52 States in 8 Months

Andi’s AWD trip

Posted in Las Vegas to El Paso, University stuff by Ulf on May 22, 2009

Hey, I told you that Andi didn’t join us on our trip to the Delicate Arch? He had other plans. There are many roads in the Arches National Park that are for four-wheel-driven cars only. Now, we had an AWD Jeep. So far so good. Andi only needed a place to go. The golden rule from old video games is: whenever you see a closed door, you have to open it. There will be something interesting behind it. Guess what Andi did?
2009-03-27_11-15-28 Springbreak 0008 Andreas
His plan was to reach one of the hills on the following picture, and things were really looking good for him! There were a few trails were other people had driven before, so Andi knew that things were safe.

2009-03-27_11-11-26 Springbreak 0006 Andreas
2009-03-27_11-07-17 Springbreak 0002 Andreas 2009-03-27_11-07-35 Springbreak 0003 Andreas

However, just before he reached the top of his hill, he got stuck. There was no chance of getting any further. He even had to put the doormats under his wheels to get enough grip for getting of the hill again…

2009-03-27_11-09-33 Springbreak 0004 Andreas 2009-03-27_11-09-48 Springbreak 0005 Andreas

In the end, Andi was really satisfied. Driving on those dirt roads is just so much fun. And he had successfully put enough dirt onto the car to make us wash it later that day…

2009-03-27_11-56-47 Springbreak 6969 Ulf 2009-03-27_18-22-22 Springbreak 7036 Ulf

You know what this story reminds me of? A very small calculation. It shows that the maximal steepness which a car can climb is quite limited. Free_Body_DiagramLet’s assume a non-interlocking connection between your wheels and the ground, and static friction. You want to drive up a hill with a steepness of α. α=0 would be a level surface, while α=π/2 would be orthogonal to the ground. Your car’s gravity force is m·g, but the normal component (which presses you to the ground) is just N=m·g·cos(α). The other fraction of the gravity force is tangential to the road, so it will try to pull you off the hill: m·g·sin(α). The only force that stops you from going downhill is the static ground friction. Its maximum value can be estimated as Ff=µ·N. Putting it all together we have m·g·sin(α)=µ·m·g·cos(α), or α=arctan(µ).

The coefficient of friction µ is roughly 0.4 on normal soil, 0.6 on tarmac and 0.8 if you drive a track vehicle (such as a tank). Now, from the formula I presented above you can easily calculate the maximum steepness α.
For µ=0.4 it is α=22°, for µ=0.6 you get α=31° and the tank will go uphill with at most 39° of steepness. That should explain why Andi couldn’t make that hill…

The numbers above do of course assume that it’s dry. Once things get wet you’ll end up with a coefficient of friction in the order of µ=0.2. That is 11° of steepness. If things get icy, then µ can become virtually 0. And if you don’t have an all-wheel-driven car, than approximately half your car’s normal force will go to non-actuated wheels. That basically means that you can divide all these values by two…

Oh, I forgot to give the good news: The calculation I presented describes only a static case. If you can get enough momentum by taking a run-up, there are no such limits :-).

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One Response

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  1. scytale said, on May 22, 2009 at 11:37 am

    Solution: http://en.wikipedia.org/wiki/Unimog ;)


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