Tailwind and speed difference



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.
Power applied is force x velocity for either a bike or an aircraft. But consider that on a bike, force is applied to a fixed road via the pedals to the rear tire. Regardless of the wind conditions, it takes the same cadence to achieve a given Vg, and the speed is always measured relative to the road. Winds only change the force required (aero drag) part of the equation, not the velocity. This is good news, because it means that with a head wind, we "only" have to pedal harder (to match aero drag force), not both harder and faster :)

OTOH, in aviation, propulsion force is applied to the air mass itself, so velocity of the air mass passing the aircraft (Vg + Vw) is the only thing that counts for determining power (at a steady altitude). As you said, the aircraft doesn't know what it's doing relative to the ground, and as a result the pilot must use nav aids or ground cues to determine groundspeed and progress. To make up for a headwind and hold a given ground speed, the aircraft has to actually fly faster, not just maintain speed like we do on the bike.
 
frenchyge said:
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? :confused:
In the example, believe the aero drag (force) on bikes 1 and 2 would be the same, since they are both working into an apparent headwind of 1 m/s. As a result, the force the rear wheel must apply to the road is identical.

What's different is that in order to apply that force, equal to the puny drag force generated by that 1m/s wind, the bike 1 (going 50 m/s) has to apply that force at 50x the velocity of bike 2. In practical terms, this means that the bike 1 rider has to be in a huge gear, while the second bike is twiddling in a tiny gear at 1 m/s. At equal cadence, this gearing disparity means the first rider has to be applying 50 times the force to the pedals to generate the same wheel force as the second rider.
 
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?
His perceived time will be roughly 1/234,567,000,000 of a microsecond faster than of someone going 51 m/s . It's elementary relativity people. :D
 
dhk2 said:
In the example, believe the aero drag (force) on bikes 1 and 2 would be the same, since they are both working into an apparent headwind of 1 m/s. As a result, the force the rear wheel must apply to the road is identical.

What's different is that in order to apply that force, equal to the puny drag force generated by that 1m/s wind, the bike 1 (going 50 m/s) has to apply that force at 50x the velocity of bike 2.
Thanks, that makes a lot of sense and explains why it appeared to me that the frame of reference was changing between the drag velocity and the propelling velocity. So, if you wanted a combined equation that was good for both cases, you'd multiply the drag force by the sum of the the body's velocity plus the velocity of the propelling mass (providing conservation of momentum).

So, Frank, what's the market for a crank-driven rear propeller that could be engaged when there's a tailwind to let a rider take advantage of the full boost of the wind? :D
 
vladav said:
<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;))
:confused::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad:GOD DAMNIT!!!!!! DID I GET THE ANSWER RIGHT OR NOT!!!!!! :mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::mad::confused:

Very truly yours,
Too uneducated to know the difference:p
 
vladav said:
DID I GET THE ANSWER RIGHT OR NOT!!!!!!
Sadly, no. Go sit in the back of the class. :p

The answer is definitely less than 10 m/s, but less than you said (and I agreed with) as well. :(
 
frenchyge said:
Thanks, that makes a lot of sense and explains why it appeared to me that the frame of reference was changing between the drag velocity and the propelling velocity. So, if you wanted a combined equation that was good for both cases, you'd multiply the drag force by the sum of the the body's velocity plus the velocity of the propelling mass (providing conservation of momentum).

So, Frank, what's the market for a crank-driven rear propeller that could be engaged when there's a tailwind to let a rider take advantage of the full boost of the wind? :D
I still don't get it. To generate 200 watts at any particular cadence requires the same amount of force on the pedals regardless of how fast the bike is going. To have the same cadence at different speeds is simply a matter of gearing.

Again, if all the other losses were zero (and the cyclist didn't know what his gearing was), I don't see how the cyclist could tell under which circumstances he was riding. 200 watts is 200 watts.
 
Fday said:
I still don't get it. To generate 200 watts at any particular cadence requires the same amount of force on the pedals regardless of how fast the bike is going. To have the same cadence at different speeds is simply a matter of gearing.
That's only true if cadence and bike speed are directly related by gear ratio (ie, non-slipping contact between the wheel and ground). Once we turn the bike into a pedal-powered zepplin then cadence, gear ratio, and speed over ground are no longer locked.

