- #1
Quinn Pochekailo
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Homework Statement
A torus has a major radius and a minor radius. When R>r by a magnitude of at least 4x, it comes to be a slim ring looking shape. When R>r by a magnitude of 1/2, the shape looks to be a donut. When R=r, the torus shape looks more like a sphere except with a small gap in the center of the shape.
I want to be able to show that volume and surface of a torus is greater than that of a sphere at the nanoscale level.
The purpose of this problem is to show that a torus at the nanoscale level can carry more particles and has a greater surface area, making this shape a more effective method than a spherical liposome.
Homework Equations
Torus:
V= 2*pi^2*R*r^2
SA = 4*pi^2*R*r
Sphere:
V= (4/3) * pi * r^2
SA = 4 * pi * r^2
The Attempt at a Solution
I started first by comparing the V and SA using R=5 and r=2.5 for simplicity and because I want the donut looking shape compared to the small ring (by using R= (1/2)r )
Final numbers for torus:
V= 616.225
SA = 492.98
Final numbers for sphere:
V = 26.17
SA = 78.5
I then calculated the magnitude of difference in V and SA, which turned out to be 23x and 6.3x respectively.
I expected to see the same results of magnitude at the nanoscale level.
Because I have to keep in mind the major radius of the torus, I made R+r = 1e-9. For the sphere it was just r = 1e-9.
Stating that R = 1/2r, I plugged that in my equation and made
(1/2r) + r = 1e-9.
r + 2r = 2e-9
3r = 2e-9
r = 6.7 e -10
R = .5 (6.7e-10) <---- R = 1/2r
R = 3.35e-10
Figuring out my values, I then used those numbers in my V and SA formulas
Torus:
V= 2.97 e -29
SA = 8.9 e-18
Sphere:
V= 4.2e-18
SA = 1.256e-17
As you can see, the calculations from the first results are very different and at the nanoscale this shows that the volume of a torus is actually less than that of a sphere. This makes sense because I am multiplying by a very small number (R) unlike that in the sphere formula.
Did I do my work wrong somewhere? Why is that at the nanoscale level the torus is less effective than the sphere?
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