All Ex wrote:I`ve got confused in my habit to think with centimeters, after all, the scale in the measurement of the width of the frame is in mm and that got me more into thinking in that way.
When you are clear on the concepts, the Photoshop ruler can also be used to make measurements directly, without counting pixels.
Here I have used Image Size to declare that the width of the image is 36 mm, equal to the sensor size.
Then I use the ruler to measure the distance between two lines that I know are 6 mm apart on the subject.
Photoshop says that the distance is 32.1 mm, and because I set the image size to be the same width as the sensor, I know that this is also 32.1 mm on the sensor.
Then 32.1 mm on sensor, divided by 6 mm on subject, gives an optical magnification of 32.1/6 = 5.35X . This is the same value, to within available precision, as the 5.338X that I got earlier by counting pixels. That's just as it should be, because under the covers all that's happening is that Photoshop is counting pixels and doing part of the calculation for me.
To be super accurate you'd need a flat, aligned subject all in focus
This is certainly true, for some definition of "super accurate".
But it turns out that
1) the
center of a line can be accurately estimated even if the edges of the line are blurred, and
2) the error due to having the target tilted with respect to the optical axis is only a factor of 1-cosine(theta), which even at 8 degrees gives less than 1% error. For smaller tilts the error is proportional to tilt squared. 5 degrees tilt gives only 0.38% error.
The combined effect of these two facts is that accurate measurements can be made even from pretty casual calibration shots, such as the one presented by All Ex.
In this whole debugging exercise, I guess the key thing was to actually
see the calibration image, instead of trusting that somebody else has done the right thing with it. Earlier, when I asked All Ex how he had made the measurement, he replied that "magnification = sensor width/ frame width I use for the calculation of the frame width a ruler which has lines every half of a mm". Those
words did not indicate any problem, but I now understand (I think!) that through confusion he was getting the wrong frame width to plug into the calculation.
--Rik