Bayway Terminalthere will be little or no flange resistance and there should not be any power issues while traveling up
Actually as many suggested, there will be flange resistance and they could be relatively large. 24" is kind of sharp. So following the suggestions from others, the frictional resistence or the rolling resistence could be large. I meaured my 3-stall hopper cars and they are about 8" long (depending on the model), and 4.2-4.5oz. If you loco have a pulling force of 4.5oz, and car weigt of 4.2oz, here is the estimate of the cars you can pull
Crr #cars
0.02 15
0.03 13
0.04 11
0.05 10
If you want to pull cars that are longer, such as intermodal cars, the number will drop quite a bit. If I take the length of 10",
0.02 12
0.03 11
0.04 9
0.05 8
Jerry
I checked again. According gregc, the effective Crr due to curve is 32/R%. So 24" curve would give you 1.33%. If I use this number
Car length #car
8" 17
10" 13
Try not to pull long cars, you might be ok.
Huh!
That string line effect always sucks doesn't it eh?
TF
Exactly!!
I tried to pull 60 cars on my layout, which is spiral with a continous grade 2-3%. I used 3 locos at the front. I had some 89' auto carrier cars near the front. They could not go through some of my curves (36"). These cars simply fell off my track because of the large side force... That made me wonder what was happening and tried to make these calculations...
LastspikemikeThis thread explains how you might estimate these effects. It also enables a better understanding of the variables which may help you solve derailments, binding, stringlining and uncoupling events.
As long as you have one curving to the right, right before you have one curving to the left. It balances itself out and you'll be OK.
Curves are utterly delightful and should not be confused with trains.
When in doubt, just add another diesel!
Bayway TerminalBased on these discussions and mathmatical analysys i think there will be little or no flange resistance
the values in the original post didn't include flange resistance
the conventional estimate for effective grade on curve is 32/rad, posted by Byron for HO model railroads
if you have a 32" curve, the effective grade is 1%.
32" curve and 2% grade, the effective grade is 3%
24" curve and 2% grade, the effective grade is 3.3%
considering that at least one measurement found rolling resistance to be approximately 2% on relatively straight and level track, ~5.3% of the total car weight is needed to pull a train up a 24" curve and 2% grade.
10 4oz cars requires 2.12 and a loco weighing ~8.5 oz
greg - Philadelphia & Reading / Reading
Coincendently i'm in the middle of planning an elevated radius for my expansion of my 3' x 14' grade level switching layout. My concept plan is for a 24 inch curve at a max 2% incline, my 6 axel diesel engines will be (double headed) and mostly be pulling 6-12 hopper or intermodel well cars at a time out to a single main track leading to a staging yard. All of the cars are weighted in accordence with NMRA standards and have metal wheel sets with kadee couplers. Based on these discussions and mathmatical analysys i think there will be little or no flange resistance and there should not be any power issues while traveling up and through the curved the incline (unless i'm wrong with my assumption) ?
Bayway Terminal NJ
I revised the formular slightly. For each loco, the force to next car is
(F-G*W-Crr*W)*A
This added to each car calculation. I also set Crr=0.02 to compare with your table.
gregcwhere do you account for the additional (flange) friction of the wheels on the curve?
mobilman44I agree its fun for some to play with stats and math formulas and matrixes, but it is not going to be an exact science in this situation.
of course it's not an exact science. but you're suggesting building the layout and then finding out the grade is too much for the loco and cars you planned to use? (of course you can shorten the train, add another loco, improve the rollability of the cars or add weight to the loco)
it's obvious that many modelers don't understand this. but it doesn't mean that a reasonable estimate can't be made by some
this is yet another aspect of the hobby.
mobilman44Remember, "one size truly fits nothing" - as in "gimme caps" and other personal items......
huh?
mobilman44 Way too many variables to set up rules for car moving capability. The only way to truly determine is to load up a train until it no longer can manage the load. And yet, that determination is only correct for those locos combined with those cars, on that section of trackage, etc., etc. I agree its fun for some to play with stats and math formulas and matrixes, but it is not going to be an exact science in this situation.
