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Consider these two situations when driving on a long straight road uphill:

  1. Starting at a high velocity $v_h$, which the car is able to maintain.
  2. Starting at a lower velocity $v_l$, and then trying to reach $v_h$ while driving uphill.

In my experience I've noticed that in case 2 it is very hard, and sometimes impossible, to reach the velocity $v_h$, even though if the car had started in that velocity it would have been able to maintain it. This observation was confirmed by another person I know.

If we do a simple analysis of the problem assuming the engine outputs some fixed power $P$, it seems that there should be no problem reaching $v_h$.

Is there something in the inner workings of the car (like the transmission or fuel injection for example) that would make it harder than expected to accelerate uphill?

For simplicity let's assume the car is always in the same gear.

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I'd say that the newtonian-gravity is not relevant. –  yohBS Feb 7 '13 at 15:16
I agree, @Qmechanic added this tag. –  Joe Feb 7 '13 at 15:32
For an infinitely long uphill you would be able to reach $v_h$ eventually. The question becomes then, why does the distance needed to reach $v_h$ increases "exponentially" with the uphill angle $\theta$. –  ja72 Feb 7 '13 at 17:38
@ja72 - I'm not sure I would be able to reach $v_h$ eventually (in the answers below there are some possible reasons why I might not be able) –  Joe Feb 7 '13 at 21:11
Here is the macro reason why it will reach $v_h$. The car will always accelerate until it reaches its terminal velocity which I assume is greater than $v_h$. Note that the top speed is less the more the incline. This is acceleration is low, but positive. The governing equation is $a=\frac{P}{m\,v}-\beta v^2-g\,\sin\theta$. –  ja72 Feb 8 '13 at 0:03

5 Answers 5

up vote 7 down vote accepted

Short, short version: It's complicated.

Slightly longer version:

Internal combustion engines have at least two relevant performance characteristics: power and torque. Furthermore the maximum attainable values for both are functions of the current engine speed (RPM).

Acceleration will cease if the current requirement for either power or torque equal the engine's maximum value at the current speed.

Both the power and torque curves (as a function of RPM) start low, rise steadily and eventually turn over and drop off. The requirements for power and torque are both monotonically increasing, which means that there must be a speed where the power requirements curve crosses the power curve. At that speed acceleration drops to zero.

Likewise, there must be a place where the torque requirement crosses the available torque and again, you can't accelerate further from there.

These two crosses can come at different engine speeds.

The result is that you can may be able to maintain a speed that you can not accelerate up to.

Full version:

The full answer to your question would require knowing the relevant curves for your engine as well as the gross vehicular weight, the current effective gear ration between the engine and the road, the slope of the road, the effective rolling friction and the car's drag coefficient. Which is why I'm not going to try to do the full version.

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I don't understand. Having fixed the gear the car is in, does the power depend on anything else except the RPM? If not then what's the meaning of saying there is a maximum attainable value of the power for a given RPM? –  Joe Feb 6 '13 at 20:02
@Joe: Every engine has a performance curve of torque vs. RPM and power vs. RPM. Both curves have peaks, after which they fall off with increasing RPM. Depending on the gear you're in, it's possible to have a stable speed (on the slope) which you might not be able to get to from a slower speed without being able to use just the right gear ratio. That's why trucks have transmissions with so many ratios. Otherwise, they could get into a climb where the lower gear would over-rev, and the next higher gear would lug. –  Mike Dunlavey Feb 6 '13 at 21:26
@Joe The power and torque also depend on the throttle position and on air flow decisions made by the computer, but if you keep the pedal hard down it will dwell near the max on both curves, so it is fair to use those values for your calculations. –  dmckee Feb 6 '13 at 22:09
I understand that this can get very complicated, but a lot of times even in complicated problems there is some simple principle that catches the main effect. @yohBS describes such a principle in his answer, but I'm not completely sure it's correct. So I'm accepting his answer for now, but I'm still open to criticism on this. –  Joe Feb 8 '13 at 9:08
@Joe I've expanded a bit on the implications of having two figures of merit. One of the big goals of engine design is to get relative flat power and torque curves, and modern cars are much better this way than old ones. –  dmckee Feb 8 '13 at 14:29

Here's my guess:

As you know, internal combustion engines burn fuel. The power output of the engine is a function of both the current RPM and the amount of fuel you inject. But there's a catch: the engine can burn a limited amount of fuel per cycle, and therefore the higher the RPM, the more fuel can be combusted. So (as @dmckee stated), at a given gear the maximum power is an increasing function of the RPM.

