Down wind faster than the wind

[Dan O.]

Humber thinks he undersrands circuits but doesn't understand frame of reference.

Here is a problem for humber in his own field: If the voltage on one side of a resistor is 10V and there is a current of 0.5A, what is the resistance?

Humber cannot solve this simple problem but will he try or will he just spin his wheels?

 

[CORed]

It's amazing. You are still asking questions about the concept of frames of reference. I say that it is not possible to reconstruct a frame of reference.
But I can let that go, and still show the idea to be wrong.

You can model the conditions of the frame you are trying to reconstruct. The treadmill is a complete failure in this respect. You don't seem to appreciate how illogical your ideas are, so you concentrate on details, as IF the treadmill were a frame of reference. Its a model, and a bad one.
Please do as I suggested. Walk from the floor onto the treadmill from the usually raised end. Your velocity wrt to the ground increases. However, as you look back, you see that the cart is still where it was, and you are in no doubt that is it stationary wrt the ground. See? It's not moving. It would be almost indistinguishable from one on the floor, right next to it.
Why does it do that? Because it essentially ignores the belt. The belt slips under it, because the friction to it is so low.

Wow! Actually, the cart remains stationary, moves "upbelt" or moves "downbelt" slower than the belt (depending on a variety of conditions) because the propeller, driven by the wheels, driven by the belt, creates thrust against the still air, in the opposite direction of the motion of the belt.
There is no automatic "force balance".

Connect it to the belt as you are when moving, and it will go backwards.

Just as, if the cart is traveling downwind on a road, and you connect it to the road, it will stop.

But that is not important, it just shows up another anomaly. It is not a frame of reference, so that really doesn't matter.

Wrong again. Two frames of reference, both valid.

So what does it say? It says that this cart, responds to being driven into a headwind (like you) by driving itself into a minimum energy state.

Actually, unless considerable effort is made to get the belt at just the right incline, the cart will either drive itself off the end against the motion of the belt, or be pushed off the other end. Of course, in the Humberverse, the cart magically balances the forces and remains stationary.

An orange stays on the belt for another reason.

I've not personally observed this phenomenon, but the most likely mechanism for it is that the weight of the orange causes the belt to sag, creating a low spot, so that the orange is continuously rolling downhill, opposing the motion of the belt. Either that, or the whole belt is moving uphill. However, I think for the orange to stay in one place for more than a few seconds, the weight-induced low spot is the most likely explanation.

My cart too. (There are many ways of doing that. Can't you think of any? No?)

Your imaginary cart? Yes there are many ways you could make it stay in one place. The mechanisms explained for the orange would probably work: Take a simple cart with low-resistance wheels, put enough weight on it to create a sag in the belt. Or put a slight incline in the belt (note that in the propeller cart tests, the incline is in the opposite direction, so this is not how they work). Heck, just build one. You might learn something (then again, if the cart doesn't work the way you expect it to, you will probably decide the cart is wrong).

In wind, if you wear a pair of skates, and launch a parachute so that it carries you down wind. Eventually, you will reach terminal velocity. Strain gauges to the parachute and in the skates, will show forces to be opposite, but at a maximum.That is the real world case for down wind travel.

Amazing! Very simple, and you still managed to get it wrong. Here is how it works in the real world (not the Humberverse). Note that for this analysis, I am assuming travel directly downwind, such that the parachute is acting as a simple drag device and is not generating lift like a sail on a tack, or a rotating propeller. The force on the parachute will be at a maximum when the stationary skater first deploys the parachute and will accelerate the skater. As the speed of the parachute and skater approach the wind speed, the force on the parachute will decrease. When the force exerted by the wind on the parachute matches the friction of the skates on the ice (very low friction, so this will be slightly less than wind speed), the skater will stop accelerating, and the force on the parachute will be at a minimum. The frictional forces on the skates increase with the skater's speed. This is not however, the primary cause of the skater reaching terminal velocity. The primary cause is the inability of the parachute to generate drag force when it is moving at the same speed as the wind.

On the treadmill, the opposite will happen. When you reach 'terminal velocity', you will be stationary as the cart is now. The gauges will show a minimum force. Opposite to the real world, except for the case where you are NOT traveling. That is the correct view of the cart or 'frame', if you must.

