Pulling power

[Geeraider]

 does anyone know the formula for pulling power for HO scale locomotives?

[rich1998]

Which particular HO locomotive? They all vary. Too general a question.

Rich

[Trainman203]

Its like trying to get a formula to see how fast you can go on a bicycle. 😂

[Terry Toenges]

Can you derive a formula from my old Bachmann HO old time 4-4-0 pulling 29 cars with a 2% grade starting about where the two red flats are in front of the green box car with yellow doors are?

[rich1998]

Yes, I remember my Spectrum 4-4-0 and 4-6-0 could d pull quite well but that was ten years ago.

Rich

[jonathan]

There is such a thing called a "tractive effort calculation". The equation is a bit long and has to do with the locomotive weight on the wheels, resistance, and so on. If u google it...

I don't know that the math would translate well down to model size. However, there is a battery operated tool you can buy, that measures how much weight a model locomotive will pull before wheel slippage. I just can't remember it's name.

Regards,

Jonathan

[Trainman203]

Of course.  The prototype locomotive manufacturers need to know so as to be able to market their product. You could call up EMD or another of the locomotive companies and tell them that you've inherited a short line, don't know anything about trains , want to buy a new engine, and ask them for the formula.

😂😂😂😂😂😂😂

[Len]

A digital fish scale that reads in ounces should show how much pull a loco has before wheel slip. A more basic rule is pulling power is equal to 25% of the loco's weight. The kicker being, how much do the cars weigh and how free rolling are they that it will be pulling?

If you want to go crazy with the math involved, this is from an old MR comment:

Caveats
If you believe you are mathematically challenged... consider stop reading now.
Make notes... I will delete this post in a few days.
Following uses information based on testing by others and my own testing. Your results may vary depending on the controlling factors of your test environment.
1 inch of elevation to flat run of 100 inches of track = 1.0% (0.01) grade.

Rolling Resistance
Rolling resistance for real railroads is about 0.2%, for model railroads it is more like 2.0% i.e. a model railroad car will roll down a 2% grade on its own. This factor must be taken into consideration when calculating tonnage ratings.
Curve Resistance
Another factor is curve resistance, which is .04% per degree of curve for the real railroads and about the same for model RRs. The big difference is model RR curves are much sharper and this resistance is a very big factor. It is equivalent to around a 30%/R where R = curve radius in inches. An 18-inch radius curve will produce about 1.7% additional resistance. Most mountain model RRs are all curves, so compensating for them usually isn't worth the time, but if you have to cheat on the ruling grade, do it on tangent track.
Locomotive Traction
An all-metal wheel model locomotive will have about a 20% adhesion factor just like the real ones. This means 20% of the locomotive weight can be turned into pulling force (locomotive weight * 20% = estimated Drawbar pull).
So how many 4-ounce cars will a 12-ounce loco pull up a 3% grade with 18" radius curves?
The equation 12 * 0.20 = ((4*N)+12) * (.03 + .02+ .017)
where 0.03 is the grade, 0.02 is the rolling resistance, and 0.017 is the curve resistance (see above)
Answer: N = 6 cars (5.955)
Use actual drawbar pull figure in place of L*E
L*E is a way to estimate drawbar pull
More accurate equation
L*E = (C*N*(G+E+R))+(L*(G+R))
L is Locomotive weight in ozs.
E is Rolling resistance factor of 0.02
C is Car weight in ozs..
N is Number of cars
G is % track grade
R is curve resistance determined by 0.3/r where r is curve radius in inches
12*0.20=(4*N*(0.03+0.02+0.017)) +(12*(0.03+0.017))
N= 7 cars (6.851)

Len

[Trainman203]

Oh man, that's good. HAWHAWHAW!

Going to create a variable-ready spreadsheet to keep track of all my 80+ engines!  Because there's not much else to do right now!

One thing is wrong with that equation, because it's old.  It says model railroad cars won't roll down a 2% grade.  That dates from pre-needle-point-axle, pre-delrin-plastic-truck days, when truck frames were die cast Zamac metal with live springs and axles were often much like the prototype.  The cars I have now, if a gnat sneezes next to one end , it'll roll.  Leveling a layout with switching and set outs is demanding.

[rich1998]

Interesting discussion. The OP has not come back yet.

Rich

[Trainman203]

No he hasn't because we probably overwhelmed him with minutiae when all he wanted was " yes here it is .... (name of formula and link)", or "no, there isn't."

[rich1998]

I have lost track of how many post I have seen over the years where many wondered why they could not figure out why their HO loco would not pull many cars, some times ten or twelve cars.

Rich

[bapguy]

Buy this: https://www.makersource.store/products/digital-pull-meter        Joe

[Hunt]

A digital fish scale that reads in ounces should show how much pull a loco has before wheel slip. A more basic rule is pulling power is equal to 25% of the loco's weight. The kicker being, how much do the cars weigh and how free rolling are they that it will be pulling?

If you want to go crazy with the math involved, this is from an old MR comment:

Caveats
If you believe you are mathematically challenged... consider stop reading now.
Make notes... I will delete this post in a few days.
Following uses information based on testing by others and my own testing. Your results may vary depending on the controlling factors of your test environment.
1 inch of elevation to flat run of 100 inches of track = 1.0% (0.01) grade.

Rolling Resistance
Rolling resistance for real railroads is about 0.2%, for model railroads it is more like 2.0% i.e. a model railroad car will roll down a 2% grade on its own. This factor must be taken into consideration when calculating tonnage ratings.
Curve Resistance
Another factor is curve resistance, which is .04% per degree of curve for the real railroads and about the same for model RRs. The big difference is model RR curves are much sharper and this resistance is a very big factor. It is equivalent to around a 30%/R where R = curve radius in inches. An 18-inch radius curve will produce about 1.7% additional resistance. Most mountain model RRs are all curves, so compensating for them usually isn't worth the time, but if you have to cheat on the ruling grade, do it on tangent track.
Locomotive Traction
An all-metal wheel model locomotive will have about a 20% adhesion factor just like the real ones. This means 20% of the locomotive weight can be turned into pulling force (locomotive weight * 20% = estimated Drawbar pull).
So how many 4-ounce cars will a 12-ounce loco pull up a 3% grade with 18" radius curves?
The equation 12 * 0.20 = ((4*N)+12) * (.03 + .02+ .017)
where 0.03 is the grade, 0.02 is the rolling resistance, and 0.017 is the curve resistance (see above)
Answer: N = 6 cars (5.955)
Use actual drawbar pull figure in place of L*E
L*E is a way to estimate drawbar pull
More accurate equation
L*E = (C*N*(G+E+R))+(L*(G+R))
L is Locomotive weight in ozs.
E is Rolling resistance factor of 0.02
C is Car weight in ozs..
N is Number of cars
G is % track grade
R is curve resistance determined by 0.3/r where r is curve radius in inches
12*0.20=(4*N*(0.03+0.02+0.017)) +(12*(0.03+0.017))
N= 7 cars (6.851)

Len

Yes Len I am old and you copied my post from a few years a go!
Hunt

[Hunt]

. . .

One thing is wrong with that equation, because it's old.  It says model railroad cars won't roll down a 2% grade.  

. . .

Trainman203,  NO it does not say a railroad cars won't roll down a 2% grade.