My essay on the maximum length of a suspension bridge concluded that the maximum-length for a steel bridge was about 2.79 miles (14,700 feet) – three times the Verrazano — based on  T1 bridge steel construction (yield strength = 100,000 psi = 100k) with no safety margin, and design where the cable makes a 30° angle with the towers, and where total span weight is twice the main cable weight. While I calculated the main span weight without calculus: from the density of the steel, 0.2833 lb/in³. I used calculus to show that the towers would have to stand 2122 feet above the roadway. This height is intermediate between the Patronas Towers (1,483 ft) and the Dubai tower (2722 ft). From the cable angle, 30° it was clear that the bridge weight must equal the maximum tension in the T1 steel. This was 100,000 psi. That is, for each square inch of bridge cable, the bridge must weigh 100,000 lbs, or 100k pounds per inch of cable cross-section.

Now, for those bothered by the need for calculus, I’d like to re-do the calculation without calculus. Instead we’ll posit a finite number of bridge sections, and we’ll find that the towers can be somewhat shorter. As an example consider the cable divided into five sections with four vertical cables between them. As shown below, there are six sections of bridge held in the middle by four vertical cables. The two end sections of bridge are held by the towers themselves. It’s seen that each vertical cable supports 1/5 of the weight of the roadway and load, and presumably represents 1/5 of the weight of the cable as well. Avoiding calculus, we’ll assign 20k to each of the four intersections, as in the figure below, with the last 10k on each side supported by the towers.

Discrete model of a long suspension bridge: 5 sections of cable; 6 sections of roadway. 100,000 psi in main cable.

The distance between each cable is 1/5 of the length we calculated before, 1/5 of 14700 ft = 2,940 ft. Since only 80% of the bridge is supported by the cable, you find that the cable angle at the towers is lower if the cable is stretched to 100k psi. As before, we will assume that the main cable is T1 bridge steel used at a tension of 100,000 psi (no margin of safety), and that the weight of the bridge + cable equals twice the weight of the main cable (2x 0.2833 lb/in³). You might want to convince yourself that 80% of a 14,700 foot bridge will weigh 80 kpsi assuming the density of the bridge cable plus roadway (and load) is 2 x 0.2833 lb/in³.

Considering the forces as vectors, and using Pythagoras’ theorem, we find the slope of the main cable at the tower equals 40/√(100² – 40²) = 0.436. This suggests an angle of 23.5°, significantly less than the slope of a 30° angle in the calculus-solution. Since the total tension is the same, the horizontal component, T°, is found to be higher: T°= √(100² -40²) = 91.65 kpsi.

Now to calculate the height of the tower from these angles and tensions. The first section of bridge, the one right next to the towers is seen to have a vertical rise that is the product of it’s slope and its length, thus 0.436 x 2940 feet = 1283 feet. The next section, the one closer to the center has half the slope but the same horizontal length. As a result it rises half as much as the previous, 642 feet. The middle section is horizontal and does not rise at all. Thus the required tower height for a five section bridge is 1283 +642 +0 = 1925 ft. This is nearly 200 feet shorter than the 2122 foot height calculated with calculus. This is a significant difference, some 20 stories.

For my amusement, I calculated again dividing the bridge cable into ten sections, with 9 vertical cables supporting 90% of the bridge weight and the rest supported by the towers. This is closer to realistic. If you repeat my calculation (not difficult) you’ll find that the tower requirement is now about 100 feet shorter than in the calculus solution. I suspect that this significant difference remains with any normal number of verticals, e.g., with 50 verticals, I suspect the towers would come out 20 feet shorter, 2 stories, and this is quite significant. I found the calculus solution somewhat easier but not vastly so, and the answers aren’t quite right. For real design, you’d want the discrete calculation, and it’s not so hard.

Here are some of the basic derivations of calculus using discrete math; all the basics are there. My problem with calculus is that it needlessly assumes continuity, when the world is not continuous. At small scales, the world is quantum, and at larger scales it’s generally fractal, or discrete. You generally want to buy a discretenumber of sandwiches, or pizzas. A city needs a discrete number of firemen.  Here, by the way, is my discrete analysis of how many police and firemen you need for a small city.  My sense is that discrete math is hardly more difficult, and a lot more flexible.

Robert Buxbaum, May 19, 2019.