2x Speed = 8x Power?

[Username1]

I’ve read that doubling your speed requires 8x the power, and I’m trying to understand why. Here’s my current understanding…

Because the speed is doubled you’re encountering (slamming into) twice as many air molecules, which on its own requires 2x the power.

Because the speed is doubled you’re slamming into the air molecules twice as hard, which on its own requires 2x the power.

So you’re encountering twice as many air molecules, and hitting them all twice as hard. These two separate doublings when both applied result in 4x power usage. Can anyone explain where the third and final doubling comes from to reach 8x power usage?

 

[DogDipstick]

That is just a "saying", and not an absolute.

It is relative to load.

The curve is an " exponential". Power vs load requirement vs friction.

It is the friction that exponentially accumulates. If the wind is what this friction comes from, then the wind is the cause. Wind can create hundreds, if not thousands, of degree temp rise from drag... enough of it. Friction.

A saying about 2x or 8x or whatever... is only that. An old saying. Applied different, for example. based on encountered speeds and load... like a bike in the wind... or a boat in the water... or a airplane in the sky, a train on a rail or a spacecraft on the way to Mars........ They all have different numbers for the equation(s). The only thing they share is a logarithmic scale... an exponential gain in the amplitude of the load ( friction) based on velocity.

Something like that.

See the little " ^2 " in this equation. That is what saps the powa. That lil old exponent. This is the root of your square.

You can read this as " Friction ( air) equals Constant K( air) times Velocity (squared)."

So you can see how velocity matters and is exponential when you divide it out.

The frictional force of air resistance ( F air) acts on the moving body.
...
Fair = CV^2

Fair depicts the force of air resistance.
C refers to the force constant.
V depicts the object's velocity.

 

[Hayds13]

double power for every 10km/h increase...

 

[Blacklite]

That is just a "saying", and not an absolute.

It is relative to load.

The curve is an " exponential". Power vs load requirement vs friction.

It is the friction that exponentially accumulates. If the wind is what this friction comes from, then the wind is the cause. Wind can create hundreds, if not thousands, of degree temp rise from drag... enough of it. Friction.

A saying about 2x or 8x or whatever... is only that. An old saying. Applied different, for example. based on encountered speeds and load... like a bike in the wind... or a boat in the water... or a airplane in the sky, a train on a rail or a spacecraft on the way to Mars........ They all have different numbers for the equation(s). The only thing they share is a logarithmic scale... an exponential gain in the amplitude of the load ( friction) based on velocity.

Something like that.

See the little " ^2 " in this equation. That is what saps the powa. That lil old exponent. This is the root of your square.

You can read this as " Friction ( air) equals Constant K( air) times Velocity (squared)."

So you can see how velocity matters and is exponential when you divide it out.

The frictional force of air resistance ( F air) acts on the moving body.
...
Fair = CV^2

Fair depicts the force of air resistance.
C refers to the force constant.
V depicts the object's velocity.

It's not exponential, it's cubic. Anything to a power is a quadratic, cubic, etc. etc. function. Exponential has the exponent that is the variable. Anyways...

For a constant force moving something at a constant velocity one can derive the following.

Knowing -
Power = work / time
Work = force * distance
For a constant velocity, Distance = velocity * time

Combining you get

Work = force * velocity * time
Power = (force * velocity * time) / time

Cancelling out the time you get Power = Force * Velocity

As the OP posted air resistance tends to scale quadratically with velocity. In other words a double of speed creates four times as much drag. Plug that back into the earlier equation.

Power required for double speed = (4 * Force) * (2 * Velocity) = 8 * Force * Velocity = 8 * Power for initial speed

So there is a cubic relationship between power and velocity.

 

[Username1]

Another example is doubling your speed requires 8x as much power. This is because air molecules are hitting you twice as hard, you are hitting twice as many air molecules - so 4x the energy - but since you are going twice as fast you put out that energy in half the time, so 8x the power.

Above is a quote I read about this subject, would you say it’s correct? For some reason my mathematically retarded brain can’t understand the 3rd doubling (highlighted in bold).

—————————————————————-

Let’s give an example using a fixed ride duration. If this is not correct please tell my why…

You ride for 1 hour using a constant 1 kW. This results in a constant speed of 50 km/h, and a distance travelled of 50 km.

You ride again for 1 hour using a constant 4 kW. This results in a constant speed of 100 km/h (double), and a distance travelled of 100 km (double).

 

[DogDipstick]

So there is a cubic relationship between power and velocity.

Hahaha Thankyou for the corrections and the good explanation.

 

[Sturmey]

Hi. My understanding is that its a mistake to treat wind resistance like friction, which is often linear. When a cyclist runs into air, he puts the air particles into motion, very much like a snooker ball striking another. There is a transfer of Kinetic energy from the cyclist to the particles of air. The Kinetic energy formulae is K.E. = 1/2 m V V (half M V squared)
So if we double our speed, the kinetic energy lost/transfer will increase by the square law to a factor of 4 but also we will bump into double the amount of air particles, so this increases our losses due to air resistance by a factor of 2 to a total factor of 8 times the velocity.
The above applies to still air (calm day). Wind in our face (for example) affects the Kinetic energy value only so wind resistance in this case is Wind resistance loss = (bike velocity + wind speed as a vector)(squared) X bike velocity.
However, its important to note that especially at lower speeds, friction and tyre rolling resistance which may act linearly play an important role, so our speed/power curve is a combination of these factors as shown below.

 

[Comrade]

its a mistake to treat wind resistance like friction

wind resistance is drag, and drag is a type of friction

20³ m/h = 8,000
40³ m/h = 64,000

64,000 / 8000 = 8x

Not rocket science.