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I hope I have phrased my question correctly.

Let's assume I have a rectangle with a width and height of X and Y.
Then I pick an aspect ratio of 0.56.
How can I calculate the size of the rectangle that bounds the original rectangle, without shrinking it?

In other words, I want to keep the original rectangle as is, and "place" it inside a canvas that obeys the aspect ratio, I'm trying to calculate the canvas size.

Any help would be highly appreciated.

Best regards,
Roi

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  • $\begingroup$ If I have understood your question. it suffices to preserve the ratio H/W=0.56. $\endgroup$ – user Jun 25 '18 at 21:05
  • $\begingroup$ @gimusi Yes. The canvas ratio should be the selected ratio. $\endgroup$ – Roi Mulia Jun 25 '18 at 21:06
  • $\begingroup$ What is your fixed dimension for H or W or for the Area? $\endgroup$ – user Jun 25 '18 at 21:07
  • $\begingroup$ Assuming you have aspect ratio is height over width, then I think you would need a rectangle of size $\max(X, Y/0.56)\times\max(Y,0.56 X)$. See if that works. $\endgroup$ – Adrian Keister Jun 25 '18 at 21:07
  • $\begingroup$ The fixed dimension is the source rectangle (I'm building an app, the source rectangle is the video rectangle which I'm trying to "fit" inside a canvas) $\endgroup$ – Roi Mulia Jun 25 '18 at 21:08
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Let $x, y$ be the dimensions of the window that you want to fit into a canvas with aspect ratio $r$. Let the desired dimensions of the canvas be $x', y'$. In all cases, we must have $y'/x' = r$. Three cases:

  • if $y/x = r$, then both dimensions fit just right: let $x' = x, y' = y$.
  • if $y/x > r$, the rectangle is tall, so we fit the y-dimension: $y' = y$ and since $y'/x' = r$, we have $y/x' = r \implies x'= y/r$.
  • if $y/x < r$, then we fit the x-dimension: $x' = x$ and since $y'/x' = r$, we have $y'/x = r \implies y' = xr$.
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  • $\begingroup$ I tried for hours to solve it, but i just saw that you said y' / x' = r, where I tried the opposite lol. Thank you sir! $\endgroup$ – Roi Mulia Jun 26 '18 at 9:42

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