User talk:Kaimbridge

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It is Friday 13 June 2008 @ 16:52 (UTC)

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[edit] Re(RD): Transverse meridional/"great-ellipse" arcradius?

Hmm...very interesting question...What is the context of the original problem? --HappyCamper 01:49, 30 October 2005 (UTC)

(move back to RD) ~Kaimbridge~ 20:36, 30 October 2005 (UTC)

[edit] Style, style, style

Hi Kaimbridge. I have one remark. According to some style rule, one should not put links in section headings, rather below in the text. For this reason I undid your change at quotient rule. I understand that sometimes links in section headings are more straightforward, but people say they don't look good there. :) (and I agree) Oleg Alexandrov (talk) 16:17, 7 December 2005 (UTC)

Okay, no problem (I'll tack it on the "see also" section). How about here, on the talk pages, such as the above heading/link? Did you see that question when I originally posted it? P=) ~Kaimbridge~ 19:51, 7 December 2005 (UTC)

[edit] Aspect ratio

I delinked "Aspect ratio" becuase disambiguation pages aren't meant to hold any information. If the Template:Planet Infobox/Earth was linking to Aspect ratio to access the dictionary defintion "The aspect ratio of a two-dimensional shape is the ratio of its longest dimension to its shortest dimension." then this is a problem because Wikipedia is not dictionary and that definiton will be removed eventually. You should only create an article Aspect ratio (ellipse) is it is more than a dictionary definition - or it will be deleted. If you want to link "Aspect ratio" to a definition, then I recommend a Wiktionary link.--Commander Keane 16:48, 10 December 2005 (UTC)

[edit] Re:question

I replied on my talk page. Sorry it took me so long, it got buried in other discussion. Oleg Alexandrov (talk) 02:08, 20 December 2005 (UTC)

I changed my mind, one argument is enough for the function. See my talk page. Oleg Alexandrov (talk) 03:02, 20 December 2005 (UTC)

[edit] Raggio mezzano

Hi,
I suspected something like that, but I wasn't sure. "mezzano" is actually derived from "mezzo", which means half, but it's a word used exclusively for pieces of forniture divided into two or more sections :-)) I'm going to reinsert your data with a fixed translation (medio (average) will do :-)) Cheers, Alfio 17:45, 29 December 2005 (UTC)

No problem——though that's why I included the "" comment next to it and link to Earth radius in the summary, just in case there was a question regarding meaning. ~Kaimbridge~ 18:42, 29 December 2005 (UTC)

[edit] Phi vs.varphi

(Moved original question from Kbolino's Talk page)

Is there a way, instead, to make phi in sans-serif to look like "regular" phi? The standard symbol for latitude is regular phi (see Vincenty's formula (PDF)). ~Kaimbridge~ 19:36, 19 February 2006 (UTC)

Not of which I am aware. It wouldn't hurt to change it back to \phi and explain in the article somewhere that the two are equivalent (the same letter, just written differently). I can do this if you'd like.—Kbolino 05:21, 20 February 2006 (UTC)

Well if you look at it in Courier text, none of them really look alike! P=) So I'd say let's put it back the way it was (it is linked to the φsymbol page). Also, I think the equations do look better with the indents. As for the "dot" vs. "times" signs (vs. no sign), I tend to go with the "times", but I think most go with no sign, so I guess the "dot" is a reasonable compromise. P=) Likewise, I prefer enclosures for all functions for consistency (e.g., "sin(P)" or, my preference, "sin{P}"), but I'll let it go here. P=) ~Kaimbridge~ 15:23, 20 February 2006 (UTC)

I tend to agree with you on function notation, but the trig functions are so common that it is customary to omit the parentheses when the argument contains a single term:

$\sin 2\theta , \quad \tan x^2 , \quad \mathrm{but} \ \csc(z + 1)$

The same is true for logarithms and hyperbolic trig. I object, however, to the use of curly braces—they imply a list, sequence, or series of terms, which is not applicable when defining the evaluation for a single, scalar value. As a result, the usage is uncommon (at best) and confusing to some. [Addendum: The cdot is purely personal, and I may have taken liberties I shouldn't have in inserting it. It is clearer to me, but not terribly common (implied multiplication being the trend) and some people don't know what it means. I also agree with the indentation: I simply changed them all to single indent for consistency.]—Kbolino 23:18, 20 February 2006 (UTC)

[edit] Modified equations in Rhumb line

I think the additional material you've put in the rhumb line article about the inverse Gudermannian function is great, but I disagree with your decision to replace all the previous equations with your new formulation. The other form, with logs, secants, and tangents, is much more common, and I think makes a great deal more sense to beginners, as most people who would read this article would know what the ln, sec, and tan functions are, without necessarily knowing about inverse hyperbolic trig functions. --Dantheox 01:32, 23 February 2006 (UTC)

Yeah, I suppose it wouldn't hurt to define the inverse gd function separately and give a tangible form of it, but I think I'll keep it as is for the tan(α) equation, since that expresses the fundamental form of the actual integration involved (i.e., why the integrand is $\frac{\Delta\lambda}{\sec(\phi)}$ and not

cos(φ)Δλ

). ~Kaimbridge~ 13:59, 23 February 2006 (UTC)

[edit] Arc (geometry) questions.

Hello Kaimbridge, thanks for the feedback. I'm sorry, though, I can't really help you out. I'm just starting to learn about the subject, and made the page primarily for disambiguation purposes. I was going to refer you to a page, but it turned out it was one you wrote, so I don't think it would have been any help ^_^ Also, most of the other pages I've been reading only deal with circular arcs, so they're no help either. Again, sorry I couldn't be more help. --Nekura 16:22, 9 June 2006 (UTC)

[edit] small details...

