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Comparison of Space City Configurations
DOI: 10.12677/JAST.2021.92004, PDF, HTML, XML, 下载: 231  浏览: 534

Abstract: The typical configurations of space city are compared, including Liu Cixin rod, Stanford ring, Bernal sphere, Liu Cixin ellipsoid and O’Neill cylinder. The stress characteristics of each configuration are analyzed, and the stress magnitudes are calculated under typical materials and sizes. The development policy of space city is summarized as follows: “from near to far, from easy to difficult, market driven, benefit first, multi-party cooperation and common development”, and the gradual implementation plan is preliminarily proposed from low orbit to high orbit, from Earth-Moon system to solar-terrestrial system. At the same time, the engineering application of rotary space solar power station can not only obtain a lot of clean energy and reduce greenhouse gas emissions, but also “shade” the earth, which has the effect of slowing down global warming.

1. 引言

2. 空间站发展

3. 刘慈欣圆杆

Figure 1. Liu Cixin rod space station

$F={\int }_{0}^{r}\rho \cdot S\text{d}r\cdot {\omega }^{2}r=\rho S{\omega }^{2}{\int }_{0}^{r}r\text{d}r=\frac{1}{2}\rho S{\omega }^{2}{r}^{2}$ (1)

${\sigma }_{0}=F/S=\frac{1}{2}\rho {\omega }^{2}{r}^{2}$ (2)

${\sigma }_{0}=\frac{1}{2}\rho {\omega }^{2}{r}^{2}=\frac{1}{2}×2.7×{10}^{3}×9.8×1000=13.23×{10}^{6}\text{ }\text{Pa}=13.23\text{\hspace{0.17em}}\text{MPa}$

${\sigma }_{e}=\frac{1.96×{10}^{5}}{3.14×{0.12}^{2}/4}\text{Pa}=17.34\text{\hspace{0.17em}}\text{MPa}$

4. 斯坦福圆环

(a) (b)

Figure 2. Stanford ring space city. (a) Integral structure; (b) Interior landscape

$\text{d}F=\rho \cdot Sr\text{d}\phi \cdot {\omega }^{2}r=\rho S{\omega }^{2}r\cdot r\text{d}\phi$ (3)

$2S\cdot {\sigma }_{0}={p}_{0}\cdot 2r$

${\sigma }_{0}={p}_{0}r/S=\rho {\omega }^{2}{r}^{2}$ (4)

${\sigma }_{0}=\rho {\omega }^{2}{r}^{2}=2.7×{10}^{3}×9.8×1000=26.46×{10}^{6}\text{ }\text{ }\text{Pa}=26.46\text{\hspace{0.17em}}\text{MPa}$

${\sigma }_{e}={p}_{e}r/S=\stackrel{¯}{m}{\omega }^{2}{r}^{2}/S$ (5)

${\sigma }_{p}^{\theta }=p\cdot D/2/t=0.1×160/2/0.199\text{\hspace{0.17em}}\text{MPa}=40.2\text{\hspace{0.17em}}\text{MPa}$

5. 伯纳尔球体

Figure 3. Bernal sphere space city

${\sigma }_{0}=\rho {\omega }^{2}{\left(r\mathrm{cos}\phi \right)}^{2}=\rho {\omega }^{2}{r}^{2}{\mathrm{cos}}^{2}\phi$ (6)

${\sigma }_{0}=2.7×{10}^{3}×9.8×1000=26.46\text{\hspace{0.17em}}\text{MPa}$

$2\pi r\cdot t\cdot {\sigma }_{p}=\pi {r}^{2}\cdot p$

${\sigma }_{p}=\frac{r}{2t}p$ (7)

${\sigma }_{e}={p}_{e}r/t=\stackrel{¯}{m}{\omega }^{2}{r}^{2}/t$ (8)

$S={\int }_{0}^{\phi }2\pi r\mathrm{cos}\phi \cdot r\text{d}\phi =2\pi {r}^{2}{\int }_{0}^{\phi }\mathrm{cos}\phi \text{d}\phi =2\pi {r}^{2}\mathrm{sin}\phi$ (9)

$\phi ={90}^{\circ }$ 时，就是上半球面面积 $2\pi {r}^{2}$，整个球体的面积为 $4\pi {r}^{2}$。如果只居住到纬度30˚，则利用面积为 $2\cdot 2\pi {r}^{2}\mathrm{sin}{30}^{\circ }=2\pi {r}^{2}$，恰好为整个球面的一半，带入数值为 $2×3.14×{1000}^{2}\text{ }\text{ }{\text{m}}^{2}=6.28\text{\hspace{0.17em}}{\text{km}}^{2}$，按人口密度8000人/km2计算，可居住约5万人。

