The force of drag is proportional to velocity?

[PhysicsLover0]

So if I double the velocity, aerodynamic drag doubles as well?

 

[PhanthomJay]

Sometimes, if the speed is relatively slow, that is correct (linear drag, F=kv). In general, however, at higher speeds, the drag goes up as the square of the velocity (quadratic drag, F=kv^2), so if you double the velocity, the drag force goes up by a factor of 4. That is why for example a very strong hurricane with 150 mph winds exerts a force on an object which is 4 times greater than the force exerted on an object by a weaker hurricane with 75 mph winds.

 

[DarioC]

Real world example. Pontiac Firebird at salt flats. Took 600 HP to go 200 mph, and 1200 hp to go 300.
DC

 

[Ranger Mike]

good one dario! takes mucho HP to move that 14.7 psi curtain of air out of the way!

 

[Researcher X]

It takes 4 times the energy to achieve twice the velocity for a given mass even in a vacuum, anyway. With drag squaring with a doubling in speed, the power requirement for overcoming that force will be 8 times when it was twice as slow! However, in reality, drag usually only makes up a proportion of the losses due to friction. One question I've often wondered about is what determines when the drag makes up the main proportion of the losses due to friction? Could terminal velocity be considered as a good marker?

For example, when terminal velocity is reached, it is because drag from the air is matching the force of gravity (or in a car the driving force). Does this mean it is producing a 9.8m/s acceleration in the opposite direction? If a car was accelerating at 20m/s^2 would the drag make up almost 50% of the losses due to friction? Can terminal velocity at a particular speed be used to calculate the losses due to drag at a particular multiple or division of that speed and then know the required power increase or decrease?

If a terminal velocity for a car when accelerated by engine power at 12m/s^2 was 50m/s would the drag force at 100m/s be 48m/s^2? If the car was producing 100hp at 50m/s, then it will need to produce nearly 800hp at 100m/s (The drag is dominant 99%)? On the other hand, below it's terminal velocity: if the car was accelerating at its maximum rate, to break a mere 25m/s; at this point the drag is only making up 1/4th of the forces slowing the car, and almost 3/4ths is just the power required to reach that velocity that quickly sans drag?

 

[D H]

It takes 4 times the energy to achieve twice the velocity for a given mass even in a vacuum, anyway.

Careful, there. This will get you into all kinds of trouble.

Suppose three people observe a 1 kg object gain a velocity of 2 m/s. Prior to the change in velocity,

• Observer #1 is at rest with respect to the object,
• Observer #2 is moving at 2 m/s wrt the object, in the direction of the new velocity, and
• Observer #3 is moving at 2 m/s wrt the object, against the new velocity.

Observer #1 will see the object gain 2 joules of energy, #2 will see the object lose 2 joules of energy, and #3 will see the object gain 6 joules of energy. They can't all be right, can they? (They are; energy is a frame-dependent quantity.)

Rockets are yet another problem.

With drag squaring with a doubling in speed, the power requirement for overcoming that force will be 8 times when it was twice as slow!

Now you are confusing the energy needed to get up to some velocity versus the energy needed to maintain that velocity. Ignoring drag, the energy required to maintain a particular velocity is *zero*.

In the case of the cars cited in post #3, most of that 300 HP (1200 HP) was expended in pushing a "14.7 psi curtain of air out of the way."