Tailwind and speed difference

rmur17 said:

the 1st example is correct but the 2nd is not ... the bicycle's frame of reference is the ground whilst clearly for the airplane or bird that makes no sense.

Anyhow, go back and look at the Martin, Coggan et al formula and simplify it to the case(s) at hand. for pure head/tail winds. I stand by P_aero (v) = 0.5*rho*CdA *Vg*(Vg+Vw)^2

You'll never win bike races if you let yourself be constrained by physics in such a conventional way 😉

Markster said:

You'll never win bike races if you let yourself be constrained by physics in such a conventional way 😉

I'm so heavy that aerodynamics don't matter anyway. give me a pure downhill course and look out 😄 .

rmur17 said:

no definitely not and even with no rolling resistance accounted for ...

I think some folks are confusing equal aero "drag" with equal aero power 🙂

let's drop Crr, drivetrain resistance - everthing except aero drag. Power for the cyclist to travel at speed Vg in headwind Vw on flat terrain will be:

P(v) = Vg x (0.5 x rho x CdA x (Vg + Vw)^2)

or to simplify P(v) = constant x Vg x (Vg+Vw)^2

So if the baseline is 200W and 6 m/s on a flat, windless course

constant = 200W / (6*(6+0)^2) or 200/6^3 or 0.926 (implying a hideous CdA BTW! ~ 1.55)

P(v) with -4 m/s tailwind will be

P(v) = 0.926*Vg*(Vg -4)^2 and for 200W solving for Vg yields 8.92 m/s not 10 m/s

If you start off with a more typical road bike CdA around 0.40, 200W with no Crr will provide a baseline speed of 9.41 m/s. Add 4m/s tailwind and the resulting speed at 200W will be 12.25 m/s or a delta of 2.85 m/s (vs 4 m/s).

Adding Crr back into the mix, road bike CdA 0.40, mass 85kg and Crr 0.004, 200W should provide 8.92 m/s. Add -4 m/s tailwind and the speed rises to 11.61 or a delta of 2.69 m/s (vs 4 m/s).

So it's not linear to start with considering aero drag/power only. You never gain the entire tailwind in bike speed. Add in Crr to the mix and you gain even less.

I think that's about it ... it's not nuclear engineering :p .

I am really confused as to what you are doing here. If you drop everything but the aerodynamic drag coefficient aren't we then talking speed over the ground of an airplane?

For the same power, doesn't a 4 m/s head wind slow the speed over the ground of an airplaine 4 m/s and a 4 m/s tailwind speed it up by the same amount?

rmur17 said:

I stand by P_aero (v) = 0.5*rho*CdA *Vg*(Vg+Vw)^2

It seems you are not the only one (http://www.mayq.com/Best_european_trips/Cycling_speed_math.htm).
OK, could you explain to a layman why it's so? It somehow makes sense to me that not the full wind speed is affecting me, I just don't have a good explanation why ;-)

rmur17 said:

I'm too old to remember where!

re that equation it's for calm conditions only - if you take 'my' equation and the factor Vg*(Vg+Vw)^2 .... set Vw to 0 and you get Vg^3 or simply v^3 as in the equation above.

Ironically, when Jim first posted about our study on a biomechanics mailing list, someone immediately pointed out that he'd made the error of using Vg^3 instead of Vg*(Vg+Vw)^2....then again, he was a mechanical engineer, not a nuclear engineer, so perhaps that explains his oversight! 😉

koger said:

It seems you are not the only one (mayq.comCycling speed math.htmOpen ↗).
OK, could you explain to a layman why it's so? It somehow makes sense to me that not the full wind speed is affecting me, I just don't have a good explanation why ;-)

Power = Force x velocity. The drag force is proportional to the square of the velocity with respect to the fluid; in other words, drag force proportional to (vg + vw)^2 to use the previous terminology. Hence, the power is proportional to vg x (vg+vw)^2.

acoggan said:

Ironically, when Jim first posted about our study on a biomechanics mailing list, someone immediately pointed out that he'd made the error of using Vg^3 instead of Vg*(Vg+Vw)^2....then again, he was a mechanical engineer, not a nuclear engineer, so perhaps that explains his oversight! 😉

In contrast, our nuclear engineer finds the equation fishy, although to be fair, he never clarified as to which equation he was referring to. 😉

koger said:

It seems you are not the only one (mayq.comCycling speed math.htmOpen ↗).
OK, could you explain to a layman why it's so? It somehow makes sense to me that not the full wind speed is affecting me, I just don't have a good explanation why ;-)

aero drag (force) is proportional to 'apparent wind' ^2 as per the Martin, Coggan et al paper. Mechanical work is simply force (in the direction of travel) x distance. Mech. power is the 1st derivative of work wrt. time or force x speed. For cyclists the reference frame is the ground!!

So the power to overcome aero drag alone is ground speed times aero drag or

Vg*(Vg+Vw)^2 for pure head/tailwinds.

for winds coming at angles other than 0/180 deg, you need to calculate the component along the direction of travel before inserting into the basic equation.

honestly, it is just high-school physics. Maybe pre-school physics these days!

rmur17 said:

the power to overcome aero drag alone is ground speed times aero drag or

Vg*(Vg+Vw)^2 for pure head/tailwinds.

for winds coming at angles other than 0/180 deg, you need to calculate the component along the direction of travel before inserting into the basic equation.

You also need to use the appropriate CdA for the resultant yaw angle.

Fday said:

I am really confused as to what you are doing here. If you drop everything but the aerodynamic drag coefficient aren't we then talking speed over the ground of an airplane?

