Solar Impulse: around the world without fuel to promote clean technologies.

1. 1/10
In this worksheet, students will get to know some units of
measurement that are less common in their daily lives, but that
are routinely used in aviation:degrees for latitude and longitude,
nautical miles for distances, and feet for altitude.
It will give them a chance to discover maps through a pilot’s
eyes. They will also find out how pilots prepare their flights,
how they keep track of where they are, and what types of maps
they use.
And finally, they will be working with scales, altitudes,
longitudes and latitudes, giving them ample opportunity to
practice converting units and using proportionality laws.
This worksheet will further offer a short historical overview of
maps,as well as a research activity in the“Brainteaser”section.
Project: EPFL | dgeo | Solar Impulse
Writing: Angélique Durussel
Graphic design: Anne-Sylvie Borter, Repro – EPFL Print Center
Project follow-up: Yolande Berga
MODELS AND MAPS
“Scale that!”

2. Concepts covered
Mathematics :
• Specialized units (feet, units for latitude
and longitude, nautical miles)
• Unit conversions
• Proportions
• Map scales
Duration of the activity
Introduction and discussion: 1 period
Exercises: 2 periods
Brainteaser: 1 à 2 periods
Delving deeper: 1 period
* Disciplines spécifiques à la scolarité vaudoise
OCOM : options de compétences orientées métiers
OS MEP : option spécifique mathématiques et physique

3. MODELS AND MAPS - GUIDE 3/10
HOW DO PILOTS KNOW WHERE THEY ARE IN FLIGHT?
While some animals are born with an internal navigation system, humans are not (See “MAGNETISM”
worksheet). Instead, they depend on representations of the world, or maps, to avoid getting lost when
traveling from place to place. For over 2000 years, people have been drawing and improving maps.
Today, GPS systems have replaced paper maps, and can be found in airplanes, trains, boats, and even
in our cellphones.
This is a good opportunity to talk about how GPS systems work. Below is a simplified summary, based
on which you can discuss the technology with your students.
Global Position System
There are about 30 satellites that are used by the global positioning system (GPS). They orbit the earth
at an altitude of 20,200 km. The satellites emit waves that travel at the speed of light (300,000 km/s) and
are detected by GPS devices. Each satellite and each receiver
are equipped with clocks. Using these clocks, it is possible to
measure exactly how long it takes for the waves to travel from
the emitter to the receiver.
This way the distance between the satellite and the receiver
can be calculated: distance = speed · time
One of the challenges lies in synchronizing the clocks in the
satellite and the receiver.
Each receiver communicates with multiple satellites. It takes at
least three for the position of the receiver to be determined ac-
curately. A fourth satellite is necessary to determine the altitude.
Some satellites spin around the earth in the same amount of time that it takes the earth to spin around
its axis. Seen from earth, they look motionless. They are said to be geostationary. They orbit the earth
at an altitude of 36,000 km, and are used mainly for meteorology and military communication. They can
also be used to relay communications to space missions. There are so many geostationary satellites in
orbit that there is a risk that they could collide!
The satellites used in by the GPS system are not geostationary. They do not stay above the same re-
gion all the time.
This animation illustrates how the satellites orbit the earth:
http://commons.wikimedia.org/wiki/File:ConstellationGPS.gif
Satellite 1
Satellite 2
d1
d2
d3
Satellite 3
P

4. 4/10 MODELS AND MAPS - GUIDE
LES CARTES
Idea: This topic can be covered at the same time as the Great Discoveries, a topic that is discussed in
10th grade history lessons.
LATITUDE AND LONGITUDE
GPSs provide information on the latitude and longitude of the
receiver. Well before they were invented, sailors knew their
latitude thanks to the North Star. By lining up an astrolabe
with the North Star, they were able to calculate the latitude. Of
course they were only able to do so at night.
It is much more difficult to determine longitude, as this requires
a precise method of measuring time. In the past, the captain
had to know exactly how much time had passed between the
moment the sun passed the zenith at his current location and
the moment it passed it in the city he had departed from earlier,
of which he knew the longitude. This allowed him to calculate
the east-west distance he had traveled. It was only in 1725,
when John Harrison invented the marine chronometer, the
first of its kind that remained accurate on boats, that it became
possible to correctly determine longitude at sea. Until then,
sailors simply didn’t know their east-west position, and often,
poor estimates of their longitude had fatal consequences.
Rama (CC-BY-SA)
Astrolabe
First marine chronometer from G. Harrisson
Pupils can test the effect of modifying the latitude and longi-
tude in real time using the following animation:
www.edumedia-sciences.com/fr/a718-longitude-
latitude
edumédia, Longitude - Latitude
Idea: Synchronize this topic with the geography class.
ALTITUDE
Extension: A foot is divided into inches (1 foot = 12 inches). This could be an opportunity to do exer-
cises using fractions.

