# CA HSR beyond 220 MPH



## Joel N. Weber II (Feb 14, 2010)

When the Boston and Providence railroad was built, I suspect its designers weren't planning for the day some parts of it would support 150 MPH operation, as the Acela achieves on that right of way today.

Does the California HSR alignment have any long straight sections that have the potential to exceed 220 MPH once the equipment becomes available?

If so, is the spacing between the two tracks being made larger than is strictly required for 220 MPH operation so as to allow for future speed increases?

Will the catenary be able to handle trains running at 350 MPH? When the French were first building 186 MPH trains, they picked a catenary design that works at nearly double the design speed for the route. Would following the same pattern mean the Californians should pick a catenary design capable of 400 MPH?


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## Rumpled (Feb 16, 2010)

I can't answer all those question; but the specific route and right of way have not been picked. The Central Valley lends itself to some very long straight sections and I'd imagine that technology and $ will be the speed limiters - not curves.

I joke that I-5 requires no turns for 200 miles, and it's not too far from the truth.


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## George Harris (Feb 17, 2010)

I think most of this is relatively public. At least it is bounced around quite a lot in this part of the world.

Track spacing is supposed to be at 16.5 feet where ever practical, 15.0 feet otherwise. To give a relationship, that used in most parts of the world is 4.3 meters, 4.5 meters and 5.0 meters. Those amount to roughly 14.1 feet, 14.8 feet, and 16.4 feet. Also for comparison, a lot of the northeast corridor is at 13.0 feet and run at 135 mph. Thanks to 130 mph being considered a step into the unknown at the time (early 1960's) much of the Japanese Shinkansen is at 4.3 m track centers, and is currently operated at 280 km/h = 174 mph, which is about as fast as practical given the terrain, curves, and station spacing.

In general, the California line is supposed to be designed to permit 250 mph operation where at all practical. That means most of the Central Valley for sure.

Catenary is the least of the problems. It is relatively cheap to modify if already a constant tension system. At some point there is supposed to be a limit, at least in theory, but I don't know where it is, and don't know why it is supposed to be a limit, Electrical capacity of the system might become a problem.


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## Joel N. Weber II (Mar 6, 2010)

Is there any publicly available estimate of the top speed that would be achievable on long sections of tangent track with 16.5 foot spacing, and/or of what spacing would be required for 500 MPH operation?

Are the Shinkansen sets the same width as the Acelas? It seems to me that narrower trainsets might reduce the track center spacing required for a particular speed (though the shape may also have some more subtle impact on the spacing required).

For all that catenary may be relatively cheap to modify, availability of money to replace it seems to be the thing limiting the Acela to 135 MPH south of New York City. (I recognize that the existing catenary there is fixed tension, although I believe I've also seen claims that replacing the wire while maintaining a fixed tension system could allow the Acelas to run at 150 MPH in some sections there, and of course there is also the issue that upgrading the 135 MPH segments to 150 MPH would have a fairly minor impact on total travel time.)

If the wattage that can be pulled from the catenary by a single train becomes a limiting factor, it might be possible to put batteries in each trainset. I'm thinking of a system where a train that had slowed down to 250 MPH to get through a curve would get all of the power it needed for the traction motors and HEP loads from the catenary, and would use whatever excess capacity was available to charge the batteries from the catenary. While braking, all of the regenerative braking energy would go into the batteries, and whatever power was available from the catenary not needed for HEP would also go into the batteries. And at the maximum speed, either the power for the traction motors and HEP loads would simultaneously come from both the catenary and the batteries, or perhaps just from the batteries if it turned out to be possible to keep the batteries adequately charged even when the pantograph was lowered for the fastest parts of the route. The cost of lithium ion batteries may end up coming down a lot in the next ten years.


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## George Harris (Mar 9, 2010)

Joel N. Weber II said:


> Is there any publicly available estimate of the top speed that would be achievable on long sections of tangent track with 16.5 foot spacing, and/or of what spacing would be required for 500 MPH operation?
> Are the Shinkansen sets the same width as the Acelas? It seems to me that narrower trainsets might reduce the track center spacing required for a particular speed (though the shape may also have some more subtle impact on the spacing required).


Don't know about spacing for 500 mph, or top speed that could be run on 16.5 foot track centers.

Shinkansen trains are wider than Acelas. Shinkanset trains are 3380 mm = 11'-1" wide, and that is over the body. No extended grabiron or any other projections beyond that dimension.


