Find the distance (departure) to the nearest whole mile between initial position 50° 00' N 02° 00' W and 50° 00' N 01° 26' W
+26388 Find the distance (departure) to the nearest whole mile between
initial position 50° 00' N 02° 00' W
and 50° 00' N 01° 26' W
\(\begin{array}{lrcll} \text{Set Latitude } & lat_1 &=& 50^\circ \\ \text{Set Longitude } & lon_1 &=& -2^\circ \\ \text{Set Latitude } & lat_2 &=& 50^\circ \\ \text{Set Longitude } & lon_2 &=& -1^\circ 26' = -\frac{43}{30}^\circ \\ \end{array} \)
\(\begin{array}{lcll} cos(g) = cos(90^\circ - lat1) \cdot cos(90^\circ - lat2) + sin(90^\circ - lat1) \cdot sin(90^\circ - lat2) \cdot cos(lon2 - lon1) \\ \text{with g: Great circle distance} \\\\ \text{Since } cos(90° - a) = sin(a) \text{ and } sin(90° - a) = cos(a) \text{ applies, the formula simplifies to: }\\ cos(g) = sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon2 - lon1) \\\\ dist = 6378.388 * \pi / 180^\circ \cdot g \\ \text{with dist: distance in km }\\ \end{array} \)
\(\begin{array}{rcll} cos(g) &=& sin(50^\circ) * sin(50^\circ) + cos(50^\circ) * cos(50^\circ) * cos(-1^\circ 26' - (-2^\circ)) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-1^\circ 26' +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-\frac{43}{30}^\circ +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(\frac{17}{30}^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * cos(0.56666666667^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * 0.99995109238 \\ cos(g) &=& 0.99997979255 \\ g &=& \arccos{(0.99997979255)} \\ g &=& 0.36424544098^\circ \\\\ dist &=& 6378.388 * \pi / 180^\circ \cdot 0.36424544098^\circ \\ dist &=& 40.5492126920\ km \\ dist &=& \frac{40.5492126920\ km}{ 1.609344\frac{km}{mi}} = 25.1961126347\ mi\\ \end{array}\)
+118658 Find the distance (departure) to the nearest whole mile between initial position 50° 00' N 02° 00' W and 50° 00' N 01° 26' W '
I am working on the Earth being a sphere with a radius of 3959 miles
These two places both have a latitude of 50 degrees north and the longitudes are different by 24 degrees.
I need to work out what the radius of the earth is on the minor circle of 50 degrees north.
\(sin(90-50)=\frac{r}{3959}\\ 3959sin(40)=r\\ r=2544.8miles\\ \\~\\ distance=\frac{24}{360}*2*\pi*2544.8 = 1066\;miles\)
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+118658 Yes Chris wouild be right I read the quesiton incorrectly.
Find the distance (departure) to the nearest whole mile between initial position 50° 00' N 02° 00' W and 50° 00' N 01° 26' W
I read the second one as 26 degrees west INSTEAD OF 1 degree and 26 minutes west.
I will do the last bit a again. I am assumingthat the earth is a perfect sphere - which it isn't.
And that the 2 points are both at sea level.
The difference in longitude is 34 seconds.
\(distance = \frac{34}{360*60}*2*\pi*2544.8=25 miles\)
Now wolfram alpha and I agree
+129839 According to WolframAlpha ≈ 25.24 miles = 25 miles
http://www.wolframalpha.com/input/?i=distance+between+50%C2%B0+00%27+N+02%C2%B0+00%27+W+and+50%C2%B0+00%27+N+01%C2%B026%27+W
+37095 ...and here is an online calculator for such things:
40.5 km = 25.17 statute miles is what it answers.... (21.87 nm)
http://www.movable-type.co.uk/scripts/latlong.html
+26388 Find the distance (departure) to the nearest whole mile between
initial position 50° 00' N 02° 00' W
and 50° 00' N 01° 26' W
\(\begin{array}{lrcll} \text{Set Latitude } & lat_1 &=& 50^\circ \\ \text{Set Longitude } & lon_1 &=& -2^\circ \\ \text{Set Latitude } & lat_2 &=& 50^\circ \\ \text{Set Longitude } & lon_2 &=& -1^\circ 26' = -\frac{43}{30}^\circ \\ \end{array} \)
\(\begin{array}{lcll} cos(g) = cos(90^\circ - lat1) \cdot cos(90^\circ - lat2) + sin(90^\circ - lat1) \cdot sin(90^\circ - lat2) \cdot cos(lon2 - lon1) \\ \text{with g: Great circle distance} \\\\ \text{Since } cos(90° - a) = sin(a) \text{ and } sin(90° - a) = cos(a) \text{ applies, the formula simplifies to: }\\ cos(g) = sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon2 - lon1) \\\\ dist = 6378.388 * \pi / 180^\circ \cdot g \\ \text{with dist: distance in km }\\ \end{array} \)
\(\begin{array}{rcll} cos(g) &=& sin(50^\circ) * sin(50^\circ) + cos(50^\circ) * cos(50^\circ) * cos(-1^\circ 26' - (-2^\circ)) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-1^\circ 26' +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-\frac{43}{30}^\circ +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(\frac{17}{30}^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * cos(0.56666666667^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * 0.99995109238 \\ cos(g) &=& 0.99997979255 \\ g &=& \arccos{(0.99997979255)} \\ g &=& 0.36424544098^\circ \\\\ dist &=& 6378.388 * \pi / 180^\circ \cdot 0.36424544098^\circ \\ dist &=& 40.5492126920\ km \\ dist &=& \frac{40.5492126920\ km}{ 1.609344\frac{km}{mi}} = 25.1961126347\ mi\\ \end{array}\)
