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WHAT
I LE
ARNED FROM
CREATIN
G AN A
DVANCED
TRIG
CLA
SS: PART
2
DR
. K
AT
I E C
ER
RO
NE
TH
E U
NI V
ER
SI T
Y O
F A
KR
ON
CO
LLE
GE
OF
AP
PLI E
D S
CI E
NC
E A
ND
TE
CH
NO
LOG
Y
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BACKGROUNDTechnical College
Our programs
Accreditation
Professional Exams
Replaced Tech Calc II
Advanced Trig
Advanced Topics
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THE ADVANCED TRIG COURSE1. Circles and Circular Curves : Arcs and central angles;
Chords and segments, Secant and tangent lines,, Perpendicular bisectors; Lengths of tangent lines, chords, curves, external distances and middle ordinates; Circular curve computation
2. Parabolic Curves: Slope of a line (grade or gradient); Distance of a line; Points of vertical curvature, intersection, and tangency; Tangent elevations; Basic form of a parabola; Finding the external distance of a vertical curve
3. Spherical Trigonometry: Spherical triangles, Interior and dihedral angles; Sine formulas for spherical triangles; Cosine formulas for sides of spherical triangles; Cosine formulas for angles of spherical triangles; Applications of spherical triangles
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VERTIC
AL CURVES
AKA Par
abol
ic C
urves
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PARABOLIC CURVESGiven: focal length f
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PARABOLIC CURVES
• Point of Vertical Curvature (PVC): the beginning of the arc
• Point of Vertical Tangency (PVT):The end of the arc
• Point of Vertical Intersection (PVI): The point where the two tangents intersect
• Length of the Chord (L): The length from PVC to PVT
PVTPVC
PVI
L
𝑔1 𝑔2
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PARABOLIC CURVESGiven:
PVTPVC
PVI
L
𝑔1 𝑔2
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PARABOLIC CURVESGiven:
PVTPVC
PVI
L
𝑔1
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PARABOLIC CURVESGiven:
Let x = 0
PVTPVC
PVI
L
𝑔1
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PARABOLIC CURVESA +1.500% grade meets a -2.250% grade at station 36+50 (3650 ft) , elevation 452.00 ft. A vertical curve of length 600 ft. (6 stations) will be used.
PVTPVC
PVI = 452 ft.
L = 6
1.5 -2.25
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PARABOLIC CURVESTURNING POINT
PVTPVC
PVI = 452 ft.
L
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PARABOLIC CURVESTURNING POINTA +1.500% grade meets a -2.250% grade at station 36+50 (3650 ft) , elevation 452.00 ft. A vertical curve of length 600 ft. (6 stations) will be used.
PVTPVC
PVI = 452 ft.
L = 6
-1.5 2.25
OR
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HORIZONTA
L CURVES
AKA Circ
ular C
urves
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WHY CIRCLES INSTEAD OF PARABOLAS?• Passenger comfort
• Side friction • Safety
• Point-mass friction analysis works for passenger cars (but not tractor trailers)• Increased safety and comfort with super-elevation
A historical and literature review of horizontal curve design by Fitzpatrick, K and Kahl, K
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CIRCULAR CURVES
• Point of Curvature (PC): the beginning of the arc
• Point of Tangency (PT):The end of the arc
• Point of Intersection (PI): The point where the two tangents intersect
• Length of the long chord (C): The length from PC to PT
PTPC
PI
C
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CIRCULAR CURVES
• Tangent distance (T): The distance from PI to PC or from PI to PT
• Deflection Angle(Δ): The central angle of the angle at the Point of Intersection (PI)
PTPC
PI
C
T T
RR
• Length of the Curve (L): the arclength from PC to PT• Radius (R): Radius of the circle• Degree of a Curve (D): the central angle that subtends a 100
foot arc
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CIRCULAR CURVESGiven D and Δ, find R. PTPC
PI
C
T T
RR
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CIRCULAR CURVESLength of the Long Chord (C): The length from PC to PTGiven R and Δ
PTPC
Δ/2C/2
R
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CIRCULAR CURVESTangent distance (T): The distance from PI to PC or from PI to PTGiven R and Δ, find T.
