I dont think the weigh descends in his hypothetical, it stays at 3 oclock which takes more and more power to maintain at higher wheel speeds.

I just thought that a hamster in a wheel is a good analogy. If the wheel has .01 Nm of friction torque for example, he is going to be burning a ton more power to keep the wheel at 100 RPM vs 10 RPM

JYL, you brilliantly solved a stunt man physics question for me a few years back. Here: http://forums.pelicanparts.com/off-topic-discussions/574805-stuntman-physics-question.html

I'm a dumbass but I "think" what you're asking is akin to why can an Olympic diver do flips and such once he leaves the high board, he has no external forces applied to allow him to do those flips. The answer is he is changing his arc and thus gravity allows for rotation.

If you vaguely remember Newton's laws, ever learned those thermowhatchacallit laws and have read the wiki page on Einstein, you know enough to tell him his theory doesn't hold water, or hamsters. No deep thinking needed, so save those brain cells for your beer tonight.

JR

You guys are helping me with the intuition.

Here's where my hangup has been. Suppose I have a spaceship of mass M with a rocket drive that produces a constant force F. It is far from any planet or sun, so there is no friction opposing its movement. Force F accelerates the spaceship to velocity V at time T1, then a while later it has accelerated to velocity 10*V at time T2. At time T1 the power of the drive is F x V. At time T2 the power is F*10*V. I was/am having trouble understanding intuitively why the constant force is producing more and more power.

Mathematically I think I get it. At time T the ship has kinetic energy KE = 0.5*M*V^2. After a small increment of time dT, at time T+dT it has KE = 0.5*M*(V+F/M*dT)^2. The change in kinetic energy is 0.5*M*[2*V*F/M*dT + (F/M*dT)^2]. The average power during dT is change in kinetic energy divided by dT, or 0.5*M*2*V*F/M*dT/dT + 0.5*M*(F/M)^2*dT). dT/dT = 1 in the first term, and as dT goes to zero, the second term goes to zero, so the instantaneous power P = 0.5*M*2*V*F/M = F*V.

Right?. That's question 1.

(This illustrates, at an elementary level, why I abandoned my physics major and switched to a math major, back in school. I could do the math but lacked intuition for the physics. What good is a physicist who can't intuitively grasp power?. But that was many decades ago and now I don't grasp any of it.)

So here is my further question 2. I assumed that, to hold the 2 kg weight at 3 oclock, the mechanism in the rim has to deliver a constant force of 20 N to the weight, thus a constant reaction force of 20 N is being applied to the wheel. That's why I thought of this device as equivalent to a pulley with a 2 kg hanging weight. Is that correct?. Or does the weight need increasing force to hold it at 3 oclock, as the wheel rotational speed increases?

Finally, question 3. It seems this would be a cheating bicycle motor (yup, that's the intended hypothetical purpose) that would be of very little help at low speeds (when it is delivering very little power), would have very little ability to help the rider make a sharp acceleration at any speed (because its maximum force is quite small), but would be very helpful if the rider is trying to maintain a high constant speed (when it is delivering high power). Is that correct?

Oh, as a practical matter I think the concept is a non starter because (1) there has to be some friction of the weight inside the channel and (2) eventually the battery will run out so nothing will be holding the weight at 3 oclock. Then there will be a 2 kg weight rotating with the wheel. I haven't calculated the centripetal force of a 2 kg weight rotating at radius 35 cm and 300 rpm (40 km/h).

Okay, it is F = M*R*(2*pi/T)^2 where T is the time required for 1 rotation. At 300 rpm, T = 0.2 s. So F = 2*0.35*(2*3.14/0.2)^2 = 690 N. I would think the bike would ride badly, if it were even rideable. With every wheel rotation, he would experience near weightlessness. Not good for traction. At higher speeds, he would be lifted off the ground, or at least his rear wheel would be.

