Dimensions of the structure
A) Form.
De-sideribus should have a shape that allows it to contain the entire solar system up to, at least, Pluto. At first glance one might think that the shape of the sphere is the most suitable, but one must take into account the strong eccentricity of Pluto's orbit and its inclination with respect to the plane of the Sun. To avoid the hypothetical waste of materials caused by useless empty spaces the most suitable form is that of the ellipsoid:
B) Dimensions.
The ellipsoid, as a three-dimensional development of two ellipses that meet perpendicularly, is characterized by three semi-axes: rx, ry, rz. To know the dimensions of the megastructure it is necessary to find those of the drive shafts. For those coplanar to the plane of the Sun we choose to use the two semi-axes of the Kuiper belt, the external asteroid belt, in addition to the orbit of Pluto, this because, assuming we build a structure that excludes them, there would be the risk that the bodies present in the fascia collide with the surface of the ellipsoid.
For a purely conceptual matter, all the values we are going to find are approximated by excess, imagining that we leave an empty margin space between the end of the structure and the chosen reference point:
1ua150,000,000 km, Earth-Sun distance
Kuiper band semimajor axis = 55 au — > rx > 55 au
Semi minor axis kuiper = 39.5 au —> ry > 39.5 au
As for the semi-axis rz, it is necessary to deal with the problem of the inclination of Pluto's orbit with respect to the Sun:
As can be seen from the image above, Pluto is the planet that most of all deviates from the coplanar reality of the solar system;
its orbit has an inclination of about 17.13°.
To correctly calculate the length that rz should have (and avoid that our favorite dwarf planet crashes into a not very credible mega-structure), let's represent the problem on the Cartesian axes:
Thanks to the sine and cosine definitions, and knowing both the angle a (which is the inclination of Pluto's orbit) and the distance od (Pluto's aphelion), we are able to find the lengths of the segments db and ob
od = 49.31 AU
a= 17.13826°
db = od x sin(a)= 49.31 x sin(17.13826°) = 14.68 au
ob = od x cos(a)=49.31 x cos(17.13826°) = 47.60 au
By adding the length of rx, and obtaining from these values the coordinates of any point of the ellipse with semiaxes rx and rz, through the definition of the ellipse itself and its equation, it is possible to obtain the value of rz:
oc= rx= 56 au
d= (47.60;14.68)
x 2 /rx 2 + y 2 /rz 2 = 1 — > 47.60 ²/56² + 14.68²/rz²=1 — >rz= 27.86728 au
Ultimately, the three semiaxes have values:
rx > 55 au = 56 au
ry >39.5 ua = 40 ua
rz > 27.86728 au = 28 au
We also calculate volume and surface area:
Ellipsoid volume= 4/3 π X x X y X z= 262.720,92164 ua³
Surface area = 13,990.87803 ua ²
Where a=rx, b=ry, c=rz, p=1.6, given as a constant to be used in calculating the surface area of ellipsoids
To give an idea of the dimensions that our structure would have with respect to the solar system, we add the link for the geogebra page in which we have represented in scale the three major ellipses of the ellipsoid, the orbits of the planets and of Pluto, and the dimensions of the Sun (all orbits have been flattened in the plane of the Sun, except Pluto): https://www.geogebra.org/classic/cwpqzfx6