PHP calculate percent that an image resolution has over another's image resolution

I would like to know how can I tell what percent does one image's resolution represents from another image's resolution. So I have $resolution1 = '480x210' and $resolution2 = '720x480' . I want to find out how much $resolution1 represents, in percentage, from $resolution2. I am trying to make a business card print preview and want to show the client how small is his uploaded image in comparison with the needed size. So I got a blank proper sized div and I will render a thumbnail of the uploaded image in it but the thumbnail is made at X% 开发者_运维技巧its size (this is what I need to know) so that I can represent how much space on the card would his image take.

I am using PHP and imagemagick binaries with exec(). Any ideas ?

Thanks.

Soimething like this should work, if not it will help you get closer to your goal:

//Get the x / y of both  resolutions
list($ax,$ay) = explode("x",'480x210');
list($bx,$by) = explode("x",'720x480');

//Get the difference of both resolutions
$calc_x = ($ax / $ay);
$calc_y = ($bx / $by);

//Calculate the increase
$increase = ($calc_x / $calc_y) * 100;

im not the best with maths but i think this is how you calculate the increase, the result of the above is 152.38095238095 which is 152.38% increase

$resolution1width / $resolution2width = $returnwidthpercent;

$resolution1height / $resolution2height = $returnhieghtpercent;

div width = rendered image width + (rendered image width * $returnwidthpercent);

and the same for height. I think that should work.