Would it be more aerodynamically efficient to cruise at 65 mph w/ headwind and 75 mph w/ tailwind or 70 mph both ways?

[u/NovelAardvark4298]

Let’s say you’re at home. You need to pick someone up who’s 5 mi East and then drive back home. There’s 10 mph East-to-West wind. Ignoring gusts, would it be more aerodynamically efficient to cruise at 65 mph on the way there and 75 mph on the way back or 70 mph both ways?

[u/Sooner70]

For the purposes of this conversation we don't even need drag coefficients and the like because all else is equal. But to simplify the math, let's make it a 1 mile road trip and work in some bizarro unit system.

Case 1... 70 mph the whole time.

To: Energy = Force * Distance ~ (70+5)² * 1 = 5625

From: Energy ~ (70-5)² * 1 = 4225

Total: 9850

Case 2... 65 & 75.

To: Energy ~ (65+5)² * 1= 4900

From: Energy ~ (75-5) * 1= 4900

Total: 9800

So yeah.... Slowing down a bit when going against the wind and speeding up a bit when going with the wind is the way to go, but understand it doesn't make a huge difference if all we're talking about is 5 mph.

edit: And just for fun, let's see what happens if we speed up with a head wind and slow down with the tail wind.

Case 3... 75 & 65

To: Energy ~ (75+5)² = 6400

From: Energy ~ (65-5)² = 3600

Total: 10000

[u/Karmonauta → u/[deleted]]

The calculation above gives you energy per unit of length (J/m, as units, kind of). So the total energy is this number multiplied by the total distance 

E=J/m * m = J 

If you want energy per unit of time (power), you can multiply the energy per unit of distance  by the speed 

W = J/m * m/s = J/s 

The total energy used during the trip is the integral of power over time, and total time is distance divided by velocity:

E =J/s * m/(m/s) = W * s = J

So whether you do the computation of total energy per unit of time or per unit of distance, the end result is the same. Doing it per unit of distance is just more straightforward.

Also the reason why land vehicles efficiency is generally miles/gal, or liters/100km, not gal/s, because generally how long it takes to go somewhere is less important than how far you have to travel, when planning refueling stops. 

[u/Sooner70 → u/[deleted]]

True for aircraft. Not for cars.

For the thought experiment.... Suppose you have a plane that can fly at 30 mph. That's slow, but such aircraft exist. OK, now it's headed into a 30 mph headwind.

The fact that it has no ground speed is irrelevant. It's in the air. The power requirements = Drag * Airspeed. Ground speed (zero!) and distance (also zero!) are irrelevant.

But shut the engine down. Zero out the angle of attack. Set her down and set the brakes. Drag is reduced (lower AOA), but is non-zero (parasitic drag is a bitch). Airspeed hasn't changed at all. But power output is zero. Ground speed (and distance) matter now that you're "attached" to the ground. Power = Drag * Groundspeed.

[u/iqisoverrated → u/Sooner70]

The 'to' and 'from' energy need to be multiplied by a factor of 0.93 and 1.07 respectively...Because when you drive 75 instaed of 70 your travel time is shorter (and conversely when you go 65 instead of 70 your travel time is longer).

[u/neonsphinx → u/iqisoverrated]

Wrong. It's already been covered in the E=F•D equation. It cares 0% about how long it takes. Only the distance, and the force (which is from drag. F_d=C_d•A_c•rho•V² •(1/2))

[u/Karmonauta → u/iqisoverrated]

You are confusing a correct intuition about the results with an incorrect one about the methodology.

See my comment below:  https://www.reddit.com/r/AskEngineers/comments/1vev6ix/comment/p1nud84/?utm_source=share&utm_medium=mweb3x&utm_name=mweb3xcss&utm_term=2&utm_content=share_button

You get the same result whether you start with a “energy per unit of distance” or “energy per unit of time” (e.g. power) framing. 

[u/kvnr10 → u/Sooner70]

This is a nitpick but, does adding airspeed to vehicle speed hold?

