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Fuel efficiency in headwinds


jaylw314

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Fair warning, there are geek alerts ahead...

So on the toilet, last night, I was wondering what the best airspeed would be if you have a headwind.  I recalled some old wives tales about flying into headwinds that go along the lines of "if you go too slow, you'll spend more time fighting headwinds, so you'll actually use less fuel if you fly faster."

So I pulled out the paper and pen.  I assumed the following:

  • Fuel flow is proportional to power
  • Power to overcome drag varies to the square of velocity (laminar flow) or the cube of velocity (turbulent flow)

After crunching some equations, it turns out, if you assume laminar flow, the minimum fuel used for any trip was if you flew at two times the headwind, and assuming turbulent flow, the best speed was 1.5 times the headwind.  Figuring real life is somewhere between the two, that suggests the best speed is about 80% more than the headwind your fighting.

Obviously, this does not make sense with headwinds of 5 knots.  The numbers probably get all wonky if they are too low.  But if you have a ridiculous headwind of 50 knots, the numbers I get suggest a speed of about 90 knots would use the minimum fuel for any distance.

If I've done the numbers correctly, this would contradict the idea that flying faster in a headwind would use less fuel in any realistic circumstances.  Flying faster would use more fuel in anything less than hurricane strength winds.

Has anyone heard of this or crunched the numbers before?

 

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From:

https://www.aopa.org/news-and-media/all-news/2017/august/pilot/efficiency-slow-burn

To determine an airplane’s Carson speed, multiply its best glide speed by 1.32 (to increase it 32 percent). Going back to the Citabria, a best glide speed of 78 mph times 1.32 equals 103 mph. So 103 mph is the optimum airspeed—and flying at the highest possible altitude at which your Citabria can deliver 103 mph IAS will get the best result in terms of true airspeed and fuel consumption.

Also, headwinds matter here, too, so the Carson speed changes with headwinds and tailwinds. If the headwind or tailwind component is 10 knots or less, don’t bother. It’s academic. In stronger winds, however, the Carson speed increases about one-third of the headwind component, and it decreases by the same fraction in a tailwind.

For math majors, there’s a vast repository of complex formulas to back this stuff up. For the rest of us, best glide plus 32 percent gets excellent fuel efficiency at a reasonable speed.

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I never heard of laminar flow drag formula being different (square vs cube), I thought it just resulted in a lower drag coefficient (Cd)?

Cd does vary based on Reynolds number but pretty sure not exponentially.

BTW, Mooney J has a Cd of 0.017, lowest of production planes, there are probably lower experimentals now days.

 

 

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I've done the analysis using numbers from my POH.  The old rule about speeding up if you are flying into headwinds is only true if you normally fly at L/D max (about 90 knots IAS or so in my J).  Since none of us do that, we usually fly closer to 125 - 145 IAS in cruise, if we want to save fuel we actually need to slow down to get closer to L/D max.  Try around 45% power or so.  Once the headwinds get up around 50 knots or so then it starts paying to use more power.

If I'm worried about running out of fuel because headwinds are much stronger than forecast, rather than slow down a lot, I'd probably land short, buy more gas, and then press on at normal speeds.

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5 hours ago, teejayevans said:

I never heard of laminar flow drag formula being different (square vs cube), I thought it just resulted in a lower drag coefficient (Cd)?

Cd does vary based on Reynolds number but pretty sure not exponentially.

BTW, Mooney J has a Cd of 0.017, lowest of production planes, probably lower experimentals now days.

 

From Wikipedia on "Drag (physics)":

Drag force is proportional to the velocity for a laminar flow and the squared velocity for a turbulent flow. Even though the ultimate cause of a drag is viscous friction, the turbulent drag is independent of viscosity.[4]

The source listed is ^ G. Falkovich (2011). Fluid Mechanics (A short course for physicists). Cambridge University Press. ISBN 978-1-107-00575-4.

So power would be proportional to the square and cube, respectively. 

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2 hours ago, Bob - S50 said:

I've done the analysis using numbers from my POH.  The old rule about speeding up if you are flying into headwinds is only true if you normally fly at L/D max (about 90 knots IAS or so in my J).  Since none of us do that, we usually fly closer to 125 - 145 IAS in cruise, if we want to save fuel we actually need to slow down to get closer to L/D max.  Try around 45% power or so.  Once the headwinds get up around 50 knots or so then it starts paying to use more power.

If I'm worried about running out of fuel because headwinds are much stronger than forecast, rather than slow down a lot, I'd probably land short, buy more gas, and then press on at normal speeds.

