# 2.0 Adhesion Limits and their Effect On Performance and Efficiency

**FORWARD**

The use of the term “adhesion” in the context of steel wheel on steel rail transport is an inaccurate use of that term. Pieces of steel with ordinary surface finishes do not adhere to each other in an empirical sense. When one steel piece is supported by another, both having relatively clean surfaces, a force is required to bring the supported piece into motion. This is a result of there being friction between the surfaces of the two pieces. The ratio between the weight of the supported piece and the force required to initiate its motion is the static friction coefficient, in the rail transport industry somewhat misnamed as the “coefficient of adhesion”. In the material following the second term will be used due to its general familiarity.

In rail transport applications all the horizontal force generated at the wheel-rail interface must be kept within the limits of the adhesion coefficient as exceeding that force will cause sliding between the wheel and rail contact surfaces, an undesired mechanical condition which will not be discussed further herein except as to how train performance is affected.

The adhesion coefficient is a number capable of considerable variation in the rail transport environment, and any deviation from a clean, burnished surface on either of the contacting surfaces will affect it in a negative way. This is a complex subject in its own right and in the following it is assumed that perfect surface conditions exist.

Further – In the following discussion, general “rules of thumb” are used which are sufficiently accurate to illustrate the concepts involved. An engineering-level analysis requires more precise numbers. For example, wind resistance has been ignored.**ABOUT BEING ADHESION LIMITED AND ENERGY EFFICIENT**

The basic physical relationship being used below is:

Horsepower (HP) = Tractive Effort (TE, lbs) x Speed (mph, ft/sec.).

A 10 car train of 60 ton coaches weighs 600 tons plus loco of 120 tons = 720 tons.

The loco weighs 30 tons/axle; using the reasonable adhesion level of 25 percent, the maximum drawbar pull (TE) will be 30 x 0.25 = 7.5 tons per axle, and with 4 axles = 60,000 pounds (30 tons). The percent weight on powered axles is 120/720 = 16.66 percent.

A 1 mile per hour per second (mphps) acceleration rate requires TE at 100 lb/ton. 720 tons requires 72,000 lb of TE, 36 tons.

Thus the acceleration rate can be no more than 30/36 mphps = 0.833 mphps.

To produce this 0.833 rate at 40 mph (40/60 x 88 = 58.66 ft/sec.) the loco has to produce 58.66 x 30 tons x 2000 lb/ton = 3,520,000 ft-lb/sec. Since 1 hp = 550 ft-lb/sec, the HP output at 40 mph is 3,520,000/550 = 6,400. To maintain the rate to 50 mph requires producing 8,000 HP. Which gives some idea as to what an ALP 46 loco might do, ie, 50 mph in 50/0.833 = 60 seconds.

To achieve the same rate at the same speed on a train consist of 60 ton trailer and 75 ton motor car, (M-T consist) having total weight of 135 tons, what is needed is 135 tons/720 tons x 8,000 = 1,500 hp, which is 1,500/4 or 375 hp/axle. Which is a practical value from a mechanical and electrical standpoint.

The TE required to produce a rate of 0.833 mphps on a weight of 135 tons is 100 x 0.833 x 135 = 11,246 lb.

By setting the consists to have equal acceleration at a specific speed both consists will have the same HP per ton, - 11.1111, - and performance above 50 mph will be identical. It is below 50 mph that differences in performance can occur due to differences in percentage weight on powered axles.

What is known as Base Speed is the speed at which power is being supplied to the motors at a value equal to their rating. This speed will be a function of the acceleration rate (TE/ton) being produced. Below Base Speed the consist is in constant TE/ton (constant acceleration rate) mode. Above Base Speed, the acceleration rate (TE/ton) declines with increasing speed. In the example above, the 8000 HP loco is reaching Base Speed at 50 mph.

