tag:blogger.com,1999:blog-8419444332771213285.post6414651823945738985..comments2014-08-27T19:38:33.258-07:00Comments on Caltrain HSR Compatibility Blog: The Top 10 Worst CurvesClemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comBlogger42125tag:blogger.com,1999:blog-8419444332771213285.post-7270606837451721152009-07-28T00:41:46.989-07:002009-07-28T00:41:46.989-07:00Regarding the San Bruno curve, is something alread...Regarding the San Bruno curve, is something already being built on the empty lot (ex lumberyard)?bffnnnhttp://www.blogger.com/profile/13655373429763259297noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-59685596456054710832009-04-14T18:48:00.000-07:002009-04-14T18:48:00.000-07:00The curve in San Bruno is 60mph, not 65mph, just a...The curve in San Bruno is 60mph, not 65mph, just a note.fjcnoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-43396213645987071702009-04-09T18:21:00.000-07:002009-04-09T18:21:00.000-07:00All curves. The gray bars show the curve-constrai...All curves. The gray bars show the curve-constrained speed limit, which (if lower than track speed) the train must slow down to meet.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-69453948511333955902009-04-09T17:08:00.000-07:002009-04-09T17:08:00.000-07:00What is up with that profile from the CHSRA? The ...What is up with that profile from the CHSRA? The parts I'm looking at are:<BR/>(1) The south end. What causes the immense speed restrictions there? Not curves, surely!<BR/>(2) The north end. Curves? Tunnels?neroden@gmailhttp://www.blogger.com/profile/07475686367097445497noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-76679827639787190762009-02-22T19:00:00.000-08:002009-02-22T19:00:00.000-08:00I think Amtrak wanted to get 9" for Acela Express,...<I>I think Amtrak wanted to get 9" for Acela Express, but I don't know if they ever did. They may be limited by the infamous "trainset too wide because of ADA compliant toilet" issue or the infamous "trainset weighs a gazillion tons" issue.</I><BR/><BR/>The Northeast Corridor has high level platforms at many stations, so I would hazard a guess that Acela trains are the same width as everything else on the NEC. <BR/><BR/>From what I've read about tilt restrictions it has something to do with the local track being too close to the express track. Wouldn't want to blow the windows out of the Metro North local as Acela speeds through Cos Cob would you? Or shear the top off both... <BR/><BR/>Non tilt equipment shares the track, Regional and commuter and in places freight, so there is a limit to the amount of super elevation you can add to the track even if your tilt equipment can take advantage of it.Adirondackernoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-32934025605237245862009-02-05T17:31:00.000-08:002009-02-05T17:31:00.000-08:00800 foot spirals sound a bit too long. If the run...800 foot spirals sound a bit too long. If the runout is 50 mm per second, the train velocity 125 mph (55 m/s), the superelevation is 175 mm, then you only need 650 foot spirals. (sorry for the mixed units)<BR/><BR/>Only half of that spiral extends beyond the ends of the equivalent circular arc. In other words, you can reverse a curve in a run length of about 750 feet. It's a bit tight, but it just might barely fit at Hayward Park.<BR/><BR/>The lateral offset due to the spiral is minimal (less than a meter) and does not significantly affect how much land is needed inside the curve.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-14217540477260066112009-02-05T14:37:00.000-08:002009-02-05T14:37:00.000-08:00Can we all please leave behind the 1800s-era degre...Can we all <B>please</B> leave behind the <A HREF="http://en.wikipedia.org/wiki/Mars_Climate_Orbiter" REL="nofollow">1800s-era</A> degrees (and feet)?<BR/><BR/>What part of "radius" is so terribly difficult for Olde Tyme PTG/HBTB/PCJPB/PBQD pretend-engineers (and their CAD systems) to understand, exactly?<BR/><BR/>If we simply got rid of <B>everybody</B> connected to this project whose first language is English (or, shudder, American) we'd be several decades, at least, ahead of where we find ourselves.<BR/><BR/>One "degree" = 1746.4m<BR/>Deal with it.