http://www.youtube.com/watch?v=XhgHrDbN4EU
Enjoy!
-Crandell
Looks like shes what she does best Moving a train at speed LOL Larry
selector http://www.youtube.com/watch?v=XhgHrDbN4EU Enjoy! -Crandell
Crandell
The Challengers were rated 8000 tons on Archer grade and this train was close to that weight. After topping Archer, 70 mph running was the order of the day. I would have to believe the Challenger could handle more tons over the hill today with the roller bearing cars.
CZ
I've watched this example many times but I still have yet to get any data on Archer Hill. If the train weighs 7,657 tons and is 8,900 ft long (WB Video liner notes), and 3985 is doing 35 mph (obvious fact? from the video, or is it 30?), the grade couldn't be all that steep. I suspect that the train is very long and not too heavy on a tons/ft basis. Also the entire train is not entirely on the worst part of the grade at the same time. The grade profile should be readily available from a UP track chart. Does anyone have one for reference purposes that they can post here?
feltonhill...and 3985 is doing 35 mph (obvious fact? from the video, or is it 30?)..,
When the engine runs abreast of the camera, I would estimate its speed, based on driver revolutions per minute, to be near 45 mph, perhaps a bit more. I agree...it just doesn't seem to be working that hard. Lots of sharp stack talk, sure, but it isn't down to the 20 mph laboured drag that it should be with all the weight. Still, though, darned impressive.
feltonhilldata on Archer Hill.
For those who don't have that, the profile looks about like this:
Train is all on near-level track at Cheyenne, 6059 ft elevation or so
Eastward, they almost immediately start down, bottoming out at 5947 ft elev after 3-1/2 miles. Then almost immediately onto the upgrade that the chart shows as 2.7 miles of 0.7% compensated. The elevation gain is 89 ft and curvature is 98 deg, so average compensated grade is 0.65% if the grade really is 2.7 miles long. Then immediately downgrade at 0.6-0.7%-- so the center of mass of the train never reaches the summit elev 6036 at Archer.
The camera is west of the summit (see the half-mile marker on the pole?) and their speed would continue to drop until the engine was well east of the summit, so 45 mph when the engine passed the camera isn't as informative as we'd like.
As I recall there's no reason to think their speed thru the sag would exceed 60 mph even if they're full throttle all the way down the hill, and the curves aren't a problem at that speed, so for all we know they were full throttle all the way from Cheyenne to passing Archer, assuming the tonnage is correct.
Edit: as the engine passes the camera I count about 25 driver rotations in 9.4 seconds, which would be 33 mph. Even that isn't going to be any snap; until I do the calculation I'm still predicting full throttle was needed to get that at that point.
Hmm.. I counted 36 revs in 10 seconds, which comes out to 44 mph if I have done the math correctly.
feltonhill I've watched this example many times but I still have yet to get any data on Archer Hill. If the train weighs 7,657 tons and is 8,900 ft long (WB Video liner notes), and 3985 is doing 35 mph (obvious fact? from the video, or is it 30?), the grade couldn't be all that steep. I suspect that the train is very long and not too heavy on a tons/ft basis. Also the entire train is not entirely on the worst part of the grade at the same time. The grade profile should be readily available from a UP track chart. Does anyone have one for reference purposes that they can post here?
Thanks much for the additional info on the line's profile. It looks like the key to the observed performance is that Archer is a momentum grade in the middle of generally falling grades eastbound.
I assume that the locomotive has the throttle wide open, and is producing as much horsepower as it possibly can. Granted, you don’t have the stack talk inherent with a slow, hard pull, but pulling at that higher speed, surely indicates that the engine is doing a lot of work as well. I particularly like the jet engine sound as the locomotive passes the camera, and then how that sound is broken up by the heat waves as the engine heads into the distance. How many horsepower could that engine produce?
3985 could probably produce about 4,600 to 4,700 DBHP at about 40-45 mph (rear of tender, total evaporation about 92,000 lbs/hr) anytime in over-the-road service. If pushed a bit, it could probably get to at least 5,100 to 5,200 DBHP at 40-45 mph (total evaporation about 104,000 lbs/hr). The absolute maximum output would be determined by what limit was reached first - grate, evaporation, drafting, or conversion (cylinders, valves and running gear determine the latter). 3985 also has an exhaust steam injector instead of a conventional feedwater heater, and this exacts a small penalty on its boiler capacity.
Knowing the grade, curvature, train length and weight, observed speed and acceleration/deceleration rate, it's possible to estimate the DBHP 3985 is developing when it passes the camera. My math skills are not sufficient to bring that off, however!
In a way, what do all the numbers matter? It's one of the all-time great runbys!! Feeble old steam locos? Baloney!
feltonhillIn a way, what do all the numbers matter? It's one of the all-time great runbys!! Feeble old steam locos? Baloney
This, of course, was why I posted it. It's a great shot. I am glad to learn more about the environs...the camera angle belies the reality that the average for the first 30 minutes or more westward from Cheyenne is flat-to-descending.