If you must write "antiln" instead of the standard "exp", I think your TeX code should say \opperatorname{antiln}, instead of the more complicated sequence you used. Is there a reason to prefer "antiln" to "exp"? Michael Hardy 21:57, 9 June 2006 (UTC)

No, you're right——I know "exp" is used in programming, but I wasn't sure about formal writing (most web references seem to formally present it as $\operatorname{e}^V\,\!$ ), rather than $\exp(V)\,\!$ , and I either didn't think to check, or overlooked it, on the TeX Formulapage. P=( ~Kaimbridge~ 13:44, 10 June 2006 (UTC)

... and if you use "exp", remember that in TeX it has a backslash: \exp Michael Hardy 22:00, 9 June 2006 (UTC)

[edit] Re: "Oblateness constant"

Haha, my lack of knowledge is showing up. I was working on Wikipedia:WikiProject Missing encyclopedic articles and 'oblateness constant' showed up. I attempted to define it on oblate and redirected the original, but it could do with a professional's view! Please do change it. Cheers akaDruid 15:26, 23 June 2006 (UTC)

[edit] Earth radius

Done :)

You can do whatever you want with the article now, I'm done with it. --Michael Retriever 17:16, 3 August 2006 (UTC)

[edit] Earth radius and finding the exact perimeter of an ellipse

Dear Kaimbridge,

The means available in the Earth radius article weren't satisfactory enough for me, that's why I searched an ideal Earth radius given that its true equation were x^2/a^2 + y^2/a^2 + z^2/b^2 = 1. Nevertheless, I suspected that the result would be equal or nearly equal to the quadratic mean, as I must say I was quite impressed to see that there is such a powerful yet simple averaging technique for three-dimensional cartesian spaces (I didn't know it before :P ).

First off, isn't it "x^2/a^2 + y^2/b^2 + z^2/b^2" rather than "x^2/a^2 + y^2/a^2 + z^2/b^2", since the equation of an ellipse is "x^2/a^2 + y^2/b^2= 1", the equator is a great circle and the north and south polar radii can be different?

How you represent it is completely up to you. When I play with equations in a three-dimensional cartesian space I imagine this isometric view

where b axis (polar radius) would correspond with Z; and a axis (equatorial radius) would correspond with both X and Y. A more familiar isometric view for you may be

where the equation is x^2/a^2 + y^2/b^2 + z^2/a^2 = 1. If you like to use x^2/a^2 + y^2/b^2 + z^2/b^2=1 then as I see it you would be playing with a view like this

where b would correspond to the equatorial radius and a to the polar radius (I am implying that when you imagine the Earth, you imagine it with its north pole looking upwards :P ). Also, it would be very kind of you to tell me which view they use in formal maths books when making three-dimensional graphics.

But you are still saying that two of the axes are equatorial and only one is polar——I'm saying shouldn't two be polar and one equatorial? As for which view books use for graphics, I have no idea (try googling for different images.) P=(

Isn't the Earth like this?

If it's not, then my radius mean and the quadratic mean are incorrect.

About the z matter and other possible functions for ellipse-related matters, I'm not fond of trigonometric functions. That explains why I used that function instead of the ones you propose :) What I think is important to mention however is that the function you used to calculate the perimeter/longitude of the Earth's great ellipse in the Meridional Earth radius is an approximation, and not the exact precise function (I think you already know this :P ). The true integral function is this one; or the integral you showed yourself?, which is a rather nicer one actually.

Of course ${}_{\left[\frac{a^{1.5}+b^{1.5}}{2}\right]^{1/1.5}}\,\!$ is an approximation——that's why I used "≈"! P=) And, yes, the complete EI2K equals M's in the complete case (i.e., 0 → 90°), as does the integral for the radius (again, in the complete case), as well even a few others: The thing is, in less than the complete case, only M's is valid!

Well, the thing is that for what I know, all integral functions which cannot be integrated with regular techniques can be translated into infinite polynomial summations (that's what the elonged S symbol in integrals means after all), and that's what I just used in my calculus. The z function I used is here, but there are many other summations which can be used in an equally powerful manner, take for example Euler's early one.

a^2 = (n+1)b^2; (a^2/b^2) - 1 = n; n = 0.0067395962783419227214499918750432

After that, you calculate the infinite series (I did it until - (273922023375/106542032486400) (n^8) = 1.0944023775784746584284715852771e-20 using the Microsoft Calculator :P ), multiply it with 4(1/2)(π)(b) = (2)(π)(b), and you get that the Earth's great ellipse longitude is 40007861.930850134920636564989507. I did the same with z (until + (654729075/3715891200)^2 · e^20/19 = 2.9540033942026385623871168669082e-25) and I got that the ellipse longitude is 40007861.930850134920638871082041. So as you can see, calculating the exact length of an ellipse's longitude is possible, it all depends on how many terms you use in the infinite summations. Muir's and Ramanujan's approximations are, after all, fairly precise approximations.

Right, what you are talking about are integrals' binomial series expansions. I much prefer trapezoidal quadrature for just immediate approximation of an integrand's average ("divided difference") and Gaussian quadrature for the "precise" average. Oh, and PLEASSSSSSSSSE tell me you are not doing this, key-by-key on a calculator (MS or otherwise)!!?!!?!!P=) If you are, you need to get UBasic——AND I MEAN YESTERDAY!!!!! P=) P=)

Yes I am!! Yes I am!! :P

There is also another matter which I want to discuss, about your Meridional Earth radius mean. When you calculate this mean, you're making the average radius of the Earth where (longitude of the great ellipse)/(2π)= Mr as I understand it. But this is of no use when you consider that the Earth is three-dimensional and not two-dimensional, is it? This mean is to find a radius of a supposed circular shape of the great ellipse, and not to find a radius of a supposed shperical shape of the ellipsoid. I'm not sure if I make myself clear, this is one of these things which is simpler to explain with a drawing; if you don't understand me please do tell.--Michael Retriever 13:36, 5 August 2006 (UTC)

I totally agree (YOU were the one saying in your first edit that 6,367.4489..——Mr——is "the one that you would get supposing that the Earth is a perfect spheroid/ellipsoid"). You're using $\sqrt{a\cdot Mr}\,\!$ as the approximation, in the same way that $\sqrt{ab}\,\!$ is an approximation of Mr——That's why I say Qr is the ellipsoidal quadratic mean (the elliptical quadratic mean would be ${}_{\sqrt{\frac{a^2+b^2}{2}}}\,\!$ , which would be the approximation fir Mr).

BTW, what is your age and math level status——I can't tell if you're still just a pimple faced kid in school, a retired old geiser, or something in between?!? P=) P=) P=) ~Kaimbridge~ 15:48, 5 August 2006 (UTC)

Not pimpled, finished college, little formal mathematical formation, found a punctual problem when trying to know the exact distance between me and someone else in the world, grew an obsessive interest in finding a solution for it in a matter of days, found a satisfactory answer contrasting different sources.

SAME HERE!!! I started about exactly 20 years ago!! P=) Your sources wouldn't be Vincenty's and Sodano's formulae, would it?

Nope xD xD. Sources are Wikipedia, my college notes on geometry, and questions I ask to a professor from time to time.

