spherical trigonometry
DESCRIPTION
Presents mathematics and history of spherical trigonometry. Since most of the figures are not uploaded I recommend to see this presentation on my website at http://www.solohermelin.com.at Math folder. Please send comments and suggestions for improvements to [email protected]. Thanks.TRANSCRIPT
1
Spherical Trigonometry
SOLO HERMELIN
Updated 12.11.12http://www.solohermelin.com
2
SOLO
TABLE OF CONTENT Spherical Trigonometry
History
Sine and Cosine Laws in Spherical Triangles
Half Angles FormulasDelambre-Gauss Equations
Napier’s Equations
References
Applications: Flight on Earth Great Circles
3
SOLO
A1
B1
C1
B
a
bc
C
A
History Spherical triangles were studied by early Greek mathematicians such as Menelaus of Alexandria, who wrote a book on spherical triangles called “Sphaerica “and developed Menelaus' Theorem. E. S. Kennedy, however, points out that while it was possible in ancient mathematics to compute the magnitudes of a spherical figure, in principle, by use of the table of chords and Menelaus' theorem, the application of the theorem to spherical problems was very difficult in practice
Further advances were made in the Islamic world. In order to observe holy days of the Islamic calendar for which timings were determined by phases of the moon, astronomers initially used Menalaus' method to calculate the place of the moon and stars, though this method proved to be clumsy and difficult. It involved setting up two intersecting right triangles; by applying Menelaus' theorem it was possible to solve one of the six sides, but only if the other five sides were known. To tell the time from the sun's altitude, for instance, repeated applications of Menelaus' theorem were required. For medieval Islamic astronomers, there was an obvious challenge to find a simpler trigonometric method.
Spherical Trigonometry
4
SOLO
History (continue - 1)
In the early 9th century, the Persian mathematician Muhammad ibn Mūsā al-Khwārizmi was a pioneer in spherical trigonometry and wrote a treatise on the subject. In the 10th century, another Persian mathematician, Abū al-Wafā' al-Būzjāni, established the angle addition formulas, e.g. for sin(a + b), and discovered the sine formula for spherical trigonometry
Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī[
(c. 780 – c. 850)
Abū al-Wafāʾ, Muḥammad ibn Muḥammad ibn Yaḥyā ibn Ismāʿīl
ibn al-ʿAbbās al-Būzjānī (940 –998)
Here, a, b, and c are the angles at the centre of the sphere subtended by the three sides of the triangle, and α, β, and γ are the angles between the sides, where angle α is opposite the side which subtends angle a, and so forth
A1
B1
C1
B
a
bc
C
A
Spherical Trigonometry
5
SOLO
History (continue - 2)
A1
B1
C1
B
a
bc
C
A
Al-Jayyani (989–1079), an Arabic mathematician in the Islamic Iberian Peninsula, wrote what some consider the first treatise on spherical trigonometry, circa 1060, entitled The book of unknown arcs of a sphere,[6] in which spherical trigonometry was brought into its modern form. Al-Jayyani's book "contains formulae for right-angle triangles, the general law of sines and the solution of a spherical triangle by means of the polar triangle". This treatise later had a "strong influence on European mathematics", and his "definition of ratios as numbers" and "method of solving a spherical triangle when all sides are unknown" are likely to have influenced Regiomontanus.
In the 13th century, Persian mathematician Nasīr al-Dīn al-Tūsī was the first to treat trigonometry as a mathematical discipline independent from astronomy, and he further developed spherical trigonometry, bringing it to its present form.[7] He listed the six distinct cases of a right-angled triangle in spherical trigonometry. In his On the Sector Figure, he also stated the law of sines for plane and spherical triangles, and discovered the law of tangents for spherical triangles
Khawaja Muhammad ibn Muhammad
ibn Hasan Tūsī (1201 -1274 )
Abū ʿAbd Allāh Muḥammad ibn Muʿādh al-Jayyānī[1]
(989– 1079
Spherical Trigonometry
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SOLO
Assume three points on a unit radius sphere, defined by the vectors
CBA 1,1,1
A1
B1
C1
B
a
bc
C
A
Figure 1: Spherical Triangle
The three great circles passing trough those three points define a spherical triangle with
- three spherical triangle verticesCBA ,,
- three spherical triangle side anglescba ˆ,ˆˆ
- three spherical triangle angles defined by the angles between the tangents to the great circles at the vertices.
