Download 4-2-Track-Geometry-Part-1-with-annotations-v2

4.2
Track Geometry
4.2
1.
2.
3.
4.
Learning Outcomes
Identify elements of a simple road curve
Compute design values related to horizontal alignment of simple curves
Identify elements of a modern road curve
Compute design values related to the horizontal alignment of modern road
curves
5. Define and explain the purpose of road superelevation
6. Compute design values related to road superelevation
Railway curve
• Modern road curves are composed of a
simple curve and transition curves at
the ends (also called spiral curve)
• The purpose of a transition curve is to
to gradually increase the curvature as
the vehicle enters the curve
• Why? An abrupt change in curvature
will also cause abrupt application of
centrifugal force on the wheels.
• A railway curve has to be provided also
with superelevation to provide the
necessary centripetal force for the
curvature
Banking / leaning / superelevation / railway cant
Photo by: Science Photo Library
Simple curve (Circular curve)
Simple curve (Circular curve)
Sample problems – simple curve
1. A railway track needs to change direction from 𝑁 5°22′13" 𝐸 to S 78°06′ 52" 𝐸.
According to the design speed, the radius of curvature should be 125.53 m. If a
simple curve is to be provided
a) What will be its length?
b) How far from the point of intersection should the curve start?
Problem #1 Solution
a) What will be its length? (𝐿𝑐 )
𝐼
𝑃𝑇
𝑃𝐼
S 78°06′ 52" 𝐸
𝐿𝑐 = 𝑅𝜃
𝐼 = (90 − 5°22′13") + (90 − 78°06′ 52")
𝐿𝑐
𝑅 = 125.53 𝑚
𝑃𝐶
𝑳𝒄 = ___________𝒎
𝑂
𝐼
𝑇
𝑃𝐼
𝐼 = __________________rad
𝐿𝑐 = 125.53 𝐼
𝐼
𝑁 5°22′13" 𝐸
𝑃𝑇
b) How far from the point of intersection should the curve
start? (Length of Tangent)
𝑇
𝑅 = 125.53 𝑚
𝑃𝐶
(rad)
𝐼
2
𝑇 = 𝑅 tan
𝐼
2
𝑻 = _____________𝒎
𝑂
Sample problems – simple curve
2. An engineer needs to design a simple curve for a change in road direction from
𝑁 42°10′ 39" 𝑊 to S 38°19′ 46" 𝑊 and has to stake on the ground points along the
curve using a total station (for the construction). According to the design speed, the
radius of curvature should be 100.0 m.
a) Starting from the PI, how far should he move to get to station PC?
b) He needs to stake out 10 equidistant points along the curve from station PC
(excluding PC, last point is PT). What are the offset angles and chord distances
he needs to stake out?
c) To check his work, he transfers back to PI, and bisects the angle between PC and
PT to locate the midpoint of the curve. What distance should M be from PI?
Problem #2 Solution
𝑃𝐼
𝐼
a) Starting from the PI, how far should he move to get to station
PC? (Length of Tangent)
𝑇
𝑃𝐶
𝑃𝑇
𝑁 42°10′ 39" 𝑊
𝐼
2
′
S 38°19 46" 𝑊
𝑅 = 100.0 𝑚
𝐼 = 90 − 42°10′ 39" + 90 − 38°19′ 46"
𝐼 = _________________rad
𝐼
𝑇 = 𝑅 tan
2
𝑻 = _____________𝒎
b) He needs to stake out 10 equidistant points along the curve
from station PC (excluding PC, last point is PT). What are the
offset angles and chord distances he needs to stake out?
For S1:
𝐼
𝜃=
= __________rad
10
𝐼
𝛼=
𝑆1
𝐿𝑆1
𝜃
𝜃
2
𝑃𝐶
𝑳𝑺𝟏
𝜃
𝛼 = = _____________rad
2
𝜃
= 2𝑅 sin
= _________________𝒎
2
Problem #2 Solution
c) To check his work, he transfers back to PI, and bisects the
angle between PC and PT to locate the midpoint of the curve.
What distance should M be from PI?
𝑃𝐼
𝐼
𝐸
𝑀
𝑃𝑇
′
S 38°19 46" 𝑊
𝑇
𝑃𝐶
𝐸=
𝑁 42°10′ 39" 𝑊
𝐼
2
𝑂
𝑅 = 100.0 𝑚
𝑇
𝐼
sin 2
− 𝑅 = _______________𝒎
Sample problems – simple curve
3. A simple curve has a radius of 125 m. What is its equivalent degree of curvature in
English and SI systems?
Solution:
a) English system: (1 𝑠𝑡𝑎𝑡𝑖𝑜𝑛 = 100 𝑓𝑡)
𝑆 = 1 𝑠𝑡𝑎𝑡𝑖𝑜𝑛
𝑅 = 125 𝑚 = 410.1 𝑓𝑡
𝑆 = 𝑅𝜃,
𝜃=
𝑆
𝑅
=
100
410.1
𝑅
𝑫 = 𝜃 = 0.243843 rad = 𝟏𝟑. 𝟗𝟕𝟏°
𝐷
b) SI system: (1 𝑠𝑡𝑎𝑡𝑖𝑜𝑛 = 20 𝑚)
𝑅 = 125 𝑚
𝑆 = 𝑅𝜃,
𝜃=
𝑆
𝑅
=
20
100
𝑫 = 𝜃 = 0.2 rad = 𝟏𝟏. 𝟒𝟓𝟗°
Compound curve
Reverse curve