Latitude and Longitude
A flight leaves Los Angeles at 9:00 p.m. on Tuesday. It lasts 11 hours and lands in Tokyo at 1:00 a.m. on Thursday. Eleven hours after Tuesday night should still be Wednesday. Where did the extra day go?
To answer that, you need to know how we locate points on a sphere. When your phone's GPS shows your position, it gives two numbers: a latitude and a longitude. They work like the coordinates $(x, y)$ on a plane, except that the "sheet of paper" is a sphere, so the coordinates are angles.
Meridians and parallels
Picture the Earth as a sphere that spins around an axis through the North and South Poles.
-
Meridians are half-circles that run from pole to pole. The reference meridian passes through Greenwich, England. It is the Prime Meridian, or $0^\circ$.
-
Parallels are circles perpendicular to the axis. The biggest one is the Equator ($0^\circ$). It splits the Earth into the Northern and Southern Hemispheres.
Points on the same parallel have the same latitude. Points on the same meridian have the same longitude.
Move the sliders and see where a point lands:
Imagine you are a beginner cartographer. On your first map, you must locate the most important reference lines of the Earth. Look at the drawing and match each letter…
Imagine you are a beginner cartographer. On your first map, you must locate the most important reference lines of the Earth. Look at the drawing and match each letter with its corresponding name:
-
Equator.
-
Greenwich Meridian.
-
North Pole.
-
Any parallel.
Two ships, Alfa and Beta, are sailing in the Atlantic Ocean. Their coordinates are: Ship Alfa: $20^\circ$ N, $30^\circ$ W Ship Beta: $20^\circ$ N, $60^\circ$ W Answer…
Two ships, Alfa and Beta, are sailing in the Atlantic Ocean. Their coordinates are:
-
Ship Alfa: $20^\circ$ N, $30^\circ$ W
-
Ship Beta: $20^\circ$ N, $60^\circ$ W
Answer with reasoning:
- a) Are they on the same parallel or on the same meridian?
- b) Which of the two ships is closer to the Greenwich Meridian?
Coordinates on the globe
Dallas is at about $33^\circ$ N, $97^\circ$ W. Latitude comes first, and the letter tells you the hemisphere. Some problems abbreviate West as O instead of W. It still means West.
Degrees and minutes. One degree is split into 60 minutes: $1^\circ = 60'$. So $96^\circ 48' = 96^\circ + \tfrac{48}{60}{}^\circ = 96.8^\circ$.
On a flat map (a planisphere), the horizontal grid lines are parallels and the vertical lines are meridians. Read the latitude from the horizontal lines and the longitude from the vertical ones. If the point sits between two lines, estimate how far along it is.
State in which hemisphere (North or South for latitude, East or West for longitude) the following geographical points are located: Point A: $34^\circ$ N, $120^\circ$ E…
State in which hemisphere (North or South for latitude, East or West for longitude) the following geographical points are located:
-
Point A: $34^\circ$ N, $120^\circ$ E
-
Point B: $12^\circ$ S, $45^\circ$ W
-
Point C: $89^\circ$ N, $5^\circ$ W
-
Point D: $40^\circ$ S, $170^\circ$ E
An explorer has mistakenly written down the following coordinates in his diary. Explain why these coordinates are geographically impossible: a) Latitude: $105^\circ$ S,…
An explorer has mistakenly written down the following coordinates in his diary. Explain why these coordinates are geographically impossible:
- a) Latitude: $105^\circ$ S, Longitude: $20^\circ$ E.
- b) Latitude: $45^\circ$ N, Longitude: $210^\circ$ W.
Look at the following flat representation (planisphere) of an ocean area. Identify the coordinates (Latitude and Longitude) of point $P$ marked with a star.
Look at the following flat representation (planisphere) of an ocean area. Identify the coordinates (Latitude and Longitude) of point $P$ marked with a star.
Antipodes
The antipode of a point is the point on the exact opposite side of the Earth. A line through the point and the center of the Earth comes out at the antipode.