Fday said:
Again, if all the other losses were zero (and the cyclist didn't know what his gearing was), I don't see how the cyclist could tell under which circumstances he was riding. 200 watts is 200 watts.
He wouldn't know the difference, but there's also no guarantee that he'd be making any speed over ground (he could even be moving backwards, for all his effort).

I think the Law of Conservation of Momentum is what really makes the conceptual difference between the 2 cases. For a rider in non-slipping contact with the earth, the tremendous mass of the earth means that the amount of backward motion generated by the forward propulsion of the bike is so imperceptibly small that it can be ignored (unnecessary if the frame of reference is placed on the earth). A rider who has slippage at the tire, or is using air mass for propulsion cannot make the same assumptions about how much forward thrust he experiences or how his wheel speed and forward speed relate.
 
Fday said:
I still don't get it. To generate 200 watts at any particular cadence requires the same amount of force on the pedals regardless of how fast the bike is going. To have the same cadence at different speeds is simply a matter of gearing.

Again, if all the other losses were zero (and the cyclist didn't know what his gearing was), I don't see how the cyclist could tell under which circumstances he was riding. 200 watts is 200 watts.
200 W is certainly 200 W, but I'm not sure what that statement means. Forget cadence and gearing; at any given road speed 200 W corresponds to a certain force which is being applied to the rear wheel. Remember, watts is just force x velocity, where the force is in units of newtons, and velocity is in meters/sec.

If a bike is traveling 10 m/sec (22.5 mph) , and the rider is putting out 200 W, then the force to the road from the rear tire is 20 nts, or ~ 2 kg force, or ~ 4.5 lbs-force (neglecting small drivetrain losses).
 
Fday said:
I still don't get it. To generate 200 watts at any particular cadence requires the same amount of force on the pedals regardless of how fast the bike is going. To have the same cadence at different speeds is simply a matter of gearing.

Again, if all the other losses were zero (and the cyclist didn't know what his gearing was), I don't see how the cyclist could tell under which circumstances he was riding. 200 watts is 200 watts.
Cadence translates to speed through the gear ratio. It is a simple mechanical system, and so this always has to be satisfied. However, the force required to maintain the same cadence (i.e. same speed assuming the same gear ratio) varies on factors such as aerodynamics, rolling resistance, etc. (for example, if you stop pedaling, the bike will slow down faster if there is a headwind). Hence, the power output required to maintain the same speed will also vary depending on these conditions.
 
TheDarkLord said:
Cadence translates to speed through the gear ratio. It is a simple mechanical system, and so this always has to be satisfied. However, the force required to maintain the same cadence (i.e. same speed assuming the same gear ratio) varies on factors such as aerodynamics, rolling resistance, etc. (for example, if you stop pedaling, the bike will slow down faster if there is a headwind). Hence, the power output required to maintain the same speed will also vary depending on these conditions.
The scenario we are discussing involves where all other losses are zero. In this instance the bike will "slow" at the exact same rate whether it is going 1 mph into a 49 mph headwind or 50 mph into a zero wind.
 
dhk2 said:
200 W is certainly 200 W, but I'm not sure what that statement means. Forget cadence and gearing; at any given road speed 200 W corresponds to a certain force which is being applied to the rear wheel. Remember, watts is just force x velocity, where the force is in units of newtons, and velocity is in meters/sec.

If a bike is traveling 10 m/sec (22.5 mph) , and the rider is putting out 200 W, then the force to the road from the rear tire is 20 nts, or ~ 2 kg force, or ~ 4.5 lbs-force (neglecting small drivetrain losses).
the scenario we have been discussing ignores all losses except for wind resistance. While the force of the tire on the road may be different under the two scenarios, the force of the feet on the pedals would be identical if the cadence were the same. The rider can only feel the force of his feet on the pedals. I don't believe the rider could discern the difference.
 
Fday said:
The scenario we are discussing involves where all other losses are zero. In this instance the bike will "slow" at the exact same rate whether it is going 1 mph into a 49 mph headwind or 50 mph into a zero wind.
really ... hmmm. .... slow relative to what reference frame? I think the baseline kinetic energies in both cases you list suggest something quite different given a std. reference frame ... or maybe I misunderstand your point ....
 
rmur17 said:
really ... hmmm. .... slow relative to what reference frame? I think the baseline kinetic energies in both cases you list suggest something quite different given a std. reference frame ... or maybe I misunderstand your point ....
Well, relative to the bicyclist, the forces would be the same. So, the deceleration would be the same except one would end up at a zero speed in relationship to the ground and the other would end up at a speed of -49 mph. Except, as I was thinking out on my bike ride this afternoon, as I often do, I have come up with a reason why the bicycle and the airplane in this scenario are different.

