Nanomedicine, Volume I: Basic Capabilities
© 1999 Robert A. Freitas Jr. All Rights Reserved.
Robert A. Freitas Jr., Nanomedicine, Volume I: Basic Capabilities, Landes Bioscience, Georgetown, TX, 1999
5.2.1 Free-Floating Solitary Nanodevices
Free-floating simple nanorobots intended solely as materials delivery or storage devices, or used as omnidirectional communications, navigational or control transponders operating independently in vivo, have no preferred orientation. Thus these nanorobots may be spherically symmetric with rigid surfaces. Spherical particles produce the smallest possible increment in blood viscosity as they tumble, in part because spheres offer the smallest possible interaction surface per unit volume of any geometrical shape. Such simple nanomachines are the kind most likely to be deployed in the greatest numbers, and their reduced surface area minimizes potential biocompatibility problems.
These free-floating nanodevices must have ready access to all tissues via blood vessels. Since they are nonmotile machines, to avoid getting stuck they must not physically extend wider across their longest axis than the width of human capillaries, which average 8 microns in diameter but may be as narrow as 4 microns (Section 8.2.1.2).
Consider a nanodevice of fixed surface area An, total volume Vn, and longest transdevice diameter Ln. Then:
A. For a spherical device of radius r, then Ln = 2 r, An = 4 p r2, and Vn = (4/3) p r3.
B. For a prolate spheroidal (football-shaped) device of length 2a, width 2b, and eccentricity e = (a2 - b2)1/2 / a, then An = 2 p b2 + 2 p a b (arcsin(e) / e) and Vn = (4/3) p a b2. For an oblate spheroidal device, An = 2 p a2 + p b2 (ln({1+e}/{1-e}) / e) and Vn = (4/3) p a2 b. In both cases, Ln = 2a and the maximum enclosed volume per unit area occurs at a = b (e.g., a sphere).
C. For a right circular disk or cylindrical device of radius r and height h, then Ln = (h2 + 4r2)1/2, An = 2 p r (h + r), and Vn = p r2 h. Maximum enclosed volume per unit area occurs at h = 21/2 r.
D. For a right circular conical device of radius r and height h, then Ln = 2r for h < 31/2 r (but Ln = (h2 + r2)1/2 if h > 31/2 r), An = p ( r2 + r (h2 + r2)1/2), and Vn = p r2 h / 3. Maximum enclosed volume per unit area occurs at h = 31/2 r.
E. For a right triangular prismatic device (Fig. 5.4) with three equal sides of length s and height h, then Ln = (h2 + s2)1/2, An = 3 h s + 31/2 s2/ 2, and Vn = 31/2 s2 h / 4. Maximum enclosed volume per unit area occurs at h = s / 21/2.
F. For a cubical device with three equal sides of length s, then Ln = (31/2) s, An = 6 s2, and Vn = s3.
G. For a right square prismatic device with two equal sides of length s and height h, then Ln = (h2 + 2 s2)1/2, An = 2 s2 + 4 h s, and Vn = h s2. Maximum enclosed volume per unit area occurs at h = s.
H. For a right hexagonal prismatic device (Fig. 5.4) with six equal sides of length s and height h, then Ln = (h2 + 4 s2)1/2, An = 6 s (h + 31/2 s / 2), and Vn = 271/2 s2 h / 2. Maximum enclosed volume per unit area occurs at h = 21/2 s.
I. For a truncated octahedral device (Fig. 5.5) of edge s, then Ln = (101/2) s, An = (6 + 4321/2) s2, and Vn = (1281/2) s3.
J. For a rhombic dodecahedron1101 (Fig. 5.6) of edge s, then Ln = (481/2 / 3) s, An ~ (11.3137) s2, and Vn ~ (3.0792) s3.
K. For a nonregular octahedron (Fig. 5.8) of equatorial edge s and vertex edge (31/2 / 2) s, then Ln = (21/2) s, An = (81/2) s2, and Vn = (1/3) s3.
L. For a regular octahedron1101 (Fig. 5.9) of edge s, then Ln = (21/2) s, An = (121/2)s2, and Vn = (21/2 / 3) s3.
Table 5.1, computed using the above relations, confirms that spheres offer the greatest storage volume per unit surface area and thus are the most efficient shape for this application.
Last updated on 17 February 2003