hardtop makes u faster?
01-30-2009, 08:08 AM
#36
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Originally Posted by ace123' date='Jan 29 2009, 10:13 PM
keep in mind the difference will be very small until at least freeway speeds. this will NOT change how quickly you get to 60, and it won't change your 1/4 mile time significantly either.
but yes, better aero = faster when drag starts to become more significant.
in rough terms, drag force increases with the square of your vehicle's speed, so if you know your drag force at 50mph:
your drag at 50 mph is a certain amount of lbs
your drag at 70 mph is 2x that amount of lbs
your drag at 100mph is 4x the amount of lbs at 50
your drag at 140mph is 8x the amount of lbs at 50
but yes, better aero = faster when drag starts to become more significant.
in rough terms, drag force increases with the square of your vehicle's speed, so if you know your drag force at 50mph:
your drag at 50 mph is a certain amount of lbs
your drag at 70 mph is 2x that amount of lbs
your drag at 100mph is 4x the amount of lbs at 50
your drag at 140mph is 8x the amount of lbs at 50
So yeah, the drag goes with the square of speed.
But the power required to overcome that drag = drag * speed
So the power required actually goes as the CUBE of speed. But the engine power available is basically fixed. And your ability to accelerate depends on the net available power. So at high speeds, drag is even more important than is implied by the speed^2 relationship.
01-30-2009, 09:15 AM
#39
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Originally Posted by mikegarrison' date='Jan 30 2009, 12:08 PM
Drag = 1/2 * air_density * speed^2 * drag_coefficient * reference area
So yeah, the drag goes with the square of speed.
But the power required to overcome that drag = drag * speed
So the power required actually goes as the CUBE of speed. But the engine power available is basically fixed. And your ability to accelerate depends on the net available power. So at high speeds, drag is even more important than is implied by the speed^2 relationship.
So yeah, the drag goes with the square of speed.
But the power required to overcome that drag = drag * speed
So the power required actually goes as the CUBE of speed. But the engine power available is basically fixed. And your ability to accelerate depends on the net available power. So at high speeds, drag is even more important than is implied by the speed^2 relationship.
The extra weight of the soft top and motors means nothing when it comes to top speed. 
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