How to map a point in $720 \times 1280$ rectangle to a point in $1080 \times 1080$?

Question

How to map a point $(x,y)$ in $720\times1280$ rectangle to a point $(x_1,y_1)$ in a square $1080\times1080$?

I try with the following formula, but seems there's something wrong.

Let $y_{offset} = \frac{1280 - 720}{2}$. \begin{align} x_1 &= x \frac{1080}{720}\\ y_1 &= y \frac{1080}{720} - y_{offset} \end{align}

The above formula works if the square is $1440\times1440$ but $1080\times1080$.

demo

Answer 1

If your goal is to make it so the rectangle of dimensions $(w, h)$, transformed, is entirely contained within the square of dimensions $(w_1, h_1)$, you should multiply by $$S = \min\left(\frac{h_1}{h}, \frac{w_1}{w}\right)$$

which makes the offsets

$$x_{offset} = \frac{w_1 - S \times w}{2}$$

and similarly for $y$.

In this case, $S = \frac{1080}{1280} = \frac{27}{32}$, and the offsets are $x_{offset} = \frac{1080 - \frac{27}{32} \times 1280}{2}=\frac{1080-1080}{2}=0$ and $y_{offset} = \frac{1080 - \frac{27}{32} \times 720}{2}=\frac{1080-607.5}{2}=236.25$.

If you instead want the image to cover the square, use $\max$ instead of $\min$ above.

Your final equation here then is

$$x_1 = \frac{27}{32}x$$ $$y_1 = \frac{27}{32}y + 236.25$$

Answer 2

For your specific case, the range of $y$ must be reduced from $1280$ to $1080$. To maintain the aspect ratio, the $x$ range must be reduced by the same ratio. You can just map $(x,y)$ to $(\frac {1080 x}{1280},\frac {1080 y}{1280})$. That will leave the image along one side of the target. As the $x$ direction of the image is small, you might want to add half the gap to the $x$ coordinates. This is $\frac 12(1080-\frac {1080 \cdot 720}{1280})=236.25$ which will center the image in $x$