[RESOLVED] Size Constraints ASAP!

I have images that I want to display on a web page. I have user defined width and height limitations. If the original width of the image is > the user defined width or the original height of the image is > the user defined height then I need to shrink the image. The problem is that I need to shrink the image and maintain the aspect ratio. What is the formula to "fit" an image into a designated space while keeping its aspect ratio.

For example if MaxWidth = 10 and MaxHeight = 5 and the image is 12 x 6, I would want the image to be 10 x 5.

For example if MaxWidth = 10 and MaxHeight = 5 and the image is 6 x 12, I would want the image to be 2.5 x 5.

Keep in mind there are cases where the one dimension may fit, but then the other dimension is still too large which would require the image to be made smaller.

PLEASE HELP!!

Think about what you are doing. How do you know which one you want to be the defining dimension? What you would do is scale the width in order to match the height available. If it is smaller than the allowed width then you have the answer. If it is greater than the allowed width, then you need to scale the other way.

VB Code:

NewWidth = OldWidth x NewAllowedHeight/OldHeight
 
 If NewWidth <= NewAllowedWidth Then
  NewWidth = NewWidth
  NewHeight = NewAllowedHeight
 Else
  NewHeight = OldHeight x NewAllowedWidth/OldWidth
  NewWidth = NewAllowedWidth
 End If

That's it.

zaza

Yeah I had an algorithm going, but I was off by just a bit. Here's what I am going with:

VB Code:

Dim dOriginalRatio As Double
    Dim dDestinationRatio As Double
 
    'get the original ratio
    dOriginalRatio = lOriginalWidth / lOriginalHeight
    
    'get the destination ratio
    dDestinationRatio = lNewWidth / lNewHeight
    
    'calculate destination width
    lDestinationWidth = IIf(dDestinationRatio < dOriginalRatio, lNewWidth, lNewHeight * dOriginalRatio)
    
    'calculate destination height
    lDestinationHeight = IIf(dDestinationRatio < dOriginalRatio, lNewWidth / dOriginalRatio, lNewHeight)

Thanks for your reply!