Note that with correct analysis of the "skater going downwind" the "real world" and "treadmill" cases are identical. So that we are not comparing apples to oranges, we will assume the treadmill is long enough to allow the parachute and skater become (nearly) stationary and substitute rollerskates (or maybe lollerskates) for ice skates. Ice doesn't bend too well, so it's kind of hard to make a treadmill out of it. Here is what will happen(from the "ground" frame of reference): The skater moving with the treadmill belt deploys the parachute. The parachute, catching the stationary air, slows down the skater The force will be at its maximum when the chute is first deployed and will decrease as the skater's speed decreases. As the skater slows down to a speed where the drag from the parachute matches the rolling resistance of his skates, which will be almost stationary, but moving slowly in the direction of motion of the belt the force on the parachute will reach a minimum, just like in the "real world'. Note that the rolling resistance of the skates will be at its maximum when the skater is at "terminal velocity" (almost stationary relative to the ground). This is not, however, the primary cause for the skater failing to come to a complete stop (relative to the ground). The primary cause is the inability of the parachute to generate any drag when it is not moving.

QED. The frames of reference are equivalent. Yet another epic fail by Humber.

Imagine a small battery powered wind cart, where a motor drives a propellor. What do you think that will do on the treadmill?

The cart it will move relative to the air in the direction the propeller drives it, with almost no effect (assuming good wheels) from the treadmill. What will it do in the Humberverse?

 

[jjcote]

Hang on, I'm about to disagree with someboby other than humber!

When the force exerted by the wind on the parachute matches the friction of the skates on the ice (very low friction, so this will be slightly less than wind speed), the skater will stop accelerating, and the force on the parachute will be at a minimum. The frictional forces on the skates increase with the skater's speed. This is not however, the primary cause of the skater reaching terminal velocity. The primary cause is the inability of the parachute to generate drag force when it is moving at the same speed as the wind.

Careful! There's an important error here that is actually critical to the operation of the cart. The reason why the force is small at this point isn't because the parachute can't work well at wind speed, it's because the drag of the skates is low. Imagine a pretty big chute (and let's make it more of an umbrella, so that we can eliminate the detail of it collapsing). It gets you up to almost wind speed, and the force is low. But let the bearings in your skates dry out, and it will keep pulling you along at (almost) the same speed, even if it takes a lot more force to do so. To think about it more clearly, don't even worry about what's on the end of the rope for a moment -- parachute, umbrella, whatever. It's just something that wants to travel along at a fixed speed, and it will deliver more force (and do more work per unit time) if that's what's required. Not so obvious? Look up and notice that I've replace the umbrella with a locomotive. If the umbrella is large enough compared to you, it will behave just like a locomotive. The amount of power available even at wind speed is large.

 

[jjcote]

If the belt of the treadmill is not moving, then the wind speed is zero. Wind speed for the cart on the treadmill is the speed of the air with respect to the surface of the belt, just as on the ground it is the speed of the air with respect to the surface of the road.

I don't know. It seems from the video that the profile above the belt is OK, but going the wrong way. It's a headwind. This seems to fall almost to zero at prop height, but then at that point, there should be a tailwind.

Whoa, whoa, whoa, WHAT?! There may be a far more basic lack of understanding going on that I had realized. You think that there's a headwind? What in the world is indicating a headwind to you? I have a notion that you have a total misconception of what the treadmill is being used for.

Are you under the impression that the treadmill is being used in an attempt to move the air in the room around? That it's some kind of inefficient linear fan or something? That isn't what's going on. The air in the room is assumed to be totally still (and the streamer is used to test that). For the purposes of the test, the only things that matter are the air and the surface of the belt, because those are the only things that the cart interacts with once the hand lets go of it. From the perspective of the camera, an object sitting on the belt moves to the left, and thus feels a wind from the left (a tailwind). The cart is moving forward on the belt (toward the right of the picture) as fast as that tailwind, so it's feeling no wind (just as an object on the road moving at windspeed feels no wind). Do you have some different misconception of what's going on here?

