19/03/2025
Engine braking for hill descent
It has long been recommended practice to adopt a low gear on descent of steep hills, a process known as engine braking. In classic cars fitted with drum brakes this is deemed essential otherwise the brakes may overheat, greatly reducing their effectiveness. But to what extent is this still true for a modern car fitted with disc brakes? In this article I attempt a rough order of magnitude answer to this question.
Conventional brakes operate by converting energy into heat. Stopping a car from speed converts kinetic energy, the energy of motion, to heat, while descending a hill converts potential energy to heat. (Electric cars use regenerative braking where much of the energy is converted back into electrical energy that is stored in the battery - a much more 21st century approach - but that topic is beyond the scope of this article. We are just concerned with so called friction brakes.)
Kinetic and potential energy are different manifestation of the same thing and they are interchangeable. An excellent example of this is a pendulum where the energy is all potential at the ends of the swing, all kinetic in the middle, and a varying mix at all points in between. The total energy, potential plus kinetic, is constant.
As mentioned earlier the main limitation of friction brakes is that they can get too hot! The discs and pads have a maximum temperature beyond which their effectiveness diminishes. Engine braking helps by diverting some of the heat away from the brakes to the engine which is equipped with a radiator and fans to dissipate excess heat.
Is there a simple way of putting some numbers on the capacity of friction brakes without resorting to materials science? Well, I think there is based on practical experience on IAM skills days and track days. (IAM skills days have been described in previous articles in this magazine. Briefly they are an opportunity to explore the handling characteristics fo your car in a safe environment - Thruxton circuit.)
First let’s introduce the simple equations that allow us to calculate the potential energy (PE) and kinetic energy (KE) of a vehicle.
PE = mgh 1) KE = 0.5mv2 2) where
m is the mass of the vehicle in kilogram (kg) - I’ll assume 1,500 kg
g is the acceleration due to gravity, generally taken to be 9.81 m/s/s
v is velocity in metres per second (m/s)
h is the height of the descent in metres (m)
KE and PE are in units of Joules (energy)
It might be helpful to remind you that a Joule per second is one Watt, or putting it another way one Watt for a duration of one second is a Joule.
Using equation 1) above we can easily calculate the energy released during a hill descent, initially without distinguishing between friction and engine braking. From how high a hill are we going to descend? Let me suggest three examples that we should consider.
Leith Hill (294m)
Ben Nevis (1345m)
Sierra Nevada (3353m)
Leith Hill is the highest point in Surrey, Ben Nevis the highest point in the UK and the Sierra Nevada has the highest road in Europe. (I’ve been up the Sierra Nevada and recommend it if you’re ever in southern Spain near Granada. You can’t get the the actual peak because it’s a nature reserve, but the public can drive to more than 10,000 ft.)
For simplicity I assume that you descend from the peak down to sea level on a constant downward slope of 10 degrees, which is pretty steep. Starting with Leith Hill equation 1) tells us that the energy to be dissipated is 4.3 MJ (4.3 million Joules). Over what period is this energy dissipated? On a 10 degree slope the distance travelled along the road will be 1693 m. At a speed of say 30 mph this will take 126 seconds. It follows that the average power is 34 kW, an enormously high value! For comparison a domestic electric fire will be no more than 3 kW. It should be stressed that the 4.3 MJ of energy is unavoidable (due to the laws of physics) and the only way of reducing the 34 kW is by slowing down.
Doing the same sums for Ben Nevis and Sierra Nevada produce the following results.
See the first table below
Most striking is the fact that the power dissipation required is the same in all cases. The reason is simply that for the assumed fixed slope of 10 deg the time to descend the hill is directly proportional to its height - you have proportionately longer time to dissipate the heat produced. Similarly reducing the speed of decent reduces the power dissipation required. For example reducing speed from 30 to 25 mph reduces the power from 34 kW to 29 kW.
At this point we can return to the question of how much of this power is dissipated in the friction brakes and how much by engine braking. Before proceeding further let me first explain, for those who may not be aware, how engine braking works. Internal combustion engines these days are so called four-stroke, because there are four strokes of each piston in a cycle, namely induction, compression, power and exhaust.
induction - open the inlet valve(s) and suck in the fuel air mixture
compression - close all the valves and compress the mixture at the top of the cylinder
power - ignite the mixture and force the piston down the cylinder
exhaust - open the exhaust valve(s) and let the piston push out the burnt mixture
In a petrol engine the majority of engine braking is produced due to the vacuum produced in the inlet manifold when the throttle is closed (i.e. when your foot is off the accelerator). There’s no air available to be sucked in and so during the induction stroke the piston has to work against a near vacuum above it. The compression stroke has little effect as not much air gets into the engine to be compressed and in any case any compressed air pushes back during the following downward stroke. Engine braking in a diesel engine is very different. There is no throttle hence no inlet vacuum. As in a petrol engine, energy stored in the compression stroke is returned during the “power” stroke, so in its standard form a diesel engine produces very little engine braking. As a result of this what follows only applies to petrol engines.(Many HGVs have special mechanism called a compression release brake which alters exhaust valve operation so that the engine acts as power absorbing air compressor.)