Way too many variables to set up rules for car moving capability. The only way to truly determine is to load up a train until it no longer can manage the load. And yet, that determination is only correct for those locos combined with those cars, on that section of trackage, etc., etc.
I agree its fun for some to play with stats and math formulas and matrixes, but it is not going to be an exact science in this situation.
Rich
Alton Junction
Remember, "one size truly fits nothing" - as in "gimme caps" and other personal items......
ENJOY !
Mobilman44
Living in southeast Texas, formerly modeling the "postwar" Santa Fe and Illinois Central
hjQi gregc but weren't you trying to account for tighter radii and coupler angles? Jerry
gregc but weren't you trying to account for tighter radii and coupler angles?
where do you account for the additional (flange) friction of the wheels on the curve?
Crr increases on curves, by 1% on a 32" radius, from 2 to 3%
I am not yet but quite look forward to it!
selectorI'm retired and getting ready to plant seeds for germination in my greenhouse.
This is great to know!
LastspikemikeWe have found that splitting two diesel locomotives into a lead and a mid train power works better than double heading. Looks better too in our tiny world.
If the two locos' speed match perfectly, I would say yes. This can maximize the usage of the power...
LastspikemikeHow about putting both locomotives in the center of the train, pulling the back half and pushing the front half? Net gain?
gregcbut weren't you trying to account for tighter radii and coupler angles?
Yes, that is correct. I think the reason is that your calculation used a fix number (effective grade) to account for the curves. Mine depends on the radius and the lenght of a car. The fixed number you used might be an average...
nealknowsevery freight car must match in weight, height, length and who knows what else has to be taken into consideration.
Neal,
This may not be necessary. The length is more of the average length. The longer the individual car is, the more force lost we would have.
The real layout could be more complicated, I agree. My calculations are just to provide a little bit help (if at all).
BTW: it will be great if you can test my calculations on your layout. Mine is also with 2%-3% grade helix, but is a mix of straight and curved...
doctorwaynef that locomotive can't do the required work, you add another one, either with it or on the rear of the train or somewhere in the middle. I've done this with all of my locomotives (I call the results "tonnage ratings").
Practically, I may do the same thing! But thinking of these is another fun thing to me in model railroading...
Definitely not a math class major, that's for sure.
However, I think what Jerry is doing and posting for us can help some people, if that's their thing.
My take is based on his numbers, if I read it correctly, that every freight car must match in weight, height, length and who knows what else has to be taken into consideration. I don't think most of us (there may be some) have perfect, matching, rolling and more freight or passenger cars.
I think in the modeling world, we have to look at our own layouts to see how we can make trains run at certain lengths, eliminating drags, coupler issues and more.
On my layout, I have a 2 track helix with a 2% grade that is 95% the same all the way around the 3 1/2 turns. Depending on the engine and cars, I can pull with a pair of matching GP38-2 engines and 16 freight cars of various lengths up my helix. I've done more, but added a helper at the end.
Are there many home layouts that run 100 cars or more on a grade or helix? I can possibly see this helpful for a club layout.
Interesting topic....
Neal
selectorWe need a vector analysis to see why stringlining takes place
free body diagram
riogrande5761 John-NYBW This is too much like math class for me. Here here. After I finished a year of Calculus and Linear Algebra in college, that was enough for me!
I agree. If you want to pull a particular train on a straight or curved track, and/or up a grade, you simply put a locomotive on it, and see how it performs. If it does the job, write down the details of the test, so you can repeat it at any time.
If that locomotive can't do the required work, you add another one, either with it or on the rear of the train or somewhere in the middle. I've done this with all of my locomotives (I call the results "tonnage ratings"). My cars aren't all the same length, nor the same weight...some open cars, without loads, weigh only 2oz., but with a full load (usually loose material) they're just slightly over 8oz.For me, it's a lot more fun doing the experimental work in determining the amount of power needed to move the train, rather than sitting around with a pencil and trying to make calculations where car weights and lengths are all the same, because for most of us they're not....just like the real ones.
