Eratosthenes and Indirect Measurement - Furman...

Eratosthenes and Indirect MeasurementMathematics 15: Lecture 5

Dan Sloughter

Furman University

September 20, 2006

Dan Sloughter (Furman University) Eratosthenes and Indirect Measurement September 20, 2006 1 / 15

Eratosthenes

I 275 B.C. - 194 B.C.

I Director of the library at AlexandriaI Among his many accomplishments:

I Maps of the known worldI An improved calendar (365 days to the year, plus an extra day every

fourth year)I An ingeniously simple way for estimating the circumference of the

earth.

Eratosthenes

I 275 B.C. - 194 B.C.

I Director of the library at Alexandria

I Among his many accomplishments:

earth.

Eratosthenes

I 275 B.C. - 194 B.C.

earth.

Eratosthenes

I 275 B.C. - 194 B.C.

I Maps of the known world

I An improved calendar (365 days to the year, plus an extra day everyfourth year)

I An ingeniously simple way for estimating the circumference of theearth.

Eratosthenes

I 275 B.C. - 194 B.C.

fourth year)

I An ingeniously simple way for estimating the circumference of theearth.

Eratosthenes

I 275 B.C. - 194 B.C.

earth.

Measuring the earth

I Eratosthenes learned that at noon on the summer solstice, a sun dialin Syene would not cast a shadow: the suns’s rays are perpendicularto the earth.

I At noon in Alexandria on the summer solstice, a sun dial would cast ashadow: the rays from the sun would make an angle of 1

50 of a circle,or 7.2◦, with the vertical.

I Moreover, Alexandria and Syene lie on the same line of longitude,with Alexandria 5000 stadia north of Syene.

Alexandria

7.25000 stadia

Measuring the earth

Alexandria

7.25000 stadia

Measuring the earth

Alexandria

7.25000 stadia

Measuring the earth (cont’d)

I Assuming that the light rays reaching Syene and Alexandria areessentially parallel, and using a basic result from geometry (a linecrossing two parallel lines makes equal angles), it follows that theangle formed by Alexandria, the center of the earth, and Syene is also150 of a circle.

I Hence, if C is the circumference of the earth,

I Thus C = 250, 000 stadia, between 23, 911 and 33, 381 miles.

Similar triangles

I Thales (624 B.C. - 547 B.C.) supposedly found the height of theGreat Pyramid as follows.

I First, he noticed that a stick 10 feet tall casts a shadow 16 feet longat the same time that the Great Pyramid casts a shadow 770 feetlong.

I He then reasoned that the triangles formed by the pyramid and itsshadow and the stick and its shadow were similar triangles (that is,the two triangles have equal angles).

I Fact from geomtry: the ratios of the sides of similar triangles areequal.

Similar triangles

Similar triangles (cont’d)

I Hence if h is the height of the pyramid, we have

and so

h =10× 770

16= 481.25 feet.

The ideas of trigonometry

I Trigonometry starts from two basic facts:

I If two triangles are similar, then the ratios of corresponding sides arethe same.

I If two right triangles have a pair of congruent acute angles, then thetriangles are similar.

I Trigonometry starts from two basic facts:I If two triangles are similar, then the ratios of corresponding sides are

the same.

I If two right triangles have a pair of congruent acute angles, then thetriangles are similar.

I Trigonometry starts from two basic facts:I If two triangles are similar, then the ratios of corresponding sides are

the same.I If two right triangles have a pair of congruent acute angles, then the

triangles are similar.

The ideas of trigonometry (cont’d)

I In particular, if 4ABC is a right triangle with right angle at C , thenwe define:

I Sine of ∠A = sin∠A =BC

I Cosine of ∠A = cos ∠A =AC

I Tangent of ∠A = tan ∠A =BC

I Hipparchus (180? B.C. - 125? B.C.), amongst others, realized thatthe values of these ratios for different angles could be compiled intables and used to compute distances which were otherwiseinaccessible.

Example

I Suppose that at a distance of 120 feet from a building we measurethe angle between ground level and the line of sight to the top of thebuilding, finding it to be 35◦.

I We then know that if h is the height of the building,

120= tan 35◦.

I Using a calculator (or Google), we find that tan 35◦ ≈ 0.7002, and so

h = 120× 0.7002 ≈ 84 feet.

Example

120= tan 35◦.

h = 120× 0.7002 ≈ 84 feet.

Example

120= tan 35◦.

h = 120× 0.7002 ≈ 84 feet.

ExampleI From the top of a 3 mile high mountain, we notice that the angle

between the vertical and our line of sight to the horizon (at sea level)is 87.77◦.

I If r is the radius of the earth, and recalling that a line tangent to acircle is perpendicular to a radius of the circle, we have

r + 3= sin 87.77◦.

ExampleI From the top of a 3 mile high mountain, we notice that the angle

between the vertical and our line of sight to the horizon (at sea level)is 87.77◦.

I If r is the radius of the earth, and recalling that a line tangent to acircle is perpendicular to a radius of the circle, we have

r + 3= sin 87.77◦.

Example (cont’d)

I Hence

r = (r + 3) sin 87.77◦ = r sin 87.77◦ + 3 sin 87.77◦,

and so

r =3 sin 87.77◦

1− sin 87.77◦.

I Using a calculator to find sin 87.77◦ ≈ 0.99924, we have

r ≈ (3)(0.99924)

1− 0.99924≈ 3944 miles.

Example (cont’d)

I Hence

r = (r + 3) sin 87.77◦ = r sin 87.77◦ + 3 sin 87.77◦,

and so

r =3 sin 87.77◦

1− sin 87.77◦.

I Using a calculator to find sin 87.77◦ ≈ 0.99924, we have

r ≈ (3)(0.99924)

1− 0.99924≈ 3944 miles.

Problems

1. A tree casts a 45 foot shadow at the same time that a 5 foot polecasts a 12 foot shadow. How tall is the tree?

2. At a distance of 100 feet from a building, it is found that the anglemade between the ground and the line of sight to the top of thebuilding is 50◦. How tall is the building?

3. If a person who is 6 feet tall stands at the sea shore and looks to thehorizon, how far away is the horizon? Assume that the radius of theearth is 4000 miles.

4. Standing on top of a 4 mile high mountain on the planet Rigel 7, Mr.Spock finds the angle between his line of sight to the horizon (whichis at sea level) and the vertical is 88◦. What is the radius of Rigel 7?

Problems (cont’d)

5. Standing on the surface of the earth, the angle between the line ofsight to the top of the moon and the line of sight to the bottom ofthe moon is 0.5◦. Given that the distance from the earth to the moonis 240, 000 miles, find the radius of the moon.

6. Twice a month the earth, moon, and sun form a right triangle withthe moon at the right angle. At this time the angle between the lineof sight to the moon and the line of sight to the sun can bemeasured, and is found to be 89.85◦. Given that the distance fromthe earth to the moon is 240, 000 miles, find the distance from theearth to the sun.

7. The angle in the previous problem is very hard to measure.Aristarchus estimated the angle to be 87◦. How would this changethe calculation of the distance from the earth to the sun?

Problems (cont’d)

8. Viewed from the earth, the sun and moon appear to be almost thesame size. Use this fact to find the radius of the sun, given that theradius of the moon is 1060 miles, the distance from the earth to themoon is 240, 000 miles, and the distance from the earth to the sun is92, 000, 000 miles.

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