19 March 2010

Roller Coaster Physics

Roller coaster rides, and avoiding them on the peninsula rail corridor, is a recurrent concern that recently came up in a discussion of tunnels. Millbrae councilwoman Gina Papan claimed that an underground station in Millbrae would cinch a tunnel through Burlingame because "you can't have a roller coaster from Millbrae to Burlingame."

Let's bust this myth with a quick look at the physics and specifications of HSR track alignment.

Discomfort arises when there is a curve in the tracks, because passengers conform to Newton's first law: they tend to keep going straight. The centripetal force imparted by the train causes them to follow the curve, and sometimes to spill their drink. In horizontal curves (left or right), engineers can use superelevation, a simple trick that harnesses gravity to provide the centripetal force and smooth the ride. Vertical curves (up and down) are a different matter: passengers must bear the full brunt of the centripetal force.

The centripetal force is perceived as a vertical acceleration, such as you might feel when riding an elevator, and goes as velocity squared divided by radius.

The recently published HSR design standards (specifically, TM 2.1.2 section 6.1.6.2 and TM 1.1.6 section 6.1.7, which trace to the AREMA manual, Chapter 5, Part 3.6) describe the design limits placed on vertical acceleration: typically just 2 to 3 percent of gravity, obviously much less than would ever be experienced on a roller coaster. Given this acceleration limit and the planned operating speed of 200 km/h (125 mph), the references above contain the following constraints:








































ParameterDesiredLimitExceptional
Passenger-Only 125 mph Vertical Curve Length*
840 ft
650 ft
420 ft
Freight 75 mph Vertical Curve Length* (sag)
2000 ft
1200 ft
1200 ft
Freight 75 mph Vertical Curve Length* (crest)
1500 ft
1200 ft
1200 ft
Passenger-Only Maximum Grade
1.0 %
1.7 %
3.0 %
Freight Maximum Grade
1.0 %
1.0 %
2.0 %

* per percent of grade change

What do these mean? To find out, it's helpful to look at a picture of what a vertical track profile looks like when you need to transition from one elevation (or depth) to another. The figure at right shows the basic anatomy of a vertical transition. Passenger discomfort (if any) occurs only in the curved portions at the beginning and end of the transition; the straight ramp in between is not perceived as having any less comfort than flat and straight track. Simply lengthening this ramp will increase the overall height of the transition. Assembling the above specifications into actual ramp lengths, we can calculate a useful metric: the total length of a vertical transition, depending on how much rise is required.












































































Transition RiseTrain TypeDesiredLimitExceptional
15 ft (Ground Level to Elevated)Passenger Only2340 ft
2140 ft
1830 ft
Passenger + Freight2830 ft
2600 ft
2600 ft
30 ft (Trench to Ground Level)Passenger Only3840 ft
3300 ft
2750 ft
Passenger + Freight4730 ft
4200 ft
4170 ft
45 ft (Trench to Elevated)Passenger Only5340 ft
4200 ft
3470 ft
Passenger + Freight6250 ft
5700 ft


5370 ft

90 ft (Tunnel to Ground Level)Passenger Only9840 ft

6850 ft
5100 ft
Passenger + Freight10750 ft
10200 ft
8100 ft
105 ft (Tunnel to Elevated)Passenger Only11340 ft
7740 ft
5600 ft
Passenger + Freight12250 ft
11700 ft
8850 ft


Keep in mind that even the "exceptional" values would not be anywhere close to a roller coaster ride: the acceleration would be only about 4% of gravity. Roller coasters routinely exceed 100% of gravity.

What immediately jumps out from the table is that the trickle of peninsula freight trains make these transitions much longer and potentially much more community-disruptive than passenger-only infrastructure.

As for Gina Papan's notion of Burlingame getting a tunnel as a consequence of a hypothetical underground Millbrae HSR station: it would take less than a mile for tracks to rise up to an elevated structure that clears Broadway in Burlingame, with the utmost passenger comfort.

Myth Busted.

17 comments:

  1. What's the formula for converting X meters per percent of grade change to a curve radius of R meters?

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  2. Why is the freight percentage of gravity lower ... is this about bulk freight (as opposed to multi-mode container) settling and shifting, or about heavy freight suspensions and safe operations?

    Note that if they persist with the FFSS configuration, the long transition for the "slow" pair could be put to use for a center-line transition for express passenger trains. At those transitions and a 2.5% grade for the express pair, a centerline transfer can drop while the HSR pair rise, gaining clearance under the HSR corridor to skew toward the slow track, rising at the 1% grade. Once it can clear under the slow track, it can pass under and rise between.

