Original post by MissoniMissoni
At 90 mph on cruise control over a 200 mile journey ?
Whatver MPG you get will depend on the car you're driving.
Original post by Tappouni
Whatver MPG you get will depend on the car you're driving.
Sorry it certainly would
I drive a 2006 BMW 320d
Original post by MissoniMissoni
Sorry it certainly would
I drive a 2006 BMW 320d
I drive a 2006 BMW 320d
Googling it says the car averages 46.1 MPG, but at 90 mph (assuming you don't slow down/speed up much), I'd guess perhaps somewhere around the 40 MPG mark
(bearing in the mind the car probably isn't quite operating as efficiently as when it came out of the factory)Original post by Tappouni
Googling it says the car averages 46.1 MPG, but at 90 mph (assuming you don't slow down/speed up much), I'd guess perhaps somewhere around the 40 MPG mark (bearing in the mind the car probably isn't quite operating as efficiently as when it came out of the factory)
(bearing in the mind the car probably isn't quite operating as efficiently as when it came out of the factory)I know it's rated at 46.1 MPG, but I would have thought on the motorway at a constant speed I should get a high MPG.
On the motorway you are at a constant speed so no need to accelerate and brake etc.
Original post by MissoniMissoni
I know it's rated at 46.1 MPG, but I would have thought on the motorway at a constant speed I should get a high MPG.
On the motorway you are at a constant speed so no need to accelerate and brake etc.
On the motorway you are at a constant speed so no need to accelerate and brake etc.
I know, but 90 you'll be working the engine harder than at 60 or 70
Original post by Tappouni
I know, but 90 you'll be working the engine harder than at 60 or 70
My car has a top speed of nearly 140 mph
So 90 is still only about 65% of top speed.
I think I'd have to be doing 120+ to see a big drop in economy
Original post by MissoniMissoni
My car has a top speed of nearly 140 mph
So 90 is still only about 65% of top speed.
I think I'd have to be doing 120+ to see a big drop in economy
So 90 is still only about 65% of top speed.
I think I'd have to be doing 120+ to see a big drop in economy
100+ you'll start noticing a big drop, but actually with a 6 speed I suppose you could achieve 50 mpg
Original post by Tappouni
100+ you'll start noticing a big drop, but actually with a 6 speed I suppose you could achieve 50 mpg
It's not the number of gears, but rather the length of the top gear that matters, although I get your point.
Straight off Wiki...
The power required to overcome the aerodynamic drag is given by:
P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
The power required to overcome the aerodynamic drag is given by:
P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
Straight off Wiki...
The power required to overcome the aerodynamic drag is given by:
P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
The power required to overcome the aerodynamic drag is given by:
P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d
Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
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