 

[spork]

Whoa, whoa, whoa, WHAT?! There may be a far more basic lack of understanding going on that I had realized. You think that there's a headwind? What in the world is indicating a headwind to you? I have a notion that you have a total misconception of what the treadmill is being used for.

Yo - be patient there JJ. humber is going to prove to us exactly why the treadmill has no relevance here. Just give him a chance. He's only been at it for 70 pages.

 

[jjcote]

If I put a treadmill on the floor and walk to it and then step on it, so as to face down the belt, I will feel a wind against my face. However, I feel the same wind when walking to the treadmill, but I just don't notice it. Effectively, I am just walking faster when on the belt.
An accelerometer will tell me that it is the belt that is moving and not the ground. I find that all windspeed changes correlate with a change in acceleration. This will confirm my that I am on a belt. Knowing that, all the anomalies are explained, and inform me of the true nature of my 'frame'

Are you aware that when you step onto a moving belt, you are no longer in the same inertial frame that you started in? The reference is not allowed to accelerate.

The accelerometer can't tell you what frame is "stationary". If you step from the ground onto a moving belt it will give you a result, but if you step from a belt moving the other way onto the ground it will give the same result. If there are a bunch of treadmills (running a different speeds and directions) lined up in the gym and you keep stepping from belt to belt, working your way across the room, you can't tell from the accelerometer when you have reached a belt that's turned off unless you know what the belt you started on was doing.

 

[jjcote]

OK slight change of plan. I will do that. I will show you how to do that very thing. You can make it yourself, if you like, but that should not be necessary, unless you have the mindset of Creationist.
Let this be clear. I cannot make anything, and I do not have to. I am not going to. Do I have to repeat that?

I'll be interested to read your description. Of course, if I build it and it fails to perform as you predict, you'll have to be prepared to explain why.

(Unless your "steel bar connected to the ground" is your proposal. The only things that the experimental apparatus is allowed to interact with are the belt and the air.)

 

[CORed]

Hang on, I'm about to disagree with someboby other than humber!



Careful! There's an important error here that is actually critical to the operation of the cart. The reason why the force is small at this point isn't because the parachute can't work well at wind speed, it's because the drag of the skates is low. Imagine a pretty big chute (and let's make it more of an umbrella, so that we can eliminate the detail of it collapsing). It gets you up to almost wind speed, and the force is low. But let the bearings in your skates dry out, and it will keep pulling you along at (almost) the same speed, even if it takes a lot more force to do so. To think about it more clearly, don't even worry about what's on the end of the rope for a moment -- parachute, umbrella, whatever. It's just something that wants to travel along at a fixed speed, and it will deliver more force (and do more work per unit time) if that's what's required. Not so obvious? Look up and notice that I've replace the umbrella with a locomotive. If the umbrella is large enough compared to you, it will behave just like a locomotive. The amount of power available even at wind speed is large.

Okay, a good point. Given a sufficiently large "sail", you can generate a huge amount of downwind force at a speed very close to wind speed. However, as I understand it, as the speed approaches the wind speed, the force is going to be reduced until it matches the resistance. This only applies to something that is moving directly downwind. In the case of a sail, airfoil or propeller blade moving at an angle to the wind, angled so that it is generating lift with a downwind component, net motion downwind faster than the wind is possible.

However, for the directly downwind, simple drag device, be it sail, parachute of umbrella, I believe that my point that the downwind force is reduced as the device approaches wind speed is correct.

 

[mender]

Careful! There's an important error here that is actually critical to the operation of the cart. The reason why the force is small at this point isn't because the parachute can't work well at wind speed, it's because the drag of the skates is low. Imagine a pretty big chute (and let's make it more of an umbrella, so that we can eliminate the detail of it collapsing). It gets you up to almost wind speed, and the force is low. But let the bearings in your skates dry out, and it will keep pulling you along at (almost) the same speed, even if it takes a lot more force to do so. To think about it more clearly, don't even worry about what's on the end of the rope for a moment -- parachute, umbrella, whatever. It's just something that wants to travel along at a fixed speed, and it will deliver more force (and do more work per unit time) if that's what's required. Not so obvious? Look up and notice that I've replace the umbrella with a locomotive. If the umbrella is large enough compared to you, it will behave just like a locomotive. The amount of power available even at wind speed is large.

