Search
Random Quotation
For all their wealth of content, for all the sum of history and social institution invested in them, music, mathematics, and chess are resplendently useless (applied mathematics is a higher plumbing, a kind of music for the police band). They are metaphysically trivial, irresponsible. They refuse to relate outward, to take reality for arbiter. This is the source of their witchery.
The American Mathematical Monthly, v. 101, no. 9, November, 1994.
Loci: Convergence
show printer friendly
send to a friend
James Gregory and the Pappus-Guldin Theorem
More on the Trunk and the Cylinder
If AB and EF are two planar figures with centers of gravity a and e, respectively, and which generate cylinders with the same height, we saw previously that
{ rev(AB) \over rev(EF)} = { trunk(AB) \over trunk(EF)} { circum(AB) \over circum(EF)}
\eqalign{ { rev(AB) \over rev(EF)} &= { trunk(AB) \over trunk(EF)} { circum(AB) \over circum(EF)} \cr &= { trunk(AB) \over cyl(AB)}{ cyl(AB) \over cyl(EF)} {cyl(EF) \over trunk(EF)} { circum(AB) \over circum(EF)} \cr &= {circum(a) \over circum(AB)} { cyl(AB) \over cyl(EF)} { circum(EF) \over circum(e)} { circum(AB) \over circum(EF)} \cr &= {circum(a) \over circum(e)}{ cyl(AB) \over cyl(EF)} \cr &= {circum(a) \over circum(e)} { area(AB) \over area(EF)} \cr }
Notice that this is Pappus' version of the Pappus-Guldin theorem. The problem is that this proof only holds if both of the figures AB and EF are symmetric about an axis perpendicular to the axis of rotation. But it is not difficult to drop this requirement. If AB is not symmetric around such an axis, it can be reflected around an axis perpendicular to the axis of rotation to generate a figure HI consisting of two copies of AB which is symmetric around an axis.
Archimedes' result shows that the center of gravity h of HI will be the midpoint of the segment connecting the centers of gravity of the two reflected copies of AB. Since the axis of reflection is perpendicular to the axis of rotation, the centers of gravity of the two reflected copies of AB will be the same distance from the axis of rotation and hence the line connecting them will be parallel to the axis of rotation. In other words, the center of gravity of HI will be the same distance from the axis as the center of gravity a of AB, or circum(a) = circum(h). Note also that area(HI) = 2 area(AB) and rev(HI) = 2 rev(AB) since HI is twice the size of AB. Similar arguments apply to a symmetric figure JK with center of gravity j constructed in the same way from EF. So the result for symmetric figures implies
{2 \,rev(AB) \over 2\, rev(EF)} = { rev(HI) \over rev(JK)} = {circum(h) \over circum(j)} {area(HI) \over area(JK)} = {circum(a) \over circum(e)}{ 2\, area(AB) \over 2\, area(EF) }
Next page >> Gregory's Proof Revealed
Pages: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
Leahy, Andrew, "James Gregory and the Pappus-Guldin Theorem," Loci (January 2009), DOI: 10.4169/loci003262