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Rolling Resistance for Large Trucks |
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| May4-12, 11:08 AM | #1 |
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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?? |
| May4-12, 04:16 PM | #2 |
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Looks about right to me:
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: |
| May16-12, 11:13 PM | #3 |
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Air resistance isn't linear in the sense we think it is.
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| May16-12, 11:35 PM | #4 |
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Rolling Resistance for Large Trucks |
| May17-12, 01:40 PM | #5 |
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| May17-12, 03:11 PM | #6 |
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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. |
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