The difference is wind shear. The bicyclist is in contact with the ground. Wind shear dictates that that a 50 mph wind in the face of a rider would have a speed of zero at the ground. So, the total wind resistance would be less compared to an airplane, where the wind would be the same at all points because it is not close to the ground. But, a bicyclist riding in zero wind is similar to the airplane because the wind he sees is the same at the feet as it is at the head. So, a bicyclist would be slowed less than an airplane in a headwind and helped less than an airplane with a tailwind.
 
Frank, here's the best I can do:

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For a case with perfectly efficient paddles which claw their way through the moving airstream without slipping:

2d9bvyo.jpg




As you can see, the wheel *has* to spin slower, since the moving airstream is pushing the paddles forward nearly as fast as the hub is moving. This may not seem important, but it is....

16a8pjc.jpg


The propulsion force equals the drag, but acts at the surface of the wheel, producing a counter-torque which must be overcome by the rider. The power that the rider must produce is equal to the torque x angular velocity of the wheel. Since both have equal torque, the relative power requirement is 50x greater in the first case since the wheel speed is 50x greater, as dhk2 explained. Good luck, buddy. :)
 
Fday said:
the scenario we have been discussing ignores all losses except for wind resistance. While the force of the tire on the road may be different under the two scenarios, the force of the feet on the pedals would be identical if the cadence were the same. The rider can only feel the force of his feet on the pedals. I don't believe the rider could discern the difference.
Not sure what the scenario we're discussing is, but the concept is the same. Power required = Force x Velocity. The force needed at the rear wheel to go 1 mph into a 49 mph headwind is the same as the force needed to go 50 mph in calm conditions. The basic equation F = 1/2 x rho x v^2 x cdA applies to both, where v=50 mph.

If you accept that the force at the tire must be the same, the power equation then dictates that power is 50 times greater for the bike traveling 50 mph vs the 1 mph bike. To generate 50 times the power, it follows that the force applied to the pedals must be 50 times higher for the 50 mph bike vs the 1 mph bike at the same cadence.

If this result is surprising, think about the gear ratio required at 50 mph vs 1 mph at the same cadence...it's 50 times higher, meaning the force at the rear wheel is 50 times less for a given pedal force. With the same amount of aero drag pushing you back, you've got to pedal a gear 50 times higher and push it 50 times harder.
 
frenchyge, our posts crossed, but believe you've got it :) Equal torque, but at 50 times the wheel angular velocity, means 50 times the power is being applied to go 50 mph vs 1 mph, and that can only be accomplished by pushing the pedals 50 times harder (at the same cadence).

What's different is of course the gearing. The big gears needed to go 50 mph reduce the torque applied to the crank by a huge ratio, while the tiny gears needed to maintain cadence at 1 mph multiply the crank torque to the wheel significantly.
 
frenchyge said:
The propulsion force equals the drag, but acts at the surface of the wheel, producing a counter-torque which must be overcome by the rider. The power that the rider must produce is equal to the torque x angular velocity of the wheel. Since both have equal torque, the relative power requirement is 50x greater in the first case since the wheel speed is 50x greater, as dhk2 explained. Good luck, buddy. :)
HMMMMMM. I am going to have to cogitate on this a bit. :)
 
dhk2 said:
frenchyge, our posts crossed, but believe you've got it :) Equal torque, but at 50 times the wheel angular velocity, means 50 times the power is being applied to go 50 mph vs 1 mph, and that can only be accomplished by pushing the pedals 50 times harder (at the same cadence).

What's different is of course the gearing. The big gears needed to go 50 mph reduce the torque applied to the crank by a huge ratio, while the tiny gears needed to maintain cadence at 1 mph multiply the crank torque to the wheel significantly.
What is a problem for me here is the answer has to be the same regardless of the frame of reference it is solved from. If we agree that the energy loss here is entirely from the wind resistance then the energy loss going to heat is the same for the rider going 1 mph into a 49 mph headwind and the rider going 50 mph into zero wind. Then the rider must simply put in the power lost to maintain speed. I am having trouble immediatly understanding what the speed of the massless wheel has to do with anything. But, as I said above, I will cogitate on this awhile and see if I can understand your viewpoint or find the flaw in it.
 

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