there have been a few ...... here
Power required goes up at the cube of change in speed, to double your speed you need 8x the power.
cube root(1.25) = 1.07721735, so a 25% increase in power would net you about 7.7% increase in speed.
That assumes you are going fast enough that aerodynamic drag is the dominant limit on your speed.
However if you are a 50cc motorcycle trying for a 33 mph speed record, rolling resistance is as much a factor as aerodynamic drag. Generally speaking aerodrag is insignificant (for our purposes) at speeds below 45-50 mph. On flat ground it only takes about 12-15 hp to cruise at 60 mph in a full sized car, and much of that is mechanical losses in rolling resistance and power train losses.
When you say 25% reduction in aero drag are you talking about a 25% reduction in frontal area, 25% reduction in CD (coefficient of drag) or a 25% reduction in the actual drag force?
The drag force varies directly with either CD or frontal area, reduce either by 25% and you get a 25% reduction in drag force. Or you can get a little of each. A common metric is to multiply the frontal area x the CD, as that number changes your drag force will change in the same proportion.
power = force x distance.
At a given speed (ie the same distance per second) the change in force will be identically the same as an equal change in power. By definition they are interchangable. But since drag increases at the square of the speed (velocity for the precise), drag changes much faster at higher speeds than at lower speeds.
As mentioned above a change in aerodrag (frontal area or CD) will hardly be noticed if the terminal speed of the car is under 50 mph, and will be very important at higher speeds.
To return to your question if by 25% reduction in drag, whether you reduce frontal area by 25% or CD by 25% or "drag force" by 25% through a combination of both you will get the same result.
Due to the definition of the relationship of power and force, there is no difference between changing power or drag force. An equal change in either will result in the same change in speed, "if your speed is fast enough that aerodrag completely dominates all other forces like rolling resistance".
what you mean by force in this definition has to do with the aerodynamics right ?
correct drag is a force that is acting opposite to the direction you are trying to go.
Quote
your saying there is no difference between changing power or the drag force . and that an equal change would result in the same speed
but the faster you go does that definition change or is changing the power or drag force still going to be equal.
No the definition does not change at any speed. What changes is that at low speed other forces are far greater than the aerodynamic drag, so even large changes in aero drag at low speed will not make any noticeable distance.
If it takes 16 hp to go 60 mph and 14 hp of that is used to over come rolling resistance and transmission loss, cutting aerodrag in 1/2 will hardly make a difference. If on the other hand the same car is going 206 mph and needs 850 hp to go that fast, it is now spending only 40-50 hp to overcome rolling resistance and transmission losses but using up 800 hp to over come air drag. If you cut its airdrag in half, it now has 400 hp extra available to go faster.
Larry
