# Would a “lifting body” design be useful for a train car?

Lifting Body

The idea of a lifting body design is to shape a vehicle's body in such a way as to produce lift without wings. Research has shown that this can be an efficient method of reducing drag while still providing lift.

This has typically done for aircraft or spacecraft:

Could a similar approach help to make trains more efficient?

Passenger trains already look streamlined and aerodynamic:

Freight trains do not:

Air resistance is not the only form of resistance that trains must overcome. They must also overcome the resistance of their wheels on the track. This is where I would think that the energy savings from a lifting body design would come from. Any additional lift created by the lifting body design would reduce the friction between the wheels and rail, thus saving energy.

Train wheels have flanges on them, so they don't need traction to steer. The cars also don't need traction on the rails since the driving wheels are only on the engines.

Could a lifting body create enough lift at typical train speeds to make a noticeable difference?

• You would still need friction in order to be able to break. As far as I know, thrust is "concentrated" in the locomotive(s), but breaking capability is distributed across all cars in the train. – Nick Alexeev Feb 26 '15 at 3:09
• For a lifting body substantial enough to create a noticeable effect, I would think that derailment could be an issue. – HDE 226868 Feb 26 '15 at 3:48
• Note that "typical train speeds" varies hugely by train, track and cargo. Inclined curves call for greatly reduced speeds to prevent, e.g., spilling 20 thousand gallons of herbicide into a major waterway. I think quantifying those "typical speeds" is the first order of business as it may answer your question right off the bat. – Air Feb 26 '15 at 5:11
• Also, a number of modern passenger trains are electric multiple units, in which the driving wheels are spread throughout the train. Even with passenger trains, a lifting body is often a bad thing. – cpast Feb 26 '15 at 10:01
• How would you go about shaping a train to provide lift? The forward facing area is tiny, and the width available for winglets is tiny. – Jon of All Trades Feb 26 '15 at 20:27

I won't say it's not possible to make a noticeable difference. But I would say that it's pretty darn unlikely.

The lift and drag forces on any body generally depend on the velocity of the body ($v$), the density of the fluid ($\rho$), the area of the object ($A$) and a dimensionless coefficient ($C_L$ or $C_D$). The applicable area can be a bit fuzzy in terms of whether you're talking about frontal area or planform area, but for a train this is going to basically be the area in the plane normal to the direction of travel (given that most of the train cars are simply in the wake of the locomotive). The coefficients can vary a bit with velocity due to turbulence, but will generally be between 0 and 2. The functional form is:

$$F_L = 1/2\rho A C_L v^2$$

Crudely estimating a freight train as being $4\, m$ tall and wide, moving at $16 \, m/s$ ($57 \, kph$). This gives us that lift will be around $2500 N$ for the entire train (not enough to lift an ordinary automobile). This calculation is not meant to be extremely precise, but even it's off by a factor of ten, lift would be minimal compared to the overall weight of the vehicle.

Given that, you might consider the merits of adding additional aerodynamic elements. Here you get into a tradeoff between generating lift (to reduce downward force on the wheels) and generating drag. While lift-to-drag ratios can get high (~50 in good cases) the rolling resistance of train wheels is actually very low ($C_{rr} \sim 0.00035$). So although the lift will reduce the losses due to friction, it will increase the losses due to drag more.

So, reducing drag is good but providing lift is probably not worth it. In fact, reducing drag at the cost of adding some weight could be beneficial because wheels are so very good at bearing that load efficiently.

N.B. I consulted Wikipedia and engineering toolbox for $C_{rr}$ data. I could not verify sources for railroad wheels, but was able to confirm that values for road tires were correct via "Fundamentals of Vehicle Dynamics" by Gillespie. Any ideas on verification would be welcome.

• Good to see the basic drag equation surface. A most useful formula in all sorts of area despite its simplicity. Add another factor of V and you get total power loss due to drag. Here =~ 40 kW (2kN x 16 m/s). That suggests that unlike small (wrt train) motor vehicles where it predominates, here the windage is a minor loss factor for a train. BUT getting weight (mg) off the wheels is as you note going to take lift per carriage. Interference amongst numerous serially adjacent wagons makes this "very very very hard". .... – Russell McMahon Feb 26 '15 at 5:46
• .... Looking at aircraft wing loadings suggests that even if clear air could be provided the tunnels would be interesting in appearance :-). – Russell McMahon Feb 26 '15 at 5:46
• What may be an interesting revelation for you (it was for me when I derived it long ago) or may be more obvious than it was for me. | For a rectangular cross section (to make life easy) A and trains velocity V, take the volume of air that the train sweeps aside in one second (A x V). Accelerate the mass involved (A x V x Rho) from 0 to V in the time available to get it out of the trains way. Calculate the power required to do this. Far out (I thought). Compressibility and drag coefficient and all the magic numbers complicate this but the basic result is gratifying. – Russell McMahon Feb 26 '15 at 5:50
• Just a quick note, even if your figure of 0.00035 is correct, trains have hundreds if not thousands of wheels. I think the typical car has 8 wheels, and there can be over 100 cars per train. For freight, the number of wheels per car may be even greater. So the incentive for reducing rolling friction to increase cruise efficiency is still there. But I have to agree with your analysis: lifting force, either from body or wings, is not viable for a train. I may post my own answer to add to this. – DrZ214 Feb 22 '16 at 22:31

You don't want to decrease the downward force on the wheels because that is your primary means of slowing down in an emergency. It already takes a while to stop a train, don't make it any longer. Remember that kinetic friction (wheels locked) is proportional to the downwards force on the wheel and the brakes are spread along to each car.

Train wheels have flanges on them, so they don't need traction to steer. The cars also don't need traction on the rails since the driving wheels are only on the engines.

They do need traction, wheels are slightly conical with the radius at the outside being larger than the inside and they are both coupled to the axle. This way when the train is off-center or in a turn the outside wheel has a larger effective radius thus will turn it back to the center of the track.