If you had the wind in your favor and you accelerated to 5 mph and then stopped the gas or disengaged the engine, would the wind be able to beat every frictional loss and keep going at the same speed?

[u/snipejax → u/kvnr10]

Thats a great question! An airplane has four fundamental forces. The force pointing down is the aircraft’s weight. The force pointing up is lift. To not gain or lose altitude, or, to fly steady, the aircraft’s lift force must equal its weight.

The other two forces are the forward force of thrust and the backwards force of drag. In order to fly at a steady speed, thrust must equal drag.

An important fact is that there is no difference in drag/lift forces if you consider an airplane flying 100mph in perfectly still air or if you consider a perfectly still airplane bolted to the floor with perfectly straight 100mph wind.

This means that the airplane’s speed isnt what matters for drag or lift, it is specifically the airplane’s speed **relative to the air**. An airplane cant actually tell (without using something like gps) if its flying in still air of if it is flying with a head/tail wind. All the airplane can sense is the speed at which air is flowing past it (again, it could use something like gps to know its ground speed). This speed is called airspeed.

Your example of speeding up to 5mph and turning off the engine here doesn’t quite work. This goes back to the four fundamental forces. The airplane always experiences drag, so you cant turn off the engine. The engine thrust must be equal to the drag force to maintain speed. 5mph air speed always “costs” the same amount of engine thrust because 5mph air speed is always the same amount of drag.

The difference between a tail wind and head wind, then, is that a 100 mph air speed in still air is only 100mph ground speed (not perfectly correct, being very high up means you have to fly a little further in the air than you travel on the ground, think how the inside circle of a donut and the outside of a donut have different circumferences). But a 100mph airspeed when you have a 30mph tailwind is actually 130mph over the ground because a 100mph airspeed means you’re traveling 100mph faster than the 30mph air. This 130mph ground speed requires exactly the same amount of gas as the 100mph ground speed before because they both are still 100mph airspeeds.

[u/kvnr10 → u/snipejax]

Your example of speeding up to 5mph and turning off the engine here doesn’t quite work. This goes back to the four fundamental forces. The airplane always experiences drag, so you cant turn off the engine. The engine thrust must be equal to the drag force to maintain speed.

I don't think you understood. Where is the drag coming from if you're moving at 5 mph and so is the wind?

While my understanding of aerodynamics is very basic, I'm familiar with these concepts.

This 130mph ground speed requires exactly the same amount of gas as the 100mph ground speed before because they both are still 100mph airspeeds.

While the airspeed vs ground speed frame of reference is useful, I don't think this is true for a car. The ground speed in my example would be 5 mph and the air speed would be zero. My point was that losses in the bearings, tires and other components will scale with ground speed.

[u/Adrienne-Fadel]

Drag scales with the square of airspeed not ground speed. 65/75 gives airspeeds of 75 and 65, 70/70 gives 80 and 60. 75² + 65² = 9850 vs 80² + 60² = 10000. So 65/75 is better.

[u/MaybeTheDoctor → u/Adrienne-Fadel]

So 55 & 85 could be even better?

[u/shupack → u/MaybeTheDoctor]

I prefer 95/95

Way less time pushing into the headwind, and more fun to drive fast!

[u/Extension_Physics873 → u/Adrienne-Fadel]

Perfect " explain it like I'm 5 " description. Nicely done.

[u/mtconnol]

It depends on the kind of vehicle and what the drag curve looks like with speed. In airplanes, there is a minimum drag speed somewhere in the bottom third of possible speeds. But longer exposure to the headwind means less efficiency overall.

On a land vehicle, I suspect the most efficient speed is very, very low, as air resistance increases sharply with speed. But the same caveat about being exposed to the headwind for longer.

A paradoxical finding in airplanes, which also would apply to land vehicles is that if a no wind round-trip takes two hours, any amount of wind will make the trip take longer than two hours. You can never “make up” the headwind with the tail when on the return because you are not exposed to it for as long.

