Speed, energy, power and air resistance

[Patrick Pihl]

What is wrong here. If I double speed drag will be 4 times higher. The engine needs to put out approximately 4 times more power. Energy consumption will be 4 time higher for a specific time period. However if it is a fixed distance I will arrive in half the time an the energy consumption will therefore only double. Now if I look on the formula E=FxS (energy=Force x Distance) the force will be 4 times higher and therefore also the total energy needed. //Thanks

 

[russ_watters]

What is wrong here. If I double speed draft will be 4 times higher. The engine needs to put out approximately 4 times more power. Energy consumption will be 4 time higher for a specific time period.

No, power (energy over a specific time) is 8x higher. Drag is 4x higher, and speed is 2x higher (or time is 1/2 as long).

 

[Patrick Pihl]

No, power (energy over a specific time) is 8x higher. Drag is 4x higher, and speed is 2x higher (or time is 1/2 as long).

Well, then it make sense. Not very intuitive (to me) that power increase that way in relation to drag.

 

[dean barry]

the aerodynamic drag force is proportional to the square of the speed
power required is : ( aerodynamic drag force overcome (+ rolling resistance if you want) ) * speed

if fuel consumption is energy consumed, is this power * time ?

 

[russ_watters]

the aerodynamic drag force is proportional to the square of the speed
power required is : ( aerodynamic drag force overcome (+ rolling resistance if you want) ) * speed

if fuel consumption is energy consumed, is this power * time ?

That's correct...if you also include engine efficiency.