Dot Physics

I ride my bike and mostly the wind makes me unhappy. On a very few days the wind is with me on the way to work and then changes so that it is with me again. But most days the wind is fairly constant. So, if the wind is constant then shouldn’t everything even out? (Even Stephen).

Assumptions:

Let me start with the assumption that I (a mere mortal) can output at a constant power (but not 57,000 Watts like some people). I will also assume an air resistance force that is proportional to the square of the relative air speed. Here is a diagram.

i-8e13039f7f5c5feb649a6354a5f14f0a-2010-02-26_bike_2.jpg

A couple of quick things to point out. First, the net force on the bike is the zero vector. This is because it is traveling at a constant speed. I am really not going to have to worry about the vertical forces on the bike – they don’t really do anything (Yes, I know I should have draw two forces for the upward force from the ground, one on each tire). The friction force is essentially from me (the rider). This includes the internal friction from the gears and stuff. There are two velocities. The vbike is the velocity of the bike relative to the ground. The vair-rel is the velocity of the air relative to the bike. This second velocity is what will be used in the air resistance force. If there is no wind, vair-rel = – vbike. If there is wind (say vwind) then:

i-5bd7b77b0d67eb292bf3622d17ddd3b7-2010-02-26_relative_1.jpg

Maybe I should have started with vbike-air instead of vair-bike – especially since the sign doesn’t really matter. (here is a tutorial on relative velocities) So, in terms of my original stuff and the velocity of the wind (which is the velocity of the air with respect to the ground):

i-16d21564ca9764d63e81c61ecd2053b2-2010-02-26_la_te_xi_t_1_3.jpg

Just a quick check: if vwind = 0 m/s, then vair-rel = – vbike. If I am riding at the same speed as the wind (and in the direction of the wind) then the relative air speed would be zero (vector). Good enough for me.

Back to power

I am going to call the power output (including the internal losses in the bike due to friction and stuff) Prider. But, what I need is a connection between this and the frictional force that pushes the bike. So, suppose the bike moves a distance s. What work would this friction force do on the bike if I consider the bike to be a particle?

i-d3d27bf9065e395eb02b3d8f16051f26-2010-02-26_la_te_xi_t_1_4.jpg

Since the friction force and the displacement are in the same direction, the work is positive. If I want the power (and I do) then I can write: (because I am lazy, I am going to write Ff for the friction force instead of Ffriction – also it is because I am really thinking of this as the force the rider exerts on a point-particle system)

i-db97a2c932ef92d6bc771df7b52d2091-2010-02-27_la_te_xi_t_1_5.jpg

If the biker is riding at a constant speed, then the air resistance force is equal in magnitude to the “friction” force. I am using the following model for the magnitude of the air resistance force.

i-64a6c3f8ab8251475cc3c4ed574e3ac5-2010-02-27_la_te_xi_t_1_6.jpg

Since the density of air, the cross-sectional area and the drag coefficient are constant, I replaced all those with the constant K. Since the air resistance is equal to the frictional force:

i-8aaf0335a6c2a3e7dd29497b3354df3f-2010-02-27_la_te_xi_t_1_7.jpg

I need the vair-rel in terms of the wind speed. So:

i-0b3bebfb9c9880af83601ec92673c597-2010-02-27_la_te_xi_t_1_18.jpg

The wind speed and the velocity of the bike are both vectors – obviously, it matters if you are riding in the same direction or opposite direction as the wind. But, this is really a 1-dimensional problem, so I can take the horizontal components of these vectors and make it look like:

i-23d7407a3e65390ee044ce710b894b5a-2010-02-27_la_te_xi_t_1_19.jpg

So the sign on these velocity components matters. Also, I got rid of the absolute value signs since I am squaring that. The wind can be positive (tail wind) or negative (head wind). It seems like this will work. Now, what I really want is to solve for the velocity of the bike in terms of the wind and the power from the biker.

i-1ccc0e1879d7983f65a691118d611528-2010-02-27_la_te_xi_t_1_20.jpg

This is a 3rd order polynomial for vbike – and you know what? Cubic equations kind of suck to deal with. Instead of dealing this symbolically, I will go ahead and determine some values for these constants.

