Based on the pixels per "whatever" that's dumped into a d-slr's sensor. If I were to roll back the 60d to shoot at 10mp. Would this give better results in high iso? Not saying I'm unhappy with the results at 18mp.
I don't think so. The noise would simply be scaled down along with the rest of the image. I believe the reason for the lower resolution option is to conserve storage space.
Strangely, the answer is "yes, it will reduce noise." That's the good news. The bad news is, not all that much. Let's do a thought exercise. Let's say that, due to random noise, the value of each pixel can vary by up to +/- 8. If we reduce the size of the photo by a factor of two in each direction, then every four pixels in the original image become one pixel in the final image. The noise will tend to average out, because equal numbers of the original pixels had brighter-than-average and darker-than-average values.
Mathematically, noise is reduced by the square root of the number of pixels averaged. So for instance, if you averaged four pixels, noise is reduced by a ratio of 1/[sqrt(4)] = 0.5 or 50%. For a reduction from 18Mpx to 10Mpx, noise is reduced by 1/[sqrt(18/10)] = 0.745 (74.5% of original value, a 24.5% reduction).
You didn't expect an actual mathematical answer, did you?
Strangely, the answer is "yes, it will reduce noise." That's the good news. The bad news is, not all that much. Let's do a thought exercise. Let's say that, due to random noise, the value of each pixel can vary by up to +/- 8. If we reduce the size of the photo by a factor of two in each direction, then every four pixels in the original image become one pixel in the final image. The noise will tend to average out, because equal numbers of the original pixels had brighter-than-average and darker-than-average values.
Mathematically, noise is reduced by the square root of the number of pixels averaged. So for instance, if you averaged four pixels, noise is reduced by a ratio of 1/[sqrt(4)] = 0.5 or 50%. For a reduction from 18Mpx to 10Mpx, noise is reduced by 1/[sqrt(18/10)] = 0.745 (74.5% of original value, a 24.5% reduction).
You didn't expect an actual mathematical answer, did you?
Strangely, the answer is "yes, it will reduce noise." That's the good news. The bad news is, not all that much. Let's do a thought exercise. Let's say that, due to random noise, the value of each pixel can vary by up to +/- 8. If we reduce the size of the photo by a factor of two in each direction, then every four pixels in the original image become one pixel in the final image. The noise will tend to average out, because equal numbers of the original pixels had brighter-than-average and darker-than-average values.
Mathematically, noise is reduced by the square root of the number of pixels averaged. So for instance, if you averaged four pixels, noise is reduced by a ratio of 1/[sqrt(4)] = 0.5 or 50%. For a reduction from 18Mpx to 10Mpx, noise is reduced by 1/[sqrt(18/10)] = 0.745 (74.5% of original value, a 24.5% reduction).
You didn't expect an actual mathematical answer, did you?
LOL kirbic I most definitely did not expect such an in-depth mathematical answer. I appreciate the work put into it though :) I'll experiment with it this wkd. Anything to get out of upgrading my harddrives.
