The TIF image is 44x34.

Trace 40(2007-04-11 08:06:18.574): img.scaledWidth() 8800.0
Trace 40(2007-04-11 08:06:18.574): img.scaledHeight() 6800.0
Trace 40(2007-04-11 08:06:18.590): img.width() 8800.0
Trace 40(2007-04-11 08:06:18.590): img.height() 6800.0
Trace 40(2007-04-11 08:06:18.590): img.getDpiX() 200
Trace 40(2007-04-11 08:06:18.590): img.getDpiY() 200
Trace 40(2007-04-11 08:06:18.590): img.scaledWidth()/img.getDpiX() 44.0
Trace 40(2007-04-11 08:06:18.590): img.scaledHeight()/img.getDpiY() 34.0

I was using:
            float iWidth = img.scaledWidth();
                (float) ( (img.scaledWidth() /
                           ( ( (img.getDpiX() == 0) ? 200 :
img.getDpiX())))
                         * 72);
And this worked fine.

Then I was trying to solve the issue where this value is larger than 200
inches (not this particular image), so tried setting the UserUnit to 200
and that is when I got into trouble.

So setting the UserUnit has no effect on the viewable PDF page?
And if it should, what should I set the UserUnit to in this case?

Thanks again in advance.

-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Bruno Lowagie (iText)
Sent: Wednesday, April 11, 2007 7:39 AM
To: Post all your questions about iText here
Subject: Re: [iText-questions] What is the maximum page size for PDF
iniText?

Estepp, Dan wrote:
> Thank you.
> 
> I now set the Userunit to 200 and x/y to 8800/6800,
 > thus should be 44 inches by 34 inches.

What kind of math are you using?

200 * 8800 = 1,760,000
1,760,000 / 72 = 24444,45

200 * 6800 = 1,360,000
1,360,000 / 72 = 18,888,89

> But when I view the image using 7.0.9 Adobe Reader, the Document
Properties show:
> "Page Size: 24,444.45 x 18,888.89 in"

Why, this is correct.
There still are certainties in life!
br,
Bruno

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