PTPC
PI
T T
RR
Δ/2
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CIRCULAR CURVESGiven D and Δ, find Length of the Curve (L): the arclength from PC to PTPTPC
PI
C
T T
RR
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CIRCULAR CURVES• External Distance (E): The
distance from the Point of Intersection to the middle of the curve
• Middle Ordinate (M): the length of the ordinate from the middle of the long chord to the middle of the arc
PTPC
PI
C
T T
RR
E
M
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CIRCULAR CURVESGiven R and Δ, find E. PTPC
PI
C/2
T T
RR
E
Δ/2
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CIRCULAR CURVESGiven D and Δ
PTPC
PI
C/2
T T
RR
M
Δ/2𝑎
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…AND N
OW F
OR SOME
SPHERIC
AL TRIG
Attribution: Peter Mercator
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MODERN APPLICATIONS
• Navigation
• Astronomy
• Geodesy
• GPS
• Satellite communication
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I GOT TO READ SOME RATHER OLD BOOKS
• Spherical Trigonometry with Naval and Military Applications (1942) by Kells, Kern and Bland
• Trigonometry Refresher (1946) by Klaf
• Sphere, Spheroid and Projections for Surveyors (1980) by Jackson
… And a little more recent one
• Heavenly Mathematics (2012) by Van Brummelen
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A LITTLE HISTORY
• Ancient Greece• Menelaus of Alexandria (70 – 140 CE)
• Ancient Persia• Abū Sahl al-Qūhī (10th century)
"Gravure originale du compas parfait par Abū Sahl al-Qūhī" (Engraving of al-Quhi's perfect compass to draw conic sections) by Abū Sahl al-Qūhī - Persian
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JOHN NAPIER (1550 – 1617)
• Mirifici Logarithmorum Canonis Descriptio (Description of the Wonderful Rule of Logarithms) in 1614
• Discussed logarithms
• Established Napier’s Rules of Circular Parts
1) Sine of the middle angle = Product of the tangents of the adjacent angles
2) Sine of an angle = Product of the cosines of the opposite angles
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THE MEDIEVAL METHOD AND PRAYING TO MECCA
The Law of Sines
OR
A
C
B
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NAVAL APPLICATIONS OD THE 19TH CENTURY
The Law of Cosines for Sides
The Law of Cosines for Angles A
C
B
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.
b = 90˚ - 41.4177˚ = 48.5823˚
c = 90˚ - 29.9933˚ = 60.0067˚
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.
A = 90.2581° - 81.8497° = 8.4084˚
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.b = 48.5823˚c = 60.0067˚A = 8.4084˚
The Law of Cosines for Sides
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.
The Law of Cosines for Sides
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.b = 48.5823˚c = 60.0067˚A = 8.4084˚a = 13.2934˚
The Law of Sines
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.b = 48.5823˚c = 60.0067˚A = 8.4084˚a = 13.2934˚
The Law of Sines
C = 33.4208˚
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DISTANCE AND BEARING
Suppose you want to find the distance and bearing from Cleveland Hopkins International Airport (41.4117° N, 81.8497° W) and Louis Armstrong New Orleans International Airport (29.9933° N, 90.2581° W). Find the distance and bearing between these two airports.b = 48.5823˚c = 60.0067˚A = 8.4084˚a = 13.2934˚
The Law of Sines
B = 28.4832˚
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VERIFICATION
a = 13.2934˚ = 919.53 miles = 1479.84 km
C = 33.4208˚ + 180˚ = 213.4208˚ SW
B = 28.4832˚+180˚ = 208.4832˚ NE
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INCREASED TEST SCORES
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KATIE
CER
RONE
The
Univer
sity
of A
kron
Colle
ge of
Applie
d Sci
ence
and T
echnol
ogy