I'm not following. Airflow over the car is the same, 60mph+25mph headwind is the same as 85mph into dead calm. So it's the same drag force. Which thus would require the same power to overcome.
How much does a 25 mph headwind matter?
- Thread starter Hermann
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Yep. always better ( time/fuel ) when less wind when doing return flights.
When headwinds are strong, dispatch will flight plan tracks to avoid the strong Jetstream heading westbound ( longer distance but less HW aka min time track ) then plan for us to follow the Jetstream as much as possible on the way back, eastbound. Minimum time tracks versus the actual distance flown.
Yes, same drag force. Not more power, because power is drag times speed.
Speed as defined how?
Road speed or airspeed?
My impression is that aero drag is the single most significant factor in fuel consumption at higher speeds.
dnewton3
Staff member
As many have said, the 25mph headwind will have a variable effect depending on your vehicle speed. Think of it as a contributor to the net speed calculation.
60mph vehicle speed + 25mph is a big headwind; equivalent to 85mph.
10mph vehicle speed + 25mph isn't nearly as difficult to overcome; 35mph equivalent.
Think of this as a "net speed" equivalent value.
Drag force (wind resistance) increases with the square of the speed. If you double the speed magnitude, you quadruple the force required to overcome that increase. It takes 4x more force to overcome drag at 80mph as it does at 40mph. (To be honest, it's not a flat slope line, but a curved partial parabola. But for a quick generic reference, you can think of it this way.) https://www.engineeringtoolbox.com/wind-load-d_1775.html
But don't confuse force with work or power; they can be related, but not they are not the same.
Force is a product of mass multiplied by acceleration. (We generally talk of force in units of pounds in the US)
Work is a product of force multiplied by distance; W = F x D (We will use lbs for force and miles for distance)
Power is a product of work per unit of time. P = W / T (We will use an hour as the standard of time) (one HP = 550 ft-lb/sec, and to convert miles to feet and hours to seconds, there is a factor-labeling constant of 1.467).
Illustrative example ... (we'll make some generic presumptions for the sake of simplicity; assume this is a "simple" physics example where air density, etc are all normalized, rolling resistance of the tires are ignored, no headwinds, etc.)
40 mph net speed example
Let's say you have a car which takes 25 lbs of force to drive through the air at 40mph.
And, consider you want to drive a distance of 40 miles.
The "work" would be 25 lbs x 40 miles = 1000 lb-miles.
The "power" would be measured in an hour's time; 40 miles driven divided by 40 mph is 1 hour resultant elapsed duration.
so the power requirement is thus:
P = W / time or 1000 lb-miles divided by one hour.
P = 1000 lb-miles per hour.
80 mph net speed example
Now, if you decided to drive that same 40 miles distance but increase your speed to 80mph, this would be the result ...
drag force to overcome is now approximately 100 lbs due to doubling the vehicle speed
"work" would now be 100 lb x 40 miles = 4000 lb-miles (four times greater than the slower example)
But ... you did the job in half the time, so the formula is thus:
P = W / time or 4000 lb-miles divided by 1/2 hour (4000 lb-mi / .5 hr)
P = 8000 lb-miles per hour
Double your net speed, and your power requirement is roughly EIGHT TIMES GREATER.
Again, much depends upon nuances like air density, size of the resistant object, rolling resistance, crosswinds, etc. The chart linked above shows a more accurate curve, but the simple mind-note of 4x force and 8x power for a double of speed are easy to remember. But the general nature of wind drag is essentially as shown above. If you hold all other things constant, then power demands are prodigiously and viciously ravaged by net speed increases.
Even seemingly minor increases in net speed can have a significant effect on power consumption.
Headwinds have the same effect as speed increases. A little extra headwind makes a BIG power demand increase.
60mph vehicle speed + 25mph is a big headwind; equivalent to 85mph.
10mph vehicle speed + 25mph isn't nearly as difficult to overcome; 35mph equivalent.
Think of this as a "net speed" equivalent value.
Drag force (wind resistance) increases with the square of the speed. If you double the speed magnitude, you quadruple the force required to overcome that increase. It takes 4x more force to overcome drag at 80mph as it does at 40mph. (To be honest, it's not a flat slope line, but a curved partial parabola. But for a quick generic reference, you can think of it this way.) https://www.engineeringtoolbox.com/wind-load-d_1775.html
But don't confuse force with work or power; they can be related, but not they are not the same.
Force is a product of mass multiplied by acceleration. (We generally talk of force in units of pounds in the US)
Work is a product of force multiplied by distance; W = F x D (We will use lbs for force and miles for distance)
Power is a product of work per unit of time. P = W / T (We will use an hour as the standard of time) (one HP = 550 ft-lb/sec, and to convert miles to feet and hours to seconds, there is a factor-labeling constant of 1.467).
Illustrative example ... (we'll make some generic presumptions for the sake of simplicity; assume this is a "simple" physics example where air density, etc are all normalized, rolling resistance of the tires are ignored, no headwinds, etc.)
40 mph net speed example
Let's say you have a car which takes 25 lbs of force to drive through the air at 40mph.