And who told you I was finished after the first edit? I wasn't!! :P

That's why you shouldn't execute/save it until you are done (you ought to create a "sandbox" on your user page, such as "User:Michael_Retriever/SandBox" or "User:Michael_Retriever/WorkBench" to create your edits on——and for articles that are going to take a while, cut & paste it into a simple text file and save it on your DeskTop).

The sandbox page is a very good idea, thank you.

Hey btw you say you know how to find the length of a portion of an ellipse's perimeter? Can you tell me how to do that in a simple way? --Michael Retriever 17:15, 5 August 2006 (UTC)

I believe a portion of ellipse perimeter equals the corresponding portion of a meridian's perimeter, in which case,

$\int_{0}^{90^\circ}\!M(l)\,dl\to\int_{l_b}^{l_d}\!M(l)\,dl\,\!$

where l_b and l_d are the basal ("base") and destination latitudes.

If you are interested in my idea on how to calculate along any angled great ellipse, see second part of this discussion! P=) ~Kaimbridge~ 20:46, 5 August 2006 (UTC)

Way too much trigonometry in that page, I'll take a break and tell you things about it in some days.--Michael Retriever 22:27, 6 August 2006 (UTC)

[edit] egg in ellipsoid

Right - good point. The so-called "egg" is now vertical :-) Deuar 15:16, 23 August 2006 (UTC)

Beats me - the "heavy end down" configuration seemed more natural to me, for example, but feel free to change it if you like. Deuar 15:48, 23 August 2006 (UTC)

[edit] Earth vs. "the earth"

Hiya Kris! Regarding "E"arth vs. "e"arth, isn't a good guideline, if it is preceeded by "the", it means Earth as a common noun, in the general, planetary sense, not requiring capitalization (e.g., "but the earth's composition" = "but the planet's composition"), whereas "Earth's..." refers to Earth as a proper noun? I've found a good way to decide is replace "Earth" with either "Mars" or "Venus": Would you say "but the Mars' composition" or "but the Venus' composition"? No, you would say either "but Mars' composition" or "but the planet Venus' composition"——right? P=) ~Kaimbridge~ 14:26, 30 August 2006 (UTC)

Hi Kaimbridge, are you ready for a headache? As I understood it you can spell the name of our planet the Earth, Earth, the earth, or earth — I was just changing it all to the most common variant already in use in the article to tidy up the various alternative forms that were littered throughout. If there's a difference when you put "the" in front, then all the occurrences that already said "the Earth", of which there were quite a few when I made the edit, would need to be changed. Alternatively, at the top of the article, the two lowercase variants can be added in the brackets along with "the Earth" so that all bases are covered, and people will accept that just the capitalized variants have been employed for continuity. I personally prefer to use the capitalized variants (putting "the" in front or not depending upon my mood or obvious appropriate contexts) so as to distinguish the use of the word in reference to our planet, which surely should be a proper noun, from the use of the common nouns and their various definitions (soil, ground, plug wire, etc.). Of course the name of our planet and any of the common noun definitions can be termed absolutely or qualified with a prefixed "the" interchangeably, so I think the distinction is a reasonable one. I now have a headache, dunno about you! If it's something you feel strongly about then please make the distinction between "Earth" and "the earth" as the core variants of the term, I'm not that bothered — I just wanted to try and introduce a bit of fluency to the article, and if you change all the terms consistently I'm totally happy with that. Kris 18:53, 30 August 2006 (UTC)

[edit] All the Earth mumbo-jumbo we talked about

I've done an Excel database to solve the problem of the perimeter of an ellipse, and also a Word document explaining how to find the distance between two coordinates on Earth. You can download it at http://www.megaupload.com/?d=KAV5AZ8X. I am sure there are other more sophisticated ways to look for a solution, but I've done it according to my maths level. I'm used to cartesian coordinates, and I know very few things about differential calculus.

This solution I found does have an accuracy of millimetres. Ground elevation isn't mentioned, I believe that is elementary. Also, I want to acknowledge that, after all, I don't believe that the roEarth is perfectly spheroidal, and I think you think that also :). There probably is a more accurate Earth equation within some NASA department, describing surface irregularities, etc. I once heard that within its spheroidal shape, the Earth has a pear-shaped tendency (it is thinner at the top and fatter at the bottom). Hey but, whatever! This is what I've been able to do with the data available, and it's more than enough :P

Please do tell me if the Excel database works fine; I've never checked language compatibility before, so I don't know whether it will or it won't. I'd also be very grateful if you mention any mistake you may see within the formulas in the Word document. Thank you for the attention you've given to the matter until now :P --Michael Retriever 19:57, 3 September 2006 (UTC)

Well, it's like this: Just as you're lost with all of the trig with most geodetic formularies, I'm equally (or more) lost with the cartesian math! P=| Soooo, most of the discussion on the Word doc. is meaningless to me (and, as for the Excel database, sorry, the only database format I have is MS Works 4.0! P=)! P=(

There is something I asked before, up above, and you attempted to answer with images, but it still seems to miss the underlying question: You say that:

$E=\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}.\,\!$ Shouldn't it be $E=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{b^2},\,\!$ meaning

$\frac{x^2}{Equator^2}+\frac{y^2}{Pole^2}+\frac{z^2}{Pole^2},\,\!$ not just switching a and b (in terms of equatorial-polar assignment), since there is one equator and two (potentially different) poles, such as in this case:

???

One improvement I can suggest is to the series expansion for the perimeter of an ellipse:

You have: $z=1-{.5\choose1}^2\frac{e^2}{1}-{.5\choose2}^2\frac{e^4}{3}-{.5\choose3}^2\frac{e^6}{5}-\ldots;\,\!$

A more productive one is: $z=\frac{1+{.5\choose1}^2f'^2+{.5\choose2}^2f'^4+{.5\choose3}^2f'^6+\ldots}{1+f'};\,\!$ , where

$f'=\frac{a-b}{a+b}\quad\mathrm{(second\;flattening)}\,\!$ , ...though it would be easier to just express it in terms of angular eccentricity:

$z=\cos\!\left(\frac{O\!\!E}{2}\right)^2\left[ 1+{.5\choose1}^2\tan\!\left(\frac{O\!\!E}{2}\right)^4 +{.5\choose2}^2\tan\!\left(\frac{O\!\!E}{2}\right)^8 +{.5\choose3}^2\tan\!\left(\frac{O\!\!E}{2}\right)^{12} +\ldots\right].\,\!$ P=) P=) P=) ~Kaimbridge~ 16:28, 4 September 2006 (UTC)

(BTW, have you attempted UBasic yet? P=)

Ok, I'll tell you what we'll do. I'd like to finish with this problem by the end of the week, so I'm turning on the boost. First, let's get on with the Earth's shape confusion. Visit Explanation of the Earth's shape.