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## jis (Mar 9, 2010)

Joel N. Weber II said:


> If the wattage that can be pulled from the catenary by a single train becomes a limiting factor, it might be possible to put batteries in each trainset. I'm thinking of a system where a train that had slowed down to 250 MPH to get through a curve would get all of the power it needed for the traction motors and HEP loads from the catenary, and would use whatever excess capacity was available to charge the batteries from the catenary. While braking, all of the regenerative braking energy would go into the batteries, and whatever power was available from the catenary not needed for HEP would also go into the batteries. And at the maximum speed, either the power for the traction motors and HEP loads would simultaneously come from both the catenary and the batteries, or perhaps just from the batteries if it turned out to be possible to keep the batteries adequately charged even when the pantograph was lowered for the fastest parts of the route. The cost of lithium ion batteries may end up coming down a lot in the next ten years.


Boy, you are reeeeaallly obsessed with batteries aren't you? Why would the amount of current necessary from the catenary become an issue? Oh I forgot, we have to have a reason to bring batteries into the discussion :lol: :lol: :lol:


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## Joel N. Weber II (Mar 12, 2010)

jis said:


> Why would the amount of current necessary from the catenary become an issue?


The last sentence of one of George Harris's posts above says ``Electrical capacity of the system might become a problem.''

The Drag article in Wikipedia says that power needed to overcome wind resistance is proportional to the cube of the speed, which suggests a 500 MPH train would need 8 times the power that a 250 MPH train does for the overcoming wind resistance portion of power consumption.


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## Guest_George Harris_* (Mar 13, 2010)

Joel N. Weber II said:


> jis said:
> 
> 
> > Why would the amount of current necessary from the catenary become an issue?
> ...


Not at the computer that has my password memorized.

What I said about electrical capacity was in reference to the capacity built for 220 mph operation. Equipment and the power system are the main things I see needed to be beefed up to run faster than 220 mph. Track and structures are not, so long as the curver radii are big enough.

Wiki is wrong, at least as it applies to speed up to 220 mph more or less, and that is by experimentation, not theory.

Aerodynamic effects on trains is definitely relative to speed squared, and that appears to be true up to somewhere in the 200 mph plus region. Above that, the French may know from their high speed runs of the TGV, but that information they are treating like the secrets of their atomic bomb.

I have heard that as the speed of sound is approached ther is a V cubed factor that comes into play, but where ever it begins to come in, that speed is for sure above 200 mph.


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## Joel N. Weber II (Mar 13, 2010)

Are we correctly distinguishing between kilowatt hours of electricity required to cover a given number of miles, vs the amount of energy the train consumes in a given amount of time?

I believe what what Wikipedia is claiming is that the number of kilowatt hours required to make a trip quadruples if you double the speed (assuming that it turns out that nearly all of the energy consumed by the train goes to overcoming air resistance). That is, the amount of money that will be paid to the electric company to get from a particular point A to a particular point B should be quadrupled when you double the speed. (Ignoring the fact that the rate structure may not simply be a linear function of kilowatt hours used, anyway.)

Because you are consuming those quadrupled kilowatt hours in half the time when you travel twice as fast, the amount of energy that needs to be carried through the catenary in the space of a second (assuming no batteries) in order to overcome wind resistance increases by a factor of 8 when you double the speed.

When you talk about ``aerodynamic effects on trains'', I'm not sure if you're talking about kilowatt hours per trip, or instantaneous power consumption.

Here's Wikipedia's attempt at explaining the distinction:



> Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.


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## George Harris (Mar 16, 2010)

I don't know about what is Wikipedia, and given more solid experimentially derived infromation, I really don't care what they say.

First question for them: Are they talking about movement in air or in water? Both gases and liquids are classified as fluids, but there is considerable difference in their response. First and foremost, gases are compressible, and liquids, water at least, for all practical purposes are not compressible.

I will copy in here what I said elsewhere, as this is really the more appropriate thread anyway.

Train resistance is most simply expressed in the form of a TR = A + B V + C V^2 formula, where "V" is the speed and the A, B, and C are constants.

Formulas of this nature have been used for many years. Search "Davis Formula" for the granddaddy of all these things. V^3 terms? Not in any of them.

In general,

"A" represents starting resistance and internal machine friction, and is constant regardless of speed.

"B" represents resistance due to deflection of the track and deviations from perfect smoothness of track

"C" is primarily aerodynamic

"A" and "B" are both relative to train weight, when different lengths of trains of the same equipment are analyzed. "C" includes a front end constant plus a length factor.

For high speed trainsets operating on tracks permitting high speed operation, both the "A" and the "B" terms are relatively small, and become less and less significant as actual speed increases.

