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andw5555
Participant
April 27, 2025
Question

How distort a image

  • April 27, 2025
  • 1 reply
  • 319 views

°Hello to everyone

I need distort a image in this way:

Imagine to shoot a sobject but to a different heght from the camera for example 2 m, up

The photo will be distort, my goal would be a image to same heght from the camera.

During the shooting i can measure the angle of the camera for take the photo, after i'd like find on Photoshoot the tool for distort the image for example of 10°. Does anyone know how can i do it?

I find a lof of tools but no one like this.

1 reply

Trevor.Dennis
Community Expert
Trevor.DennisCommunity Expert
Community Expert
April 28, 2025

Do you need a method that will give you the exact change in perspective at a given angle?

Is that angle important, or does it just need to be close?

 

You could work it out using Photoshop as a very rough and redy CAD application, but it would be much easier and more accurate with a prober CAD app like Autocad or Sketchup.   I have probably got the diagram below wrong as the 10° should be the center line, and not the bottom of the projected shape.

 

If you only need a rough estimate, then Free Transform > Perspective will do it.

 

    A
    andw5555Author
    Participant
    April 28, 2025

    Thanks for the reply.

    I need a precise tool, I don't need a closer image

     

    "You could work it out using Photoshop as a very rough and redy CAD application"

    How?

    Trevor.Dennis
    Community Expert
    Trevor.DennisCommunity Expert
    Community Expert
    April 29, 2025
    quote

    "You could work it out using Photoshop as a very rough and redy CAD application"

    How?


    By @andw5555

     

    If you look at my diagram above, the 10° line is accurate done using Free Transform.

     

    If this was a projection onto a screen, then the dimension I have marked as Width would be the width of the projected image on our imaginary screen.

     

    However the height would be extended because of the angle it hits the screen.  We started with a square box because it suits the calculation.  So the aspect ratio between width and height can be applied to your situation.   Is there enough information there to do the Trig?  I don't think we have, because we don't know the angle.  That 10° doesn't help you because the angles of the right angle triangle depend on the height (hypotenuse).  

     

    It is also not clear exactly what you are trying to work out.  You'd need to provide diagrams/screen shots and a more detailed description.

     

    If you were able to draw your diagram with a CAD app, then you could accurately measure and workout the relationship between width and height, but other than that 10°, the rest is arbitrary.  It wouldn't be linear either. That aspect ratio would be different for a wider angled projection.  It's like I said before...  You'd need to split the angles between the top and bottom line and use the lens axis of this imaginary projector.  The aspect ratio would then remain linear regardless.   

    I'm old and out of my depth, so you need to provide more information, and someone like our Dave, who is cleverer than most of us, might offer a solution.

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