After you've done that, tell me if you agree with it or if you don't. We discuss, find the correct equation for the Earth's shape, and I make the appropiate modifications in my files.

Alright, I looked at it! P=)

I see what you are saying——and it makes sense——but what keeps nagging at me is the discrepancy in assignment between bi-axial and tri-axial:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=bi-axial\,\!$ (a = equatorial, b =polar);

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=tri-axial\,\!$ (a = b = equatorial, c = polar).

It's not just that b is another equatorial axis (whereas b is defined as the polar axis), but that y is assigned to the equatorial and not the polar. Now, I have seen some sources use X, Y and Z for tri-axial, instead of x, y and z——is that a proper distinction to make?

But, to bolster your view, look at Mars' radii (scroll down a bit): Here, the mean equatorial radius ≈ ½(a+b) and c ≈ the north polar radius——I would think a would equal either the mean equatorial radius or treat the equator as an ellipse and find its mean radius (i.e., using your z) and b,c would be the north, south polar radii. Actually, I would think this should properly be "quad-axial"! P=)

Regardless, if it is a(a_1), b(a_2) and c(b), that needs to be pointed out and emphasized in the various articles.

The data available of Mars in that webpage is different to the one given of the Earth in Wikipedia. It says:

Mean equatorial radius: 3396.200 ± 0.16 km (I assign to this radius letter p for our equation)
North polar radius: 376.189 ± 0.05 km (I assign to this radius letter q for our equation)

That's obviously a typo (it should be 3376.189), as confirmed here.

South polar radius: 3382.580 ± 0.05 km (I assign to this radius letter r for our equation)

I wish they gave us the varius equatorial radii they used for the mean, but anyway here's the best equation I can think for it:

They did, in the triaxial section: a = 3398.627, b = 3393.760 (since the error is ± 0.160, they rounded .1935 to .200).

Oh, now that's nice. Then it all comes down to:

$\frac{x^2}{p^2}++\frac{y^2}{p'^2}+\frac{(z+|z|)^2}{4q^2}+\frac{(z-|z|)^2}{4r^2}=1$

Where p and p' are the two different equatorial radii, supossing they are the ones that correspond to x and y axes.

And about the x y z being with this radius or with the other radius, it does not matter. What is indeed important is that if you're supposed to use two radii a with two dimensional variables, you use them with two dimensional variables; and that once you choose, you are coherent with yourself.

You can use

$\frac{x^2}{b^2}+\frac{y^2}{c^2}+\frac{z^2}{a^2}$

instead of

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$

if you like. But then, when you use a three-dimensional Cartesian coordinate (for example (1,6,3) ), you must associate each individual coordinate with its right dimensional variable ( 1 to x, 6 to y, and 3 to z ) and with its right axis in a drawing of the figure. This is easy when you only work with your own drawing and your own coordinates, but if someone who's using the second equation instead of the first one gives you a three-dimensional coordinate, and you try to look at it in your equation or in your drawing, you'll have to change the order of them. I use the setup I use because it's the one I've seen the most.

And after this... how come there is more data about Mars than about the Earth itself?

$\frac{x^2}{p^2}+\frac{y^2}{p^2}+\frac{z^2}{q^2}=1$ when

z > 0

(for north pole)

$\frac{x^2}{p^2}+\frac{y^2}{p^2}+\frac{z^2}{r^2}=1$ when

z < 0

(for south pole)

Which is equivalent to:

$\frac{x^2}{p^2}+\frac{y^2}{p^2}+\frac{(z+|z|)^2}{4q^2}+\frac{(z-|z|)^2}{4r^2}=1$

The more data for different axis they give you, the more accurate the equation is. But you cannot change its basic form because we are always using a three-dimensional Cartesian space, and what I mean by not changing its basic form is that each radius you give me must still be associated with the right dimensional variable. You can have up to an hexaxial shape, but the dimensional variables are still only three. The one for height goes with height radii, the one for width goes with width radii, and the one for depth goes with depth radii. Hee hee, and you missed the spelling there, it's quadraxial for quadrātus.

Next point: about the series expansion for the perimeter of an ellipse. What I like about the formula I use is that it is very elegant-looking for people experienced in technical drawing. When you make a building or garden plan or any drawing of mainly common geometric shapes, and you draw an ellipse in it, you mention the eccentricity of it. Everyone knows what you mean when you say "the eccentricity of this ellipse is 0.6". On the other hand, no one has a clue what you're talking about if you say "this ellipse's f' is 0.6, where f'=a-b/a+b", or "this ellipse's n is 0.6, where a^2=(n+1)b^2" (Euler's early formula). Thus, other formulas for the perimeter of an ellipse look weird, while the one that contains a simple series of divided integers and a fraction of a power of the eccentricity looks very intuitive and easy to remember.

Ah, but f' is the "second flattening" (just like e' is the "second eccentricity", ${}_{\sqrt{\frac{a^2-b^2}{b^2}}}\,\!$ ): So, visually, what does "the eccentricity of this ellipse is 0.6" mean? On the other hand, saying "the aspect ratio of this ellipse is 0.8" does have a visually tangible meaning (the ellipse's height is 80% of its width)!

The Excel database I made calculates up to 60 diminishing terms in the formula I use. For an ellipse with an eccentricity between 0 and 0.3, that is more than half the diminishing terms needed for a precision of 16 digits after the decimal separator; which is absurdly accurate for calculations with ellipses within the Earth. Probably other formulas are better for higher eccentricities, but for the problem we have at hand, they aren't useful enough to make an intuitive formula turn into something complicate. Nevertheless, as a curiosity, I'll check your f' formula along with Euler's one with ellipses of different eccentricities, to study their behaviour. The OE one I cannot test because I don't know what it means.

You certainly do!

$O\!\!E=\arccos\left(\frac{b}{a}\right);\,\!$

$e=\sin(O\!\!E);\quad e'=\tan(O\!\!E);\,\!$

$f=2\sin\left(\frac{O\!\!E}{2}\right)^2;\quad f'=\tan\left(\frac{O\!\!E}{2}\right)^2;\,\!$

I'll check that.

Last point: UBasic. I'm doing fine with the tools I have already: I have a powerful calculator, lots of paper and ink to draw and write, and some office tools in my computer (which by the way has no internet access). I'm not in real need for a maths programme, and you know I dislike formulas editors :P

I haven't tried UBasic yet. But since you insist so much, I'll try it. With one condition. You must download some spreadsheet-reading software, and try the Excel database I made. I think OpenOffice does that, but I'm not sure if there's OpenOffice for Windows (I think there is).

