schroder said:
I was saying you could disregard the tread velocity altogether, and use the floor reference only. But you are right in that the cart must be doing some work at the drive wheels to generate the torque. But this cannot be the amount of work to cancel out the motion of the tread altogether.
Stop. An amount of work doesn't cancel a motion. You constantly mix kinematical and dynamical considerations, and as I said many times before, they are distinct.
Kinematics = description of MOTION
Dynamics = description of FORCES
The link between both is through Newton's equation, which links forces to acceleration. From that, one can also derive other techniques, such as energetic considerations. But any state of motion (position/velocity) is compatible a priori with any dynamical state (set of forces).
If it were, that by definition would be at least a unity device and the prop pushes it over unity and we KNOW that cannot be happening.
You go way too fast here. There's no link between the premise and the conclusion. You are jumping to conclusions. Don't. Go 1 step at a time.
Let’s look at the grab bag numbers I picked before: tread 10 m/sec and cart 2 m/sec with respect to the floor. OK I was saying the cart does NO work on the tread, so we only need to consider the velocity wrt the floor so we have a 2 m/sec cart. That assumption disregards any work being done by the wheels to provide the Force to the propeller and so it is not correct.
Right.
You were basically saying the cart is working against the full 10 m/sec and so we can use the tread as the reference and we have a 12 m/sec cart. That is impossibly high, as it is 120% efficiency and it can’t be happening so your assumption is not correct.
Again, you go too fast. You always want to jump to your conclusion, without making a step-by-step argument.
The wheels are exerting a force on the treadmill, and the treadmill is exerting a force on the wheels. As per action = reaction, both forces are equal and opposite. This holds in any (inertial) frame, as forces are invariant under change of inertial frame. So whatever is the force that the wheels of the cart exert on the treadmill, that is that force, and that is independent of what VELOCITY the treadmill has (in no matter what frame). There is a force acting upon the treadmill. And that same (opposite) force is acting upon the wheels. In any frame.
Can you agree with that ? It is elementary mechanics.
I obviously need to take into consideration the work being done by the wheels, convert that to a velocity and add that to the velocity measured wrt the floor and will get a higher number than what I measured.
Mechanical power done on something = the velocity of that something x the force done on it, right ?
So this is dependent on the frame in which you express the velocity. So work is a quantity which is frame-dependent. We know that there is conservation of energy, but in order to do so, we need to express all work done in one and the same frame. No matter what frame. But one and the same for all contributions.
Now, if the cart is exercising, say, 12 Newton on the treadmill, then the treadmill is exercising 12 Newton (in the opposite direction) on the (wheels of) the cart.
Now, in the ground frame, the direction of these 12 Newton is opposite to its velocity (which is 10 m/s), so the treadmill is not gaining, but delivering 12 x 10 = 120 watt (our values are not realistic, but no matter). But again, this power is frame dependent: we did it in the ground frame. We'll stick to it.
Let us look at the cart. We already know that there is a force of 12 Newtons on it dragging it in the sense of the treadmill. If, by assumption, the cart is in steady state, that means that no (horizontal) force is acting net on it, so this means that the only other thing acting upon it, the air, must be pulling 12 Newton in the other direction (in the direction of the motion of the cart).
The work done by the air, however, is 0 (in the ground frame), because the air is not moving, and work = force x displacement.
So no power delivered by the air. 120 Watt delivered by the treadmill. No gain in kinetic energy anywhere (actually, there is, momentaneously, in the air stream behind the propeller, but this will eventually be dissipated).
So net power balance of the system: 120 Watt IN. This is pure dissipation.
Let us do the same thing now from the frame of the cart. In the frame of the cart, the air is coming in at 2 m/s. The treadmill is coming in at a velocity of 12 m/s.
The forces are the same (don't change under a change of reference frame, as long as they are inertial).
So in this frame, the energetic balance is:
on the air, there is a force of 12 N in the sense of the treadmill (because this is the reaction on the fact that the air is pulling on the cart with that force). The air is moving in that same direction at 2 m/s.
So in this frame, there is work done on the air, the air is receiving 2 x 12 = 24 Watt.
On the other hand, in this frame, the treadmill is running at 12 m/s. We already knew that it was undergoing a force in the opposite direction of 12 N. So the treadmill is delivering 144 W in this frame.
Net balance: the treadmill is delivering 144 W, the air is taking 24 Watt. Net goes into the system: 120 Watt. Again, dissipation.
Finally, let us look upon it from the frame of the treadmill. In this frame, the treadmill is of course at rest. The air is coming in at 10 m/s, but in the sense of the cart. The cart is moving at 12 m/s.
Here, the force on the air is 12 N, but in the opposite direction of the motion of the air. So the air is DELIVERING 12 x 10 = 120 W of power. As the treadmill is standing still, the 12 N acting upon it don't do any work. Net balance, again: 120 W in.
It is interesting to return to the frame of the cart, and look upon the energy balance inside the cart.
The wheel is turning and at its outer side which turns at 12 m/s, a force of 12 N is acting, which gives a torque on the axle. It can easily be established that this wheel is receiving 144 W. This power is transmitted through the gearing system to the air, where only 24 W is delivered to do mechanical work on the air which establishes the force of 12 N. All the rest (120 W) is dissipated. It is mainly dissipated by accelerating the air until it settles down (through turbulence), and also in the gears and wheels. The large dissipation comes from my arbitrary choice of 12 N of course.
So we are still a long way away from a potential over-unity device. We dissipate a lot of power here.
