In what way is the Riemann curvature tensor related to 'radius of curvature'?

In Misner, Thorne & Wheeler, they say, in their delightful 'word equations' that

$$\left(\frac{\mathrm{radius\,\, of \,\,curvature}}{\mathrm{of\,\, spacetime}}\right) = \left(\frac{\mathrm{typical\,\, component\,\, of\,\, Riemann\,\, tensor}}{\mathrm{as\,\, measured\,\, in \,\,a \,\,local \,\,Lorentz\,\, frame}}\right)^{-\frac{1}{2}}.$$

My question is: does this definition of radius of curvature (and others like it - where tensor are described in words) depend on the valence of the Riemann curvature tensor?

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To see that the estimate is right, just calculate the Riemann tensor for a sphere of radius $a$. You will get the Ricci scalar equal to something like $2/a^2$. The only thing you need to fix in your equation is the power. The typical component of the Riemann tensor goes like $a^{-2}$ where $a$ is the curvature radius. The power may be calculated by dimensional analysis. I guess that you – or they? – just forgot the power.
Thanks for the reply. Your comment on the difference between in/extrinsic curvature is helpful (note that the links were added by Qmechanic). Also, very insightful is the comment on the word 'typical', +1 for those. Yes, I forgot the power, I'll edit that. Although, one thing I'm still not clear on is how the valence of the Riemann tensor affects its physical meaning, or does it not? E.g. does it matter if in the above equation we use $R^{a}_{bcd}$ or $R_{abcd}$? –  user12345 Feb 16 '13 at 14:04