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-62720084004905036292009-02-05T14:05:00.000-08:002009-02-05T14:05:00.000-08:00On further reflection-- if you're trying to do 110...On further reflection-- if you're trying to do 110 mph around a 1-degree curve you might want 800-ft spirals, or thereabouts, and maybe that would be a problem at Hayward Park for a four-track line that can't swing as wide across the right-of-way as a two-track line. At San Bruno, a 1-deg curve with 800-ft spirals is 86.0 ft inside the assumed present centerline.Timnoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-56484410290875790082009-02-04T15:15:00.000-08:002009-02-04T15:15:00.000-08:00If you zoom way into the map (or better yet, open ...If you zoom way into the map (or better yet, open the KML in Google Earth), you'll see the six I-380 pillars that cause problems at San Bruno, as well as the outline of the 4-track right of way for a 110 mph (@ 10 inch equivalent cant) curve.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-82906810411211179192009-02-04T14:15:00.000-08:002009-02-04T14:15:00.000-08:00If you go to Hayward Park and take a look, I think...If you go to Hayward Park and take a look, I think you'll agree the spiral lengths aren't a problem there.<BR/><BR/>(In SP days Hayward Park was a 2-degree curve; as I recall it's now 1 deg 30 min.)<BR/><BR/>Did some calculation for San Bruno--<BR/><BR/>Assuming the total angle of the curve is 23.64 degrees, and assuming the centerline of the track is now a 3 deg 08 min curve (chord definition) with 375-ft spirals...<BR/><BR/>Then if we want a 1-degree curve with 500-ft spirals we need to shift the centerline inward by 83.1 ft at the midpoint of the curve.<BR/><BR/>(That total angle is from Google Earth-- I'm guessing it's correct within 0.1 degree or so, but can't promise.)<BR/><BR/>Next question: how much of a problem are those I-380 pillars. Too lazy to work on that yet.Timnoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-74197117150519321902009-02-04T11:36:00.000-08:002009-02-04T11:36:00.000-08:00Looking closer at the Hayward Park curve, I suspec...Looking closer at the Hayward Park curve, I suspect the problem is not the curvature of the track itself but rather the fact that it's a reverse curve (chicane). I believe this is problematic because it limits the length of the transition curve (i.e. the section between the tangent track and the circular curve).<BR/><BR/>The bottom line is that it may not be possible to fix this curve without substantial eminent domain takings to the east of the ROW, which may be why the run simulations assume that the curve is limited to 95 mph.mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-33833615810581643432009-02-03T19:56:00.000-08:002009-02-03T19:56:00.000-08:00Thanks to mike and Richard M for sharpening their ...Thanks to mike and Richard M for sharpening their pencils and doing the extra math that I didn't bother with.<BR/><BR/>It turns out that the acceleration profiles shown in the CHSRA simulation are indeed consistent with the capabilities of a standard ICE-3 high speed train, powered at just under 20 kW/tonne.<BR/><BR/>We armchair experts now all agree: the 32.5 kW/tonne label on the CHSRA plot is evidently some kind of minor typographical error, and is not indicative of unrealistic train performance assumptions, as I originally assumed.<BR/><BR/>Post has been expunged accordingly; I'm glad, because this was about curves, not timetables.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-63401172370746752942009-02-03T12:25:00.000-08:002009-02-03T12:25:00.000-08:00Even more Siemens data:0-100 km/h in 0.6 km0-200 k...<A HREF="http://republicans.transportation.house.gov/Media/File/110th/Rail/3-20-07--roundtable-Brady-siemens.pdf" REL="nofollow">Even more Siemens data:</A><BR/><BR/>0-100 km/h in 0.6 km<BR/>0-200 km/h in 3.9 km<BR/>0-300 km/h in 17.9 km<BR/><BR/>Blowing up the PDF and measuring with a very precise ruler, the simulated CHSRA train does:<BR/><BR/>0-100 km/h in 0.7 km<BR/>0-200 km/h in 4.5 km<BR/>0-300 km/h in 23.8 km<BR/><BR/>So the assumed performance is actually slightly <B>inferior</B> to an off-the-shelf ICE-3.