# Satellite pictures

Thursday 9th July 2020

Given that there are 12 inches to the foot and \(2.56cm\) to the inch, and that each letter in the BLACK LIVES MATTER slogan is \(35ft\) in height:

- Can you express the scale of the photo as a ratio?
- Can you estimate the area of the park in m\(^2\)?

The satellite that took this image is located \(695km\) above the earth. If this is the best resolution that can be taken with its camera, estimate how far away the satellite would have to be so that it could no longer read the slogan.

By printing off the page and measuring the photo (note the following measurements are taken from an A4 printed page, but regardless the answer should be the same), we get an approx. size of \(5mm\) for the height of the lettering. This implies that \(1mm\) on the photo equates to about \(7ft\). With conversion of ft to inches to mm this means \(1mm\) (image) = \(7\times12\times25.4mm\) (real) which gives an approx. ratio of \(1:2137\).

Approximating the footprint of the park to be \(4cm\times6cm\) or \(40mm\times60mm\), we have measurements of \(40\times2137\times60\times2137mm^2\) or \(85.48\times128.22m^2 = 10960\;(0.dp.)m^2\).

Answering the final question requires a lot of assumptions, and perhaps many may disagree with the following (please do so!). However, the thinking behind it is one based on pixel resolution…

Our first assumption is that if the photo was approximately \(\frac{1}{12}\) smaller (i.e. lengths approx. 3.5 times less) then we would lose resolution and be unable to read the lettering (see diagram above). Note that in the current format, we can reduce and enlarge to our hearts content and still be able to read the writing because the image isn’t losing the resolution data. (It does however when it is rasterised – i.e. when the image is reduced and re-pixelated based on its current size).

(Please feel free to disagree with this also!) Our next assumption is that the scaling of the distance is itself ‘linear’ over distance. I.e. 1 goes to \(\frac{1}{2137}\) over a distance of \(695km\), therefore over twice this distance it would go to \(\frac{1}{4274}\) etc. Therefore, since we believe that the photo can be approx. 12 times smaller, which equates to approx. 3.5 times smaller in length and width, then we could assume that the satellite could be 3.5 times further away so our camera’s limit in terms of distance to be able to read the BLM lettering would be \(695\times3.5=2432.5km\).