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標題: mass/ gravity /coefficients of friction [打印本頁]

作者: 毅字 時間: 2009-11-5 22:32 標題: mass/ gravity /coefficients of friction

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作者: 毅字 時間: 2009-11-5 22:33

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作者: 毅字 時間: 2009-11-5 22:37

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作者: 毅字 時間: 2009-11-5 22:59

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作者: yuusaku 時間: 2009-11-6 00:14

I've heard that putting some weight at the rear of a FWD would help, since most FWD weight distribution is usually light on the rear...

I personally have driven both FWD with(couple bags of sand) and without weight in trunk on snowy road...feels similar...

can anyone confirm this myth?

落雪﹐架車越重越好﹖定係越輕越好﹖
毅字發表於 2009-11-5 23:32

作者: Lik 時間: 2009-11-6 00:47

Everything else being equal (with identical weight balance, in particular), you want minimal mass, but maximum weight (b/c weight = downward force, which in turn provides friction -- which is grip).

In a typical everyday case, having less mass would make it easier to drive in snowy condition.

-Lik

作者: ricrick 時間: 2009-11-6 01:11

本帖最後由 ricrick 於 2009-11-6 02:14 編輯

recall my physics11. (if i am right) force of friction = coefficient of friction x mass x gravity x cos(θ)
the greater the force of friction is, the faster your car can stop
sub in the numbers, then you will know the result

作者: CWong 時間: 2009-11-6 03:23

都要睇架車有几長同啲重量集中係架車边个部份喎！

作者: Lik 時間: 2009-11-6 10:00

recall my physics11. (if i am right) force of friction = coefficient of friction x mass x gravity x cos(θ)
the greater the force of friction is, the faster your car can stop
sub in the numbers, then ...
ricrick 發表於 2009-11-6 01:11

Physics 11 is completely inadequate in accounting for real life situations because of all the assumptions it makes.

-Lik

作者: shutterbug 時間: 2009-11-6 12:09

Physics 11 is completely inadequate in accounting for real life situations because of all the assumptions it makes.

-Lik
Lik 發表於 2009-11-6 11:00

But I think it's good enough for what we're talking about here.

作者: shutterbug 時間: 2009-11-6 12:15

Everything else being equal (with identical weight balance, in particular), you want minimal mass, but maximum weight (b/c weight = downward force, which in turn provides friction -- which is grip).

...
Lik 發表於 2009-11-6 01:47

Using this forumla for the simplicity of our discussion, and having coefficient of friction, gravity, angle being constant:

force of friction = coefficient of friction x mass x gravity x cos(θ)

The only thing that varies is mass. More mass = more friction. More friction helps you get out of snow.

作者: ricrick 時間: 2009-11-6 12:30

o.. we have to think about velocity too.. given that the cars are running at 50km/h
suppose a heavier car needs more time to stop on a ramp due to inertia

作者: 蛋撻 時間: 2009-11-6 12:49

Yes, more mass will give you more fraction.

HOWEVER, more mass also increase your momentum while moving which equal you need more force to stop.

So in the end, a lighter car will have a tough time picking up if it is at a complete stop, but heavier car will have hell of a time trying to break to a stop.

(I'm just thinking out load using my logic... I might be wrong)

作者: shutterbug 時間: 2009-11-6 13:06

momentum = mass * velocity

so in snow it's better to have a heavy car but you drive slowly...haha

作者: Derek128 時間: 2009-11-6 15:30

eg bmw 1 man 528 (3,700 lb) vs 5 men 550 (5,000 lb), eg same winter tires, same brakes

which vehicle would have a better stopping distance on 35 degree downslope on snow?
毅字發表於 2009-11-5 23:59

bmw 528 同 bmw 550 個brake 唔會一樣掛
550 多咁多隻用同一個brake 會唔識停

作者: 毅字 時間: 2009-11-6 16:27

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作者: 毅字 時間: 2009-11-6 16:30

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作者: 毅字 時間: 2009-11-6 16:50

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作者: ricrick 時間: 2009-11-6 18:14

F=MA

作者: B2L2 時間: 2009-11-6 21:00

A car has 4 wheels and therefore the CG matter in the calculation. If we put that aside and consider our cars here only have 1 wheels and the CG is at the center of the wheel.

My answer is:
1) when the slope is below 45 degree, the heavy car stops better
2) above 45 degree, the lighter car stops better
3) when it is 90 degree, both drop at the same speed, LOL.

Things that I know:
- coefficient of friction is between the tire and road surface and it has nothing to do with mass
- magnitude of the friction force linearly depends on the normal force
- when the car is on the slope, the normal force decreased, cos(angle)

作者: sheep 時間: 2009-11-6 21:09

bring back my Physics

作者: chobit 時間: 2009-11-6 21:09

Here my 2 cents..!  With an assumption that we are talking about snow, but not ice.

When everything being equal in terms of tire size, and weight distribution..!  My guess is ligher car will be able to stop quickly as long as it has enough weight to keep the car/tires not floating on top of the snow.
One of the condition that help our car stop in the snow, is the amount of snow build-up in front of your tire as you stop, and the amount of snow build-up in real life probably wouldn't change alot with the same size tires.  Therefore, the light car will end up being stopped faster not only by the amount of friction generated by the tires, but the amount of snow build up.

One of the factor and reason why snow tires are usually purchase with width less than the OEM tires is because you don't want you car to float on top, but to cut into the snow or ice.  Therefore, wider tires would need heavier weight car to dig-in.

