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Comment by venomsnake | Hacker News Reader

At 2.6-sh second mark your car accelerates with more than 9.8m/s^2 to get to 60.

But even lower accelerations will glue you to the seat because gravity acceleration affects every part of your body. Where as the car grabs you by the butt and yanks you forward.

So - I think he has probably some vague idea of Newtonian physics but just didn't manage to produce coherent sentence.

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> But even lower accelerations will glue you to the seat because gravity acceleration affects every part of your body. Where as the car grabs you by the butt and yanks you forward.

The feeling of weight we associate with gravity is not distinguishable from elevator-style or car-style acceleration (not counting vibrations; or tiny gradients in the field). You don't feel gravity gluing you to the ground, you feel the ground continuously shoving you skyward to stop the free fall.

See: equivalence principle https://en.wikipedia.org/wiki/Equivalence_principle

How is more than one g accelleration possible? There is one g making the friction between the tyre and the road.

I think this is intuitively wrong for a vehicle with a gear wheel locked to geared track: the gears push against each other and acceleration would be possible even in a zero g environment. Now, cars don't have gear wheels, but I assume the same principle applies.

Edit - here's a more stringent discussion: http://physics.stackexchange.com/questions/75032/maximum-acc...

Cars can do more than 1G even without downforce. Sports car street tires commonly ~1.2g, race tires commonly ~1.5g+

High school friction is wrong.

Source: various of my cars (in the past, I just have lame hybrids now) and a g meter

> High school friction is wrong.

No, it's at least mostly correct. High school friction says

    F = mu * W

Where F is the output force, mu is the coefficient of friction, and W is the normal force (typically equal to the weight).

If it's failed you, it's failed in failing to mention that the tire-road interface can have a coefficient of friction greater than one, and in failing to mention that the normal force can be increased with aerodynamic downforce.

High school physics tends to claim that the coefficient can't be greater than one.

Source: note all the people here who thought it couldn't be greater than one.

How is that due to high school physics rather than naive intuition?

Mine never implied that, and I could have easily disproved it with the grippy pads on my giant calculator.

Because the coefficient of friction could be greater than 1. And the "sticking" force is proportional to the tire area. And the air pushes you down while you move.

As long as the torque is lower than friction you should be fine.

There's only one answer to that: the Lotus 78.

https://en.m.wikipedia.org/wiki/Lotus_78

I don't understand the question. The car is accelerating horizontally, the gravity acts vertically.

The cars also gets pressed down by aerodynamics.

But downforce isn't appreciable until over 50mph (80kph), short of having stupidly oversized wings (which would drastically slow your acceleration) or active downforce (fans etc).

This is actually why a lot of expensive sports cars have 0-60 times of 2.9 seconds. Unless you design specifically for it, it's difficult to generate much downforce at low speeds.

Related: if you've ever wondered why some drag cars (e.g. Funny cars) have short exhaust pipes angled up and back, now you know.

Similar to the Red Bull F1 car's exhaust-blown diffuser: http://jalopnik.com/5892403/the-exhaust-pipe-that-made-red-b...

Imagine the tires were velcro'd to the ground. Or that they had giant spikes stabbing down for traction. Clearly you could go over 1g with that, right?

The same kind of thing (not literally, but analogously) is already happening at the atomic level, giving the tires a coefficient of friction of more than 1.

Indeed. My weak Google-fu indicates that the coefficient of friction of tyres against a dry road surface is something like 0.9, so there must be some improvement in tyre technology here just to make the thing go without creating a thick black smelly line on the road.

A car doesn't drag along the road. The tyres are in relatively static contact (unless you're drifting etc.). Static and dynamic friction are quite different.

As an aside: Spinning-rust hard drives are rated in the hundreds of Gs. While the earth only pulls with 1G, if you drop the drive onto a hard surface, the sudden stop is an acceleration of hundreds of Gs, due to the very short time frame.

Acceleration of more than 1g is probably not possible for 0-60; there isn't much of that speed range that would allow for downforce (downforce would only start have a tangible effect after approximately 50mph (80kph)), and there is no other appreciable way for the car to generate more friction with the road that doesn't scale with vehicle mass.

I suspect what the OP was trying to get at is the vector addition of gravity plus the forward acceleration of the car means that the apparent scalar force feels substantially higher than normal (i.e. 1g downwards); it's just poorly worded.

No they actually mean what they write - 1g gives you 0-60 in 60*1609/3600/9.81 = 2.73 seconds. The way to get that friction is probably to use a particular combination of tires and road. It's equivalent to standing still on a 45 degree slope -- not impossible, but definitely a risk of slipping there.

but there is four tires!

and 4/4ths of a car

With rear-wheel drive, acceleration transfers most of the vehicle weight to the rear wheels. That's why funny cars do wheelies. That doesn't work so well for front-wheel drive, so those vehicles can't accelerate as quickly as rear-wheel drive vehicles. With four-wheel drive, weight transfer cuts both ways, but more rubber on the road helps, for sure. Low center of gravity helps too, I think.

With wheels on the bottom of the car acceleration transfers weight to the rear axle. The forward forces on the car body are below the CG so produce a net rearward tilting torque.

Yes. But my point is that, with front-wheel drive, that transfers weight away from the driven axle.

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