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house on the beach near Acadia National Park, Maine. Canon AE-1, film Kodak Elite Chrome

What happens if the image resolution exceeds the printer's dots per inch? For example, if an image intended to be 4x6 inches is 3600x5400 pixels, then the digital image resolution is 900 pixels per inch. If a printer can print 300 dpi, then each one inch square of image contains 900x900 pixels but must be represented by 300x300 dots of ink. The documentation of my editing software says that to represent multiple pixels by one, down-sampling is performed by taking a weighted average of neighboring pixels. Different methods give different amount of weight to the original pixel vs neighbors. -There is obviously a limit to UP-sampling, as you are adding "fake" data. Is there such a thing as too much DOWN-sampling? -There are a bunch of different algorithms in the options, like bilinear, bicubic etc. Is there a "best" method/setting to choose for reducing image size? Do these algorithms only apply to interpolation methods for INCREASING size, or do they also apply to reducing size? -if I DON'T resize the image, and just send the large file to the printer and ask it to print 4x6, does the printer have its own algorithm to downsize?

Sunset over Washington Street Bridge in Binghamton, NY, 2005. Scan of 35mm slide

View of downtown Milwaukee from Kadish Park Ilford HP5, Canon ae-1, nFD 50mm SC 1.8. Exposure not recorded.

Steve Sweringen at http://www.cameraclinicusa.com/ did a great job on restoring my newly acquired Canon FTB. He works on F1, and knows canon fd cameras very well. He is very knowledgeable and enthusiastic.

Hahaha!!!

I did a series of 40 test exposures on a nikon d3400. Test conditions: -incident and reflected metering using Gossen Luna-Pro SBC -camera in grayscale mode -sRGB color space -target uniformly lit -exposure values given in as stops over/under, with zero exposure anchored at meter reading -mean luminance (from 0-255) of image calculated as 0.2126*R+0.7152*G+0.0722*B Findings: -exposure predicted by hand-held light meter (which is within 1/3 stop of the camera's meter) results in a luminance of around 155-160 (not 128, as I would have though) -change in luminance with change in exposure is not symmetric. Increases to white much faster than decreases to black. I guess this is expected. -Below are some paired exposure values along with luminance values luminance (0-255) : stops over/under light meter exposure 30: -2.9 stops 60 : -1.9 stops 100: -1 stop 130: -0.5 stops 153: 0 stops 195: +0.5 stops 217: +1 stop 240: +1.8 stops

Oops. I made a typo. I meant: "Therefore, the slope of the curve, if linear, from log(Hm) to log(exposure where density 0.8 above Hm) = (0.9-0.1) / 1.3 = 0.6."

The above works if 0.75 is indeed the correct TRANSMISSION density on a NEGATIVE for mid-gray. I think this would be correct if 18% transmission in a negative results in 18% reflectance in the printed image, as [uSER=4774668]@alan_marcus|2[/uSER] has stated (that 18% is the anchor point). I found at least 1 other thread on photo.net in which [uSER=4774668]@alan_marcus|2[/uSER] and [uSER=2403817]@rodeo_joe|1[/uSER] argue this point vociferously. 18% Gray Card I did not come up with the equation for the relationship between iso and proper exposure for mid-gray. I found relationship stated on the website below. I only tried to figure out why it might make sense. It does, if the density of 0.75 is right, and we take the definition of speed point and use a gamma of 0.6. A Practical Guide to Using Film Characteristic Curves | Film Shooters Collective

I am sitting at work not working. So I figured out the source of the formula I cited above log (H-midgray) = log (10/ISO). It comes from Hm, the speed point, and proper developing, as described by rodeo joe, placing mid-gray at 0.75D above baseline+fog, as described by alan marcus, and an assumption of linearity in the density/exposure curve between log(Hm) and log(H-midgray) H(mid-gray): exposure needed for 0.75D above baseline+fog Hm: exposure needed for 0.1D If film properly developed, 1.3 log exposures above log(Hm) yields density of 0.9 Therefore, the slope of the curve, if linear, from log(Hm) to log(H-midgray) = (0.9-0.1) / 1.3 = 0.6 Density of 0.75 is 0.65 units above density at Hm (0.1). 0.65 occurs at 1.1 log exposures above Hm: 0.6*1.1 = 0.65 Hm= 0.8/ISO ==> log Hm = log 0.8 - log iso Log(H-midgray) = log Hm + 1.1 = log 0.8 - log ISO + 1.1 = -0.1 - log ISO + 1.1 = 1 - log ISO = log (10/ISO) So Log (H-midgray) is 1 - log(iso) = log(10/ISO) if the response curve is linear between the speed point and proper exposure for mid-gray QED!!!!!!!!!!