All these terms are derived primarily from measurements of actual train performance as determined by power draw in relation to speed and acceleration. Since proper derivation of these terms is quite expensive, the results are usually not published.

If doing this for freight, the terms can be extrodinarily complex, as A, B, and C depend upon the characteristics of each piece of equipment, and, for the "C" term its relationship to its neighbors in the makeup of the train.

Power and tractive effort: TE = P/V. Therefore, TE is infinite at a speed of zero. This obviously is not possible, and even if possible could not be applied to the rail. Generally, at low speeds TE is limited based on a permissible acceleration rate.

Adhesion has also been determined to decline with speed. Usually this is not a factor in EMU's, but can be with high powered light axle load locomotives.

Again, this factor is experimentially determined. One set of formulae is in the form of Adhesion = D / (V + E), where D and E are constants and V is the speed. Usually there are multiple values used here, one for dry rail which is effectively the maximum, another for wet rail, which is the maximum to be considered in scheduling, and a third for braking which is a worst condition situation that you had better consider in determining safe stopping distances. For the accelerating condition, the factor must be multiplied by the proportion of the train weigth on powered axles. For braking, the full weight may be used, or a reduced factor based on some ratio of failed brakes included.

Numbers: All these are in metric units, because that is the way I have dealt with these things. The numbers given are not the real ones, as these are regarded as confidential by the equipment supplier, but the are in the general range of those that are used.

For train resistance for high speed train sets, with "V" in km/h and resitance in kilonewtons, think of something on the order of: 8.00 + 0.08 V + 0.0008 V^2 for a train with a weigth of around 600 metric tonnes.

For power, think in terms of motors of around 275 to 325 kilowatts each times however many axles you feel like so long as your weigth per axle is in the range of 13 to 16 tonnes. To get TE, you must convert the "V" term from km/h to meters/second.

For adhesion, the Shinkansen braking adhesion formula was published in AREA Bulletin No.727, October 1990. It is

Adhesion = 13.6 / (V + 85) again, remember these are metric units.

Also this is a ratio, so at say 200 km/h, the Adhesion is 4.77%. If this sounds low, it is. The braking rate is somewhat higher than this, because braking is assisted by train resistance.

End of lesson.

All further work is left to the student.

Discussion of power requirements will be in a later installment.


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## jis (Mar 16, 2010)

Now if one takes a step back and consider what is the power (energy per unit time) required by the train to overcome the resistance (force), one might see where the additional O(v) comes from as explained below. The Wiki page is not incorrect in what it says. though the way it says it may leave it open to misinterpretation.

BTW, in the equation TR = A + B V + C V2, the third term is essentially Rayleigh's equation which holds true for high Raynold's Number flows (as opposed to Stokes equation which holds for low Reynold's Number flow and that one is O(v) not O(v2), where

C = rho Cd Ar / 2

where

rho = density of the fluid

Cd = drag coefficient

Ar = reference area (area of the orthographic projection of the object in the direction of motion)

Experimentally one can compute the three coefficients A, B and C by running the object at various speeds under otherwise similar conditions and recording the force required to exactly balance the drag and solving the equation for A, B and C given the values of TR which is the same as the balancing force, and V.

Both Rayleigh's Equation and Davis Formula are about force, not about energy or power.

No we know that Power P = F . v

Hence substituting for F (= TR) into the formula for TR one gets for PTR, the power needed to keep the train running at constant speed in face of the train resistance TR as

PTR = A V + B V2 + C V3

That is where we get the V3 from, the same V3 that Joel was talking about.

Sorry if I bored you all out of your skins. I can do that once I get going 

Boy all this is coming back to me from my Physics Graduate School days - all these things like Navier Stokes Equations and stuff about fluid dynamics and laminar and non-laminar flow and what nots. Good stuff! OK got to go and review this stuff in some depth.


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## DET63 (Aug 29, 2010)

Regarding issues of curvature on the proposed California HSR system, the 200+ mile stretch through the central valley shouldn't be much of an issue. However, the hills between San Francisco (and the rest of the Bay Area) and the valley, as well as the mountains between the valley and Los Angeles, will undoubtedly present a number of engineering challenges.

To be fair, though, Japan is a rather mountainous country, and yet the Japanese have been able to not only build an extensive HSR system, but were in the fact the pioneers in doing so.


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## Guest (Aug 29, 2010)

Google California High Speed Rail and you will find some web sites with considerable discussion on the alignment and other issues. Some is very speculative, some discusses plans that have been released for public discussion. keep hunting and you can find some of the Technical Memos, or excerps that were written for design guidance. The preliminary alignment developed to cross Pacheco Pass is good for 220 mph.


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