A comment upon economical driving.

[Jim314]

Originally Posted by Bill_56

I'm now thinking that the cornering car is not equivalent to the rotating body of 'O' level newtonian physics, and there is one fundamental difference...

..if I tie a weight on the end of a rope and spin it around my head, the rope comes under tension, and the tension of the rope is the force that causes the weight to follow a circular path, and so it keeps spinning without any extra work from me. The faster I spin the weight, the greater is the force and so the equations all balance at any speed.

However, when a car goes around a corner, the closest thing to the 'rope tension' above is the frictional force between the tyre and the road surface. Friction is a much more complicated force, and is limited by the tyre's 'co-efficient of friction' and thus there is a ceiling on the frictional force,. Assuming co-efficient of friction no greater than 1.0, the frictional force can never be greater than the weight (ie downwards force) of the vehicle.

A vehicle on rails, such as a roller coaster, is not bound by the frictional forces. Reactive force against the vertical wall of the rail provides the 'tension' that provides loss-less rotational movement. And that reactive force has no bounds, the faster the carriage travels, the greater is the force.

I know that's only a half baked argument but intuitively, I've now convinced myself that a freewheeling vehicle (on tyres, not rails) would never be able to complete a full circle using just it's own kinetic energy. Am I wrong? If I had a bicycle I'd try to prove/disprove it, but I don't fancy trying it in the S60.

Forty years ago I taught O- and A-level physics. You are right that lateral tyre friction is not as strong as a rope, but there is no fundamental difference in the two types of forces for the present discussion. The fact that the coefficient of friction would normally be less than 1 is not of significance here. One could imagine a type of tread with velcro bonded to it where the coefficient of friction would be greater than 1. So there is no fundamental reason why the coefficient of friction cannot be greater than 1.

In this thread people are reporting that their instantaneous mpg drops in sweeping level turns. I think this observation should be provisionarily accepted as true. (The one caveat I would have is that a sweeping turn could be associated with a slight rise in the road that might be hard to detect.) So what could be the reason why the instantaneous mpg would drop in a sweeping level turn at constant speed?

I think a drop in instantaneous mpg in a turn has to do with engineering details of the vehicle and not simple physics. In a turn the edges of all four tyres on the inside of the turn are traveling a shorter distance than the outside edges. This should cause scrubbing of the tread on the road surface which would increase rolling resistance of the tyres. This would require the engine to produce more power to keep the speed constant.

A while back I was in a tyre and alignment shop getting new tyres and an alignment for my wife's XC90. This was a large indoor space with a very clean and sticky floor. I stood in the shop on the edge of the works area to observe the tech installing the tyres and doing the alignment. As cars were being driven around I was amazed to hear loud sounds of tyre scrubbing in low speed but sharp turns. For decades I have patronized a different type of tyre shop where the vehicles are driven straight in to bays so the vehicle is not turning inside a building which retains the sound of tyre scrub.

In a turn generating 0.3 g of lateral acceleration the vehicle will lean to the outside and increase the load on the two outside tyres and decrease the load on the two inside tyres. This would widen the contact patch of the outside tyres and maybe increase scrubbing of the two outside tyres in the turn.

[Brendan W]

Originally Posted by Jim314

In a turn the edges of all four tyres on the inside of the turn are traveling a shorter distance than the outside edges. This should cause scrubbing of the tread on the road surface which would increase rolling resistance of the tyres. This would require the engine to produce more power to keep the speed constant.

Got it in one I think or at least a significant part. For the car to complete a 360 turn each tyre contact patch will have a vertical axis 360 rotation superimposed on its rolling motion. So the problem reduces to how much energy is needed to rotate a loaded wheel about a vertical(ish) axis - simple. Anyone care to guess the shape of the contact pads?

[foggyjames]

The additional factor to consider here is where the engine is most efficient. That doesn't necessarily mean when / where does it use the least fuel, but under what conditions does the work done / fuel consumed ratio work out for the best? I reckon that the ideal situation (from a fuel consumption point of view) is full-load acceleration to the middle of the engine speed range (say 4500rpm for my V70R) through a couple of gears to get up to cruising speed as quickly as possible.

Certainly lugging the engine at very low speeds for an extended period (i.e. specifically avoiding high engine speeds) is not efficient. Taking 20 seconds to reach 70mph from a sliproad without exceeding 2500rpm (when the engine doesn't make much power until 3000rpm+ - most older turbo designs, and pre-VVT multi-valve engines) is almost certainly going to use more fuel than revving it out a bit, and getting to speed in half that time.

Or, as a friend once very accurately put it, if the engine isn't accelerating at full throttle but would in a lower gear, you're wasting fuel by not changing down. Even if the engine speed is higher, you can reduce the load significantly, which makes a bigger difference.

cheers

James

[Bill_56]

Originally Posted by Jim314

Forty years ago I taught O- and A-level physics.

I've never taught professionally, but I've always felt that half the battle of teaching is to get the student's interest, maybe by giving them a real-world and interesting problem to consider. And I can't help thinking that this very problem, the behaviour of a vehicle rolling around a bend would be a perfect example. Yet I can't find any other discussions anywhere on the great Internet.

I certainly bow to Jim's greater knowledge. Though I must admit I've an uncomfortable feeling that at some time in the future, maybe hours, maybe weeks or months, I'll have a 'Eureka' moment and spot some aspect of the issue we've all been missing. Listen out!

Incidentally, lest anybody suggests this thread has gone off topic... the objective of this diversion remains to explain and quantify a fuel-consumption penalty experienced by folks living in Lincolnshire. Sounds quite reasonable, for the topic, No?

[S60D5-185]

Originally Posted by Brendan W

So the problem reduces to how much energy is needed to rotate a loaded wheel about a vertical(ish) axis - simple. Anyone care to guess the shape of the contact pads?

Energy is quantified by G x mu to the power of 7 + V squared.

On that basis with the assumption that you are using a 225/45/17 tyre correctly inflated the contact patch will be 6.22217777 inches recurring.

Hope this helps


Darryl

[Brendan W]

Cheers Darryl, no matter what I do I can't get the tyre scrubbing losses to add up to more than half a horsepower. Was expecting a bigger number.