## Shawshank Redemption – the tunnel length

(Note: Despite my best efforts, the maths formulas tend to show up awkwardly every now and then. Please let me know if you see that happening; I’ll revert to regular text based formatting.)

I watched Shawshank Redemption  again a couple of nights ago, and every time I watch it I wonder about the length of the tunnel that Andy carved out. I couldn’t hold it any more and I attempted to unravel the mystery. I did all the maths in my head while lying in bed staring at the dark, so did a lot of generous rounding offs to get nice numbers, and have reproduced the same numbers here. This is what I came up with

Andy joins the prison in 1947 and manages to escape in 1965. Let us assume that he started carving the wall from 1950 onwards, everyday. That is 365 x 15 number of days or 5475 days in all. Let us round it off to 5500.

It is shown, when Red recounts, that Andy disposes the dust in the yard. The dust doesn’t seem to be much in quantity. Let us assume it was 200 grams per day. That makes it 200 x 5500, that is 1100000 grams in all. Or 1100 Kgs.

The tunnel appears to be cylindrical (I am still amazed at the pains he took to carve a near perfect circle. May be that is what took him that long to finish). And by the glimpse that we get of it from the inside when the warden punctures it, and also by the way Andy crouches while crawling through, it appears to be about 2 feet in diameter. In metric that is 0.6 m. Using a couple of simple formulas that we learnt in school geometry and physics, we can estimate the length of the tunnel.

Density is mass divided by volume.

$\rho={m\over v}\dots(1)$

Volume of a cylinder is

$v = {{\pi\times d^2\times h}\over 4}$

where d is the diameter and h is the height. But in our case since it is length that we are after, let us use ‘L’ in place of ‘h’. That gives us.

$v={{\pi\times d^2\times L}\over 4}\dots(2)$

Using (1) and (2) we get

${\rho = {{m\times 4}\over{\pi\times d^2\times L}}\dots(3)}$

Taking pi to be 3.143 and using the other values that we have, putting them in (3) we get

$\rho = {{1100 \times 4}\over{3.143 \times 0.6 ^2 \times L}}$

or

$\rho = {4400\over{1.1315\times L}}$

that is

$\rho = {3888.71\over L}$

and after transposing L and p we get

$L = {3888.71\over\rho}\dots(4)$

If we can get the value of p for the material of the wall, finding the length is trivial. Remember, the got length will be in meters. Could the length be in hundreds of meters? Was the “wall” just an extra thick wall (since it appears that Andy was in a corner cell) or did it include a series of stacked walls? Was the exit right on the other side of Andy’s cell? Did the tunnel go in a straightish line or was it akin to mining? Did the tunnel continue into the ground?

We have no clue. At least I, who have only seen the movie and haven’t read the novel, don’t find a clue. However, some analysis of (4) is going to help us.

The density of the wall material is the most important variable. If the tunnel length is to be in hundreds or even tens of meters, the material of the wall must be not very dense. Check out this link that lists densities of over 300 materials. http://www.simetric.co.uk/si_materials.htm. We realize that any material that is fit to be made a prison wall out of has to have a density that ranges from 2000 kg/cu-m to 2500 kg/cu-m. Let us assume 2250 kg/cu-m.

With this density, from (4), we get 1.72 m, or 5.6 feet. An average man is 5 ft 10 inches tall. Andy seems a taller bloke than that and could be about 6 ft 2 inches tall. That would be 1.86 m. Huh, but we see Andy crawl out a fair distance, don’t we! Something seems amiss in our assumptions. If we look at the contributing variables, we find that in order for the length to be more, more wall material must be carved out for a longer period and/or the tunnel must be narrower. The latter is inconceivable albeit Andy’s narrow frame.

So let us update our assumption to 500 grams of dust being disposed everyday. With this we get

$L = {9721.61\over\rho}$

Plugging in the density, we end up at 4.32m or 14 feet. This implies that Andy, with his height of 1.86m, crawled 2.32 times his own length. Meaning, once he entered the tunnel and lied down such that his feet were placed flat on the imaginary door at the opening, he had to crawl one and a third more of his own length when his head would touch the imaginary door at the other end of the tunnel. (What this also implies is that with his arms outstretched, he had to crawl only just about his own body length but the crawling mechanics won’t agree with this. Just a thought anyway)

Somehow, 14 feet feels too small for our romantic idea, doesn’t it ? The only remaining factor that could fulfill our craving for a longer tunnel is time. Red tells us that Andy asked for Rita Hayworth’s poster in 1949 but we have no clue how far into ’49 did he ask. Let us assume he asked for it half way through and started his attempt immediately. With this updated time period, we have 180 days more to the already assumed 5500 days giving us 5680 days.

If we rework our equations with 500 gms of dust, carved out of the wall having density of 2250 kg/cu-m, disposed off for 5680 days, we arrive at 4.46 m for the length, or 14.6 feet, which is 2.4 times Andy’s tall frame.

You may feel this is still not good enough but there is nothing else we can do to improve this to fit our romantic idea of a long tunnel. Further, we have already been very generous in our assumptions and rounding offs, so we just have to accept it.

PS: From the in universe perspective, we have to give credit to Andy for carving out that 14.6 feet long a tunnel, tenaciously. Considering the conditions he was in, it would be as much a feat as carving a 50 feet tunnel.