Recognitions:
Gold Member

## Rolling Resistance for Large Trucks

I'm calculating the rolling resistance and air drag for a 60,000 lb truck. I used a Cd=0.7 for drag coefficient; 6 m2 for frontal area; and a rolling coefficient of Cr=.008; density of air=1.3 kg/m3.

I get a force from rolling resistance of 2142 N (Cr*mass*g). I get a force from air drag at 35 mph of 676 N (1/2*ρ*Aveh*Cd*v2).

All is well - until I calculated the deceleration of this truck based on these two forces (I.E. the transmission is in neutral and no brakes) (a=f/m). I get a deceleration of -0.011 g (-0.103 m/s2).

This decel rate yields a distance to coast to a stop (v2/2*a) of 1186 meters - nearly 3/4 of a mile and taking 152 seconds (almost 2.5 minutes). This, to me, seems unbelievable for a 60,000 truck. If I change the velocity to 60 mph I get a 1 1/2 mile coast taking about 3 minutes. Can't be - where did I go wrong??
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study

And I don't think the car was in neutral, so there was probably some engine braking involved. There is also this other real-life example:
 On a training day with negligible wind, sunswift IV entered the flat straight of the race track at 70 k.p.h. The driver then switched the motor off and coasted, using no brakes. After covering a marked kilometer, she was travelling at 50 k.p.h. What is her acceleration?
In your example, the real result would even be longer as drag will lower with speed, hence acceleration will lower as well.
 Air resistance isn't linear in the sense we think it is.

## Rolling Resistance for Large Trucks

 Quote by jimgram All is well - until I calculated the deceleration of this truck based on these two forces (I.E. the transmission is in neutral and no brakes) (a=f/m). I get a deceleration of -0.011 g (-0.103 m/s2).
As jack action observes, air resistance decreases in magnitude as the speed decreases, so it's incorrect to use a constant acceleration formula to solve this problem. The solution is a bit more complicated since it's a non-linear ODE as well: $$m\dot{v} = - b v^2 - Cmg.$$ If you solve it (it's a little tricky to do, but possible), you'll see that you get a bit more than 2.5 minutes stopping time when you take into account the reduced force with time. If you're confused, you're probably thinking it's because the truck is so massive that it should have a large resistive force against it -- and that's true. But it also has a lot of momentum, and the drag forces take a long time to erode it all. Or another way to think about it is that the mass represents resistance to acceleration, and in particular more mass makes it take longer to decelerate the truck (and this effect competes with the larger rolling resistance).

 Quote by mylarcapsrock Air resistance isn't linear in the sense we think it is.
Well, hardly a criticism, as the drag listed in the OP is indeed non-linear :)

 Quote by Steely Dan Well, hardly a criticism, as the drag listed in the OP is indeed non-linear :)
Really? Show me where it's NOT in his calculation.
 Recognitions: Gold Member I'm convinced. I think I'm just accustomed to trucks with improperly inflated tires and drivers rarely, if ever, shift to neutral and totally decouple the engine. In fact, it's more common to use engine braking. BTW - I know air drag is a function of velocity squared and while it only accounts for 24% of the total drag at 35 mph, it very rapidly diminishes, making the coast time and distance even greater. Thanks for all of the input.