I am not averse to using a "test run" of this Challenger, also on youtube under that name, to poke fun at our modern diesel fans, but on the modeller forum. Somehow, it just seems right, if perverse, to find a good video of a Challenger towing two fully set-up six-axle modern diesels across Cheyenne yard.
timz ...The elevation gain is 89 ft and curvature is 98 deg, so average compensated grade is 0.65% ...
http://www.youtube.com/watch?v=c2BoMFZcnDI&feature=related
CAZEPHYRThe Challengers were rated 8000 tons on Archer grade and this train was close to that weight.
timz [snip] Then almost immediately onto the upgrade that the chart shows as 2.7 miles of 0.7% compensated. The elevation gain is 89 ft and curvature is 98 deg, so average compensated grade is 0.65% if the grade really is 2.7 miles long. [snip]
timz, good information and to the point as always - you're one of the few that's sharper than me on these matters. However, a little clarification is needed - something's wrong with that curvature figure - should it perhaps be 0.98 degree instead ? That would be consistent with the actual grade being in the 0.62 % range (89 feet / 2.7 miles = 33 ft. per mile / 52.8 ft. per mile for each 1.00 % = 0.624 % ), and compensated up to the 0.7 % value at the rate of adding 0.05 % or so for each degree of curvature.
If a constant or average curvature, a 0.98 degree curve is a curve with a Degree of Curvature of about 0 deg. 59 mins. in civil engineering terminology, which is a radius of about 5,850 feet. (I'm not going to split hairs here over whether that's computed by the highway engineer's 'Arc' definition, or by the railroad engineer's 'Chord' definition - for such a curve, there's no practical difference.)
The other aspect is - is that a continuous curve for that whole distance ? If so, that would imply a curve that turns through a total arc or angle approaching 140 degrees of central or deflection angle = 0.98 degrees per 100 ft. chord x 2.7 miles x 5,280 ft. per mile, etc. I don't know the territory, but that seems a little odd to me. Or, if it's an average, what are the consituent curves and their relative degrees of curvature ?
None of this takes away from the essence of the analysis, conclusions, or admiration of the 3985 above - but I wanted to clarify this point, because it seems a little odd to me, that's all.
- Paul North.
Do we see the location being discussed above in the video? I see the train running through a fairly tight reverse curve on the way up the grade. I also detect what I feel must be a slippage of about 7 mph by the time the Challenger crests. The last three seconds of the video that show the tail end on the curves suggests that the consist is moving more slowly.
selector Do we see the location being discussed above in the video? I see the train running through a fairly tight reverse curve on the way up the grade. I also detect what I feel must be a slippage of about 7 mph by the time the Challenger crests. The last three seconds of the video that show the tail end on the curves suggests that the consist is moving more slowly. -Crandell
It is Archer Hill just a few miles east of Cheyenne. The diesels that pulled the train into Cheyenne were removed at Cheyenne and the 3985 pulled the train by request of the stack train company. It did slow down for sure on the grade, but take note, five GE's had pulled the train into Cheyenne from the west. This was back in 1993 I believe so those units would have been 8-40CW's most likely.
Sorry, everyone, and thanks CZ. Not my best composition skills evident there. What I meant was, based on the discussion between Timz and PDN, above, and looking at the S-curve and grade, albeit with some telephoto foreshortening when the camera looks to the meet near the top of the grade, it seems steeper than a mere .7%. Even so, my question pertains to the S-curve...would this not impose a substantial impediment to the climb up the grade, say by adding about 0.2%?
Bucyrus I assume that the locomotive has the throttle wide open, and is producing as much horsepower as it possibly can. Granted, you don’t have the stack talk inherent with a slow, hard pull, but pulling at that higher speed, surely indicates that the engine is doing a lot of work as well. . . . How many horsepower could that engine produce?
Keep in mind that HP is proportional to Tractive Effort x Speed (and if in lbs. and MPH, respectively, then divided by 375 to get the drawbar HP). So, even if the steam engine at speed is not pulling as hard, if it's going considerably faster it might still be producing more HP than at a slower speed.
For a fictitious example, if 3985 was producing 95,000 lbs. TE - per GP40-2's post on the previous page of this thread - at 20 MPH, that would be about 5,070 HP. But if at 40 MPH it was only producing 50,000 lbs. TE, the 'stack talk' might be considerably less - but the HP output would be close to 5,335. So there's not necessarily a correlation between not working the steam locomotive as hard meaning it's putting out less HP.
Crandell/ selector has a good point about the 'S' or 'reverse' curves as shown in the original video clip, as well as the evident grade. I'll force myself to review it again later on to see if I can estimate their sharpness. For sure, they are not constant, and much sharper than the 0.98 degree that I speculated above, which makes the supposed curvature resistance data in timz's post on the 1st page of this thread all the more puzzling.
