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I have a question i am stuck on in my maths class. I am going to change the numbers so i can figure it out this example myself.

Need to get the distance between the two pixels calculated.

Thanks for the help

Thats 100" diagonal and 1920 pixels across and 1080 going up.

user2860471
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3 Answers3

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Using the screen resolution, you find the ratio of width to height:

$$\frac{w}{d} = \frac{1024}{768} \implies w = \frac{1024d}{768}.$$

Then, using the pythagorean theorem, you have

$$w^2+d^2 = 100^2.$$

Using the ratio derived above, we have

$$w^2+d^2 = \left(\frac{1024}{768}\right)^2d^2+d^2 = 100^2,$$ or $$\left[\left(\frac{1024}{768}\right)^2+1\right]d^2 = 10000.$$

Solving, we find $d^2 = 3600 \implies d = 60 \textrm{in}$. Then, $w = \frac{1024}{768}\cdot 60 \textrm{in} = 80 \textrm{in}.$

From this, we conclude that each pixel is $\frac{80}{1024} \textrm{in}$ wide by $\frac{60}{768} \textrm{in}$ tall. The distance between centers of pixels can then be computed as multiples of these heights and widths, using the Pythagorean theorem again.

For instance, the distance between the lower left pixel located at (1,1) and the pixel located at (10,20) is

$$\begin{align} w &= (20-1)\cdot\frac{80}{1024} \\ h &= (10-1)\cdot\frac{60}{768} \\ d &= \sqrt{w^2+h^2}\\ &= \sqrt{\left(19\cdot\frac{80}{1024}\right)^2+\left(9\cdot\frac{60}{768}\right)^2} \\ &\approx 1.64 \textrm{in} \end{align}$$

Emily
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My interpretation of the question is to compute the distance between the centres of two pixels diagonally opposite each other. Clearly the answer should be slight smaller than 100".

Assuming each pixel is a square of side $p$, then Pythagoras gives $p = {100 \over \sqrt{1024^2+768^2}} = {100 \over 1280}$ inches.

The coordinates of the centre of the lower left pixel are $(x_1,y_1)=({1 \over 2}, {1 \over 2})p$, the coordinates of the centre of the upper right pixel are $(x_2,y_2)=(1024-{1 \over 2}, 768-{1 \over 2})p$.

Hence the distance $d$ between the centres is $d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = (\sqrt{1023^2+767^2})p$, which is $d = 100 { \sqrt{1023^2+767^2} \over \sqrt{1024^2+768^2} } \approx 99.81$ inches.

Addendum: It seems that this is not what the OP was asking.

The distance between the centres of the pixels at $(m_1,m_2)$ and $(n_1,n_2)$ (the coordinates are integers) will be given by

$(\sqrt{(n_1-m_1)^2+(n_2-m_2)^2} ) \cdot p$ inches, where $p = {100 \over 1280}$ inches.

copper.hat
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  • Well, that is why I asked if you wanted to compute the distance between the centres of the pixels. The drawing suggests that you wanted to compute the distance between the 'extreme' pixel centres. – copper.hat Jan 10 '14 at 17:27
  • Then that would be $p={100 \over 1280}$ inches above. – copper.hat Jan 10 '14 at 17:46
  • Yes. The square root term gives the distance between centres in 'integer' coordinates. Multiplying by $p$ just scales this distance appropriately. – copper.hat Jan 10 '14 at 18:38
  • No need to switch answers, glad to help in any case. – copper.hat Jan 10 '14 at 18:45
  • I switched answers because the previous answer i dont believe it gave me the distance from the centre of two pixels. – user2860471 Jan 10 '14 at 18:54
  • If you have a 200" screen then $p$ will have a different value. Using the coordinates from your previous comment with a 100" screen the distance will be $\approx 1.64$". If you have a 200" screen with the same 1024x768 format, then $p = {200 \over \sqrt{1024^2+768^2}}$ and the distance will be $\approx 3.29$". – copper.hat Jan 10 '14 at 18:57
  • Ok no i realise my confusion. Im going to ammend the last formula of your answer. Please feel free to change it, i was a bit confused and someone in the future could use this. Thanks for your help! – user2860471 Jan 10 '14 at 19:04
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Pixels on HDTVs are square, so if the distance between pixels is $1$, then the diagonal distance is $\sqrt{1920^2+1080^2}$. Therefore, the distance between pixels is $$ \frac{100''}{\sqrt{1920^2+1080^2}}\doteq0.04539'' $$ or about $1.153$ mm.

robjohn
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