From: Dave Soemarko ^lt;*fotodave@dsoemarko.us*>

Date: 09/08/04-12:33:03 PM Z

Message-id: <005701c495d2$49032470$0a808080@wds>

Date: 09/08/04-12:33:03 PM Z

Message-id: <005701c495d2$49032470$0a808080@wds>

Hi Loris,

I am back from work, and I have worked out the formula as follows:

Say we have an x-% dot (and for the derivation I will assume x-% coverage in

ink, not x-% clear area). In an area of 100, you will have x dark area and

(100-x) light area. The dark area has a density of Dmax and the light area

has a density of Dmin.

Since you are measuring print, I will use reflectance. For negative, you can

change it to transmittance. The following notations are used:

R = reflection

D = density

and we know D = log (1/R) or R = 1 / (10 ^ D)

In the dark area, the reflectance is R_dark = x * 1 / (10 * Dmax)

In the light area, the reflectance is R_light = (100 - x) * 1 / (10 * Dmin)

If you always subtract Dmin and consider Dmin to be zero, then R_light

becomes simply R_light = (100 - x)

Since this is covering the area of 100, the actual reflectance is R_act =

(R_dark + R_light) / 100

And the actual density would be D_act = log (1 / R_act)

You can write it all out and simplify the equation if you want to. Or if you

are using spreadsheet, you can use the formulas step by step as shown above.

However, after applying the formula, I found that the difference is small.

For example, if you can print using a dark ink when Dmax is 3.0, then D_act

will be 0.3006; but if you print with Dmax of 1.33, the D_act will be

0.2812. I think it is pretty neglegilble (wrong spelling, I know), so you

can probably just use the simple formula, especially since getting the 50%

area to print as 50% is not really giving you a theoretically right value.

Dave S

----- Original Message -----

From: "Loris Medici" <loris_medici@mynet.com>

To: <alt-photo-process-l@sask.usask.ca>

Sent: Wednesday, September 08, 2004 9:53 AM

Subject: RE: How to translate log density readings to percent?

*> Thanks again for the explanations. Yes I want the formula! :)
*

*>
*

*> Regards,
*

*> Loris.
*

*>
*

*> > -----Original Message-----
*

*> > From: Dave Soemarko [mailto:fotodave@dsoemarko.us]
*

*> > Sent: Wednesday, September 08, 2004 3:57 PM
*

*> > To: alt-photo-process-l@sask.usask.ca
*

*> > Subject: Re: How to translate log density readings to percent?
*

*> >
*

*> >
*

*> > ...
*

*> >
*

*> > But if want the density of 50% with Dmax of 1.33 (or
*

*> > different Dmax), I can give you the formula later. I don't
*

*> > have it written somewhere, so I need to write it out myself,
*

*> > but I need to go to work now, so I will write later.
*

*> >
*

*> >
*

*> > Dave S
*

*>
*

*>
*

*>
*

Received on Wed Sep 8 12:33:28 2004

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