If my numbers are anywhere near correct (always questionable), increasing power would consume less fuel for any given distance once headwinds got up to about 70-80 knots.  Less than that, you'll always use less fuel going slower (whether you have the time for that is a whole 'nother question).

The scenario this is relevant would be if you are on a long trip and find out you have unforeseen headwinds that you can't avoid.  If you're worried about making it to your destination, slowing down will almost always INCREASE your range, even if it takes significantly longer.

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1 hour ago, Jeff_S said:

If you did all that analysis while sitting on the toilet, your hemorrhoids would be so painful that you'd fly as fast as possible so you weren't sitting in the cockpit all that long!

:lol:

Hey, leave my bathroom habits alone! :D

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From Wikipedia on "Drag (physics)":

Keep reading...

 

Section “Drag at high velocity”

 

“...the drag equation with a constant drag coefficient gives the force experienced by an object moving through a fluid at relatively large velocity (i.e. high Reynolds number, Re > ~1000). This is also called quadratic drag.”

For air at Mooney speeds (Re >3,000,000), so power at high Reynolds numbers is proportional to velocity cube.

 

Section “Very low Reynolds numbers “

“The equation for viscous resistance or linear drag is appropriate for objects or particles moving through a fluid at relatively slow speeds where there is no turbulence (i.e. low Reynolds number Note that purely laminar flow only exists up to Re = 0.1 under this definition. In this case, the force of drag is approximately proportional to velocity.

 

BTW, Mooney has laminar wings, not a laminar fuselage, which I would guess is where most of the drag comes from.

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47 minutes ago, jaylw314 said:

Hey, leave my bathroom habits alone! :D

you started the conversation.:o

 

I fly WOT unless landing or boring holes in the sky.  When I have a head wind I want to get though it as fast as possible and yes I will try different altitudes to see if I get any better ground speed.

 

 

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23 minutes ago, teejayevans said:

Keep reading...

 

Section “Drag at high velocity”

 

“...the drag equation with a constant drag coefficient gives the force experienced by an object moving through a fluid at relatively large velocity (i.e. high Reynolds number, Re > ~1000). This is also called quadratic drag.”

For air at Mooney speeds (Re >3,000,000), so power at high Reynolds numbers is proportional to velocity cube.

 

Section “Very low Reynolds numbers “

“The equation for viscous resistance or linear drag is appropriate for objects or particles moving through a fluid at relatively slow speeds where there is no turbulence (i.e. low Reynolds number Note that purely laminar flow only exists up to Re = 0.1 under this definition. In this case, the force of drag is approximately proportional to velocity.

 

BTW, Mooney has laminar wings, not a laminar fuselage, which I would guess is where most of the drag comes from.

Er, huh??  That was my understanding, that drag at laminar flow (low Reynolds #) is proportional to velocity, and drag in turbulent flow (high Reynolds #) is proportional to velocity squared.  Power at low Reynolds # would be proportional to velocity squared, and power at high Reynolds # would be proportional to velocity cubed, right?

I figured since only the wing leading edges are laminar flow, that the power exponent would be something like 2.8 or something in real life...

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Er, huh??  That was my understanding, that drag at laminar flow (low Reynolds #) is proportional to velocity, and drag in turbulent flow (high Reynolds #) is proportional to velocity squared.  Power at low Reynolds # would be proportional to velocity squared, and power at high Reynolds # would be proportional to velocity cubed, right?

 

I figured since only the wing leading edges are laminar flow, that the power exponent would be something like 2.8 or something in real life...

Yes, but we’re talking the laminar flow contribution to overall drag is on the order a of less than 1%.

You have drag from:

Rest of the wing

Fuselage

Antennas

Cooling drag

 

It’s all baked into the Mooney’s coefficient of drag constant (Cd=0.017).

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2 minutes ago, teejayevans said:

Yes, but we’re talking the laminar flow contribution to overall drag is on the order a of less than 1%.

You have drag from:

Rest of the wing

Fuselage

Antennas

Cooling drag

 

It’s all baked into the Mooney’s coefficient of drag constant (Cd=0.017).

Is it really as small as 1%?  If that's the case, then power should be pretty much proportional to velocity cubed, in which case the minimum fuel used would be at about 1.5x the speed of any headwind.