Since the gear is fixed, the RPM is a linear function of velocity. Therefore, when you enter the slope at a low velocity, the car is at a low RPM, and the maximal power that your engine can provide is smaller than what it could provide if you'd enter the slope at a higher velocity. This is why you can not reach the terminal velocity that you could have maintained if you started with it in the first place.

BTW, what people do in such situations is kicking down a gear. This keeps you at the same velocity but at a higher RPM, thus increasing the available power that your engine can supply.

Anyway, I'd recommend that you buy a better car.

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So far this seems like the best answer –  Joe Feb 7 '13 at 15:36
"at a given RPM the maximum power is an increasing function of the RPM" Only for so long, once the engine is moving fast enough the exhaust valve opens before burning finishes and the power curve turns over. Worse, in some engines (especially those with variable valve timing, the power curve may have more than one local maximum). It's all very ugly from a pencil and paper analysis point of view. In any case, torque matters too, and similar consideration apply to the torque curve. –  dmckee Feb 7 '13 at 15:40
Also, although this may fully explain "impossible" (which, according to the OP, only happens sometimes), it doesn't seem to explain, or even allow for, "[merely] very hard". –  Glen The Udderboat Feb 7 '13 at 16:01
@dmckee I agree, but I think the car is not yet at this regime. –  yohBS Feb 7 '13 at 19:01
@Gugg In my experience, the phenomena is pretty common. Also, there are surely many other factors involved (the computer regulating fuel injection, exauhst, wind/friction dissipation...) so this simplified model does not capture the difference between "hard" and "impossible". –  yohBS Feb 7 '13 at 19:03

Here is my take on it. It is somewhat similar to what many others have said but I will try to explain in greater detail, so bear with me in terms of length.

Power from engine First let us understand power-torque relationship for an engine. The working fluid of the engine (expanding combustion gases) apply a torque on the crankshaft. This torque varies with in-cylinder pressure and is net positive (useful) only in the power stroke (one of the four strokes in a 4-stroke engine in most cars). However, let us assume that we have a average net positive torque coming from the engine. This assumption is ok since most automobiles have multiple cylinders that are phase shifted so that at any time at least one or more cylinder are in power stroke.

So lets call this torque $\tau$. The work done by the engine will be $\tau\theta$, or in other words, the average power from the engine will be: \begin{align*} &P_e= \tau_{e}N_{e}2\pi \end{align*}

Power asked for by the wheels The load on the car comes from 1) friction on the road 2) air resistance 3) $mgsin\theta$ to climb a hill at a slope of $\theta$. Since the car does not revolve around its center of mass but only translates, all the load on the car (friction, wind, gravitational body forces) can be considered as some combined torque that is overcome by the torque applied by the engine. We can further analyse this using free body diagram of wheels, but that a different discussion to be had later. Essentially for a constant velocity car climbing say fixed inclined slope and a fixed frictional/air load, the torque demand is fixed and equal to the resistive load. Lets call this $\tau_w$ (wheel). Next lets us assume we are going at a constant velocity of our choice that translates to a wheel RPM (rev/min) of $N_w$. Don't worry I will get to an accelerating car later. For now: \begin{align*} &P_w= \tau_wN_w2\pi \end{align*} Since energy is not accumulated in the car's drive-train (assuming the engine parts don't heat up much after warming up) the power produced by the engine is consumed at the wheels. Again this assumption is fairly accurate since most of the fuel energy either goes to the wheels or leaves as exhaust from the engine, other effects such as frictional heating in the transmission fluid etc are negligibly small \begin{align*} P_w&=P_e\\ \Rightarrow \tau_eN_e&=\tau_wN_w \end{align*}