5. MODELS AND MAPS - GUIDE 5/10
True False
On a map drawn to scale, the proportions are always conserved
4 cm on a map with a scale of 1: 25,000 represent 100 m on the ground
A diagram always depicts objects as being smaller than they are in reality
If 1 cm on a map represents 0.02 cm in reality, the scale of the map is 50 : 1
SCALES
It is essential that the pupils know how to convert from one unit of length to another. On maps, dis-
tances are often measured in cm, while distances outdoors are measured in m or km. There are three
common types of problems when considering scales:
*
* The correct answer is 1,000 m.
A) We know the scale (S) and the distance on the map (M), but the distance in reality (R) is unknown.
B) We know the distance in reality (R) and the scale (S) but want to know the distance on the map (M).
C) We know the distance on the map (M) and the distance in reality (R) but want to know the scale (S).
When the representation of an object is larger than the object itself, the procedure is the same (pay
attention to where you put the 1 in the proportionality table):
Scales are an excellent example of how to apply laws of proportionality. It is important to express all
values with the same units to be able to use the proportionality coefficient.
Pay attention to the units: the results are in cm and have to be converted to m or km.
?
1.5 km
2.5 km
?
3 cm
?
10 cm
3 cm
1 : 25,000
map
map
map
map
3 cm ?
? 1.5 km = 150,000 cm
2.5 km = 250,000 cm10 cm
3 cm
1
1
1
21
25,000
25,000
?
?
1
reality
reality
reality
reality
1 : 25,000
?
21 : 1
Quiz

6. 6/10 MODELS AND MAPS - GUIDE
ALL THIS IN NUMBERS...
Exercise 1
a)
Exercise 2
a) A circle has 360° = 360 ∙ 60’’ = 21600’’. The circumference of the earth is 4,000 km.
That means that 40,000 / 21,600 ≈ 1.852 km = 1 mille marin.
Exercise 3
If the students have not completed exercise 2, provide them with the value of a nautical mile = 1.852 km.
a) The distance between Paris and Brussels is 5.9 cm ∙ 4,500,000 = 26,550,000 cm = 265.5 km.
b) The latitude is 46° 50’ 35’’ N.
To transform this into decimal notation, the minutes and the seconds have to be converted into
degrees: 50 minutes / 60 ≈ 0.833 degrees
35 seconds / 3’600 ≈ 0.010 degrees
The latitude of the Payerne airdrome in decimal notation is obtained by adding these two values to
46,843° N.
c) The difference in the latitude of the cities is 47.633 – 46.843 = 0.790°.
One minute corresponds to one nautical mile.
That means that 0.79 degrees = 47,4 angular minutes, or 47.4 ∙ 1.852 km ≈ 87.8 km.
Thus, the two cities are ~ 88 km apart.
model figurine
31.6 cm 180 / 202 = 0.89 cm63.8 m = 6,380 cm 180 cm
6,380 / 31.6 = 201.9 2021 1
reality realityb)
The scale of the model is about 1: 202. The figurine of the pilot would be 0.89 cm tall.
b) We know that one nautical mile corresponds to on angular
minute and equals 1.852 km.
The 15,329 km between the two cities
equals 15,329 / 1.852 ≈ 8,277 nautical miles,
which corresponds to 8,277 angular minutes,
or 8,277 / 60 ≈ 137.9° = 137°57’.
The latitude of Eyrarbakky is said to be 63°50’. Lyddan is
therefore at 137° 54’ – 63° 50’ = 74° latitude South.
latitude Nord
latitude Sud
Eyrarbakky
Lyddan
15'329 km
63,8°
74,1°

7. MODELS AND MAPS - GUIDE 7/10
Exercise 4
a) On the plan, the line indicating the length of the
Airbus is 14.1 cm long. The goal is to find out
how long the line representing the 63.8 m wing-
span of Solar Impulse is:
Exercise 5
1 foot = 30.48 cm = 0.3048 m.
So 4,900 feet = 1,493.5 meters et 28,000 feet = 8,534.4 meters.
The difference between the two altitudes is 7,040.9 meters.
Exercise 6
1 foot = 30.48 cm = 0.3048 m.
So 5,600 feet = 1,706.9 meters
To reach 2,200 m + 400 m = 2,600 you have to gain 893.1 meters = 2,930.1 feet.
Exercise 7
On the plan, 5.9 cm (the distance between the centers of the wheels) corresponds to 650 mm = 65 cm.
The scale is about 1: 11.
The height of the tow tug is 7.6 cm on the plan, which corresponds to 83.6 cm in reality.
map map
14.1 cm 14.1 cm64.45 m = 6,445 cm 64.45 m = 6,445 cm
63.8 m = 6,380 cm 11 m = 1,100 cm6,380 ∙ 14.1 / 6,445 ≈ 13.96 cm 1,100 ∙ 14.1 / 6,445 ≈ 2.41 cm
reality reality
b) For the Cessna :
c) The ratio of the wingspans of Solar Impulse and the Airbus is 63.8 : 64.45 or 1: 1.01.
d) The ratio of the wingspans of Solar Impulse and the Cessna is 63.8 : 11 or 1: 0.17.
The dimension line for the wingspan of Solar
Impulse is 13.96 cm long.
The dimension line for the wingspan of the
Cessna measures 2.41 cm.

