The hanging cable - the intrepid part 2

I mentioned at the end of my post yesterday (see "The Hanging Cable") that I might be able to optimise the maximum height of the cable by adding a taper. The aim of today, aside from the fairly mundane past papers I had scheduled, was to model this as best I could. The result of the monster integral, which I have verified using Wolfram Alpha, is a complicated function in H_max which I don't believe can be solved by conventional algebraic means. Nevertheless, the working is fairly satisfying, if I have made it clear at all, and there are several interesting qualitative conclusions that can be drawn from the model.

It is important to note that I have not repeated the first-principles derivation of this method below. For this, see here.

The function at the bottom is equated to zero to indicate how the required root is to be found. I have entered it into Desmos graphing calculator here, where it is easy to read off the x-intercept as the value of H_max for the material. The sample data on it is for 2800 maraging steel; when a taper angle of 1° ( ≈ 0.0175^c ) is selected, a clearly too large but a good arbitrary starting figure, the graph of y = f(H_max) looks like this:

This shows that the theoretical maximum length of cable, fitting this specification, is 7288km, a pretty awesome distance. However, for a little perspective, let's consider how fat the cable will be at the very top, where it is hanging from its superlatively streadfast loop:

0.5d = Htanθ = 7.288·10⁶ · tan(0.0175^c) = 127553.0213 ≈ 128km (3sf)

If this doesn't seem unwieldy and inappropriate enough, next consider the volume and mass of this so-called 'cable':

V = ¹/₃πH_max³tan²θ = ¹/₃π · (7.288·10⁶)³ · tan²(0.0175^c) = 1.241705149·10¹⁷ ≈ 1.24·10¹⁷m³ (3sf)
m = ρV = 1.24·10¹⁷ · 8100 = 1.005781171·10²¹ ≈ 1.01·10²¹kg (3sf)

This cable therefore makes up 0.0114% of the Earth's volume, enough to excavate the grand canyon nearly 30000 times (using the common estimate of 5.45 trillion cubic yards, where 1 yard = 0.9144 metres), and 0.0169% of the Earth's mass! Now, for the worst part, consider the estimated cost of this object, with the assumption that all the forges in the world could produce enough maraging steel between them. It is difficult to find a precise figure for the price of any particular variety, but in general the world steel price is about $60 per tonne, or 6 cents per kilogram:

Cost = 0.06·m ≈ $(6.06·10¹⁹)

This cost, just over 60 quintillion US dollars or 4.21 quintillion GBP, would be enough to pay off the £1.56 trillion UK deficit nearly 27 million times!

In essence what I am trying to illustrate is that a 1 degree taper is a truly ridiculous idea for extending the snapping length of the cable, even though it does a very good job of doing so. Let's see what happens as the angle is changed...

Since I am using graphical means to solve f(H_max) = 0, I can see no way to easily find the equation for a graph of H_max against taper angle. However some empirical experimentation with Desmos shows that the smaller the taper angle, the greater the value of H_max. This is rather counterintuitive - the smaller the taper angle, the closer this model should come to the model established in the previous article.

Nevertheless I reckon I have found a possible flaw: since the cable starts with a radius of zero, having such a small taper means it is pretty much non-existent for the first few kilometres. I see this as analogous to the critical assembly of a system - just like ants can support weights disproportionate to their own mass due to their tiny size, having such a microscopic cable allows very disproportionate behaviours to occur in comparison with the macroscopic world. This idea doesn't entirely explain away the fact that the previous model completely decoupled the value of H_max from the diameter of the cable, but I'm working on that bit! Perhaps the cable could instead be modelled as a frustum, with a significant radius at the bottom?