6. 刘慈欣椭球

Figure 4. Liu Cixin ellipsoid space city

$\left\{\begin{array}{l}x=b\mathrm{cos}\phi \\ y=a\mathrm{sin}\phi \end{array}$ (10)

$\frac{\text{d}y}{\text{d}x}=\frac{a\mathrm{cos}\phi \text{d}\phi }{-b\mathrm{sin}\phi \text{d}\phi }=-\frac{a}{b}\text{ctg}\phi$ (11)

$-\frac{\text{d}x}{\text{d}y}=-\frac{-b\mathrm{sin}\phi \text{d}\phi }{a\mathrm{cos}\phi \text{d}\phi }=\frac{b}{a}\text{tg}\phi$ (12)

$\frac{b}{a}\text{tg}\phi =\text{tg}{30}^{\circ }=\frac{\sqrt{3}}{3}$

$\text{tg}\phi =\frac{a}{b}\cdot \frac{\sqrt{3}}{3}$ (13)

$L=4{\int }_{0}^{\pi /2}\sqrt{\text{d}{x}^{2}+\text{d}{y}^{2}}=4{\int }_{0}^{\pi /2}\sqrt{{b}^{2}{\mathrm{sin}}^{2}\phi +{a}^{2}{\mathrm{cos}}^{2}\phi }\text{d}\phi =4{\int }_{0}^{\pi /2}\sqrt{{a}^{2}-\left({a}^{2}-{b}^{2}\right){\mathrm{sin}}^{2}\phi }\text{d}\phi$

$L=4a{\int }_{0}^{\pi /2}\sqrt{1-\frac{{a}^{2}-{b}^{2}}{{a}^{2}}{\mathrm{sin}}^{2}\phi }\text{d}\phi =4a{\int }_{0}^{\pi /2}\sqrt{1-{e}^{2}{\mathrm{sin}}^{2}\phi }\text{d}\phi$ (14)

$L=\pi \left(a+b\right)\left(1+\frac{3h}{10+\sqrt{4-3h}}\right)$$h=\frac{{\left(a-b\right)}^{2}}{{\left(a+b\right)}^{2}}$ (15)

$b=1000\text{\hspace{0.17em}}\text{m}$$a=\sqrt{3}b=1732\text{\hspace{0.17em}}\text{m}$，计算得 $h=7-4\sqrt{3}$$L=8737.6\text{\hspace{0.17em}}\text{m}$。假设椭球体内气压为均匀，切出半个长轴椭球，根据力的平衡条件，有：

$Lt\cdot {\sigma }_{p}=\pi ab\cdot p$

${\sigma }_{p}=\frac{\pi ab}{Lt}p$ (16)

${\sigma }_{e}={p}_{e}r/t=\stackrel{¯}{m}{\omega }^{2}{r}^{2}/t$ (17)

$S={\int }_{0}^{\phi }2\pi b\mathrm{cos}\phi \cdot \sqrt{{b}^{2}{\mathrm{sin}}^{2}\phi +{a}^{2}{\mathrm{cos}}^{2}\phi }\text{d}\phi =2\pi ab{\int }_{0}^{\phi }\sqrt{\frac{{b}^{2}}{{a}^{2}}{\mathrm{sin}}^{2}\phi +{\mathrm{cos}}^{2}\phi }\text{d}\left(\mathrm{sin}\phi \right)$

$\gamma =\mathrm{sin}\phi$，有

$S=2\pi ab{\int }_{0}^{\mathrm{sin}\phi }\sqrt{\frac{{b}^{2}}{{a}^{2}}{\gamma }^{2}+\left(1-{\gamma }^{2}\right)}\text{d}\gamma =2\pi ab{\int }_{0}^{\mathrm{sin}\phi }\sqrt{1-\frac{{a}^{2}-{b}^{2}}{{a}^{2}}{\gamma }^{2}}\text{d}\gamma$

$S=2\pi ab{\int }_{0}^{\mathrm{arcsin}\left(e\mathrm{sin}\phi \right)}\frac{1}{e}{\mathrm{cos}}^{2}\beta \text{d}\beta =\frac{2\pi ab}{e}{\left[\frac{1}{2}\beta +\frac{1}{4}\mathrm{sin}2\beta \right]}_{0}^{\mathrm{arcsin}\left(e\mathrm{sin}\phi \right)}$

