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What's the Maximum Steepness a Car Can Climb?

12/27/2009 2:37 AM

A 2D physics problem. A car of unknown weight is going up a hill at some angle theta. Distance between tires are 2.90m with C.O.G situated 1.15m behind front wheel. Co-efficient of static friction between road and tires is given to be 0.7. The question posed is what the maximum angle would be that the car could climb in 3 scenarios :

  • a) Rear wheeled drive
  • b) Front wheeled drive
  • c) Both

I think it wouldn't matter and the max angle is determined by arctan(0.7) = 35 degrees. Is this right or could any of the above three scenarios give a higher angle than that? Impossible I think given that you cannot change co-efficient of friction, even though the normal force is 30% higher on front wheel than rear due to the askew weight distribution. I just want to know if I was right :)

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#1

Re: Maximum Steepness A Car Can Climb

12/27/2009 2:56 AM

The coefficient of friction is multiplied by the normal force, so even if the C.F. does not change, the exertible tangential force does change.

In this example, we don't know how high the C.G. is above the wheelbase. It is conceivable that as the car tilts back, the C.G. could favor the rear wheels over the front. In that respect, we don't have enough information.

The rear drive may gain an additional advantage: as it accelerates the car, it tends to lift the front wheels and give greater normal force to the rear wheels. But if you do a wheelie and flip over backwards, all bets are off!

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#2
In reply to #1

Re: Maximum Steepness A Car Can Climb

12/27/2009 3:10 AM

C.O.F = Frictional force / Normal Force

My thinking was that since c.o.f is constant, if normal force increases on any wheel, frictional force in tangential direction decreases or vice versa to hold the same c.o.f. So arctan(c.o.f) gives 35 degrees as the angle which I think is the max and this cannot be increased due to the vehicle employing rear wheel drive, or front wheel drive or both. If increased beyond this hill angle, the tires will slip or a moment about rear wheel can cause the car to topple. Am I right?

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#3
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Re: Maximum Steepness A Car Can Climb

12/27/2009 3:12 AM

I'm sorry. The problem has it that C.O.G is 0.6 m above the ground, 1.15 m behind the front wheel, and 1.75 m forward of the rear wheel.

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#4

Re: Maximum Steepness A Car Can Climb

12/27/2009 4:00 AM

The weight of the car is distributed between the front and rear wheels. Thus only part of the weight is available to be multiplied by the C.F. and thereby contribute tractive force. The analysis would be easy if the C.G. were at height zero (the car's weight would divide 1.15 rear versus 1.75 front over 2.90 total, in inverse proportion to the front/rear location of the C.G.) With the C.G. above ground zero, all the angles change. I'm too sleepy to tackle this right now, but I'll give it a shot tomorrow.

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#5
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Re: Maximum Steepness A Car Can Climb

12/27/2009 4:17 AM

Agreed. The normal forces hence are different for both wheels on account of the assymetrical weight distribution. But c.o.f cannot change, so if front wheel's normal force is higher, the frictional force acting on the front wheel's tire has to be higher to keep c.o.f the same right?. So either way, c.o.f does not change and max angle is arctan(Frictional force/Normal Force) = artan(0.7) = 35 degrees which is the minimum slope at which the car slides. My feeling is that this angle cannot be theoretically increased by virtue of rear or front wheel drive or both. I don't know how it would be in practice.

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#6
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Re: Maximum Steepness A Car Can Climb

12/27/2009 4:48 AM

The 0.7 has to be multiplied by the fraction of the car's weight that goes to either front/rear. For the easy case in which the C.G. is at ground level, about 40% of the weight goes to the rear wheels, and about 60% to the front. The front wheels thus can produce about 42% (0.7 x 0.6) of the car's weight as traction; the rear wheels about 28% of the car's weight. Arctan 0.42 ≈ 22.8° for the front wheels; arctan 0.28 ≈ 15.6° for the rear.

Because of the C.G. actually being upward 0.6 M, the front wheels have somewhat less advantage, and the rear wheels somewhat more. I'll try to run better numbers later....

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#15
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Re: Maximum Steepness A Car Can Climb

12/28/2009 10:54 AM

You're better than I if you can do this without a drawing! I have to see it:

This drawing was made with a 30° slope. As should be visible, at this slope more than half the weight is supported by the rear wheels.

This also sounds like a homework problem, so I won't give the complete answer, but I think the drawing should help. With a coefficient of friction of 0.7 and the CG given, there is no danger of tipping over, as long as you are going up front first. With a front wheel drive backing up the hill, it is close enough that 'gunning it' just might make it tip.

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#18
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Re: Maximum Steepness A Car Can Climb

12/28/2009 9:27 PM

Its not a homework problem. I was flipping through an old mechanics text in my library and found a problem to spend some time thinking about. You know I'm bored....

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#7

Re: Maximum Steepness A Car Can Climb

12/27/2009 4:49 AM

Take a run at it.

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#8
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Re: Maximum Steepness A Car Can Climb

12/27/2009 5:14 AM

That's no fair! With a running start you can climb quite a bit steeper of a short hill.