<BR/><BR/>Clem, it might be helpful to others to update the post, noting that although the run simulation claims 33 kW/t, the actual acceleration curve that they use is totally consistent with current conventional HSR equipment.mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-12676148490055949832009-02-03T11:38:00.000-08:002009-02-03T11:38:00.000-08:00We can use physics to calculate an upper bound on ...We can use physics to calculate an upper bound on how powerful the simulated CHSRA train could be.<BR/><BR/>We know that a TGV with 43 kW/tonne can reach 400 km/h in 22 km (this was recorded during the 1989-1990 speed trials before they opened LGV Atlantique). The TGV was 300 tonnes with 12,800 kW of power.<BR/><BR/>We also know that the simulated CHSRA train reaches ~295 km/h in 22 km. The CHSRA train is 642 tonnes with unknown power (power is claimed at 20,800 kW, but this is clearly wrong).<BR/><BR/>Assume that there is no friction/drag (we'll return to this later).<BR/>Energy = Power*Time.<BR/>Kinetic Energy = 0.5*Mass*Velocity^2.<BR/>Power = Force*Distance.<BR/>Force = Mass*Acceleration.<BR/>Acceleration = dVelocity/dTime.<BR/>Velocity = dDistance/dTime.<BR/><BR/>If Acceleration is roughly constant (it's not, but for our calculations it turns out not to make a big difference), and we have:<BR/>Velocity = Acceleration*Time<BR/>Distance = 0.5*Acceleration*Time^2<BR/><BR/>Plugging in known quantities, we have:<BR/>Distance(Time') = 22 km = 0.5*Acceleration*Time'^2<BR/>Velocity(Time') = 400 km/h = Acceleration*Time'<BR/>So Acceleration = 400 km/h / Time'.<BR/>And 22 km = 0.5*(400 km/h / Time')*Time'^2 = 200 km/h*Time'.<BR/>So 0.11 hours = Time'.<BR/>Thus we estimate that it takes 0.110 hours for the TGV to reach 400 km/h. In real life, where acceleration declines over time (as speed increases), it should take less time. And in fact, it took the actual TGV 0.104 hours to reach 400 km/h, so we're only off by 5%.<BR/><BR/>For the simulated CHSRA train, the same calculations imply 22 = 147.5 km/h*Time', so 0.149 hours = Time'. This is assuming acceleration is constant. I can't know exactly how long a CHSRA train would take to reach 295 km/h, but the estimated vs. real TGV numbers above suggest that it definitely shouldn't be less than 0.130 hours. So I'll conservatively assume 0.130 hours.<BR/><BR/>Finally, we can do comparisons. The TGV must expend a minimum of 0.5*(300 tonnes)*(400 km/h)^2 = 24 million units of energy in 0.104 hours. It thus must have at least 230 million units of power. The CHSRA train must expand a minimum of 0.5*642*295^2 = 28 million units of energy in 0.130 hours. It thus must have at least 215 million units of power.<BR/><BR/>In a dragless world, the CHRSA train is thus using no more than 93% of the TGV's power, or 0.93*12,800 = 11,900 kW. The actual power/weight ratio of the simulated CHSRA thus does not exceed 18.5 kW/tonne, which is actually less than the current ICE-3's power/weight ratio of 19.3 kW/tonne.<BR/><BR/>Of course, all of these calculations were done without taking into account drag. However, accounting for drag will only <B>reduce</B> the amount of power than the CHSRA train needs relative to the TGV. Both trains have similar drag coefficients and similar frontal areas, but the TGV is running much faster than the CHSRA train (400 km/h at exit vs 295 km/h at exit). Since drag is proportional to the square of velocity, the TGV is overcoming roughly 83% more drag than the CHSRA train when it crosses the 22 km mark. Actual energy expenditures are equal to kinetic energy plus energy dissipated as drag. Thus the ratio of actual TGV energy expenditures to actual CHSRA energy expenditures is even higher than I calculated above, which implies that the ratio of the TGV's power to the CHSRA's power must also be higher than I calculated above.<BR/><B><BR/>Bottom line, this CHSRA train is performing like an HSR train that has 17-18 kW/tonne of power. These run times are achievable with a conventional, off-the-shelf HSR trainset. You definitely don't need 33 kW/tonne to achieve them.