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Your description of method 2 makes no sense.   

All the values are simple ratio -- darkness of one ink vs darkness of another ink.
So K vs LK and LK vs LLK -- then calculate K vs LLK.
It's just simple algebra ->>    (a/b)*(b/c) = (a/c)

The trouble is that he's got it wrong.  He shows how he interpreted it 
and it's just not correct and not what the PDF says.  
Please stop saying it's an "alternate" interpretation.

Roy -- the designer and author of all this stuff

Dear Roy:

I am sorry to say that the math you suggest cannot be applied in the present case. The problem in your above equation is that the 'b' in the denominator of the first factor is not he same as the 'b' in the nominator of the second factor. You cannot simply replace a, b and c by K, LK and LLK in this equation. The reason is K, LK and LLK are names of ink channels and not numbers! But you need numbers to be inserted in your equation.

Here is the explanation. It is a bit longer than you may want, but I feel I have to do the explanation in very small steps for clarity, and clarity requires precision of expression. I hope to convince you.

When dealing with the calibration plot, we need to understand the printed step wedges as luminosity functions L(K) of the step variable K. Since we have 3 luminosity functions, each for one of the ink channels, I introduce three functions of K which I denote L_K, L_LK and L_LLK:

• L_K(K)
• L_LK(K)
• L_LLK(K)

Please note that the subscript K denotes the ink channel K and not the step variable K which is the argument in the parentheses. Now the independent variable K runs from 0 to 1 (100%=1). We want to find the two values K1 and K2 of the step variable K at which the luminosity function L_LK evaluated at K=1 equals the luminosity function L_K at some unknown K1, and at which the luminosity function L_LLK at K=1 equals the luminosity of L_K at some unknown K2. Mathematically, this means

• L_LK(1)?? = L_K(K1)???? (1)
  
• L_LLK(1) = L_K(K2)???? (2)
  

I have numbered the equations. These 2 equations relate the two luminosities L_LK and L_LLK at K=1 (=100%) (on the left side of the equations) to the luminosity L_K (right side of the equations), which serves as the common reference function (as intended). So we have to solve the two equations for the arguments K1 and K2. This can be done by using Newton's iteration (if one has explicit expressions for the continuous functions - I determined them by fitting an exponential), but it is easier to do this graphically. To this end one has to draw the graphs of the two functions on both sides of equation (1) and find their intersection point. At the intersection point the equation is satisfied and the abscissa of the intersection point yields the value K1. The same can be done for equations 2. Well, here we are done.

Here is the mistake

You claim that one arrives at the same value K2 when multiplying K1 with K3, where K3 is the step K at which the luminosity of L_LLK at K=1 equals the luminosity of L_LK. To find K3 we need to solve the equation

• L_LLK(1) = L_LK(K3)??? (3)
  

This equation can be solved graphically in the same way as above. However, for general nonlinear functions L_K(K), L_LK(K) and L_LLK(K) the claimed relation (multiplication of the fractions K1 and K3)

• K2 = K1*K3???? (4)
  

is not generally satisfied! This is my claim. For a proof I have given an example in my post /g/QuadToneRIP/message/19394 . Shilesh has given another proof in his post /g/QuadToneRIP/message/19398 .

It is interesting, however, that equation (4) is satisfied, if the 3 luminosity functions would be linear in K. (Note that I am talking here of the hypothetical case that one measures a linear behavior of the luminosity functions in the calibration plot. It should not be mixed up with the linearization process in QTR.) The linearity of L_K(k), L_LK(K) and L_LLK(K) is is a very special and hypothetical case. A physicist would call this a 'model'. It is unrealistic for inkjet printing (but it may possibly hold approximately for some type of matrix printers in which each pixel is made of many non-overlapping black squares which ultimately cover the whole pixel completely).?

Here I give the proof of equation (4) for this model. For luminosity functions which are linear in K equations (1) to (3) yield

• L_LK(1)??? = L_K(K1)??? = a - b1*K1???? (5)
  
• L_LLK(1)? = L_K(K2) ?? = a - b1*K2???? (6)
  
• L_LLK(1)? = L_LK(K3)? = a - b2*K3???? (7)
  

where 'a' is the luminosity of the paper white and b1 and b2 and are the constant negative slopes of the (now linear) functions L_K(K) and L_LK(K). Subtraction of the two last equations '(6) - (7)' yields

• 0 = (a - b1*K2)? - (a - b2*K3)

or

• K2 = (b2/b1) * K3?????? (8)
  

Now we evaluate (7) at K=1 to obtain

• L_LK(1) = a - b2

This is now inserted in (5) to obtain

• a - b2 = a - b1*K1

from which we get (b2/b1) = K1. This can now be inserted in (8) to find

• K2 = K1 * K3????? (9)
  

This proofs that the multiplication of fractions is equivalent to directly determining K2, if all luminosity functions are linear in K. But if the luminosity functions are exponential (as it seems to be the case)

• L_ink(K)? =? L_ink(K-> infty) + [a - L_ink(K-> infty)] * exp( -s_ink * K)

where L_ink(K-> infty) is the asymptotic luminosity for large K (saturation) and s_ink the decay rate of the luminosity (equivalent to the slope of the luminosity L_ink(K) at K=0), then (9) does not generally hold. The fact that (9) holds for linear functions can also be derived geometrically using the "Intercept theorem". I hope I did not commit any typos.

I am sorry to have found this misconception. I find it not only on page 4 of the calibration guide , but also in the user guide of Tom Moore. On page 15 one finds:?

" For QuadTone inks, this process is repeated for each lighter ink, comparing it to the next darker ink, calculating its density relative to that ink and then converting it to a density relative to black."?

Presumably, the misconception has not been discovered/discussed previously, because QTR is quite forgiving with respect to selecting the densities for multiple gray inks. So most practitioners may not care and follow their intuition. But you will understand that I had to disproof your statement "The trouble is that he's got it wrong".

Hendrik? --? a beginner in QTR but not a beginner in math ;-) :-)