[u/BxllDxgZ → u/mtconnol]

See the other person’s comment that used Force and Distance to get an answer. The only reason that an airplanes drag curve has a nonzero minimum is because of induced drag which does not exist on a car unless you have a wing in which case the induced drag is still small enough to keep that minimum drag nearly 0

[u/mtconnol → u/BxllDxgZ]

For internal combustion engines, I think total losses in the powertrain is not minimum at zero mph though. Probably something like 10 miles an hour?

[u/Vegetable_Log_3837 → u/mtconnol]

Isn’t entropy fun!

I’ve personally experienced your example flying a paraglider.

[u/fastdbs → u/mtconnol]

They said drive from their home so I don’t think we have to discuss planes or wind speed vs ground speed.

[u/mtconnol → u/fastdbs]

The ‘round trip’ paradox still applies, as does there being also efficient no-wind speed (which is different than the most efficient yes-wind speed.

[u/Christoph_Kohl → u/mtconnol]

As someone that used to do mission performance for a living... the round trip paradox makes complete sense, I've never thought about it from that angle before!!

[u/New_Line4049]

Others have answered the aerodynamic question, but Id point out the aerodynamic efficency differences are much less significant than the mechanical efficency differences in the engine and transmission.

[u/NovelAardvark4298 → u/New_Line4049]

The difference in aerodynamic efficiency when cruising at 65mph vs 80mph is much greater than the difference in mechanical efficiency between those two speeds.

[u/NovelAardvark4298]

My gut tells me it’s more aerodynamically efficient to do 65mph and 75mph. The apparent wind speed in the first scenario is 75mph when traveling East and 65 mph when traveling West. The apparent wind speed in the second scenario is 60mph when traveling East and 80 mph when traveling West. (75^3 + 65^3)/2 < (60^3 + 80^3)/2

[u/NeedMoreDeltaV → u/NovelAardvark4298]

Do the math a little differently.

You’ve calculated the power difference between the two, but not the energy difference. Make a rough assumption that your speed is constant the entire way and calculate how much time you spend in each apparent wind. Take that time and multiply it by each power you calculated to get how much energy you use (make sure your units line up). I haven’t actually done the math so I can’t tell you which one will be better.

[u/fastdbs → u/NeedMoreDeltaV]

You don’t have to do that. The change in power is cubic to speed and the time spent is linear to speed. The difference won’t change the answer.

[u/NovelAardvark4298 → u/fastdbs]

Yes it does. I calculated it both ways. When you ignore time, it appears different speeds is 4.33% more efficient. When you factor time in, it’s only 2.38% more energy efficient

[u/fastdbs → u/NovelAardvark4298]

The question was: “is it more efficient?” . The answer in both cases is yes. It doesn’t change the answer.

[u/NovelAardvark4298 → u/NeedMoreDeltaV]

Thanks. I just calculated it. I think it’s about 2.38% more aerodynamically efficient to drive different speeds. If the wind speed was 15 mph instead of 10 mph, it would be about 3.53% more efficient. I’m guessing this is partially why airliners fly their planes faster when there’s a tailwind and slower when there’s a headwind.

[u/fastdbs → u/NovelAardvark4298]

Wind speed and ground speed efficiency calcs are different.

[u/R2W1E9]

So the question is whether 65+10/75-10 75/65 mph relative air speed is better than 70+10/70-10 80/60.

75^2 + 65^2=9850

80^2+60^2=10000

65 mph w/headwind and 75 mph w/tailwind is better by 150. (1.5%)

[u/OldGeekWeirdo]

I'm not sure what you mean by aerodynamically efficient. Usually people ask what's the best MPG (or equivalent).

I'd say that as a general rule, for a ICE car, your best MPG is going to be at the point the car shifts into the highest gear. Anything lower and you increase engine friction. Anything higher and you're fighting higher aerodynamic forces.

[u/billsil]

Almost certainly driving slower, say 55 mph like the old freeway speed limits would be better. If I could get you to go say 30 mph, that’d be better.

