Down wind faster than the wind

[sol invictus]

Ignoring the observed tiny speed. we have two options:
The cart is at windspeed when in the hand or on the belt. ( no sensible wind over cart)
The cart is at zero velocity when in the hand or on the belt. ( no sensible wind over cart)
It's paradoxical, because it is nonsense.
This platform represents a cart that keeps itself at zero velocity regardless of windspeed.
It is understandable, but unbelievable that such a model would be proposed. That's the difficulty of understanding.
All this headbanging is astonishment at support of such a banal and absurd claim. It is not even wrong.

I really have no idea what you're trying to say. It's clear you've misunderstood something very basic, but I don't know what it is and I'm sick of trying to guess. So I give up.

 

[Thabiguy]

Ignoring the observed tiny speed. we have two options:
The cart is at windspeed when in the hand or on the belt. ( no sensible wind over cart)
The cart is at zero velocity when in the hand or on the belt. ( no sensible wind over cart)
It's paradoxical, because it is nonsense.

You're right. Now, how about this?

The cart is at windspeed when in the hand or on the belt. ( no sensible wind over cart)
The cart is at zero velocity with respect to the air when in the hand or on the belt. ( no sensible wind over cart)

 

[sol invictus]

Which is wrong from a physics standpoint. You always need a torque greater than zero to turn a wheel.

Are you really an engineer? If so, I hope (for my own safety if I ever use something you've designed) that you're familiar with the concept of "inertia", or "moment of inertia"? Or maybe that you've heard of something called Newton's First Law?

 

[humber]

Huh? Sol explicitly stated that by connecting the propeller to the wheels means a torque greater than zero is required to turn the wheels. With suitable gearing and propeller selection the net force from the treadmill on the cart and the prop can be adjusted (in theory and apparently in practice) to be in a direction which causes the cart to accelerate to above the speed of the treadmill.

Yes, huh.
The torque is the tiny difference between the drive from the wheels and the equal and opposite reaction from the propellor*. In an ideal world, these forces would be in balance, resulting in zero torque. It is fundamentally incapable of doing any work. Other ratios are possible, but that simply means that the propellor turns at a different speed in order to create the same result of standing still. It is not a model of anything other than it is. It is literal. There is no wind. The belt drives the wheels and the propellor.

*attributed to Newton or Zeus.

 

[humber]

There's no need to wonder.

JB

I wonder how it is that you can speak for these men. They do an awful lot of your talking. Could it be that you have nothing to say that is not connected to an appeal to authority?

 

[ThinAirDesigns]

techno:

Right but it needs to be done in BOTH DIRECTIONS at the same time in order to travel faster than wind. You can't do it with just a belt drive.

Techno, I've seen you state the above and your "screw" analogy numerous times here on this thread. I know why you use the screw analogy and I understand why you think it's correct -- but it's just not. I know you don't believe me and you feel you have very good reasons for that, but it's just not.

I will make one small attempt at demonstrating your fallacy.

If the prop were not connected to the wheels, you would be using your "screw" analogy just perfectly -- a prop really is nothing more than a screw after all. Wind from the front ... screw turns one way. Wind from the back ... screw turns the other way. We get it, we really do. But the prop IS geared to the wheels and in spite of every bone in your body telling you this doesn't make a difference, IT DOES!.

If you're "screw" analogy was to suddenly become valid in this application, ever sailboat on the water would have suddenly come dead in the water and would have to toss up a spinnaker and head straight downwind only.

The keel on a sailboat allows the sail to "see" wind from a direction *other than* the direction you would think. The gearing between the wheels and the prop allow the prop to "see" wind from a direction *other than* you would think.

While you will likely remain rooted in your vision that when the cart chassis transitions from 'slower than the wind' to 'faster than the wind' the propellor sees the same change, it will never be true.

You can learn *why* it's not true, or you can continue believing your "screw" analogy while sailors the world over prove you wrong.

Your choice.

 

[humber]

And you, sir?

 

[my_wan]

Perfect. I take back my criticism of the substance of your argument.

In fact, I explained with great detail why that is. You just didn't understand my explanation. Perhaps I didn't explain it clearly enough.

I'll let that slide sense the substance in this post is perfect. It even makes the source of our disagreement crystal clear and also makes your original claim very understandable. Nobody could ask for more.

Let me try again. The craft is travelling right to left, wrt ground, at 4 mph. The wind is going right to left, wrt ground, at 8 mph (and also right to left, wrt craft, at 4 mph). The prop blows the air left to right wrt the craft, at 2 mph.