And, consider you want to drive a distance of 40 miles.
The "work" would be 25 lbs x 40 miles = 1000 lb-miles.
The "power" would be measured in an hour's time; 40 miles driven divided by 40 mph is 1 hour resultant elapsed duration.
so the power requirement is thus:
P = W / time or 1000 lb-miles divided by one hour.
P = 1000 lb-miles per hour.
80 mph net speed example
Now, if you decided to drive that same 40 miles distance but increase your speed to 80mph, this would be the result ...
drag force to overcome is now approximately 100 lbs due to doubling the vehicle speed
"work" would now be 100 lb x 40 miles = 4000 lb-miles (four times greater than the slower example)
But ... you did the job in half the time, so the formula is thus:
P = W / time or 4000 lb-miles divided by 1/2 hour (4000 lb-mi / .5 hr)
P = 8000 lb-miles per hour
Double your net speed, and your power requirement is roughly EIGHT TIMES GREATER.
Again, much depends upon nuances like air density, size of the resistant object, rolling resistance, crosswinds, etc. The chart linked above shows a more accurate curve, but the simple mind-note of 4x force and 8x power for a double of speed are easy to remember. But the general nature of wind drag is essentially as shown above. If you hold all other things constant, then power demands are prodigiously and viciously ravaged by net speed increases.
Even seemingly minor increases in net speed can have a significant effect on power consumption.
Headwinds have the same effect as speed increases. A little extra headwind makes a BIG power demand increase.
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Head wind does make a difference. I make regular trips over the Pennines from Manchester to York and the prevailing wind is almost always behind me on the way out and against me on the way back. Even with modest windspeeds of 10 mph that's a 20mph differential and it typically makes a 5 mpg difference between outward and return trips.
I think the best illustration that has been given is the effects on a pedal cycle. You can really feel a head wind even when pedalling at just a few mph.
I feel a head wind on a motorcycle too as like a pedal cycle the drag coefficient is pretty appalling compared to a car so it will be impacted at lower speeds than a car.
I think the best illustration that has been given is the effects on a pedal cycle. You can really feel a head wind even when pedalling at just a few mph.
I feel a head wind on a motorcycle too as like a pedal cycle the drag coefficient is pretty appalling compared to a car so it will be impacted at lower speeds than a car.
Direct headwinds are also largely negated when following close enough behind other traffic on the highway.
I bought a pair of Ford Festivas 5 sp manuals from a guy and was driving them 50 miles one way to Champaign IL for work. Interstate 57 and wide open flat corn fields. One was really carboned up and I was driving into a 40 mph headwind. Doing OK until I tried to pass a semi and all I could do is get the nose of the car in front of his bumper and then his drag stopped me from gaining.
I had to wait until a curve with a overpass and tree line to break the wind before I could get by him. I bet he was laughing away the whole time.
I had to wait until a curve with a overpass and tree line to break the wind before I could get by him. I bet he was laughing away the whole time.
That's my exact thoughts as well. I80 is my main road trip highway so I tend to get a lot of north/south winds when driving east/west and it sucks!
Even in the logging truck, a head wind makes a big difference. Have you noticed new trucks are lacking the big external air cleaners? They are gone for fuel mileage.
Side winds push the logging truck around greatly, despite being 130,000 lbs, because of the large surface area it has.
Side winds push the logging truck around greatly, despite being 130,000 lbs, because of the large surface area it has.
Coming across the California desert on I40 from Arizona, with a 20MPH headwind pretty much straight on, vs. coming home to AZ with a tailwind, I'm always getting a 4-5 MPG differential.
Air speed (difference between air speed and car's road speed) determines drag force, other factors being equal. That drag force times road speed yields power to overcome that drag.
You last sentence is generally correct. (Tire rolling resistance is more significant than aero drag at low speed.)
That's called drafting, also known as tailgating.
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We make a road trip of about 4k miles roundtrip most summers over the mountains and into the flat lands. We pull about 30mpgs for the first 1/3 part where there's often very little wind. Then, it's a crap shoot. If the wind is helping at all, we'll continue pulling 30+mpgs going 80 for the next 600 miles. If it's into us at all, we're looking at 22ish. It absolutely eats oil if it's under that extra wind load, too. I figure it's a great way to do a 600 mile long Italian tuneup
I don't know how accurate this tool is but the ecomodder website has a calculator that is fun to play with.
Aerodynamic & rolling resistance, power & MPG calculator - EcoModder.com
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ecomodder.com
When I go to work I have a headwind most of the time (10-20 mph), but when going home at night the wind usually has died down so can't benefit from the tailwind nearly as much as I should.
Un faired motor cycles are notoriously bad in terms of drag coefficient. There was a famous test done back in the 50's (well it was famous at the time). They streamlined a 350cc Royal Enfield and did some painstaking and accurate measurements of the impact on top speed, acceleration and fuel consumption. What strikes me is that even though the faired machine weighed 15% more, low speed acceleration times improved and even at 30mph mpg improved by approx. 25%.
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