Wayne
We need a vector analysis to see why stringlining takes place after certain limits on curves of a certain radius and with the draw/resistance calculated for different weights of trailing scale cars...trailing tonnages.
And don't look at me. I'm retired and getting ready to plant seeds for germination in my greenhouse.
Jerry, thanks for taking the time it took to consider this and to post it. Nice work!
Hi Lastspikemike,
When we place two cars on a curved track, these two cars are not aligned. There is a small angle between them. We can see this more clearly if we place two passenger cars on a 24" curve. Because of this small angle, the front car cannot pull the car after with 100% of its pulling force. It actually only uses F*cos(t), where t is the angle between the two cars. For the example I gave, on a 40" radius curve and car length of 8", cos(t)=0.98.
So if we have one loco with 4oz pulling force at the front, assuming the loco is also 8" long, the force that pulls the first car is 3.92oz; we lose 0.08oz force. The force that pulls the second car is 3.92*0.98=3.84, we lose another 0.076oz, and so on.
But if we put two locos at the front, the pulling force from the locos is 8oz. The force that actually pulls the first car is 8*0.98=7.84oz, so we lose 0.16oz force. The force that pulls the second car is 7.84*0.98=7.68oz, so we lose another 0.16oz force. So the more power we put at the front, the faster we lose it (in absolute value). Because of this, one loco can pull 33 cars on 40" curved track with 2% grade; but two locos at the front can only pull 53 cars (because the force is lost so fast by its absolute value).
In prototype trains, the curve radius is so large so that we barely see any angle between two cars. So there is no such issue. This is more of a consideration in model railroad when our curves are sharper (as compared to our cars).
LastspikemikeWhy do you say it matters where in the train you pull it?
hjQiespecially when the radius becomes large.
but weren't you trying to account for tighter radii and coupler angles?
John-NYBW This is too much like math class for me.
This is too much like math class for me.
Rio Grande. The Action Road - Focus 1977-1983
gregci believe your value of 0.1% is too low and a better estimate is 2% effective grade of a curve is estimated to be 32/radius (e.g. 1.14% for a 28" rad). it also needs to account for flange resistance
I changed the Crr to 0.02, which I think might be too high for rolling. But anyway, I got the numbers that are very close to what you listed that table, especially when the radius becomes large. Great!
gregcrather hastely measured a pull of 2.4 oz for 30 cars ~4 oz each, total 120 oz. comparable to going up a 2% grade. still expected it to be a bit less. if max TE is ~20% of weight, a 12oz loco should be able to pull this train.
if max TE is ~20% of weight, a 12oz loco should be able to pull this train.
hjQiRolling resistance: Crr=0.001
i believe your value of 0.1% is too low and a better estimate is 2%
effective grade of a curve is estimated to be 32/radius (e.g. 1.14% for a 28" rad). it also needs to account for flange resistance
consider following for loco tractive force of 4oz and car weight of 4oz
rad effGr fPul #cars grade fPul #cars 0 0.00% 0.080 50.0 2.00% 0.160 25.0 28 1.14% 0.126 31.8 2.00% 0.206 19.4 32 1.00% 0.120 33.3 2.00% 0.200 20.0 36 0.89% 0.116 34.6 2.00% 0.196 20.5 40 0.80% 0.112 35.7 2.00% 0.192 20.8 50 0.64% 0.106 37.9 2.00% 0.186 21.6 60 0.53% 0.101 39.5 2.00% 0.181 22.1 100 0.32% 0.093 43.1 2.00% 0.173 23.1 200 0.16% 0.086 46.3 2.00% 0.166 24.0
This post might be long. So let me summarize first. For a loco with pulling force of 4oz, it can pull a long train (each car is 8” long, 4oz weight),
1. On a straight track with 2% grade, one loco can pull about 50 cars.
2. On a curved track (R=40”) with 2% grade, one loco can pull about 33 cars.
3. If we use two locos to pull at the front, the number of cars is about 53 cars.
4. To maximize the pulling power, we may want to put the second loco in the middle to pull 66 cars.
Now, we first consider a straight track (I did this in another post; but for completeness of this post, I include it here).
