Applications of Parabolas:

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Transcript Applications of Parabolas:

Applications of Parabolas:
Highway Overpasses using Type 1
Vertical Curves
John Catlett
Mathematics Teacher
North Star High School
What is a parabola?
• A parabola is the graph that results from an
equation of the form: y  Ax 2  Bx  C
• Parabolas are symmetric about a vertical line
known as the Axis of Symmetry.
• The Axis of Symmetry runs through the
maximum or minimum point of the parabola
which is called the Vertex.
• Examples of parabolas:
Exploration of Parabolas
• How do the coefficients A, B and C affect the
graph of the parabola?
• Let’s use Geogebra to find out:
• http://isite.lps.org/dtravis2/jcat/geogebra/par
abolaexploration.html
Parabolas and Vertical Curves
• In road design, vertical curves are designed
using parabolas. There are four types:
Focus: Type 1 Curves and
Overpasses
Design Factors
• Parabolas used in the overpass design are
based on a number of factors:
– The entrance and exit grades, g1 and g2
– The length of the vertical curve, L (NDOR uses 600
ft as the minimum L)
– The design speed (speed the road is designed for)
– The sight stopping distance, S (based on design
speed and line of sight)
– The elevation at the start of the curve, ElevBVC
Equation of the parabola:
y  Ax  Bx  C
2
A
g2  g1
,
2L
B  g1 , C  ElevBVC
Finding the Parabola for Overpass
Design
• Lets find equations to fit some existing
overpasses.
• http://isite.lps.org/dtravis2/jcat/geogebra/ove
rpassexample1.html
• For our purposes today, we will use:
– The NDOR minimum ( 600 ft ) for L
– The distance above the roadway that runs below
the overpass at the start of the curve as the
ElevBVC
(these calculations are more complex in real life)
Now lets find the equation of the parabola using
our formula, y  Ax 2  Bx  C
g2  g1
A
,
2L
B  g1 , C  ElevBVC
Example:
The entrance grade, g1, is 3%, the exit grade, g2,
Is 2%, the ElevBVC is 8 feet and the horizontal
length of the curve, L, is 900 feet.
Solution:
.02  .03
A
 0.00003 B  0.03 C  8
2  900
Equation: y  0.00003x 2  0.03x  8
Check Solution:
http://isite.lps.org/dtravis2/jcat/geogebra/overpassequationc
hecker.html
Final Thoughts
Video: http://isite.lps.org/dtravis2/jcat/
Other things to consider?