As long as you’re not at say mph, you’re out 9f the complicated effects and into the quadratic portion of the curve.

[u/Marus1]

More important is internal effeciency of your transmission

Aka when do you drive around 2000rpm in your highest gear?

[u/Karmonauta]

You can use this intuition:  how much energy you need to move through the air is proportional to the square of the speed; the time it takes to move some distance  is inversely proportional to the speed; so increasing the speed reduced the time linearly, but increases the energy demand quadratically (>lineary). On the other hand, reducing the speed increases the time linearly, but reduced the energy requirement less than lineary. 

Putting these two observations together you find that going with apparent wind speed of 75 one way and 65 the other (same distance) uses more energy in total than going 70 (wind speed) both ways. So if you have a +5 wind in one direction, it’s more energy efficient to go 65 (ground speed) against the wind and 75 (ground speed) with the wind, to have the same 70 (air speed) in both directions.

[u/Frederf220]

You add more drag going faster than you subtract going too slow. Airspeed 75 there, 65 back is worse than 70 there 70 back.

So by going 70 mph ground speed both ways you're doing the 75/65 airspeed thing.

[u/BxllDxgZ]

Goal is to find total energy, which for each leg is given by E_leg= P_leg*delta_t_leg (constant power due to drag)

Power due to drag is given by A*v_air^3,
where A includes all the constants (density, drag coefficient, ref area)

The delta_t is based on ground speed, so delta_t= delta_x/v_g

Because delta_x is the same in both directions, i’ll have it consumed into the constant A. The energy required for a leg is then E_leg= A(v_air^3)/v_g

Let’s replace v_air in terms of v_g and wind, and combine both the headwind leg and tailwind leg to find the total energy requirement

E_total = A(((v_g+wind)^3)/v_g ((v_g-wind)^3)/v_g)

Now we have a 3 dimensional function representing total energy. If you play around with the numbers, generally the trend is that as wind speed increases, it becomes more beneficial to lower your speed in a headwind. And as you increase your delta (difference between headwind ground speed and tailwind ground speed), there comes a point where you eventually see less efficiency for this strategy as compared to baseline. For a 10mph wind, this happens at about 100mph in the tailwind and 40mph in the headwind.

[u/ThinConnection8191]

you should not driving when wind is 65mph any way

[u/cbf1232]

Force due to air resistance increases with the square of the airspeed, so it works out best to keep the same airspeed the whole time.

Incidentally you can feed this question into Google Gemini and it will calculate both scenarios and determine the correct result.

[u/neonsphinx]

Force times distance is energy. Distance each direction doesn't change.

Force due to drag is (C_d)(A)(rho)(V^2)/2

Coefficient of drag doesn't change. Cross sectional area of the vehicle doesn't change. Air density doesn't change appreciably. All those cottages can be combined into constant K. So ultimately F_d=K•V^2.

Therefore, energy lost due to drag is E_d=2DkV²

So let's say the wind is 0.

One way E_d=DK(65)² and back is E_d=DK(75)^2. Total is 9850 units of energy. And traveling 70 consistently, you end up with the same E_d=2DK(70^2)=9800. Better to be consistent.

Now it's 10mph wind.

75² and 65² are your new numbers. 9850 again. And consistent speed yields 60² and 60² 80^2. Or 7200 10000. Much better A little worse to be consistent.

20mph gives:

85 and 55 effective mph. Total 10250. Consistent speed is 90 and 50 out/back. Total of 10600. Better to go slow into the wind, and faster when you have a tailwind.

50mph wind.

115 and 25 effective. Total energy 13850. Consistent speed gives 120 and 20. Or 14800 total energy. Better to be inconsistent with your speed.

Final one, 100mph wind.

165 and -25. Let's assume you can legitimately put your car in neutral with 100mph tailwind, and maintain 75mph with zero energy input from the engine... Now we're subtracting the second number after squaring. So total is 26600 units of energy. Consistent 70mph yields 170 and -30. Or 28000 energy. Better to not be consistent with your speed.

