- #1
jimgram
- 95
- 1
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??
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??
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