<p>Your model is quite clear, although the question is less so. You ask if increasing the height of the image in the background to 200% is the best way.</p>

<p>The best way for what purpose?</p>

<p>If your objective is to maintain an approximation of the original image, the number would not be 200%. </p>

<p>Consider that the red line extends below the forward lip of the shadowbox. Your measurement point should begin at the intersection of the orange triangle and the red line. From that point, the correct calculation is to use the square root of two times the original height, presuming the 45° mentioned in the question. </p>

<p>For angles other than 45° it is necessary to use a bit of trigonometry.</p>

CC BY-SA 3.0
2018-03-18T19:38:22.557Z
*Thanks!* So if the original height is 8.5 inches, and the innner-height, less the orange triangle (straight up 0deg) is 7 inches... The square root of 17 is 4.123.. I'm unclear about what to do with the 4.123, since that's shorter than the visible portion of the frame.
Regular Jo
CC BY-SA 3.0
2018-03-18T19:59:18.703Z
Square root of two is 1.414 (exponential actions take priority over multiplication) and that number times 7 is 9.89 or close enough to 10 to probably work well enough. This is presuming that there's nothing below the 7 inch mark on the original, blocked by the triangle. If the original image is 8.5 and all of it is to be visible above the triangle, the size would be 12.
fred_dot_u
CC BY-SA 3.0
2018-03-18T19:27:46.123Z
2018-03-18T19:27:46.123Z
fred_dot_u
<p>This is an old question and kind of a unique situation.  But if anyone else wanted to try this, the existing discussion doesn't really address what's required.  This is a perspective problem.</p>

<p>Your field of view is like a cone.  The farther the distance, the more you see both horizontally and vertically, and the smaller items appear.  The item up close occupies more of your visual field than the same item far away.</p>

<p>With the tilted picture, the bottom of the frame is close and it gradually gets farther away as you get to the top.  Think of the frame as a slice of your perspective.  The top of the frame will look narrower than the bottom because it is farther away.</p>

<p>How you adjust the image needs to reflect what you want to achieve with the perspective.  </p>

<ul>
<li>Do you want the image to look like a normal, vertical image despite being tilted?  If so, you would need to shape it like a wedge, getting both taller and wider moving to the top.  </li>
<li>If you want the image to look like it would if the objects in the picture were tilted away from you, don't adjust anything.  </li>
<li>If you want to accentuate the distance even more (the question mentions a mountainous landscape), you would need to shrink in both width and height going toward the top of the image.  Also, when dealing with great distances (the horizon), the shrinkage isn't linear between you and the horizon.  The closer objects are to the horizon, the smaller the increment between a given distance interval on the ground, but vertical distances are still uniform.</li>
<li>If you are starting with a photograph of mountains in the distance, that image already contains the perspective, and will look correct when viewed vertically.  Viewing it at an angle, you are introducing more perspective adjustment.  For the image to continue to look correct, you would need to unadjust for the perspective added by tilting the picture.  That is basically what's described in the first bullet.</li>
</ul>

CC BY-SA 4.0
2019-08-31T01:24:10.180Z
2019-08-31T01:24:10.180Z
fixer1234
<p>I have a sort of shadowbox frame. I want to print something (a mountainous landscape) so that it's best viewed when the frame leaned back from the observer at 45 degrees. Is stretching (to 200% height) the best way to achieve this?</p>

<p>I have a model-prop that will stand up straight (-45deg from the backdrop) in front of it.</p>

<p>I am not particularly creative, or close to well-versed in artistic terminology, which is painful when I'm trying to create something.</p>

<p>Terribly sorry for the bad demo but this is what I'm thinking of</p>

<p><a href="https://i.sstatic.net/97lBv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/97lBv.png" alt="enter image description here"></a></p>

<ul>
<li>Black is the frame</li>
<li>Red is the picture</li>
<li>Purple is the the prop</li>
<li>Orange is a triangle of wood to hold the prop up</li>
<li>And blue is the support for holding the frame up.</li>
</ul>

<printing><linear-perspective>
Printing to be viewed at 45deg
Regular Jo
# Printing to be viewed at 45deg

(Asked by: username_0 on 2018-03-18)

I have a sort of shadowbox frame. I want to print something (a mountainous landscape) so that it's best viewed when the frame leaned back from the observer at 45 degrees. Is stretching (to 200% height) the best way to achieve this?

I have a model-prop that will stand up straight (-45deg from the backdrop) in front of it.

I am not particularly creative, or close to well-versed in artistic terminology, which is painful when I'm trying to create something.

Terribly sorry for the bad demo but this is what I'm thinking of

[![enter image description here](https://i.sstatic.net/97lBv.png)](https://i.sstatic.net/97lBv.png)

* Black is the frame
* Red is the picture
* Purple is the the prop
* Orange is a triangle of wood to hold the prop up
* And blue is the support for holding the frame up.

**Tags:** printing, linear-perspective


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## 2 answers

### Answer by username_1 on 2018-03-18. Score: 2 | Accepted answer

Your model is quite clear, although the question is less so. You ask if increasing the height of the image in the background to 200% is the best way.

The best way for what purpose?

If your objective is to maintain an approximation of the original image, the number would not be 200%.

Consider that the red line extends below the forward lip of the shadowbox. Your measurement point should begin at the intersection of the orange triangle and the red line. From that point, the correct calculation is to use the square root of two times the original height, presuming the 45° mentioned in the question.

For angles other than 45° it is necessary to use a bit of trigonometry.

> **Comments:**
> **username_0**: *Thanks!* So if the original height is 8.5 inches, and the innner-height, less the orange triangle (straight up 0deg) is 7 inches... The square root of 17 is 4.123.. I'm unclear about what to do with the 4.123, since that's shorter than the visible portion of the frame.  
> **username_1**: Square root of two is 1.414 (exponential actions take priority over multiplication) and that number times 7 is 9.89 or close enough to 10 to probably work well enough. This is presuming that there's nothing below the 7 inch mark on the original, blocked by the triangle. If the original image is 8.5 and all of it is to be visible above the triangle, the size would be 12.  

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### Answer by username_2 on 2019-08-31. Score: 1

This is an old question and kind of a unique situation. But if anyone else wanted to try this, the existing discussion doesn't really address what's required. This is a perspective problem.

Your field of view is like a cone. The farther the distance, the more you see both horizontally and vertically, and the smaller items appear. The item up close occupies more of your visual field than the same item far away.

With the tilted picture, the bottom of the frame is close and it gradually gets farther away as you get to the top. Think of the frame as a slice of your perspective. The top of the frame will look narrower than the bottom because it is farther away.

How you adjust the image needs to reflect what you want to achieve with the perspective.

* Do you want the image to look like a normal, vertical image despite being tilted? If so, you would need to shape it like a wedge, getting both taller and wider moving to the top.
* If you want the image to look like it would if the objects in the picture were tilted away from you, don't adjust anything.
* If you want to accentuate the distance even more (the question mentions a mountainous landscape), you would need to shrink in both width and height going toward the top of the image. Also, when dealing with great distances (the horizon), the shrinkage isn't linear between you and the horizon. The closer objects are to the horizon, the smaller the increment between a given distance interval on the ground, but vertical distances are still uniform.
* If you are starting with a photograph of mountains in the distance, that image already contains the perspective, and will look correct when viewed vertically. Viewing it at an angle, you are introducing more perspective adjustment. For the image to continue to look correct, you would need to unadjust for the perspective added by tilting the picture. That is basically what's described in the first bullet.


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