I'm going to disagree with you on this, JJ. If the parachute is at windspeed, it can't supply any force. In order to produce a force, the parachute has to change the momentum of the air around it, and it can't do that when it is at the same speed as the air. Also, the higher the drag from the skates, the more force is required to overcome that, and the more the momentum of the affected air has to be changed to produce that force. That will result in a greater difference between the parachute speed and the air speed. A bigger parachute will increase the amount of force available and will get the system closer to the actual windspeed than a smaller parachute - but it will never quite be at rest with respect to the air.

That is a crucial point for anything powered by the wind. The amount of force available is calculated from the speed difference between the air and whatever the wind-powered device is anchored to. If the device happens to be moving but is linked to the ground, the force available is still calculated according to the difference in speed between the air and the ground. As the difference in speed between the device and the air gets larger, more energy can be harnessed from that speed difference. If that amount of energy is higher than the drag of the device when it is at the same speed as the wind, the device will accelerate to a higher speed as long as it remains linked to the air and the ground in order to harness that force. When the forces balance, the speed will be constant.

Surprisingly, humber got the treadmill example correct and the real world example wrong, as you pointed out, JJ. I'm curious to see how he's going to wiggle out of this one. Likely something about the fact that we don't understand that minimum and maximum values of something are relative and since the treadmill and the real world aren't equivalent, he can show that either one can be used to illustrate his point.

ETA: JJ, I see CORed beat me to it.

 

[sol invictus]

I'm going to disagree with you on this, JJ.

Everything CORed wrote is correct as far as I can see as well. But I think jj's point is that even though the force is at a minimum (in the process CORed described), that doesn't necessarily mean it's small. If you grab a giant umbrella on a windy it will drag you downwind almost as fast as the wind no matter how much you dig your heels in, and digging your heels in a little more will just increase the force you feel without slowing you down (wrt the ground) much at all.

 

[sol invictus]

That contradicts the treadmill. Which is why you so strenuously avoided saying it.

I have no idea what you even mean by "contradicts the treadmill", and neither does anyone else. Personally I don't spend a lot of time arguing with non-sentient objects (except on this forum).

Yes, transfer will not be 100% perhaps, but quite clearly, more. The masses are known, so we can build the buffer so the error is negligible. Same problem for the belt, though. So?

There.is.no.such.thing.as.100%. Any amount of energy can be released, from zero to infinity. The energy released depends on the relative velocity, which means if you change the velocity of the "buffer" while leaving the object alone, you change the amount of energy released. That's why your definition is nonsense - KE is not an intrinsic property, it's a frame-dependent quantity (or if one tried to use your definition, buffer-dependent).

 

[Subduction Zone]

Hello all. I have been reading with delight humberisms in this thread but it seems like everyone missed this one:

No good. Two objects side by side. I measure all possible sources of velocity and KE from here to the centre of the Universe. This is then my datum.

He did not keep this up but it seems that even using the Earth as a frame of reference was not good enough for humber. Well if he does not get the Nobel Prize for his brilliant arguing here he will definitely get it when he finds the center of the universe.

On a sad but serious note his arrogance, thickheadedness, and misuse of terms that he thinks he understands reminds me of a girl I knew who had a meth addiction. When she was under the influence she had delusions of grandeur and an attitude that matches humber. She would also try to use terms that she did not understand and if you listened to her she almost seemed to make sense sometimes.

A quick note about me I was on the anti side on the Mythbusters' forum until spork and JB made their first video. Then I was one of the first to congratulate them and own up to my mistake. Unlike humber I recognize a valid frame of reference when I see one.

 

[Subduction Zone]

I have also been waiting all day for humber's proof that the treadmill is not a valid frame of reference. It seemed he was going to do some sor of demonstration as shown by this statement to spork:

It will indeed be fatal. It's really simple, but I have no scanner to hand. By tomorrow for sure.

Oh well, just another unfulfilled promise from humber.

