Re: Can a car with a large engine drive up a hill w/ an angle greater than 45*?

Laurence,

My initial guess is to say that some cars may have the power necessary to drive up a 45 degree angle, but they probably won’t have the traction. Let’s see how good my guess is.

I’ll start with an Excursion since you mentioned that truck in your question. I looked up some basic specs on the Excursion from carpoint. An Excursion weighs 7156 lbs and with the diesel engine makes 500 ft-lbs of torque. To do this right, I also need to know the transmission gear ratios and tire diameters, but I’ll just make some reasonable assumptions. We’ll say it has a 3:1 first gear ratio, a 3:1 axle ratio, and 2ft diameter (1ft radius) tires. May not be perfect, but it will do. There’s also a handful of other assumptions I’m kind of making here, but it will do to answer your question.

Force the Excursion can produce:

500ft-lbs * 3 * 3 / 1ft = 4050 lbs of force.

Force the Excursion must produce to climb a 45 degree incline:

7156 lbs * sin(45) = 5060 lbs.

Doesn’t look like the Excursion is going to make it. It’s just too heavy. We’ll try something a little lighter and see how it does. How about an F- 250 with the same engine. Force output is the same (4050 lbs).

Force the F-150 must produce:

5603 lbs * sin(45) = 3962 lbs.

Looks like the F-250 has the strength to do it, but there’s still the traction issue. At a 45 degree angle the force you must exert to climb the hill is equal to the normal force (the weight of the vehicle against the hill) because cos(45) = sin(45). That means in order for ANY vehicle to be able to climb a 45 degree angle hill, the coefficient of friction between the tires and the ground must be greater than or equal to 1. And that just isn’t realistic. If I remember right, an vehicle with good tires on dry pavement only has a coefficient of friction of about .8. That means that even though the F-250 can produce 4050 lbs of force, it will start to slip when it’s producing greater than 3170 lbs (3962 * .8). And that’s with 4- wheel-drive. It will be even worse with 2 wheel drive, and just plain useless in a front wheel drive car.

Here’s a couple more tidbits of knowledge: The steepest angle a car with 4wheel drive and tire coefficient of friction of 0.8 could drive would be 39 degrees. Also, a 45 degree angle translates into a 100% grade. We have a 30% grade (17 degrees) here at our test track, and it’s pretty intimidating.

Your Mad Scientist,
Mike Scannell