In Appendix 1 it is shown how you can “calibrate” the engine braking levels of your car. In summary, for a given steepness of hill there will be one gear ratio at which the engine braking holds the car at a steady speed, neither slowing nor speeding up. With discrete gear ratios as in a conventional manual gearbox (as distinct from continuously variable gearbox) it may be necessary to estimate the gear ratio which matches a particular hill.
I happen to live close by an approximately 10 degree hill and I know that I can descend at a constant speed if I select 2nd gear - no friction braking at all, braking done entirely with the engine. A simple sum, shown in Appendix 1, indicates that the engine is producing a retarding force of 2,555 Newtons.
But what if I left it in a higher gear? From information in my car owners’ manual I can see that the difference between 2nd and the other gears are as shown in the adjacent table, which also shows how the retarding force of the engine is reduced in higher gears. Using this information it is possible to calculate the amount of friction braking required for a descent.
Generalising the specific examples given above to a range of slopes and available gears leads to the graph below which shows the friction braking power required for slopes between 10 and 20 degrees and gears 2 to 6. The graph applies for a speed of 30 mph, but can be scaled linearly for other speeds - twice as fast, twice the power.
See the 2nd table below, with graph
You can see that, as expected, for a 10 degree slope and 2nd gear no friction braking is required. At the other extreme in top gear on a 20 degree slope about 55kW needs to be dissipated.
Modern warning signs for steep hills indicate the approximate steepness as a percentage. The highest I recall seeing in the UK is 20% which equates to 11.5 degrees. The famous Stelvio Pass has an average slope of 7.5% with a local maximum of 13% (7.5 degrees). According to Mr Google the steepest road in Surrey is Barhatch Lane at up to 21%. All this is saying that in practice you’re unlikely to move far from the left side of this graph. Slopes of around 20 degrees are extreme and likely to be short.
We now want to compare the rigours of downhill with on track braking. I regularly participate in track days at a local circuit. From many laps I know that my brakes are satisfactory (no noticeable reduction in braking effectiveness) under the following conditions
two braking efforts per lap from 160 down to 40 mph
120 seconds lap time
laps repeated for > 20 minutes
Using equation 2) for KE reveals that the heat energy generated per lap is 7.2 MJ (about seven million Joules). This energy will be absorbed by three mechanisms namely,
friction braking,
engine braking and
air resistance
You might expect that air resistance would be significant at the speeds mentioned, but in Appendix 2 I show that in fact it is negligible. Braking is completed in 6th gear so we can use the 939 N engine braking force that was calculated in Appendix 1. The remaining braking effort must be due to the friction brake, the value we want to estimate. Appendix 3 shows the calculation which reveals that 93% of the energy is dissipated in the friction brake and hence only 7% is due to engine braking, as might be expected in top gear. This results in an average power dissipation over a lap of 56 kW in the brakes. Of course the temperature of the brakes will peak at the end of each braking event and then cool during the remainder of the lap. From the fact that the brakes can sustain this performance for over 20 minutes without noticeable loss of performance it is reasonable to conclude that they can dissipate about 56 kW of heat (by radiation, conduction and convection.)
Comparison with the graph above shows that for all reasonable slopes and for a sensible descent speed there is ample capacity in the brakes irrespective of the gear selected. From this rough order of magnitude calculation I deduce that there is no imperative to use engine braking during hill decent (at least with a petrol engine).
So after all this effort let’s summarise and draw some conclusions.
Engine braking only operates on the driven wheels, usually two, whereas of course the friction brake operates on all four. That means, contrary to what you can read on the internet, that use of the friction brake is less likely to produce a skid: the braking force is spread over four contact patches not two.
I’ve never owned a diesel car but, from what I’ve read and tried to explain above, they produce a lot less engine braking than does an equivalent capacity petrol engine. They seem to survive around the world without cooking their brakes. This tends to support the idea that engine braking isn’t essential. If anyone knows more about engine braking in diesel cars I’d like to know! Are diesels in cars modified to produce additional engine braking?
Recall that the above graph assumes the car travelling at 30 mph. Reducing speed will reduce the braking power required proportionately so if in doubt reduce your speed of descent.
Some people like to use engine braking because it reduces wear on the pads and discs. Ok, but remember you are putting additional wear on the engine due more engine revolutions and note which is cheaper to replace.
At night engine braking will reduce the amount of brake light glare for following vehicles.
You may wonder how the results presented above relate to your car. I expect that for a modern, manual gearbox, petrol engined saloon car the results will be reasonably representative.
Remember that use of engine braking is a traditional IAM technique. On your test if there’s a significant hill your examiner will expect you to demonstrate the technique.
Perhaps the real question is not whether but how much engine braking should be applied, in other words which is the best gear for a particular hill. You might conclude from the above graph that slopes of around 10 degrees (~ 20%) degrees hardly need any action. On the other hand for very steep hills, especially around 20 degrees selection of a lower gear would be very wise.
