# Tailwind and speed difference

#### koger

##### New Member
Hello

If there is no wind and I put in 200W let's my pace is 6m/s.
I then get a tailwind on 4m/s and keeping the 200W, should my pace in theory raise to 10m/s?

##### New Member
koger said:
Hello

If there is no wind and I put in 200W let's my pace is 6m/s.
I then get a tailwind on 4m/s and keeping the 200W, should my pace in theory raise to 10m/s?
<Raises hand> Oo! Oo! I know this one!

It would be (10M/s) - (increase in rolling resistance) - (possible but unlikely increase in drivetrain resistance) - (any braking)

#### frenchyge

##### New Member
Vladav's right. The aero portion of the entire overall resistance would drop because of the tailwind, and then bike speed would increase until the new overall resistance matched the previous 200w value. The new speed would be less than 10 m/s.

#### Fday

##### New Member
frenchyge said:
Vladav's right. The aero portion of the entire overall resistance would drop because of the tailwind, and then bike speed would increase until the new overall resistance matched the previous 200w value. The new speed would be less than 10 m/s.
It would be less than 10 m/s because of rolling resistance, which doesn't give a hoot about what the wind is doing.

#### rmur17

##### New Member
koger said:
Hello

If there is no wind and I put in 200W let's my pace is 6m/s.
I then get a tailwind on 4m/s and keeping the 200W, should my pace in theory raise to 10m/s?
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 .

#### luban

##### New Member
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 .
Is Crr constant for any speed?What is the formula for Crr calculation?Iam bad in fysik/and English/.

#### rmur17

##### New Member
luban said:
Is Crr constant for any speed?What is the formula for Crr calculation?Iam bad in fysik/and English/.
yes within reason of course typical Crr ranges from around 0.002 to 0.008 on the track and road to much higher for mountain bike tires and surfaces (don't have any data for that).

not for the calculation of Crr per se but the power required to overcome rolling resistance is:

P = Vg x m x g x Crr

Vg= bike ground speed in m/s
m = total mass of bike, rider and kit
g = gravitational constant ~9.81
Crr = unitless coefficient of tire/surface rolling resistance

#### koger

##### New Member
rmur17 said:
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

Thank you for your great post. I have one question though, where did you find the equation? I found this at http://en.wikipedia.org/wiki/Drag_(physics)

P(v) = 0.5*rho*v^3*A*Cd

Could you tell me what I missed?

Thanks again

#### rmur17

##### New Member
koger said:
Thank you for your great post. I have one question though, where did you find the equation? I found this at http://en.wikipedia.org/wiki/Drag_%28physics%29

P(v) = 0.5*rho*v^3*A*Cd

Could you tell me what I missed?

Thanks again
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.

Actually what I listed is a very simple version that applies only to direct headwinds and tailwinds. Add in a 'yaw angle' and the math is a little messier -- though the concepts exactly the same.

#### Yojimbo_

##### Well-Known Member
I would also like to see the derivation of that equation, because it looks fishy to me.

I think the correct equation (for wind directly behind the rider) and ignoring all effects except aerodynamic drag is:

Power = 0.5 rho V**3 A Cd (as said by another poster)

where: rho is density, A is area, Cd is drag coefficient, and V is the speed of the bike relative to the air.

(And I am a nuclear engineer).

#### TheDarkLord

##### New Member
Yojimbo_ said:
I would also like to see the derivation of that equation, because it looks fishy to me.

I think the correct equation (for wind directly behind the rider) and ignoring all effects except aerodynamic drag is:

Power = 0.5 rho V**3 A Cd (as said by another poster)

where: rho is density, A is area, Cd is drag coefficient, and V is the speed of the bike relative to the air.

(And I am a nuclear engineer).
Which equation are referring to as being fishy? So, you saying that this is nuclear engineering?

#### rmur17

##### New Member
Yojimbo_ said:
I would also like to see the derivation of that equation, because it looks fishy to me.

I think the correct equation (for wind directly behind the rider) and ignoring all effects except aerodynamic drag is:

Power = 0.5 rho V**3 A Cd (as said by another poster)

where: rho is density, A is area, Cd is drag coefficient, and V is the speed of the bike relative to the air.

(And I am a nuclear engineer).

I'm old and cranky most days and I stand by my math. Period.

Unless corrected by a true expert of course . Where are the true boffins when you need them????

Hey you're not in charge of installing backup safety pumps at Chalk River are you?

#### frenchyge

##### New Member
Yojimbo_ said:
I would also like to see the derivation of that equation, because it looks fishy to me.

I think the correct equation (for wind directly behind the rider) and ignoring all effects except aerodynamic drag is:

Power = 0.5 rho V**3 A Cd (as said by another poster)

where: rho is density, A is area, Cd is drag coefficient, and V is the speed of the bike relative to the air.
Me too. You must use the same reference point for all velocities, whether that's the ground, rider, air, or an observer. Rick's equation uses the ground as the reference point for the rider velocity Vg, and the air as the reference point for the drag force (ie, by subtracting Vg and Vwind *before* squaring the terms) calculation. If the drag force is a function of the body's speed relative to the air, then the drag power is a function of that same relative velocity.

