I have never really understood why if you have a flat out and back course with relatively windy conditions coming from a variety of directions. Why does the wind not mostly balance out such that there is net zero effect on effort required?
Looking at wind tunnel data on wheels, especially disc wheels, there can be actually negative drag. So would seem actually a positive effect on watts required when windy compared to no wind. My assumption is that this is such a little percent of the issue it does not overcome the major issue and that the riders body extra drag when windy.
The position in aero bars does seem to be optimized for wind at a 0 degrees or 12 o’clock. When wind hitting at an angle it creates lots of more drag on the body? More wind gets to the chest?
An additional factor I have never completely understood is when you have a wind directly behind you. My simple thinking is that an aero position would not be ideal because your butt and back would not catch the wind and it would be better if you sat up. I have read that the issue is if the wind is moving slower than you then you still are going into the wind. So stay in aero position. For whatever reason if I do need to sit up I do it with a rear wind if possible. Does it actually make any difference? But the entire concept is not really crystal clear in my mind.
Last question on the subject is a disc wheel. I have a completely flat disc. So I am thinking in terms only of a flat disc. But why would a disc help in windy conditions? It would seem that when the wind is hitting you at say 2 o’clock there is considerable more drag as the disc is like a sail going into the wind. But when the wind is hitting you at 4 o’clock it is helping you the exact same amount. But in both cases it is pushing you from the side which would be more inefficient than a regular spoke wheels.
I am asking because I have a race coming up and this time I have been doing some testing on bike setup and watts required for a specific speed. But conditions here in the summer are almost no wind. The race I will be doing it will be windy. I would like to make a WAG (Wild Ass Guess) on a percent of additional effort that will be required for the same speed. I am also thinking if there is any differences I would make in my setup for the race because of wind. I only have a flat disc but have considered purchasing a toroidal disc.
I really, really appreciate any help on the subject. I did do some searching on the forum for about 5 - 10 minutes and did not find any obvious posts. But if there is a link that better explains what is going on I would appreciate a link.
It hurts you more than it helps you b/c you’re making the wind “more” than what it is on the way out by riding into at X mph (so basically add to that the fact that it’s also coming at you at Y mph), and on the way back you’re making the wind “less” by riding away from it at Z mph, while it’s still only blowing at Y mph. Maybe someone can explain this more scientifically than I have (if it’s really true)
You ride INTO the wind for longer (cuz it’s slower) than you ride WITH the wind.
Just like with hills - you climb for longer than you descend, hence more watts required for the same avg speed.
The windier/steeper, the greater the watts needed to overcome it, overall.
You seriously don’t get this concept? It ain’t rocket surgery.
A disk can be faster on windy days since it can act as a sail - yes, very much like on a sailboat.
When are sailboats faster, on calm days or windy days?
Aero postition is faster, even with a honkin tailwind.
Actually the explanation by Dawgtown makes sense to me. It is not something I had read before.
The disc wheel having less drag than a box rim when the wind is between 1 and 3 o’clock (net result when factoring bike speed and wind angle) still is tough for me to understand. Originally sail boats could not sail into a headwind. Modifications were made to the sail in later years to make it work. But you can’t move a disc wheel it is in a fixed position unlike a sail.
I am still having difficulty understanding wind tunnel data with a disc wheel such as what HED publishes. The drag numbers for a disc are better than a box rim when I would think the disc would actually create a lot more drag. There is an explanation on the site about stall. Basically where the wind will not come off the wheel cleanly any longer. I am guessing this behavior is what allows the disc wheel to not create more drag than a box rim. That the wind behaves with the disc in a manner much better than a box rim and it is not as simple as thinking about a solid board into a head wind is a lot harder to hold than a screen. I realize it is not dead straight headwind with a disc ever.
What I would like to find is wind tunnel data comparing a completely flat disc compared to a toroidal shape at different yaw angles and the difference in behavior.
Wind resistance goes up with the square of the bikes velocity, relative to the wind. So going into the wind you are riding for a longer time and the wind resistance is greater.
Going with the wind you are riding for a shorter time against much less resistance.
Or look at it this way suppose you can ride a 40 k tt in an hour with 0 wind. If the wind speed in your face is 40 mph you would have a hard time making any progess It could take much longer than an hour to even go 1/2 way. Coming back you’d fly but if it take 61 minutes to ride 20 km you aren’t getting back in -1 minutes.
Because power, for the aero term, is proportional to Vt^2*Vg, where Vt is “true air speed” and Vg is what you see on the speedo. Vt is Vt=Vg +/- Vw, where Vw is the wind speed. When you include rolling resistance, you just find that as Vw deviates from 0, it takes more average power to get the same overall time.
Actually, the answer is completely the second part of your answer and nothing to do with the first part. While wind resistance does go up with the square of the bikes velocity, the rider essentially will see the same wind resistance each way.
For example: I head out into a 5mph wind at 20mph. When I head back I have a 5mph tailwind and I ride 30mph. Both rides give me a 25mph wind in my face so the wind resistance is the same.
The difference, like you said, is that I spend much more time at my slower (20mph) “half” than my faster (30mph) “half” (as compared to simply going 25mph the whole way with no wind).
“For example: I head out into a 5mph wind at 20mph. When I head back I have a 5mph tailwind and I ride 30mph. Both rides give me a 25mph wind in my face so the wind resistance is the same.”
A 5 mph wind doesn’t slow you down 5 mph and it doesn’t speed you up 5 mph.
“For example: I head out into a 5mph wind at 20mph. When I head back I have a 5mph tailwind and I ride 30mph. Both rides give me a 25mph wind in my face so the wind resistance is the same.”
A 5 mph wind doesn’t slow you down 5 mph and it doesn’t speed you up 5 mph.
Styrrell
yeah but a 5 mph wind will slow you down some fraction of 5 mph and will speed you up the same fraction of 5 mph
Because power, for the aero term, is proportional to Vt^2*Vg, where Vt is “true air speed” and Vg is what you see on the speedo. Vt is Vt=Vg +/- Vw, where Vw is the wind speed. When you include rolling resistance, you just find that as Vw deviates from 0, it takes more average power to get the same overall
time.
That is absolutely correct. If you have a straight out and back course with a headwind on the way out and a tailwind on the way back, it will be slower than on a calm day.
Here is an easy way to think about it. Take an airplane going in a straight line from A to B then B to A. Assume the same power setting to achieve the same true air speed in both directions, let’s use 100MPH. Point A to B is 100 miles. The wind from A to B is 20mph headwind, then a 20mph tailwind from B to A. With a true airspeed or 100mph we will have a ground speed of 80mph from A to B. The time to get from A to B will be 1H 15min. From B to A at the same true airspeed of 100mph, same power setting, your ground speed will be 120 and the time from B to A will be 50 min. Total trip time 2H 5Min. On a calm day total trip time would have been 2 H 0Min. Extra 5min total trip time with the wind conditions. The only way to get the trip time back to 2 hours would be to increase the power on one of the directions. You do not make up the lost time on the tailwind section, especially adding rolling resistance.
Like Tigermilks example above, this same math from an airplane would apply to a bicycle with the exception of the rolling resistance of course.
