I'm reading the derivation of the equation of the hanging catenary from Wikipedia:


In the same article the parabolic equation governing the shape of the main cable of a suspension bridge is discussed, and its equation is obtained as following:

$$y={\frac {w}{2T_{0}}}x^{2}+\beta$$

Here $w$ is the density of the deck and $T_{0}$ is the horizontal tension on the cable at lowest point.

Let's say I'm designing a suspension bridge with the length and density of the deck and its other features are known. I would be interested in knowing the equation of the main cable to calculate the length of the secondary cables (the vertical ones). But how do I get $T_{0}$?

  • $\begingroup$ Are you sure that parabolic isn't just a first-order approximation to the catenary? $\endgroup$ Aug 22, 2018 at 18:02
  • $\begingroup$ @CarlWitthoft If the deck weight is far larger than the weight of the cable, then the parabola is a better fit; I explore the transition here. $\endgroup$ Aug 22, 2018 at 18:17
  • $\begingroup$ @Chemomechanics interesting.... I would have guessed that the deck weight could be treated as a series of masses attached to the support cable at specified locations. Does this mean a non-uniform chain fails to fit to a catenary? $\endgroup$ Aug 22, 2018 at 18:19
  • 1
    $\begingroup$ @CarlWitthoft You're exactly right, but those specific locations for the case of a massive deck are per unit horizontal distance rather than per unit distance along the cable; the difference is important unless the sag is very small. Yes, the general non-uniform chain assumes a more complex shape than a catenary. $\endgroup$ Aug 22, 2018 at 18:30

1 Answer 1


Since $x=0$ marks the midpoint of the parabola, let $y(0)=0$ at this point to eliminate $\beta$. Then, we only need to know the sag $s$ of the cable (which is typically set during the design process). The sag is equivalent to the value of $y$ at one of the symmetric towers (where $x=L/2$, where $L$ is the span or horizontal distance between the towers).

Thus, if we know the sag $s$ and span $L$ (and the distributed weight $w$), then we can immediately obtain $T_0$ from $y=\frac{w}{2T_0}x^2$.

If we know the length $l$ of the cable but not the sag, then we can use the formula for the length of a parabola $$l=\sqrt{\left(\frac{L}{2}\right)^2+4s^2}+\frac{L^2}{8s}\sinh^{-1}\left(\frac{4s}{L}\right)$$ to obtain the sag $s$. (We'll probably have to solve this equation numerically.)

  • $\begingroup$ By "sag", do you mean the height of the parabola? But the height at a point is not something I know, this is what I'm trying to find out. $\endgroup$
    – S. Rotos
    Aug 22, 2018 at 20:22
  • $\begingroup$ Yes; if you were to invert the parabolic shape, the so-called sag would become the height. If you don't know the sag, the cable length, or the tension, then the problem is ill-defined. $\endgroup$ Aug 22, 2018 at 20:28
  • $\begingroup$ Ill defined how? Sorry, English is not my first language. $\endgroup$
    – S. Rotos
    Aug 22, 2018 at 20:45
  • $\begingroup$ Ah, sorry. It means that you have more variables than equations—that multiple combinations of sag and tension could be compatible with what you know about the span length and the deck mass. Also known as underdetermined. But the sag/height of the bridge is usually known/set during the design process. Then the tension is calculated, and the appropriate length of cable is spun with cross-sectional dimensions that are suitable to bear the calculated tension. Edited the answer to clarify. $\endgroup$ Aug 22, 2018 at 21:06
  • $\begingroup$ Ah, now I understand. So to design a bridge, I plug in the density of the deck, I choose some acceptable tension on the cable (within its limits) and then solve for the height of the tower? Apparently the cable is in higher tension when the tower is lower? Also a final question: Are the different parts of the cable in different tension? For some reason I initially thought the tension would be the same at every point.. Is the tension highest in the lowest point, $T_0$? $\endgroup$
    – S. Rotos
    Aug 23, 2018 at 6:51

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