# Wind Force is Equal?

by 909kidd
 P: 262 This is not a simple question at all. Flow and turbulence can be very complicated, in addition we’re dealing with a human body hence our picture becomes very clouded. In cases such as this one I’ll try to reduce the variables as much as possible. So I imagine comparing the energy needed for my car to be parked up in a 25 kph wind and then the same car to drive 25 kph on a windless day. I’d be p*** off if I had to refill fuel just for standing still. So what’s the difference if the forces are the same? I’d imagine that if we have to ignore tire friction, engine heat etc then when driving, the fuel energy gets transferred to higher air vortices in front and behind the car. This higher turbulence will eventually end up in a higher air KE, ie heat up the surrounding air. I cannot see how mass comes into this, unless you mean the mass of air involved. Edit: I just realised I’ve left out another subtle but important factor. When I’m driving the car it’s the car which is causing the turbulence, when parked the wind provides the energy.
P: 261
 Quote by rcgldr Same impulse equals same change in momentum, so from the riders frame of reference, on the windless day the earth surface speed increased from 25 kph to 25+ΔV kph, and on the windy day from 5 kph to 5 + ΔV kph. Assuming the earth only rotates faster due to surface speed increase, the energy change is relative to 1/5 m (V12 - V02) = 1/5 m (50 x ΔV) on windless day and = 1/5 m (10 x ΔV) on windy day. So there is 5x energy added to the earth on the windless day versus windy day (if the force x time are the same).
Didn't quite catch that. How did you get that the energy change is relative to that?
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P: 7,176
 Quote by rcgldr Same impulse equals same change in momentum, so from the riders frame of reference, on the windless day the earth surface speed increased from 25 kph to 25+ΔV kph, and on the windy day from 5 kph to 5 + ΔV kph. Assuming the earth only rotates faster due to surface speed increase, the energy change is relative to 1/5 m (V12 - V02) = 1/5 m (50 x ΔV) on windless day and = 1/5 m (10 x ΔV) on windy day. So there is 5x energy added to the earth on the windless day versus windy day (if the force x time are the same).
 Quote by chingel Didn't quite catch that. How did you get that the energy change is relative to that?
Assume that all the energy change in this situation is angular, and v is surface velocity of earth relative to rider, and that earth is a solid uniform sphere.

E = 1/2 I ω2
for solid uniform sphere: I = 2/5 m r2
ω = v/r
E = 1/2 (2/5 m r2) (v/r)2 = 1/5 m v2

The increase in energy = (1/5 m) ((v + Δv)2 - v2) = (1/5 m) (v2 + 2 v Δv + Δv2 - v2) ~= 1/5 m 2 v Δv
 P: 261 OK, I follow that now, the earth seems to have more energy because the rider is going faster, but according to a different frame from the rider's frame, let's say in a frame originally at rest to the earth before the bicycle started riding, where does the energy go?
P: 4,213
 Quote by chingel let's say in a frame originally at rest to the earth before the bicycle started riding, where does the energy go?
Into the air. On a windless day he definitively adds KE to the air. By going slowly upwind he eventually even removes KE from the air. But he cannot use that energy. It is converted into turbulence.
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P: 7,176
 Quote by chingel In a frame originally at rest to the earth before the bicycle started riding, where does the energy go?
Keep in mind that the rider is generating 5 times the power into the system on the windless day versus the windy day, the force at the rear tire is the same, but the speed on the windless day is 5x that of the windy day. Also the frame of reference initially at rest with the earth, is inertial and doesn't accelerate with the surface of the earth.

So the impulse to the earth is the same, resulting in the same increase in velocity and energy. On the windless day, energy is added to the air by the rider. On the windy day, energy is extracted from the air (and transferred to the earth). Total energy of this system is conserved (assuming no losses), so after 1 hour, the total kinetic energy is increased by the decrease in rider potential energy, which is 5x more on the windless day. In this frame, the total energy change of the earth is the same on both days. On the windless day, the rider provides all of the energy added to the earth and wind. On the windy day, the rider supplies only a fraction of the energy added to the earth and the slowing the wind provides most of the energy added to the earth.
Thanks
PF Gold
P: 12,256
 Quote by A.T. Into the air. On a windless day he definitively adds KE to the air. By going slowly upwind he eventually even removes KE from the air. But he cannot use that energy. It is converted into turbulence.
To say that he removes KE is to ignore the fact that the mean velocity of the air molecules is not changed - but this is only locally for a short time.Any interaction with the air is, essentially, a loss mechanism - to my mind.
 P: 3 Power is defined as force times velocity, it states in the problem that the force is the same for both situations, so therefore power is greater when the bike is going 25 km/h than when the bike is going 5 km/h. Energy is equal to power times time when power is constant (which i think is assumed for this problem) and the problem states that they both are riding for one hour, so the bike that has the higher power, the one that's going 25 km/h, also uses more energy in the time frame. I don't understand why there's such a debate over this :/ (this is my first post :D)
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P: 7,176
 Quote by sophiecentaur To say that he removes KE is to ignore the fact that the mean velocity of the air molecules is not changed but this is only locally for a short time.
Forces don't vanish, so the forward force exerted by the rider to the air results in some net residual velocity, probably in the form of an impulse that ends up being dispersed into an ever increasing volume of air, but the magnitude and direction of that impulse will remain the same until opposed by some other force.