Paul_D_North_Jr something's wrong with that curvature figure
To compute an average compensated grade over the 2.7-mile climb, we don't need to know how sharp the curves are (assuming we're making the usual assumption that curve resistance is directly proportional to curve sharpness). If we make the usual assumption that a 1-degree curve is equivalent to 0.04% upgrade, and sharper curves proportionately more, then 98 degrees of total angle is equivalent to 3.92 feet of altitude gain, which we add to the actual altitude gain to get the average compensated grade over the 2.7 miles.
selector it seems steeper than a mere .7%.
selector the S-curve...would this not impose a substantial impediment to the climb up the grade, say by adding about 0.2%
Paul_D_North_Jr For a fictitious example, if 3985 was producing 95,000 lbs. TE - per GP40-2's post on the previous page of this thread - at 20 MPH, that would be about 5,070 HP. But if at 40 MPH it was only producing 50,000 lbs. TE, the 'stack talk' might be considerably less - but the HP output would be close to 5,335. ...
For a fictitious example, if 3985 was producing 95,000 lbs. TE - per GP40-2's post on the previous page of this thread - at 20 MPH, that would be about 5,070 HP. But if at 40 MPH it was only producing 50,000 lbs. TE, the 'stack talk' might be considerably less - but the HP output would be close to 5,335. ...
CAZEPHYR.It did slow down for sure on the grade, but take note, five GE's had pulled the train into Cheyenne from the west. This was back in 1993 I believe so those units would have been 8-40CW's most likely. CZ
.It did slow down for sure on the grade, but take note, five GE's had pulled the train into Cheyenne from the west. This was back in 1993 I believe so those units would have been 8-40CW's most likely.
jr 611 good video but have you seen N&W #1218 + 100 Coal Hoppers (1987)
For Comparison, and because you mentioned and I had not see it; I looked it up and linked it here;
http://www.youtube.com/watch?v=pl-QVxHIq6o
Edit:
http://www.youtube.com/watch?v=so7-Fu2psjc&feature=related
Here is a video of the 1218 in Apr 87. What is amazing is the size of the passenger train (I'm guessing; about 20 or so cars), Maybe some one can get a better count. Anyway there is a segment of the 1218 slipping on Christianburg Hill near Arthur,Va. Slipped 3 times in starting that train on the grade...Anyone provide what percentage that grade is? Thanks!
It seems you have copied the same URL. ?
samfp1943 And just for good measure Two Y's hauling 174 loads, Awsome!
And just for good measure Two Y's hauling 174 loads, Awsome!
Your link was incorrect. Perhaps you meant this link:
http://www.youtube.com/watch?v=zV8rA3UE-lc&feature=related
Hey feltonhill, any idea where this was filmed at? Also, can you estimate the weight of the train based on the capacity of the N&W coal cars at that time?
Most of the action was filmed on Blue Ridge grade east of Boaz (the natives tell me it's pronounced boze, not bow-az) siding near Vinton, VA. The sample on youtube was made as the train approached "photographers bridge" about halfway up the hill. The bridge is long gone and the area has grown over considerably since 1957. I just bought the DVD in Roanoke last weekend, and it's probably one of the best I've ever seen.
A train weight of about 11,000 to 12,000 trailing tons sticks in my mind if three loco's were used, but I'm not sure.
While there is a mathematical approximation between horsepower and tractive effort, it does not hold in real life actual performance.
Tractive Effort, more correctly known as Starting Tractive Effort, does not remain constant, but immediately drops off for most engines after about 10 mph.
Locomotive horsepower curves are bell shaped, and for steamers, generally increasing as speed increases (and different from diesel horsepower curves).
That's why diesels can start a train they cannot keep moving on some grades (and stall), and steamers can pull more train than they can start (helpers were used on long trains on the PRR just to get them rolling).
The Union Pacific 4-12-2 produced 4917 horsepower at 37 mph as originally tested during 1926.
The Union Pacific Challenger was designed to exceed that power output and run as fast as 90 mph.
Kratville in The Challenger Locomotives says they were designing for about 5500 horsepower. IIRC, the actual drawbar horsepower was said to be in the 5200 horsepower range.
Starting tractive effort was not important to the Union Pacific. Instead, horsepower at speed was--and they really wanted high speed, because they were in a race with Santa Fe and others to see who could move freight the fastest.
John
Also--today's diesels are using computerized wheelslip control to maintain a much more constant (and higher) rate of adhesion between the steel wheel and the rail than what steam power could achieve. As much as I love steam power, there is no real comparison to today's diesels. They're just different animals.
Also--it was customary on the old railroad plans to quote degrees of curvature. This really had to be degrees of central angle, and not the civil engineering definition of degree of curve.
As already pointed out above, the math just doesn't work out--those old track charts are stating degrees of central angle.
John (civil engineer)
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