Are Sensor Pixels Square?

Started Oct 9, 2012 | Discussions thread
Stephen Barrett
Regular MemberPosts: 495Gear list
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I Was Wrong
In reply to Draek, Oct 10, 2012

The tentative conclusions in my previous posting are incorrect.  The macro magnification close to the lens-barrel is extremely sensitive to distance and focusing.  In my photos of the clear plastic ruler, the horizontal measurements were made on different photos.  The following photo (Macro with object touching the lens) is used to determine horizontal and vertical image size in a single photo:

Calibration of SX30 Macro Magnification with Object Touching the Lens

(Small size for posting)

The "object" diameter at the outer edge of the radiating lines is 9.90 +/- 0.03 mm both vertically and horizontally.  I used the Canon Image Browser with the "Image Viewer" at its maximum size, which was a frame of 265 mm x 199 mm on my monitor, in order to calculate image size and magnification.
The Screen-Image Diameter at the outer edge of the radiating lines was 99.5 mm +/- 0.3 mm in both the horizontal and vertical directions, so the horizontal and vertical magnifications are the same within better than 1%.
Since the image is not distorted, it appears that the sensor of the 6.17 mm x 4.55 mm is not completely used because it does not have a 4:3 ratio.  The Canon specs,

http://www.canon.ca/inetCA/arcproducts?m=gp&pid=4812#_030
say there are 14.5 total Megapixels and 14.1 Effective Megapixels on the sensor.
Assuming that the total area of 6.17 mm x 4.55 mm = 28.0735 mm^2 corresponds to 14.5 Total Megapixels and that the "Effective Area" corresponds to 14.1 Effective Megapixels:
Effective Area = (14.1 /14.5) x 28.0735 mm^2 = 27.299 mm^2
                    = 6.033 mm x 4.525 mm effective width and height
                       actually in use on the sensor contributing to the final image.
If we use 4320 x 3240 pixels = 14.0 Megapixels, the effective area becomes
Effective Area = (14.0 /14.5) x 28.0735 mm^2 = 27.1054 mm^2
                     = 6.01 mm x 4.51 mm  (Effective width & height with 4:3 ratio)
Using these numbers, the vertical and horizontal macro magnifications for the photo above are:
Horizontal Magnification = (99.5 / 265) x 6.01 mm / 9.9 mm = 0.228 (for object touching the lens)
Vertical Magnification     = (99.5 / 199) x 4.51 mm / 9.9 mm = 0.228
(The numbers in brackets are the on-screen measurements of image size and frame size, made on my computer monitor.)

So, using "Effective Sensor Size" of 6.01 mm x 4.51 mm, the image magnification on the sensor is less than when the sensor dimensions from the specs are used, and the discrepancy between horizontal and vertical magnifications has been removed.
Does this look reasonable?

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