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Why is headwind more punishing then the benefit of tailwind?(14 posts)

Why is headwind more punishing then the benefit of tailwind?glia
Nov 8, 2003 6:46 PM
Ever did a ride or time trial where you go for 10 miles into the wind the turn around and go back with tailwind. In the end, I always end up taking longer with wind then without. Yet, 10 miles tailwind and 10 miles headwind should cancel each other out. Doesn't for me. Any thoughts on this? Why is headwind more punishing then the benefit of tailwind?
... nature never gives anything for free...Akirasho
Nov 8, 2003 6:56 PM
... that's why... if possible, I plan my rides for a return tail...

In general, the energy expended fighting a headwind is never compensated for by the adjoining tail (factoring in your fatigue)... and to make matters worse, I'm sure you've experienced the head to head where, at the turnaround point, the wind changes directions!!!

Be the bike.
Get your zen onwooglin
Nov 8, 2003 7:11 PM
When you're riding at 20mph with a 10mph tailwind, is it really helping you? Or is it just hindering you less?
..... oohhmmmmmmmmmmm... (nm)Akirasho
Nov 8, 2003 7:17 PM
Be the bike.
Get rid of all those duplicate letterswooglin
Nov 8, 2003 7:33 PM
and you've got my middle name. No, really.

PS--I think its oooommmmmmmm ;)
It's simpleiktome
Nov 8, 2003 9:02 PM
you travel with the wind for a shorter amount of time than you travel against it. it's the same thing with hills. every minute the wind slows you down is another minute to go slow and lose time. but every minute you go faster is one less minute to go fast. so if you travel 5 mph into the wind for 10 miles, you have 2 hrs to go slow. but coming back at 15 mph for 10 miles is only 40 minutes to go fast. you "lose" 1 hr 20 minutes.
A train leaves Chicago traveling at 67.9 km/h... YIKES!!! nmBowWow
Nov 8, 2003 9:07 PM
very simple math problem. (long workout)Qubeley
Nov 9, 2003 9:03 AM
Problem 1: Why doesn't wind cancel each other out?

Say without wind, you are doing 20mph.
Without wind: 10mi/20mph + 10/20mph = 1 hour
suppose the wind affect your speed equally with or against you, say 2mph.
With wind: 10mi/18mph + 10/22mph = 1.01 hour
Now, I am sure you know the assumption isn't true, the wind slow you a lot more than it help you, so the actual formula is more like this:
10mph/17mph + 10/22mph = 1.04 hour. see?

Problem 2: Why does wind slow you more than it help you?

When the wind is behind you,(unless the wind is faster than you speed) it merely means less air in front you that you have to work against. 20mph bike into 18mph wind, you are facing 2mph relative speed(still headwind technically). Still gotta work, plus road resistance.
If the wind is faster than you from behind, 22mph wind pushing 20mph bike, now we talking about possibility of going forever without ever pedal at all.(in theory at least, if 2mph relative speed is enough to overcome road resistance.)
Now here's the no fun part: 20mph into 10mph wind, much less wind than previous two case. However, relative speed between you and wind is 30mph! compare this to 2mph in the previous case. No need for me to say how much more work this is. For comparision, no wind in the sense of this problem will be 20mph relative speed. Growth of wind resistance is not linear, rather sqaure of the speed.
So the formula in problem 1 might even look like:
10mi/15mph + 10mi/22mph = 1.12 hour

Hope this is clear..
very simple math problem. (long workout)asgelle
Nov 9, 2003 1:08 PM
Now, I am sure you know the assumption isn't true, the wind slow you a lot more than it help you, so the actual formula is more like this:

I don't understand this. The rider has power to overcome 20 mph of wind drag. Therefore with a 2 mph headwind, a relative wind speed of 20 mph results in a land speed of 18 mph. Similarly, a 2 mph tailwind and a 20 mph relative wind speed, gives a 22 mph speed over the ground. Power to overcome wind resistance depends only on the relative velocity of the rider and the surrounding air; 20 mph relative wind speed is the same regardless of the magnitude of the wind or it's direction.

For the headwind to hurt you more than the tailwind helps, there must be some other speed related factor to slow you down more at the higher speed with the tailwindwind than the lower speed of the headwind. Assuming level terrain, this leaves only rolling resistance. From analytic cycling, rolling resistance makes up about 10% of the total power requirement. So the difference from rolling resistance would be no greater than (22/20)x0.1 or 1 % of the total power for the tailwind case. Since apeed goes as the cube root of power this means a decrease in speed of 0.3% for the tailwind section from lower rolling resistance.
By your accounting the formula should read
10mi/(18x(1-0.003)) + 10mi/(22x(1+0.003)) = 1.009

Your explanation for problem 2 is totally beyond me. Simply put, air molecules are moving with an average speed (the wind speed), they collide with bike and rider molecules moving at the riders speed. The force is proportional to the difference between the two numbers. A rider produces enough power to overcome some relative wind speed, RW = RS-W, where RW is the relative wind speed, RS is the riders speed over the ground, and W is the wind speed, or RS=RW+W. That is the rider's speed equals the relative wind speed he has enough poewr to overcome plus the movment of the air around him (W). Another way to look at it is to consider the inertial reference frame of the moving wind.
Physics 101C-40
Nov 9, 2003 3:34 PM
No one here ever took college physics I see.

The explanation is far simpler than anyone has posted.

When traveling into the wind you add the bike's speed to the speed of the headwind to get the relative windspeed. For example, riding at 15mph into a 20mph headwind results in a 35 mph wind resistance. Ride away from the same headwind at 15 mph and the relative windspeed would be -5 mph or in other words, it would be providing 5mph of "wind assist", compared to the 35mph of "wind resistance" going the opposite direction. Obviously, 5mph of wind assistance is not going to increase speed as much as the 35 mph wind resistance reduced speed.
Physics 101asgelle
Nov 9, 2003 4:54 PM
There is no wind resistance and wind assitance from headwind or tailwind; there is only aerodynamic drag from the relative wind velocity (this is a vector quantity). For the simple case of wind directly ahead and behind the rider the wind speed can be added to or subtracted from the rider's speed. For your example, with the headwind, the rider is producing power to overcome the aero drag from 35 mph (15 rider +20 headwind). In your example, the rider continues to ride at 15 mph with the tailwind. Now the rider needs to produce power to overcome -5 mph of aero drag (15 rider - 20 tailwind), in other words there is no aero drag, and the wind is assiting the rider in overcoming other forces (rolling resistance, gravity, etc). However, this is not a fair comparison, The fair way to compare the two cases is to say that the rider can produce enough power to overcome 35 mph aero drag (based on the headwind case), and this power would allow the rider to ride 55 mph at the same power, (35 relative wind =55 rider - 20 tailwind).
Physics 101asgelle
Nov 9, 2003 5:07 PM
"No one here ever took college physics I see."

That's pretty funny coming from someone who refers to "wind resistance" (the proper term is aerodynamic drag) and "wind assist" (the proper term is still aerodynamic drag, how about that).

I can't speak for anyone else here, but if you want, you can do a lit search on A.S. Geller. If you want to discuss bicycle aerodynamics off-line, I'd be glad to dicuss this in whatever detail you like.
Missing the point ;)wooglin
Nov 9, 2003 5:21 PM
After giving this more thought, its clear everyone is missing the point. The reason a tailwind doesn't help as much as a headwind hinders is because we're more aerodynamic from the rear than from the front.
And that's also why..............Mike Tea
Nov 9, 2003 6:28 PM
......we have a headwind almost everywhere we go - if we're doing 25mph with a 15mph tailwind we're feeing a 10mph headwind.