    Each centerline cross-over requires a footprint of five tracks wide for a sufficient width of the transition, but would be four tracks wide at the overpass/underpass.

    The reverse would seem to work for a trench, though the cross-over would need to be completed before entering a tunnel.

    This is obviously substantially more expensive per crossover than the direct switch of the FSSF layout, and so would not be very well suited to an Express/Ltd/Local service structure, but it could work for an Express / Semi-Express / Local service structure with only four to six cross-overs.

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  3. radius = 100 * L where L is the run length per percent of grade change.

    The passenger values in TM 1.1.6 translate to 25 km, 20 km and 13 km... which are incredibly generous (and inconsistent) with 17 km, 11.5 km and 7.5 km in TM 2.1.2.

    I've been using 10 km, on the tight side. But that's to minimize impacts.

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  4. "What's the formula for converting X meters per percent of grade change to a curve radius of R meters?"

    Here's the German textbook answer because that's what I have in front of me. (I haven't checked exactly what Clem used. Also, I'm slightly glossing over the differences between the conventional German design parameters and the European norm EN 13803):

    Vertical radius design goal
    r_a = 0.4 v^2
    where v is in kmh and r_a in m.
    Minimum without a design exception
    r_a = 0.25 v^2 with r_a >= 2000m.
    Absolute exceptional value
    r_a = 0.16 v^2 (crests), 0.13 v^2 (sags) (Absolute max 3000m and 2000m respectively when there are turnouts in vertical curves.)

    Parabolic transitions between steady grade and vertical curves. Between grades of s_1 and s_2 (specified here in percent grades, even though Yurpeens use promille in this context) the length of such a transition will be
    l_a = r_a * (s_1 + s_2) / 100
    and the height gained in the transition
    y_e = r*a * ((s_1 + s_2) / 200) ^2

    So for example to transition from flat to a 2% grade with a (cost inflating and completely ridiculous for the SF peninsula) design speed of 200kmh one has
    Default vertical curve radius r_a = 16000m
    Transition length l_a = 480m
    Transition rise y_e = 3.6m

    So to get up to the 15 foot (4.6m) "ground to elevated" level a 3% grade would never be achieved using the design default vertical curve (since 3.6 > 4.6/2)

    Using the tighter, but still nothing like "roller coaster" discretionary design limit r_a = 0.25 v^2 = 10000m, one would have instead l_a = 300m, y_e = 2.25m, meaning that over a total run of 602.4m (1976 feet) a transition from flat to rising to 3% (constant for just 2.4m!) back to flat again would gain 4.572m (15 feet).

    If I typed everything in correctly, that is...


    Anyway, the idea of putting anything underground anywhere south of 16th Street in San Francisco is simply bat shit insane (to use the approved European standards body terminology), with a Design Goal of Maximizing Contractor Profits and Absolutely Nothing Else.

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  5. Adirondacker1280020 March, 2010 13:57

    Anyway, the idea of putting anything underground anywhere south of 16th Street in San Francisco is simply bat shit insane

    But the catenary is going to so visually intrusive. . . it's going to just ruin this view in Palo Alto

    And the lack of horns, bells and diesel engines is going to be replaced with the whooshing of very quiet electric trains.

    .... tunnels, the only solution.

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  6. @Bruce, if I had to guess, the AREMA freight radii are probably gentler (especially in sag) so that extremely high drawbar forces will not pull with an upward component... there's probably also a desire to reduce slack action.

    As for overtakes in FFSS, you can kiss them good bye. No fancy saut-de-moutons will be built for Caltrain, if only because they shouldn't be.

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  7. Richard, you're mentioning promilles but your second set of equations is for percent. In addition, the vertical arc has no transition, it's quadratic parabola with vertical axis. See here.

    There's something wrong with your actual numbers:
    R = 0,4V^2 = 0,4 * 200^2 = 16000 m
    t_z = 0,5 * 16000 * (20 - 0) / 1000 = 160 m
    y_v = 0,5 * 160^2 / 16000 = 0.8 m
    y_zl = 160 * 20/1000 = 3.2 m

    R = 0,25V^2 = 0,25 * 200^2 = 10000 m
    t_z = 0,5 * 10000 * (20 - 0) / 1000 = 100 m
    y_v = 0,5 * 100^2 / 10000 = 0.5 m
    y_zl = 100 * 20 / 1000 = 2 m


    Clem, there's minimum distance of grade intersections:
    L_n >= 4V >= 200 m
    L_n = 4*200 = 800 m > 200 m
    So if you stick to european standards, the minimum ramp length is t_z1 + 800 m + t_z2.