ˆ,ˆˆ
The extreme cases are when the three vertices are on the same great circle:
• When exists one diameter on the great circle such that all three vertices are on the same side, then
180ˆ&0ˆˆ • When doesn’t exists one diameter on the great circle such that all three vertices are on the same side
180ˆˆˆ Therefore 570ˆˆˆ180
Spherical Trigonometry
Sine and Cosine Laws in Spherical Triangles
7
SOLO
A1
B1
C1
B
a
bc
C
A
c
BAˆsin
11
b
CAˆsin
11
Those angles can be defined using unit vectors as follows
CBA 1,1,1
180ˆ,ˆ,ˆ0
11ˆsin&11ˆcos
11ˆsin&11ˆcos
11ˆsin&11ˆcos
cba
BAcBAc
ACbACb
CBaCBa
0ˆsin&0ˆsin&0ˆsin
ˆsinˆsin
1111ˆsin&
ˆsinˆsin
1111ˆcos
ˆsinˆsin
1111ˆsin&
ˆsinˆsin
1111ˆcos
ˆsinˆsin
1111ˆsin&
ˆsinˆsin
1111ˆcos
cba
ab
BCAC
ab
BCAC
ca
ABCB
ca
ABCB
bc
CABA
bc
CABA
180ˆ,ˆ,ˆ0
1
2
Spherical TrigonometrySine and Cosine Laws in Spherical Triangles
8
SOLO
A1
B1
C1
B
a
bc
C
A
We will use the following vector identities
wvuvwuwvu
vuwuwvwvu
to perform the following computations
3
4
abcBCACBACCBCACBCACab
cabABCBACBBABCBABCBca
bcaCABACBAACABACABAbc
ˆcosˆcosˆcos1111111111111111ˆsinˆsinˆcos
ˆcosˆcosˆcos1111111111111111ˆsinˆsinˆcos
ˆcosˆcosˆcos1111111111111111ˆsinˆsinˆcos
1
1
1
5
We obtained the Laws of Cosines for Spherical Triangle Sides
6
Spherical TrigonometrySine and Cosine Laws in Spherical Triangles
ˆcosˆsinˆsinˆcosˆcosˆcos
ˆcosˆsinˆsinˆcosˆcosˆcos
ˆcosˆsinˆsinˆcosˆcosˆcos
babac
cacab
cbcba
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SOLO
A1
B1
C1
B
a
bc
C
A
ab
abc
ca
cab
bc
bca
ˆsinˆsin
ˆcosˆcosˆcosˆcos
ˆsinˆsin
ˆcosˆcosˆcosˆcos
ˆsinˆsin
ˆcosˆcosˆcosˆcos
6
Let compute
ab
abccba
ab
abcabcab
ab
abcabc
ab
abc
ˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcosˆcos1ˆcos1
ˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆsinˆsin
ˆcosˆcosˆcos1ˆcos1ˆsin
22
222
22
22222
22
2222
22
from which
abccbaab ˆcosˆcosˆcos2ˆcosˆcosˆcos1ˆsinˆsinˆsin 222 7
The plus sign is used since, by definition, 0ˆsin,ˆsin,ˆsin ab
Spherical TrigonometrySine and Cosine Laws in Spherical Triangles
10
SOLO
A1
B1
C1
B
a
bc
C
A
ˆsin,ˆsin,ˆsinFrom the definitions of, given by (1) and (2), we have
BACACACCBACBCACab
ACBABCBBACBABCBca
CBACABAACBACABAbc