F=Crr*N*W=0.001*1000*W=4oz
So one loco can pull it.
For a track with a 2% grade, the total pulling force F is
F=Crr*N*W+G*1000*W=4+0.02*1000*4=84oz
The second term is due to weight contribution to direction along the track. In this case, we would need 21 locos. We also see on a 2% grade track, the rolling resistance is not that important. The main resistance comes from the gravity (or weight) contribution along the track slope, i.e., we need to lift the train higher, (or increase the potential energy as DrW used in his approach).
Now we consider the curved track. To make my calculation easier, I only consider one loco. Also, since Crr is not a major contributor on the graded track, I will ignore it. But it can certainly be added.
So now, a loco can pull 50 cars on a 2% graded straight track. When the track is curved, my key assumption is that the pulling force is lost slightly because the cars are not aligned straight.
For a track with radius of R=40”, filled with cars of L=8”, the number of cars on one lap of the track is
C=2*3.14*R/L=2*3.14*40/8=31.5
Then the angle between the two adjacent cars is
t=360/C=11.4 degree
Since a car is only pulled by the force along its direction, the pulling force from the previous car to the next car is reduced because of this angle. So if the pulling force after the m-th car is Fm, the pulling force on the m-1 car is
P= Fm* cos(t)= Fm*A
A=cos(11.4)=0.98027
So only 98% of the force from the previous car is transmitted to pull the car immediately after the previous car. Certainly, A depends on the car length. The shorter the car, the larger A is (smaller force loss), vise versa. For example, for L=4”, A=0.995; for L-12”, A=0.955.
Now, consider our loco. It has a pulling force of f=4oz.
The pulling force on the first car is
F1=f*A=4*0.98=3.920z
The pulling force on the second car after pulling the first car is
F2=(F1-G*W)*A= (4*0.98-0.02*4)*0.98=3.76oz
The pulling force on the third car is
F3=(F2-G*W)*A= (3.76-0.02*4)*0.98=3.60oz
We can use an excel sheet to do the calculation easily. When the pulling force is smaller than the force required to pull one car, we reach the maximum number of cars N_max. In the case discussed above (L=8”, W=4oz, G=0.02, R=40”, Crr=0.0), N_max=33. So we lose 17 cars due to the curved track.
From the above calculation, we can see the most important contributing factors to the pulling force drop are curve radius and car lengths.
If we maintain other factors the same and change the radius, i.e.,
L=8”, W=4oz, G=0.02, Crr=0.0
R N_max
28” 25
32” 28
36” 31
40” 33
50” 37
60” 40
100” 45
200” 48
If we maintain other factors the same and change the car length, i.e.,
W=4oz, G=0.02, R=40”, Crr=0.0
L N_max
4” 43
6” 38
8” 33
10” 28
12” 24
14” 21
Slope certainly matters,
L=8”, W=4oz, R=40”, Crr=0.0
G N_max
0.01 53
0.02 33
0.03 24
0.04 19
0.05 16
Crr can be included by slightly change the formula but seems to have less effects.
L=8”, W=4oz, G=0.02, R=40”
Crr N_max
0.000 33
0.001 32
0.002 31
0.003 30
0.004 29
0.005 28
Of course, increasing Crr may increase the pulling force of a loco, which may increase the number of cars.
Last thing, since the force drops by percentage, the larger the pulling force at the beginning, the larger the drop. For example, if we put two locos (8oz total) at the front, these two can only pull 53 cars, not 33*3=66 cars. This is because we have more force loss at the front cars. So the best strategy is to put one loco at the front, and one loco in the middle. This can maximize the number of cars.