In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2], which doesn't even converge for h < 2 r. The area element for the general x,y,z parametric surface is just Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]] (I don't recall this from 18.03.) E.g., for the torus with radii r and R, In[116]:= % /. {x :> (Cos[#2]*(R + r*Sin[#1]) &), y :> (Sin[#2]*(R + r*Sin[#1]) &), z :> (r*Cos[#1] &)} Out[116]= Norm[Cross[({Cos[#2] (R + r Sin[#1]) & , Sin[#2] (R + r Sin[#1]) & , (1,0) r Cos[#1] & }) , ({Cos[#2] (R + r Sin[#1]) & , (0,1) Sin[#2] (R + r Sin[#1]) & , r Cos[#1] & }) ]] In[117]:= % /. f_[L_List] :> f /@ L /. Abs[w_] -> w Out[117]= Sqrt[Power[-(-r Sin[#1] & ) (Cos[#2] (R + r Sin[#1]) & ) + (0 & ) (r Cos[#1] Sin[#2] & ), 2] + Power[-(0 & ) (r Cos[#1] Cos[#2] & ) + (-r Sin[#1] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2] + Power[(r Cos[#1] Cos[#2] & ) (Cos[#2] (R + r Sin[#1]) & ) - (r Cos[#1] Sin[#2] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2]] In[119]:= % /. (f_ &) :> (f &)[t, u] Out[119]= 2 2 2 2 Sqrt[r Cos[u] Sin[t] (R + r Sin[t]) + 2 2 2 2 r Sin[t] (R + r Sin[t]) Sin[u] + 2 2 2 (r Cos[t] Cos[u] (R + r Sin[t]) + r Cos[t] (R + r Sin[t]) Sin[u] ) ] In[120]:= FullSimplify[%] Out[120]= 2 2 Sqrt[r (R + r Sin[t]) ] In[125]:= Assuming[R > r > 0, Integrate[%120, {t, 0, 2*Pi}, {u, 0, 2*Pi}]] Out[125]= 2 4 Pi r R Did Archimedes know this? --rwg
Somewhat apropos ellipsoid geodesics, On Sat, Apr 24, 2010 at 5:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get
pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2],
which doesn't even converge for h < 2 r.
The area element for the general x,y,z parametric surface is just
Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]]
(I don't recall this from 18.03.)
[...] Neil has hanging above his computer a lamellation of not-so-little laminated cards full of those good old geometry formulas, and some not-so-good ones! One clearly states (and illustrates) that the side area of a skew circular cylinder is just s/h times the side area of an unskewed cylinder, where h is the height and s is then slant height! It goes on to claim this is true for *arbitrary* (noncircular) cylinders! For the circular case, radius r, height h, slant height s, Mma and I get 2*r*(s*EllipticE[1 - h^2/s^2] + h*EllipticE[1 - s^2/h^2]). For a slant of 45 degrees, this gives (h*r*(Gamma[-(1/4)]^2 + 4*Gamma[1/4]^2))/(4*Sqrt[2*Pi]) ~7.6404 h r, compared to 2 pi h r for unslanted, and compared to 2 sqrt(2) pi h r ~8.88577 h r for that bogus card! I wonder how many of these cards decorate the walls of Boeing and Lockheed. --rwg
One clearly states (and illustrates) that the side area of a skew circular cylinder is just s/h times the side area of an unskewed cylinder, where h is the height and s is then slant height! It goes on to claim this is true for *arbitrary* (noncircular) cylinders!
It is the supremum for arbitrary cylinders; however, no cylinders actually attain this value.
I wonder how many of these cards decorate the walls of Boeing and Lockheed.
(More worryingly, I wonder how many of these cards were used by the team designing Boeing and Lockheed. : ) Sincerely, Adam P. Goucher
Neil's cheater cards again: Copyright 1984 by Papertech Marketing Group (Canada), printed in Canada, Science Editor C. Bello, M.A. Sc.; Design Editor S. M. Cali. SPHERES AND SPHERIODS (sic) Gives area of spheroids (needlessly separating prolate and oblate cases, perhaps to avoid imaginary/imaginary) in terms of eccentricity, for which they give a formula with units of sqrt(length) ! The oblate case has a completely undefined "n" which should have been pi. This is a clearcut attempt to sabotage American industry and education. --rwg Using the correct eccentricity formula, the oblate case actually agrees with Macsyma's geofuncs formula, 2 * %pi * b * (((a * asinh(sqrt(((b^2)/(a^2)) - 1)))/(sqrt(((b^2)/(a^2)) - 1))) + b) (Henry's beloved arcsinh again) and the prolate case agrees also, modulo a major bout of Mma's FullSimplify. --Bill On Mon, Jun 27, 2011 at 2:32 AM, Bill Gosper <billgosper@gmail.com> wrote:
Somewhat apropos ellipsoid geodesics,
On Sat, Apr 24, 2010 at 5:45 PM, Bill Gosper <billgosper@gmail.com> wrote:
In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get
pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2],
which doesn't even converge for h < 2 r.
The area element for the general x,y,z parametric surface is just
Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]]
(I don't recall this from 18.03.)
[...]
Neil has hanging above his computer a lamellation of not-so-little laminated cards full of those good old geometry formulas, and some not-so-good ones! One clearly states (and illustrates) that the side area of a skew circular cylinder is just s/h times the side area of an unskewed cylinder, where h is the height and s is then slant height! It goes on to claim this is true for *arbitrary* (noncircular) cylinders! For the circular case, radius r, height h, slant height s, Mma and I get
2*r*(s*EllipticE[1 - h^2/s^2] + h*EllipticE[1 - s^2/h^2]).
For a slant of 45 degrees, this gives
(h*r*(Gamma[-(1/4)]^2 + 4*Gamma[1/4]^2))/(4*Sqrt[2*Pi])
~7.6404 h r, compared to 2 pi h r for unslanted, and compared to 2 sqrt(2) pi h r ~8.88577 h r for that bogus card! I wonder how many of these cards decorate the walls of Boeing and Lockheed. --rwg
participants (2)
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Adam P. Goucher -
Bill Gosper