what i learned from creating an advanced trig class dr. katie cerrone the university of akron...
TRANSCRIPT
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WHAT
I LE
ARNED FROM
CREATIN
G AN A
DVANCED
TRIG
CLA
SS
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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PARABOLIC CURVESGiven: focal length f
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PARABOLIC CURVES
• Point of Curvature (PC): the beginning of the arc
• Point of Intersection (PI): The point where the two tangents intersect
• Point of Tangency (PT):The end of the arc
• Length of the Chord (L): The length from PC to PT
PTPC
PI
L
𝑔1 𝑔2
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PARABOLIC CURVESGiven:
PTPC
PI
L
𝑔1 𝑔2
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PARABOLIC CURVESGiven:
PTPC
PI
L
𝑔1
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PARABOLIC CURVESGiven:
Let x = 0
PTPC
PI
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.
PTPC
PI = 452 ft.
L = 6
-1.5 2.25
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PARABOLIC CURVESTURNING POINT
PTPC
PI = 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.
PTPC
PI = 452 ft.
L = 6
-1.5 2.25
OR
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CIRCULAR CURVES
• Point of Curvature (PC): the beginning of the arc
• Point of Intersection (PI): The point where the two tangents intersect
• Point of Tangency (PT):The end of the arc
• Length of the long chord (L): The length from PC to PT
PTPC
PI
L
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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
L
T T
RR
• Length of the Curve (C): 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
L
T T
RR
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CIRCULAR CURVESLength of the Curve (L): The arclength from PC to PTGiven R and Δ
PTPC
Δ/2L/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 C. PTPC
PI
L
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
L
T T
RR
E
M
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CIRCULAR CURVESGiven R and Δ, find E. PTPC
PI
L/2
T T
RR
E
Δ/2
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CIRCULAR CURVESGiven D and Δ
PTPC
PI
L/2
T T
RR
M
Δ/2𝑎
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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