Test embedding - please don't vote : Custom Object : SummitPost

Disclaimer

If you stumble across this test object, know that this is where I fiddle around with embedded maps (and some other stuff). I've added some explanation, primarily in order to remind me what I'm trying to do, but also to enable others that might be curious to see it. If you see something wrong, or have remarks or questions about the subject, by all means, let me know.

As this is a test object. I reserve the right to delete or modify it at any time. Also, while I'll try to keep the HTML code displayed below in line with the actual code, I might make a mistake and forget to update something. Therefore, use this at your own peril.

Embedding Maps

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The Equation

D = f(Φ,Z) =

\sqrt{\frac{a^{4} \cos^{2} Φ + b^{4} \sin^{2} Φ}{a^{2} \cos^{2} Φ + b^{2} \sin^{2} Φ} + 2 Z \sqrt{a^{2} \cos^{2} Φ + b^{2} \sin^{2} Φ} + Z^{2}}

The Derivation

Note: This derivation assumes your browser correctly shows the square root symbols. If it does not, possibly try a different browser.

The oblate spheroid shape of the Earth is not as flattened as the above figure would suggest and value e is small, placing ray R_a close to C and more closely parallel to ray R_g. The squashed look is provided here for clarity.

The true distance D is as shown. This is because the latitude that gives the position of a person or object on the Earth is the astronomical latitude, alat. This is a ray that extends outward from the Earth's surface at a right angle to that point. Because the Earth is an ellipsoid from the Equator to the pole, a ray that is perpendicular cannot pass through the exact Center of the Earth except at two latitudes (0^o at the Equator and 90^o at the poles). Since elevations above sea level are measured directly up from the surface, they must therefore be parallel to ray R_a.

The geocentric latitude, glat, on the other hand, defines the angle of ray R_g that extends from the Center of the Earth to the same point on the surface.

General Trigonometric Equations

D² = L_d² + H_d²     (Eq. A1)

H_d = (R_a + Z)sinΦ

H_r = R_gsinθ = R_asinΦ ► R_a = R_gsinθ/sinΦ     (Eq. A2)

L_g = R_gcosθ
L_a = R_acosΦ
e = L_g – L_a = R_gcosθ - R_acosΦ

L_az = (R_a + Z)cosΦ
L_d = L_az + e

a²tanθ = b²tanΦ ► tanΦ = a²/b²tanθ ► Φ = tan^-1(a²/b²tanθ) ► θ = tan^-1(b²/a²tanΦ)    (Eq. A3)

These are relationships between geocentric latitude, glat (θ), and astronomical latitude, alat (Φ). See here. Note geodetic latitude is synonymous with astronomical latitude.

Also,

R_g =

\sqrt{\frac{a^{4} \cos^{2} Φ + b^{4} \sin^{2} Φ}{a^{2} \cos^{2} Φ + b^{2} \sin^{2} Φ}} = \sqrt{expr1}

(Eq. A4)

See here.

Combining Equations

L_d = L_az + e = (R_a + Z)cosΦ + R_gcosθ - R_acosΦ = R_gcosθ + ZcosΦ = cosθ

\sqrt{expr1}

+ ZcosΦ

L_d² = (cosθ

\sqrt{expr1}

+ ZcosΦ)(cosθ

\sqrt{expr1}

+ ZcosΦ) = cos²θ[expr1] + 2ZcosθcosΦ

\sqrt{expr1}

+ Z²cos²Φ

H_d = (R_a + Z)sinΦ = R_gsinθ + ZsinΦ = sinθ

\sqrt{expr1}

+ ZsinΦ

H_d² = (sinθ

\sqrt{expr1}

+ ZsinΦ)(sinθ

\sqrt{expr1}

+ ZsinΦ) = sin²θ[expr1] + 2ZsinθsinΦ

\sqrt{expr1}

+ Z²sin²Φ

D² = L_d² + H_d² = [expr1](cos²θ + sin²θ) + 2Z

\sqrt{expr1}

(cosθcosΦ + sinθsinΦ) + Z²(cos²Φ + sin²Φ)
= [expr1] + 2Z

\sqrt{expr1}

cos(θ – Φ) + Z²

D = f(θ,Φ,Z) =

\sqrt{[expr1] + 2 Z \sqrt{expr1} \cos (θ - Φ) + Z^{2}}

(Eq. A5a)

Fully Combined Form for Distance

Substituting in Eq. A3 where θ = tan^-1(b²/a²tanΦ):

D = f(Φ,Z) =

\sqrt{[expr1] + 2 Z \sqrt{expr1} \cos (\tan^{-1} (b^{2} / a^{2} \tan Φ) - Φ) + Z^{2}}

(Eq. A5b)

The second term of Eq. A5b can be further simplified to remove the contraction cos(θ – Φ), resulting in the following:

D = f(Φ,Z) =

\sqrt{\frac{a^{4} \cos^{2} Φ + b^{4} \sin^{2} Φ}{a^{2} \cos^{2} Φ + b^{2} \sin^{2} Φ} + 2 Z \sqrt{a^{2} \cos^{2} Φ + b^{2} \sin^{2} Φ} + Z^{2}}

(Eq. A5c)

This is the final equation.
The distance to the Center of the Earth from any point, like a summit, is a function of only latitude, Φ, and elevation, Z.

Here is the expected result when Z = 0: