Take the 2-minute tour ×

'tis the season as they say!

It seems to me obvious that it's better to drive in high gear on snowy roads to reduce the torque.

However, there are completely opposite advices being given on different sites:

Who is right?

You can use a simplified model of the car:

  • the torque of the engine is directly controlled through a smooth limited function ($0\leq T\leq T_{max}$ ) whose time derivative is also limited.
  • torque is transmitted through a choice of gears to the wheels (ratios are givens)
  • the wheels need not slip on the road ($\mu$ is given, circumference is given)


  1. What is the relationship between gear ratio, torque function and car acceleration?
  2. Is it true that there is a preferred gear ratio (given a set) that always maximises acceleration, independent of the choice of function? If so, which?
  3. What is the best choice of function and gear to maximise acceleration?
share|improve this question
For those not terribly experienced (like me) low gear inhibits the temptation to go too fast. –  dmckee Dec 19 '10 at 23:03
Use public transportation I say. –  Cem Dec 19 '10 at 23:08
Have you tried public transportation in England with snow? You haven't! Reason: there is basically NO public transportation with snow... it shuts down! ;-) –  Sklivvz Dec 19 '10 at 23:11
I don't see this getting answers that are likely to add any value to a physics Q&A site. (It would be perfectly on topic at a transportation or auto care SE) If you edit it to get at the underlying physics in more detail, then perhaps it could be reopened. –  David Z Dec 20 '10 at 2:16
-1. Even though I sympathize with your automotive dilemma. –  user346 Dec 20 '10 at 8:42

5 Answers 5

I used to be really really good at this(but now I live in California and my skill has gone totally stale). I think it depends on the sort of snow we are talking about. If it is packed or ice then the ability to control torque, especially when starting from a dead stop, is important. In really deep snow, torque (and maintaining momentum) is the key. I used to drive my old Z71 pickup in as much a 3/4 meter of unplowed (uphill too), and compound low plus a heavy foot was essential for that.

Of course the real key to safety is stopping. That requires practice. And keeping in mind that you want the wheels in rolling contact, i.e. you want to keep any braking/steering forces below the value where they break free and can slide sideways. Being able to feather both the accelerator and the braking in response to whether the wheels have traction or not is key. The feel for that can vary greatly by vehicle model. My old landcruiser was terrible that way. You became aware of a skid by looking out the window, whereas with my even older Subaru my brake foot somehow knew if the tires had traction.

share|improve this answer
What is "compound low"? –  Mark C Dec 20 '10 at 2:17
Most 4WD (Im not sure about all wheel drive) vehicles have a transfer case, which usually has settings for 2WD neutral 4HI and 4Lo. 4Hi is 4wheels, with the same gear ratios as 2WD. 4HI goes through another gear reduction (usually around 2 to 2.5 times). In 4Hi even at the highest gear speed max is probably only 25-30mph, and the lowest gear goes only to single digits. Huge torque is available. –  Omega Centauri Dec 29 '10 at 15:05
The benefit of compound low in this case is likely related more to the limitation on wheel speed than to the high torque. –  Colin K Nov 23 '11 at 14:27

I have driven enough cars on ice ponds to have a good feel for it. Most of my driving was done with a 1974 Datsun pickup with a clutch. You simply start in low and give it just enough gas to not break free, sensing that point with engine tone. The breakaway spot is sensed by mashing and letting up on the accelerator pedal. You must deliberately and continuously cause a breakaway condition every two or three seconds so that you know how hard you can push it. You work on up through the gears, shifting higher as soon as possible. Higher gears are better because the wheels will not break away as easily. The maximum acceleration will be at a torque just below breakaway. I drove a K-Whopper cab over with a 21 speed Spicer for awhile. It is all in that sweet spot.

share|improve this answer

On snowy roads you have reduced friction (duh!) which translates to reduced available traction on the driving wheels. Driving in 1st gear is a bad idea, because with just a little extra throttle you can exceed traction (due to the torque multiplication on the wheels). Driving in 5th gear is a bad idea also, because you have to maintain higher speeds (>35 mph typically) for the engine not to stall. So the answer is in-between.

In order to maintain control of the car and dampen the gas pedal response as much as possible, I would recommend driving in whatever gear keeps the engine happy at about 2000-2500 rpm (gas engines, lower for diesels).

The trick is keep the momentum going on slopes (no acceleration / deceleration ) and brake early before you start cornering. Do *not trail brake (apply brakes and steering at the same time). If you feel the driving wheels spinning let go of the gas slowly and wait for the car to settle.

share|improve this answer

After the edit Lets get some nomenclature out of the way. Tire radius is $R$, total gear ratio is $\gamma$ such that the engine rotational speed $\omega_E$ is a function of tire speed $\omega_E = \gamma \omega_T$ and the vehicle speed is $v=\omega_T R$. The maximum traction (force) applied on the car allowed by friction is $F_{max}=\mu\, m_T\, g$ where $m_T$ is the total mass supported by the driving wheels ($m_T = m_{car}$ for AWD, and $m_T = 0.6 m_{car}$ for typical FWD).

So at any given car speed $v$ the engine speed is $\omega_E = \gamma\, \frac{v}{R}$ while the engine is making $T$ torque. The torque at the wheels is $T_{wheel} = \gamma\, T$ with forward thrust of $F = \gamma \frac{T}{R}$.

Check that power is conserved through the gears, $F\,v = T\,\omega_E$.

If thrust is equal traction, then $F=F_{max}$ or $ \gamma \frac{T}{R} = \mu\, m_T\, g $ which is solved for torque as $T =\mu \dfrac{m_T\,g\,R}{\gamma}$.

So the more friction, mass, or tire radius you have the more torque you can apply. The more gear ratio you have the less torque. Note that with this notation 1st gear has the highest $\gamma$. This is found by multiplying the final drive ratio (like 3.2:1) to the first gear ratio (like 3.75:1).

It all makes sense.

NOTE: the reason I went to trouble to establish the engine speed first is because typically the max. available torque is a function of engine speed. Also the max. engine speeds limits what is the lowest gear available at any speed.

share|improve this answer

The issue is that the available friction coefficient is highest when the tire is not sliding. Once it starts sliding, it has friction, but less than when it was not. For example, once the tire starts sliding, the heat of sliding friction melts a thin layer of water, which acts as a lubricant.

The advantage of using a higher gear is it is not quite as easy to break the friction. You don't want your wheels spinning or sliding, even a little.

Ignorance of this is comical in a city like Atlanta, where people are so proud of their muscle-cars. They get a couple inches of snow, and suddenly everybody's in the ditch.

share|improve this answer

Your Answer


