I just did, and then showed you that all of the other components of the van are dispensable while preserving the dynamics of this system.
I'm not sure why you say this, but it is in fact true.
What I am saying is the the reasoning that is used to conclude "zero differential velocity" is windspeed can also be used to conclude that the cart is motionless in still air.
I see nothing to distinguish the two. You say that the belt makes the difference, I say not. The belt is in no way the equivalent of wind.
A couple of things: The first syllogism, though indeed invalid, has nothing to do with the equivalence of treadmill and cart at windspeed.
The specified case is not an example of the abstracted syllogism; It's imprecisely formulated, and still has nothing to do with the matter at hand.
For instance, the second premise is just a special case of the first, namely for windspeed equals zero.
This makes the syllogism valid and the conclusion true, though trivial, since it is just a reformulation of the one premise.
I say it relies upon both. The
only characteristic that you have used is to equate apparent velocities.
Also, "no differential wind at wind speed" has not been demonstrated for the cart. The treadmill says nothing about the cart's ability to get there.
Also, swapping symbols or objects to undermine my syllogisms, just invites me to modify them in response.
ETA:
What I do find odd, is that if it does work when still, how could moving it around add any new information?
It seems to be only symbolic. The treadmill should have handles, so it can be carried around like the Ark of the Covenant.
What we should actually be interested in, however, are the relative speeds between the air, the cart, and the surface. Do these relative speeds differ between the treadmill and a cart moving over a road at windspeed? Let's see:
Outside:
Vcart - Vair = Vwind - Vwind = 0
Vcart - Vroad = Vwind - 0 = Vwind
Vair - Vroad = Vwind - 0 = Vwind
Inside:
Vcart - Vair = 0 - 0 = 0
Vcart - Vroad = 0 - Vtreadmill = - Vtreadmill
Vair - Vroad = 0 - Vtreadmill = - Vtreadmill
This means that for any Vtreadmill = - Vwind , the two situations are identical. That is, for every wind speed, we can set a treadmill speed that results in equivalent dynamics.
Like I said before, this ignores differences in the boundary layer flows, but does not affect the working principle of the cart. Both the treadmill and the road develop a similar boundary layer flow, and the bigger the treadmill, the more alike the two will become.
Of course. Arithmetic subtraction of vectors. Perhaps you are unaware that is what is thought to be my "problem". I can't get my head around such an everyday occurrence. This sort of abstraction is done every time you model something in your mind and turn it around. There is no need to consider that point any further.
But equivalence guarantees "no difference", of course. It all boils down to solely how good your model is. No equivalence necessary. It is redundant.
Most use wind tunnels to model the wind, but you say a belt is enough.
I say not even close.
This is the problem:
It ignores the momentum exchange that accompanies real air travel.
The cart (clearly) does not pick up enough momentum to allow it to remain in the 'still' air. Even allowing for a special case, the reaction to being pushed backwards should generate a significant force. It does not.
Without that, it cannot be said to be a model of dynamic system in any useful sense. Move the treadmill 1m to the right. Attach a 1m drive shaft to the motor and use an idler wheel to turn the cart wheels. Wind now? Cover the treadmill with a cardboard box, allowing an small port for the drive shaft. In the wind now? It is no model.
This is not what we are saying. In similarly short wording I would say:
Air moves over road
Velocity is relative.
Road moving in still air is the same as air moving over road.
And the modelling consists of:
Belt moving in still air is equivalent to road moving in still air.
I will tell you no such thing. In fact, I'm trying to tell you that they are all equivalent.
You can't, that is what equivalence means. I'm still not sure what your problem is.
What provides the energy is, as sol invictus pointed out, the difference in speed between the surface and the air. In the treadmill this difference is maintained by the treadmill motor; outside, this difference is maintained by the wind.
Taking a mathematical perspective, which is all that is being done, is not the same as creating an environment viewed from that perspective.
A system of inquiry should be useful. If I sit on the missile, or on the deck of the battleship, your view says that I can't know if I am traveling or stationary in either case. Great "I can't know" . I'll take two.
Something that vague is not likely to be useful.
Equivalency, says that I cannot expect to gain any advantage by changing my viewpoint because all things are equivalent from that view. Therefore abstraction is redundant.
One thing that it does allow is to argue for a belt being a road, on the basis of equivalent velocity alone.
Is this a valid simplification of the principle of equivalence? The lack of torque in the cart indicates not. That is an outright failure of the model.
Adequate power cannot come from the motor, because there no return path to couple it to the propeller. There must be a suitable path to carry the force.
There is no
momentum carrier, that is the equivalent of the wind.
The cart runs on "leakage momentum" to ground, which it why is is so feeble. This is not a model of a cart in wind, of any sort
.