I have MS Works 4.0, which has spreadsheet and database modules, as well as word processor (I just hardly use any of them)——they just aren't compatible with Excel (though I tried looking at it with WordPad and, through the formatting garbage, what I did see I wasn't too familiar with, as is the case with most of your "perimeter" paper——other than the series expansion part! P=)

As for UBasic, it is not some math program like Mathematica or Maple, it is just an old fashioned BASIC (like was popular back in the "Apple IIe"/"Commodore 64" days!), that gives you a blank screen and you either create programs (e.g., "10 PRINT "Let's see if this works!?!":End"), or you can just use it in direct mode (e.g., "e=.6:?1-e^2/4-e^4*3/64-e^6*5/256-e^8*175/16384-e^10*441/65536") like a fancy calculator, and even create small programs with functions and use them in direct mode!

See you in a while. --Michael Retriever 01:07, 6 September 2006 (UTC)

Now this is weird: Ellipse perimeters. For some reason, it seems that the formula I suggested is more accurate than the other two... --Michael Retriever 11:51, 6 September 2006 (UTC)

That's because you goofed on the coefficients for the f' form:

${.5\choose 1}^2=\frac{1}{4}\,\!$ (okay), ${.5\choose2}^2=\frac{1}{64}\ne\frac{1}{16}=\frac{1}{4^2}\,\!$ , ${.5\choose3}^2=\frac{1}{256}\ne\frac{1}{36}=\frac{1}{6^2}\,\!$ , ${.5\choose4}^2=\frac{25}{16384}\ne\frac{1}{64}=\frac{1}{8^2}\,\!$ , ${.5\choose5}^2=\frac{49}{65536}\ne\frac{1}{100}=\frac{1}{10^2}\,\!$ , etc.——you are familiar with binomial coefficients, aren't you (they certainly provide a concise, formulatic notation! P=)? But don't feel bad, I goofed on giving your equation in BC form: It should be,

$z=1-1{.5\choose1}^2e^2-3{.5\choose2}^2e^4-5{.5\choose3}^2e^6-\ldots;\,\!$

Now see, if you had UBasic, you could have a simple BC function and use that in direct mode:

    99 End
    100.BC(XP,RG):V_bc=1:For TN=1 To RG:V_bc=V_bc*(1+XP-TN)/TN:Next:Return(V_bc)

then ("?" = "print"),

    e=.6:z=1:for I=1 to 10:z=z+(2*I-1)*.BC(.5,I)^2*e^(2*I):?(2*I-1)*.bc(.5,I)^2*e^(2*I):Next:?:?"z =";z

And, the best part is, even in direct mode (as long as it is still on the screen), you don't have to retype everything!!

To get different decimal precisions, change the "point" ("4" is default, equalling about 19 decimal places; I like 7 <33 places> as a base value):

    10 Point 7

As for the third, "Euler's early one", I don't know what you are referring to (An approximation? Another series expansion?). ~Kaimbridge~ 23:14, 6 September 2006 (UTC)

Oh so UBasic is one of them many tools for programming in Basic. I used VisualBasic for a short while (some years ago?), and I'm supposed to know Java; so if I'm not using them it's simply because I'm too lazy. Also, I'm supposed to know binomial expansions because it's in Computer Science's 1st year things-to-do list (thing which I'm supposed to be studying), but then you can call me lazy again even if I'm not supposed to be. I'll check the formulas, and I'll add the OE one. The Euler's early formula is the one I mentioned in our previous conversation (press me pleease!) --Michael Retriever 12:40, 7 September 2006 (UTC)

Alright then, n = e'² = tan(OE)²!

These three series expansions (f' , e, e' ) are just expansions of different forms of the same integrand:

$e:\sqrt{1-(\sin(P)e)^2}=\sqrt{1-(\sin(P)\sin(O\!\!E))^2};\,\!$

$e':\frac{b}{a}\sqrt{(\cos(P)e')^2+1}=\cos(O\!\!E)\sqrt{\cos(P)\tan(O\!\!E))^2+1};\,\!$

$f':\frac{\sqrt{1+2\cos(2P)f'+f'^2}}{1+f'}=\cos\left(\frac{O\!\!E}{2}\right)^2\sqrt{1+2\cos(2P)\tan\left(\frac{O\!\!E}{2}\right)^2+\tan\left(\frac{O\!\!E}{2}\right)^4}\,\!$

$x\,\!$

Thus all of that mathematical rigamarole in the Euler article reduces to just another form of the integrand used in the elliptic integral of the 2nd kind! P=)

Wait... I know how to get the binomial expansions for your formula because you say the formula I suggested is equivalent to

$z=1-1{.5\choose1}^2e^2-3{.5\choose2}^2e^4-5{.5\choose3}^2e^6-\ldots;\,\!$

thus

$1{.5\choose1}^2=\frac{1}{2} \cdot \frac{1}{1}$ , $3{.5\choose2}^2=\frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{1}{3}$ , etc.

No, look again——you didn't square the BCs! P=/

Ouch I forgot :P But I did use it correctly in the new study for the different formulas. I told you I hate formulas editors!! And specially this Wikipedia one, it's like... clumsy.

then I can check the formulas in the perimeters study I showed you. But I have no clue how you do that, since k>n in the binomial expansions. --Michael Retriever 13:00, 7 September 2006 (UTC)

Because n is not a whole, positive number (in this case, it's .5, a decimal), the numerator never lands on zero (thus k can be ∞). ~Kaimbridge~ 15:18, 7 September 2006 (UTC)

I've tested again the formulas, and the results are quite pretty. 10 terms in your formula are more powerful than 80 in mine! But the OE formula doesn't seem to work. You sure I have to multiply by cos((OE/2)^2) ? Ellipse perimeters. --Michael Retriever 19:40, 7 September 2006 (UTC)

Well, you still have the BCs written out wrong for those——you're still not squaring!!!——but you are doing them right! P=)

As for the OEs, it seems we've been experiencing a major expressive miscommunication: I'm of the school that (e.g.) $\sin(P)\sin(P)=\sin(P)^2, \mathrm{\;NOT\;} \sin^2(P)\,\!$ , as those are the only functions that do that——plus, does $\cos^{-1}(P)=\sec(P) \mathrm{\;or\;} \arccos(P)\,\!$ ?