Now lets say the car just got on the slope and you want to maintian the same velocity as before so you slam the accelerator here is what happens. The new load is some $\tau_w$ and the speed you want is $N_w$ so you are asking the engine to deliver $N_w\tau_w$. If you are in a fixed gear $N_e = GN_w$ where $G$ is the gear ratio. Hence the engine has to deliver a torque of \begin{align*} \tau_{e,\; desired} = \frac{\tau_wN_w}{GN_w} \end{align*} So, not only you have a desired power, you also desire fixed engine RPM (or at least hope to stabilize at). Essentially you are asking for a desired torque.

Let us see what the engine can give us

Engine Load-Map
A typical engine map looks like the figure below (hand drawn so excuse me for wobbly curves). For now ignore the red circles $A$ and $B$. Torque and power vs. Engine RPM Torque comes from the work output given by the combustion gases. So the max torque curve (for any RPM) is when you are putting most fuel (diesel engine) or least throttle (gasoline engine). The curve drawn in the figure is the max torque for each speed. Even this curve has a peak value at some RPM. As you change engine speed initially you are doing good, i.e., increasing air flow speeds into the engine (that allows higher mass of air into cylinder due to greater rate of suction/pressure drop) allowing higher thermodynamic work output and higher torque, but after a certain speed the engine breathing efficiency (volumetric efficiency) goes down, then the cylinder is gasping for air (usually the flow chokes in the intake valve I think). So at high speed the volumetric efficiency goes down,there is less air, you can burn less fuel (even if press full throttle) to keep emissions within limits, and torque from the engine goes down. The power on the other hand keeps going up because it is a product of speed and torque. The increase in speed means more power strokes per unit time even if each stroke gives less torque. So the power peaks almost at max engine RPM.

Now the problem Lets study what happens when you are trying to climb a hill. Let's say you started climbing the slope and the desired engine speed (for a fixed vehicle velocity you want to hold constant and fixed gear) corresponds to red circle A. You press the gas pedal the engine gives torque $\tau_A$. If the load is such that $\tau_A<\tau_{desired}$ the vehicle will decelerate and $N_e$ will go down. This will make the $\tau_A$ to go down further (you move left on the torque curve) and you will not be able to keep speeds up. In this situation usually you will gear shift to allow engine to rev up more relative to the wheels, but if you are at constant gear you will not be able to accelerate.

Instead if you were at red circle $B$ and $\tau_B<\tau_{desired}$ at first the engine will try to slow down but its torque output will go up now the engine will stabilize to your desired acceleration. So if you started at the correct side of the torque curve speed-wise you can accelerate to your desired velocity.

Gear changes only allow us to jump to the right engine speed to be able to pull this off. Of course this is assuming that your engine is sufficiently powered in the first place. Otherwise you will go over the hump backwards and then decelerate further. So the power has to work out!

Non-issues Most cars of today have a fairly powerful engine and are heavy enough that minor things like fuel tank weight, whether the engine is in the front or back, etc are not an issue. A typical sedan is like 1250-1500 kg, only five heavy ppl and a full trunk of luggage can seriously load the car, not its own peripherals. Furthermore engine electronics (with electronic fuel injection, high pressure fuel rail etc) precision solenoid injectors, etc., are robust enough that fuel supply system is never a limiting factor. Engine peripherals are not that underpowered or lossy. Of course you never have a vapor-lock in fuel lines.

It is just a torque-vs-rpm issue, that comes from how well we can breathe and do combustion. Even throttling loses in gasoline engines have been reduced with better acoustically-tuned intake manifolds, refined valve timing, etc. Torque-curves are becoming flatter and flatter. Also one rarely ever goes to max-power situations (around 6000 rpm for typical sedan engine).