<BR/><BR/>Even more evidence? <A HREF="http://www.siemens.pl/upload/images/TS-Velaro%20CN.pdf" REL="nofollow">Siemens claims 0-200 km/h in 148 seconds for the Velaro (i.e. ICE-3)</A>. The simulated CHSRA train reaches 200 km/h in roughly 4.7 km, which implies a few seconds under 169 seconds (169 seconds is calculated assuming that acceleration is constant). The performance profile sure looks a lot like a conventional ICE-3.<BR/></B>mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-67648379857958238172009-02-03T00:28:00.000-08:002009-02-03T00:28:00.000-08:00Also note that it takes this mythical ICE-3 over 1...Also note that it takes this mythical ICE-3 over 1.5 km to get from 0 to 160 kph. I've been on Amtrak trains with only 13-14 kW/t of power that have gone from 0 to 160 kph in roughly that distance (2 km at most). There is no way that these simulations actually correspond to 33 kW/t of power.mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-52839827772829177732009-02-03T00:19:00.000-08:002009-02-03T00:19:00.000-08:00At 32.52 kW/tonne, they're assuming an almost 21 M...At 32.52 kW/tonne, they're assuming an almost 21 MW of total power. That train is going to need to raise all four pans in order not to fry the pan and wire. I'm not sure how many such beasts the Caltrain electrification system will be able to support simultaneously. I'm guessing somewhere between zero and one per substation.<BR/>As for the souped-up TGV, what they usually end up with is using higher supply voltage and generally pushing the traction package to its limits, and removing as many intermediate cars as possible. In fact, the latest test run had two locomotives pulling what amounted to a three-car MU with half its axles powered. I don't think this sort of thing is really representative of anything at all.arcadyhttp://www.blogger.com/profile/06394805356595604336noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-62884726520528804202009-02-03T00:01:00.000-08:002009-02-03T00:01:00.000-08:00Something is weird about the "super-powered" ICE-3...Something is weird about the "super-powered" ICE-3. It sure doesn't accelerate like it has 33 kW/t.<BR/><BR/>For example, in the run time simulation it takes 51 km for the ICE-3 to accelerate from 0 to 330 kph on level track. In real life (1989-1990 test runs), a souped-up TGV accelerated from 0 to 300 kph in 11 km, 0 to 350 kph in 16 km, 0 to 400 kph in 22 km, and 0 to 440 kph in 33 km (and this was on a very slight uphill grade). The TGV did have 43 kW/t, but a mere 30% better power-to-weight ratio shouldn't generate this kind of discrepancy...the TGV made it to 350 kph in less than one-third the distance that it takes the supposedly super-powered ICE-3 to get to 330 kph!mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-24280694427642091382009-02-02T23:37:00.000-08:002009-02-02T23:37:00.000-08:00Doing some back-of-the-envelope calculations using...Doing some back-of-the-envelope calculations using your graph (based on the official graph) here are some numbers I calculated. My calculations are from the start of breaking, through the turn, and to the end of acceleration. I don't take 2-turn combos into account, so my seconds lost are vs an ideal running of 200kph the entire distance (some of which might be part of another turn).<BR/><BR/>Hayward park: 178kph avg, 3.2km, 7 sec lost<BR/>Millbare: 175kph avg, 3.6km, 9 sec lost<BR/>San Bruno: 153kph avg, 6.6km, 35 sec lost<BR/>Sierra Point: 161kph avg, 4.9km, 21 sec lost<BR/><BR/>Can someone double check my math?Peterhttp://www.blogger.com/profile/12497947698006908034noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-50467726421709453012009-02-02T23:08:00.000-08:002009-02-02T23:08:00.000-08:00I object in the strongest terms to the term "Prius...I object in the strongest terms to the term "Prius brake"! Electric traction has had regenerative braking for many decades, with the recovered energy being returned back to the grid. If anything, the brake on the Prius should be called the "trolley brake" or something.arcadyhttp://www.blogger.com/profile/06394805356595604336noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-38102647721276589532009-02-02T22:55:00.000-08:002009-02-02T22:55:00.000-08:00Ahh, hold the presses. I hadn't noticed the littl...Ahh, hold the presses. I hadn't noticed the little "32.52 kW/tonne"... their simulated train is <I>ridiculously</I> over-powered.