I have been reading the original Paul Birch paper published in JBIS 1982. I have been using this copy found here.

In particular I am wondering about generalizing the ideas, like for a different planet or body what would the forces look like? In short, I wondered what a chandelier city could look like in a hypothetical scenario.

In section 3 of the paper Birch presents an equation and then breaks the terms into constituents which are then brought back together on the next page.

$F_T$ is defined elsewhere and is the force applied at a tether 'anchor' I suppose, a location where the ring is directed into another orbit and the momentum transferred to holding a tether aloft. Measured in newtons.

$\Delta \Theta$ is the angle change at one of the anchor stations

This is the base equation. Over the next two pages he develops the terms and then is able to rewrite $F_T$ in terms of $\Delta H$, the height difference in the perigee of the new orbit and $\alpha$, the angle of the orbit of the ring after exiting a skyhook station. I have attached a screen shot of this final equation.

He provides 'Table 4-Permissable Skyhook Weights' with a set of values and the resultant $F_T$ weight of a skyhook tether in N (newtons).

I then have been setting up my work thusly and having some trouble with the equation:

$m$ = given in table footnote as $2.5 \pi * 10^3$

$g$ = acceleration due to gravity $9.8 m/s^2$

$R$ = radius of earth at equator of $6378$ km

$\Delta H$ = as given in the table is $\frac{\pi}{\alpha}$ and provided as $2$ for the first example.

$$ F_T = \frac{4mgR^2 * \frac{\Delta H}{\alpha}}{(R+H)(R+H+\Delta H*(1-\frac{2}{\alpha ^2})}$$

I am having trouble replicating the data in the table. Using the values there I have interpreted $\frac{\Delta H}{\alpha}$ to be $\frac{3}{2}$ and when that did not produce the expected result I tried to use $\frac{\pi}{\alpha}$. I am not sure if this is the error in the calculation.

From the table I am expecting a result for $H=300$, $\Delta H=3$, $\frac{\pi}{\alpha}$ of $540000000 * 10^{-9}$ or $0.54$. The result I am getting is substantially different.

Is anyone familiar with the math behind the concept to offer a hand to help me understand where I've gone wrong?

  • $\begingroup$ I love the cover art (first page of PDF) $\endgroup$
    – uhoh
    Jan 25, 2022 at 4:32
  • 1
    $\begingroup$ Please edit your question and supply the actual substantially different value you are getting so that someone can help. $\endgroup$ Jan 25, 2022 at 12:50
  • $\begingroup$ @OrganicMarble Thank you. I should've made things more clear in my question that I wasn't even sure I had the right values as inputs. From looking at the response I see that I was incorrect on two values. $\endgroup$
    – b00klegger
    Jan 25, 2022 at 22:36

1 Answer 1


The table should be understood as follows: the first data column (0.54, 1.08, ...) contains the values of $F_T\cdot 10^{-9}$ in Newtons for different values of $\pi/\alpha$ (2, 4, ...) in the case where $\Delta H$ is 3 km; the second column (0.54, 1.09, ...), for the case where $\Delta H$ is 30 km, etc.

For example, if we take $H = 300~\text{km}$, $\Delta H = 3~\text {km}$ and $\alpha = \pi/2$, then $$ F_T = \frac{4~\cdot~ 2.5\pi\cdot10^3~\text{kg m}^{-1}~\cdot~9.8~\text{m s}^{-2}~\cdot~ (6371000~\text{m})^2~\cdot~ \frac{3000~\text{m}}{\pi/2}}{6821000~\text{m}~\cdot~(6821000~\text{m} ~+~3000~\text{m}~\cdot~(1 - \frac{8}{\pi^2}))} \approx 5.4\cdot 10^8~\text{N}, $$ or $F_T\cdot 10^{-9}$ is $0.54~\text{N}$.

  • $\begingroup$ This is a perfect explanation. I see that I was incorrect in my understanding of the $\frac{\pi}{\alpha}$ terms in the equation. $\endgroup$
    – b00klegger
    Jan 25, 2022 at 22:50
  • $\begingroup$ Would I be correct in thinking that the $\frac{\pi}{\alpha}$ term is the number of radians between the previous orbit modified by the 'station' and the perigee of this new orbit? $\endgroup$
    – b00klegger
    Jan 25, 2022 at 22:58
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    $\begingroup$ @b00klegger If I understand Fig. 6 correctly, $\alpha$ is the angle (in radians) between a station and an adjacent orbit's perigee, when seen from the Earth's center. If perigees are located at midpoints between two stations, then $\frac{\pi}{\alpha}$ is the total number of stations along the Earth's circumference. $\endgroup$
    – Litho
    Jan 26, 2022 at 8:47

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