Quote:
Originally Posted by *GhostRide37  yeah, you said "probly" (probably) dumping more gas at 1hr half speed. with you saying "probably", ill take my chances and get there half hour earlier. unless you're 100 % sure...
and air resistance is an entire different story. if you were in an airless vacuum doing 80 at low rpm and doing 30 at high rpm, youd probably use more gas doing 30.
*but anyway, before people get carried away here...much of what i said is just me and my sarcastically extremist views for shock value......but in all seriousness.....if you drive slow but do gas-brake-gas-brake....or drive fast and just dont brake until your there....youll use the same amount of gas (if not less for not breaking and accelerating) |
I'll get carried away i guess:
Basic thermal eff.
Nth= Wout / Qin
Where Nth= engine efficiency Wout= energy from force out
Qin= energy in (fuel)
Basic translation is Efficiency equals the % of heat transformed into work (energy).
Engine eff. is roughly 25% due to losses in the engine (engine heat, friction, work used to turn engine equipment)
This simply means that out of 100% energy that fuel contains (Kinetic energy), only 25% is actually utilized to create work (move your car).
At high rpms your engine needs to works harder to create the same out of Wout at a lower rpm due to higher friction losses, more engine heat, etc.
Basically the total Wout will be less because your Wlost is higher at high engine speed.
Havent you seen that old school quaker state motor oil commercial with Dennis Leary?
HEAT is bad for engines, and in thermal efficiency terms you see bigger losses the harder the engine runs and the hotter it gets
Nth=Wout/Qin
When engine runs at high rpm, Wloss increase. Which means Wout is less.
For example
Nth1=Wout/Qin @ 65mph
Nth2=1/2Wout/Qin @120mph
Nth1 > Nth2