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Feb 5, 2015 at 6:15 comment added geometrian For a real world example of this, many view $2\pi$ as a bletcherous fossil from the past.
Feb 3, 2015 at 10:12 comment added Joce It is not true that the final relation you have is not used. It is used by many physicists working with objects which have a variable volume (can be inflated/deflated) but a fixed surface area (inextensible material is forming their surface). That's a good approximation of what a ball is. $V_{sphere}$ in that case is the maximum volume you can get for such an object of given area. So yes, formulae with non-integer exponets are in use, no question about that. Why are their in less frequent use is the question: one answer is yours (not as convenient), and you can read mine on linearity.
Feb 3, 2015 at 5:56 comment added KCd @R.. The diameter, yes, but not the radius (directly, as I wrote). Even though a diameter can be measured, nevertheless we still most often give mathematical formulas related to a sphere in terms of its radius rather than its diameter.
Feb 3, 2015 at 5:01 comment added R.. GitHub STOP HELPING ICE @KCd: My calipers beg to differ. The diameter is trivial to measure, and radius has a natural and simple relationship to diameter, of course.
Feb 3, 2015 at 4:11 comment added KCd A sphere's radius is not easy to measure. If you are given a solid steel ball or a marble, you would be unable to measure the radius directly. A more relevant reason for describing spheres in terms of the radius comes from the standard equation of a sphere: $x^2 + y^2 + z^2 = R^2$. Whether or not the radius of a sphere is easily accessible to measurement (often it is not), in mathematical formulas it is a convenient way to distinguish one sphere from another (with the same center).
Feb 3, 2015 at 0:34 comment added Jeppe Stig Nielsen Another example would be Kepler's third law which we might write $T^2 = K r^3$. We could use fractions, $r = K' T^{2/3}$, but that looks uglier.
Feb 2, 2015 at 16:17 history answered Inquisitive CC BY-SA 3.0