Tailwind and speed difference

rmur17

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

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Fday

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.

patrick_

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

Fday

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

frenchyge

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.

Fday

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.

frenchyge

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?

Fday

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.

patrick_

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.

frenchyge

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?

dhk2

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.

dhk2

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.

Piotr

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.

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frenchyge

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?