# Lighter-Than-Air-Bridge II: Tethered Gondolas

Would something like this be feasible? Span of bridge is approximately 5km. Height above canyon floor is approximately 800 meters (1/2 mile). Bridge structurally close to neutral. Tethered dirigibles carry extra load. Tube trams carry passengers. Structure is flexible, actively conformal and segmented. Site is fairly sheltered. One end is urban area. Other end is wilderness resort / preserve. Climate mild - like Pacific Northwest possibly. Atmosphere somewhat more dense than earth. Winds tend to flow mostly down-canyon or up canyon, perpendicular to structure. I am an artist, not an engineer or anything.

• What holds the bridge up, and what are the "gondolas?" – 2012rcampion Mar 31 '15 at 22:54
• The bridge is supported by the cables. So it is a sort of suspension bridge. The gondolas are the blimps that carry heavier loads. They are travelling back and forth along a sort of track. I am working on a better drawing. – clarkland Mar 31 '15 at 22:58
• In your new drawing you might want to use the same language as in your question text ("tethered dirigibles"), which would make it totally clear what they are. Also clarify whether the balloons support the bridge, vice-versa, or neither. Another question, what is meant by "structurally neutral?" Other than those points I think I understand what you're getting at. – 2012rcampion Mar 31 '15 at 23:16
• By structurally neutral I mean that the bridge by itself is self-supporting. But the balloons support any extra weight that would be transporting a load across the bridge. So the balloons do not support the bridge, only the added loads. – clarkland Mar 31 '15 at 23:26

A self-supported cable of constant density forms a curve known as a catenary, which is close to a parabola. (In fact a parabola is formed when the weight is distributed evenly horizontally like a suspension bridge, instead of evenly along the cable.)

I won't go into all the mechanics, but there are two important equations we need. First, the shape of the caternary is described by:

$$y=a\left[\cosh\left(\frac x a\right)-1\right]$$

Where $a$ is some characteristic length. We can describe the tension $T$ in the cable as:

$$T=\lambda ga\cosh\left(\frac x a\right)=\lambda g(y+a)$$

Where $\lambda$ is the cable's mass per unit length, and $g$ is the gravitational acceleration.

All we need to know is how wide the span is and how much the cable sags. In the best-case scenario, the cable will be allowed to sag all the way to the bottom of the canyon; we can use this as a lower bound on the required material strength.

Assuming that $y=800~\text{m}$ and $x=2.5~\text{km}$ (and implicitly $g=9.8~\text{m}/\text{s}^2$), we get:

$$a = 4.03~\text{km} \\ \frac T \lambda = 47\,400~\text{m}^2/\text{s}^2=47.4~\text{kN}\cdot\text{m}/\text{kg}$$

The ratio $T/\lambda$ is the required specific strength of our bridge material (I'll use the symbol $\varsigma$ to denote it). At first this looks pretty good, since steel has a strength of around $250~\text{kN}\cdot\text{m}/\text{kg}$. However, this doesn't account for the whole bridge, only the cables.

We can figure out the payload fraction $\zeta$ (the fraction of the weight of our bridge that is not cable) with this formula:

$$\zeta = 1-\text{SF}\times\frac{\varsigma_\text{requirement}}{\varsigma_\text{material}}$$

Note $\text{SF}$, the safety factor. It ensures that we have some margin, so we don't fall apart at the slightest touch. Based on some quick research, typical actual safety factors for bridges are around 4, but I'll use a design SF of 5. The maximum payload fraction for steel cables over this canyon is therefore

$$\zeta = 1-5\times\frac{47~\text{kN}\cdot\text{m}/\text{kg}}{250~\text{kN}\cdot\text{m}/\text{kg}}=6.3\%$$

This means that pretty much all of the weight of the bridge will have to be dedicated to the supporting cable(s).

Plotting the shape of the bridge reveals another problem: the slope at the ends is pretty steep.

Assuming a more "reasonable" value for the sag of $125~\text{m}$ (obtained by limiting the grade at the ends to 10%) gives us a shape like this:

Which is probably more like what you had in mind. However, in order to raise the bridge we have to increase the tension. The numbers we get now are:

$$a = 25~\text{km} \\ \varsigma = 247~\text{kN}\cdot\text{m}/\text{kg} \\ \zeta_\text{steel} < 0$$

The required strength is just about the limiting strength of a steel cable. While a steel cable could support itself, it would do so just barely. Any nick in the cable, or a strong gust of wind, and it would snap.

However, Kevlar has 10× the strength-to-weight ratio of steel, and is not an implausible material to use: it's a manmade polymer, and common production processes can spin strands of unlimited length. You could lay the cable in the same manner used to lay the steel cables for the Golden Gate bridge, but spinning the Kevlar strands as you go.

$$\varsigma_\text{Kevlar} = 2514~\text{kN}\cdot\text{m}/\text{kg} \\ \zeta_\text{Kevlar} = 51\%$$

With a conservative safety factor, the total weight of the bridge, cars, and passengers/cargo should be no more than that of the cables themselves. I would suggest a thin, aerodynamic, flexible composite shell surrounding the cables and tram cars to keep live loads from the wind low, and to keep the cables spaced correctly. You wouldn't be able to have a hard, rigid deck like modern bridges, so think of it like a high-tech rope bridge.

As for the balloons, I'm guessing they're essentially like modern blimps, but tethered to the bridge so that they can simply pull themselves along. Although this is more energy-efficient, it won't be much faster since you don't want to load the bridge too much. Plus you will still have to give the balloons maneuvering capabilities to avoid each other when they pass in opposite directions (or build two parallel bridges spaced a couple hundred meters apart). Unless you need high volume or high speed, the bridge won't really add much to the balloons operating by themselves as aerial ferries.

So yes, such a bridge is possible, as long as you can use advanced materials (and are careful to avoid aeroelastic flutter!).

• Interesting question; I haven't had to pull out the catenary equations since freshman statics! – 2012rcampion Apr 1 '15 at 2:11
• I'm not sure why you wouldn't just scrap the balloon support and just get a stronger cable. The cable needs to be a bit stronger than minimum to hold strain from winds, and the force pulling/pushing the gondola. Or if you have great, safe lighter than air craft, why tether them to the ground? – Oldcat Apr 1 '15 at 19:10
• @Oldcat I assumed no support from the balloons (as per the question), and included a safety factor in my calculations. However, now that you mentioned it, I actually looked up safety factors for real suspension bridges, which are higher than I originally thought, so I'll go ahead and redo the numbers. (I'm used to the aerospace industry where SF is commonly 1.25 to 1.5 due to tight controls.) – 2012rcampion Apr 1 '15 at 19:27
• What a great answer. Thanks much. I am not wedded to the blimps. – clarkland Apr 6 '15 at 20:33