<BR/><BR/>German ICE 3 - 19.6 kW/tonne<BR/>French AGV - 22 kW/tonne<BR/>Japanese Series 700 - 20.4 kW/tonne<BR/>CHSRA "magic train" - 32.5 kW/tonne<BR/><BR/>@James - the run time is in the CHSRA document. This is a speed vs. distance plot, so the area under the curve doesn't tell you much about time savings.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-46903759887486715692009-02-02T22:46:00.000-08:002009-02-02T22:46:00.000-08:00So the run time is the inverse of the area under t...So the run time is the inverse of the area under the curve? Then is is a matter of finding the ratio the area with and without the reductions to find the time and cost savings.Jameshttp://www.blogger.com/profile/17296808260419563238noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-1527684818436690142009-02-02T22:14:00.000-08:002009-02-02T22:14:00.000-08:00I just updated the post with a CHSRA run time simu...I just updated the post with a CHSRA run time simulation, which reveals many of the assumptions made about curves on the peninsula. Hat tip to Richard M for bringing my attention to this.Clemhttp://www.blogger.com/profile/01374282217135682245noreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-57435034588345520182009-02-02T17:28:00.000-08:002009-02-02T17:28:00.000-08:00It's actually less than 47 miles from end of track...It's actually less than 47 miles from end of track at 4th and Townsend to San Jose-- 46.75 plus or minus 0.1 would be a fair guess. UP changed the milepost at San Jose.<BR/><BR/>The Bayshore curve used to be 1 degree, but when Caltrain four-tracked they inserted a short straight for the switches-- so as I recall the curve is now 1 deg 30 min. Similarly, the Lawrence curve is a bit sharper than it was in SP days.<BR/><BR/>I suppose they'll remove that newish offramp overpass south of the 101 overpass at Sierra Point? They can add two tracks outside the piers, but they'd be kinky.Timnoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-21756984579938001782009-02-02T16:39:00.000-08:002009-02-02T16:39:00.000-08:00One question Clem left unanswered is whether the 3...One question Clem left unanswered is whether the 31 minute SF-SJ express is feasible.<BR/><BR/>Under the assumption that Sierra Pt is 85 mph (5" cant deficiency), San Bruno is 75 mph (5" cant deficiency), and all the "low-hanging fruit" curves are improved as Clem suggests, here is what I see:<BR/><BR/>Transbay to Diridon = 48.5 miles (current route is 47.5 miles)<BR/><BR/>Transbay to curve at (current) MP 0.7 = 1.7 miles @ 30 mph = 3.4 minutes<BR/>MP 0.7 to MP 1.2 @ 50 mph = 0.6 minutes<BR/>MP 1.2 to MP 2 @ 90 mph (acceleration phase) = 0.5 minutes<BR/>MP 2 to MP 45 @ 125 mph (cruise phase) = 21 minutes<BR/>MP 45 to MP 47 @ 60 mph (deceleration phase) = 2 minutes<BR/>MP 47 to MP 47.5 @ 30 mph (terminal approach) = 1 minute<BR/><BR/>+ 31 second penalty for San Bruno curve and 21 second penalty for Sierra Point curve<BR/><BR/>Total running time = 3.4 + 0.6 + 0.5 + 21 + 2 + 1 + 0.55 + 0.35 = 29.4 minutes.<BR/><BR/>Add in a couple minutes of dwell time at Diradon Station, and it looks like 31 minutes is achievable even without the removal for the Sierra Point and San Bruno curves (though it could still be worth it to remove the San Bruno curve)mikenoreply@blogger.comtag:blogger.com,1999:blog-8419444332771213285.post-75348481107056883892009-02-02T15:41:00.000-08:002009-02-02T15:41:00.000-08:00Rafael: I like your idea of getting Caltrans DOR i...Rafael: I like your idea of getting Caltrans DOR in on this. They at least have some experience with building rail projects, albeit fairly small and isolated ones mostly done by other railroads. As for Sierra Point, the limitation is not the lagoon, it's the configuration of the mountain and the 101 overpass. Any modifications there will have to involve massive earthworks and the reconstruction of the highway. Given that not too far north you have Bayshore and the tunnels, where I doubt you'd be going faster than 80 anyway, I just don't think it's worth it, at least compared to all the other much easier realignments that can be done elsewhere.arcadyhttp://www.blogger.com/profile/06394805356595604336noreply@blogger.com