 

[davefoc]

I'm going to disagree with you on this, JJ. If the parachute is at windspeed, it can't supply any force. In order to produce a force, the parachute has to change the momentum of the air around it, and it can't do that when it is at the same speed as the air. Also, the higher the drag from the skates, the more force is required to overcome that, and the more the momentum of the affected air has to be changed to produce that force. That will result in a greater difference between the parachute speed and the air speed. A bigger parachute will increase the amount of force available and will get the system closer to the actual windspeed than a smaller parachute - but it will never quite be at rest with respect to the air.

That is a crucial point for anything powered by the wind. The amount of force available is calculated from the speed difference between the air and whatever the wind-powered device is anchored to. If the device happens to be moving but is linked to the ground, the force available is still calculated according to the difference in speed between the air and the ground. As the difference in speed between the device and the air gets larger, more energy can be harnessed from that speed difference. If that amount of energy is higher than the drag of the device when it is at the same speed as the wind, the device will accelerate to a higher speed as long as it remains in linked to the air and the ground. When the forces balance, the speed will be constant.

ETA: I see CORed beat me to it.

It seems like some people here understand what is going on better than I do. I have been trying to work it out for a few days without success. So I'd like to talk about it a bit.

First. I believe the source of power for the device is derived from the rotation of the axle the wheels are on. I was happy with that as the total explanation for awhile, but then I got to thinking about that.

If a device derives energy from the rotation of the axle, the axle will be slowed. And thus the cart will be slowed per force to less than the wind speed if the power is just dissipated away in a way other than making the cart go faster.

But if that power is used to drive the wheels faster the best that can be hoped for is that speed that was lost by the addition of the power extraction device can be regained if everything works at 100% efficiency and without friction losses.

The obviousness of this idea is, I think, what makes the DWFTTW cart seem at first implausible.

But clearly the propeller makes the situation much more complex. The propeller adds drag so while the cart is going slower than the wind the propeller increases the amount of power available to the generator attached to the axle. This seems like what mender was alluding to above.

Cool, but so far all we've got is the idea that a bigger spinnaker can make the cart go downwind faster but not faster than the wind.

But the propeller is complicated because it is also providing thrust but if that was all that was going on we still couldn't go faster than the wind because the drag that was helping overcome the power sucked up by the generator attached to the wheels can't produce any power when the cart gets to wind speed to provide power for the thrust device.

And the thrust produced by the propeller can not produce more power than can be extracted by the generator (see thought experiment below). So the result of the power balance where the thrust produced is constrained by the amount of power extracted from the rotation of the axle is that the cart can't go faster than the wind.

So the total answer seems to involve understanding the complex behavior of the propeller with respect to all this. I think the answer is related to what I thought the solution was in the first place. The propeller changes the air flow in front of itself enough that the apparent down wind speed the cart sees is somewhat faster than the actual down wind speed.

There are confounding factors that make it difficult for a person with my level of understanding to predict what would happen right around the transition to greater that down wind speed. As the cart speeds up there is more power available from the generator. With a low friction, lightweight cart a speed very close to the down wind speed will be obtained even if a small load is added to the axle. So when the cart is moving very close to the down wind speed there's going to be a significant power source available. If that power can be used for thrust, the cart will go faster but as the cart exceeds wind speed it seems like that power won't be available, unless there is some magic in the propeller (and I think there is).

I'm sure that people have posted very good descriptions of what is going on here that address my confusion. The problem is that I couldn't easily easily find those descriptions buried in the 60 odd pages of this thread. Any references to the previous posts will be appreciated.

Thought experiment:
If the power obtained from the rotation of the axle is used to power a mass chucker (100% efficient ion cannon, rail gun, etc.) instead of the propeller would the cart exceed the down wind speed?

Even assuming 100% efficiency, I suspect that the answer is no.

As an aside, I've been very impressed with the videos and demos of spork and others, but if they'd really like to impress me I'd like to see the ion cannon experiment done next. Thank you in advance.

 

[humber]

Seriously Ross, you can't be saying there's anything seemingly intelligent about this troll - can you!? The guy knows a few words, but has yet to put any of them together into a remotely meaningful sentence, and he has the logic and attention span of a rutabaga.

The lack of any supporting argument from in this debate shows that you cannot think on your feet, but when given time, it is best that you stay silent.