(As a check on the above calculation, using the M-T set numbers, that rate (0.833) at 50 mph (73.33 ft/sec) from TE of 11,246 lb is work at the rate of 73.33 x 11,246 = 824,669 ft-lb/sec. HP is 824,669/550 = 1,499.4 hp, as above.) Because of the greater percentage of weight on powered axles, which is 75/135 = 55.5 percent for the M-T consist, (vs 16.66 % on the loco hauled train), at lower speeds the TE/ton can be made higher. The 25 percent adhesion maximum acceleration rate possible for the M-T consist will be 0.555 x 22 x 0.25 = 3 mphps.

Overall, performance is determined by HP/ton. For the loco hauled train it is 8000/720 = 11.11 HP/ton. For the M-T set, it is 1500/135 = 11.11. The values are the same because we have given both consists the same acceleration value at the same speed. So, overall upper speed range performance is comparable, but the M-T set has the advantage of much higher initial rate capability (3 vs 0.833 mphps), and so has a quicker time to 50 mph. (calculating this is more complex, as the relationships are not linear.)

The loco train, being adhesion (weight on drivers) limited will be restrained in rate (0.833), therefore will have a constant rate (TE/ton) to 50 mph, where it will be developing its rated HP of 8000, after which the rate will start decreasing so that the product of rate times speed will decline from 8000 hp. In traction jargon we say that 50 mph is the Base Speed above which HP output will decline. Base Speed is a variable value inversely related to rate, ie, a lower acceleration rate (TE/ton) will produce a higher Base Speed, and vice versa. But it will always occur at the speed at which the traction system is developing its rated horsepower. **(1)**

The M-T set will be free to accelerate at 3 mphps until it reaches a speed where it is developing its rated 1500 HP, after which it also will accelerate at lower rates with speed until it is at 0.833 at 50 mph, above which it will have a performance curve the same as the loco hauled train.

The constant rate of 3 mphps can be maintained to the speed where total TE times speed equals 1,500 hp. A 3 rate on 135 tons requires 300 lb per ton of TE = 135 x 300 = 40,500 lb TE. Since HP = ft/sec x TE, we can solve for speed. 1 hp = 550 ft-lb/sec. 1,500 hp = 825,000 ft-lb/sec. TE is 40,500 lb. Then base speed is 825,000/40,500 = 20.37 ft/sec, = 13.8 mph. This sounds low, but the M-T consist gets there in 5 seconds, including jerk limited build up. (approx.)

At this speed the transition from constant rate to declining rate (TE/ton) and HP occurs. But at all speeds below 50 mph the M-T consist will have a higher rate of acceleration.

Example - The M-T consist acceleration rate at 25 mph would be lower. 25 mph = 36.67 ft/sec. HP = 1,500, = 825,000 ft-lb/sec. TE = 825,000/36.67 = 22,498 lb.

Rate = 22,498lb/135 = 166.6 lb/ton = 1.66 mphps, still twice the rate of the loco hauled train.

The above is based on each type of train having a fixed HP/ton rating. Because the M-T consist is not adhesion limited, it is possible to push the traction equipment and thus maintain higher rates to higher speeds, so the above is understated. That can't be done with a loco hauled train because it would result in wheel spin. One could maintain a 25 percent adhesion demand all the way to 50 mph , but that would require far more HP. It might be possible to build a M-T set with 500 hp per axle, but given space limitations........

For simplicity of comparison, in the above both types of train were given the same level of power in HP per ton of 1500/135 = 11.11. For comparison below are some other approximate values of HP/ton:

Birney streetcar - 50/6.5 = 7.7

Standard 1920s streetcar - 140/20 = 7.0

PCC streetcar - 220/17 = 12.9

Typical NY IRT subway car, 1970s - 400/35 = 11.4

Typical North Shore Line car - 560/50 = 11.2

P & W bullet car – 400/26.2 = 15.23

Silverliner MU cars, 1960s - 600/52 = 11.5

Norristown line N5 car - 820/38 = 21.6

Dynamic braking brings some other considerations into play. As a ballpark number, the added work that dynamic braking imposes on the traction motors when providing full service braking requires that they have more HP rating than required for the acceleration achieved. That's because HP is a rate of doing work, in this case converting electrical energy into the energy of motion (kinetic) and (vice versa) kinetic energy into kilowatts. In the process of doing so, some of the energy becomes heat in the windings of the motor. The amount of HP output that the motor can produce either mechanically or electrically while staying within the temperature rating of the motor's insulation - which in turn is determined by the ability of the machine to dispose of the heat generated - determines the HP rating of the motor. The degree to which motor HP needs to be increased when dynamic braking is added will depend upon station spacing. The longer the station spacing the more heat is removed from the motor before it is again required to handle maximum power. In streetcar service with 10 stops per mile, motor HP rating needs to be increased by 20 or more percent. The term used in this respect is the Duty Cycle imposed on the motor.

Caveat - In the following the contributions of air resistance and mechanical losses to brake rate are disregarded in the interest of simplicity. As a rule of thumb, these factors can contribute up to 10 percent to brake rate, with the maximum occurring at high speed on single car trains and decreasing with longer trains and lower speeds. Also, the discussion is predicated on employment of electrically powered trains.

In accelerating the traction system produces a mechanical output. In braking the traction system produces an electrical output. The relationship between the two types of output is that

1 HP can produce 550 ft-lb of work in a second, such as lifting 550 lb one foot in one second (or any other combination of pounds and feet whose product is 550), and that a motor will require supply of 746 Watts of energy to do that work in one second. The power used will be 746 Watt-seconds. Similarly, generating 746 Watts of energy per second will require a 1 HP machine.

When dynamic braking is employed to produce the same brake rate as the acceleration rate the motors have to produce power at levels higher than their one-hour rating. **(2)**

For example, using the numbers from above, if it takes a 1,500 hp traction system to produce a 3 mphps brake rate at 13.8 mph, it takes a 3000 HP generating capability to produce that brake rate at 27.6 mph. (a two to one ratio is a practical limit) Above that speed the dynamic brake rate has to be tapered lower to stay within the traction equipment's voltage capabilities. Therefore at twice the speed, the rate will be half, ie, at 27.6 x 2 = 55.2 mph the rate will be 1.5 mphps. If it's desired to maintain a constant 3 mphps brake rate above 27.6 mph it is necessary to add in some friction brake. If a lower full service brake rate is acceptable, such as 2 mphps, there will be no need to add friction brake unless speed is above 41.66 mph. From an efficiency standpoint, it will be possible to put more energy back into the supply line with the lower rate as the friction brake will be less used, wasting motion energy as heat, and the power system will have greater acceptance of a lower value of regeneration; both lead to greater energy efficiency.

The greater HP/ton capability the train possess, the greater will be its ability to use almost maintenance-free dynamic braking. And the greater will be its ability to return power to the supply system during busy periods when the degree of acceptance is highest. However, if the installed HP is provided in a way that adhesion limitations cap the amount of regenerated power that can be produced, then equivalent regeneration efficiency can only be produced by utilizing a brake rate no higher than the maximum acceleration rate, with high speed limitations as discussed above.

In overly simple terms, higher energy efficiency occurs with lower schedule performance, and vice versa. Performance that achieves the minimum running times that attract passengers and provides for maximum train throughput at peak periods is at odds with maximum energy efficiency unless there is a high degree of regeneration capability. A systems approach is needed to determine which is the most important parameter.

For high speed trains making widely spaced station stops the percent of weight on driven axles has much less importance. At high speed most of the power output of the powered axles is used to overcome air resistance. This energy loss cannot be recovered, therefore regeneration capability is of much less importance and percent of weight on powered axles needs to be no more than is required to start a chosen maximum train weight on the maximum grade of the route under the lowest expected adhesion coefficient conditions.

