[scribus] Gravity of images in image frames
gregp_ky at yahoo.com
Wed Jan 25 03:54:16 UTC 2012
On 01/24/2012 10:27 PM, Gregory Pittman wrote:
> On 01/24/2012 08:00 PM, Hadmut Danisch wrote:
>> when creating image frames, either in free scaling or scaling to frame
>> size, the image seems to be always anchored to the top left corner.
>> Is there a menu or setting to have an image anchored to the other
>> corners or to have it vertically or horizontally centered in its frame?
> For one thing, you have to think through what you are asking. If we
> might think of images having some absolute dimensions of themselves, the
> DPI would be unsuitable for most purposes, so you must decide what DPI
> you wish to have for your image, then consider how to center.
> Let's imagine you wish to import an image to a frame of a certain size,
> at a certain scaling, then center it.
> Make your frame, import the image, then adjust scaling as desired (you
> will need to use Free Scaling of course), but leave the image with its
> X-Pos, Y-Pos (as shown in the Image tab of Properties) at 0, 0. Now open
> the Multiple Duplicate dialog, use the By Number of Copies tab, select 1
> copy, select Shift Created Items By, and leave the Horizontal and
> Vertical Shifts at 0 each, then click OK. Your copy will be right over
> your original. Select this, and from Context menu select Adjust Frame to
> Image -- this size of this frame will be the size of the image at the
> scaling you have chosen. Now push this down a level below the original.
> In X,Y,Z you can get the Width and Height of this adjusted frame. let's
> imagine these values are 897.76 and 602.64 pts. Let's imagine that the
> width and height of the original are 398.54 and 314.42. Now, select your
> original, small frame and go to Image tab of Properties, and in the
> X-Pos spinbox enter -843.21/2 + 398.54/2,
typo here, should be: -897.76/2 + 398.54/2
and in the Y-Pos -602.64/2 +
> 314.42/2. This will center the middle of the image horizontally and
> vertically with the original frame.
> This description may sound complicated but conceptually it's rather
> simple (I think) -- you are shifting the midpoint of the image to the
> upper left of the original frame, then adding half the width and height
> of the frame.
> Again, I would repeat that this kind of operation seems unlikely to be
> in much use once you consider the scaling variability.
And if you wanted to have the lower right corner of the image at the
lower right corner of the frame, in this example above, enter X-Pos =
-897.76+398.54 and for Y-Pos = -602.64+314.42.
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