I hope I have addressed all issues still unanswered. Like I said I am not saying anything new just explaining it in more detail. There is something about invariance to inverting gravity. That is just a matter of decreasing load. If you come down a hill very fast and you don't brake, well you will gain speed. You will step off the gas, now your engine torque will come down from the max torque curve. Then it depends on how the engine speed torque stabilizes on a lower torque curve (i.e., if you in highest gear and your load requirement is very low) technically your consuming very little fuel in the engine but it is so fast that it will try to run away (highest rpm) but I am sure there are ways to prevent that, and lose power in engine friction (essentially high speed engine braking) but in most cases you will probably brake much much earlier!

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I was looking at my hand-drawn power curve and I realized it looks very linear at low speeds. This not true since both torque is not constant with speed. So use the figure only to understand what goes up and what goes and don't worry about the slopes being right! –  Sankaran Feb 13 '13 at 18:38
Nice. I hadn't noticed this answer until just a few minutes ago. –  dmckee Nov 18 '13 at 22:26

It takes force (power from the engine) to accelerate to reach the higher speed. F=ma. When maintaining the desired speed, no additional power is needed from the engine (other than to overcome drag from air resistance.)

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But the power required to maintained $v_h$ is higher than the one required for $v_l$, so if the engine outputs that power when it's going $v_l$ it already contains the additional power which is supposed to make it accelerate to $v_h$. –  Joe Feb 6 '13 at 19:18
His question is why can some car maintain a speed $v_1$ up a hill if they hit it fast enough, but if they hit the bottom more slowly can't rise above some other speed $v_2 < v_1$. (And I've had a couple of cars for which this was true). –  dmckee Feb 8 '13 at 14:32

As the velocity increases, power has to stay the same, so the acceleration must decrease. Here's my analysis:

So, say you're on some slope, we can say $c$ meters vertical rise for every $b$ meters traveled along the road.

You travel at speed $v$ meters per second, along the road, so you travel $v \frac{c}{b}$ meters per second vertically. Potential energy is just $mgh$, or in other words, you have an energy of $mg$ joules, per meter. If we multiply the vertical speed with the potential energy required per meter, we get a power of $mgv\frac{c}{b}$ joules per second. If the car is accelerating, that's not the only power the engine has to fill, it also has to fill the increase in kinetic energy. Kinetic energy is $\frac{1}{2} m v^2$. Differentiating with respect to time, we find that the power needed is $\frac{1}{2} m 2 v \frac{dv}{dt}=mva$, where $a$ is the acceleration of the car. So the total power that the engine has to exert is $mva+mgv\frac{c}{b}=mv(a+g\frac{c}{b})$ joules per second.

So, to plug in numbers, lets say the car weighs $1500$kg, is travelling at $18$ meters per second (40mph), and isn't accelerating at all. Lets say there's a $20^\circ$ grade, so that $\frac{c}{b}=\sin(20^\circ)$, and of course $g=9.81$ meters per second per second. Then the formula gives that the power needed is $90590.9$ joules per second, or 121 horsepower. (that's a pretty steep grade though)

If you're going at $30$ on the same hill, but are accelerating at $2$ meters per second per second, you get $144591$ joules per second, or $194$ horsepower.

and the formula shows that as the velocity increases, for power to stay the same, $a$ must also decrease. Depending on what values you plug in, you can definitely switch it up, and wind up with a case where the power needed to accelerate is LESS than the power needed to maintain your current speed. but this depends on which numbers you plug in. It is still the case that to maintain constant power, your acceleration must decrease as your velocity increases.

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You're right of course! But I think what I addressed in the post (even though it's simple kinematics) helps clarify a reason it would be difficult, though not impossible (the equation doesn't imply it would ever be impossible w/ constant $P$). ("sometimes impossible" was the question, so, this would cover the other cases, right? So it's relevant!) –  NeuroFuzzy Feb 8 '13 at 15:38

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