All you need to know.

W = F X D

Work (energy consumed) = Force (aerodynamic parasite drag) X Distance traveled.

Aerodynamic parasite drag force increases as the Square of the increase in speed.

From 55 mph to 80 mph the speed increase is 1.45. 1.45 squared is 2.11

Over DOUBLE THE ENERGY USED!

Conversely slowing from 80 to 55 (55/80 = .6875 Square this = .473 used)

cuts the energy, fuel burned to less than 1/2.

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).