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Navigational systems are extremely useful in our society today. Although not directly connected to our school environment, it is used in the real world to locate different vessels both on and off land. These systems can be used to determine the least possible time from departure to destination, which are extremely important for search and rescue missions. For example, a call of emergency is made and the caller is stranded on a mountain or in the ocean. I thought about how the search and rescue team navigated and what math they used to find the fastest route to the victim.
Even during my childhood, I have always liked being up in the air and on the water.
I would always enjoy long plane flights, and stress-relieving cruises through the vast ocean. I joined sea cadets when I was 12 years old. I was nervous at first however soon after I found it quite interesting and extremely beneficial to my knowledge as well as my discipline.
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Although I am no longer a sea cadet, my passion for ships and other vessels has not disappeared.
I constantly ask myself how control towers were able to guide planes and ships so perfectly. I realize we have advanced technology that allows us humans to navigate, however in the past there were no technology to aid people, therefore they would rely solely on the knowledge of the navigators. I took an interest in this topic as my passion for these methods of transportation have never faltered.
I decided to do research and show how the math we have learned have been implemented in navigating and navigational systems.
In the past, the primary mode of transportation was boats. Cities were based around bodies of water, everything revolved around that body of water: transportation of people, the trading of goods and supplies, etc. Before modern technology and advancements, it was extremely hard, almost impossible to navigate at sea. However, the sea was essential for many industries such as fishing, and transportation, thus having a navigation system would change the world.
The first advancement in navigation was using the position of the Sun and the stars. It was discovered that the patterns depended upon where you were on the Earth. By working out a coordinate system for the Earth (latitude and longitude), it was evident that the latitude could be determined by measuring the angle of the Sun at noon above the horizon. This required a good knowledge of angles and trigonometry. The angle itself could be measured using a sextant, again using ideas from trigonometry. An example using a diagram is listed below:
Ex: Find the distance “x”
Using a sextant we can determine that the angle of elevation is 30°, and the object we are looking at, a lighthouse for example, is 100m. Using trigonometry, a sailor would be able to find the distance from their location to the object.
Solution: tan 30°= 100/x
x = 100/tan 30°
≈ 173 m
As mentioned above, navigation systems are extremely useful for search and rescue missions. In a real world problem we can use trigonometry and bearings to help us determine the shortest possible flight path. Search and rescue missions are often in remote areas with the closest help tens or even hundreds of kilometers away therefore require the fastest and most efficient transportation possible (ex: helicopters, planes, speedboats).
An example question would be:
Some hikers are stuck on a mountain and call for a rescue team. The base (A) determines the bearing from the caller (C) to the base is 54°. The distance between them is 50 km, and the base decides to send a helicopter. The wind speed is 20 knots blowing on a bearing of 180°, and the aircraft speed is 100 knots. Solve for the ground speed and the bearing the helicopter has to fly at.
First we solve for the ground speed (b). This we can use the cosine law:
c2 = a2 + b2 - 2ab cosC
1002 = 202 + b2 - 2(20)(b) cos126
0 = b2 - 2(20)(b) cos126 + 202 - 1002
0 = b2 + 23.5b - 9600
b = -23.5(-23.5)2 - 4(1)(-9600)2(1)
= -23.5 197.36324382
86.9knots (correct) -110(inadmissible because speed cannot be negative)
86.9 knots 44.7 m/s
Knowing this we can use sine law to find angle B because we have side b, angle C and side c.
sin Bb = sin Cc
sin B86.9=sin 126100
sin B =86.9 (sin 126)100
B =sin-186.9 (sin 126)100
B 44.7 °
Therefore, if the aircraft takes a bearing of 44.7° with a speed of 100 knots, and the wind blows towards the south with a speed of 20 knots, the ground speed of the aircraft becomes 86.7 knots and the bearing becomes 54°.
Trigonometry can also be used to find one’s positioning when out at sea. One can find their positioning using a sextant and a navigator’s book. First, they would locate a star in the sky, and the time it is. In this case, using a sextant, the angle from the sailor’s location to the star at 50°. These two pieces of information are necessary to find positioning. Next, the book (HO-229) will indicate that at 6:00 pm star A would be located at a position, in this case directly over the sailor’s head.
We can now use trigonometry to find the distance from the sailor’s location to the point directly below the star using the tangent because you have the angle and the location (longitude and latitude). The sailor’s location could be anywhere on the circle, any point on the circle would be looking at the same star at 50°. Therefore, using a second star, star B, and repeating the same method, we can create another circle, or arc. These two arcs intersect at a specific point, the point of intersection, and is called a fix. To be more accurate, one could use another star but two is enough to pinpoint the location of the sailor.
The development of spherical trigonometry was needed to solve triangles on the surface of the Earth that resulted from the navigational measurements. Tables were used alongside the ephemerides (tables of the location of the Sun, planets and other stars for frequent time intervals every day throughout the year). This is an example of a nautical table from the 18th century, these included extreme amounts of calculations which would have been done by human computers today.
Spherical trigonometry is used for terrestrial navigation.
This special case of spherical triangle consists of two random locations L0 (Lat0, Lon0) and L1 (Lat1, Lon1), and the North Pole (NP) as the vertex points. These three sides are great-circle segments meaning the length of the diameter of the sphere. The angles A0 and A1 are the angles of the segment from L0 to L1 that intersect with the meridians (half of a great circle, ended by either the North Pole or the South Pole) in L0 and L1.
The third angle (North Pole) is the difference of the Longitude of the two locations (Lon1-Lon0). Since the North Pole is one of the vertex points, and two of the sides are meridian segments and thus great-circle segments, the third side D is the great-circle distance between L0 and L1. The meridian segments of the triangle are the complementary angles of the Latitudes of the positions L0 and L1: 90° - Lat 0/1.
With this information, the two principal navigational problems can be solved with this spherical triangle arrangement:
The 'distance problem': If the locations L0 and L1 are known, the distance of side D and the angles A0 and A1 (true bearings) can be calculated.
The 'destination problem': If the location L0 (departure) and the distance D and the angle A0 are known, the position of the destination L1 can be calculated.
The 'Distance Problem': finding distance D and true bearings A0 and A1
To find the distance D we must use Supplemental Cosine Rules, used specifically for spherical trigonometry. The rule is:
cos A = cos B cos C + sin B sin C cos a
cos B = cos C cos A + sin C sin A cos b
cos C = cos A cos B + sin A sin B cos c
So we can apply this to find distance D:
cosD = cos(90°- Lat0)cos(90°- Lat1) + sin(90°- Lat0)sin(90°- Lat1)cos(Lon1 - Lon0)
***Let x and y be any real number***
***cos(90°- x) = sin (x)***
cosD = sin(Lat0)sin(Lat1) + cos(Lat0)cos(Lat1)cos(Lon1-Lon0)
D[°] = acos[sin(Lat0)*sin(Lat1) + cos(Lat0)*cos(Lat1)*cos(Lon1-Lon0)]
This identity gives the 'angular' distance D between the locations L0 and L1 in degrees. The distance expressed in nautical miles is obtained by the fact that along a great circle, one minute-of-arc corresponds to one nautical mile. So each degree of angular distance corresponds to 60 nautical miles: D[Nm] = 60 * acos[sin(Lat0)*sin(Lat1) + cos(Lat0)*cos(Lat1)*cos(Lon1-Lon0)]
Trigonometry in Navigation: From Ancient Techniques to Modern Search and Rescue. (2024, Feb 15). Retrieved from https://studymoose.com/document/trigonometry-in-navigation-from-ancient-techniques-to-modern-search-and-rescue
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