EGR1301_FALL2015_Earth1_to_Earth2_150903.docx
Azimuth
angle
North Pole
(Latitude = 90° N)
(1)
(2)
Greenwich
Lat. 51.50° N
(3)
Point P on
Earth’s surface
Prime Meridian
(Longitude = 0°)
Zenith
angle
Earth radius R = 6371 km
θ = 90° − latitude N
φ = longitude E
x-y is the plane of the equator
Equator
(Latitude = 0°)
Waco latitude = 31.56°N, longitude = −97.20°E,
Moscow latitude = 55.70°N, longitude = 37.50°E
aR
To a person standing on the
Earth’s surface at point P,
 aR points straight up,
aφ
aθ


𝐴𝜋𝑟 2
𝐋𝐞𝐧𝐠𝐭𝐡 𝐑𝐜𝐨𝐬(𝛉)
P
𝐋𝐞𝐧𝐠𝐭𝐡 𝑅𝐬𝐢𝐧(𝛉)𝐬𝐢𝐧(𝛟)
𝐴𝜋𝑟 2
Spherical coordinate unit vectors at point P are
aR = sin(θ) cos(φ) aX + sin(θ) sin(φ) aY + cos(θ) aZ
(4)
aθ = cos(θ) cos(φ) aX + cos(θ) sin(φ) aY − sin(θ) aZ
(5)
aφ = −sin(φ) aX + cos(φ) aY
(6)
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aθ points south.
aφ points east.
EGR1301_FALL2015_Earth1_to_Earth2_150903.docx
Problem 1. Obtain the vector in x,y,z coordinates that points from Waco to Moscow
To obtain the x,y,z vector from Waco to Moscow, first compute x,y,z coordinates at Waco and Moscow
using (1), (2), (3).
x,y,z coordinates at Waco
x = 6371 sin(90° – 31.56°) cos(–97.20°)
= 6371 (0.8521) (–0.1253) = –680.2 km
x,y,z coordinates at Moscow
x = 6371 sin(90° – 55.70°) cos(37.50°)
= 6371 (0.5635) (0.7933) = 2848.0 km
y = 6371 sin(90° – 31.56°) sin(–97.20°)
= 6371 (0.8521) (–0.9921) = –5385.8 km
y = 6371 sin(90° – 55.70°) sin(37.50°)
= 6371 (0.5635) (0.6088) = 2185.6 km
z = 6371 cos(90° – 31.56°)
= 6371 (0.5234) = 3334.6 km
So,
z = 6371 cos(90° – 55.70°)
= 6371 (0.8583) = 5263.1 km
So,
Rw(Center of Earth to Waco),
Rm (Center of Earth to Moscow),
Rw = –680.2 aX – 5385.8 aY + 3334.6 aZ km
and from sqrt(sum of squares),
Rm = 2848.0 aX + 2185.6 aY + 5263.1 aZ km
and from sqrt(sum of squares),
Mag(Rw) = 6371 km (checks out)
Mag(Rm) = 6371 km (checks out)
Now, starting at the Center of Earth, Rw + Rwm – Rm = 0, where Rwm is a vector from Waco to Moscow.
Rewriting, Rwm = Rm – Rw, so
Rwm = (2848.0 – (−680.2)) aX + (2185.6 –(−5385.8)) aY + (5263.1 − 3334.6) aZ km.
Collecting terms yields the x,y,z vector from Waco to Moscow,
Rwm = 3528.2 aX + 7571.4 aY + 1928.5 aZ km (corresponds to 8572.9 km straight line distance
through Earth.
Problem 2. Convert the x,y,z vector from Problem 1 to spherical coordinates and determine the
compass direction at Waco that points toward Moscow. Then, determine Moscow’s tilt angle below the
horizon as seen at Waco.
To obtain the compass angle heading at Waco, take the dot product of Rwm in x,y,z with the three
spherical unit vectors at Waco (expressed in x,y,z). The three components of the dot product and their
meanings are
Rwm ● aR,w gives the component of Rwm in the aR,w direction (corresponds to straight up
at Waco),
Rwm ● aθ,w gives the component of Rwm in the aθ,w direction (at Waco, points toward the
southern horizon),
Rwm ● aφ,w gives the component of Rwm in the in the aφ,w direction (at Waco, points toward the
eastern horizon).
