The power required to overcome the aerodynamic drag is given by:
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
Source: Wikipedia
Many thanks to Jim Messina for sharing the following in depth analysis.
Air forces (parasite drag in pounds) goes up as the SQUARE of the increase in speed.
Work(energy used) = Force (parasite drag) x Distance traveled General Physics definition
Doubling speed, increases drag and energy burned by 4.
From 50 mph to 100 mph....100 ÷ 50 = 2 increase in speed. Square this(2= 4), energy burned is 4 times more, because the Force of air drag only, has gone up 4 times. 400%
From 55mph to 70mph....70 ÷ 55 = 1.2727 increase in speed. Square this (1.2727=1.62), 1.62 or 62% more drag to overcome, 62% more energy burned.
From 55 to 80 (the speed many drive on Interstates). Speed increase 1.45, square this (1.45 = for 2.11 increase in drag. Over double the energy used!
NOT included: rolling friction of tires (soft is bad or firm is better), interference drag between belly of the car and road nor power train friction. This treatment is to show speed and aerodynamic drag matters significantly. Not opinion but simple general physics and math.
CONTINUING
Horsepower Required goes up as the CUBE of the increase in speed.
Power = Force x Speed. General Physics definition
Derivation: Power = Force (of Drag) X Speed.
Power = (Drag increases as Speed² )X Speed -----> Speed³
Power ~ Speed³
Doubling the speed increases the power required by a factor of 8
From 50 mph to 100 mph needs 8 times the power (rate of doing work)
I.e. how rapidly the energy is used to cover the distance).
Simply 100 ÷ 50 = 2 2³ = 8
From 55 to 70 requires 2.06 times the power at 55, over double the power!
(Simply 70 ÷ 55 is 1.2727 1.2727³ = 2.06)
From 55 to 80 requires 3.1 times the power at 55.
(Simply 80 ÷ 55 is 1.45. 1.45³ = 3.1)
The typical coefficient of drag on an aerodynamically clean car is ~ .3
Honda Insight is one of the best .25
Porsche Cayman .29
Toyota Prius at .26
Hummer at .57
Obtain the Equivalent Front Plate area by multiplying the actual front plate area X the coefficient of drag. e.g. Insight 20.45 ft² frontal area X .25 = 5.11 ft² Equivalent.
Hummer 26.3 ft² equivalent (46.1 ft² frontal area X .57 = 26.3 ft² Equivalent)
http://www.insightcentral.net/encyclopedia/enaero.html
Obtain the air pressure at a speed in question.
In FLYING magazine July 2008, Peter Garrison “Technicalities” shows “q” pressure at 100mph is 25.7 #/ ft²
Insight 5.11 ft² X 25.7 #/ ft² = 131.4 # from frontal plate drag (at 100 mph)
Hummer. 26.3 X 25.7 #/ ft² = 675 # from frontal plate drag (at 100 mph).
Converting drag to horsepower using the "familiar" 1 HP = 550 ft-#/sec and the "familiar" 60 mph = 88ft/sec : 100 mph is 147 ft/sec
Recalling
Power = Force (parasite drag) X Speed
HP = Drag(#) X Speed (ft/sec) ÷ by 550 ft-#/sec
Insight at 100 mph
35.1 HP = (131.4 X 147) ÷ 550
Insight at 50 mph power required is 4.38 HP
50 ÷ 100 = ½ (½)³ = 1/8 (35.1 ÷ 8)
Hummer at 100 mph
180.4 HP = (675 X 147) ÷ 550
Hummer at 50mph power required is 22.5 HP
50 ÷ 100 = ½ (½)³ = 1/8 (180.4 ÷ 8)
This does not consider anything but flat plate aerodynamic drag resulting from different speeds. The weight, tire friction,
power train losses are in addition to the air drag).