Let me start with the case of no wind. My brother cycles a lot and has a PowerTap. He estimates that I would be at about 200 watts going about 20 mph (9 m/s). So, from this I can get a value for Ffriction which will give me the value of Fair. What I really want is K:

i-9bfe52d8aedd90765991e0b6d0b86e40-2010-02-27_la_te_xi_t_1_21.jpg

Now for the fun stuff. I need to solve that cubic equation for different values of the wind speed. Here is a method that I am going to use. Now for a graph. This is the bike rider speed as a function of wind speed (I randomly chose to go from wind speed of -15 m/s to +15 m/s where the +15 means the wind is in the same direction as the rider). One more note – 15 m/s is really fast (over 30 mph). You probably shouldn’t ride your bike if it is this windy outside.

i-e7b85716ab1a8163df0d9755cd58e04e-2010-02-28_riderspeed_1png.jpg

Remember my initial point (I know it was quite some time ago) – the wind has more of a negative impact than a positive one. Let me plot the magnitude of the change in rider speed due to the wind.

i-5733bc10cb9fd71665cb321d1d925852-2010-02-28_untitled.jpg

In terms of speed adjustment, you can see I was wrong. What was I thinking? Take a look at, say, 8 m/s wind. If it is going with the rider then it will increase the rider’s speed by about 6 m/s. If it is going against the rider, it will decrease the rider’s speed by only a little over 4 m/s. I am not sure I have a good explanation for why this is the case – so instead I will make another argument to show that I am correct.

Suppose this is a round trip and the wind is constant in magnitude and direction for the full round trip. Then, I will go faster when going with the wind and slower against the wind. How about I calculate the time for a round trip for 5 km one way with different wind speeds?

i-9716d65ec807bc3bf44b1f4657a3e417-2010-02-28_roundtripbikepng.jpg

See. So even though a rider may ‘gain’ more speed with the wind, the trip takes longer. Really, this is a classic intro-physics problem (but usually with an air plane where the speed difference with and against the wind is the same). The answer is that it takes longer with the wind because the rider will spend more time on the slow part than on the fast part. This means the average speed is not something close to the speed with zero wind.

One more thing – how fast would the wind have to be to not be able to go at all?

i-dabdee39bf9342bdebad7c1bf4e48ffc-2010-02-28_hurricaneridepng.jpg

From this plot, even in a 90 mph wind you would still be going forward (though not really fast). I am not going to put too much weight on this calculation because I know some weird things can happen with cubic equations when the sign of the result changes.

One more thing

Here is an online bike calculator. You enter parameters like your power and stuff about your bike and it tells you your speed.

Main entry continued

Comments

  1. #1 Connie
    March 1, 2010

    Last season I started referring to headwinds as my training partner “Wendy”.

  2. #2 natural cynic
    March 1, 2010

    You have only elucidated one of the perceptible properties of the cycling universe: There’s always going to be more headwinds than tailwinds. This goes along with property #2: there will always be more uphills than downhills.

    And as a corollary of the first: when your ground speed matches the tailwind speed you are most likely to pass gas.

  3. #3 Patrick
    March 2, 2010

    I don’t know know who natural cynic is, but he is very clever and hilarious! Anyway, one thing to consider would be the differing force exerted by the biker against the wind vs the biker with the wind. If I’m biking against the wind, I’m going to be exerting a near constant force to combat the wind in a nearly uniform manner. But when the wind is on your side, you tend to relax. You don’t exert a constant force over a given distance as you would when you were combatting the wind. You peddle a little, then coast… applying no force. So… over a given distance less force is applied by the biker going with the wind (i.e. less power output). If I’m right, this presents a problem that can’t be easily estimated. How much force does one exert into the wind at different windspeeds/velocities? I don’t know. I suspect a graph of speed vs. wind speed would be nearly identical for no wind at all vs with the with the wind bikers though.

  4. #4 Patrick
    March 2, 2010

    Let me rephrase that last sentence. What I’m trying to say is that your graph of wind speed vs rider speed should level off very quickly. That is to say the rider’s speed will not increase exponentially as the wind speed increases. If that were the case, a 10-15 m/s tailwind would have all bikers souring down the sidewalk at (very roughly) 200m/s!!

  5. #5 Patrick
    March 2, 2010

    Nevermind, I was calculating a non-zero net force on the bike… Stupid mistake!

  6. #6 Rob Monkey
    March 4, 2010

    This is totally nitpicking, but wouldn’t the wind resistance be greater with a headwind because of the shape of the rider? I think a tailwind would slide over your convex back, but a headwind would be going into your chest, which is in a sort of concave orientation. Would that affect the net force on the rider?

  7. #7 Rhett Allain
    March 4, 2010

    @Rob,

    Good point – I didn’t even think about that. I guess I could fix this by saying “assume a spherical rider”

    Oh – also, as long as the relative air speed is going into the rider, the CA term will be the same.

  8. #8 DCH
    April 20, 2010

    Very nice analysis: Now I fully understand the effect of wind.

    About those hills:

    If I ride 10Km on flat, using the same constant power (same assumption for your wind analysis) I assume by near similar logic my time for 10Km with included hills (lets assume there is no wind in the mountains) will not be as good?

    Or will it? Intuative assumptions don’t always give the same results as mathematics…

  9. #9 Rhett Allain
    April 20, 2010

    @DCH,

    I think the time will be longer – you will spend more time going UP hills at a slow speed than going down at a faster speed. That is my first guess, but maybe I should do the calculations in another blog post – good idea!