111111111111111ˆsinˆsinˆsin
111111111111111ˆsinˆsinˆsin
111111111111111ˆsinˆsinˆsin
0
0
0
But since
BACACBCBA 111111111
we have
abccba
abcabc
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsin
222
by dividing this equation by we obtain the Law of Sines for Spherical Triangle
bca ˆsinˆsinˆsin
cba
abccba
cba ˆsinˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆsin
ˆsinˆsin
ˆsinˆsin
ˆsin 222
8
9
Spherical TrigonometrySine and Cosine Laws in Spherical Triangles
11
SOLO
A1
B1
C1
B
a
bc
C
A
10
Let compute the expression
bca
bcacbaa
bca
bcacbaa
bca
bcacaabbcbcaaa
ab
abc
ca
cab
bc
bca
ˆsinˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcos1ˆcos
ˆsinˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆsinˆcos
ˆsinˆsinˆsin
ˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆcosˆsinˆcosˆsin
ˆsinˆsin
ˆcosˆcosˆcosˆsinˆsin
ˆcosˆcosˆcosˆsinˆsin
ˆcosˆcosˆcosˆcosˆcosˆcos
2
222
2
222
2
22222
But from (8)
abcaabccba ˆsinˆsinˆsinˆsinˆsinˆsinˆcosˆcosˆcos2ˆcosˆcosˆcos1 222
Substituting this in (10) gives ˆsinˆsinˆcosˆcosˆcosˆcos a
Finally we can write the Laws of Cosines for Spherical Triangle Angles.
11
Spherical TrigonometrySine and Cosine Laws in Spherical Triangles
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c
b
a
ˆcosˆsinˆsinˆcosˆcosˆcos
ˆcosˆsinˆsinˆcosˆcosˆcos
ˆcosˆsinˆsinˆcosˆcosˆcos
12
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas
From (6)
bc
bcaˆsinˆsin
ˆcosˆcosˆcosˆcos
Using the half angle formulas from trigonometry, we have
bc
bcba
ccba
bc
cbacba
bc
acb
bc
bcabc
ˆsinˆsin
ˆ2
ˆˆˆsinˆ
2
ˆˆˆsin
ˆsinˆsin2
2
ˆˆˆsin
2
ˆˆˆsin2
ˆsinˆsin2
ˆcosˆˆcos
ˆsinˆsin2
ˆcosˆcosˆcosˆsinˆsin
2
ˆcos1
2
ˆsin 2
If we define2
ˆˆˆ:ˆ
cbap
we obtain bc
bpcpˆsinˆsin
ˆˆsinˆˆsin
2
ˆsin
12
Spherical Trigonometry
13
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas (continue – 1)
13
In the same way
bc
app
bc
acbcba
bc
cba
bc
bcabc
ˆsinˆsin
ˆˆsinˆsin
ˆsinˆsin2
2
ˆˆˆsin
2
ˆˆˆsin2
ˆsinˆsin2
ˆˆcosˆcos
ˆsinˆsin2
ˆcosˆcosˆcosˆsinˆsin
2
ˆcos1
2
ˆcos2
bc
appˆsinˆsin
ˆˆsinˆsin
2
ˆcos
or
From (12) and (13)
app
bpcpˆˆsinˆsin
ˆˆsinˆˆsin
2
ˆtan
14
2
ˆˆˆ:ˆ
cbap
Spherical Trigonometry
14
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas (continue – 2)and
bc
cpbpappˆsinˆsin
ˆˆsinˆˆsinˆˆsinˆsin
2
ˆcos
2
ˆsin2ˆsin
2
ˆˆˆ:ˆ
cbap
from this equation and (8) we obtain
abccba
cpbpappbc
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆˆsinˆˆsinˆˆsinˆsinˆsinˆsinˆsin
222
15
Spherical Trigonometry
15
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas (continue – 3)In the same way from (11)
16
ˆsinˆsin
ˆcosˆcosˆcosˆcos
a
we can write
ˆsinˆsin
ˆ2
ˆˆˆcos
2
ˆˆˆcos
ˆsinˆsin2
2
ˆˆˆcos
2
ˆˆˆcos2
ˆsinˆsin2
ˆˆcosˆcosˆsinˆsin2
ˆcosˆcosˆcosˆsinˆsin
2
ˆcos1
2
ˆsin 2
aa
If we define2
ˆˆˆ:ˆ
we obtain
ˆsinˆsin
ˆˆcosˆcos
2
ˆsin
a17
Spherical Trigonometry
16
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas (continue – 4)Also
18
ˆsinˆsin
ˆˆcosˆˆcosˆsinˆsin2
2
ˆˆˆcos
2
ˆˆˆcos2
ˆsinˆsin2
ˆˆcosˆcosˆsinˆsin2
ˆcosˆcosˆcosˆsinˆsin
2
ˆcos1
2
ˆcos2
aa
from which
ˆsinˆsin
ˆˆcosˆˆcos
2
ˆcos
a
From (17) and (18)
ˆˆcosˆˆcos
ˆˆcosˆcos
2
ˆtan
a
19
Spherical Trigonometry
17
SOLO
Half Angles Formulas (continue – 5)
Summarize
Spherical Trigonometry
ˆsinˆsin
ˆˆcosˆˆcos
2
ˆcos
a
ˆˆcosˆˆcos
ˆˆcosˆcos
2
ˆtan
a
ˆsinˆsin
ˆˆcosˆcos
2
ˆsin
a
bc
appˆsinˆsin
ˆˆsinˆsin
2
ˆcos
app
bpcpˆˆsinˆsin
ˆˆsinˆˆsin
2
ˆtan
bc
bpcpˆsinˆsin
ˆˆsinˆˆsin
2
ˆsin
ˆsinˆsin
ˆˆcosˆˆcos
2
ˆcos
b
ˆˆcosˆˆcos
ˆˆcosˆcos
2
ˆtan
b
ˆsinˆsin
ˆˆcosˆcos
2
ˆsin
b
cpp
bpapˆˆsinˆsin
ˆˆsinˆˆsin
2
ˆtan
ˆsinˆsin
ˆˆcosˆˆcos
2
ˆcos
c
ˆˆcosˆˆcos
ˆˆcosˆcos
2
ˆtan
c
ˆsinˆsin
ˆˆcosˆcos
2
ˆsin
c
bpp
apcpˆˆsinˆsin
ˆˆsinˆˆsin
2
ˆtan
ac
bppˆsinˆsin
ˆˆsinˆsin
2
ˆcos
ac
apcpˆsinˆsin
ˆˆsinˆˆsin
2
ˆsin
ba
cppˆsinˆsin
ˆˆsinˆsin
2
ˆcos
ba
bpapˆsinˆsin
ˆˆsinˆˆsin
2
ˆsin
2
ˆˆˆ:ˆ
2
ˆˆˆ:ˆ
cbap
18
SOLO
A1
B1
C1
B
a
bc
C
A
Half Angles Formulas (continue – 6)And
20
ˆsinˆsin
ˆˆcosˆˆcosˆˆcosˆcos
2
ˆcos
2
ˆsin2ˆsin
aaa
ˆˆcosˆˆcosˆˆcosˆcosˆsinˆsinˆsin a
or
By cyclic substitution of the angles we obtain
ˆˆcosˆˆcosˆˆcosˆcos
ˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsinˆsin
cba
ˆsinˆsinˆsinLet divide by this equation to recover the Law of Sines (9)
cba
cpbpapp
cba
abccba
cba
ˆsinˆsinˆsin
ˆˆsinˆˆsinˆˆsinˆsin
ˆsinˆsinˆsin
ˆcosˆcosˆcos2ˆcosˆcosˆcos1
ˆˆcosˆˆcosˆˆcosˆcos
ˆsinˆsinˆsinˆsin
ˆsinˆsin
ˆsinˆsin
ˆsin
152229
21
22
Spherical Trigonometry
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19
SOLO
A1
B1
C1
B
a
bc
C
A
Delambre-Gauss Equations
Jean Baptiste Joseph chevalier Delambre
(1749 –1822)
Using (12) and (13) we can write
2
ˆsin
2
ˆcos
2
ˆcos
2
ˆsin
2
ˆˆsin
ca
cpap
cb
app
ca
bpp
bc
bpcpˆsinˆsin
ˆˆsinˆˆsin
ˆsinˆsin
ˆˆsinˆsinˆsinˆsin
ˆˆsinˆsinˆsinˆsin
ˆˆsinˆˆsin
2
ˆˆsin
2
ˆˆˆcos2
2
ˆˆcos
2
ˆˆˆsin2
2
ˆcos
ˆsin
1ˆˆsinˆˆsin
ˆsinˆsin
ˆˆsinˆsinˆsin
1
babap
babap
capbp
ba
cpp
c
2
ˆcos
2
ˆsin
2
ˆˆsin
2
ˆcos
2
ˆcos
2
ˆˆcos
2
ˆˆsin
2
ˆcos2
2
ˆˆcos
2
ˆsin2
2
ˆcos
2
ˆcos
2
ˆsin2
1
c
ba
c
ba
bac
bac
cc
Spherical Trigonometry
Carl Friedrich Gauss
(1777 – 1855)
20
SOLO
A1
B1
C1
B
a
bc
C
A
Delambre-Gauss Equations (continue – 1)
Jean Baptiste Joseph chevalier Delambre
(1749 –1822)
Therefore
2
ˆcos
2
ˆˆcos
2
ˆcos
2
ˆˆsin
bac
23
2
ˆcos
2
ˆˆsin
2
ˆsin
2
ˆˆsin
bac
24
Spherical Trigonometry
Carl Friedrich Gauss
(1777 – 1855)
The Delambre-Gauss Formulas were first discovered by Delambre in 1807 (published in 1809) and rediscovered independently by Gauss.
Delambre: Director of Paris Observatory (1804 – 1822)Gauss : Director of Gőttingen Observatory (1807 – 1855)
21
SOLO
A1
B1
C1
B
a
bc
C
A
Delambre-Gauss Equations (continue – 2)
Jean Baptiste Joseph chevalier Delambre
(1749 –1822)
25
Also
2
ˆsin
2
ˆsin
2
ˆcos
2
ˆcos
2
ˆˆcos
ca
cpap
bc
bpcp
ca
bpp
cb
appˆsinˆsin
ˆˆsinˆˆsinˆsinˆsin
ˆˆsinˆˆsinˆsinˆsin
ˆˆsinˆsin
ˆsinˆsin
ˆˆsinˆsin
2
ˆˆsin
2
ˆcos2
2
ˆˆcos
2
ˆsin2
2
ˆsin
ˆsin
1ˆˆsinˆsin
ˆsinˆsin
ˆˆsinˆˆsinˆsin
1
cp
c
cp
c
ccpp
ba
apbp
c
2
ˆsin
2
ˆsin
2
ˆˆsin
2
ˆsin
2
ˆcos
2
ˆˆcos
2
ˆˆsin
2
ˆcos2
2
ˆˆcos
2
ˆsin2
2
ˆsin
2
ˆcos
2
ˆsin2
1
c
ba
c
ba
bac
bac
cc
Therefore
2
ˆsin
2
ˆˆcos
2
ˆcos
2
ˆˆcos
bac
2
ˆsin
2
ˆˆsin
2
ˆsin
2
ˆˆcos
bac
26
Spherical Trigonometry
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Carl Friedrich Gauss
(1777 – 1855)
22
SOLO
A1
B1
C1
B
a
bc
C
A
Napier’s Equations [4]
John Napier of Merchiston (1550 –1617)
Divide (23) by (25)2
ˆcot
2
ˆˆcos
2
ˆˆcos
2
ˆˆtan
ba
ba
to obtain
28
Divide (24) by (26)2
ˆcot
2
ˆˆsin
2
ˆˆsin
2
ˆˆtan
ba
ba
to obtain
27
Spherical Trigonometry
2
ˆˆsin
2
ˆˆsin
2
ˆcot
2
ˆˆtan
ba
ba
2
ˆˆcos
2
ˆˆcos
2
ˆcot
2
ˆˆtan
ba
ba
23
SOLO
A1
B1
C1
B
a
bc
C
A
Napier’s Equations (continue)
