Module 1-E
Railway Alignment Design and
Geometry – AREMA/FRA Basis
Alignment Examples
© AREMA & APTA- 2015
Reference Sources
• The tables used in the following examples come
from mixed sources to obtain clearest presentation
of data.
• Similar tables can be obtained from the FRA in the
track safety standards.
http://www.gpo.gov/fdsys/granule/CFR-2011title49-vol4/CFR-2011-title49-vol4-part213
© AREMA & APTA- 2015
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Reference Sources
• Additionally CSX, NS, UP, BNSF publish tables in
their Grade Separation Design guidelines or
Industrial sidetrack documents available from their
websites.
http://www.nscorp.com/content/dam/nscorp/ship/shippingtools/Public_Projects_Manual.pdf
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References – Curve Ea Table 1.5” Eu
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References – Curve Ea Table 3” Eu
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References – Minimum Spiral Lengths
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References – Minimum Element Lengths
• Minimum length of any track element - 100 feet
• Preferred length of any track element - 3 times the
velocity in mph. Example Track speed 60 mph,
preferred length = 3 X 60 = 180’
Note: These guidelines are for educational purposes only, actual requirements vary
drastically between railroads and transit agencies
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Example 1 – Allowable Speeds
SC 11+55
TS 10+00
SP-1
C-1
Dc = 2d 15’
CS 15+00
SP-2
ST 16+55
Ea = 2 ½”
A. How fast can a freight train operate through this segment @ Eu = 1 ½”? ______
B. How fast can a passenger train operate through this segment @ Eu = 3” ? ______
Note: Round speed to nearest lower multiple of 5 mph.
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Example 1 – Answers
SC 11+55
TS 10+00
SP-1
C-1
Dc = 2d 15’
CS 15+00
SP-2
ST 16+55
Ea = 2 ½”
A. How fast can a freight train operate through this segment @ Eu = 1 ½”? __50 mph__
B. How fast can a passenger train operate through this segment @ Eu = 3” ? __50 mph__
Note: Round speed to nearest lower multiple of 5 mph.
© AREMA & APTA- 2015
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Example 1a – Solution 1 of 3
A. How fast can a freight train operate through this segment @ Eu = 1 ½”? __50 mph__
Steps
1. Check the Curve’s Ea.
Using the Curve Ea Table 1 ½” Eu as shown the following pages
a. Enter the Table with the Dc of 2d 15’
b. Drop down to the Ea of 2 ½”
c. Read across to the speed 50 mph
d. Round to the nearest lower 5 mph multiple – 50 mph
2. Check Spiral #1 Length
Using the Minimum Spiral Length table as shown the following pages
a. Enter the table with the 50 mph and read the ½” runoff in 31 feet
b. 2 ½” runoff @ ½” per 31 feet = (2.5” / 0.5”) X 31 feet = 5 X 31 = 155’ Minimum
c. SC – TS = Spiral Length = Ls = 11+55 – 10+00 = 1+55 = 155’ >= 155’ Min. OK @ 50 mph
3. Check Spiral #2 Length
a. Ls = ST – CS = 15+55 – 14+00 = 1+55 = 155’ >= 155’ Min.
OK @ 50 mph
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Example 1a – Solution 2 of 3
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Example 1a – Solution 3 of 3
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Example 1b – Solution 1 of 3
B.
How fast can a passenger train operate through this segment @ Eu = 3” ? __50 mph__
Steps
1.
Check the Curve’s Ea.
Using the Curve Ea Table 3” Eu as shown the following pages
a. Enter the Table with the Dc of 2d 15’
b. Intersect with the Ea of 2 ½”
c. Read across to the speed 59 mph
d. Round to the nearest lower 5 mph multiple – 55 mph
2.
Check Spiral #1 Length
Using the Minimum Spiral Length table as shown the following pages
a. Enter the table with the 55 mph and read the ½” runoff in 39 feet
b. 2 ½” runoff @ ½” per 39 feet = (2.5” / 0.5”) X 39 feet = 5 X 39 = 195’ Minimum
c. SC – TS = Spiral Length = Ls = 11+55 – 10+00 = 1+55 = 155’ << 195’ Min. FAIL
d. Lower Speed to 55 mph and recheck spiral
e. 2 ½” runoff @ ½” per 31 feet = (2.5” / 0.5”) X 31 feet = 5 X 31 = 155’ Minimum
f. SC – TS = Spiral Length = Ls = 11+55 – 10+00 = 1+55 = 155’ >= 155’ Min. OK @ 50 mph
3.
Check Spiral #2 Length
a. Same as Spiral #1
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Example 1b – Solution 2 of 3
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Example 1b – Solution 3 of 3
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Example 2 – Spiral Manipulations
C-1
SC 11+55
TS 10+00
SP-1
Dc = 2d 15’
CS 15+00
SP-2
ST 16+55
Ea = 3 ¼ ”
If the spirals on Curve #1 were increased in length to allow a passenger of at least
60 mph, what would the approximate stations of the TS and SC of Spiral #1 be?
TS ≅ _____________
SC ≅ _______________
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Example 2 – Answers
SC 11+55
TS 10+00
SP-1
C-1
Dc = 2d 15’
CS 15+00
SP-2
ST 16+55
Ea = 3 ¼ ”
If the spirals on Curve #1 were increased in length to allow a passenger of at least
60 mph, what would the approximate stations of the TS and SC of Spiral #1 be?