$S=\frac{\pi ab}{e}\left(\mathrm{arcsin}\left(e\mathrm{sin}\phi \right)+e\mathrm{sin}\phi \sqrt{1-{e}^{2}{\mathrm{sin}}^{2}\phi }\right)$ (18)

$S=\frac{\pi ab}{e}\left(\mathrm{arcsin}\left(e\right)+e\sqrt{1-{e}^{2}}\right)=\pi {b}^{2}+\pi ab\cdot \mathrm{arcsin}\left(e\right)/e$ (19)

$b=1000\text{\hspace{0.17em}}\text{m}$$a=\sqrt{3}b=1732\text{\hspace{0.17em}}\text{m}$$\phi ={45}^{\circ }$，代入有 $e=\sqrt{2/3}$$e\mathrm{sin}\phi =\sqrt{1/3}$，整个椭球体表面积为 $2\left(3.14+3.14×3/\sqrt{2}×\mathrm{arcsin}\sqrt{2/3}\right)×{1000}^{2}\text{ }\text{ }{\text{m}}^{2}=19.01\text{\hspace{0.17em}}{\text{km}}^{\text{2}}$，坡度较缓的宜居面积为 $2\left(3.14+3.14×3/\sqrt{2}×\mathrm{arcsin}\sqrt{1/3}\right)×{1000}^{2}\text{ }\text{ }{\text{m}}^{2}=14.48\text{\hspace{0.17em}}{\text{km}}^{\text{2}}$，利用率为76%。按人口密度8000人/km2计算，可居住 $14.48×8000=115840$ 人，即超过11万人。

$b=1000\text{\hspace{0.17em}}\text{m}$$a=3b=3000\text{\hspace{0.17em}}\text{m}$$\phi ={60}^{\circ }$，代入有 $e=\sqrt{8/9}$$e\mathrm{sin}\phi =\sqrt{2/3}$，整个椭球体表面积为 $2\left(3.14+3.14×9/2\sqrt{2}×\mathrm{arcsin}\sqrt{8/9}\right)×{1000}^{2}=30.88\text{\hspace{0.17em}}{\text{km}}^{\text{2}}$，宜居面积为 $2\left(3.14×3/2+3.14×9/2\sqrt{2}×\mathrm{arcsin}\sqrt{2/3}\right)×{1000}^{2}\text{ }\text{ }{\text{m}}^{2}=28.51\text{\hspace{0.17em}}{\text{km}}^{2}$，利用率为92%。按人口密度8000人/km2计算，可居住 $28.51×8000=228080$ 人，即超过22万人。

7. 奥尼尔圆筒

Figure 5. O’Neill cylinder space city

$2t\cdot \sigma =2r\cdot \left({p}_{0}+\rho t{\omega }^{2}r+\stackrel{¯}{m}{\omega }^{2}r\right)$

$\sigma =\frac{r}{t}{p}_{0}+\rho {\omega }^{2}{r}^{2}+\frac{r}{t}\stackrel{¯}{m}{\omega }^{2}r$ (20)

8. 太空城市建造

Figure 6. Combination configuration of space city

Figure 7. Sketch map of Lagrange points

1) 在低轨道建设第五代空间站，测试模拟重力的实际效果并不断完善，探索空间站旋转启停技术和与之配套的停靠补给技术，积累人类在其中较长期生活的健康保持技术。

2) 在中轨道建设小型斯坦福圆环太空城市，完善太空焊接组装施工技术，开发太空旅游观光等商业价值。可在中高轨道建造多个旋转展开的中小型太阳能薄膜发电站，并将清洁能源输送回地球。

3) 在高轨道建设大型斯坦福圆环太空城市，探索基于太阳光压的轨道维持技术，拓展人类生存空间，并将之作为火箭燃料补给的停靠点，打造为深空探测前哨，不断发展太空农业和太空医疗等功能。

4) 建设月球生产和生活基地，开发月球各类资源，冶炼生产构筑太空城市的建筑材料，建立高效电磁推进装置以节省运送费用，不断积累人类长期驻留地外空间的各项技术。

5) 利用月球基地，在地月拉格朗日L4、L5点建设大型太空城市，可作为重要的天文观测基地和太空生产基地。采用积极的生态循环方法，不断完善太空城市封闭的生态系统维持技术。

6) 在日地拉格朗日L1点建设可容纳上百万人口的太空城市群，可作为探测金星的优良基地。建造巨型太阳能薄膜发电站，可在获取清洁能源的同时为地球打伞“遮阳”，以减缓地球气候变暖。

9. 结语

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