1. WHAT I LEARNED FROM CREATING AN ADVANCED TRIG CLASS: PART 2 DR. KATIE CERRONE THE UNIVERSITY OF AKRON COLLEGE OF APPLIED SCIENCE AND TECHNOLOGY

2. BACKGROUND Technical College Our programs Accreditation Professional Exams Replaced Tech Calc II Advanced Trig Advanced Topics

3. THE ADVANCED TRIG COURSE 1. 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

4. VERTICAL CURVES AKA Parabolic Curves

5. PARABOLIC CURVES

6. 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 PVT PVC PVI L

7. PARABOLIC CURVES PVT PVC PVI L

8. PARABOLIC CURVES PVT PVC PVI L

9. PARABOLIC CURVES PVT PVC PVI L

10. PARABOLIC CURVES PVT PVC PVI = 452 ft. L = 6 1.5-2.25

11. PARABOLIC CURVES TURNING POINT PVT PVC PVI = 452 ft. L

12. PARABOLIC CURVES TURNING POINT PVT PVC PVI = 452 ft. L = 6 -1.52.25

13. HORIZONTAL CURVES AKA Circular Curves

14. 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

15. 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 PT PC PI C

16. 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) PT PC 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

17. CIRCULAR CURVES PT PC PI C T T RR

18. CIRCULAR CURVES PT PC Δ/2 C/2 R

19. CIRCULAR CURVES PT PC PI T T RR Δ/2

20. CIRCULAR CURVES PT PC PI C T T RR

21. 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 PT PC PI C T T RR E M

22. CIRCULAR CURVES PT PC PI C/2 T T RR E Δ/2

23. CIRCULAR CURVES PT PC PI C/2 T T RR M Δ/2

24. …AND NOW FOR SOME SPHERICAL TRIG Attribution: Peter Mercator

25. MODERN APPLICATIONS Navigation Astronomy Geodesy GPS Satellite communication

26. 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

27. A LITTLE HISTORY Ancient Greece Menelaus of Alexandria (70 – 140 CE) Ancient Persia Ab ū Sahl al-Q ū h ī (10 th 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

28. 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

29. THE MEDIEVAL METHOD AND PRAYING TO MECCA A C B

30. NAVAL APPLICATIONS OD THE 19 TH CENTURY A C B

31. 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.

32. 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˚

33. 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˚

34. 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˚

35. 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.

36. 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˚

37. 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˚

38. 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˚

39. VERIFICATION a = 13.2934˚ = 919.53 miles = 1479.84 km C = 33.4208˚ + 180˚ = 213.4208˚ SW B = 28.4832˚+180˚ = 208.4832˚ NE

40. INCREASED TEST SCORES

41. KATIE CERRONE kc24@uakron.edu The University of Akron College of Applied Science and Technology