Just compare the e, e'²/n and f' values with the OE function counterparts——they should always equal (for cos((OE/2)^2), I have NO IDEA what you are doing there, as the values you give don't match either way! P=|). ~Kaimbridge~ 16:04, 8 September 2006 (UTC)

As of 19:30, YAH STILL GOT IT WRONG:

${.5\choose1}^2=\left(\frac{.5}{1}\right)^2=\frac{1}{4};\quad {.5\choose2}^2=\left(\frac{.5\cdot-.5}{2\cdot1}\right)^2=\frac{1}{64};\,\!$

${.5\choose3}^2=\left(\frac{.5\cdot-.5\cdot-1.5}{3\cdot2\cdot1}\right)^2=\frac{1}{256}.\,\!$

So, the coefficients for the e and e' models are: $1{.5\choose1}^2=\frac{1}{4};\quad 3{.5\choose2}^2=\frac{3}{64};\quad 5{.5\choose3}^2=\frac{5}{256}.\,\!$

Now, to further confuse you, for each of those, there is another series, based on M (as given in Earth radius): Where your "z" = V^.5, M = cos(OE)²V^^-1.5, so, then the only difference is the BC is based on -1.5 rather than .5,

${-1.5\choose1}^2=\left(\frac{-1.5}{1}\right)^2=\frac{9}{4};\quad {-1.5\choose2}^2=\left(\frac{-1.5\cdot-2.5}{2\cdot1}\right)^2=\frac{225}{64};\,\!$

${-1.5\choose3}^2=\left(\frac{-1.5\cdot-2.5\cdot-3.5}{3\cdot2\cdot1}\right)^2=\frac{1225}{256}.\,\!$

and everything is multiplied by cos(OE)² (and cos(OE/2)² becomes sec(OE/2)⁶)! ~Kaimbridge~ 19:37, 8 September 2006 (UTC)

Sorry for answering this late, I didn't have access to the internet until now. Anyway, I believe the work is done: we've seen the Earth's equation, I found a way to find the distance between two coordinates on Earth, and you've found a damn good formula for the perimeter of an ellipse. There are only some minor details left I want to discuss:

Why the heck do we have more information about Mars' axes than about the Earth's axes? Seems contradictory to me...

What's contradictory? Mars has four different axes (two unique a's and b's), whereas Earth just has two different axes (an a, and a b equal to both poles).

Is the n formula for the perimeter of an ellipse really that bad? (Different formulas for the perimeter of an ellipse.)

It's not that it's "bad", it's just the least efficient: Since n = tan(OE)², e = sin(OE) and f' = tan(OE/2)^2, f' < e < n, therefore the f' series will converge before the e, which will converge before the n series——in fact, since n = tan(OE) and a binomial series doesn't converge if the parameter (here, tan(OE)) is ≥ 1 and tan(45°) = 1, if the aspect ratio (cos(OE)) is ≤ .7071, that series won't converge. Besides the new equations (based on -1.5, rather than .5) I gave above, there is one other important series expansion, the reverse convergence model, using cos(OE) as the parameter. This one uses a second term factor, involving a BC, permutation and ln:

$C_1=1+P(2,2)\ln\!\left(\frac{4}{\cos(O\!\!E)}\right) -{2\choose2}^{-1}; \,\!$

$C_2=1+P(4,2)\ln\!\left(\frac{4}{\cos(O\!\!E)}\right) -{2\choose2}^{-1} -{4\choose2}^{-1}; \,\!$

$C_3=1+P(6,2)\ln\!\left(\frac{4}{\cos(O\!\!E)}\right) -{2\choose2}^{-1} -{4\choose2}^{-1} -{6\choose2}^{-1}; \,\!$

$z=\frac{2}{\pi}[1 +{.5\choose1}^2C_1\cos(O\!\!E)^2 +{.5\choose2}^2C_2\cos(O\!\!E)^4 +{.5\choose3}^2C_3\cos(O\!\!E)^6+\ldots];\,\!$

With this model, covergence is quicker with greater ellipticity!!! Unfortunately, I don't know what the integrand this particular model is based on (or the equivalent one for tan((90°-OE)/2)^2!). P=(

I've taken out the chart for the OE formula; even though you said both f' and OE formulas were equivalent, I didn't realise it until I saw the results on the charts. After that, I thought it was kind of silly to put them as different formulas.

Over all, after seeing your way of understanding ellipse-related matters, I must say I was kind of disturbed to see that trigonometric functions are really that powerful. I've always felt a stitch in my brain when thinking about them; school teachers are not a great example for rigurosity (e.g.):

(1) The equation of a circle is x^2+y^2=r^2, where x is the distance from the point to the origin
in the x axis, and y is the distance from the point to the origin in the y axis.

Great, interpretation of a graphic through elementary algebraic expressions.

(2) The equation of a circle is (cos(α)·r)^2+(sin(α)·r)^2=r^2, where α is the angle between the
x axis and the line that goes from the origin to the point.

Uhm, okay, you describe an equation with two auxiliar functions, cos(α) and sin(α). And the
equations for those functions are?

The functions cos(α) and sin(α) are trigonometric functions.

Which means? ...

It means that to know the result of their equations you must use a calculator.

¬¬'

And thus, I never understood how they really work, because no one has ever cared to define them to me. Even more incredible is the fact that you get the same answer when asking some university professors! :/

What do you expect——most are probably written by university professors! P=)

But trigonometric functions are powerful and of great utility, so I'll care to study them carefully for some time, until I get to use them with ease.

Explore them with UBasic! P=) P=) P=)

It was a pleasure discussing the ellipse matter with you. You'll probably want to make some clean-up now in your discussion section! :P --Michael Retriever 13:26, 12 September 2006 (UTC)

One little (big?) nitpick I have regarding your perimeter page: What do (you think) you mean by "measures the distance between two points in the ellipse's quarter of a perimeter every 0.01/0.001 degrees"? These series terms don't measure different segments of the perimeter; what they do is add more and more ellipticity to the average radius. Consider the "e" based series: The first term is "1", giving the equatorial radius as the mean elliptical radius, then ¼e² is subtracted from the elliptic factor, with subsequent terms also subtracted. Or, another way to look at it: If all of these different series converge at different terms, then how can a given term in each of these different series (giving different values) measure a particular segment of the perimeter (especially since it takes longer——i.e., more terms——to converge, as the ellipticity grows)?

Also, I reverted the Qr value in Earth radius, to keep it consistant with the rest of the article. ~Kaimbridge~ 16:16, 13 September 2006 (UTC)

The polygonal paths 1 and 2 are just two more measures I made for the ellipses' perimeters, and they have nothing to do with the infinite binomial expansions. But hey, you poking my article? I'll poke yours then!

