[OpenLayers-Users] How to calculate Bounds for Image Layer Example?

[Ilya Braude]

Andreas Hocevar wrote:
>> Problem is how to calculate OpenLayers.Bounds?
>> Simply replacing
>>    new OpenLayers.Bounds(-180, -88.759, 180, 88.759),
>> by
>>   new OpenLayers.Bounds(-20037508, -20037508, 20037508, 20037508.34)
>> did not work.
>> (It just blows up a single pixel to cover the whole map-div.)
>>     
>
> The easiest for image layers that do not have to align with a map is
> to set the bounds according to the image size. So let's assume you
> have an image, like the one in the example, with is 580 pixels wide
> and 288 pixels high, then the best approach is to set the bounds to 0,
> 0, 580, 288.
>   
I have a similar use case where I do this in order to keep the proper 
aspect ratio for the image. However, is there a way to have the bounds 
be calculated automatically without me having to know the image size a 
priori?  For example, if the image URL is dynamic or user-submitted.

The only way I can think to do this (and I haven't tested this) is to 
have the browser pre-load the image in a new <img/> tag, read the size 
attributes from it, and then use those values for creating the image 
layer.  In this case, how do I know that the image has been downloaded 
for me to read those attributes?

Something like:

var img = new Image();

img.src = "http://path.to/img/url";

...

var bounds = new OpenLayers.Bounds(0, 0, img.width, img.height);



Is there a way that OpenLayers can handle this if I don't care about the 
extent of the image layer, just that it keeps it's aspect ratio without 
me having to specify the correct (in terms of aspect ratio) bounds?

>   
>> So how to calculate OpenLayers.Bounds to display the image at its original
>> 100% size?
>>     
>
> See above.
>   
I assume that by "above" you mean using the pixel size of the image for 
the bounds?  Would it make sense that this would be the default behavior 
for an image layer and implemented by OL directly? 

Thanks,
Ilya