Oh, I had looked for [Adobe Reader in the Microsoft Store] but I did not find it.

Get Adobe Acrobat Reader DC from the Microsoft Store

Hello Joe

Thank you.

The Windows 10S computer that I am using at present will not accept Adobe Reader. I do not want to unlock the S on this computer, though I have done so on my new Windows 11 computer as I wanted to load Affinity programs for which I bought licences some time ago. At present I am keeping that computer for preparing artwork.

I think Edge to view a PDF document is by default, I don't think that I set it up.

Interesting William as I didn't know about the 'S' mode. I just had a look and adobe reader is available in the microsoft store if you wanted to download it and use it in 'S' mode.

Thank you.

The Windows 10S computer that I am using at present will not accept Adobe Reader. I do not want to unlock the S on this computer, though I have done so on my new Windows 11 computer as I wanted to load Affinity programs for which I bought licences some time ago. At present I am keeping that computer for preparing artwork.

I think Edge to view a PDF document is by default, I don't think that I set it up.

I am using the Microsoft Edge browser.

Something that I noticed today, which may already be known to some readers, is that if one clicks on the attached file and then displays it in Microsoft Edge, then if one clicks on the diagonal two-headed arrow that is second from the right in the Microsoft Edge header, then one gets a full screen display of the PDF document.

Press on the Escape key on the keyboard (often in the upper left corner) to leave full screen display mode.

Alfred wrote:

You can change the shape of the decay curve by altering the parameters of the gradient.

Is that by changing the opacity from the 20% that I used?

Or something else?

William

You can alter the other parameters, too. When you choose the Fill Tool, the control line has stops for the different colours and opacities along the gradient. You can add or remove stops, and you can reposition the stops or the ‘midpoints’ between pairs of stops (positioned exactly halfway by default).

You can change the shape of the decay curve by altering the parameters of the gradient.

Is that by changing the opacity from the 20% that I used?

Or something else?

William

So it seems that the colours move from the background colour towards the yellow colour of the bands as a sort of chunky exponential decay.

You can change the shape of the decay curve by altering the parameters of the gradient.

(128, 128, 255)

The five colours of the bands have (r, g, b) values as follows.

(153, 153, 204)

(173, 173, 163)

(189, 189, 130)

(202, 202, 104)

(213, 213, 83)

heading towards (255, 255, 0)

So it seems that the colours move from the background colour towards the yellow colour of the bands as a sort of chunky exponential decay.

William

Do you know how each of the red, green, and blue channels is calculated when opacity of a layer is used please?

I am wondering if it is, if opacity is expressed as a value p in the range 0.0 to 1.0, if there are n layers, as

x = p * x(i) + (1 - p) * x(i - 1)

Alas, I cannot use square brackets, as the forum software treats them as markup.

…

Using Paint to look at what is the colour gives (153, 153, 204), so it looks like my guess at the formula might be correct, at least for just one semi-opaque layer upon an opaque layer.

I think your formula is indeed correct, unless we’re both missing something!

As for using square brackets, you need to ‘escape’ them by using {code} markup via the ‘angle brackets’ button on the toolbar above the message text area.

x = p * x[i] + (1 - p) * x[i - 1]

I am wondering if it is, if opacity is expressed as a value p in the range 0.0 to 1.0, if there are n layers, as

x = p * x(i) + (1 - p) * x(i - 1)

Alas, I cannot use square brackets, as the forum software treats them as markup.

and recursively back to the background colour x[0]..

In my picture, a few posts back, (r, g, b) for the blue background is (128, 128, 255) at 100% opacity.

Then each of the five yellow areas have (r, g, b) as (255, 255, 0) at 20% opacity, so p=0.2.

So if my guess at the formula is correct, then

(r, g, b) for the single yellow layer on the blue would be about (153, 153, 204)

Using Paint to look at what is the colour gives (153, 153, 204), so it looks like my guess at the formula might be correct, at least for just one semi-opaque layer upon an opaque layer.

William