I do see how you got got your numbers now and why. I qualified my original claim you objected to with, "When the craft first takes off" for a reason. In the first video 4 mph is most of the ground wind speed and certainly can't be defined as "when the craft first takes off". By arbitrarily picking the prop wind speed at some arbitrary craft speed (such as 2 mph when the craft is at 4 mph) you can point the prop wind speed either direction you want wrt ground. Your chosen numbers would give a prop wind speed of 2 mph wrt the ground in the same direction as the craft.

If craft speed is 4 mph (far too fast to be when the craft first takes off) then:

From above wrt the ground (ground= 0mph wrt the ground).
Definitions:
Right to left is -mph wrt the ground.
Left to right is +mph wrt the ground.
A switch from + to - or - to + is a direction reversal wrt ground.
Sentence by sentence in the same order you stated above.
Craft = -4 mph wrt ground, right to left.
Wind = +8 mph wrt ground, left to right.
Prop air = +2 wrt craft but craft = -1 wrt ground therefore:
Prop air wrt ground = -4+2 = -2 mph wrt ground (same direction as as craft wrt ground).

What happens when we use these same numbers except with a craft speed much closer to "when the craft first takes off" as I specified, such as 1 mph wrt ground. Now the prop air speed wrt ground is 1 mph opposite the direction of the craft.

Craft = -1 mph wrt ground, right to left.
Wind = +8 mph wrt ground, left to right.
Prop air = +2 wrt craft but craft = -1 wrt ground therefore:
Prop air wrt ground = -1+2 = +1 mph wrt ground.

This is exactly why I specified "when the craft first takes off" in the quote of mine you chose to take issue with. To see it in context here is where you objected:

When the craft first takes off the prop is moving air in the opposite direction of the wind.

With respect to the device, yes. With respect to the ground, no. The air pushed back is moving in the opposite direction of the air wrt device, and in the same direction as the air wrt ground.

If you don't realize this, then you don't understand how the device works.

What direction is the air blown by the propeller travelling, wrt ground? In the same direction as the air, or the opposite direction of the air?

When the craft first takes off it is in the opposite direction of the wind, as I originally stated and will show why. Now since gearing can be set so that either one of us can be right about the craft who is right concerning the craft given sporks specifications? We know that the gear ratio is 1 to 1 in rpm. So this same question becomes how does a propeller turning the same rpm as the wheels blow air faster than the wheels are moving the craft?

This one is easy. The prop has a much larger radius than the wheels. This means the tips of the prop has to have a higher angular velocity than the edges of the wheels on the road, even at the same rpm. Any part of the prop with a greater radius of the wheels must eject air faster than the wheels are moving the craft. My statement therefore stands.

Does this make statement number 2 true, or false?

Given the design specifications of the craft statement number 2 remains false. Of course you could change sporks specifications (gear ratio, wheel size, prop pitch, etc.) to make it true.

Does this clarify what I mean? If not, what is still unclear to you?

Yes, perfectly. I even understand now how you did the vectors to reasonably lead you to that claim.

ETA: I'm sorry, I mistyped one of the directions when I was typing this post. The error is now corrected. If you have read it before, please read the post again.

Not a problem. The above issue with gearing ratio changing the physics of the crafts operation is an excellent example of why "idealized" models are so often useless. I try real hard to qualify my statements for the specific characteristics of the real model under actual operational conditions. It is impossible to avoid all mistakes.

 

[ThinAirDesigns]

Arguing over inertial frames of reference in this simple application is a silly as arguing over the basic laws of gravity.

I drop a brick on my foot -- it hurts. I switch frames of reference -- nothing changes.

If that's an appeal to authority -- I claim guilty.

JB

I wonder how it is that you can speak for these men. They do an awful lot of your talking. Could it be that you have nothing to say that is not connected to an appeal to authority?

 

[mhaze]

Here's a simple test - Put machine on treadmill, use fish scale and measure stationary thrust. Do not allow machine to move in xy during test.

Might be easiest to suspend machine between two fish scales on a tight string.

 

[ThinAirDesigns]

Here's a simple test - Put machine on treadmill, use fish scale and measure stationary thrust. Do not allow machine to move in xy during test.

EXCELLENT!! Someone has suggested a test. As it turns out, we do tests. We do tests on request.

We'll set it up and post it.

mhaze, any special concerns or requirements you wish to see us meet during the test?

Thank GOD someone wants to actually *do* something.