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EGR1301_FALL2015_Earth1_to_Earth2_150903.docx
For convenience, Equations (4), (5), (6) are repeated here:
aR = sin(θ) cos(φ) aX + sin(θ) sin(φ) aY + cos(θ) aZ
(4)
aθ =cos(θ) cos(φ) aX + cos(θ) sin(φ) aY − sin(θ) aZ
(5)
aφ = −sin(φ) aX + cos(φ) aY.
(6)
The three dot product components become
Rwm ● aR is (Rwm,x) (aR,x) + (Rwm,y) (aR,y) + (Rwm,z) (aR,z). Upward,
Rwm ● aθ is (Rwm,x) (aθ,x) + (Rwm,y) (aθ,y) + (Rwm,z) (aθ,z). Southern,
Rwm ● aφ is (Rwm,x) (aφ,x) + (Rwm,y) (aφ,y) + (Rwm,z) (aφ,z). Eastern,
where the spherical unit vector are for Waco. Substituting in (4),(5),(6) yields
Rwm ● aR = Rwm,x sin(θ) cos(φ) + Rwm,y sin(θ) sin(φ) + Rwm,z cos(θ). Upward.
Rwm ● aθ = Rwm,x cos(θ) cos(φ) + Rwm,y cos(θ) sin(φ) − Rwm,z sin(θ). Southern.
Rwm ● aφ = −Rwm,x sin(φ) + Rwm,y cos(φ) . Eastern.
Substituting in θ and φ for Waco, the vector toward Moscow has components
Pointing Up,
3528.2 (0.8523) (–0.1253) + 7571.4 (0.8523) (–0.9921) + 1928.5 (0.5234) km
= –376.7 – 6402.1 + 1009.4 = –5769.4 aR km.
Pointing Toward the Southern Horizon,
3528.2 (0.5231) (–0.1253) + 7571.4 (0.5231) (–0.9921) – 1928.5 (0.8523) km
= –231.3 – 3929.3 – 1643.7 = –5804.3 aθ km.
Pointing Toward the Eastern Horizon,
–3528.2 (–0.9921) + 7571.4 (–0.1253) km = 3500.3 – 948.7 = 2551.6 aφ km
Mag check = sqrt(5769.4^2 + 5804.3^2 + 2551.6^2) = 8572 km (checks).
Referring to the Figures on the Next Page:
Azimuth angle = arctan(east/north) = arctan(east/(–south)) = arctan(2551.6/(-5798.3)) = 23.8⁰. Azimuth
angle is compass angle but without magnetic declination factored in. At Waco, compass points 4.5⁰ east
of north. Thus, the observed compass angle is 23.8 – 4.5 =19.3⁰.
Zenith angle = arccos(up/Mag) = arccos(-5769.4/8572) = 132.3⁰. Zenith angle = 0 corresponds to straight
up, 90 degrees is tangent to the Earth, and 180 degrees is straight down. Thus, Moscow is 132.3 – 90 =
42.3⁰ degrees below the horizon.
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EGR1301_FALL2015_Earth1_to_Earth2_150903.docx
True North at Waco
Moscow
– aθ
Standing on the Earth at Waco, Looking Straight Down. Align this
graph with true north, and stand on it. A step along the dashed
line is a step toward Moscow. If you used compass north instead
of true north, rotate the paper counterclockwise by 4.5° .
23.8°
5804
km
north
The projection of the Moscow vector tangent to the Earth’s surface
at Waco is sqrt(2552^2 + 5804^2) = 6340 km long.
Waco
aφ
2552 km east
Waco
6340 km tangent to Earth at Waco
Side View of Earth. The vertical
cross section contains Waco,
Moscow, and center of Earth. You
can see that the Moscow vector
points at an angle that is
arctan(5769 / 6340) = 42.3⁰ degrees
below the horizon at Waco. The
distance along the Earth’s surface to
Moscow is nearly one-fourth the
circumference of the Earth.
42.3°
5769 km
toward
center of
Earth
8572 km
Earth radius
6371 km
– aR
Moscow
Center of
Earth
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EGR1301_FALL2015_Earth1_to_Earth2_150903.docx
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