Life Below the Equator or Splitting the Third Dimension
So I ran the colour test and got some good news and bad news. First, the bad news. As indicated below I got A LOT of white space at the top of the chart. More than usual on the "B" chart (the blue one) and not as bad on the "A" chart (the green) one. The HSL Array gets you in the neighbourhood lets say, but doesn't always knock on the door for you. Luckily, the good news, printing the ChartThrob 101 Step Chart knocks on the door -- the gradations are much finer -- it pin points the starting block quite accurately. At this point you could just ignore everything I've written below and let the curve fix everything for you. Or, you can do what I did and go back to the original coloured charts and measure the colour of the block where density starts. You have to print a second set of charts but at least you know you're starting at right density --- but read on....
View attachment 86
On the left:
"B" -- Hue 240, Saturation 34, Brightness 66. (RGB112,112,169)
On the right:
"A" -- Hue 120, Saturation 53, Brightness 100. (RGB 120,255,120)
Above are the returned values. At first the results of "B" really baffled me, "how could I be so far off?" Then I went back and measured the colour where density actually started -- What!? I was no longer on the "skin" of the RGB double cone, I was inside the cone. Check it out: HSB :240,34,66 (RGB 112,112,169) is IN not on the cone. The reason: the HSL Array (most density wedges in fact) measures density on the outside only -- remember the shape of the model -- two ice cream cones stuck big end to big end.
View attachment 89
When you pick a colour on or above the equator (green line) as the negative reduces the saturation it follows on the skin until it reaches white, BUT if you're below the equator (the blue line) it takes the shortest path to white short-cutting through the model! (Remember no matter what colour you pick it always moves toward the north as it desaturates.)
Imagine it this way...you just landed on the planet HSL-Array in the Chaos system and you've landed at the North pole of this planet, where everything's white (the south pole is completely black BTW), and you're carrying a laser pointer so you can signal the ship. It's a very strange planet in that it is the shape of two cones stuck together and it's also translucent gelatin in the core. You decide to go exploring and start walking toward the equator. Once and a while you point your laser back at your comrades waiting at the space ship, they lift their glasses to you and smile back. The laser is always going straight back to the ship through the planet's atmosphere up to the point that you start walking south of the equator. Lucky for you though the planet is translucent so even though you're now standing 23.5 degrees south of the equator on the tropic of Capricorn (i.e. colour "B") your friends can still see you and your laser signal because it's able to cut right through the gelatin (unhardened by glyoxal, an in joke) like substance and still puts a shiny red dot on your space ship.
OK, so you're asking why don't a see a little bit of colour at 0 per cent on the ChartThrob grid? The answer is, sometimes you will and sometimes you won't depending on if you over or underestimate which HSL value you start with. If you did manage to catch a piece of colour it might fade out to white and then fade back in again before it started its progression towards white. It kind of makes me think the core of the model is denser than the skin all things being equal, hard to say without plotting out more points.
So what have we learned. One, using colour at or above the equator (i.e. "A") of the model probably gets you closer to where you want to be, initially. Two, while path "B" is perfectly valid option what you see on the HSL Array doesn't really reflect the gradation you will end up with because you're no longer on the surface.
I counted about 31 discernable steps in the Chartthrob test of "A" and about the same in "B" so as far as one colour giving you an advantage in tonality over the other, I don't see it. To me, the obvious choice, if it is available, is "A". In fact maybe a wedge based on the hue of 120 degrees would be useful in this case. A single green with fine gradations which runs from the equator to the North pole. Ho ho ho!
~m