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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
BACKGROUNDTechnical College
Our programs
Accreditation
Professional Exams
Replaced Tech Calc II
Advanced Trig
Advanced Topics
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
VERTIC
AL CURVES
AKA Par
abol
ic C
urves
PARABOLIC CURVESGiven: focal length f
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
PARABOLIC CURVESGiven:
PVTPVC
PVI
L
𝑔1 𝑔2
PARABOLIC CURVESGiven:
PVTPVC
PVI
L
𝑔1
PARABOLIC CURVESGiven:
Let x = 0
PVTPVC
PVI
L
𝑔1
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
PARABOLIC CURVESTURNING POINT
PVTPVC
PVI = 452 ft.
L
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
HORIZONTA
L CURVES
AKA Circ
ular C
urves
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
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
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
CIRCULAR CURVESGiven D and Δ, find R. PTPC
PI
C
T T
RR
CIRCULAR CURVESLength of the Long Chord (C): The length from PC to PTGiven R and Δ
PTPC
Δ/2C/2
R
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
CIRCULAR CURVESGiven D and Δ, find Length of the Curve (L): the arclength from PC to PTPTPC
PI
C
T T
RR
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
CIRCULAR CURVESGiven R and Δ, find E. PTPC
PI
C/2
T T
RR
E
Δ/2
CIRCULAR CURVESGiven D and Δ
PTPC
PI
C/2
T T
RR
M
Δ/2𝑎
…AND N
OW F
OR SOME
SPHERIC
AL TRIG
Attribution: Peter Mercator
MODERN APPLICATIONS
• Navigation
• Astronomy
• Geodesy
• GPS
• Satellite communication
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
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
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
THE MEDIEVAL METHOD AND PRAYING TO MECCA
The Law of Sines
OR
A
C
B
NAVAL APPLICATIONS OD THE 19TH CENTURY
The Law of Cosines for Sides
The Law of Cosines for Angles A
C
B
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.
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˚
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˚
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
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
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
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˚
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˚
VERIFICATION
a = 13.2934˚ = 919.53 miles = 1479.84 km
C = 33.4208˚ + 180˚ = 213.4208˚ SW
B = 28.4832˚+180˚ = 208.4832˚ NE
INCREASED TEST SCORES
KATIE
CER
RONE
kc24@uakron.edu
The
Univer
sity
of A
kron
Colle
ge of
Applie
d Sci
ence
and T
echnol
ogy
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