why do engines have so much unused horsepower?

Question

I've looked at the power rating of an engine used in a Humvee (5500 lbs) and was surprised it's only ~190 horsepower, which is exceeded by many 4 door sedans.

So an obvious question is why doesn't my Camry SE (3300 lbs) burn more gas than a Humvee and I think it's just because during regular use, it doesn't use anywhere near 250 HP.

Someone on the physics stack forums gave me an estimate that it only takes ~35 HP for an average car to go 50 mph on the highway.

So I have 2 questions:

1. Can someone speculate what the torque vs RPM graphs of a sedan engine and a Humvee engine look like and what the typical operating points are? It seems the Humvee engine would be operating much closer towards its peak power while spinning slower, but at much higher torque?

2. If the peak power is never used for passenger vehicles, why don't they use a smaller engine and save some money?

Answer 1

The root problem here is that you are conflating several different terms. See Wikipedia for a calculation of horsepower from torque (tau) and rpm (f in this equation):

If you assumed a flat torque curve, you can see that peak horsepower would continue to increase with RPM. In fact, if you wanted to increase your marketing horsepower for a new vehicle, you'd just need to raise the rev limiter and keep torque from dropping faster than one foot-pound / rpm. Not only that, you can also see that the scalar values for power and torque (i.e., without their units) will always be equal at 5252 rpms.

1.Can someone speculate what the torque vs RPM graphs of a sedan engine and a Humvee engine look like and what the typical operating points are? It seems the Humvee engine would be operating much closer towards its peak power while spinning slower, but at much higher torque?

As previously remarked, the diesel in the HMMWV has a high peak torque at low rpms but the torque drops off rapidly, resulting in a peak horsepower at lower rpms than the Camry (190 hp @ 3,400 rpm / 380 lbf·ft @ 1,700 rpm cited on the Wikipedia page). The normally aspirated Camry has a much flatter torque curve with a torque peak closer to 4700 rpm and power peak at 6200 rpm (making some assumptions that we're discussing a fairly recent Camry SE with the 2GR-FE engine).

If you go back to the original equation, that means that the first derivative of the Camry's torque curve is zero at 4700 rpm and negative one at 6200 rpm (where torque's rate of decrease finally overwhelms the linear increase of rpm).

2.If the peak power is never used for passenger vehicles, why don't they use a smaller engine and save some money?

They do. You just didn't buy that one.

Civilian engine and automotive design is driven by certain marketing points. In this case, the Camry has a broad torque and power curves so there's no need to shift frequently (more than two shifts to 60 mph affects that critical 0-60 marketing number). Military transports have entirely different mission parameters including such factors as ease of maintenance, weapon systems, not getting blown up and so forth.

If economy is your key factor, you can choose a smaller engine. Every manufacturer is offering a variety of displacements and engine technologies now. The current Camry SE offers a smaller inline four cylinder, a four cylinder hybrid and a larger V6.

Regardless, it's nearly impossible for a gasoline powered engine to beat the miles per gallon of a turbo diesel.

Answer 2

Your question has a few unconnected points in it:

• as jmosrt253 said, a Humvee engine is a large diesel engine - which means its torque is low down in the revs range
• Humvee engines, like many large diesels, are not highly tuned. They are built to be robust and survive in extreme conditions
• up to a point, horsepower is irrelevant for speeds. It is relevant for acceleration. Pretty much all cars these days can do 50mph, but do you want the car to be able to get there in 3 seconds or 12 seconds. (As speeds get higher you do need horsepower to offset drag, which is why only the most powerful cars reach 200mph)
• peak power is used for passenger vehicles when you need it, but when you don't need it, you take your foot off the accelerator and are instantly saving money by sending less fuel to the engine. Some modern engines also stop providing fuel to half the cylinders when cruising

The real driver for engine size has always been the cost of fuel, so in the US, where fuel is cheap, fuel consumption was not an issue - hence large, inefficient engines. Compare this with the UK, Europe and Japan, where fuel is very expensive and you find manufacturers making highly tuned, highly efficient, small engines.

I couldn't find a power/rpm graph for the Humvee, but there are any number online for ordinary cars.

To answer your question "Why doesn't my Camry burn more fuel than a Humvee":

The Camry engine is a far more efficient engine - you are getting much higher horsepower from much less fuel.

Answer 3

Why have extra engine power? Well if 35HP is required to do 50 MPH and that's all you have, it will take a very long time to accelerate to that point - possibly several minutes.

Also, people like the extra power for the sportiness of the car - quick acceleration when you want it, and not having to press the throttle down much to go up hills etc. You'll find that the Camry will actually be using a very significant amount of the available HP. I'd say at least 50% at times (e.g. accelerating up a hill with a full car), and depending on the driver, closer to 100% is not unheard of (e.g. overtaking).

Why is the Camry more efficient? People seem to be missing one of the key points - the Humvee is one metric tonne heavier (so much more rolling friction), and has the aerodynamics of a brick. To put numbers to it, Wikipedia cites the coefficient of drag of a Hummer H2 as 0.57, and a Toyota Camry 2007 at 0.27. So it's more than twice the drag coefficient, and that's excluding the larger cross-sectional area on Humvees. And if you have a gunner up top, it would be even worse!

So if you put the Humvee engine in the Camry, you would see a massive fuel consumption decrease compared to the original Humvee.

Answer 4

Excuse me for answering my own question, but the answer is so clear to me now at least for non-electric motors. They have so much unused power because at idle speeds, you get only a fraction of the advertised power. For the 2.4L Camry engine at idle (1000 RPM), it only puts out 30 HP, and only gets to 140 HP at 5000 RPM (about 30 MPH based on acceleration test dashboard videos).

The V8 diesel reaches its full 200 HP at only 2500 RPM, so that explains why it doesn't use an over sized engine like on gasoline vehicles to compensate for the low HP at low RPMs.

To see how much HP you need if your engine had a perfectly flat power curve, I looked at an acceleration test for a 2008 Toyota Camry LE 2.4L, 158 hp, which gets goes from 0 to 60 mph in 8.5s. For a perfect constant power engine, velocity = sqrt(2* power * time / mass), so you would only need an 86 HP engine to achieve the same acceleration.

But now that raises a new question. If electric motors have such a good power band, then again, why so much HP? Also, why is the torque flat around 0 RPM instead of going to infinity like on a perfect engine. Is it something to do with the magnets being saturated?

Here are the torque vs RPM graphs for

1. 2.4L Camry 2009

2. V8 6.5L diesel (Hummer H1)

3. Tesla Model S traction motor (found somewhere on cleantechnica.com)

Update: As DucatiKiller pointed out, I didn't consider engine weight or size, which might explain why some types of engines have such a high horse power since if the fixed weight/size cost of an engine (e.g. a 1/2 HP gasoline model airplane engine) is small or the marginal cost to increase the HP is small, then it doesn't cost much extra to make an overpowered engine.

Here are estimates of the power to weight ratios of the various engines:

1. 2.4L Camry 2009

   158 HP / 121kg = 1.3 HP/kg

2. Detroit Diesel 6.5L diesel

   200 HP / 750 lbs = 0.59 HP/kg

3. Tesla Model S traction motor

   400 HP / 70 lbs = 12 HP/kg

   with batteries or else wouldn't be fair:

   400 HP / (70 lbs + 540 kg) = 0.70 HP/kg