Drive Your Car Like a Rocket?

[Whirligig Girl]

AFAIK, the reason you wanba get to orbit as fast as possible is that Isp is lower in atmo.

No, it's drag.

[Brethern]

I can't comment on rockets because I don't know but cars are a different story.

Engines produce torque. For diesels it's typically between 1200 - 1800 RPM. Gasoline I believe is around 3000RPM. Going higher or lower than those numbers means you start losing efficiency. So if you can keep your RPM up the speed isn't an issue.

[Right]

Wow, definitely a rich mixture of answers. Thank you all for the enlightening arguments.

Perhaps a better way of asking the question is this: 1. are you fighting less gravity by going up a hill faster? 2. If so, how fast is best?

It may also be useful to think about it from the opposite direction: in order to capture as much gravity as possible from a hill would you want to move down it slowly, or quickly? The opposite should be true of the opposite answer.

In reality, I suspect that most of you are right to say that there is a sweet spot, determined by many factors; road quality, car's age, engine/transmission efficiency, altitude, even wind velocity.

That said, I just wanted to reiterate a couple (possibly unrealistic) assumptions for the purpose of isolating the concept we're talking about.

1. Linear ratio of fuel consumption to acceleration. (e.g. 50% acceleration costs 1x fuel & 100% acceleration costs 2x fuel)

This means going half the speed takes less fuel, but it also means spending fuel for twice as long. In a simplified theorical picture, moving a car along a flat surface from A to B will cost the same amount of fuel moving at all speeds. (discounting aerodynamics)

2. Total friction is equal (e.g. Moving at 30mph you lose 30 units of energy to friction, moving at 60mph you lose 60 units of energy to friction. This means from A to B, the lose of energy to friction will be equal at all speeds)

Gravity and aerodynamics are the two primary considerations. Gravity is a force over time, which means exposing yourself to it for as short a time as possible ensures you are encountering less of it. (This means going up a hill fast) Drag however should place an upper limit on the optimal speed.

On the note of aerodynamics, I would like to argue what I feel is an important point. On flat ground, terminal velocity is 0 m/s. This is because while gravity is exerting a force on you, it is not literally accelerating you (I know I have to be careful with that word in this context). As an example, anyone who has flown jets or SSTOs in KSP with mechjeb might have noticed mechjeb only slows your vessel if your vertical velocity reaches terminal velocity. This also explains why rockets with lower T/W ratios start their gravity turns later, and vice versa.

So...if you were moving up a 30° incline, the terminal velocity would be multiplied by sin(30°), or half of its normal value. Terminal velocity on earth of a car falling might be 90mph and thus its optimal speed on a 30° incline would be ~45mph. BUT, recall that this is 45mph of vertical velocity. At a 30° incline, one would have to travel 90mph to get that much vertical velocity. (Well that was a waste of time... Still useful demonstration)

The curious revelation about this is that terminal velocity for a car at a 30° incline and terminal velocity for an ascending an aircraft (shaped like a car, weighing as much as a car, and with the same drag as a car) at a 30° incline are very different; double in fact. (Because half of the gravity the car is encountering is defeated by the ground.) It is notable the the more the incline increases, the more the terminal velocity will increase for a car.

Critique away my logic!

Edited February 12, 2014 by Right

[magnemoe]

I always thought of it like this:

- Vehicle speed scales linearly with engine rpm.

- Fuel usage scales fairly linearly with engine rpm in a frictionless environment (like when you run the engine on a test stand).

- Combined friction from wheels, aerodynamic drag, slip etc. increase exponentially with vehicle speed.

- Going uphill while keeping fuel input constant slows you down.

Therefore:

- Maintaining speed while going uphill requires more fuel input.

- Going faster, regardless of uphill or not, worsens your gas mileage.

Finally leads to:

- Going uphill fast is less fuel efficient than going uphill slowly.

This is true, only time you earn on going faster is coming down to an bottom before going up next hill, still here you have an pretty low maximum speed because of drag.

Again another exception, icy hills might have minimum speeds as you don't have enough traction in the hill to maintain speed.

[Seret]

That said, I just wanted to reiterate a couple (possibly unrealistic) assumptions for the purpose of isolating the concept we're talking about.