So, honeslty. back to the real life question.., the tires would have to fit in the equation for an answer..!

As for the brake...!  Bigger size brake probably wouldn't make any difference at all, since any cars out there would brake with enough braking power to overcome the amount of friction available.
When make a car brake better in these condition beside driver skills, is how well the brake force are distributed to each of the wheel to maximize braking power..kind of like those high end active yaw ..traction control stuffs...!

作者: sheep 時間: 2009-11-6 21:14

0甘
呔氣要打漲0的
定係要淋0的?

作者: tofu 時間: 2009-11-6 22:42

"""Everything else being equal (with identical weight balance, in particular), you want minimal mass, but maximum weight"""

i don't get it. if everything else is being equal, how can min mass with max weight

作者: ricrick 時間: 2009-11-7 00:02

各位同學﹐要考究GE係多出左一千三百磅GE負荷﹐ 加埋本身部車GE時速五十的恆衝
沖力﹐再加落斜個角度GE地心吸力。到底邊LUN部車會停得好過對方。如果輕1300磅
果架車停得比重1300磅果架好﹐冬天擠重野在車尾箱果班朴 ...
毅字發表於 2009-11-6 17:50

條formula真係吾識寫

吾好諗落斜果d野，但同一道理，如果行得好快，輕車應該停得快過重車
如果加重架車，會對於起步同慢速停車有利

總之速度越快，重車越差

作者: B2L2 時間: 2009-11-7 00:13

條formula真係吾識寫

吾好諗落斜果d野，但同一道理，如果行得好快，輕車應該停得快過重車
如果加重架車，會對於起步同慢速停車有利

總之速度越快，重車越差 ...
ricrick 發表於 2009-11-7 01:02

On a flat surface, if both car could brake at the threshold braking point, then two car could brake in the same distance. The limiting factor is the friction.

On the heavy car, you just need to brake harder to reach the threshold. So heavy car with bigger brake could stop in short distance too.

作者: ricrick 時間: 2009-11-7 00:25

樓主話乜都一樣，淨係重量吾同啊

作者: 毅字 時間: 2009-11-7 00:58

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作者: chobit 時間: 2009-11-8 01:04

Larger brake will not help your car to stop faster, because all brake can probably generate enough braking power to break the traction between your car and the road (Snow).

In terms of putting extra weight on the car.  First of all, it's a common practice for the following type car.

RWD cars and 4WD trucks. like someone else suggest, it's about starting your car, more than stopping your car.
It's common to find most RWD are front heavy, and you do need the extra weight at the back to help your car move forward.  Same thing with many 4WD trucks.  Light-weight in the rear + the 4WD system doesn't generate any power to the front unless the rear is not spinning..!  (I go 4 wheeling, so I know quite well about 4x4 locker, center diff... etc..)

As for FWD, since they are front engine with front driver, you pretty much have all the weight needed to keep the tires on the ground, therefore extra weight at the back won't do you any goods.

Back to the question about braking.  It doesn't matter if you are driving FWD,RWD, AWD, 4WD..!  They all brake with 4 wheels/tires/brakes.  The following is how extra weight at the back may help you brake better in the snow.

This is a made up assumption to help us understand this better..

Say this FWD weight 2000lbs, the weight distributions for the front is 1300lbs, back is 700lbs.  During braking, more weight is transfer to the front from the back.  Say, 1500lbs / 500lbs.  During max braking, your front tires are braking at it's max, however, because of the weight distribution, your front brakes/tires are at 100% stopping power, but the back may only at 50%.

But adding more weight to the back, you would need extra braking power to stop the car, but since the weight is added to the back.  The weight may put out 100% braking front & back.  The extra stopping power allows you to stop an extra 400lbs for the same distance, and the extra weight you add to the back is only 100lbs.  Now your car can stop faster, because the gain in traction is more than enough to overcome the extra weight added.

EL

On a flat surface, if both car could brake at the threshold braking point, then two car could brake in the same distance. The limiting factor is the friction.

On the heavy car, you just need to ...
B2L2 發表於 2009-11-7 00:13

作者: BarF1 時間: 2009-11-8 10:13

本帖最後由 BarF1 於 2009-11-8 11:23 編輯

落雪﹐架車越重越好﹖定係越輕越好﹖
毅字發表於 2009-11-5 23:32

This is a grossly simplify problem.
You can probably solve by linear optimization.

More weight = more normal force = more frictional force through tires = better deceleration rate
Less weight = less mass to decelerate

So you just have to find the intersection where the two graph meet up.
Just use 1D kinematics formulas.

This ignores pretty much everthing in real life; weight transfer, polar moment around CG, variable conditions ...

Putting heavy things in the trunk also help reduce polar moment in pitch and yaw, so your car becomes less twitchy as well ...

作者: B2L2 時間: 2009-11-10 10:23

樓主話乜都一樣，淨係重量吾同啊
ricrick 發表於 2009-11-7 01:25

Yeap, there is nothing wrong. On the exact same car, when you have a single driver. You could be using 50% of your braking power to reach the threshold point. When you have 4 passengers and max load, you could be using %70 of your braking power to reach the threshold point. Note that you have more mass to decelerate, but you having max out your braking power yet. Both able to stop at the same distance in two cases.

Now on snow, you will no way to max out your braking power before the tire starts to skid.

Agree with BarF1 on the calculation.

作者: Monkii 時間: 2009-11-18 14:42

May be off topic...but I believe the best practical answer is get some good snow tires.

The result is much more significant from a change in the coefficient of friction than a change in the mass.

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