Upwind/Downwind Not Equal?
 

Was reading a book written by a flight navigator and he made the point that the speed gained by a tailwind does not equal the speed lost by a headwind. Wonder if that is true of cycling also?
When trying to figure average speed and cycling out and back against and with a wind factor what is the effect of loss? Seems this factor would be enhanced by the posture of a cyclist also where the wind at his/her back is meeting a more aerodynamic shape. Even in a low crouched posture there is the drag on a shape that is essentially concave. Any thoughts?

It is true. Air resistance is a function of your velocity squared relative to standing air. Going into the wind increases your velocity relative to the air. A tailwind simply reduces your velocity relative to the air.

You also have to remember that if you're measuring speed, then the key element is time, not distance. So don't think that if you went 15 mph on the way out and then 25 mph on the way back that your average would be 20. You spent less time at 25 than at 15 so your real "average" is the harmonic mean, not a simple average.

^^^.... just as, for example, if you go up a 2-mile hill at 2 mph, and then back down it at 60 mph, your average speed is not 31 mph. The total time is an hour and two minutes to go 4 miles, which means a smidge under 4 mph.

People call wind an endless hill sometimes, jokingly, and basically, it is. Well, a kind of wiggly hill.

The thing that gets me every time here on the Pacifica coast is that I can be riding along, not feeling any wind at my back on the way south, and don't even notice any trees or bushes swaying in the wind...but then when I turn around and start heading back north to go home, wham! there it is, slowing me up like two flat tires and a 100 lb. load, reducing what was a nice fast ride to a 12mph suffer-fest!

when i try to figure out my average speed of cycling i usually look at the little computer on my handlebars. also, i don't really care, i just like riding my bicycle.

i would like to add that when discussing airpseed, i know the air speed of a swallow, especially a european swallow, depends on if it is laden or not.

hope this helps.

It must be true because when it gets windy here, 20mph and up, my times are slower. One other thing to think about is that even a side wind is tiring. I am constantly battling the side winds to keep myself upright and following my line. So, wind from the front and both sides is going to tire me and slow me down, while only the tail wind is going to speed me up.

Additionally, let's say I have a 20 mph tail wind. The advantage essentially ends after I reach 21 mph. Okay, not entirely true, but you get what I am saying. While if I have a 20 mph headwind, regardless of speed, that headwind is affecting me negatively.

Further.....having a headwind on a hill is doubly tiring. Because we are not machines, but humans, at some point all that additional effort is going to add up and slow you down for the duration of the ride. If that 10% incline feels like a 20% incline because of a headwind, and you have gearing that essentially requires you to go no slower than 6 mph, then you will be forced to expend more energy than you typically do to keep from falling over.

How'd aircraft get pulled into this? We are talking about bikes, right? Bikes are propelled by a tire rolling against the ground, not by a propeller pushing air.

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This is somewhat incorrect. Again, it is a function of your velocity squared relative to the air. If the air is moving (i.e. wind), then you must take that into account. A 20 mph headwind has you starting at 20 mph relative to the air while a 20 mph tailwind starts you at -20 mph. The negative airspeed will actually act as an assisting force up until you reach 0 mph, after which you will be back to 0 (relative to the air). However, it still helps at this point, since you are at a lower relative velocity.

I think he deleted his post?

The actual equation is:

Drag force = 0.5 * Adpv^2
A = Cross-sectional area
d = drag coefficient
p = density of air (perhaps the air mass ratio he was thinking of?)
v = velocity relative to the air

Well, an extreme case -- suppose your plane flies 100 mph, and you've got a 100 mph headwind. It would just hover -- it would take forever to reach it's destination, so your average speed would be zero.
Yes, but for a slightly different reason. If you do 20 mph with no wind, with a 20 mph headwind you'd do like 10 mph or so, and with a 20 mph tail wind you'd do 28 mph or so. But even if you did 30 mph with the tailwind, you'd spend more time doing 10 mph than 30 mph, so your average would be lower than 20 mph.

Hills work the same way, by the way -- yes, you store energy as you go up the hill, but burn it up really fast as you go fast downhill, and when you average equal distances worth of slow speeds plus high speeds -- the average is closer to slow than high.

Want to get places quickly? Move somewhere flat without wind.

I think there is an echo in here...

Same thing here in Ventura, especially when riding along PCH. It is so weird to be riding along and then when I turn for the return leg of the ride I get smacked on the face by a strong headwind...not fun.

air speed like going against the tide , the speed thru the air or water may be X,
but physically you can be going nowhere or backwards.

on a bike you may just fall over
or get off and walk,
when the headwind is too strong to proceed thru.

friends tell tales of crawling on hands and knees to get to the car,
on the southern Irish Coast , in winter storms.

Er, hills don't work the same way. Assuming ideal and symmetrical conditions, the work spent going up a hill will be regained on the way down. That's conservation of energy.

Found poetry.

Well, what we're talking about here is speed. And if you're trying for a certain average mph over the course of a ride, then no, the damage done to that average by climbing up the hill will not be regained on the downhill. Because, as explained, one spends much less time on the downhill. Now, if your "symmetrical condition" is defined as "the same elapsed time" then sure. But uphills and downhills are more usually of similar distances, not similar travel times on a bicycle.

FWIW, I remember a story maybe 20 years ago (possibly in Bicycling back when it used to be a real magazine) that analyzed this. Don't recall all the details, but they looked at riding positions, bike speed relative to wind speed and other factors, and there was a graph showing that only winds within 45 degrees of the bike's direction of travel actually aided the cyclist. There was a small zone where there was no effect, but basically about 270 degrees counted as a headwind.
It didn't make complete sense to me then, and I'm not sure it does now. But that's what they claimed.

Read my post. That is exactly what I said and explained why that is the case.

That is why I said

We've all had the experience on an out-and-back ride, of riding into a headwind thinking, "man...well, at least I'll have a tailwind going back!". But then, when you turn around, it still feels like you have a headwind. But the leaves are still blowing in the same direction they were blowing before you turned around. It seems to make no sense, but it does make sense.

Wind is wiggly and shifty. Real hills stand still.

The answer is perfectly simple. There is no such thing as a tailwind.

Oh, yeah, right. Next you're gonna tell me there's no such thing as Santa Claus. Well, riddle me this, Batman, Who's been giving me all those presents for all these years, and who's been eating the cookies?

:)

It's interesting to compare bikes to airplanes; an airplane takes longer to get from point A to point B with a wind directly from the side, because it flies crabbed with a slower groundspeed than airspeed. The bike has the benefit of tires to keep it on a direct path, but that has to come at a cost, and I guess it's more tire deformation.

When driving a car with a strong crosswind, I notice the steering wheel needs to be turned into the wind. That would add tire drag, it's probably the same kind of thing going on.

Plus it is more work.

I'm pretty sure that the tires are actually still pointed straight. You're simply applying a force on the wheel to counter the force that the wind transfers to your steering wheel since it's trying to turn your car. Having the tires not pointed straight would imply that there was slipping going on, and this would actually cause a great deal of wear over a short distance.

You might be right. But I'm still pretty sure that the side load on the tire would cause more tread distortion, and more drag. That performance loss might be small enough not to notice, but significant enough for these forum theoretical discussions.