Consider the case where a rider is standing stationary in the face of a 10mph wind. The rider is still expending energy to resist the force of the wind, despite the fact that she is not moving relative to the ground. There is still a frictional energy transfer taking place because of the relative motion.

If you use (Vg + Vw) in place of the first Vg in Rick's equation, you do indeed get 10m/s for Vg in the example of a -4m/s wind, since the example neglects the other sources of friction.

#### Alex Simmons

##### Member
Yojimbo_ said:
I would also like to see the derivation of that equation, because it looks fishy to me.
Here's a clue to where it was sourced:

#### rmur17

##### New Member
frenchyge said:
Me too. You must use the same reference point for all velocities, whether that's the ground, rider, air, or an observer. Rick's equation uses the ground as the reference point for the rider velocity Vg, and the air as the reference point for the drag force (ie, by subtracting Vg and Vwind *before* squaring the terms) calculation. If the drag force is a function of the body's speed relative to the air, then the drag power is a function of that same relative velocity.

nope the ground is the reference frame.

Consider the case where a rider is standing stationary in the face of a 10mph wind. The rider is still expending energy to resist the force of the wind, despite the fact that she is not moving relative to the ground. There is still a frictional energy transfer taking place because of the relative motion.

the rider (unless swaying to and fro) performs no mechanical work while stationary and thus the power is zero.

If you use (Vg + Vw) in place of the first Vg in Rick's equation, you do indeed get 10m/s for Vg in the example of a -4m/s wind, since the example neglects the other sources of friction.
oh dear. try again ! look up the definition of mechanical work !!!

on a simply practical basis - haven't you folks noticed that headwinds don't slow one down proportionally to headwind speed and the opposite for tailwinds? Does a say 10 m/s headwind stop you folks in your tracks? !!!

I'm honestly shocked at the lack of understanding here ...

#### frenchyge

##### New Member
rmur17 said:
look up the definition of mechanical work !!!
Aero drag is the shear force between the fluid and the body. The drag is the frictional force parallel to the flow and the distance is the length of the body over which it flows. Makes no difference whether the object is moving through stationary air or the air is moving against a stationary body -- only the frame of reference has changed.

rmur17 said:
on a simply practical basis - haven't you folks noticed that headwinds don't slow one down proportionally to headwind speed and the opposite for tailwinds?
By 'proportionally', you don't mean linearly, do you? I don't believe anyone has stated that the aero drag power is other than a cubic function, and that the total power includes that plus some other factors.

#### rmur17

##### New Member
frenchyge said:
Vladav's right. The aero portion of the entire overall resistance would drop because of the tailwind, and then bike speed would increase until the new overall resistance matched the previous 200w value. The new speed would be less than 10 m/s.
so you stand by this statement that the only reason the bike speed won't reach 10 m/s is because of the increase in the force/power to overcome Crr?

Go into analytic cycling or another proven bike speed/power calculator, set Crr to zero and run thru some examples.

#### frenchyge

##### New Member
rmur17 said:
so you stand by this statement that the only reason the bike speed won't reach 10 m/s is because of the increase in the force/power to overcome Crr?

Go into analytic cycling or another proven bike speed/power calculator, set Crr to zero and run thru some examples.
The difficulty I'm having is believing that a cyclist without rolling resistance who is experiencing a 1 m/s relative wind is expending 50 times more power while riding at 50 m/s than at 1 m/s (assuming the wind is much stronger in the second case so that the relative wind remains 1 m/s in her face).

Additionally, I have difficulty believing that a bird or plane flying against a strong headwind but making no speed over ground is doing no work and expending no power.

Your equation above would indicate that both of those examples are true, and maybe they are. I'll have to think about it some more once the post-race delirium wears off.

#### phantoj

##### New Member
Isn't the idea that drag force is proportional to relative windspeed squared... and power is speed relative to ground x drag force.

Power is relative to groundspeed because the power is being applied to the ground. Right?

#### rmur17

##### New Member
frenchyge said:
The difficulty I'm having is believing that a cyclist without rolling resistance who is experiencing a 1 m/s relative wind is expending 50 times more power while riding at 50 m/s than at 1 m/s (assuming the wind is much stronger in the second case so that the relative wind remains 1 m/s in her face).

Additionally, I have difficulty believing that a bird or plane flying against a strong headwind but making no speed over ground is doing no work and expending no power.

Your equation above would indicate that both of those examples are true, and maybe they are. I'll have to think about it some more once the post-race delirium wears off.
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

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