Er, I don't know if you forgot to put a smiley at the end, or I ruffled your feathers... (if I did, none was intended! P=(

In terms of the polygonal paths, the way it is worded and the term chart is presented in two sections, one could think you mean the first section equals "path 1" and the second, "path 2".

1. It's not $e, e', e'', e''', e'''', ...\,$ it's $e, e^I, e^{II}, e^{III}, e^{IV}, ...\,$

Well, in all of the sources I've seen (including journal published articles) it is ', '' or ", ''', etc.——e.g., see [1] or [2] P=)

You've always seen $f, f', f'', f''' ...\,$ because the articles you visit don't do further than $f'''\,$ . But those $'\,$ are supposed to be Roman numbers; and more specifically, they're usually written in lower case letters (though I find that a bit aesthetic). Check the English equivalent for "derivada enésima" ( $f^{n}\,$ ) on Google , or look at the first page I found on the subject in Google Spain.

I agree with the $f^{(>3)}\,\!$ , but for ≤ 3, I believe——at least in recent times——LaGrange's "prime" (rather than Roman) notation is de facto (though it could be a regional/cultural thing).

Most practical. I don't write $f^{I}\,$ myself unless calculating $f^{IV}\,$ or higher. Just confirming that you know the formal expression. f prima, f secunda, f tertia, f quarta... I love Latin :)

2. If you want to keep clarity within your articles, I am sure there is no problem if you use $\alpha\,$ for naming angles. Greek lower case letters starting from $\alpha, \beta, \gamma, \,$ and onwards are normally used within equations to define an angle. Using $O\!\!E\,$ instead of $\alpha\,$ DOES make it confusing, and it's the reason why I didn't know what you where talking about when you mentioned a function that depends on $O\!\!E\,$ .

Like most notations, since there is usually variation between different sciences (and even different papers for the same subject), what's important is that the writer clearly defines what means what and that the reader doesn't "read" more into it than intended! P=)

Fair enough. But proof-reading won't be enough then, and expect the first reaction to be "what are you talking about?". At least for people with a similar level of maths than me, which isn't as good as I wish at the moment.

It is obvious that even though $e\,$ is normally 2.7182..., if you define $e\,$ as eccentricity, and you give the equation for eccentricity, $e\,$ regarding an article about ellipses won't be $e\,$ for 2.7182.... The same applies for $\alpha\,$ . It is obvious that even though $\alpha\,$ is normally (within your regular mathematical environment) the horizontal angle in an angular coordinate on Earth, if you define $\alpha\,$ as angular eccentricity, $\alpha\,$ regarding an article about angular eccentricities won't be $\alpha\,$ for the horizontal component of an angular coordinate on Earth.

Right, but if $\alpha\,\!$ is already defined in a given paper as "azimuth" (as is the case in most geodetic papers), something has to give (besides, in TeX, $\alpha\,\!$ looks awfully like $a\,\!$ ! P=)

And, while defining e as eccentricity may typically suffice, there may be situations where either e (2.7182...) may be used in conjuction with e (sin(OE)), or e (sin(OE)) may be used as a base or exponent (such as in finding conformal latitude) which may lead the reader to question whether e still means eccentricity or 2.7182... in that case. Hey, if one sees $\pi\,\!$ , there is generally only one meaning, 3.141592653589793.... Since angular eccentricity hasn't really caught on (yet? P=) and "OE" isn't established as a conceptual constant or variable (and, symbolically, you have a circle, "O", being squashed by eccentricity/ellipticity, "E"), as long as it is clearly defined, it makes an ideal variable. Plus, as it can be expressed plain textually as either "Œ" or "OE", it can be directly used (unlike "α") as a variable in any program. Heh, originally I denoted it as "Oz", "imaginary angle of oblateness"! P=)

I tell you, I'm of the group that agrees with writing trigonometric functions ALWAYS with a Greek lower case letter; since they are functions which always depend on an angle, not on a physical magnitude with a true unit of measure. Unlike distance or mass or volume (which have their own units), an angle is the relation between two equal physical magnitudes (distances), so the two units of measure cancel out themselves in the calculation of an angle. In order for this to be always present, angles deserve to stay with weird characters of their own in equations. You'll never see a considerable maths article using $x\,$ or $P\,$ for an angle. Because of the fact that I prefer to use always a Greek lower case letter for angles, I am reticent to read $O\!\!E$ instead of $\alpha\,$ .

Well then, how about $\varepsilon\,\!$ ( $\epsilon\,\!$ looks sawed-off)?——or even $o\!\varepsilon\,\!$ ?! P=)

Or, seriously, $\alpha\!\varepsilon\,\!$ ("angular eccentricity")??

Seeing there is more than one article using $e\,\!$ for eccentricity, I think $\varepsilon\,\!$ is most suited for the angular version of it. --Michael Retriever 18:22, 17 September 2006 (UTC)

3. When defining a function/equation that depends on one sole variable, if you're not being too rigurous, you normally give to that variable the name $x\,$ . So if we're discussing trigonometric functions just like, you know, saying $\cos\,$ of whatever, it is usual to say $\cos x\,$ . It is not usual to say $\cos P\,$ .

I just prefer P, as in "point", rather than the "abscissal" x. P=)

I like cookies :)

4. You are right when you say that $\cos (x)^2 = (\cos (x))^2 =\cos^2 (x)\,$ , but I'm of the school that you should always write either $(\cos (x))^2\,$ or $\cos^2 (x)\,$ , because $\cos (x)^2\,$ can lead to confusions.

I guess we just went to different schools! P=) P=) P=)

That makes me be a bit more sure about your photograph being a true photograph of yourself. I've never seen your face before, not at school either! :P

Oh yeah, that's me alright! P=)

As for schools, I know we didn't go to the same school, since——as Three Dog Night sang it——♪♫ I've Never Been To Spain ♫♪. P=) P=) P=)

And now, another doubt. Going back to the discussion about the ellipse's perimeter page we both love so much; you see the second mathematical expression? Isn't it blacker than the rest, and also a bit blurred? Do you know how to change that? The same happens for b/a = cos α and α = cos^-1 (b/a).

No, I don't notice any difference——could it be your browser or a monitor setting?

BTW, did you try the other series expansions I added above (particularly the "reverse convergence" model)?

No, I haven't tried them. Will I try them? Probably. Maybe next month; I'm off ellipses for a while.