 

[humber]

More great vanishing tricks from the humber:

Even a flywheel look like magic to the less advanced.

Humber - why? What in god's name is the possible connection? Extreme example? You said that the cart on the treadmill does what it does (i.e. goes forward) because it gets rid of as much KE as it takes from the tread, and that that is all the prop was doing, just balancing that energy input by getting rid of it. You then said that a flywheel would do the same thing, and later repeated that it was no more than a flywheel on wheels.

NO that is an analogy to show that the translation from windspeed to zero ( as seen on the treadmill) is mathematically a reflection, and so unreal ( like everything is the wrong way around, to save you a wikki)
That last point indicates that 'cart' is no only a singular noun for you, but unique.

If it were fastened solid to a bar, it can't move forwards or backwards, can it, and no comparison can be made between the action of the prop and that of a flywheel.

Strange, Modified tried, (while brave Sir Spork ran away).
He says that the bar is equivalent to an enormous mass traveling at windspeed. This suggests to me that no connection can be forged between being stationary on the belt and windspeed, unless there can be enormous flying masses at the ready to paste over the holes in the hypothesis.

You would have a point if you properly joined the dots. There are no real framer effects in the treadmill, for or against, just coincidences. This is hardly surprising that they should existy given tht the laws of nature are consistent, but you need to know which are applicable to a given situation,
I need to connect them to the treadmill to develop the argument, so that you may understand. There are other reading, and they may be better able to follow it than you can.

 

[humber]

Humber thinks he undersrands circuits but doesn't understand frame of reference.

Here is a problem for humber in his own field: If the voltage on one side of a resistor is 10V and there is a current of 0.5A, what is the resistance?

Humber cannot solve this simple problem but will he try or will he just spin his wheels?

Several errors
1: Not my really field
2: Voltage not on one side. Presume,'across'
3. Shocking laxity. DC? AC? RMS? Linear resistor or otherwise?

It's 20 ohms, ( for you) but then, for example, if it's very small, it may not work as you expect. You need to be more specific. A Bit like defining all complex dynamics problems as free bodies in space.


Sometimes need more than simple formula to understand.

 

[sol invictus]

2: Voltage not on one side. Presume,'across'

That was the whole point, humber. As usual, it's gone whizzing far over your head.

Absolute voltages are meaningless. Only voltage differences mean anything, just like velocities.

 

[humber]

Wow! Actually, the cart remains stationary, moves "upbelt" or moves "downbelt" slower than the belt (depending on a variety of conditions) because the propeller, driven by the wheels, driven by the belt, creates thrust against the still air, in the opposite direction of the motion of the belt.
There is no automatic "force balance".

Flat out wrong. Demonstrate otherwise.

Just as, if the cart is traveling downwind on a road, and you connect it to the road, it will stop.
Wrong again. Two frames of reference, both valid.

No contradictions may exit between frames.

Actually, unless considerable effort is made to get the belt at just the right incline, the cart will either drive itself off the end against the motion of the belt, or be pushed off the other end. Of course, in the Humberverse, the cart magically balances the forces and remains stationary.

Balance seem to be heap big magic to you. Try thinking, and not following.

I've not personally observed this phenomenon, but the most likely mechanism for it is that the weight of the orange causes the belt to sag, creating a low spot, so that the orange is continuously rolling downhill, opposing the motion of the belt. Either that, or the whole belt is moving uphill. However, I think for the orange to stay in one place for more than a few seconds, the weight-induced low spot is the most likely explanation.

You think know a lot about what you have not seen. I observed it out of interest. I worked in an orchard after school. I have curiosity.
For cleanliness, the floor of the belt is flat stainless steel. Alos, if rolled is will resume its activity elsewhere. The belt is not inclined. They will do this for a long time, so that accretion rings form The orange is not a unique case, in fact quite common. Apples too. All the rest, wrong, wrong, wrong. Cherries can climb perhaps a foot up an apparently smooth and motionless inclined plane. How do you think they do that? Oh, I forgot.