Divide (24) by (23)
to obtain
29
Divide (26) by (25)
to obtain
30
2
ˆˆtan
2
ˆtan
2
ˆˆsin
2
ˆˆsin bac
2
ˆˆtan
2
ˆtan
2
ˆˆcos
2
ˆˆcos bac
Spherical Trigonometry
John Napier of Merchiston (1550 –1617)
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2
ˆˆsin
2
ˆˆsin
2
ˆtan
2
ˆˆtan
c
ba
2
ˆˆcos
2
ˆˆcos
2
ˆtan
2
ˆˆtan
c
ba
24
SOLO
Napier’s Rules for Right Angles Spherical Triangles
Let γ = 90 T , i,e, a Right Angle Spherical Triangle.Napier arranged the quantities as in the Right Circle . Any of t6he parts of this Circle is called a Middle Part, the two neighboring parts are called Adjacent Parts, and the two remaining parts are Opposite Parts.
ˆcosˆcosˆtanˆtanˆsin
ˆcosˆcosˆtanˆtanˆsin
ˆcosˆcosˆtanˆtanˆsin
ˆcosˆcosˆtanˆtanˆsin
ˆcosˆcosˆtanˆtanˆsin
coaccobco
bacococco
bcoccoaco
coccocoab
ccococoba
Spherical Trigonometry
I
Ecuator
1R
2R
Ex
Ey
Ez
1
2
TrajectoryGreat Circcle
1 2
A
B'B
1R
2R1
2
O
A
B
Ca
b
c
Earth Center
North Pole
90ˆ
b a
coco
cco ˆ
co = complement
ccco
co
co
ˆ90ˆ
ˆ90ˆ
ˆ90ˆ
Napier’s Rules for Right Angles Spherical Triangles are
• The sine of any Middle Part equals the product of the tangents of the Adjacent Parts.
• The sine of any Middle Part equals the product of the cosines of the Opposite Parts.
25
SOLO
Napier’s Rules for Right Angles Spherical Triangles
Let γ = 90 T , i,e, a Right Angle Spherical Triangle.Napier arranged the quantities as in the Right Circle . Any of t6he parts of this Circle is called a Middle Part, the two neighboring parts are called Adjacent Parts, and the two remaining parts are Opposite Parts.
ˆsinˆcosˆcotˆtanˆcos
ˆcosˆcosˆcotˆcotˆcos
ˆcosˆsinˆcotˆtanˆcos
ˆsinˆsinˆcotˆtanˆsin
ˆsinˆsinˆcotˆtanˆsin
acb
bac
bca
cab
cba
Spherical Trigonometry
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I
Ecuator
1R
2R
Ex
Ey
Ez
1
2
TrajectoryGreat Circcle
1 2
A
B'B
1R
2R1
2
O
A
B
Ca
b
c
Earth Center
North Pole
90ˆ
b a
coco
cco ˆ
co = complement
ccco
co
co
ˆ90ˆ
ˆ90ˆ
ˆ90ˆ
Napier’s Rules for Right Angles Spherical Triangles are
• The sine of any Middle Part equals the product of the tangents of the Adjacent Parts.
• The sine of any Middle Part equals the product of the cosines of the Opposite Parts.
