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12/08/2011 11:59:59 PM · #1 |
| 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. |
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12/09/2011 12:39:36 AM · #2 |
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.
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12/09/2011 09:03:26 AM · #3 |
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?
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12/09/2011 09:17:05 AM · #4 |
Originally posted by kirbic:
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? |
That sounds bout right.... |
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12/10/2011 12:24:02 AM · #5 |
Originally posted by kirbic:
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.
Edit to add: Thanks
Message edited by author 2011-12-10 00:24:49. |
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