Campag4life

Thought I would posit a question to those that have thought about it. This question is part comedy but nonetheless rooted in reality aka aerodynamics...

When wind is say 10-20 mph and a route is out and back...my route is generally 10 miles + North and then 10+ miles South return. When the wind is dead downwind one direction and a pure headwind back, how much does average mph vary for the round trip compared to the same route on a dead calm day?

In my experience, my average aggregate speed out and back always seems to be faster on a rare calm day. How about you guys? Always killer to ride into a stiff wind and seemingly not offset by favoring downwind in terms of increased speed keeping net average speed the same compared to a zero wind day.

I try to dissect why. Headwinds as most know can be brutal on a bike no matter how low we try to get. Also, the benefit down wind never seems to equal the detriment in terms of reduced speed due to air drag into the wind. Or perhaps it does or close...but I tend to ride the wind and not put out the same watts when riding down wind...but like many pedal like hell into a wind to try and keep my speed up.

Anybody do any testing with a power meter?

Personal experience?
thanks

Smokehouse

I have the same issue where I live and I've done the same thing more times than I can count. Out with/against a strong wind, back with the opposite. I do not have a power meter but I can say that I often average out better when there's no wind in my face. I won't pic up enough extra mph with the wind to combat the mph loss when riding against it. That doesn't even factor in the irritation of riding for an extended period of time directly into a strong wind...which can be maddening.

HTupolev

It shouldn't. Wind drag isn't linear with your speed relative to the air, it's more like ~quadratic. The penalty of riding into a headwind should be far worse than the benefit of having the same wind on your tail.

Not to mention that wind can also do things like mess with handling, particularly when it's somewhat from the side.

tiiger

I somehow have a headwind on the ride to work AND on the way home.

it's a conspiracy.

Dan333SP

Every summer I go to the Outer Banks, where the only cycling is north/south on a beach road with 0 elevation change. It's a good test case for what you're referring to, because the wind tends to blow steadily without too many gusts and is, for whatever reason, rarely a pure crosswind, it's either a head or a tailwind.

I don't have a PM, unfortunately, so I can't share any data there. I do notice that, if I'm just going out to ride at a steady pace for 1-2 hours, my average speed will generally be higher if I ride out with the tailwind and ride back with the headwind rather than the reverse.

It's definitely a mental thing. If I have a strong tailwind when I start out, I push harder because it's fun to cruise at 25-30, and then when I turn around I get angry that I'm fighting to hold 16-18 so I tend to continue to push myself.

If I do the opposite, I burn myself out grinding into the headwind until I turn around, and then I take it easier on the way back so the overall average would be less.

I'd say I would be faster over that same route with 0 wind, but it's never not windy out there so I can't really say.

aubiecat

If I'm riding and there is wind it's in my face 99% of the time.
I had a tailwind once, it was fun while it lasted.

blackdogmax

Yes the penalty against the wind is bigger than the benefit of the wind at your back. You lose an amount of time going out. Coming back you have less time to make up time since you are traveling faster.

gsa103

This. At ~200W you should be cruising at approximately ~20 mph. In zero wind, a 10-mi each way should take ~1 hr.

With a 10 mph wind, same 10 mi each way.
Head wind = 14 mph, 43 min
Tail wind = 26 mph, ~23 min
Total time = 66 min (roughly 10% slower)
(Bike Calculator)

FBinNY 

It's fairly easy once you understand the basics. But the net effect also depends on your speed.

Understand that the wind drag (resistance) you're working against is proportional to the Square of your wind speed.

So assume for simple math that you're riding 10 miles at 15mph against a 10mph wind.

On the way out, your wind speed is 25mph, so the drag is 625units. Coming back your wind speed drops to 5mph, so the drag is a tiny 25units. That's an average of 325units out and back.

Compare that to the 125units of drag of doing the same ride in still air.

Once you have the basics, you can factor distance to calculate the energy used, or speed to calculate the power needed to maintain that in either direction.

So the key to efficient (least energy used) riding is to try to keep wind speed as constant as possible, and fight headwinds less, while using high pears and your power to maximize the benefits of tailwinds.

BTW - all the above assumes flat terrain and that wind is the only source of drag, but that's not unreasonable since non-wind drag is the same in both directions.

gsa103

I feel your pain. Usually no wind on the ride in, ~10-15mph headwind on the return. My commute times are actually much faster when I'm riding well after dark, just because the wind has died down.

Wheever

In my experience, tailwinds are a myth! I go out, there's a headwind; I go back, there's a headwind.

rpenmanparker 

Excellent analysis. My average speeds definitely demonstrate this phenomenon. Much better on a calm day.

PepeM

That's basically it. You spend more time going slower than you do going faster.

Edit: supposed to quote blackdogmax.

kkapdolee

Out and back ride in the wind has always been slower than that with no wind for me. I think this is because of the aerodynamic drag function where the drag increases proportionately to speed squared.

According to Wikipedia,



is the density of the fluid,[SUP][10][/SUP] is the speed of the object relative to the fluid, is the cross sectional area, and is the drag coefficient – a dimensionless number.
https://en.wikipedia.org/wiki/Drag_(physics)

If I've done my math correctly, here's what I think is happening.

