# Driving 55 and Math | Last Updated: July 14, 2020

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). An ex-salesman of industrial equipment, Shawn used to drive nearly 60K miles a year just commuting to clients. He also has a little project Miata build going on the side. Safe to say, Shawn has slain a few tires in his days. He knows all about horrid road-noise, hydroplaning risks, and how much damage a bad alignment can do to your wallet. He enjoys helping us out and Chris always values his opinion when designing something new for the website.