As I read my own question, it occurs to me that it cannot be answered as asked. , It depends on the idiosyncratic response curve of the sensor (film, digital, whatever). For example, some slide films will become almost pure white (luminance 255) with 3 stops of over-exposure. Some other films or sensors may have 9 stops of latitude between what is rendered as pure white vs pure black.

not sure if this is the right forum, but lots of knowledgeable people here, so here goes: As I understand it, luminance of an image on screen varies from 0 (pure black) to 255 (pure white). If the image is color, a formula such as the one below can be used : RGB Luminance value = 0.3 R + 0.59 G + 0.11 B Lets say im using RGB color space. Each color can take on a value 0-255 If I were to properly expose a picture of a grey card, I think it would end up as RGB (130,131,130). My question is the following: If I were to change the exposure of the gray card up or down by x number of stops, can we say what the new RGB value would be? Is there a mathematical conversion for various shades of gray, to convert form stops of exposure to lightness? Thanks.

Got it. Thank you. I've seen another formula for iso (may be used in color neg film? Or maybe differing standards over time?) Iso=10/H(mid) where H(mid) is the exposure that gives properly exposed middle gray. Thus for iso 100, H is 0.1 so log H is -1. The formula based on speed point is new to me. Thanks for sharing.

Oops, meant -1.7 or -1.6, which would be 2 or 2.1 on ilford scale

Thanks for the thoughtful reply. How you determined that hp5's speed point is has an absolute (log) exposure value of - 2.7? From there, i understand that we're making a relative value of 1 = - 2.7. And if the absolute exposure for iso 400 for middle gray is about - 1.8 or- 1.7, then yes, that's 1 or 1.1 x units from the speed point. So about 3.5 stops.

Looking at several different films, it appears that absolute log exposure = - 1 results in a density of 1 for iso 100 SLIDE films. This is the case for kodak e100, fuji velvia 100, and fuji provia 100f. Not the case for color c41 or b&w print films. Maybe because the film's inherent exposure density is different. So my question still remains regarding a film like hp5, which gives its curve in relative log exposure: if a gray card is exposed properly for the iso marked on box, and we develop according to some preset standard, where do we end up on the curve? Forgive me for ignorance regarding "standard" developing as I send my film out

I read many previous posts and could not find a clear answer to my question, although I'm sure someone will find it somewhere, so please excuse. For clarity, I understand logs, fstops, etc. A film like Fuji Velvia gives the x-axis for its characteristic curve in ABSOLUTE terms. Velvia 100 https://asset.fujifilm.com/master/emea/files/2020-10/2f3c7f90a0b0c6e605e84f98b7d489c2/films_velvia-100_datasheet_01.pdf Velvia 50 https://www.ishootfujifilm.com/uploads/VELVIA%2050%20Data%20Guide.pdf For velvia 100, an absolute exposure of log(lux*sec)= -1 results in a density of 1. For velvia 50, the exposure resulting in the same density is approximately -0.7, which is approximately 1 stop more of exposure. Which makes perfect sense. Some films, such as ilford HP5 400, give the x axis as relative log. In either case, I can figure out the approximate exposure latitude of the film in stops, but what about latitude to over vs under exposure? Where is the anchor point for exposure or density, relative to "correct" exposure? Based on the table provided in this website, if (log) exposure is given in absolute terms, -1 yields the correct density for middle gray for iso 100 film. -1.3 for 200, etc. Thus, one can figure out how much under or over exposure can occur before the density stops changing linearly. It may be asymmetric, which is why knowing the "anchor point" is useful. https://www.filmshooterscollective.com/analog-film-photography-blog/a-practical-guide-to-using-film-characteristic-curves-12-25 What about "relative" log exposure? I've seen some kodak reference sheets refer to "log H ref." Thanks all.

This is an incredible portrait. Story?

magnifier s works super well. really helpful. thanks

thanks all. I bought the magnifier S. Coming in the mail tomorrow. Going to try some macro shots with my AE-1 and Kodak Elite Chrome.

Sunrise at Hoan Bridge in Milwaukee, on the shores of Lake Michigan. Canon AE-1, Canon nfd 28/2.8, fuji reala 100

Thank you Niels!

Have been doing macro photography with my ae-1, using extension tubes. As my vision gets worse, getting harder to focus very precisely. Anyone have experience with the Magnifier S? There are some available pretty cheap on ebay. If this would help with focusing, can someone give some guidance on how to mount/attach it? My camera's eyepiece has a rubber cup for glasses. Would I need to change out the the eye piece? Thanks. SS

crap. i did not mean to post it 6 times. very sorry. someon eplese let me know how to delete. I have reported myself.