Hi Gerhard,
OK I understand, and I see the difficulty in making a call on this.
So my gut feel would be to run with the lower -r, perhaps 0.5.
Cheers...
Milton Taylor wrote:The parameter controls the trade-off between smoothness and fitting accuracy.Hi Graeme,
I'm interested in finding out more about the -r option. I'm remaking my film scanner profile, and this switch seems to have quite a dramatic effect on the results.Leaving it off, i.e. default value, gives the best results in terms of avg and peak dE. Going from -r 0.5 to -r 0.75 and the error doubles. At -r 1.0 it's even worse.This depends mostly on your particular device and your particular measurements. If your device and your measurements were perfectly noise-free and perfectly accurate, and if the reproducibility of your device would be perfect, then indeed rather a low r (resulting in a low dE fitting error) would be desired.
You indicated that the default was too low, so why select a higher -r value if it gives a profile that is apparently not as good?
Just for info, in my case, -qh gave these results: r = 0.25 avg dE=0.38 peak 2.8 r = 0.50 avg dE=0.48 peak 3.7 r = 0.75 avg dE=1.0 peak 7.8 r = 1.0 avg dE=1.6 peak 9.5
Which is the 'right' value of -r to use?
However, when fitting noisy data (which are "contaminated" with random measurement errors, device reproducibility errors, etc), then an allegedly good fit (low dE) may be in fact not as good as it seems, and just fit the noise in the measurements, while a somewhat smoother profile with a larger dE may indeed predict the actual average device behaviour more accurately.
Since typical real-world devices and measurements are never perfect, your r=0.75 (or maybe even the 1.0) result looks more plausible to me than your r=0.25 result (which I'd rather expect to overfit the data).
On the other hand, if you would create a profile from a ti3 file which has been generated by averaging 16 (independent) measurement sets, then for this profile I would rather expect a fitting error like you got it with r=0.25, since averaging 16 independent measurements reduces random noise by a factor of 4 = sqrt(16).
Regards, GerhardCheers, Milt
0.75 (or maybe even the 1.0) result looks more plausible to me than your r=0.25 result (which I'd rather expect to overfit the data). <br> <br> On the other hand, if you would create a profile from a ti3 file which has been generated by averaging 16 (independent) measurement sets, then for this profile I would rather expect a fitting error like you got it with r=0.25, since averaging 16 independent measurements reduces random noise by a factor of 4 = sqrt(16). <br> <br> Regards, <br> Gerhard <br> <blockquote type="cite">Cheers, <br> Milt <br> </blockquote> <br> <br> </blockquote> </body> </html>