Let's say your regular out and back ride is 20 miles each way.
Assuming all else equal: your cross sectional area remains the same (your body volume doesn't change, you wear the same clothes, use the same bike) , air density remains the same, etc.,
on a no wind day, say you ride 20 mph each way ending the trip in 2 hours with an average mph of 20.
On a wind day, say your speed going out was 15 mph. In order to achieve the same average speed for the entire trip, you would have to average 30 mph on the way back.

2 hrs = 20 miles / x mph out + 20 miles / y mph back
No wind day x = 20, y = 20 and total trip hour = 2
Wind day if x = 15 then y has to equal 30 for total trip hour to equal 2 and to have the same average speed.

Let's say on this particular windy day the wind is blowing from north to south at 5 miles an hour. On your way out, you can average 15 mph with the same power output as you could average 20 mph on a no wind day because your v relative to the ground on a wind day is 15 mph. e.g. 15 (your speed) + 5 (wind blowing towards you) = 20 (your speed relative to the fluid, in this case air).

On your way back, if you generate the same power output, you would average 25 mph. 25 (your speed) + (-5 (wind speed on your back)) = 20 ( your speed relative to the fluid).
Averaging 15 mph out and 25 mph back for 20 miles each takes [20 / 15 + 20 / 25 = 2 hours and 8 minutes ]
40 mile distance traveled over 128 minutes is 18.75 mph.
Thus it is slower than the average mph you can produce with the same power output on a no wind day which is 20 mph.

Wind speed doesn't have to be 5. You can use any positive number for the wind and your speed relative to the ground will always be slower with the same power on a windy day compared to a non windy day.

FBinNY 

That's usually true, and not an illusion for most people in most places. A decent rider maintains speeds between 15 and 20mph or higher. 15mph is pretty decent wind on most days, so you're usually riding faster than the wind, meaning even with a tail wind, you're riding into a head wind.

Add to that the reality that you'll ride faster with a tailwind. As a very rough guestimate, you'll probably ride at least fast enough to produce a net headwind of 10mph (if you have the gearing, because that's what it takes for the effort to feel normal.

So headwind or tailwind, you'll alwaus be riding into the wind. The difference is the gear you use, and your ground speed.

PepeM

That calculation you just did does not take the quadratic nature of drag into consideration. Which is ok, even if drag was linear, out and back would lead to lower average velocity than no wind day.

kkapdolee

Hmm.. yeah. Now that you mentioned, I realize that in my example, in order to use the same power output to overcome the same drag force, I used the same velocity relative to air and didn't end up demonstrating how drag force being proportional to speed squared affects average speed. Nevertheless, it still shows that average speed on a wind day will be lower than non wind day with same power.

On second thought, as you said "even if drag was linear, out and back would lead to lower average velocity than no wind day", the fact that drag force is proportional to speed squared doesn't actually have anything to do with lower average speed on wind days. hmm..

MikeyBoyAz

Where I live it isn't a conspiracy, it's common knowledge. The wind lightly flows up the canyon in the evening and oozes out during the morning... if you take the road up the canyon into a headwind, you will take the headwind back down the canyon in the evening.

FBinNY 

I also sail, and there's a similar effect so it's all upwind, all day.

Campag4life

Believe the out and back speed difference isn't the same guys. In other words for the same wind speed, the amount of increase in drag is not equal to the decrease in drag with downwind.
This I believe is due to how the drag coef. is affected. Into the wind, even leaned over body has a higher drag coef. compared to reversing the bike in a wind tunnel with the same wind speed. This means the body has a higher drag coef. into the wind than downwind. Its further complicated because a rider is always riding into the wind on some level.

So I believe the math is more complex than described because the drag coef. isn't the same for the same rider into the wind compared to downwind. Just some further thoughts. Thanks for all the great comments and math back up.

caloso

This.

Dan333SP

Everyone is talking about drag and time spent in headwinds, and while that's all true, this discussion isn't the full story unless we're assuming someone is holding the same power throughout their ride, and has the same drag coefficient (so in the drops the whole time with the same arm angle or whatever).

Like I said before, a lot of our anecdotal evidence about winds is a mental thing. I definitely push myself more throughout a ride when I start with a tailwind, and there's the added bonus of always going further than I intended when I set out.

caloso

I've done long intervals (such as 2x20's) on an out and back course in the wind. It's much easier to keep the target wattage going into the wind than with a tailwind. I think mostly it's psychological.

Campag4life

No it isn't for the reason I explained with due respect to FB. The reason is..when the wind is at a rider's back the force of the wind is acting on a smaller drag coef....a convex surface versus a direct headwind where windspeed + bike speed is acting on a concave surface resulting in a higher drag coef. Putting it in more layman's term. The body from behind is less of a sail then a wind into the body. The body natively catches more air when wind is into it.

FBinNY 

There's the physics, and there's what people do under various conditions.

As I said earlier, most people will ride at a speed that the winds allow, and so will ride much faster down wind. Plus people aren't machines and tend to tire when exerting at any level.

So if one starts out in a headwind, and turns around when he reaches the halfway point in time, he'll ride home faster and get home sooner than he expects. OTOH if he starts out down wind, he'll fly down the course, and if he doesn't plan carefully may find himself too far out and run out of his allotted time, and maybe endurance trying to get home.

Divers are very well aware of this issue because we have no slack in our schedules. So divers always start out up current and turn around when they've used half their dive time. Then the current takes them home sooner, rather than later. That's important because later isn't an option.