If the longitude has minutes, write $180^\circ$ as $179^\circ 60'$ before subtracting. For example, $180^\circ - 10^\circ 20' = 179^\circ 60' - 10^\circ 20' = 169^\circ 40'$.
If you travel by airplane from Dallas ($40^\circ$ N, $3^\circ$ W) to a city that is exactly at its antipodes (the opposite point on the Earth's sphere): a) What would…
If you travel by airplane from Dallas ($40^\circ$ N, $3^\circ$ W) to a city that is exactly at its antipodes (the opposite point on the Earth's sphere):
- a) What would the latitude of that point be? (Same value but opposite direction).
- b) If the longitude of a point and its antipode must add up to $180^\circ$ and be in opposite hemispheres, what would the longitude of that point be?
The antipodes of a place are said to be the diametrically opposite point on the Earth's surface. If the coordinates of your city are $40^\circ$ $25'$ N latitude and…
The antipodes of a place are said to be the diametrically opposite point on the Earth's surface. If the coordinates of your city are $40^\circ$ $25'$ N latitude and $3^\circ$ $41'$ W longitude:
- a) What would the latitude of your antipodes be?
- b) What would the longitude of your antipodes be?
(Hint: For the latitude, change the hemisphere; for the longitude, find the meridian that completes $180^\circ$ and change the E/W hemisphere).
Distance along a meridian
Two cities on the same meridian are joined by an arc of a great circle. A full circle around the Earth through the poles has length $2\pi R$. With $R = 6{,}371$ mi, that is $2\pi(6{,}371) \approx 40{,}030$ mi. A central angle of $\alpha$ degrees covers $\tfrac{\alpha}{360}$ of the circle.
Why does this work? One degree of arc is $\tfrac{1}{360}$ of the whole circle. So $\tfrac{40{,}030}{360} \approx 111.2$ mi per degree.
Nautical miles. Sailors measure along meridians in minutes of arc. One nautical mile is the length of a meridian arc of $1'$, which is $\tfrac{1}{60}$ of a degree. So the number of nautical miles along a meridian is the number of minutes of latitude: $\alpha^\circ \times 60$.
Two cities, Villa Sol and Villa Luna, are located on the same meridian. If Villa Sol has a latitude of $45^\circ$ N and Villa Luna has a latitude of $20^\circ$ S, how…
Two cities, Villa Sol and Villa Luna, are located on the same meridian. If Villa Sol has a latitude of $45^\circ$ N and Villa Luna has a latitude of $20^\circ$ S, how many degrees of meridian arc separate them?
A humpback whale migrates following exactly the same meridian from a feeding ground located at $65^\circ$ N latitude to a breeding ground at $15^\circ$ S. If the mean…
A humpback whale migrates following exactly the same meridian from a feeding ground located at $65^\circ$ N latitude to a breeding ground at $15^\circ$ S. If the mean radius of the Earth is considered to be $6\,371$ miles, calculate the total distance in miles that the whale travels on its outward journey.
In navigation, the "nautical mile" is used, which is defined as the length of a meridian arc of $1'$ (one minute of a degree). a) Calculate how many miles a nautical…
In navigation, the "nautical mile" is used, which is defined as the length of a meridian arc of $1'$ (one minute of a degree).
- a) Calculate how many miles a nautical mile is equivalent to (Radius of the Earth = $6\,371$ miles).
- b) If a sailboat has sailed along a meridian from a latitude of $30^\circ$ N to $35^\circ$ N, how many nautical miles has it traveled?
Distance along a parallel
A parallel is a circle, but not a great circle. Its radius $r$ is smaller than $R$. At latitude $\alpha$, the radius of the parallel is the leg of a right triangle with hypotenuse $R$:
The triangle explains the formula. In it, the angle at the center is $\alpha$, the hypotenuse is $R$, and the side next to the angle is $r$. So $\cos\alpha = \tfrac{r}{R}$. At the Equator, $\alpha = 0^\circ$ and $\cos 0^\circ = 1$, so $r = R$. At the pole, $\cos 90^\circ = 0$, so the parallel is just a point. Make sure your calculator is in degree mode.
