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