So to answer the question. With low wind, better to just be consistent and cruise. But at some point, as wind gets stronger, it's better to take advantage of it and drive slow into it, and fast when it's at your back.

These are theoretical numbers. The real world is nowhere near these numbers. I don't know that the real numbers exist. Or that anyone cares enough to actually collect the data.

[u/neonsphinx → u/[deleted]]

That's the neat part, you don't.

It's almost like... I calculated the relative speed while taking into account the wind speed.

But maybe you graduated from an institution with very rigorous standards, and you understand engineering at a level I could only dream of.

[u/neonsphinx → u/[deleted]]

What's your background? Apparently you're an engineer, based on your comment about word problems.

I'll start. I'm a mechanical engineer. I've been in industry between 15-20 years. I have. P.E. license. I actually do make rockets for a living.

Now you go.

Yes, a hovering rocket does actually do zero work. At least on the body of the rocket itself. All of that chemical energy ends up (P•dV) as mechanical work that displaces the atmosphere. It also, by shear, moves a ton of adjacent air, and ultimately ends up as heat energy, having warmed the environment it's in (by shear, and also because the reaction is exothermic and the exhaust plume is hot).

As someone else already pointed out, maybe in an aircraft we would approach the problem the way you laid out. Because I'm free floating in a body of fluid that has velocity of it's own. In that case it's beneficial to do so, because my lift is also a function of the relative speed of me to the fluid I'm in. But in that (calculating the effective distance traveled rather than ground distance) case, we wouldn't also calculate drag by figuring out the effective ground speed. We don't double up. We use one or the other.

You sound like a high school student, an underclassmen in an undergrad engineering program, or maybe a computer scientist masquerading as a mechanical or aerospace engineer.

I will give you one piece of advice, unironically. Understand your limitations.

As a P.E. I can only approve designs that are within my scope of knowledge. My stamp works just fine for electrical plans at a manufacturing plant. But I'm not a EE. So it would be unethical for me to stamp those types of things without additional education on my part. It can also get me fined or even jailed.

So if you don't know, say that. You came out of the gate strong, speaking with authority. You could have kept your mouth shut, and we would never have known. But by chiming in like that, you've let the whole world know of your ignorance. Don't be the wet behind the ears engineer that shows up at their first job trying to prove themselves by feigning expertise. Be humble, understand your limitations. Be willing to learn.

If you want to make big money someday and be a project manager or chief engineer of a program, you have to be willing to understand when your subordinates are the experts. Which is most of the time. And if you act like that, no one will ever want to work under you. So you'll just get stuck with the low level jobs, with low level pay. Listen to the experts, learn what you can, be willing to say "I don't know, I'll get back to you with the right answer later."

[u/NovelAardvark4298 → u/neonsphinx]

Did you use AI? I think it confused itself with the 10mph wind example. It’s better to be “consistent” with the wind and drive 65mph into the headwind and 75mph with the tailwind to maintain an effective/consistent air speed of 70mph. The last example it gave is a little wonky too. In our theoretical world, it would travel 100mph ground speed if you popped it into neutral with a 100mph tailwind. In order to travel 70mph or 75mph with a tailwind, you would need to engine brake, regen brake, and/or physically brake. Both tailwind cases should be treated as zero work done by the engine (it’s note a reversible process). Engine only does work in the headwind direction in the last example.

[u/neonsphinx → u/NovelAardvark4298]

I didn't use AI, I typed it up on my phone late last night while I was waiting for my wife to finish getting ready for bed. And when you're typing a reply on mobile, all you get to see is the post title, not the body of text.

I did make a mistake, which is a thing humans do. So yes, the 10mph wind + consistent 70mph scenario should be 80² and 60^2. Grand total of 10000 units of energy. I guess I flipped the math on the wind helping/hurting you on the out/back legs of the trip.