For a school project, you need to calculate the length of the $40^\circ$ N parallel of latitude. Knowing that the Earth's radius is $R \approx 6\,371$ miles, calculate:…
For a school project, you need to calculate the length of the $40^\circ$ N parallel of latitude. Knowing that the Earth's radius is $R \approx 6\,371$ miles, calculate:
- a) The radius of the circle formed by this parallel (use the expression $r = R \cdot \cos(\text{latitud})$).
- b) The total length (perimeter) of this parallel.
Two cities, P and Q, are located on the same meridian. City P is at $42^\circ$ N and city Q is at $15^\circ$ N. Another city, R, is located on the same parallel as Q,…
Two cities, P and Q, are located on the same meridian. City P is at $42^\circ$ N and city Q is at $15^\circ$ N. Another city, R, is located on the same parallel as Q, but at a distance of $20^\circ$ of longitude to the East.
Calculate, using $R = 6\,371$ miles:
- a) The distance between P and Q along the meridian.
- b) The distance between Q and R along the parallel.
Which of the two distances is greater?
Two cities, A and B, are located on the same parallel at a latitude of $45^\circ \text{ N}$. City A has a longitude of $20^\circ \text{ E}$ and city B a longitude of…
Two cities, A and B, are located on the same parallel at a latitude of $45^\circ \text{ N}$. City A has a longitude of $20^\circ \text{ E}$ and city B a longitude of $70^\circ \text{ E}$.
Knowing that the radius of the Earth is approximately $6371 \text{ mi}$:
-
Calculate the radius of the $45^\circ \text{ N}$ parallel (Remember that the radius of a parallel at a latitude $\alpha$ is calculated as $r = R \cdot \cos \alpha$. For $45^\circ$, you can use $0,707$ as an approximation of the cosine).
-
Determine the distance in miles that separates both cities following the arc of the parallel.
Longitude differences and shortest routes
Longitude difference.
-
Both points in the same hemisphere (both E or both W): subtract.
-
One E and one W: add. If the sum is more than $180^\circ$, the shortest way goes across the $180^\circ$ meridian, so use $360^\circ - \text{sum}$.
For example, $150^\circ$ E and $140^\circ$ W add up to $290^\circ$. The shortest difference is $360^\circ - 290^\circ = 70^\circ$.
Shortest route. The shortest path between two points on a sphere is an arc of a great circle. Meridians (with their opposite halves) and the Equator are great circles. Other parallels are not. If two points have the same longitude, the shortest route follows the meridian. If two points are on the same parallel other than the Equator, the shortest route is not along that parallel.
From distance back to angle. If you know the distance, divide by $111.2$ mi per degree on a great circle to get the angle. On the Equator, longitude behaves like latitude on a meridian.
Travel time. Use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$. A knot is one nautical mile per hour. Use the conversion to mph that the problem gives.
Two oceanographic research ships are on the Equator. Ship A is at a longitude of $12^\circ$ E and ship B is at a distance of $2\,500$ miles to the West of ship A.…
Two oceanographic research ships are on the Equator. Ship A is at a longitude of $12^\circ$ E and ship B is at a distance of $2\,500$ miles to the West of ship A. Determine the exact longitude of ship B, knowing that the radius of the Earth is $6\,371$ miles.
Calculate the shortest distance between two points, P and Q, located on the Earth's surface if their coordinates are: P: $(40^\circ \text{ N}, 10^\circ \text{ O})$ Q:…
Calculate the shortest distance between two points, P and Q, located on the Earth's surface if their coordinates are:
-
P: $(40^\circ \text{ N}, 10^\circ \text{ O})$
-
Q: $(40^\circ \text{ S}, 10^\circ \text{ O})$
Consider the Earth to be a perfect sphere of radius $6371 \text{ mi}$. Express the result in miles. Does the traveler move along a meridian or a parallel? Justify your answer.
A submarine travels along the seabed from a point X $(0^\circ, 175^\circ \text{ E})$ to a point Y $(0^\circ, 170^\circ \text{ O})$. 1. How many degrees of longitude has…
A submarine travels along the seabed from a point X $(0^\circ, 175^\circ \text{ E})$ to a point Y $(0^\circ, 170^\circ \text{ O})$.
