2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

1
$\begingroup$

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 $0.54~\text{N}$.

$\endgroup$
3
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.