2. Cartesian Coordinates for Waco and Moscow (Normalized to Earth Radius)

_12_EGR1301_Fall2014_Earth1_to_Earth2_140919.docx

1. Relationship Between Cartesian and Spherical Coordinates

2. Cartesian Coordinates for Waco and Moscow (normalized to Earth radius)

WacoMoscow

x = sin(58.44) = –0.1068x = sin(34.30)cos(37.50°) = 0.4470

y = sin(58.44)sin(–97.20) = –0.8454y = sin(34.30)sin(37.50°) = 0.3431

z = cos(58.44) = 0.5234z = cos(34.30) = 0.8261

Rwaco(Center of Earth to Waco),Rmoscow(Center of Earth to Moscow),

= –0.1068ax–0.8454ay + 0.5234az km= 0.4470az+0.3431ay + 0.8261az m

Rwaco,mag = 0.9999Earth radii (checks out)Rmoscow,mag = 0.9996 Earth radii (checks out)

Let Rwm be a vector from Waco to Moscow. Starting at the center of Earth and returning to the center of the Earth, Rwaco + Rwm –Rmoscow = 0 because the net distance is zero. So,

Rwm = Rmoscow – Rwaco, yielding

Rwm = Rwm,xax + Rwm,yay + Rwm,zaz

= (0.4470 + 0.1068)ax + (0.3431 +0.8454)ay + (0.8261− 0.5234)azEarth radii,

= 0.5538 ax + 1.1885 ay + 0.3027 azEarth radii.

Mag = 1.345 Earth radii, corresponding to 8570 km.

Now, to get the compass angle heading at Waco, convert Rwm to local spherical coordinates at Waco. Then, take the dot product of Rwm with Waco spherical unit vectors. Dot product works with vectors whose coordinate references have the same rotations. Dot product is ordinarily used with Cartesian (xyz) coordinates and works as follows:

The projection of vector (Aax + B ay + C az) onto vector (D az + Eay + Faz) is AD + BE + CF. To calculate a dot product, both vectors must be expressed in the same coordinate system. For example, the projection of Cartesian vectorRwm on a unit vector in the r direction, abbreviated as ar, requires that ar be converted to Cartesian coordinates. That conversion is shown on the previous figure to be

ar = cos(Θ) az + sin(Θ)•[cos(Φ) ax + sin(Φ) ay].

When rearranged in xyz order, ar is

ar = sin(Θ)•cos(Φ) ax+ sin(Θ)•sin(Φ) ay + cos(Θ) az,

soRwm●ar(straight up component) becomes

(Rwm,xax + Rwm,yay + Rwm,zaz) ● (sin(Θ)cos(Φ) ax+ sin(Θ)sin(Φ) ay + cos(Θ) az)

= Rwm,xsin(Θ)cos(Φ) + Rwm,ysin(Θ)sin(Φ) +Rwm,zcos(Θ),

andRwm●aΘ(due south component) becomes

(Rwm,xax + Rwm,yay + Rwm,zaz) ● (cos(Θ)cos(Φ) ax + cos(Θ)sin(Φ) ay − sin(Θ) az)

= Rwm,xcos(Θ)cos(Φ) + Rwm,ycos(Θ)sin(Φ) −Rwm,zsin(Θ),

andRwm●aΦ(due east component) becomes

(Rwm,xax + Rwm,yay + Rwm,zaz) ● (−sin(Φ) ax + cos(Φ) ay)

= −Rwm,xsin(Φ) + Rwm,ycos(Φ)

3. Angles at Waco, Looking Toward Moscow

In thear(up) direction at Waco,

= Rwm,xsin(Θ)cos(Φ) + Rwm,ysin(Θ)sin(Φ) +Rwm,zcos(Θ),

= –376.7 – 6402.1 + 1009.4 = –5769.4 km

In theaΘ (south) direction at Waco,

= Rwm,xcos(Θ)cos(Φ) + Rwm,ycos(Θ)sin(Φ) −Rwm,zsin(Θ),

= –231.3 – 3929.3– 1643.7 = –5804.3 km

In theaΦ (east) direction at Waco,

= −Rwm,xsin(Φ) + Rwm,ycos(Φ)

= 3500.3 – 948.7 = 2551.6 km

Mag check = sqrt(5769.4^2 + 5804.3^2 + 2551.6^2) = 8572 km (checks).

True Azimuth Angle at Waco, Looking Toward Moscow

Atan(east/north) = atan(east/-south) = atan(2551.6/-5798.3) = 23.8⁰.

Compass Angle at Waco, Looking Toward Moscow

True azimuth does not include the effect of magnetic declination. At Waco, a compass points 4⁰ east of north, so the compass angletoward Moscow is(23.8⁰ + 4⁰) = 27.8⁰.

Zenith Angle at Waco, Looking Toward Moscow

= acos(up/Mag) = acos(-5769.4/8572) = 132.3⁰. Zenith angle = 0 corresponds to straight up, 180 degrees is straight down. The horizon is at zenith angle = 90. So, Moscow is (132.3⁰ – 90⁰) = 42.3⁰ below the horizon.

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