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Cal video... - Page 2

post #31 of 54
So for example, in the following graph, generated by my last precalibration run in HCFR, here is what you see:




The white curve refers to the reference gamma. I would have thought this would be a completely flat line. I have no idea why it changes slightly. Shouldn't this be a flat line?

The cyan curve refers to the average gamma (in my case, 2.2). This is perfectly flat, which makes sense. The average is the average - it cannot change over individual points.

The yellow curve is what I've been asking about. You can see that 10% luminance, it's around 2.1, but by 20%, it's around 2.2
post #32 of 54
The reference gamma you see there is ITU-R BT.1886. You can change this in HCFR preferences to a flat line at a gamma of your choosing.
post #33 of 54
In other words, the yellow curve is my actual measured gamma as calculated using the BT.1886 equation then? So it's not a "target curve"
post #34 of 54
The yellow curve is your actual measured gamma, the white curve is the BT.1886 reference. The yellow curve has no equation applied to it in that sense, it wouldn't change if you changed reference gamma to something linear. After measuring it you would try to match it as closely as possible to whatever reference you've selected. That is only possible with 10pt controls.
post #35 of 54
sorry, I meant white curve not yellow.

But if the white curve is a reference line, shouldn't it be flat? The BT.1886 equation has a constant exponent. Its power relationship does not change as a function of luminance.
post #36 of 54
While the equation does have a constant exponent, it also has a variable in it (input signal level). Obviously the higher the signal level, the higher the solution for the equation, so that's definitely BT.1886. For BT.1886 gamma on the low end is supposed to be lower than on the high end. If you change BT.1886 for normal gamma in HCFR, the reference (white line) is going to be straight.

For more see here:
http://www.itu.int/dms_pubrec/itu-r/rec/bt/R-REC-BT.1886-0-201103-I!!PDF-E.pdf
post #37 of 54
This is the BT.1886 document.

Here is a snapshot of the equations (I used them to create the curves in my earlier figure)



Notice that the power exponent is a constant - it does not change as a function of V (and is recommended to be 2.4).

I'm assuming the reference line is referring to the value of the exponent, in which case it should be a straight line.

Perhaps the reference line isn't a target reference line, but rather a visual indication of how the actual measured curve matches the BT.1886 equation, at various luminances. If the reference line is flat at 2.4, it would mean that my measured curve matches BT.1886 @ 2.4 perfectly.
post #38 of 54
Quote:
Originally Posted by monvo View Post

While the equation does have a constant exponent, it also has a variable in it (input signal level). Obviously the higher the signal level, the higher the solution for the equation, so that's definitely BT.1886. For BT.1886 gamma on the low end is supposed to be lower than on the high end. If you change BT.1886 for normal gamma in HCFR, the reference (white line) is going to be straight.

For more see here:
http://www.itu.int/dms_pubrec/itu-r/rec/bt/R-REC-BT.1886-0-201103-I!!PDF-E.pdf

sorry didn't notice ur edit until after I posted.

Yes, of course the higher the signal level, the higher the solution for the equation. Any monotonically rising function will have that property.

Perhaps there's a misunderstanding of what is meant by the term "gamma" here. I'm taking gamma to mean "exponent".
post #39 of 54
Quote:
Originally Posted by spacediver View Post

sorry didn't notice ur edit until after I posted.

Yes, of course the higher the signal level, the higher the solution for the equation. Any monotonically rising function will have that property.

Perhaps there's a misunderstanding of what is meant by the term "gamma" here. I'm taking gamma to mean "exponent".

Most people use gamma to refer to the transfer function as a whole or the log(Yn)/log(stim) value of a specific point.

Gamma in the mathematical realm does refer to an exponent, which is why people like Charles Poynton talk about the curve as an Electro Optical Transfer Function or EOTF. But every control in every display calls it gamma or a gamma function.
Edited by sotti - 10/3/13 at 11:01am
post #40 of 54
With all the PC applications today and the smart TVs, isn't it time for an on-line calilbration service? Customers can buy or rent the tools they would need and someone on the other end can do a diagnostic and adjustment on the fly.
post #41 of 54
I can't remember what was said in that video but I wonder if the black levels/horizontal bars match the blackness of the tv frame sitting in a fairly dim room.
post #42 of 54
Quote:
Originally Posted by sotti View Post

Most people use gamma to refer to the transfer function as a whole or the log(Yn)/log(stim) value of a specific point.

Gamma in the mathematical realm does refer to an exponent, which is why people like Charles Poynton talk about the curve as an Electro Optical Transfer Function or EOTF. But every control in every display calls it gamma or a gamma function.

By Yn, do you mean the luminance value on the y axis?

