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I am not a mathematician by any stretch but have this real world problem that I need help with.

So imagine a set of steps that have a tread length of 12" and a riser height of 6". There are 30 steps in total.

Your fifty feet away from the steps and you printed out a large image such as your face. You cut it into strips and apply it to each riser to form a very large portrait from your vantage point. You realize that the edges look pretty jaggy as it moves back 12" on each step. The solution would be to increase the size of each strip by a certain amount to make the edges of your face line up to create the illusion from your 50' vantage point.

I thought it would be a standard percentage from step to step but when I work it out it doesn't work that simply unfortunately. My initial belief was print the first step at say 100%. The second at 106% and the each strip would increase 106% from the previous one all the way up so the edges matched. When I drew it out the image got progressively larger as it went up each step.

Any help would be greatly appreciated.

Thanks

Grant

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possibly helpful: instructables.com/id/… –  vadim123 Jun 5 '13 at 18:30

2 Answers 2

The strips are at distance 50 feet plus $n$ times 12 inches (starting from $n=0$). The scaling for step $n$ should therefore be proportional to $\frac1{50+n}$, that is $\frac{5000}{50+n}\%$ for step $n$ (if I remember correctly that one foot equals twelve inches).

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+1 for simplicity; note, however, that this neglects the distance in the vertical dimension; the last riser is $79$ feet away as the crow flies, but $\sqrt{79^2 + 14.5^2}$ "as the eye sees", assuming the person stands level with the bottom step. The general formula is then $1/\sqrt{ (50+n)^2 + (n/2)^2 }$. –  Douglas B. Staple Jun 5 '13 at 20:10
    
And by "stands level with the bottom step", I mean "lays down with their head on the ground"... It'd be better to use $1/\sqrt{ (50+n)^2 + (n/2 - h)^2 }$, where $h$ is the target audiences' height in feet (e.g. $h=6$ for me). –  Douglas B. Staple Jun 5 '13 at 20:21

The scale (blow-up factor) on each step is proportional to the distance from the viewer. To a good approximation, we can use the distance from the eye to the top of the riser (for simplicity).

Let $\ell$ be the tread length, $h$ the riser height, $e$ the eye height above the ground, and $x + \ell$ the horizontal distance from eye to the base of the first stair. I have chosen to name things this way, in which we imagine a "$0$th stair" that extends on the ground a distance $\ell$, to make formulas simpler.

Then, the distance from your eye to the corner of the $n$th stair is $$ d_n = \sqrt{(x + n \ell)^2 + (e - n h)^2}, $$ and so the scaling factor (compared to the first stair) is $$ s_n = \frac{d_n}{d_1} = \frac{\sqrt{(x + n \ell)^2 + (e - n h)^2}}{\sqrt{(x + \ell)^2 + (e - h)^2}} = \sqrt{\frac{(x + n \ell)^2 + (e - n h)^2}{(x + \ell)^2 + (e - h)^2}}. $$

With your numbers, assuming eyeheight is $5$ feet: $(\ell = 1, h = 0.5, e = 5, x = 49.5)$, the absolute scale of the $n$th stair works out to $$ s_n = \sqrt{\frac{9901 + 376n + 5n^2}{10282}}. $$

Here's a table of these numbers. The relative scale column indicates the factor by which you scale up each step from the previous one. $$ \begin{array}{lll} n & \textrm{abs. scale} & \textrm{rel. scale} \\ \hline 1 & 1. & \text{} \\ 2 & 1.01884 & 1.01884 \\ 3 & 1.0378 & 1.01861 \\ 4 & 1.05688 & 1.01839 \\ 5 & 1.07608 & 1.01816 \\ 6 & 1.09538 & 1.01794 \\ 7 & 1.11479 & 1.01772 \\ 8 & 1.13429 & 1.01749 \\ 9 & 1.15389 & 1.01727 \\ 10 & 1.17357 & 1.01706 \\ 11 & 1.19333 & 1.01684 \\ 12 & 1.21318 & 1.01663 \\ 13 & 1.23309 & 1.01642 \\ 14 & 1.25308 & 1.01621 \\ 15 & 1.27314 & 1.01601 \\ 16 & 1.29327 & 1.01581 \\ 17 & 1.31345 & 1.01561 \\ 18 & 1.33369 & 1.01541 \\ 19 & 1.35399 & 1.01522 \\ 20 & 1.37435 & 1.01503 \\ 21 & 1.39475 & 1.01485 \\ 22 & 1.41521 & 1.01467 \\ 23 & 1.43571 & 1.01449 \\ 24 & 1.45626 & 1.01431 \\ 25 & 1.47685 & 1.01414 \\ 26 & 1.49749 & 1.01397 \\ 27 & 1.51816 & 1.01381 \\ 28 & 1.53887 & 1.01364 \\ 29 & 1.55962 & 1.01348 \\ 30 & 1.5804 & 1.01333 \end{array} $$

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