-
How many degrees of longitude has it traveled if it followed the shortest route along the Equator?
-
If the submarine travels at a constant speed of $20 \text{ nudos}$ ($1 \text{ nudo} \approx 1,85 \text{ mph}$), how long will it take to complete the journey? (Radius of the Earth: $6371 \text{ mi}$).
-
Was there a change of date during the journey? Explain why.
Time zones
The Earth turns $360^\circ$ in 24 hours, so it turns $360^\circ \div 24 = 15^\circ$ every hour. That is why the world is divided into 24 time zones, each $15^\circ$ wide. The Sun rises first in the east, so it is later to the east and earlier to the west.
Theoretical zones. Zone number $n$ is centered on the meridian $15n^\circ$. It covers $15n^\circ \pm 7.5^\circ$, and $n$ is the UTC offset (positive east, negative west). To find the zone, divide the longitude by $15$ and round to the nearest whole number. Use $+$ for east and $-$ for west.
We know that the Earth takes 24 hours to complete a $360^\circ$ rotation on its axis. a) How many degrees of longitude correspond to a one-hour difference? b) If it is…
We know that the Earth takes 24 hours to complete a $360^\circ$ rotation on its axis.
- a) How many degrees of longitude correspond to a one-hour difference?
- b) If it is 12:00 noon in a city located at $0^\circ$ longitude (Greenwich), what time will it be in a city located at $30^\circ$ E?
A group of friends in Dallas (longitude $96^\circ$ $48'$ W) wants to watch a live space rocket launch from Cape Canaveral (longitude $80^\circ$ $36'$ W). If the launch…
A group of friends in Dallas (longitude $96^\circ$ $48'$ W) wants to watch a live space rocket launch from Cape Canaveral (longitude $80^\circ$ $36'$ W). If the launch is scheduled for 14:30 local time in Florida, and knowing that each $15^\circ$ time zone is equivalent to a theoretical one-hour difference, at what time should they tune into the live stream in Texas? Use the theoretical time zone diagram to carry out the calculation.
Flights and the date line
When a trip crosses time zones, the calendar date can change too. The safest method is to go through UTC (Greenwich time):
-
Convert the departure time to UTC. Subtract the zone number: UTC = local time $-$ (zone number). For west zones, that means adding hours.
-
Add the flight duration to get the UTC time at landing.
-
Convert to local time at the destination: local = UTC + (zone number).
This method handles the International Date Line (near the $180^\circ$ meridian) automatically, as long as you use $+$ for east and $-$ for west. Crossing the line eastward, you get a day back. Crossing westward, you jump a day ahead.
Back to the start: the day was not lost. Los Angeles is 17 hours behind Tokyo. The 11-hour flight plus that 17-hour gap adds up to 28 hours, which is more than one full day.
A reconnaissance plane takes off from a base located at coordinates $(30^\circ \text{ N}, 15^\circ \text{ O})$ at 08:00 local time. It flies directly East following the…
A reconnaissance plane takes off from a base located at coordinates $(30^\circ \text{ N}, 15^\circ \text{ O})$ at 08:00 local time. It flies directly East following the same parallel until it reaches a city located at $60^\circ$ longitude East. The flight has a total duration of 9 hours.
At what local time of the destination city will the plane land? Take into account the change of time zones during the journey (consider theoretical time zones of $15^\circ$ each).
An airplane takes off from Tokyo ($35^\circ$ N, $139^\circ$ E) at 08:00 on Monday bound for San Francisco ($37^\circ$ N, $122^\circ$ W). If the flight lasts 10 hours,…
An airplane takes off from Tokyo ($35^\circ$ N, $139^\circ$ E) at 08:00 on Monday bound for San Francisco ($37^\circ$ N, $122^\circ$ W). If the flight lasts 10 hours, what time and day will it be in San Francisco when the airplane lands? Take into account the date change when crossing the Greenwich antimeridian ($180^\circ$).
