Speed into wind and with the wind
 

Hi,

Not sure this is the best forum but it is the most technical.

Anyway today I rode on the flat into a 20mph+ headwind doing about 10mph.
So aero drag is equivalent to doing 30mph and should totally dominate.

Fact is I can't remotely do 30mph on the flat no wind, so how can I do 10mph
into a 20mph wind ? Coming back the other way with the wind I think I hit
25mph maximum on the flat, how does that work with a 20mph tailwind ?

Before you say the wind simply wasn't that strong I can assure you it was.

rgds, sreten.

First of all, since you're only doing 10mph ground speed you've probably shifted down so were using a lower gear and therefore had enough torque. In many ways it's comparable to climbing a long shallow grade. You might also have made it easier on yourself by assuming a lower profile than normal to cut wind drag. So while you were pedaling hard enough to go 30mph (wind speed) you're only covering 1/3rd the distance.

The reverse happens on the downwind leg, at 25mph you only had an effective wind speed of 5mph, allowing you to fly along in a much higher gear than you could sustain in still air.

This effect is the main reason that I set my bikes up so my "cruising" gear is 2 down from high. That leaves me roughly 20% "overdrive" for downgrades and tailwinds.

If you were riding on a treadmill running at 20MPH, that logic would make sense.

Wind drag is not ground speed.

Hi,

Gearing into the wind was 42 front and mainly
24 or 22 rear.Coming back with, 52 front and
mainly 20 or 18, (though I hit top speed 16).

rgds, sreten.

Your post was also moved to the Folder subforum, so I'm copying my reply there to this subforum:

Aero drag goes as the square of wind velocity and power is proportional to the drag times the ground velocity. So in a 20 mph headwind the required power is proportional to 30^2 x 20 or 18000. That's the same as if you were riding 26.2 mph with no wind at all (26.2^3 = 18000). Of course that's assuming that aero drag was the only force working against you (no rolling resistance, gradient, mech. inefficiency) so in practice your equivalent speed without wind would be a little less - maybe 25 mph or so.

If you can't go close to 25 mph using the same level of effort then I do question whether the wind was really that strong. Keep in mind that the wind speed down near the ground is usually significantly reduced. So your wheels, legs, and lower part of the bike may have been seeing a wind speed of less than 20 mph even if your face was getting the full 20 mph headwind.

sreten, I've wondered the same thing. As I understand it, your assumption is correct - air resistance is the only thing keeping a cyclist from going 100mph (if we had gearing to do so, of course). Bearing drag and rolling resistance are fairly minimal in comparison. There's no need to get into squared drag physics - the issue is simply the speed of the rider relative to the speed of the wind (assuming the wind is parallel to the rider).

I would think the same thing that you do: riding 10mph into a 20mph headwind should be roughly equivalent to riding 30mph on a windless day (assuming flat road in each case). However, as you note, riding 10mph into a 20mph headwind seems easier in experience. I don't have an answer for why this is. My best guess is that the actual headwind isn't consistent in terms of molecules - or as prathmann says, the effective wind speed is notably slower as you get closer to the ground, lowering the drag on the bike.

have never seen a flat road,or no wind so I don't pay attention to the speed,(don't have one of those either)-just ride and enjoy every minute of it!

Hi,

Duplicated posts is a pain.

However here is my take on trying to work it out :

In a 20mph wind if you stop and sit there that is
clearly nothing like doing 20mph with no wind.

Rolling resistances at low speeds into the wind will
be much less than with higher speeds with the wind.

Still I can't get my head around why I can do 10mph
into a 20mph headwind, but only 25mph same tailwind.

(I will work it out eventually, physics is physics)

rgds, sreten.

I can be wrong, but rolling resistance depends on the elasticity of the tire(s) and the road, dismissing the road's elasticity for the moment the tire's elasticity will remain constant at any speed. The road's elasticity can change, for example when traversing a patch of loam from and to an asphalt surface.

Brad

Exactly. As I stated before, if we neglect all forces except air resistance then:

F (i.e. the drag force) = Cd x Area x (air speed)^2

Power = F x v (i.e. ground speed)

So if you're sitting still, then your ground speed is zero and it takes no power to stay in place even if there's a substantial wind (in a really strong wind there will be some internal muscle energy losses to overcome just to hold on - but no actual external work is done in a physics sense).

So from the standpoint of power required, there's a big difference between sitting still in a 20 mph wind vs. moving at 10 mph into a 10 mph wind.

For your original example I worked out the math in the previous post - going 10 mph into a 20 mph headwind is about equivalent in power required to going 26.2 mph with no wind (assuming level ground and no other energy losses). So that's part of the answer. The other two parts are that the wind speed close to the ground was probably less than 20 mph and that there are other drag forces (rolling resistance and mech. losses) in addition to air drag.

Your math was wrong. 10mph into a 20mph headwind is closer to 20.8 with no wind.

prathmann, I'm fully with you on the calculation of F (drag force), but why would power depend on ground speed? When I think about power, it's driving the continual acceleration required to maintain speed against resistance (some rolling resistance and bearing drag, but mainly air resistance) which is trying to decelerate the rider. I'm not saying you're wrong (and your calculations are consistent with our experience), but theoretically, I still don't understand why ground speed is part of the multiplier for calculating power. Can you explain the rationale here a bit more?

