User talk:Rmo13

From Wikipedia, the free encyclopedia

You have new messages (last change).

Thanks Thanks for the article on Jerzy Petersburski. I was about to write it myself... Halibutt 13:28, 10 February 2006 (UTC)

Hi there! I replied to your questions at Talk:Rzeczpospolita. BTW, you might want to sign your posts next time you make a comment at the talk page. To do this just write four tildes at the end (~~~~) and the wiki will convert that to your name and exact date. Halibutt 18:06, 22 February 2006 (UTC)

[edit] Tide

Dear Rmo13, At Tide, you added:

Atmospheric tides are the dominant dynmics from about 80 km to 120 km where the molecular density is too small, and are both gravitational and thermal in origin.

Hmm? Too small for what? -- JEBrown87544 15:51, 21 March 2007 (UTC)

[edit] Earth radius; oblique curvature

Hiya Rmo13!
Nice addition to Earth radius! P=)
I particularly like your touching on the rhumb/Pythagorean relationship to radius of curvature——though there does need to be a distinction made between spherical ("globoidal") azimuth, denoting here as \widehat{\alpha}\,\!, and the (local) elliptical/geodetic azimuth, \tilde{\alpha}(\phi)\,\!:

R_c=\frac{{}_{1}}{\frac{\cos(\tilde{\alpha}(\phi))^2}{M}+\frac{\sin(\tilde{\alpha}(\phi))^2}{N}}=\frac{(M\cos(\widehat{\alpha}))^2+(N\sin(\widehat{\alpha}))^2}{M\cos(\widehat{\alpha})^2+N\sin(\widehat{\alpha})^2},\,\!

as \tilde{\alpha}(\phi) =\arctan\left(\frac{N(\alpha)\sin(\widehat{\alpha})}{M(\alpha)\cos(\widehat{\alpha})}\right) =\arctan\left(\frac{N}{M}\tan(\widehat{\alpha})\right).\,\!


In your relating D\cos\widehat{\alpha}\approx Md\phi\,\! and D\sin\widehat{\alpha}\approx N\cos\phi d\lambda\,\!, you miss a more profound relationship:

\begin{align}D&\approx\sqrt{(Md\phi)^2+(N\cos(\phi)d\lambda)^2},\\ &=\sqrt{(M\cos(\widehat{\alpha}))^2+(N\sin(\widehat{\alpha}))^2}\sqrt{(d\phi)^2+(\cos(\phi)d\lambda)^2},\\ &=\frac{\sqrt{(d\phi)^2+(\cos(\phi)d\lambda)^2}}{\sqrt{(\frac{\cos(\tilde{\alpha}(\phi))}{M})^2+(\frac{\sin(\tilde{\alpha}(\phi))}{N}})^2},\\ &=\widehat{\overline{O}}(\widehat{\alpha},\phi)\sqrt{(d\phi)^2+(\cos(\phi)d\lambda)^2},\\ &=\tilde{\overline{O}}(\tilde{\alpha}(\phi),\phi)\sqrt{(d\phi)^2+(\cos(\phi)d\lambda)^2};\end{align};\,\!

Thus,

\begin{align}R_o=O&=\widehat{\overline{O}}(\widehat{\alpha},\phi)=\tilde{\overline{O}}(\tilde{\alpha}(\phi),\phi),\\ &=\sqrt{(M\cos(\widehat{\alpha}))^2+(N\sin(\widehat{\alpha}))^2} =\frac{1}{\sqrt{(\frac{\cos(\tilde{\alpha}(\phi))}{M})^2+(\frac{\sin(\tilde{\alpha}(\phi))}{N}})^2};\end{align}\,\!

With those relationships in mind, would O be considered the "oblique radius of curvature" (i.e., radius of plane curve arc, or "radius of arc", or "arcradius"), and the curvature, itself, "oblique curvature" (i.e., plane curve arc), as north-south curvature is "meridional curvature" and east-west curvature is "normal curvature", with their respective radii, M and N? ~Kaimbridge~19:40, 28 March 2007 (UTC)

Retrieved from "http://en.wikipedia.org../../../r/m/o/User_talk%7ERmo13_f25d.html"