My concern about exponent to the velocity is that that determines how drag relates to changing velocity.  The drag coefficient would only tell you how much drag you would have for different shapes at the SAME velocity

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Is it really as small as 1%?  If that's the case, then power should be pretty much proportional to velocity cubed, in which case the minimum fuel used would be at about 1.5x the speed of any headwind.
My concern about exponent to the velocity is that that determines how drag relates to changing velocity.  The drag coefficient would only tell you how much drag you would have for different shapes at the SAME velocity

Drag coefficient is velocity independent (dimensionless). Except at very low speeds and when approaching supersonic speeds.But at the speeds we’re going, it’s constant.
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3 minutes ago, teejayevans said:


Drag coefficient is velocity independent (dimensionless). Except at very low speeds and when approaching supersonic speeds.But at the speeds we’re going, it’s constant.

Hm, I was talking about how the force of drag, not the drag coefficient, varies with velocity.  yes, the drag coefficient would not change.

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I was forced with this decision last spring flying to the Texas Mooney gathering.  It was an approaching cold front.  Heading north directly into the wind.  In a sail boat your best speed is with a slight wrinkle in the forward part of the Jib.  Which is about 30 degrees.  So would it be better to tack up wind?

Or the other wind was since it was a front.  There is warm air going above cold air.   Should I be in the warm air riding the wave over the cold air?

I think Paul went high (darn turbo) and was coming at a different angle and had much better ground speed.

 

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Shouldn't your fuel efficiency, based on true airspeed,  be the same regardless of headwind?  Your fuel efficiency based on ground speed will certainly be affected.  I will burn the same gph wether or not I am in a head wind.  If I go 360 nautical miles at 120knots in a 60 knot headwind burning 6.5gph with a GS of 60 knots I will burn 39 gallons.  If I go on the same trip burning 10.4gph at a true airspeed of 154knots with a groundspeed of 94knots I will use 39.83 gallons and get there faster.  These numbers are off the top of my head but seem doable +- a few% either way.  The first trip takes six hours while the higher power setting takes 3.83 hours. This example is a huge time savings.

This stuff is certainly thought provoking.

If you do the first trip with no headwind you'll burn only 19.5 gallons in 3 hours.  The second one will burn 24.11 gallons in 2 hours and 20 minutes or 2.33 hours.  In this case the first trip with no headwind is 23.6% more efficient in fuel while only losing 40 minutes in time. 

  

 

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5 minutes ago, INA201 said:

Shouldn't your fuel efficiency, based on true airspeed,  be the same regardless of headwind?  Your fuel efficiency based on ground speed will certainly be affected.  I will burn the same gph wether or not I am in a head wind.  If I go 360 nautical miles at 120knots in a 60 knot headwind burning 6.5gph with a GS of 60 knots I will burn 39 gallons.  If I go on the same trip burning 10.4gph at a true airspeed of 154knots with a groundspeed of 94knots I will use 39.83 gallons and get there faster.  These numbers are off the top of my head but seem doable +- a few% either way.  The first trip takes six hours while the higher power setting takes 3.83 hours. This example is a huge time savings.

This stuff is certainly thought provoking.

If you do the first trip with no headwind you'll burn only 19.5 gallons in 3 hours.  The second one will burn 24.11 gallons in 2 hours and 20 minutes or 2.33 hours.  In this case the first trip with no headwind is 23.6% more efficient in fuel while only losing 40 minutes in time. 

I was talking about fuel consumption for a fixed ground distance, rather than fuel consumption per hour.  My thinking was

  • Fuel consumption = fuel flow x time
  • Fuel flow = Power = Drag x airspeed (or airspeed ^2)
  • Time = distance / (airspeed - headwind)

Like I said, you will certainly save time (much time) by going faster into a headwind, but you will (almost) always use less fuel by going slower (unless you're fighting headwinds that are 100 knot headwinds in a Mooney).  What I haven't done yet is graph fuel consumption as a function of airspeed to see how much of a difference it makes, e.g. do you just save a tiny bit of fuel or a significant amount?  The only data point I do know is that if your airspeed is equal to the headwind, your fuel consumption would be infinite since you'd never get there!

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This only applies if you normally cruise at less than full power. For example for a low wind condition I will aim for 145ktas on 8.5gph lop . That’s 17nmpg.

With a 20kt tailwind, I’d opt to go 140ktas on 8.0gph because it would be 19.4nmpg on 8.5gph but 20nmpg for just a few knots slower.

On the other hand, I’ll speed up in a headwind up to 75% power on 10.0gph (more lily 70% on 9.0 for most headwinds). At about 155ktas, that’s 15.5nmpg. 