The energy required to move something against a resisting force is equal to the size of the force times the distance that the object is moved. The simplest example is lifting a weight - say you lift a 10 lb weight up a 100' cliff. The energy required is then 1000 ft-lbs. Similarly, the energy required to overcome air resistance would be the drag force due to the air times the distance you are moving against that force. Power is energy per unit time, so the power needed is the drag force times the distance traveled divided by the time. But distance divided by time is just your ground speed. So the power is the drag force times your ground speed.

And yes, greg is correct the speed should have been 20.8 mph w/o wind as equivalent to 10 mph into a 20 mph headwind. (I was looking at 20 mph into a 10 mph wind.)

It seems most things have been covered by the excellent analysis of prathmann. I can think of one more thing. An exact head- or tailwind is unlikely. As soon as the wind is coming from the side, you are losing speed by lateral turbulence.

I once read a study that compared speeds on a closed loop trajectory on windy and windless days and the conclusion was that no wind is always better for the average speed, which bears out what you seem to have noticed.

Right. Force needed to over come drag goes as the square of road speed. So, power needed to overcome drag goes as the cube of road speed.

Not quite, Drag is proportional to the square of wind speed not ground speed.

What everyone here is forgetting is that in the example given of riding 10mph against a 20mph wind is that the total distance covered is 1/3rd what would be ridden if doing 30mph is still air. I tried o give Sreten the clue in my first post (#2). So while the resistance is comparable to 30mph, the work done is one third.

So using 10mph as a base line, and discounting mechanical and tire resistance for the moment) the drag in a 20mph headwind would be 9 times what it would be in still air (3x air speed, squared). But since the distance is only 1/3rd we divide that by 3 and can compare it to 3x drag, and take the square root to convert air drag back to speed and find that riding 10mph against a 20mph wind is comparable to riding 17.3mph in still air.

If Sreten was able to ride 10mph against a 20mph wind, I'll venture that he'd be able to cruise along at 17mph or so in still air at a similar level of effort.

BTW- I once (and that was enough) had to ride against 40mph winds, and it was possible but was like climbing a mountain. But there's no way I can ride anywhere near 40mph.

Hi,

Pretty much bang on the money, though the 17mph or so cruising on the
flat no wind feels easier due to their being no wind gusting. One of my
most tiring rides was into a very gusty head/crosswind, very hard work,
a lot of effort* to keep balanced, and difficult to be in the right gear.

No doubt the best conditions are no wind, unless it suits the course,
headwind downhill, tailwind uphill, and all cross winds slow you down.

rgds, sreten.

* Or continual tension in the body, doing something it usually can
do quite relaxed normally, and is not used to continual adjustment.

The obvious thing I missed is that at 10mph into a 20 mph wind
with the low gearing I'm producing enough torque to overcome the
30mph effective drag, I haven't a hope in hell of producing that sort
of torque with a gearing that would allow me to do 30mph in still air.

This thread is mildly confusing in the sense that I feel like I'm in math class and got bored. With that said, in a nutshell, if there is a 20mph tailwind vs a 20mph headwind, do they even out or will you still exert much more going into the headwind than going with the tail wind?

It just seems like that when I ride on a windy day, i'm much more tired than the same ride on a still day but I would have thought that the going against vs going with would have evened out but it doesn't seem like that.

Hi,

Generally wind always makes things harder.

They don't even out, the fastest on a flat course there and back
is always no wind. All wind slows you down on a flat course.

But if you have a course where there and back is uphill / downhill
then a suitable wind can make you faster than no wind at all.

rgds, sreten.

You never braak even when wind is involved. Since wind resistance is proportional to the square of the (wind) speed the handicap of a headwind where your speed added is always much larger then the benefit gained when your speed is subtracted.

Skipping the details, imagine you ride 15mph up and down in a 5mph wind. In still air at 15mph the drag factor would be 225x. Going upwind it jumps to 400x (a big penalty for only 5mph wind speed). When you reverse and go downwind the effective speed drops to 10mph for a drag factor of 100x.

So upwind you're penalized by 175 force units, but downwind you only get back 125.

That's in a perfect world, but as anybody who spends much time riding in wind can tell you, the world is anything but perfect. So after you fight a headwind all morning, and get psyched for that easy ride home, the wind also reverses an it's another tough stretch up wind. Nobody likes hills but at least they play fair and stay put.

No, we didn't forget that, in fact Prathmann stated it explicitly in posts 5 and 13 in this thread.

BTW your maths is wrong, Prathmann's second answer (20.8) is correct, = cube root (30 x 30 x 10)

Let's say you are fighting a headwind on the way out and average 10 MPH. On the way back, the tail wind helps you average 20 MPH. It will take you twice as long on the first leg as it will on the second so you will spend twice as long fighting the wind as you will having it help you.

Doh! I wasn't paying attention. My bad. In still wind, power to overcome drag goes as the cube of the road (hence wind) speed. If there is wind, it gets slightly more arithmetically complicated.

That's OK. I ran the problem in my head over breakfast coffee, and dropped a factor, so while my instincts were right, my math was off and my answer was close, but no cigar.