<Raises hand> Oo! Oo! I know this one!koger said:Hello
I'm thinking about drag resistance.
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?
It would be less than 10 m/s because of rolling resistance, which doesn't give a hoot about what the wind is doing.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.
no definitely not and even with no rolling resistance accounted for ...koger said:Hello
I'm thinking about drag resistance.
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?
Is Crr constant for any speed?What is the formula for Crr calculation?Iam bad in fysik/and English/.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 .
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).luban said:Is Crr constant for any speed?What is the formula for Crr calculation?Iam bad in fysik/and English/.
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
I'm too old to remember where!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
Which equation are referring to as being fishy? So, you saying that this is nuclear engineering?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).
Canadian P.Eng here graduated with a 'specialization' in Electrical in '86.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).
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.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.
Here's a clue to where it was sourced:Yojimbo_ said:I would also like to see the derivation of that equation, because it looks fishy to me.
oh dear. try again ! look up the definition of mechanical work !!!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.
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:look up the definition of mechanical work !!!
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 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?
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?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.
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).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 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.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.
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