On the opacity of a constituency

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Here may be the question, the channel in the white representative of the selection, Black is not a constituency, then what other colors represent?

In the course #01 we have learned that the channel contains only grayscale, there is no such color as Red green. In addition to white and black, the other is a transition between white and black color. We have also learned that in a separate RGB channel grayscale, the white represents the full glow, the black represents the total no light, in other words, white represents a fully saturated state, and black represents a state of complete absence.

Now, for the alpha channel generated by the storage selection that we have previously contacted, consider that the selected area after the selection is stored in white in the channel, and that the selection is black. In fact, it can be seen as: white represents the "full saturation" state of the selection, while black represents the "no" state of the selection. The transition grey between black and white can also be viewed as a transition from "full saturation" to "no" in a constituency.

In addition, the gray color unit is K, White is K0, pure black is K100. K the larger the color of the more black, the smaller the number of color more white.

Now remove the front S1 and ALPHA1 channels by dragging and dropping the channels onto the Trash button, as shown below.

Then, like the following figure, create a rectangular selection and store it as a channel. Switch to a separate display, pull up the information palette 〖f8〗, move the mouse to the white square, see the color value of K0. If the information palette does not display a K value, click the Cyan Circle and select "Actual Color" from the pop-up menu.

Now use the Shape tool 〖u or Shift U〗, select the rectangle, and use the third way of drawing, as shown in the following figure of the mouse. The third way is to populate pixels (the difference in three ways will be introduced later). Note that this must be the right choice. There are other options in the public column that can be referenced in the following illustration.

Open the Color palette 〖f6〗 and select k49%. Draw a rectangle of approximately the same size on the right side of the white rectangle, and you will see the effect of the following left image. Then draw one with k50%. Finally, draw one with k51%. Figure three below.

So in this channel, we have four rectangles. Review their colors: The first color of the rectangle is K0, which is pure white, the second is K49, the third is K50, and the fourth is K51.

The next three colors look the same, because they have a small difference in grayscale. Now put this channel into the selection after the effect of the following figure.

Found a strange phenomenon, in the channel there are clearly 4 squares, there should be 4 rectangular constituencies only right, why now only three?

Now we go back to normal RGB mode 〖ctrl ~〗, click on the Layer palette

button to create a new layer, and the layer selection stays on the new layer. The following is the left figure. Then use Black to populate the 〖d〗〖alt Delete〗 and cancel the selection 〖ctrl D〗. You'll see the effect of the following right image.

Now that we see the effect is more puzzling, after the fill, there are 4 rectangles on the screen, just like the four rectangles we draw in the channel.

Why do you see only three dotted boxes when you load a selection? And why is it that only one of the four rectangles is black and the other three are gray, do you just fill it with a grayscale color?

Let's look at the second question first. We clicked on the front eye sign of the Layer Palette background layer (below the cyan circle), which hides the background layer, as shown below. There are many gray and white squares in the image, which is the way Photoshop represents the transparent part, because the phenomenon of transparency in the display is actually not perfect expression, because the monitor can not be transparent. So Photoshop is the way to show transparency. Note that white is not confused with transparency, white is a color, and transparency is without any color. The two are conceptually completely different.

From the above, we see that the leftmost square is completely opaque, while the remaining three squares have a translucent effect.

So it appears gray, not because it's filled with gray, but because it's filled with translucent black, plus a white background that looks like it's gray.

If the background color is different, the colors presented will be different. To prove this, we first display the background layer (the same way that the layer is displayed and the hidden layer is clicked in the same position). Then pull out the information palette 〖f8〗, move the mouse to the rightmost square, the following left figure, see RGB values equal, indicating that this is a grayscale color.

Then fill the background layer into a green (after the Layer palette select the background layer fill, pay attention not to fill the wrong layer), and then look at the RGB value, you will see changes, the following right figure, is no longer a grayscale color. This shows that the translucent part of the image and background color will have a blending effect.

Now we are dealing with a very important question: the constituency is not necessarily a complete choice. The same is the range within the dotted box, but there may be different choice of transparency.

The choice of transparency can not be directly identified with the eyes, only after the selection has been filled and other operations will show the translucent effect. In addition to padding, the effect will be different if the selection is used for color adjustment. The following figure uses the same color adjustment for two different transparency selections. You can see that the right side of the adjustment effect is weaker than the left. This is because the selection of the right region is less transparent, such as the region's "power" is relatively weak, it has a relatively small impact.

As we can see from the above, the pure white part of the alpha channel is "fully saturated" after it is converted to a selection, which is "half saturated" after the gray is converted to a selection. As for black, it is "totally without".

As we said earlier, the white representation in the alpha channel is the choice, and the black represents the option.

In fact the correct statement should be: White (K0) part of the representative is the full choice. The Black (K100) section represents a complete selection. The rest of the gray according to Gray, each representing a different degree of choice.

OK, now it's time to answer the question of why the last rectangle doesn't have a dotted box.

First to review the color of 4 squares, from left to right is: k0,k49,k50,k51. According to the previous law, K0 is white, and white represents a complete choice, so can be seen as the K0 of the choice is 100%, the same K100 choice is 0%, is nothing the state of choice. Then the choice degree of K49 is the 51%,K50 choice degree is the 50%,k51 Choice degree is 49%. The relationship is shown below.

In Photoshop, if the selection is less than 50%, the boundaries of the selection will not be visible. is not to see the flow of the dotted line. So only the first three of the four rectangles showed a flowing dotted line, and the fourth one didn't. Because the fourth choice is less than 50%.

Remember that the percentage of K and the percentage of selection is exactly the opposite. The higher the K value, the lower the degree of selection. K The highest there is no choice at all. The lower the value of the K, the higher the selection. K at the lowest time represents a complete choice.

But even in sight, constituencies are still there. As with the previous fill, color adjustment, and so on, you will see the effect.

If the transparency is less than 50% anywhere in a selection, the dotted box will not be visible across the entire selection. Now, with a k51% large rectangle, cover the original four rectangles in the channel. The following is the left figure. Click this channel to load the selection. See a Photoshop pop-up warning. The following is the right figure.

Close this warning box, back to normal RGB mode 〖ctrl ~〗, fill a color or color adjustment, you will see the effect of the selection.

Note, therefore, that the flow-dotted border of the photohsop selection does not necessarily represent the entire range of the selection. Even if there is no dotted line at all, there may be a constituency.

If you just load a selection that doesn't see a border, you can't smear it outside the selection by using the drawing tool. This is because once a selection is established, basically all operations are only valid for the selection. This time, because the constituency does not show the flow of dashed lines, it will create confusion.

The solution is also very simple, remove the selection 〖ctrl D〗 can be arbitrarily applied. If you encounter similar problems in the future, you can try to solve this problem.