For the same power, doesn't a 4 m/s head wind slow the speed over the ground of an airplaine 4 m/s and a 4 m/s tailwind speed it up by the same amount?

1. I agree with your 2nd statement: for an airplane
2. A bicycle is not an airplane
3. Tell me about the drag and power to propel characteristics of nuclear subs.

acoggan said:

You also need to use the appropriate CdA for the resultant yaw angle.

oh that's for the advanced class Andy. This is "b" level physics 🙂

rmur17 said:

1. I agree with your 2nd statement: for an airplane
2. A bicycle is not an airplane
3. Tell me about the drag and power to propel characteristics of nuclear subs.

I understand a bicycle is not an airplaine. Now, what is different in solving the problem from assuming an airplane and a bicycle where every component except aerodynamic drag has been eliminated?

A nuclear submarine going at 5 knots in the gulf stream with a "tail current" of 1 knot is moving 6 knots over the ocean floor. A submarine behaves as an airplane in this regards.

Fday said:

Now, what is different in solving the problem from assuming an airplane and a bicycle where every component except aerodynamic drag has been eliminated?

Propulsive force is between bicycle and ground in one case and between airplane and air in the other.

patrick_ said:

Propulsive force is between bicycle and ground in one case and between airplane and air in the other.

How can the bicycle know where the propulsive force is generated if all other factors except wind resistance have been eliminated?

Fday said:

How can the bicycle know where the propulsive force is generated if all other factors except wind resistance have been eliminated?

I've wondered the same thing, and I'm thinking it's because we've eliminated the *rolling* resistance (ie, tire flex, etc.), but not the static friction that keeps the wheel from slipping against the road during motion. I tried to envision bicycles on rollers or a moving treadmill, but what I really need is a set of tires with zero rolling resistance so I can go outside and convince myself. 🙂

Edit: I'm pretty sure that a bicycle on ice reacts to the wind exactly the same way that an airplane or bird would, as absurd as that example would be.

frenchyge said:

I've wondered the same thing, and I'm thinking it's because we've eliminated the *rolling* resistance (ie, tire flex, etc.), but not the static friction that keeps the wheel from slipping against the road during motion. I tried to envision bicycles on rollers or a moving treadmill, but what I really need is a set of tires with zero rolling resistance so I can go outside and convince myself. 🙂

Edit: I'm pretty sure that a bicycle on ice reacts to the wind exactly the same way that an airplane or bird would, as absurd as that example would be.

Or a bicycle on an air hockey board. :-)

I could see there might be a difference with acceleration, perhaps, compared to an airplane. But, once up to the speed of the tailwind, the bike would require zero power input to maintain that speed if all other losses were zero. It would behave as a helium balloon as far as the wind were concerned and the bike rider would think he were on a bike at rest in zero wind. And, as it went faster, the only resistance it would see would be air resistance. It would behave just like an airplane as far as I can see.

Fday said:

Or a bicycle on an air hockey board. :-)

I could see there might be a difference with acceleration, perhaps, compared to an airplane. But, once up to the speed of the tailwind, the bike would require zero power input to maintain that speed if all other losses were zero. It would behave as a helium balloon as far as the wind were concerned and the bike rider would think he were on a bike at rest in zero wind. And, as it went faster, the only resistance it would see would be air resistance. It would behave just like an airplane as far as I can see.

But now, the bicycle on the road requires 50x more power to propel itself 50 m/s against a 1 m/s relative wind than the bicycle on the air hockey board (say, it's propelled magnetically or perhaps using a warp-drive.... 😉 ). So, the question still in my mind is: if that's not due to rolling resistance, what is the cause?

frenchyge said:

But now, the bicycle on the road requires 50x more power to propel itself 50 m/s against a 1 m/s relative wind than the bicycle on the air hockey board (say, it's propelled magnetically or perhaps using a warp-drive.... 😉 ). So, the question still in my mind is: if that's not due to rolling resistance, what is the cause?

Since there are no resistances on the road, how can the bike rider know the difference between riding the bike 50 m/s against a 1 m/s head wind and riding 51 m/s into a zero head wind without a speedometer telling him how fast he is going over the ground? It is the same issue the plane has. The plane only knows how hard it is working and what the airspeed is. It cannot know its ground speed without looking or somehow measuring it.

Fday said:

how can the bike rider know the difference between riding the bike 50 m/s against a 1 m/s head wind and riding 51 m/s into a zero head wind without a speedometer telling him how fast he is going over the ground?

It requires x Newton force relative to the ground to keep speed constant, at a constant cadence you need a different gearing and a different force at the pedals.

Fday said:

Since there are no resistances on the road, how can the bike rider know the difference between riding the bike 50 m/s against a 1 m/s head wind and riding 51 m/s into a zero head wind without a speedometer telling him how fast he is going over the ground? It is the same issue the plane has. The plane only knows how hard it is working and what the airspeed is. It cannot know its ground speed without looking or somehow measuring it.

Yeah, I'm with you on all that, and I'm going the next step back towards the *correct* equations presented earlier.

In a 49 m/s tailwind, Bike 1 has Crr=0 (rigid steel tires, or something), but still rolls on the ground and requires 50x more power to maintain a speed of 50 m/s over ground vs Bike 2 which has no resistance between the wheel and ground whatsoever (essentially a hovercraft with aero drag only).

So, my question back to the others is: if it's not due to the rolling resistance of Bike 1, why is the power requirement so dramatically higher? What else is different between the 2 bikes? 😕