"For an 80 car, 7200 ton train with two six axles, here’s the air resistance as a % of total train resistance (traction HP required to move train):
on the level:
10 mph - 17% (143 HP)
30 mph - 46% (1466 HP)
50 mph 60% (5162 HP)
70 mph 68% (12418 HP)"
What formula and assumptions produce those figures? Do they really say the total train resistance at 10 mph is less than 0.7 pounds per ton?
(If we pretend for a minute that those horsepowers are at the rail, rather than traction HP, then to get them the train resistance (in pounds) would have to be A + BV + CV^2 where V is speed in miles/hr and
A = 1669.27
B = 277.1875
C = 9.276
And of course those coefficients would actually have to be a bit lower, since the figures are traction HP. So where did those come from?)
The horsepowers you gave can’t be based on the actual Davis formula; you remember his constant term (what I called A) was 1.3 pounds per ton, plus 29 pounds per axle. Your spreadsheet doesn’t actually use any Davis coefficients, does it? Which is why I was wondering where its coefficients did come from.
This aparently supports the move to 286K equipment. There’s less resistance per ton of train (Fuel Savings!) This makes sense. The railroad will be using fewer axles to move the same weight - less flange friction against the rail.
Are these formulas for roller bearing or friction (journal) bearings? The original Davis formula was, I assume, for friction bearings, and I haven’t seen public sources on roller bearings.
If, and only if, the cross section and drag resistance is identical between the 263 k and 286 k would there be there a slight 2% advantage to the HAL cars in terms of resistance for the same train weight, notwithstanding 7% fewer axles. But, those numbers would be necessary to know to make the conclusion you did and those numbers would have to show identical dimensions to the two sizes. The move to HALs is generally justified in consulting studies by a variety of other factors.
I don’t have that info, but I sincerely doubt it would be in the 1970 car & loco book. With the heavier axle loading on the larger capacity car I am skeptical that it would have the identical moving and air drag coefficients as the smaller capacity car.
In general, however, it is true that the efficiency gains in overall motion resistance go to the heavier, not the lighter, cars. Not exactly the argument we heard on the other thread. Railroads continue to try and generate efficiency gains by overall heavier, not lighter, carloads. The tare is heavier, the heavier overall carload is heavier.
Well, that’s not what it says on page 842. And it makes no sense. (1.3+0.29) = 1.59 every time. Why would they complicate the equation by writing (1.3+0.29)/W instead of 1.59/W?
According to the MR December, 2003 article (page 75) on freight trucks, roller bearing trucks became widely used on freight cars in the late 1950s and 60s. They were required equipment on cars built after 1966. Solid bearing trucks were banned from interchange after 1980. Since these trucks were rated in the 70 to 125 ton range, the number of solid bearing trucks in use in 1970 depended on how many 50 ton capacity freight cars were still around at that time.
seems to have an interest in resistance formulas and offers up some coefficients.
He gives a “passenger car” formula of
R=1.3 + 29/W + 0.03V + (0.041/(WN))V²
where 1.3 lbs/ton is a constant resistance, 29/W is a resistance that depends on axle loading W (for a 29 ton axle, 29/W = 1 lbs/ton), 0.03V is a “viscous term” (he gives .03 instead of .045, and the .041/(WN) times V^2 term is aero drag.
You don’t lump the 1.3 and 29/0.29 coefficients together – the 1.3 is a constant drag while the 29 is a coefficient of a drag that is higher for lightly-loaded axles.
As to the aero term (this is the aero drag per ton for a car weighing amount WN where W is the axle load and N is the number of axles on the car – this is seemingly odd, but you need to divide the aero drag by the weight WN to get the aero drag per ton of car). This drag comes from assuming a coef of drag times frontal area of 20 ft sq. Since a car has about 140 sq ft frontal area, this optimistic drag coefficient must be taking into account the drafting effect of cars that trail one another.
A couple of the studies done, which describe AAR’s wind tunnel studies – as opposed to “real world” studies – specifically relating to intermodal are located at these links:
“During the 1980s, a number of studies focused on technologies to reduce train resistance and therefore fuel costs (7, 8). Aerodynamic drag was known to be a major component of total tractive resistance, particularly at higher speeds, so the Association of American Railroads (AAR) supported research on wind tunnel testing of rail equipment, including large-scale intermodal car models (9, 10). The results were used to develop the aerodynamic subroutine of the AAR’s train energy model (TEM) (11).”
Air drag is so significant with intermodal, that the study recommends adding – yes, adding – empty containers because notwithstanding the extra dead weight – the airflow improvement more than compensated for the extra weight by lowered resistance and added fuel savings. This is of course contrary to what some have argued. These studies, however, precisely define just how important airflow is compared to higher or lower weight in the case of complicated airflow situations such as intermodal.
I appreciate the common sense of air flow and wind tunnel studies, first, and only then doing field application, rather than spending millions of dollars to do it backwards and probably wrong. In my eight years with an engineering R&D project/agency with the U.S. Government, I participated in many a wind tunnel study and aerodynamic drag analysis, my office being just two doors down from the wind tunnel control room office in the facility where I worked.
Thanks for posting the spreadsheet. I gather you were trying to use Davis resistance in the spreadsheet (except for increased air resistance) but it looks like you left out the 29 pounds per axle. Then the figures you gave earlier (quoted at the beginning of this thread) don’t agree with the spreadsheet, or with Davis.
I think you’ll make fewer mistakes if you rewrite the Davis formula to give train resistance directly, instead of pounds per ton. Using your example, a train consisting of 7200 trailing tons in 80 four-axle cars, pulled by two 199.8-ton six-axle units, has a total resistance (in pounds) on level track of A + BV + CV^2 where V is speed in miles/hour and
A = (1.3 times total tonnage) + (29 times total axle count) =19507.48
B = (0.045 times trailing tons) + (0.03 times loco tons) = 335.988
and C depends on the type of car, unlike A and B, so I won’t bother trying to figure it.
In other words, total train resistance starts out at 19507 pounds at speed zero and climbs from there. Your spreadsheet doesn’t agree with that, does it? And your figures from the beginning of the thread are even lower.