1. Linear ratio of fuel consumption to acceleration. (e.g. 50% acceleration costs 1x fuel & 100% acceleration costs 2x fuel)

This means going half the speed takes less fuel, but it also means spending fuel for twice as long. In a simplified theorical picture, moving a car along a flat surface from A to B will cost the same amount of fuel moving at all speeds. (discounting aerodynamics)

Nope, the relationship between RPM, acceleration and fuel consumption isn't linear. If it was you wouldn't need gears. There is an optimum RPM, revving above this will consume a lot more fuel for minimal increase in power. Also, you can't discount aerodynamics, it's too important.

2. Total friction is equal (e.g. Moving at 30mph you lose 30 units of energy to friction, moving at 60mph you lose 60 units of energy to friction. This means from A to B, the lose of energy to friction will be equal at all speeds)

Again, I'm afraid you can't simplify this to a linear relationship. There are several sources of drag on a car (aerodynamics, rolling resistance, friction in the drive train, etc). Some of these are linear, some aren't, and it all depends on speed. At low speeds things like rolling resistance are dominant, but as speed increases the aerodynamic forces dominate, and these are very much non-linear.

Going as fast as you can up a hill to reduce the time that the gravity vector is supplementing drag will use more fuel than travelling at the optimum speed for that vehicle. The bottom line is keeping your RPM in the right zone. Since there is more effective drag on the car when going up a hill you won't be able to maintain the same speed you would on the flat without increasing your RPM. So the most fuel efficient speed for climbing a hill will be somewhat lower than it would be on the flat.

In general though, rockets and cars are different machines, draw the FBD for each and you'll see that different forces are acting in different ways.

[adammada]

First about "sweet spot speed" - it depends on engine RPM AND gear. If engine is most efficient at 2000 RPM then it will be different speed on 1-st gear and on 6-th gear. And of course aerodynamics. (you will for sure be more efficient at 2-gear than 1-gear, because air resistance is still small, but you can be more efficient at 5-th gear than 6-th, because even you are driving faster with the same RPM, you need more fuel to keep that RPM).

I also heard that most efficent for car engine is not too drive with constant speed. Most efficient way (used in contest for low fuel usage) is to accelerate hard then let car slow down, because engine is more efficient while accelerating, and doesn't use fuel at all while decelerating (engine breaking). Its not health for engine, but its most fuel efficient. Just like ecodriving - it says "accelerate hard, but change gear to higher as fast as you can" - it will lower fuel consumption, but will wear your engine faster.

And now back to the hill problem - for sure you can't drive too slow. Its just like rocket - you need to use part of your engine force to fight gravity. Rocket with TWR 1.01 will not be efficient. Also driving car 1 km/h uphill while pushing pedal so much that on flat you would go 31km/h means that you are wasting 30km/h way too long.

You can pretty easly compute what additional force works on your car while driving uphill, you just need to know how steep hill is, and use sin function. Normally this force is neutral because works vertically and you go horizontally. But while going uphill it is like rocket fighting gravity. Rocket stops fighting when it will reach orbit. Car will stop fighting gravity when it will reach top of the hill.

Does anybody here was ever driving uphill on bike (without gears)? Does going slow make it easier? As far as I remember its best to start fast and go up as fast as you can, not go slowly...

[adammada]

Also, about going downhill.

If you drop red ball it will reach faster finish line on green(longer) track than on black track.

[Right]

Nope, the relationship between RPM, acceleration and fuel consumption isn't linear. If it was you wouldn't need gears. There is an optimum RPM, revving above this will consume a lot more fuel for minimal increase in power. Also, you can't discount aerodynamics, it's too important.

Again, I'm afraid you can't simplify this to a linear relationship. There are several sources of drag on a car (aerodynamics, rolling resistance, friction in the drive train, etc). Some of these are linear, some aren't, and it all depends on speed. At low speeds things like rolling resistance are dominant, but as speed increases the aerodynamic forces dominate, and these are very much non-linear.

You're welcome to decline that these assumptions are realistic. I've already pointed that out. But they are, never the less, assumptions. The point is to work through in a simplified context. Ceteris paribus - second paragraph.