Your imaginary cart? Yes there are many ways you could make it stay in one place. The mechanisms explained for the orange would probably work: Take a simple cart with low-resistance wheels, put enough weight on it to create a sag in the belt. Or put a slight incline in the belt (note that in the propeller cart tests, the incline is in the opposite direction, so this is not how they work). Heck, just build one. You might learn something (then again, if the cart doesn't work the way you expect it to, you will probably decide the cart is wrong).

No belt sag required. Wow, he almost thought it...low resistance.
though it does show that you are willing to accept any thing that stays on the belt as being windspeed. Can I sell you som real estate? You make up both our minds, OK? You build one, but mind the sharps.

Amazing! Very simple, and you still managed to get it wrong. Here is how it works in the real world (not the Humberverse). Note that for this analysis, I am assuming travel directly downwind, such that the parachute is acting as a simple drag device and is not generating lift like a sail on a tack, or a rotating propeller. The force on the parachute will be at a maximum when the stationary skater first deploys the parachute and will accelerate the skater. As the speed of the parachute and skater approach the wind speed, the force on the parachute will decrease

Risible. You are using "feels like" as an instrument. Folk physics. (51%, remember). Wow! tragic error.
It's the fact that the forces are not only equal but opposite, but maximised. Forces are maximised not minimised at terminal velocity. I put a gauge in the skates, so that you wouldn't fall into your own trap. Oh, boy, Spork has it easy!

When the force exerted by the wind on the parachute matches the friction of the skates on the ice (very low friction, so this will be slightly less than wind speed), the skater will stop accelerating, and the force on the parachute will be at a minimum.

Actually they are roller skates, ( hence wheels) but even so, why does the skater not continue to accelerate?

The frictional forces on the skates increase with the skater's speed. This is not however, the primary cause of the skater reaching terminal velocity.
The primary cause is the inability of the parachute to generate drag force when it is moving at the same speed as the wind,

Another schoolboy error. Make it an aircraft. Make it a car....
And no, parachutes do not behave like that. They can be designed to be like a 'constant' force or 'constant' velocity.

Note that with correct analysis of the "skater going downwind" the "real world" and "treadmill" cases are identical.

That's wrong. I will ignore the rest.

So that we are not comparing apples to oranges, we will assume the treadmill is long enough to allow the parachute and skater become (nearly) stationary and substitute rollerskates (or maybe lollerskates) for ice skates. Ice doesn't bend too well, so it's kind of hard to make a treadmill out of it. Here is what will happen(from the "ground" frame of reference): The skater moving with the treadmill belt deploys the parachute. The parachute, catching the stationary air, slows down the skater The force will be at its maximum when the chute is first deployed and will decrease as the skater's speed decreases. As the skater slows down to a speed where the drag from the parachute matches the rolling resistance of his skates, which will be almost stationary, but moving slowly in the direction of motion of the belt the force on the parachute will reach a minimum, just like in the "real world'. Note that the rolling resistance of the skates will be at its maximum when the skater is at "terminal velocity" (almost stationary relative to the ground). This is not, however, the primary cause for the skater failing to come to a complete stop (relative to the ground). The primary cause is the inability of the parachute to generate any drag when it is not moving.

Quite wrong. A canard.

QED. The frames of reference are equivalent. Yet another epic fail by Humber.

You say, that forces are mimimum at terminal velocity. All your arguments are either based upon that false assumption, or other matters equally so.

QED. You have embarrassed yourself.

 

[humber]

That was the whole point, humber. As usual, it's gone whizzing far over your head.

Absolute voltages are meaningless. Only voltage differences mean anything, just like velocities.

This is really too good to let go. You missed the point at least twice.
Dan_O's questions are always loaded, you see. Notice that? Simple arithmetic, for someone in "my field". ROLF

The analog of voltage is velocity. What happens when you apply the same velocity 'across' an object. Ohh nothing it the same humber oh the same!!
You forgot that like voltages, forces can be common-mode. Any differences becomes a 'frame' eh?

I said, if the resistance is very small, even that may not apply. Small in value, or small in size? The other 'end' may also find its own conductive path.

Anyway, now that you agree my cart/radar/buffer means that I can be quite certain that the cart is moving wrt the ground, and that I can measure the resultant KE, you cannot apply your counter-argument to the treadmill.