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#9

Re: Maximum Steepness A Car Can Climb

12/27/2009 9:02 AM

Agree with all comments. I\d just propose one worst case as the limit: Calculate the angle required to transfer all the normal force to the rear axle. This is going to be the maximum. In this scenario, really will not matter if there\s front axle transmission or not. And - of course - if it tilts back, this is another thing... In the real life, as you should preview it, probably you\ll find that a 4-traction vehicle could push you a little further.

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#10

Re: Maximum Steepness A Car Can Climb

12/27/2009 11:37 AM

A better question would be, "What is the steepest angle, at which, that the car would hold from slipping down the hill?". That gets us out from underneath the standing-start, overcoming-inertia, flipping-over-backwards problem.

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#11

Re: Maximum Steepness A Car Can Climb

12/27/2009 1:50 PM

A life time of living out in the country has shown me that rear wheel drive climbs hills much better than front wheel drive. And all wheel drive is even better than rear wheel drive.

If you cant get up a hill with a front wheel drive car you just turn it around and go up it backwards! Believe me it works!

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#12
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Re: Maximum Steepness A Car Can Climb

12/27/2009 2:03 PM

That's very interesting. I have wondered whether car engines are better at rear drive pushing up a hill than a front drive pulling.

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#13
In reply to #11

Maximum Steepness A Car Can Climb

12/27/2009 11:41 PM

It works because the reduction in the reverse gear is higher than the 1st gear. This is true for rear wheel drive vehicles also. i.e., irrespective of whether it is front wheel drive or rear wheel drive, the gradeability of a vehicle is better in reverse gear.

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#20
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Re: Maximum Steepness A Car Can Climb

12/29/2009 11:46 PM

Try it! Gear reduction is not the problem. Traction is. Try it with a tractor as well. A rear wheel drive tractor will go up a hill better forward than when going backwards even though the weighted end of a tractor is the back tires. They still don't go up as steep a hill in reverse as well they can when going forward. Even if you reverse the tractor tire tread pattern they still cant climb up backwards as well as forward.

I have pushed snow for 20 years with common rear wheel drive tractors and when ever you cant back up a slope or a ditch you just turn around and go forward and you have a much better chance of crawling out going that way. Its similar as to sliding into the ditch with a front wheel drive car. Instead of trying to turn around and drive out forward simply backing up has better traction. Going forward with front wheel drive only works with momentum and some places you just cant safely turn around a get a run at the slope.

Like I said I grew up in the country and we live in a valley also. I have at one time or another had to climb a hill with every type of machine and vehicle we have ever had and I know from experience a front wheel drive car can drive out of a steep ditch or up a steeper hill better backing up than going forward. Just the same as a rear wheel drive tractor can drive up a steeper slope going forward rather than backward.

Even an unloaded rear wheel drive pickup can go up a steeper slope forward even though the weight is on the front!

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#14

Re: Maximum Steepness A Car Can Climb

12/28/2009 12:24 AM

I started to do a table on this, but it won't be accurate unless I account for the wheel radius....

Just a quick preliminary result: at about 30 degree or so of slope, the load will be divided about equally between the front and rear wheels. Less slope, the front wheels can give better traction; greater, the rear wheels. (Assuming either can be driven.)

Of course, if the car is rear drive, the front wheels are useless. If front drive, the rear wheels are useless. Under all front drive circumstances, putting the front wheels downhill and going in reverse will improve traction.

Dunedin NZ Baldwin St. 35%

Canton St Pittsburgh 37% (shorter length) (20.3 degrees)

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#16

Re: What's the Maximum Steepness a Car Can Climb?

12/28/2009 2:49 PM

And, there are at least three very short sections of street and one alley that are steeper than Canton!

However, back to your question.

Both wheels = 35°

Rear wheels = 17.98º

Front wheels = 20.25º

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#17
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Re: What's the Maximum Steepness a Car Can Climb?

12/28/2009 9:22 PM

TVP45 :

Those are interesting conclusions. It would be great if you could go through the math in brief with me. Note that others too might be reading this and it would be great help to all to know this stuff.

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#19
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Re: What's the Maximum Steepness a Car Can Climb?

12/29/2009 8:27 AM

OK. Leaving out units of distance since it is scale invariant.

Do a free body diagram (a rectangle with little points on each end. Show the COM at 1.75,0.6.

Draw all the forces. At the COM: Straight down is mgcosΘ; straight to the left is mgsinΘ.

At the place where the right point touches the level (road). Straight up is (1.75mgcosΘ - 0.6mgsinΘ)/2.9 {from torques about left hand point}.

At the left point: straight up is (1.15mg cosΘ + 0.6mgsinΘ)/2.9; to the right is 0.7[(1.75mgcosΘ - 0.6mgsinΘ)/2.9 ) + (1.15mg cosΘ + 0.6mgsinΘ)/2.9)].

Since every term contains mg, drop it from all. Use the entire bracket [] for both wheels, the left term for rear wheel drive, the right term for front wheel drive. I assume the car is travelling uphill to the right. Solve for Θ.

Happy New Year.

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