But now compare 145ktas@8.5gph vs 155ktas@10.0gph as a headwind is added on a 1000nm trip:

0kts headwind - 20nmpg, 145ktas, 145kgs, 6.9hrs or 15.5nmpg, 155ktas, 155kgs, 6.5hrs

10kts headwind - 15.9nmpg, 145ktas, 135kgs, 7.4hrs or 14.5nmpg, 155ktas, 145kgs, 6.9hrs

20kts headwind - 14.7nmpg, 145ktas, 125kgs, 8.0hrs or 13.5nmpg, 155ktas, 135kgs, 7.4hrs

50kts headwind - 11.2nmpg, 145ktas, 95kgs, 10.5hrs or 10.5nmpg, 155ktas, 105kgs, 9.5hrs

So as you see, without a headwind adding 10knots of speed only saves 0.4 hours. However, as the headwind approaches 50 knots, the time savings can be an entire hour!!!!! Fuel flow difference is the same in either case! 1.5gph extra. But that extra fuel spent goes a longer way in a headwind.

Lets now compare the cost to benefit at say $5 a gallon (the difference is even more pronounced if you have to buy expensive fuel):

At no wind $293 vs $320. $37 to save 0.4 hours.

At 50kts wind $446 vs $475. $29 to save 1.0 hours. Depending on what you spend on oil, maintenance, engine time, it practically pays for itself just to save the tach time on the engine.

Meanwhile in the no wind condition, that $37 to save 0.4hours translates into $93/hour of savings. So my point is that it’s not an old wives tale. If you want to do more flying for the same fuel spending each year but you like to get there pretty quickly, then fly near Carson’s Speed in low/no wind conditions, even slower in a tailwind, and speed up (but only while staying on the LOP side) into a headwind. Your extra fuel spent goes a much longer way in a headwind than it does in calm conditions. 

 

Edited by 201er
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2 hours ago, 201er said:

This only applies if you normally cruise at less than full power. For example for a low wind condition I will aim for 145ktas on 8.5gph lop . That’s 17nmpg.

With a 20kt tailwind, I’d opt to go 140ktas on 8.0gph because it would be 19.4nmpg on 8.5gph but 20nmpg for just a few knots slower.

On the other hand, I’ll speed up in a headwind up to 75% power on 10.0gph (more lily 70% on 9.0 for most headwinds). At about 155ktas, that’s 15.5nmpg. 

But now compare 145ktas@8.5gph vs 155ktas@10.0gph as a headwind is added on a 1000nm trip:

0kts headwind - 20nmpg, 145ktas, 145kgs, 6.9hrs or 15.5nmpg, 155ktas, 155kgs, 6.5hrs

10kts headwind - 15.9nmpg, 145ktas, 135kgs, 7.4hrs or 14.5nmpg, 155ktas, 145kgs, 6.9hrs

20kts headwind - 14.7nmpg, 145ktas, 125kgs, 8.0hrs or 13.5nmpg, 155ktas, 135kgs, 7.4hrs

50kts headwind - 11.2nmpg, 145ktas, 95kgs, 10.5hrs or 10.5nmpg, 155ktas, 105kgs, 9.5hrs

So as you see, without a headwind adding 10knots of speed only saves 0.4 hours. However, as the headwind approaches 50 knots, the time savings can be an entire hour!!!!! Fuel flow difference is the same in either case! 1.5gph extra. But that extra fuel spent goes a longer way in a headwind.

Lets now compare the cost to benefit at say $5 a gallon (the difference is even more pronounced if you have to buy expensive fuel):

At no wind $293 vs $320. $37 to save 0.4 hours.

At 50kts wind $446 vs $475. $29 to save 1.0 hours. Depending on what you spend on oil, maintenance, engine time, it practically pays for itself just to save the tach time on the engine.

Meanwhile in the no wind condition, that $37 to save 0.4hours translates into $93/hour of savings. So my point is that it’s not an old wives tale. If you want to do more flying for the same fuel spending each year but you like to get there pretty quickly, then fly near Carson’s Speed in low/no wind conditions, even slower in a tailwind, and speed up (but only while staying on the LOP side) into a headwind. Your extra fuel spent goes a much longer way in a headwind than it does in calm conditions. 

 

Ooooh, nice number work!  I can see you like working with real numbers instead of theory! :P

I hadn't thought about working out what winds do to the property of "airspeed/power".  Supposedly, this value is maximized at Carson's Speed.  Now that I've written it down that way, if Power = "motive force (or drag force)" x airspeed, then Carson's Speed is the speed that maximizes 1 / "motive force".

I'll have to think about that on the toilet tonight! :D

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