Euler s rule

The closure of the graphite sheets can be explained by the substitution of pentagons for hexagons in the nanotube sheets. Six pentagons are necessary to close a tube (and Euler s Rule is not violated). Hexagon... 

The existence of other hydrate structures is also suggested by the striking analogy of hydrate cavities to the large Buckminsterfullerene family of carbon cavities. Both types of cavities obey Euler s Rule cavities have exactly 12 pentagonal faces and any number of hexagonal faces, except one. Hydrates have the additional restriction that cavities should fill space continuously without excessive strain on the hydrogen bonds. 

Vickery and Taylor (81) used a Naphtali-Sandholm method containing all of the MESH equations and variables [M2C + 3) equations] with the variables represented by x. H is the Jacobian from the Naphtali-Sandholm method solution of the known problem, G(x) = 0, This is numerically integrated from t = 0 to t - 1, finding a H, at each Step and updating H when the solution is reached at each step, With Hj. and H, known, dxjdt is solved, and with step size t, a new set of values for the independent variables x is found by Euler s rule... 

Adams-Moulton Family of Methods One-step Adams-Moulton, backward Euler s rule... 

Equation [26] is simplified by applying Euler s rule, then separating the real and imaginary parts, so Eq. [22] becomes the solution for the quasi steady-state case... 

Here,/, v, and e are the numbers of the faces, vertices, and edges, respectively, and/ is equal to p+ h. Since each edge shares two polygons, we have the relationship 2e = 5p+ 6h. We also have the relationship 3v = 5p+ 6h because each vertex shares three polygons. Thus, we obtain the relationshipp = 6(/+ v — e), which givesp = 12, and V = 20+ 2h from Euler s rule. Namely, p must always be 12, and fullerenes must have even numbers of carbon atoms. If = 0, u is 20. This is the smallest fullerene... 

Now recall Euler s rule that e = cos(0) + i sin(0) and apply it to both terms. We do not know Cl or C2 but there would be a cos( ) part from both terms and a sin() part from each term. Let the coefficients of these two parts be new unknown constants so we have... 

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