Quote Originally Posted by Photo Engineer View Post
Retro...

The sheets can be used that way. The vertical height of the curve is the response in LogE. The "Y" axis is thus the speed at that wavelength. However, the speed is the integral of the entire curve. And so the speed that you measure at one wavelength is infinitely small regardless of the "Y" value unless combined with all of the other values.

This all is related to calculus, I'm afraid. And, the curves have to be smoothed out because they are bumpy due to the dyes themselves.

It is easier for a B&W film than for a color film, due to overlap. Thus, some green light produces red exposure. How do you estimate that? You need to do spectral curves of each dye formed at each wavelength and then begin to calculate what must be done to improve color reproduction.

To do that, you must also include a factor for the human eye vs the measuring system, and you must therefore consider the half band width of the imaging dyes.

If I can find one, I might post a typical picture used for this type of measurement.

PE
Photo Engineer, thanks for the response,

I actually am familiar with calculus, so feel free to discuss more in terms of it, I actually prefer more technical explanations.

Could you work out an example? Lets say I have a spectral power distribution where the spectral irradiance is as follows: between 450 and 451 nm, the spectral irradiance is 7 erg cm^-2 s^-1 nm^-1 and at every other wavelength the spectral irradiance is zero. What would the exposure be for the blue layer in lux-seconds for Kodachrome 25? I would like to see an example worked out so I can find the error I think is present in my calculations. You can just give me the approximate value (no need to actually integrate over that 1nm range, just approximate using a riemann sum of 1 rectangle to keep calculations simple)

I noticed that the exposure in the spectral sensitivity curves for kodachrome 25 is a radiant exposure
but the exposure on the H-D curve is a luminous exposure. Ive accounted for that in my conversion by involving the luminous efficiency function, I think this may be where my mistake lies, theres a detail(s) Im failing to account for.