Good, go get a copy of UBasic and get comfortable with it! P=)

Oh! and sin^-1 = arcsin, cos^-1 = arccos, tan^-1 = arctan. --Michael Retriever 17:53, 13 September 2006 (UTC)

Yes, I realize you meant that, I'm just saying that func^-1 can be interpreted as either arcfunc or 1/func! P=)

You checked if arcsin x =/= 1/sin x? Who knows, I wouldn't be surprised if it turned out to be equal :P

Again, if I did ruffle your feathers, it wasn't intentional. P=)

(And, BTW, you're not the first to say that my notation is——shall we say——a bit "challenging"! P=| P=)

~Kaimbridge~ 21:57, 13 September 2006 (UTC)

I forgot to put a smiley :) Your poking was just an excuse to mention a few things I wanted to say to you sooner or later. --Michael Retriever 09:34, 14 September 2006 (UTC)

No problem. P=) ~Kaimbridge~ 16:46, 14 September 2006 (UTC)

[edit] Binomial coefficients

I think I understand now. Mind take a look? :) Binomial coefficients with real non-natural numbers --Michael Retriever 19:09, 21 September 2006 (UTC)

From just the casual glance I've given so far, it looks okay (one suggestion: For continuity's sake, rather than presenting the square root exponent as "1/2", just keep it as ".5"). One related concept you might want to look up is Pochhammer symbol, as that is really the BC's numerator. Also, one special BC is

${-1\choose RG}\frac{-1\cdot-2\cdot-3\cdot-4\cdot-5\ldots-RG}{1\cdot 2\cdot 3\cdot 4\cdot 5\ldots RG}= -1^{RG}\,\!$ !!!

~Kaimbridge~ 19:32, 23 September 2006 (UTC)

[edit] Ellipses and Ellipsoids

Hi Kaimbridge, wow - those are extended talks higher up on this page! Regarding planetary bodies, the equal axes are almost always the equatorial ones as you say, at least when you've got an oblate spheroid like the Earth etc. However check out (136108) 2003 EL61 which is thought to be prolate for reasons connected to its fast rotation (it's weird). Now, you mentioned that

oblate is a>c and prolate is a<c,

It is really just a matter of what definitions of a,b,c, you take, but it is obviously inconsistent with the present statement a>=b>=c in ellipsoid!

I'm not sure what you mean is inconsistent:

If it is assumed a ≥ b ≥ c, then, when:

Meaning: The equatorial x's a is ≥ to y's b, both of which are ≥ to polar z's c, right?

a = b = c, it is the aforementioned sphere;
a = b > c, the ellipsoid is an oblate spheroid (disk-shaped);

Meaning: Equatorial = a = b, polar = c; equatorial > polar;

a = b < c, the ellipsoid is a prolate spheroid (cigar-shaped);

Only this one. If a≥c, then you can't have a<c (also, if b≥c, you can't have b<c). Deuar 20:12, 23 October 2006 (UTC)

Ahh——DUH P=)——yeah, now I see what you mean! Okay, I fixed up (for now) the opening paragraphs. ~Kaimbridge~ 15:37, 24 October 2006 (UTC)

;-) Deuar 21:20, 24 October 2006 (UTC)

Meaning: Equatorial = a = b, polar = c; equatorial < polar;

a > b > c, the ellipsoid is scalene.

Meaning: The equatorial x's a is > y's b, both of which are > polar z's c, right?

That's presented right, isn't it? ~Kaimbridge~ 19:11, 23 October 2006 (UTC)

I'm afraid i'm not well versed in what are the conventions with x,y,z labeling in cartography, spheroids, etc, so I can't say anything wise there, although I suspect the conventions may differ depending on the exact field. aargh! Good luck with the edit... :-) Deuar 19:02, 22 October 2006 (UTC)

The "dimensions" of "EL61" are given as 1960×1518×996: If you measure the image on the screen, it appears what you see is 1960×1518, the two equatorial axes——thus you are looking down on "EL61" (i.e., the polar perspective), not at it (i.e., facing the equator)——which means it's triaxial, not prolate (at least as it is dimensionally defined). ~Kaimbridge~ 17:22, 23 October 2006 (UTC)

Right - good point. Even more interesting :-) Deuar 17:30, 23 October 2006 (UTC)

[edit] Your signature

Hi Kaimbridge: please remove your image from your signature. It is against WP:SIG#Images. You will see various reasons for this on that page. Thanks! —Mets501 (talk) 21:28, 13 December 2006 (UTC)

[edit] Re: Bad BOT archive

Thanks. (Further reply on my talk page.) —Steve Summit (talk) 01:07, 5 August 2007 (UTC)

[edit] Earth Radius

Hi, thanks for the note - I've been behind in my watchlist for a while :p Before I comment, I need to find some time (sigh) to sit back and read your explanations carefully. Cheers, Deuar 10:15, 26 October 2007 (UTC)

[edit] Ellissoide

Hi! It looks like you modified a page on it.wiki, replacing different letters with the same symbol $o\!\varepsilon$ . I reverted your edit. I inform you that such an action could be looked as vandalic. In case the edit was wrongly attributed, I beg your pardon for the inconvenience. You can contact me here. Thank you --Zio illy (talk) 22:11, 1 March 2008 (UTC)

Oh, now I see what you meant to do; I misunderstood your summary to the edit. Still, it seems to me that œ is but one possible notation (see de:Ellipsoid, nl:Ellipsoïde, pl:Elipsoida, pt:Elipsóide, and th:ทรงรี). As far as I know, it is not the main one in italian culture. Off Topic: I would advice againts the use textrm in a math environment (except in TeX). ^^ --Zio illy (talk) 23:10, 1 March 2008 (UTC)

[edit] Old question answered

You asked at Help talk:Displaying a formula how one requests a new feature. Since you asked in December I figured you may no longer be watching that page so I have come here to let you know that the place to go is the Mediawiki bugzilla and Wikipedia:Bug reports will explain how you do it. Asking questions on talk pages often does not get results depending on how well frequented they are. It is usually better to ask at WP:Helpdesk who will direct you to the right place if they do not know themselves. SpinningSpark 19:12, 26 April 2008 (UTC)

[edit] Latitude

You may think you reverted me, but you were too late! I had already undone all of my edits.

I had always assumed that the latitude everyone uses is the geocentric latitude, which I why I thought the surface distances were wrong. Turns out, without knowing it, everything I've ever done has been geodetic. Oh well.Rracecarr (talk) 18:15, 11 June 2008 (UTC)