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May 31

Meridional Radius of Curvature

Hi y'all.

$M(\phi )d\phi =\frac{(ab{)}^2}{((a\cos \phi {)}^2+(b\sin \phi {)}^2{)}^{\frac{3}{2}}}d\phi =\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}d\beta =??(\beta )d\beta .$

(φ, β = geodetic, reduced latitudes)

If $\frac{(ab{)}^2}{((a\cos \phi {)}^2+(b\sin \phi {)}^2{)}^{\frac{3}{2}}}=M(\phi )$ equals the "meridional radius of curvature", then what does

$\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ equal ("reduced meridional radius of curvature"?) and what is its symbol (_rM(β)? )?

--2601:19C:4A01:7057:4C27:AD22:B7E2:D04A (talk) 15:35, 31 May 2024 (UTC)

I cannot relate the quantity $\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ to a radius of curvature. It is the speed of a particle moving along the meridian for $\dot{\beta }=1.$

For the radius of curvature of the meridian at reduced latitude $\beta$ I find

$\frac{((a\sin \beta {)}^2+(b\cos \beta {)}^2{)}^{\frac{3}{2}}}{ab}.$

As far as I know there is no standardized symbol for this. I don't think that the notation $M(\phi )$ is common either.  --Lambiam 20:18, 31 May 2024 (UTC)

Of course $M(\phi )$ and $\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ are valid (though here written with e and e' instead of a and b), but β is not given its own integral identity, even though $\frac{(ab{)}^2}{((a\cos {\phi }_n{)}^2+(b\sin {\phi }_n{)}^2{)}^{\frac{3}{2}}}\ne \sqrt{(a\sin {\beta }_n{)}^2+(b\cos {\beta }_n{)}^2}$. 2601:19C:4A01:3561:4C27:AD22:B7E2:D04A (talk) 03:23, 1 June 2024 (UTC)

I don't understand what it means that $\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ is "valid".

The angles $\beta$ and $\phi$ are related by

$a\tan \beta =b\tan \phi .$

Here is a numeric example, randomly generated:

$a$ = 239.2188308713, $b$ = 192.1989786957

$\beta$ = 1.3880315979, $\phi$ = 1.4233752785

Then

$\frac{(ab{)}^2}{((a\cos \phi {)}^2+(b\sin \phi {)}^2{)}^{\frac{3}{2}}}$ = 292.5274922901

 

$\frac{((a\sin \beta {)}^2+(b\cos \beta {)}^2{)}^{\frac{3}{2}}}{ab}$  = 292.5274922901

This is not a numerical coincidence. For comparison,

$\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ = 237.8141595361.

 --Lambiam 09:28, 1 June 2024 (UTC)

Right, $\frac{(ab{)}^2}{((a\cos \phi {)}^2+(b\sin \phi {)}^2{)}^{\frac{3}{2}}}=\frac{((a\sin \beta {)}^2+(b\cos \beta {)}^2{)}^{\frac{3}{2}}}{ab}$ but $\frac{(ab{)}^2}{((a\cos \phi {)}^2+(b\sin \phi {)}^2{)}^{\frac{3}{2}}}d\phi =\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}d\beta \ne \frac{((a\sin \beta {)}^2+(b\cos \beta {)}^2{)}^{\frac{3}{2}}}{ab}d\beta$!

So what is $\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$, which equals $\frac{ab}{\sqrt{((a\cos \phi {)}^2+(b\sin \phi {)}^2)}}$?--2601:19C:4A01:650:1123:BA2C:D056:629 (talk) 15:00, 1 June 2024 (UTC)

Writing $R$ for the meridional radius of curvature, a variable that depends on $\beta$ (or, equivalently, on $\phi$), we have:

$\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}=R\frac{d\phi }{d\beta }.$

This is the tangential speed of a particle moving along the meridian when $\beta =t+{\beta }_0,$ in which case the rhs equals $R\dot{\phi }.$  --Lambiam 17:30, 1 June 2024 (UTC)

Okay, so you are saying $\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}$ is the variable for tangential speed (let's call it "S") and using the chain rule:  M(φ) = S(β(φ))β'(φ) and S(β) = M(φ(β))φ'(β), therefore M(φ)dφ = S(β)dβ.

But:  $M(0)=\frac{b^2}{a}$ and $M(90)=\frac{a^2}{b}$ , while $S(0)=b$ and $S(90)=a$, so S is a radius, not speed (I know, speed here is a calculus thing, not literally "speed", but still) and I should point out S(90-β) = R(β), geocentric radius! --2601:19C:4A01:650:19F3:4CE1:97CE:10D5 (talk) 19:09, 1 June 2024 (UTC)

I also just figured out $\frac{a}{b}S(\beta )=N(\phi )=\frac{a^2}{\sqrt{(a\cos \phi {)}^2+(b\sin \phi {)}^2}}$, the prime vertical radius of curvature and conversely, of course, $\frac{b}{a}N(\phi )=S(\beta )$.  --2601:19C:4A01:6E9F:1937:5ABC:70DF:B9B3 (talk) 15:38, 3 June 2024 (UTC)

My $0.02: $\phi$ plays a favored role in defining the radius of curvature because this latitude defines the normal vector to the meridian and so is directly related to the definition of curvature. The corresponding expression in terms of $\beta$ is useful in carrying out integrals but I don't think it's necessary to invent a name for the integrand. cffk (talk) 19:53, 3 June 2024 (UTC)

Technically, isn't "S" the integrand for geodetic distance (just here focused on the north-south meridian distance, either geodetically or as the parametric version of the plane Pythagorean distance), with respect to σ rather than β? I've seen some articles define the geodetic "s" as the spacetime variable, itself!  --2601:19C:4A01:C40C:A8A9:9580:7416:38DE (talk) 19:29, 4 June 2024 (UTC)

While the terms meridian, latitude and geodesic suggest a problem in spheroidal geometry, everything going on here can mathematically be seen as taking place on a good old planar ellipse. The formula

$\underset{{\beta }_1}{\overset{{\beta }_1}{\int }}\sqrt{(a\sin \beta {)}^2+(b\cos \beta {)}^2}d\beta$

gives the elliptic arc length between two points on an ellipse in the standard parametric representation $(x,y)=(a\cos \beta ,b\sin \beta )$ using $\beta$ as the name of the parameter. It is easily seen to be equivalent to the formula given at Template:Slink.  --Lambiam 04:43, 5 June 2024 (UTC)