View attachment 7346
I decided to do a pixel count to get a better estimate of the two planes. Attached above is the picture I used with pixels marked off so that you can dispute my methods and do your own estimates if you so choose. Each dot was one pixel, and each black line represents 5. I zoomed in on MS paint to mark each pixel and to figure out where the plane ended and the asphalt began.
I first started off with the J-15, which should share the same length as the Su-33, 21.15 m. Counting from the tip of the nose to the tip of the tail, the J-15 was about 44 pixels long. Divided by 21.15 I got roughly .48 meters per pixel (rounded). This seemed pretty close to 0.5 meters per pixel, so I decided to round the J-15's length up to 22 meters (divided 44) in order to get .5 meters per pixel, for the sake of this exercise.
The J-20 seemed to be about 43 pixels long. What was clear when zoomed in was that the J-20 was slightly shorter than the J-15. If we extrapolate this out using the resolution of .48 meters per pixel estimate the J-20 was 20.64 meters long. If we assume a resolution of .5 meters per pixel then the J-20 should be about 21.5 meters long.
It made sense to round the pixels up to .5 meters because what I've heard is that the resolution of satellite photos are often set to very clean ratios for the sake of getting a clean measure of distance and sizes. Now, by all reasons, given the J-15's real length, assuming it is the same as the Su-33's, the J-15 should have been 42-43 pixels long (and closer to 42) if the resolution of the picture is indeed meant to be .5 meters a pixel. One reason the count could have been higher is due to a slight angle of the plane's position relative to the orthogonal plane of the camera, which would have registered either or both the nose or/and the tail to an additional pixel each (their pixel being a combination of that and everything else in that .5 meter square) when being caught by the camera. (This is the peril of using a low resolution photo to do estimations).
What is clear though is that the J-20 seems to be tilted in a similar manner. Therefore, the size difference of about 1 pixel should remain roughly the same (might be slightly bigger, but in an exercise with this high of an error the difference should almost be negligent), which means that the J-20 is about half a meter shorter than the J-15. Thus assuming a resolution of .5 meters a pixel, corrected for the tilt the J-20 should be about 41-42 pixels long (probably closer to 41), and therefore 20.5-21 meters long.
I decided to do a pixel count to get a better estimate of the two planes. Attached above is the picture I used with pixels marked off so that you can dispute my methods and do your own estimates if you so choose. Each dot was one pixel, and each black line represents 5. I zoomed in on MS paint to mark each pixel and to figure out where the plane ended and the asphalt began.
I first started off with the J-15, which should share the same length as the Su-33, 21.15 m. Counting from the tip of the nose to the tip of the tail, the J-15 was about 44 pixels long. Divided by 21.15 I got roughly .48 meters per pixel (rounded). This seemed pretty close to 0.5 meters per pixel, so I decided to round the J-15's length up to 22 meters (divided 44) in order to get .5 meters per pixel, for the sake of this exercise.
The J-20 seemed to be about 43 pixels long. What was clear when zoomed in was that the J-20 was slightly shorter than the J-15. If we extrapolate this out using the resolution of .48 meters per pixel estimate the J-20 was 20.64 meters long. If we assume a resolution of .5 meters per pixel then the J-20 should be about 21.5 meters long.
It made sense to round the pixels up to .5 meters because what I've heard is that the resolution of satellite photos are often set to very clean ratios for the sake of getting a clean measure of distance and sizes. Now, by all reasons, given the J-15's real length, assuming it is the same as the Su-33's, the J-15 should have been 42-43 pixels long (and closer to 42) if the resolution of the picture is indeed meant to be .5 meters a pixel. One reason the count could have been higher is due to a slight angle of the plane's position relative to the orthogonal plane of the camera, which would have registered either or both the nose or/and the tail to an additional pixel each (their pixel being a combination of that and everything else in that .5 meter square) when being caught by the camera. (This is the peril of using a low resolution photo to do estimations).
What is clear though is that the J-20 seems to be tilted in a similar manner. Therefore, the size difference of about 1 pixel should remain roughly the same (might be slightly bigger, but in an exercise with this high of an error the difference should almost be negligent), which means that the J-20 is about half a meter shorter than the J-15. Thus assuming a resolution of .5 meters a pixel, corrected for the tilt the J-20 should be about 41-42 pixels long (probably closer to 41), and therefore 20.5-21 meters long.
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