Edited February 12, 2014 by Right

[Seret]

I also heard that most efficent for car engine is not too drive with constant speed. Most efficient way (used in contest for low fuel usage) is to accelerate hard then let car slow down, because engine is more efficient while accelerating, and doesn't use fuel at all while decelerating (engine breaking).

That's a technique called pulse and glide. It works best in something like a hybrid, where you can let the car glide with the engine off for the second part of the maneuver. I do it all the time, it does work. You don't want to accelerate TOO hard, but you don't want to be too much of a granny either.

You're welcome to decline that these assumptions are realistic. I've already pointed that out. But they are, never the less, assumptions. The point is to work through in a simplified context. Ceteris paribus - second paragraph.

Simplifying assumptions are great, but you've got to be careful with them. If you choose simplifications that cause your model to act significantly differently from reality then the model becomes a bit pointless.

[adammada]

That's a technique called pulse and glide. It works best in something like a hybrid, where you can let the car glide with the engine off for the second part of the maneuver. .

Yes, but lets forget about hybrid and about restarting engine during driving - both are rather impossible for most drivers. But if you do not shut down your engine during "glide" its better not to be in "neutral".

When you are in neutral engine uses fuel to spin at idle RPM (around 900 usually). But if you are not in neutral while driving and not pressing gas, you are doing "engine breaking", which means that engine is not using fuel at all, because engine is propelled by wheels. You are loosing speed faster than in neutral (engine breakes), but you do not use fuel at all.

Most opinions i've read says that it is more fuel efficient to use "engine breaking" and not use fuel at all, than glide longer but with engine on idle.

[drakesdoom]

Yes, but lets forget about hybrid and about restarting engine during driving - both are rather impossible for most drivers. But if you do not shut down your engine during "glide" its better not to be in "neutral".

When you are in neutral engine uses fuel to spin at idle RPM (around 900 usually). But if you are not in neutral while driving and not pressing gas, you are doing "engine breaking", which means that engine is not using fuel at all, because engine is propelled by wheels. You are loosing speed faster than in neutral (engine breakes), but you do not use fuel at all.

Most opinions i've read says that it is more fuel efficient to use "engine breaking" and not use fuel at all, than glide longer but with engine on idle.

I am unaware of IC engine in factory cars that cuts fuel under engine breaking conditions. Unless you have evidence otherwise I believe that engines continue to use idle levels of fuel.

[Seret]

I am unaware of IC engine in factory cars that cuts fuel under engine breaking conditions. Unless you have evidence otherwise I believe that engines continue to use idle levels of fuel.

Some hybrids definitely do, and I wouldn't be surprised if some of the stop-start ones did either. But generally you're right, they will run at idle. The glide part of the cycle should be done in neutral, which makes it a lot less practical in conventional cars IMO.

[K^2]

I am unaware of IC engine in factory cars that cuts fuel under engine breaking conditions. Unless you have evidence otherwise I believe that engines continue to use idle levels of fuel.

I used to drive a '98 Tiburon that did that. If it sensed engine brake, it would cut fuel and open up a bypass valve. The downside is that you couldn't actually engine brake with that car, because the engine did not have to work against the manifold pressure, but on the plus side, it didn't use up any fuel when coasting downhill on high gear.

I wouldn't be surprised if a lot of cars cut fuel supply when engine braking. This is a challenge with a carburetor engine, but when you have a computer-driven injector, it's a few extra line of code that significantly improve mileage.

[Kerbin Dallas Multipass]

http://en.wikipedia.org/wiki/Fuel_economy-maximizing_behaviors#Pulse_and_Glide

Most modern petrol vehicles cut off the fuel supply completely when coasting (over-running) in gear

http://en.wikipedia.org/wiki/Engine_braking#Legal_implications

most modern engines don't use any fuel while engine braking which helps reduce fuel consumption.[citation needed] This is known as DFCO or Deceleration Fuel Cut-Off

This is being done since the 1980s in gasoline injection engines and even earlier in diesel engines.

[Seret]

Well, you can tell I went straight from carburetors to hybrids...

[drakesdoom]

Ahh, that explains it. Me over here with my carb on 4.3L v6 from '93.