Understanding the Maths Behind Space Elevators

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SUMMARY

This discussion focuses on the mathematical principles behind space elevators, specifically the calculation of the volume of a frustum of a cone. The volume formula for a cone, V = (1/3)Ah, is applied to derive the volume of the frustum by subtracting the volume of the missing top section. Additionally, the discussion highlights the method of using a single integral to calculate the volume through the surface of revolution, emphasizing the importance of finding the right boundary equation and integrating over the y-axis.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of volume of revolution.
  • Knowledge of geometric shapes, particularly cones and frustums.
  • Basic algebra skills for manipulating equations.
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  • Study the concept of volume of revolution in calculus.
  • Learn about frustums and their properties in geometry.
  • Explore integration techniques for calculating volumes, including definite integrals.
  • Investigate the applications of space elevator designs in aerospace engineering.
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Students and professionals in mathematics, physics, and engineering, particularly those interested in aerospace applications and the theoretical foundations of space elevators.

RooftopDuvet
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I'm doing some research on space elevators and have found a site (http://www.zadar.net/space-elevator/#transverse") which gives some insight into the maths behind the elevator cable. If you click on the link and scroll down, up, whichever direction to equation (2) and a picture of a blue trapezium the guy has found the difference in the volume of the tapering cable with what looks like volume of revolution, but I can't seem to figure out what he's done. He says that it's 'easy to show...'. Am I missing something? I could use a little help. thanks.
 
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Quick way to show this: You are dealing with finding the volume of a frustum of a cone. The volume [itex]V[/itex] of a cone is [itex]V=\frac{1}{3}Ah[/itex], where [itex]A[/itex] is the area of the base and [itex]h[/itex] is the height.

You can get the volume of the frustum illustrated in the figure by repeatedly applying the above formula and subtracting volumes, as follows:

[tex]V=V_{frustum+missing top}-V_{missing top}[/tex]

Long way to show this: You can calculate the volume with a single integral because the frustum is a surface of revolution. Find an equation for the right boundary of the figure (it's a straight line segment, so that's easy) and solve it for x. Then revolve that line segment about the y-axis and write down an expression for the differential volume. Then integrate over y, and you should get the same expression.
 

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