The final result is
26
Navigation SOLO
I
Ecuator
1R
2R11,
Ex
Ey
Ez
1
222 ,
,
12 TrajectoryGreat Circcle
1 2
0
Flight on Earth Great Circles
The Shortest Flight Path between two points 1 and 2 on the Earth is on the Great Circles (centered at Earth Center) passing through those points.
The Great Circle Distance between two points 1 and 2 is ρ.The average Radius on the Great Circle is a = (R1+R2)/2
a kmNmNma 852.11deg/76.60/
1
2111 ,, R
222 ,, RR – radiusϕ - Latitudeλ - Longitude
27
Navigation SOLO
I
Ecuator
1R
2R11,
Ex
Ey
Ez
1
222 ,
,
12 TrajectoryGreat Circcle
1 2
0
1
2
Flight on Earth Great Circles
The Great Circle Distance between two points 1 and 2 is ρ.
a
1
2111 ,, R
222 ,, RR – radiusϕ - Latitudeλ - Longitude
212121 cos90sin90sin90cos90cos
/coscos
a
From the Law of Cosines for Spherical Triangles
or
212121 coscoscossinsin/cos a
2121211 coscoscossinsincos a
The Initial Heading Angle ψ0 can be obtained using theLaw of Cosines for Spherical Triangles as follows
a
a
/sincos
/cossinsincos
1
120
2
222
22221
coscoscossinsin1cos
coscoscossinsinsinsincos
The Heading Angle ψ from the Present Position (R,ϕ,λ) to Destination Point (R2,ϕ2,λ2)
28
Navigation SOLO
I
Ecuator
1R
2R11,
Ex
Ey
Ez
1
222 ,
,
12 TrajectoryGreat Circcle
1 2
0
1
2
Flight on Earth Great Circles
The Distance on the Great Circle between two points 1 and 2 is ρ.
1
2111 ,, R
222 ,, RR – radiusϕ - Latitudeλ - Longitude
The Time required to travel along the Great Circle between points 1 and 2 is given by
22
2121211 coscoscossinsincos
yxHoriz
HorizHoriz
VVV
V
a
Vt
2121211 coscoscossinsincos a
29
Navigation SOLO
I
Ecuator
1R
2R
Ex
Ey
Ez
1
2
TrajectoryGreat Circcle
1 2
1R
2R1
2
O
A
B
Ca
b
c
c
Earth Center
North Pole
A
B'B90
P
Flight on Earth Great Circles
1
2111 ,, R
222 ,, R
If the Aircraft flies with an Heading Error Δψ we want to calculate the Down Range Error Xd and Cross Range Error Yd, in the Spherical Triangle APB.
R – radiusϕ - Latitudeλ - Longitude
Using the Law of Cosines for Spherical Triangle APB we have
aaYd /sin
90sin
/sin
sin
2/sin/sin
/cos/cos/cos0ˆcos 21
90ˆ RRa
aYaX
aYaXaP
dd
ddP
Using the Law of Sines for Spherical Triangle APB we have
aY
aaX
dd /cos
/coscos 1
sin/sinsin 1 aaYd
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References
http://en.wikipedia.org/wiki/
[1] Lass H, “Vector and Tensor Analysis”, McGraw-Hill, 1950, pp.25-27
[2] Wertz, J. editor, “Spacecraft Attitude Determination andControl”,
Appendix A, D.Reidel Publishing Company, 1985[3] “ASM Handbook of Engineering Mathematics”, American Society for
Metals, 1983, pp.86-89
[4] Tuma, J.J., “Engineering Mathematics Handbook”, 3 Edition, McGraw-Hill, pp.34-35
Spherical Trigonometry
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http://mathworld.wolfram.com/SphericalTrigonometry.html
April 13, 2023 31
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TechnionIsraeli Institute of Technology
1964 – 1968 BSc EE1968 – 1971 MSc EE
Israeli Air Force1970 – 1974
RAFAELIsraeli Armament Development Authority
1974 – 2013
Stanford University1983 – 1986 PhD AA