, it doesn't much matter who made the lens or what the maximum focal length was, in every case the measured maximum focal length is less than the marked maximum focal length, and the average difference is about 5%. The difference is always on the short side. I didn't find any examples where the actual focal length of a telephoto was longer than marked!"
-Bob Atkins, discussing the focal lengths of SLR lenses
Focus Breathing (Focal length variation with focus distance)- Bob Atkins Photography
Focus Breathing is a change in focal length, angle of view and/or magnification (depending on your viewpoint) as the focus distance of a lens is changed. Usually it manifests itself as a decrease in focal length/magnification (or an increase in the angle of view) as a lens is focused closer
www.bobatkins.com
Bob Atkins describes the methods that he uses to measure the focal lengths, his "hard way" and his "easy way". His "easy way" is to set the focus to infinity and use:
f = Image size / [2 tan( object angle / 2)].
This is very similar to method E that I used above in Equation 5, except that, with an SLR lens, Atkins was able to make a direct measurement of image size, whereas we are unable to remove our PowerShot lenses to get a direct measurement of the image size. Nevertheless, if SLR telephoto lenses have actual focal lengths that are on average 5% smaller than specified values, it seems likely that this practice may also apply to PowerShot lenses. For example, the maximum equivalent focal length of the SX50 might be only 95% x 1200 mm = 1140 mm.
If f were to be reduced by 5% in my calculations, the magnification value in Equation 2 and the frame width in Equation 2 would also be reduced by 5% and the discrepancy for closeup lenses would vanish. Likewise, for the camera alone, the frame width calculated in equation 6 would also be reduced by 5% and that discrepancy would also vanish.
Conclusion
Use of the nominal value f = 150.5 mm at infinite focus on my SX30 leads to frame sizes on the sensor that are about 5% too large to fit on the typical 1/2.3" sensor.
- One possibility is that Canon uses 1/2.3" sensors that are 5% larger than typical.
- A second possibility is that Canon uses the typical size sensor together with the area-based definition of crop factor and also uses an actual focal length that is about 5% smaller than the nominal value.
- A third possibility is some combination of the first two, such as a 2% larger sensor size with the diagonal-based definition of crop factor (C above) and focal lengths that are 3% smaller than nominal values.
- A fourth possibility is that I have made multiple blunders.
But nice analysis...