Polyhedra, Euler's formula, and surface area
Some problems in this section use solids. Here are the tools you need.
How to count. Add up the sides of all faces. Each edge is shared by 2 faces, so $E = \tfrac{\text{total sides}}{2}$. If every vertex has the same number $k$ of faces meeting there, then $V = \tfrac{\text{total sides}}{k}$.
Cutting off a vertex (truncating). When you slice off a corner where $k$ edges meet, you create a new face with $k$ sides. Each original face gains sides, and the cut face is a new regular polygon.
Regular polygon areas (side $s$):
-
Square: $s^2$.
-
Equilateral triangle: $\tfrac{\sqrt{3}}{4}s^2$.
-
Regular hexagon: $\tfrac{3\sqrt{3}}{2}s^2$.
A regular hexagon is made of 6 equilateral triangles. The side of the hexagon equals the radius of the circumscribed circle. The apothem is the height of one triangle, so $a = \tfrac{\sqrt{3}}{2}s$. Also, area $= \tfrac12 \cdot \text{apothem} \cdot \text{perimeter}$.
Prisms. Lateral area $=$ (perimeter of the base) $\times$ height. Total surface area $=$ lateral area $+$ 2 bases.
A truncated octahedron is constructed from a regular octahedron with an edge of $12 \text{ cm}$. To do this, each of the six vertices is cut off so that small regular…
A truncated octahedron is constructed from a regular octahedron with an edge of $12 \text{ cm}$. To do this, each of the six vertices is cut off so that small regular square-based pyramids are removed, turning each original triangular face into a regular hexagon.
-
Determine the number of faces, vertices and edges of the resulting polyhedron and check if it satisfies Euler's formula.
-
If the edge of the new polyhedron (that of the resulting hexagon) is $4 \text{ cm}$, calculate the total surface area of the truncated octahedron.
A water tank has the shape of a right regular hexagonal prism. It is known that the radius of the circumscribed circle of the hexagonal base measures $4 \text{ m}$ and…
A water tank has the shape of a right regular hexagonal prism. It is known that the radius of the circumscribed circle of the hexagonal base measures $4 \text{ m}$ and that the height of the prism is three times the apothem of the base.
Calculate the cost of waterproofing the side walls and the bottom (bottom base) if the price of the material is $12,50 \text{ \$/m}^2$. Round the result to two decimal places.
Spheres, cones, and frustums
A frustum is a cone with its tip sliced off. Its volume is the big cone minus the small cone you removed. Similar triangles give the height of the missing tip.
Combined solids. Find each volume separately and add (or subtract). When the problem says to take $\pi \approx 3.14$, use it at the end. Science and lab problems often use metric units, where $1 \text{ mL} = 1 \text{ cm}^3$. In customary units, $1 \text{ ft}^3 \approx 7.48$ gallons.
Comparing volumes. To prove a relationship, write both volumes using the same variable and divide. Things usually cancel.
A perfume company designs a bottle in the shape of an inverted frustum of a cone topped by a hemisphere that fits perfectly on the top base. The design data is as…
A perfume company designs a bottle in the shape of an inverted frustum of a cone topped by a hemisphere that fits perfectly on the top base. The design data is as follows:
-
Height of the frustum of a cone: $8 \text{ cm}$.
-
Radius of the bottom base (where it rests): $3 \text{ cm}$.
-
Radius of the top base (where the hemisphere fits): $6 \text{ cm}$.
Calculate the total volume of the bottle and determine if it is able to hold a standard $330 \text{ ml}$ bottle of essence. (Take $\pi \approx 3,14$).
There is a cylinder of radius $R$ and height $2R$. A sphere of radius $R$ is placed inside it, fitting tangentially with the walls and bases of the cylinder. 1. Prove…
There is a cylinder of radius $R$ and height $2R$. A sphere of radius $R$ is placed inside it, fitting tangentially with the walls and bases of the cylinder.
-
Prove analytically the relationship between the volume of the sphere and the volume of the cylinder.
-
If the volume of the cylinder is $54 \text{ cm}^3$, what volume of air is left free inside the cylinder once the sphere is inserted?