The amount of resistance you get on one straight should be made up for by the amount of help you get on the other so why are mile times changed for wind?
The amount of resistance you get on one straight should be made up for by the amount of help you get on the other so why are mile times changed for wind?
If your running 4:10 pace and there's a 30 mph wind the tailwind isn't going to help you fully recover. Also what if the wind is blowing side/side vs head/tail.
probably retarded wrote:
The amount of resistance you get on one straight should be made up for by the amount of help you get on the other so why are mile times changed for wind?
Supposedly, studies have shown that the amount of resistance you get from running into the wind is far greater than the aid you get while the wind is at your back.
Running into the wind will create a lot more resistance versus any momentum from running with it.
ttc wrote:
Running into the wind will create a lot more resistance versus any momentum from running with it.
Sorry 'bringer of news'. I basically repeated what you said.
There's a simpler explanation, folks... When you're fighting a headwind on one straightaway, it takes you longer to run that straightaway (assuming s constant effort), so you're being slowed down for a longer period of time. When you're being helped by a tailwind on the opposite straightaway, you finish that straightaway more quickly, so you're being helped for a shorter period of time. End result: If there's wind during a track race, you spend more time with a head wind than tail wind. Thus, wind slows you down.
ttc was on to something. But it is more than that, and simple.
If you are running 5 min/mile then you are running at 12 mph. If you are running into a headwind that is blowing 20 mph then the net wind speed is equal to a 32 mph wind. When you are running with the wind then the net wind speed is 8 mph.
Hope that helps.
trainerman wrote:If you are running 5 min/mile then you are running at 12 mph. If you are running into a headwind that is blowing 20 mph then the net wind speed is equal to a 32 mph wind. When you are running with the wind then the net wind speed is 8 mph.
Hope that helps.
not necessarily so simple....you have to compare those to your resistance in still air...
let's say, heading north with wind, you have 32 mph negative wind....heading north in still air you have 12 mph negative wind....difference of negative 20 mph
heading south with wind, you have 8 positive wind, and in still air, you have 12 mph negative wind...difference of 20 positive mph
the difference works out the same...i also don't think the time spent in the wind idea works out because it's the same distance covered...
i suspect it has a lot do with breaking up your rhythm, and that the body burns more energy running against a wind than with it, not necessarily that it slows you down more...
after two hard laps into the wind, you ahve little energy left
Resistance or drag caused by the wind is proportional to the square of the velocity. When you're running 10mph in no wind you are actually creating a 10mph headwind. If you are running into a real 10mph wind, the effective headwind increases from 10mph to 20mph. If it's at your back you go from a 10mph headwind to a 0mph headwind.
Because the resistance is proportional to the square of the velocity there is a bigger difference between 10 and 20mph than between 0 and 10mph, so you lose more in the headwind than you gain with it at your back.
Have you guys never taken physics? When two forces oppose each other, if one force increases (your speed) the opposing force increases by the same ratio (the wind). Therefore, the faster you run the more wind you deal with.
Going with the wind, you are not opposing a force so the wind is not nearly as effective.
this is the correct explanation.
when something hits you, the amount it will slow you down depends on its momemtum, p.
momentum is mass * velocity
p = mv
in this case, you are being hit constantly by the wind, so to figure out how much it slows you down, you need to know not total momentum but the rate of momentum, i.e. momentum of the wind per unit time.
p/t = (mv)/t = m/t*v = wind resistance
the mass of wind per unit time that hits you is the density of wind, d, times the wind's velocity
m/t = dv
putting it all together
p/t = m/t*v=d*v*v = d*v^2 = wind resistance
that is why wind resistance goes with velocity squared
(p.s. technically, it should say d*v^2 is proportional to wind resistance, but i can't make the little jesus fish symbol with my keyboard)
also, technically i should have written
resistance = dp/dt = dm/dt*v+m*dv/dt
but dv/dt = 0
Also, another way of thinking of this: Ever hit a headwind so hard you practically stopped in your tracks? Now, ever have the wind at your back and suddenly you're effortlessly being pushed along running 4:00 mile pace? I can honestly say I don't think I've ever had the wind "push" me along in the least. That alone should help you realize that the net result is a slower time.
part II
why the fact wind resistance = v^2 makes wind a hindrance
let's say you run 100m down the staightaway, then turn around and run back. you run 15mph
on a calm day, you experience wind resistance at 15mph = 15^2* 2 straightaways = 450 units of wind resistance
with a 10mph wind, when you run with the wind you have 5mph wind = 25 units wind resistance. when you run against the wind you have 25mph wind = 625 units wind resistance = 650 units wind resistance total
in general, if you run x mph and wind speed is y, then on a calm day you would have 2x^2 units of wind resistance and on a windy day (x+y)^2+(x-y)^2 = 2x^2 + 2y^2 units wind resistance. the difference is always y^2.
in other words, if you run at the same pace, total wind resistance will always be greater when there is a wind. you could try to counter this a little by running slower into the wind and faster with it at your back, but you won't be able to negate the effect completely
oops. difference is always 2y^2
zzzzzzzz wrote:
http://sheldonbrown.com/brandt/wind.html
Actually, I think there is a mistake in the analysis at this link. I don't think a crosswind will result in an increase in drag except to the extent that it increases turbulence.
this all sounds like a bad word problem!! ugh!
What is the airspeed velocity of an unladen swallow?
Think of it this way, guys. It's like hills in a xc race. On a flat xc course, you can run a lot faster than on a hilly xc course. Even if the hilly course ends where it starts(which would mean you come down as much as you go up), the hilly course will be tougher/slower because of the fact that you have to deal with the resistance of hills for certain periods of time compared to a flat course where you never have to deal with that resistance. Same on the track with wind. If there is wind, you go against it at some point every lap. Even if it helps you for a certain period of time every lap, you will still run slower on a windy day because of the fact that you have resistance for a certain amount of time compared to on a calm day where there is no resistance ever during the race. Hills on a xc course, wind on a track, I think it makes sense.