    Radii for freight can be so generous because:
    - coupler construction: you can't do nothing about that if it is the case)
    - radial forces & slack action: you can limit them easily by limiting maximum length of train.

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  8. I can see the percentages of gravity being lower for freight because an equal increase in terms of percentage of gravity for freight as for passenger trains would mean the the freight trains would still be exerting a much greater increase in weight on the tracks and track bed.

    That would be the same way airplanes are weight-limited (at least one reason), because the wings on a heavier plane have to generate more lift, causing greater stress on the airframe.

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  9. Anyway, the idea of putting anything underground anywhere south of 16th Street in San Francisco is simply bat shit insane (to use the approved European standards body terminology), with a Design Goal of Maximizing Contractor Profits and Absolutely Nothing Else.

    Well I agree with the bat shit insane part, but the Maximizing Contractor Profits part is incomplete - it would also Minimize Visual Intrusion (good conceptually at any rate) and Maximize Costs (bad, and probably why NIMBY's really want it - costs skyrocket, project dies on the vine. Status quo, baby!).

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  10. For a post with physics in the title, the first part of this is a bit misleading.

    From the first sentence of the quoted wikipedia link: "Centripetal force is a force that makes a body follow a curved path." For the earth circling the sun, the centripetal force is provided by gravitational attraction. For the case of a train going through a turn, the centripetal force is the force of the rails pushing against the train, as per Newton's Laws.

    You can argue that the force exerted by the rails on the cars is passed along to the passengers in the car, but this perceived force has another name: centrifugal force, and it is this force that's responsible for knocking over any drinks (certainly not "the centripetal force imparted by the train").

    Of course, you can always consult xkcd.

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  11. Though not on this topic, I have been asked about electric/electronic interference from the 25Kv system about to span the Caltrain corridor in the next several years.

    The questioner wondered about not only IT hardware, but stuff like pacemakers, etc.

    Is it a proximity issue?

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  12. Given that 25kV is in use all over the world without any ill effects, I am guessing that it is in fact a proximity issue. I've never even heard of a person with a pacemaker standing on a platform served with 25kV OCS having any problems.

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  13. Maritn,

    If I was you, I would be more worried about the 110kv+ power/transmission lines in the cities instead of the 25Kv train lines

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  14. @Steven: As observed from an inertial reference frame, the centripetal force, if not imparted by gravity (superelevation), is imparted to the drink by its container. The centrifugal vs. centripetal issue is irrelevant; it is a matter of intertial vs. accelerated reference frame and does not change the outcome--a spilled drink. I stand 100% by the physical explanation.

    @Martin: Eric is correct; think of electrification as a simple and minor extension of the power grid. I'd be far more worried about the emissions from variable-frequency traction electronics in the train itself, and even that doesn't seem to set off pacemakers. Generally, all this equipment is tested for compliance with (and certified to) government electromagnetic emissions standards. Our very own FRA has already done studies on European HSR... including the TGV, which has a 5 megawatt, 25 kV power main on the roof of the train itself, just feet from passenger's heads. Google it.

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  15. @Martin: distance from the source plays the dominant role in EMF exposure, and drops by the inverse square of the distance (e.g. field at 2 feet is 1/4 as strong as at 1 foot). 25kv/60hz electrification on a nearby rail line does not hold a candle to what most people are exposed to in their homes, in their beds, with 110v/60hz power lines all throughout their walls, ceilings, etc. Electric clock radio on the bedstand? Electric blankets? Hair dryers? While at much lower voltage, all expose people to much higher EMFs due to proximity than a 25kv line dozens or even hundreds of feet away.

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  16. Technically 25kv isn't even considered "high voltage" it's "medium voltage".

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  17. "So for example to transition from flat to a 2% grade with a (cost inflating and completely ridiculous for the SF peninsula) design speed of 200kmh one has
    Default vertical curve radius r_a = 16000m
    Transition length l_a = 480m
    Transition rise y_e = 3.6m"

    A 16000-meter-radius vertical curve from zero to 2% would be 320 meters long. You're saying a 200-km/hr railroad needs to replace that with... two 480-meter transitions?

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