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If you look at the earth as it is, the longitudes
intersect the latitudes at 90o only at the equator. So the distance
to be traversed will be either one-fourth the circumference of the
earth or three-fourths the circumference of the earth.
Assuming that the earth is a perfect sphere,
the same works with London as the pole. Using spherical trigonometry,
the points B & C can be either of the following
Point B - at 26 deg 6' 18.591" N and
127 deg 56' 54.905" E, which would be a point in the island
of Okinawa
and
Point C - at 26 deg 6' 18.591" N and
128 deg 9' 30.905" W, which would be a point south-west of
Santa Barbara, California in the Pacific Ocean
OR
Point B - at 26 deg 6' 18.591" S and
52 deg 3' 5.095" W, which would be near Linha São Sebastião,
Francisco Beltrão-PR, Paraná, Brazil
and
Point C - at 26 deg 6' 18.591" S and
51 deg 50' 29.095" E, which would be a point south-east of
Madagascar in the Indian Ocean.
Venky
From some math or physics class or puzzle
site we have all heard about the triangle with three right angles
- because it is drawn on a sphere. Here it is again, but not so
simply. From the question, we can in fact deduce that we have such
a triangle. Let's call the first segment of the trip c (since it
is opposite the angle at point C). The second segment, a, is defined
as the same length as c. The law of sines for triangles on the sphere
says sin(a)/sin(A) = sin(c)/sin(C),
so the angle of the trip at A (London) must also be a right angle.
Similarly, the segment of the trip from C
to London must be the same length as the other two segments.
Now that that's out of the way, where are
B and C? Let's start with the example three-right-angle triangle
from all those homework questions, the one with one vertex at the
north pole and the other two on the equator at 0 deg and 90 deg
W. Since the Earth is a sphere (of course it isn't exactly, but
I'm never going to solve this without assuming it is), we can slide
that triangle around until one vertex is in London. That still leaves
us one degree of freedom (i.e. it can be rotated around London)
so we put one segment at a bearing of 45 deg from London.
We know the lengths of the segments are one
quarter the circumference of the Earth (obvious from the initial
triangle). This allows us to draw another triangle ABN, where N
is the North Pole, and use the law of cosines on a sphere:
cos(a) = cos(b)cos(n)+sin(b)sin(n)cos(A)
In radians, n = c = pi/2 , A = pi/4 and b
= (90-latitude_of_London)*pi/180 = 0.67195.
Since sin(n) = 1 and cos(n) = 0, cos(a) = sin(b)cos(A)
and a = 1.1487 radians = 65.82 deg. This puts B at latitude 24.18N.
A calculation for N (two solutions for inverse sine, pick the right
one!) puts B at latitude 128.18E. Symmetry about the prime meridian
(since the original bearing of 45 degrees bisects the 90 degree
angle of the triangle at London) puts C at 24.18N, 128.18W.
We can double check the maths with the triangle
BCN, we know b = c = 1.1487 (it was 'a' just a moment ago, but our
angles have changed name) and angle N = 1.7738 (the longitude between
B and C). This tells us that the length of the segment BC is indeed
90 degrees, give or take rounding errors, or one fourth the circumference
of the Earth.
So in all, your trip has taken you over Scandinavia,
Russia and China to point A off the east coast of Taiwan, over the
Pacific to point B off the west coast of Mexico, and back over Mexico,
the U.S., my house (go ahead, prove me wrong :-) ), and the North
Atlantic home again.
Quite the personal transport you have, Martin!
Zack
Okinawa, Japan.
This is a somewhat ambiguous question, since
"bearings" are quite different entities depending on whether
you are dealing in Cartesian or spherical geometry. For this question,
I will assume spherical geometry, where a "straight line"
is the shortest distance across the sphere (i.e. a line following
a "great circle", whose center is at the center of the
earth) and a "bearing" is the initial compass direction
defining the circle. This is a bit counterintuitive when thinking
in two dimensions, since the compass bearing will typically change
throughout the trip.
In order to meet the criteria of the given
trip (two 90 degree turns that return you to your starting point)
the distance traveled from point A to point B will be 1/4 the circumference
of the earth (~10,000 km). To see why this is true, imagine the
world tilted so that London is actually the North Pole. A trip on
any bearing will proceed down a line of longitude (i.e. due South).
After 10,000 km we would be at the equator, so a 90 degree left
turn would send us east along the equator. (Note that longitude
lines and the equator are all "great circles"). After
10,000km (1/4 of the way) around the equator, another left turn
would send us due north again, back to the pole.
Now that we can visualize the trip, we need
to do the same thing only from where London really is, and starting
off at 45 degrees. Sparing you the math, a 10,000km great circle
trip bearing at 45 degrees would take you around 26.40N 127.80E,
which is on the island of Okinawa. You would arrive on a bearing
of about 150 degrees, so a left turn would send you off again on
a bearing of 60 degrees. Another 10,000km along this arc takes you
to around 26.81N 127.90W, which is in the Pacific Ocean (between
southern California and Hawaii). You would arrive on a bearing of
120 degrees. Another left turn would send you off at 30 degrees,
which brings you back to London.
Pete Mitchell
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