 Quote by sophiecentaur Any interaction with the air is, essentially, a loss mechanism - to my mind.
From the frame that is initially at rest with the surface of the earth, on the windy day, the wind is allowing the rider to generate the same force at 1/5th of the speed of the windless day, only requiring 1/5th the power from the rider while the earth still ends up with the same amount of energy on both the windless and windy days. So on the windy day, slowing down the wind provides most of the energy being transferred to the earth.

Simpler still, change the situation on the windy day so that the rider just stands in place without any movement relative to the earth's surface (zero energy generated by the rider). In this case the earth still gains energy, and all of the energy gained by the earth is due to the wind being slowed down. Momentum is conserved, so any gain in momentum of earth is offset by a loss in momentum of the wind.
 P: 261 OK thank you I follow that now.
 Mentor P: 15,202 A bit late, but just in case anyone reading some of the posts in this thread starts to wonder whether some PF mentors are complete idiots, let me reassure you. I am a complete idiot. At least in this thread.
P: 132
 Quote by D H A bit late, but just in case anyone reading some of the posts in this thread starts to wonder whether some PF mentors are complete idiots, let me reassure you. I am a complete idiot. At least in this thread.
It's reassuring to see that a PF mentor can get something wrong, but it's even more reassuring to see that a PF mentor can admit a mistake with so much candour. Thanks.
 P: 143 The scenario allegedly provided by a Ph.D. was very poorly articulated therefore, it has resulted in a myriad of answers based upon individual impressions of that which was meant by the professor’s use of the term “effort”. “Effort” is a relative term to humans, as one individual may deem a required effort as considerable whereas another may deem the same requirement somewhat insignificant. Humans possess various states of physical conditioning, some with seemingly unending stamina, others with exceptional strength, while others are the 90 pound weaklings that sway by manner of a stiff breeze. When we think of “effort”, our impression is typically that of an “overall effect” therefore it takes in one’s impression of how hard it was to pedal (the force required) as well as one’s impression of how draining it was (energy required per a distance traversed) or (energy required per pedaling for a given time duration), none of which was specified by the professor. The professor receives an ‘F’ for his poorly articulated scenario. As an ardent bicyclist and having just recently pedaled 100 miles on this past March 14, 2012 (half of which was into the wind as I headed westward), I can convey from firsthand experience that pedaling into the wind is far more draining to traverse a given distance compared to traversing the same distance via still air and conversely, a tailwind on the return trip most definitely reduces the energy requirement to traverse the same distance. Since all bicycle “rides” involve traversing some manner of distance––typically a predefined favorite route––one is not interested in how much energy is used to pedal against the wind at a lower speed rather, they are only interested in the amount of energy required to traverse the desired distance by manner of pedaling. Conversely, only those on stationary bicycle trainers are concerned with the amount of energy required while pedaling, as all the pedaling in the world leaves them right where they started. LOL
P: 132
 Quote by Gnosis When we think of “effort”, our impression is typically that of an “overall effect” therefore it takes in one’s impression of how hard it was to pedal (the force required) as well as one’s impression of how draining it was (energy required per a distance traversed) or (energy required per pedaling for a given time duration), none of which was specified by the professor. The professor receives an ‘F’ for his poorly articulated scenario.
It's true that "effort" might mean either force applied or work done (i.e. energy expended). Since the time duration was specified (one hour in both cases), it's reasonable to conclude that "effort" here means work done.

 Quote by Gnosis As an ardent bicyclist and having just recently pedaled 100 miles on this past March 14, 2012 (half of which was into the wind as I headed westward), I can convey from firsthand experience that pedaling into the wind is far more draining to traverse a given distance compared to traversing the same distance via still air and conversely, a tailwind on the return trip most definitely reduces the energy requirement to traverse the same distance.
I'm sure all will agree that going against the wind requires more power than going at the same speed with no wind: in the first case, the resistance from the air is greater. In the given question this is not the case: pedalling for an hour at 5 km/h does not require as much energy as pedalling for an hour at 25 km/h, if in both cases the air offers the same resistance.
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P: 7,176
 Quote by Gnosis I can convey from firsthand experience that pedaling into the wind is far more draining to traverse a given distance compared to traversing the same distance via still air.
Except in this case the time factor is the same and the distance traversed on the windy day is 1/5th that of the windless day.

The force from the wind could be replaced by going up hill with a tail wind going the same speed as the rider. The "windless" day would correspond to going 25km up the hill at 25 kph for one hour, while the "windy" day would correspond to going 5 km up the hill at 5 kph for one hour (probably using lower gearing).

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