Because if I plot log(Yn)/log(stim) of a BT.1886 curve (where Yn = luminance), I get nothing that looks like the white reference curve in the previous image.
post #43 of 54
Quote:
Originally Posted by spacediver View Post

By Yn, do you mean the luminance value on the y axis?

Because if I plot log(Yn)/log(stim) of a BT.1886 curve (where Yn = luminance), I get nothing that looks like the white reference curve in the previous image.

x,y are chromaticity coordinates, Y is the luminance. Yn is normalized luminance (Y/YMax) typically reported 0-1.
Stim would be the stimulus percent also 0-1.

So to calculate Yn for a given stim at a gamma of 2.2 you get Yn = stim^2.2, the formula above unwinds that so you can calculate the exponent for that particular point.
post #44 of 54
Ok I'm still not getting the same curve as that white reference curve.


On the left is a perfect BT.1886 function with an exponent of 2.4, and a black level of 10% of the white level.

On the right is a plot of log(Yn)/log(Stim).


post #45 of 54
On Sam e450 here I tried too have a bright gamma like I could on crts,but it does not work because the brightest (usually little)white patches are often torch white and there's nothing you can do to change it except turn contrast way down but that destroys the picture.I'm now using gamma -3 as it helps with the crappy wash out with bright picture levels,and turning color up to counter dark picture,colors is around 55.brightness now 55.The gamma seems more stable,luminance levels look better,contrast ratio is better..Im not noticing much metallic silver look either.
Edited by Vic12345 - 10/3/13 at 4:47pm
post #46 of 54
Have you viewed The Dark Knight Rises with those settings? The Bat-Pod tunnel scene, 45 - 47 minutes in when the guy runs off the bike. Gamma 2.4 makes the floor look grungy/noisy.
post #47 of 54
I don't watch many movies.more bright sports like hockey,football where the picture is bright and brings out flaws with this tv....I lowered flesh control -4. Too.May have too have the grayscale really hooked up well with the tv too use dark gamma..try raising color.

Edit -3 gamma is too dark.now trying -2.
Edited by Vic12345 - 10/4/13 at 1:02am
post #48 of 54
Quote:
Originally Posted by spacediver View Post

Ok I'm still not getting the same curve as that white reference curve.


On the left is a perfect BT.1886 function with an exponent of 2.4, and a black level of 10% of the white level.

On the right is a plot of log(Yn)/log(Stim).



That might be correct for a display with 10:1 contrast ratio, isn't really realistic as even very mediocre projectors do better than 400:1.

Most displays are around 1000:1, with high contrast displays being 5000:1 or more. I don't know what the black level used in HCFR is, but I'm sure it's something more realistic.
post #49 of 54
Very cool. Looks much more like the reference curve with a 10,000:1 contrast ratio:



edit: I get what gamma means now (in this context). If you take an actual curve, and you take any given point along the stim axis, you can ask the following question: Given the obtained value for luminance at that point, what exponent would have been necessary to produce that luminance value in a pure power law relationship where Luminance = Input^exponent.

Without a black offset, and without any contrast scaling, the BT.1886 equation reverts to a pure power law relationship, and the Gamma would be a flat 2.4 (or whatever value you chose).
Edited by spacediver - 10/3/13 at 8:01pm
post #50 of 54
Quote:
Originally Posted by spacediver View Post

edit: I get what gamma means now (in this context). If you take an actual curve, and you take any given point along the stim axis, you can ask the following question: Given the obtained value for luminance at that point, what exponent would have been necessary to produce that luminance value in a pure power law relationship where Luminance = Input^exponent.

Without a black offset, and without any contrast scaling, the BT.1886 equation reverts to a pure power law relationship, and the Gamma would be a flat 2.4 (or whatever value you chose).

Yep you've got it now.
post #51 of 54
Quote:
Originally Posted by Pres2play View Post

With all the PC applications today and the smart TVs, isn't it time for an on-line calilbration service? Customers can buy or rent the tools they would need and someone on the other end can do a diagnostic and adjustment on the fly.

Answered in your thread about the same subject.
post #52 of 54
Bump in case someone wants to see a new OLED and a
disk cal and then meter/software cal

The green difference fron the minolta and the xrite spectro is curious.
any info about which meter was more accurate?
post #53 of 54
Quote:
Originally Posted by sotti View Post

That might be correct for a display with 10:1 contrast ratio, isn't really realistic as even very mediocre projectors do better than 400:1.

Most displays are around 1000:1, with high contrast displays being 5000:1 or more. I don't know what the black level used in HCFR is, but I'm sure it's something more realistic.

HCFR uses the measured white and black levels to create the target BT. 1886 curve in the plot, so it will change depending on the display being measured.
post #54 of 54
thanks Zoyd, good to know.
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