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Road Cycling “It is by riding a bicycle that you learn the contours of a country best, since you have to sweat up the hills and coast down them. Thus you remember them as they actually are, while in a motor car only a high hill impresses you, and you have no such accurate remembrance of country you have driven through as you gain by riding a bicycle.” -- Ernest Hemingway

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Old 08-26-13, 10:23 AM   #1
profjmb
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physics of crosswinds?

Yesterday with mild crosswinds I felt (and rode) slower. Why? (In my head or real?)

Back in the day, in resolving vectors, I don't think that a crosswind would exert force against my forward progress that should require increased energy. But that was a simple, probably simplistic, model.

Why would a crosswind make me work harder to obtain the same speed?
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Old 08-26-13, 10:27 AM   #2
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The short answer is: any wind not within 15 degrees of being a perfect tailwind hurts you.

The longer answer involves your speed, apparent wind speed, stall angles, and lots of reading at Slowtwitch.
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Old 08-26-13, 10:32 AM   #3
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I feel a Sail Bike posting coming on.
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Old 08-26-13, 12:32 PM   #4
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You don't use the forces exerted by the separate wind vectors, you have to use the vector sum of the forward motion (rearward wind) and crosswind. Then apply the force of that wind vector. Since force is proportional to the square of the wind speed, the crosswind factor gets "locked into" the value so to speak. The compnent in the opposite to the dorection to travel is greater than just from headwind. A simple example: 2 mph forward motion, 2 mph perfect crosswind so the resulting wind is 2 X root 2 at 45 deg to forward direction. Since force is proportional to the square of the wind speed, the total force of the wind on the bike and rider is proportional to 8. The component of that force pushing opposite to the direction of travel is 8 divided by root 2 or 4 X root 2. But if there were no crosswind, the force would be proportional only to the headwind component = 2 squared or 4. So the actual impeding force would be a factor of root 2 times the headwind force. That is the effect of the crosswind. Simple, huh? Sorry to be so obtuse but I am typing on my phone. Equations aren't so easy.
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Old 08-26-13, 12:42 PM   #5
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^ agree with the above, although it's probably over-simplified somewhat.
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Old 08-26-13, 07:48 PM   #6
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^ agree with the above, although it's probably over-simplified somewhat.
And a little hard to parse, but the gist is there.
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Old 08-26-13, 08:52 PM   #7
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Parker’s explanation is correct and understandable, but I’m sitting at a desktop, so lets do a simple example using numbers from a 3-4-5 triangle so we can do the math in our head. Some paragraphs will add to the clarity.

Wind drag is proportional to the square of the wind speed. This example is simplistic by assuming a constant drag factor in any direction and ignoring the drag factor by using an arbitrary value of 1.
Bike forward progress is 4, side wind is 3, the apparent wind is the vector sum and turns out to be the hypotenuse of our sample triangle or a magnitude of 5. Force is 5^2 or 25 applied at an angle of about 37 degrees IIRC.

But all we care about is the opposing force, or that portion of the force in the direction of travel. That is cos theta (4 / 5) X the force of the wind in the direction of the apparent wind. or .8 X 25 = 20.
Without the side wind, the apparent wind is simply the riding speed and drag is 4^2 = 16.

It is the fact that drag is proportional to the square of the speed that creates a non-linear drag function.

The value of the neutral wind angle that neither hinders nor helps will change with the wind speed and rider speed, but 15 degrees is in the ball park for a typical example

Last edited by rdtindsm; 08-26-13 at 08:57 PM.
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Old 08-26-13, 09:48 PM   #8
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Quote:
Originally Posted by Bah Humbug View Post
The short answer is: any wind not within 15 degrees of being a perfect tailwind hurts you.

The longer answer involves your speed, apparent wind speed, stall angles, and lots of reading at Slowtwitch where people believe in 20 degree yaw angles.
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Old 08-26-13, 09:50 PM   #9
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The best technical response I can offer is "crosswinds suck"...ride harder or find a fool...er um meant more fit person to break the wind for you...
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Old 08-26-13, 10:49 PM   #10
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The other thing is that your swept area is greater with an apparent wind on your forward quarter than with the wind dead ahead. "Getting aero" doesn't help as much as it does with a headwind or still air. This is very noticeable on our tandem. With apparent wind on the forward quarter, the stoker is no longer hidden behind the captain. We hate crosswinds.
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Old 08-27-13, 05:42 AM   #11
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The short answer is: any wind not within 15 degrees of being a perfect tailwind hurts you.
The actual angle is around 60 degrees, not 15.

For wind speeds greater than ground speed the angle is less than 60 degrees, for wind speeds less than ground speed the angle is greater than 60 degrees. For a wind 15 degrees off a tailwind to have a net negative effect the wind speed needs to be nearly double your ground speed.

The above ignores the effect of change of apparent area with aspect and any increase in rolling resistance from steering into the side wind.
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Old 08-27-13, 08:19 AM   #12
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The actual angle is around 60 degrees, not 15.

For wind speeds greater than ground speed the angle is less than 60 degrees, for wind speeds less than ground speed the angle is greater than 60 degrees. For a wind 15 degrees off a tailwind to have a net negative effect the wind speed needs to be nearly double your ground speed.

The above ignores the effect of change of apparent area with aspect and any increase in rolling resistance from steering into the side wind.
I agree...I'll take a 45 degree tailwind any day.
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Old 08-27-13, 02:11 PM   #13
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wut
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Old 08-28-13, 10:02 AM   #14
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Originally Posted by rpenmanparker View Post
You don't use the forces exerted by the separate wind vectors, you have to use the vector sum of the forward motion (rearward wind) and crosswind. Then apply the force of that wind vector. Since force is proportional to the square of the wind speed, the crosswind factor gets "locked into" the value so to speak. The compnent in the opposite to the dorection to travel is greater than just from headwind. A simple example: 2 mph forward motion, 2 mph perfect crosswind so the resulting wind is 2 X root 2 at 45 deg to forward direction. Since force is proportional to the square of the wind speed, the total force of the wind on the bike and rider is proportional to 8. The component of that force pushing opposite to the direction of travel is 8 divided by root 2 or 4 X root 2. But if there were no crosswind, the force would be proportional only to the headwind component = 2 squared or 4. So the actual impeding force would be a factor of root 2 times the headwind force. That is the effect of the crosswind. Simple, huh? Sorry to be so obtuse but I am typing on my phone. Equations aren't so easy.
That's more confusing that those "If a train leaves Chicago traveling 150MPH" word problems from high school
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Old 08-28-13, 10:03 AM   #15
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The best technical response I can offer is "crosswinds suck"...ride harder or find a fool...er um meant more fit person to break the wind for you...
Breaking wind while pulling someone is not a cool thing to do
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Old 08-28-13, 10:08 AM   #16
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Breaking wind while pulling someone is not a cool thing to do
Waving negates that...
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Old 08-28-13, 10:09 AM   #17
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That's more confusing that those "If a train leaves Chicago traveling 150MPH" word problems from high school
I'm surprised you noticed. Seriously, if you diagram it on paper, it will make sense.
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