How Much Axle Torque Is Needed to Prevent a Car from Rolling Downhill?

[danja]

Alright, I've re run this numerous times in my head, and on paper. It seems I must be missing something stupid here. Bottom line is, I am trying to calculate the amount of axle torque (post differential) required to stop a 1500 KG car from rolling down a 20 degree hill. Road to axle radius is assumed to be 8 inches. No, it's not a homework question =D though it does sound like one.

Anyways, here is the issue. I have arbitrarily picked 20 degrees since most roads don't slope higher than that. Since this is the torque to hold the car still, the angular acceleration is zero along with drag and the like.

Now, summing the torques will give the general equation (neglecting drag and friction):

Taxle = m*g*r*sin(hillangle) + m*r^2*(alpha)

In this case:

Taxle = 1500*9.8*0.2*sin(20) + 0

Or: Taxle = 1021.63 Nm (753.55 lbft)

Am I to believe that this is really correct?? Given that an average car has less than 200 Nm peak torque (at the wheels) this seems very off. Now what did I miss? This is driving me crazy!

 

[russ_watters]

Welcome to PF.

Are you sure that 200 Nm is the torque at the wheels, not the torque at the crankshaft? The difference would be that you multiply the torque at the crankshaft by the gear ratio to get the torque at the wheels. I've never done a dyno test, but I would think they are calibrated to remove wheel size and gear ratio from the results so you don't have people screwing with the test by selecting a different gear.

 

[Dr.D]

Your calculation looks OK; I don't know where you got the 200 Nm peak torque value from, but there does seem to be a conflict. A 20 deg incline is very, very steep, more than I think I have ever seen.

 

[danja]

Welcome to PF.

Are you sure that 200 Nm is the torque at the wheels, not the torque at the crankshaft? The difference would be that you multiply the torque at the crankshaft by the gear ratio to get the torque at the wheels. I've never done a dyno test, but I would think they are calibrated to remove wheel size and gear ratio from the results so you don't have people screwing with the test by selecting a different gear.

I had some trouble finding OEM curves, but for example here is the dyno from a stock older Integra. Little less than 1500 Kg, but the point still holds.

http://image.importtuner.com/f/1684..._z+1993_acura_integra_gs+stock_dyno_chart.jpg

Max Tq of 110 lbft (150 Nm). And as far as I know, dyno's measure wheel torque, and calculate wheel power from that, so the gearing/wheel radius has already been taken into account with that.

I must have something wrong, or am missing something, but I simply can't see what it is.

 

[danja]

Your calculation looks OK; I don't know where you got the 200 Nm peak torque value from, but there does seem to be a conflict. A 20 deg incline is very, very steep, more than I think I have ever seen.

I took it as the torque you might expect an average 1500 Kg car to reach at peak torque. It's not really based off a specific car, but I can tell you a lot of smaller cars don't go over that from the factory (as you can see from the integra, it is around 2/3 of that).

 

[danja]

Welcome to PF.

Are you sure that 200 Nm is the torque at the wheels, not the torque at the crankshaft? The difference would be that you multiply the torque at the crankshaft by the gear ratio to get the torque at the wheels. I've never done a dyno test, but I would think they are calibrated to remove wheel size and gear ratio from the results so you don't have people screwing with the test by selecting a different gear.

Alright, sorry to re-quote, but it seems you might be right. I thought the wheel torque dyno's measured referred to the torque output from the wheel in a specific gear. Apparently, that's not what it means. Wheel torque is the engine torque derived from force measured at the wheels, so it is just engine torque including frictional losses. Now it all makes sense since putting that 150 Nm through gearing easily drives it up the hill!

Here is actual wheel thrust for a generic vehicle. Friend of mine just found it.

http://wahiduddin.net/race/images/thrust1.gif

Here, 1600 lbf of thrust (at the wheel) is about 7120 N, which far exceeds the 1022 N required to hold the car from movement.

So, dumb question has been resolved. If you ask me, wheel torque should be renamed to estimated torque. =)

 

[Bob S]

I believe your eqn T_axle = m*g*r*sin(hillangle) is ok, as long as r is the wheel radius. Your differential is a torque multiplier (divider), as is your transmission. So using m = 1500 Kg, g = 9.81 m/sec², R= 0.3 m, hillangle = 10 deg, differential = 4:1, and trannie = 1.3:1 gives a flywheel torque of 147 Newton-meters.

 

[danja]

I believe your eqn T_axle = m*g*r*sin(hillangle) is ok, as long as r is the wheel radius. Your differential is a torque multiplier (divider), as is your transmission. So using m = 1500 Kg, g = 9.81 m/sec², R= 0.3 m, hillangle = 10 deg, differential = 4:1, and trannie = 1.3:1 gives a flywheel torque of 147 Newton-meters.

Yep, thanks for the verification. Glad to have sorted out my terminology now! Also, I think it's tranny, not trannie.. =p

Actually the reason I asked this question is I am working on a Simulink model of a motorized bicycle I plan to build, and when I put in a simulated hill to check the response, it threw everything off, but now I know why haha.

Not quite a car, but I figured a more general question was the way to go.