For trains operating on relatively short station spacing regeneration will be a much more useful capability. To provide a maximum recovery of motion (kinetic) energy and a minimum waste of energy as heat during braking the percentage of weight on powered axles needs to be much higher and higher installed HP/ton traction equipment is required.

While it is possible to equip both types of trains with the same braking capabilities in terms of mphps rates, for a given HP/ton level of performance the consist with the greatest percent of weight on powered axles will have the greater regenerative capability. But the station spacing will be the determining factor as to whether that capability is fully utilized. (The assumption is that the propensity of the electrical supply system to accept regenerated power is the same for both cases, which is not always true and cannot be ignored in any evaluation.) A final conclusion as to what is the optimum consist configuration should consider all the factors involved, with the individual factors weighted in accordance with the primary goals of the rolling stock application.

R. E. Jackson, September 31, 2014

**Notes****(1)** With DC motors base speed is the value achieved while maintaining 100 percent field strength. When base speed has been reached -which is the speed at which the motors are receiving full line voltage -the motor's HP output can be increased by weakening the motor's field strength by use of field taps or field shunts. Instantaneously the motor current increases. Since output power is always the input power less the losses -which are constant -the increased input power (Volts (constant) x Amps) must result in increased output power and greater acceleration. HOWEVER, to subject the motor's commutator to such a surge of current would make the motor very prone to flash-over as commutators have a current versus speed limitation. Therefore following achievement of base speed motor current is allowed to decline to what is considered a safe value at which field strength reduction can be applied. At all speeds above base speed the weakened field motor will produce more HP than a full field motor can, producing a higher balancing speed for the vehicle, balancing speed being the speed at which motor HP output balances the drag of wind, grade, rolling resistance, etc. With AC induction motors the commutator limitations are replaced by what is known as the slip curve limitation. In order to produce torque (which becomes TE) the rotor must rotate at a speed below that of the magnetic field rotating within the stator. (as generated by the inverter providing the stator with electrical current) The frequency of the AC currents generated in the rotor because of the “slip” must be held within limits beyond which the torque decreases rather than increases. There is no “weak field” condition as created in DC machines; the software in the inverter control must adhere to the motor's characteristics once the motor is receiving its maximum AC voltage as it advances stator frequency to achieve higher speeds.**(2)** The higher power output in dynamic braking is achieved by increasing generated voltage, as a current-only increase would impose greater heat losses on the motor and require the brushes and commutator be sized for the greater current, which is counter to making the motor small and light. The technique used with compact series DC motors is to construct a 600 V motor that is able to provide the desired TE and HP at 300 V when accelerating and can reliably generate 600 V in braking. Two motors, permanently series-connected, will generate up to 1200 volts. Because of this, regeneration to the power system is not possible, and the generated power is disposed of as heat in a resistor. Braking using 600 V motors has been used in a number of circuit configurations. The motors must still generate the needed power either through higher voltage or current, requiring a physically larger motor, which is at odds with the goal of producing a car with a minimum of floor height. Use of a single large motor driving both axles of a truck has been a successful way of accomplishing this, albeit with some weight penalty. DC motors whose input and output are controlled by power electronics, as with a chopper, can be either 300 or 600 V rated machines, and capable of generating voltages as above, but because the chopper controls the motor(s) voltage in braking, it is possible to deliver the generated power back into the power system at a voltage level safe for all users. Motors will be connected in parallel, rather than in series due to the line voltage limitation.

With AC induction or permanent magnet motors one builds a machine and a four-quadrant (DC-AC & AC-DC) inverter with the ability to deliver the HP and TE needed for both acceleration and braking. The basic principle involved is that when an AC motor is fed a frequency corresponding to a speed lower than that existing the motor will generate power. The inverter will control motor voltage such that it can be returned to the supply. AC motors do not have the current limitations at high speeds that DC motors have, so can be smaller in size and weight.

As with DC motors, AC motor circuits contain resistors that can dispose of generated power as heat when the power system cannot accept all or a part of it.