The idea is a battery in the wheel that powers this mechanism. What the mechanism is, no idea. Probably something like a solenoid. But it could be a hamster on EPO.

maybe if you see another example, which is even more counter

intuitive because the force is going down while the power is going up.

here are the Torque and HP curves for a popular 911 motor

torque is the angular analog of force applied in your example and power of course is the analog of power in your example and rpm is the analog of linear velocity in your example

Note that the torque(force) is decreasing from ~5k rpm yet the power is increasing from that point to ~6200rpm. Why? because the hp is the product in the mathmatical sense of the decreasing force(torque) and increasing rate(rpm)
http://forums.pelicanparts.com/uploa...1455666865.jpg

BTW there are many concepts in Physics that go against intuition, among the most bizarre at the macro scale is exhibited by a simple toy, a spinning top.

That's not a Carrera, 210 HP is Honda minivan type numbers :-)

Where'd you get those extra three horsepower? Mine only has 207!!!

Watched this on TV tonight- reminded me of this thread:
Fast forward to about 4:00 in:

<iframe width="420" height="315" src="https://www.youtube.com/embed/vU2FDhXJe2U" frameborder="0" allowfullscreen></iframe>

Again, this doesn't really seem like a power problem. Maybe a torque or kinetic energy problem but not really a power problem. Again though, I really don't understand the concept. But never forget that there is no free lunch. Energy output will never equal energy input in a system because there are always losses due to efficiency. No system is 100% efficient, except the gerarator.;)

The reason I'm talking power is because in cycling, that's the common language. It takes pedaling at 250 watts to the crank, to ride 20 mph on flat ground in calm air. So when I saw the claim that this drive will produce 200 watts at 20 mph, that was interesting - if true, the drive could propel a bike at 20 mph with the rider barely exerting himself.

Plus, as established here, I don't understand power.

Re question 1: I believe the amount of rocket fuel required to maintain constant thrust will increase linearly with the rockets velocity.

The thrust is proportional to the mass flow rate of the exhaust times the change in momentum. It becomes more difficult to change the moment (out the back of rocket) when it already has so much forward velocity

Said more simply is in general a prolonged rocket burn is probably not a constant force event

Hey, I noticed something else - tell me if this sounds correct.

If this drive produces constant force (20 N) all the time that it is turned "ON", then it produces constant acceleration all the time. Whether you want it or not. When you are coasting fast down a steep incline, when you are holding position amidst other riders in the peloton, when you are trying to slow down for a turn - you don't want acceleration, but you will get it anyway. It is like a car with a stuck throttle. The only way to stop the constant 20 N is to turn the drive off. But when the drive is turned off, the wheel is very imbalanced as discussed above.

Doesn't seem that far fetched to me as 2kg is not a tiny mass and 10m/s is a decent clip.

I don't think you could fit your device in a road bike wheel easily though. Also it could produce peak torque more than the gravity limit, because the device could accelerate the mass backwards in theory to produce an inertial torque on the wheel. This would not be practical for very long though.

No matter how your drive is configured it needs to have a very "rigid" feedback control loop, it could theoretically have a cost mode where it rides at the bottom or a brake mode where it rides on the other side of the wheel

Re q2: this is the same hold up you have originally. The force is constant but it is applied at increasing speeds so requires more power to maintain as speed increases.

The 20N force is approximately equal to what a 2kg mass would exert on a string if static, however I think thinking about the static condition may be tripping you up

^^^ time to draw those force diagrams!!!

Thanks aaschen. I think I'm getting it. A little.

My conclusion is that, in theory, this motor would work.

In practice . . .

You seem to ignore the concept that power is a derived quantity, it is the mathematical product of force and velocity. If either of those change, power changes as a consequence.

when pedaling an input force is applied to the crank, the force is changed by the mechanical properties of the drivetrain to a different force as a thrust at the rear wheel. The other input variable is the cadence(speed at which the force is applied) this is also altered by the mechanical properties of the drivetrain. Opposing those 2 input variables is friction and aero which also have a velocity dependence. If the mathematical product of the inputs exceeds opposing mathematical product there is acceleration, if the opposing variables exceed the input there is deceleration. There is constant speed when the input and opposing power are equal.

1 N at the pedal 2 1 rpm, is much less power than 1N at the pedal at 100rpm. There will be 0 friction and aero resistance at 0 mph, both will rise w/ speed