TS ≅ __9+50.5__
SC ≅ ____12+04.5____
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Example 2 – Solution 1 of 3
What would the approximate stations of the TS and SC of Spiral #1 be?
Steps
1. Check the Curve’s Ea.
Using the Curve Ea Table 3” Eu as shown the following pages
a. Enter the Table with the Dc of 2d 15’
b. Drop down to the Ea of 3 ¼ ”
c. Read across to the speed 63 mph
d. Round to the nearest lower 5 mph multiple – 60 mph
2. Determine Spiral #1 Length
Using the Minimum Spiral Length table as shown the following pages
a. Enter the table with the 60 mph and read the ½” runoff in 39 feet
b. 3 ¼ ” runoff @ ½” per 39 feet = (3.25” / 0.5”) X 39 feet = 6.5 X 39 = 253.5’ Minimum, SAY Ls = 254’
c. The length of the spiral is split approximately 50/50 around the perpendicular to the center of the circular curve C #1
d. Use existing spiral TS and SC to set approximate location of the perpendicular
e.
f.
i.
11+55 – 10+00 = 155’
ii.
10+00 + (155/2) = 10+00 + 77.5 ≅ 10+77.5
TS ≅ 10+77.5 – (254/2) = 10+77.5 – 127 = 9+50.5
SC ≅ TS + Ls = 9+50.5 + 254 = 12+04.5
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Example 2 – Solution 2 of 3
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Example 1b – Solution 3 of 3
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Example 3 – Element Length Control
C-1
Dc = 2d 15’
Ea = 3 ¼ ”
Existing Freight Speed 55 mph
CS 25+00
SP-1
ST 26+55
C-2
Dc = 2d 15’
Ea = 3 ¼ ”
TS 29+30
SP-2
SC 30+85
If the spirals on Curves #1 & #2 were increased in length to allow a passenger of at least
60 mph, would the connecting tangent length be adequate for Desirable / Minimum
Criteria?
A. For Freight ___________ / ___________
B. For Passenger ___________ / __________
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Example 3 – Answer
C-1
Dc = 2d 15’
Ea = 3 ¼ ”
Existing Freight Speed 55 mph
CS 25+00
SP-1
ST 26+55
C-2
Dc = 2d 15’
Ea = 3 ¼ ”
TS 29+30
SP-2
SC 30+85
If the spirals on Curves #1 & #2 were increased in length to allow a passenger of at least
60 mph, would the connecting tangent length be adequate for Desirable / Minimum
Criteria?
A. For Freight ____YES____ / _____ YES ___
B. For Passenger ____NO____ / ____ YES ____
© AREMA & APTA- 2015
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Example 2 – Solution 1 of 4
Would the connecting tangent length be adequate for Desirable / Minimum Criteria?
A. For Freight __YES_ /__ YES _ B. For Passenger __ NO___ / ___ YES __
Steps
1.
Check the Curve’s Ea.
Using the Curve Ea Table 3” Eu as shown the following pages
a. Enter the Table with the Dc of 2d 15’
b. Drop down to the Ea of 3 ¼ ”
c. Read across to the speed 63 mph
d. Round to the nearest lower 5 mph multiple – 60 mph
2.
Determine Spiral #1 Length
Using the Minimum Spiral Length table as shown the following pages
a. Enter the table with the 60 mph and read the ½” runoff in 39 feet
b. 3 ¼ ” runoff @ ½” per 39 feet = (3.25” / 0.5”) X 39 feet = 6.5 X 39 = 253.5’ Minimum, SAY Ls = 254’
c. The length of the spiral is split approximately 50/50 around the perpendicular to the center of the circular curve C #1
d. Use existing spiral TS and SC to set approximate location of the perpendicular
e.
f.
i.
26+55 – 25+00 = 155’
ii.
25+00 + (155/2) = 25+00 + 77.5 ≅ 25+77.5 29+30 + (155/2) = 29+30 + 77.5 ≅ 30+07.5
30+85 – 29+30 = 155’
ST1 ≅ 25+77.5 + (254/2) = 25+77.5 + 127 = 27+04.5
TS2 ≅ 30+07.5 - (254/2) = 30+07.5 - 127 = 28+80.5
© AREMA & APTA- 2015
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Example 2 – Solution 2 of 4
Steps - continued
3. Check the Tangent Length - Freight
Using the Element Length Criteria
a. Desirable Length = 3V = 3(55) = 165’
b. Actual Length = 28+80.5 – 27+04.5 = 176’ > 165’ OK
4. Check the Tangent Length – Passenger
Using the Element Length Criteria
a. Desirable Length = 3V = 3(60) = 180’
b. Actual Length = 28+80.5 – 27+04.5 = 176’ < 180’ FAIL
c. Minimum Length 176’ >> 100’ OK
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Example 2 – Solution 3 of 4
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Example 1b – Solution 4 of 4
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Presentation Author
Michael Loehr
Practice Leader Rail & Transit – Americas - Civil
Transportation Business Group
CH2M HILL
8720 Stony Point Parkway, Suite 110
Richmond, VA 23235
Mobile 570.575.4692
[email protected]
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