JB

 

[humber]

You're right. Now, how about this?

What is the one thing that connects all together. No wind. There is no "equivalence" what you are seeing is a toy on a belt, doing what a toy would do if it were on a belt.

It is not a cart in wind, it is a cart on a belt in still air.
The monumental conceit, is that dead physicists are invoked to suggest that it otherwise. Why? Because no living physicist would support the idea, once he realised tghe difference between what is said to be presented, and what is presented.
I have challenged Spork to publish his calculations at on the Physics Forum. He doesn't seem at all keen.

 

[sol invictus]

The torque is the tiny difference between the drive from the wheels and the equal and opposite reaction from the propellor*. In an ideal world, these forces would be in balance, resulting in zero torque. It is fundamentally incapable of doing any work. Other ratios are possible, but that simply means that the propellor turns at a different speed in order to create the same result of standing still.

I see - they're always in balance. How odd... would they remain in balance if I aimed the propeller up? What if I applied a resistive brake to the wheels - will the cart remain stationary and its wheels spin at the same rate in that case too?

Rather than continue repeating yourself, why don't you show us why those forces are always in balance?

 

[humber]

EXCELLENT!! Someone has suggested a test. As it turns out, we do tests. We do tests on request.

We'll set it up and post it.

mhaze, any special concerns or requirements you wish to see us meet during the test?

Thank GOD someone wants to actually *do* something.

JB

No need to bother. You can see from the video #5. It is quite clear that the cart can be moved very easily. There is no appreciable torque.
This has been mentioned several times, but we have yet to hear an explanation from Newton/Einstein/Galileo's clairvoyant.

 

[my_wan]

Assuming a constant coefficient of drag, the resistance increases with the square of velocity. It's the power required that increases with the cube of velocity.

Yes but you have a set power available for the crafts use under a given wind speed and cross-section. Power is then the quantity you design for. Also, the coefficient of drag is not constant given the operational specs of this craft.

 

[spork]

http://www.youtube.com/watch?v=Y15NnGZIBuM

 

[Thabiguy]

my_wan,

I'm genuinely pleased that we've been able to clear up some misunderstandings and move forward. Now, there's one significant thing to resolve.

What happens when we use these same numbers except with a craft speed much closer to "when the craft first takes off" as I specified, such as 1 mph wrt ground. Now the prop air speed wrt ground is 1 mph opposite the direction of the craft.

Craft = -1 mph wrt ground, right to left.
Wind = +8 mph wrt ground, left to right.
Prop air = +2 wrt craft but craft = -1 wrt ground therefore:
Prop air wrt ground = -1+2 = +1 mph wrt ground.

I underlined the problem here. Correctly, the situation would be this:

Craft = -1 mph wrt ground, right to left.
Wind = +8 mph wrt ground, left to right.
Prop air = +0.5 wrt craft but craft = -1 wrt ground therefore:
Prop air wrt ground = -1+0.5 = -0.5 mph wrt ground.

The important thing to realize here that the prop air speed, wrt craft, is linearly dependent on the ground speed wrt craft (i.e. the propeller will slow down if the cart slows down wrt ground). It can't possibly stay constant, such as +2.

Are you with me so far?

(P.S.: I'll be AFK for a couple hours.)

 

[humber]

Remember Mr Newton "for every action there is an equal and opposite reaction"
Changing the orientation of the propellor, will simply change the amount of drag, and therefore the amount of energy absorbed from the belt.
The two forces will always be in balance, not only because of that law, but that the sum of the two forces is that derived form the belt.
If I were to replace the propellor with a sail, it would do much the same. I can make the cart move forward or backwards according to how much drag I allow the sail to create.
It is held in placeon the belt by drag created by the propellor. The propellor is not a motive force, but a dissipator of energy. It is too busy keeping itelf on the belt that it does not have any time to do any other work.
A lame duck, backed by hubris and bravado.

 

[spork]

Yes but you have a set power available for the crafts use under a given wind speed and cross-section. Power is then the quantity you design for. Also, the coefficient of drag is not constant given the operational specs of this craft.

Nevertheless, I was commenting on your assertion that drag is proportional to the cube of speed. It's not.

 

[humber]

Arguing over inertial frames of reference in this simple application is a silly as arguing over the basic laws of gravity.

I drop a brick on my foot -- it hurts. I switch frames of reference -- nothing changes.

If that's an appeal to authority -- I claim guilty.

JB

Yes, all bricks have the same mass to you. That last bit sums up your knowledge of inertial frames.