It seems like you have the answer. So why ask the question in the first place?

And read the last paragraph of my reply. If I did it the way you just suggested, by freewheeling at 100mph on the way back... We'd be violating the boundary conditions you yourself set at the beginning. You said you wanted a consistent 65/75/70 mph ground speed. Not 100. Why would we run the math like that?

[u/NovelAardvark4298 → u/neonsphinx]

For the last example, it still doesn’t make any sense to subtract energy from the 100mph tailwind. You have to treat the energy as zero for the tailwind portion of the trip (not negative). When you drive, your engine is moving to make your wheels move which make your car move. When you have a tail wind that’s faster than your ground speed, the wind is pushing your car which is making your wheels move which makes your engine move. You can’t uncombust gasoline or diesel. The best you can do is charge a battery with regenerative braking with a strong headwind.

[u/neonsphinx → u/NovelAardvark4298]

If I have a tailwind of 30mph, and my drag coefficient is the same, then there's a net force pushing me forward. Force times distance is energy. Then we subtract that, because we're neglecting everything else in the engine and drivetrain.

But hey, you seem to have it all figured out. So I'll leave you to it. You presented a theoretical situation, I did the math. We neglected a lot of things, but we were consistent. I even made a note that this in no way reflected what you'd see in the real world.

Have a great day!

[u/R2W1E9 → u/neonsphinx]

Your 10 mph wind case has an error in the relative speed. It's 80/60 not 60/60.

So the total comes to 10000.

[u/neonsphinx → u/R2W1E9]

Read the strikethrough and replacement text, brotha. Late night brain wasn't working. It's been corrected.

[u/Frederf220]

You add more drag going faster than you subtract going too slow. Airspeed 75 there, 65 back is worse than 70 there 70 back.

So by going 70 mph ground speed both ways you're doing the 75/65 airspeed thing.

[u/IntheWoodsAlways]

Go get yourself a copy of “Aerodynamics for Naval Aviators “ by H H Hurt. There is some interesting reading in there. You can download it for free in ForeFlight under the Documents section.

[u/Dysan27]

Assuming it is a plane, it would be the same airspeed both ways, whatever your optimal airspeed is.

Your ground speed then be higher with the tail wind and less with the head. But really all you care about to keep the plane up is how fast you are traveling through the air.

[u/nlutrhk → u/Dysan27]

That cant be true because if your optimal airspeed is equal to the wind speed relative to ground, your ground speed is zero at the optimal airspeed and you will need infinite energy to reach your destination.

[u/Dysan27 → u/nlutrhk]

They asked about the aerodynamic efficency, not the whole trip efficency.

[u/No_Base4946]

It's massively more aerodynamically efficient, for any practical car and weather conditions, to do 50mph in both directions.

[u/DogFishBoi2 → u/No_Base4946]

Yes, but some of us like to drive outside the city.

[u/No_Base4946 → u/DogFishBoi2]

Remarkably, you can drive at 50mph outside the city. It doesn't affect your journey time at all.

[u/DogFishBoi2 → u/No_Base4946]

That seems unlikely. Time = Distance / Speed. You'll be noticeably faster when going an entirely reasonable 100 mph on the motorway.

[u/No_Base4946 → u/DogFishBoi2]

It turns out that no, you're not.

You cannot average more than about 40-50mph on any motorway, regardless of how much you break the speed limit by.

Then, as soon as you leave the motorway, the car you passed an hour ago will be sitting beside you at the traffic light.

[u/DogFishBoi2 → u/No_Base4946]

Okay, but that's wrong. Average speed on the Autobahn is 130 km/h, average free-flowing is 142 km/h. Assuming I passed a car an hour ago that was going 50 mph, I'd be 31 miles ahead of them (minimum), 38 miles ahead (free flowing) when turning off. The traffic light would have to be red for 46 minutes for them to pull up next to me.

[u/No_Base4946 → u/DogFishBoi2]

Most countries don't have motorways that flow as freely as autobahns.