8. 8/10 MODELS AND MAPS - GUIDE
BRAINTEASER
a) a) The first step to finding the cities on the map is de-
termining where the plane will be at 6 o’clock. On the
map, draw a line connecting Le Russey and Morteau,
and mark the middle. Then, draw a straight line from
Payerne through the mark, and extend it by the same
proportion as on the map. This lets us deduce that the
plane will be above Rougemont at 7 o’clock.
By drawing the lines on a road map with the right angles
and respecting proportionality, we can determine that
the plane will be above Grostenquin at 9 o’clock, and
above Saarlouis at 10 o’clock. At noon, the plane will be
a few km north of Marche-en-Famenne.
b) 1. The distances have to be calculated using the numbers indicated on the map.
Between Metz and Grostenquin : 43 + 16 = 59 km.
Given that a car travels at about 60 km/h on a sec-
ondary road, this distance would take about 1 hour to
covern.
Between Grostenquin and Saarlouis, the distance
is approximately: ~2 + 17 + 3 + 25 + ~3 = 50 km.
This corresponds to a 50 min trip.
Between Metz and Saarlouis : 27 + 26 + ~3 = 56 km,
a little under 1 hour.
In conclusion, if Jean wants to see the plane at both locations, he had better be quick! It takes
50 minutes to get from Grostenquin to Saarlouis. If there is a lot of traffic or if he is stopped at
the border, he will miss the second sighting!
2. The best solution would be to go close to St-Avold. From Metz, the distance is about 44 km
(about a 45 minute trip), and the plane passes there at about 9:30 a.m.
3. If Jean decides to go to St-Avold, he would have to leave at about 8:30 a.m. to have some extra
margin. He would be back at around 11 o’clock. He would have plenty of time to watch the plane
fly by.
6:00
7:00

9. MODELS AND MAPS - GUIDE 9/10
c) To find out the how far the plane has to fly to go from Payerne to Brus-
sels, you first have to determine the scale of the map, then that of the
flight chart.
The scale of the map can be determined by measuring a straight stretch
of road on the map. For example, the road from Moudon to Payerne is
1.6 cm long on the map and corresponds to 20 km. The scale of the map is therefore 1: 1,250,000.
On the map, a straight line from Payerne to Brussels is about 38.5 cm long, corresponding to a
distance of about 480 km in reality (38.5 ∙ 1,250,000 = 48,125,000 cm).
This value can be used to determine the scale of the copy. On the copy, the line from Payerne to
Brussels is about 19.2 cm long, which means that the scale is a 1: 2,500,000.
Next, all of the straight flight segments drawn on the flight chart have to be summed up. They add
up to: 3.5 + 6.9 + 1.6 + 2.3 + 3.6 + 3.7 = 21.6 cm, which corresponds to a distance of about 540 km.
This is the distance the plane covered from 5:00 a.m. to 1:00 p.m., so in 8 hours. The plane was
flying at an average speed of 67.5 km/h.
d) If the plane continues to fly at 67.5 km/h for 7 more hours, it will cover an additional 472.5 km around
Brussels airspace before being able to land.
e) A 600 km round trip with 15 hours time is perfectly feasible. With an average speed of 100 km/h,
it would take the driver 6 hours. Even with traffic jams and stops, he would have enough margin to
make it home on time.

10. 10/10 MODELS AND MAPS - GUIDE
DELVING DEEPER…
a) The three sections of the wings are shaded in gray, the cockpit and the hull are in red, the horizontal
stabilizers are in blue and the vertical stabilizer is in green:
b) On the representation below, the green piece (vertical stabilizer) is seen from above.
The length of the wings measured on the plans is shorter than their actual length, as the wing seg-
ments are loaded onto the Boeing 747 diagonally. This can also be neglected in the exercise. The
answer of part d) can be used by more advanced students to draw a sketch to scale.
c) The result is 8,6 cm + 8,5 cm + 8,5 cm. This adds up to a total of 25.6 cm if we consider that the
three segments are connected end to end in reality.
These 25,6 cm correspond to the wingspan of the plane: 63.8 meters. The scale of the plan is there-
fore approximately 1 : 250.
The scale that is indicated on the plan is 1 : 150. This means that the original plan has been reduced.
Extension: you can ask students by what factor the original plan was reduced.
The answer is 150 : 250 = 0.6 = 60%.
d) If the scale of 1: 150 indicated on the map corresponds to a scale of 1: 250 in reality,
it means that the scale of 1: 75 for the cross section is actually 1: 125
(because 250 : 150 = 1.666, and 75 · 1.666 = 125)
In reality, the 2.6 cm width of each wing corresponds to about 3.25 meters.