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Down wind faster than the wind

Or, is my math wrong?
Any reply appreciated.
I hate to walk around with a delusion in my head.
 
quarky said:
Or, is my math wrong?
Any reply appreciated.
I hate to walk around with a delusion in my head.
Click to expand...
There was no math in that post. It was in your head, so it is somewhat difficult to decide how correct it was.

As for very high speeds..

You get more air resistance (proportional to the square of the speed of the cart through the air.), so my WAG would be that the process would be less efficient, and you would end up with a lower multiple of the wind speed.

That had a little math in it. :)
 
Ririon said:
There was no math in that post. It was in your head, so it is somewhat difficult to decide how correct it was.

As for very high speeds..

You get more air resistance (proportional to the square of the speed of the cart through the air.), so my WAG would be that the process would be less efficient, and you would end up with a lower multiple of the wind speed.

That had a little math in it. :)
Click to expand...

Thanks.

My unspoken math was the power/speed ratio (cubic), against the squared losses.
 
quarky said:
Still not sure if its wrong-think. I'm missing something.
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If I read you correctly, you are saying that the efficiency of the propeller is (roughly) proportional to the cube of the speed. We seem to agree that the losses to air resistance is (roughly) proportional to the square of the speed. (Not getting into the finer points of which speeds we are talking about right now.) I'd like to see an explanation for the cube factor. I think that is where the dog i buried. (Is that an expression in English?)
 
The question asked is rather an interesting one, and I don't have the answer.

What can be said is that a craft speed between 0 and wind ground speed you actually gain energy, by the wind/ground speed difference. This is because the craft has a tail wind up to that speed, with a wheel coupled to the prop so the effective power doesn't fall off as the craft approaches wind ground speed. The effective tail wind even increases with speed up to ground wind speed, because the prop air speed is counter to the wind.

Once the craft exceeds wind ground speed, it begins expending that energy differential against a head wind. It doesn't expend it linearly because the increased speed also increases the prop exhaust speed. Presuming a simplistic conservation effect I presume this is why it exceeds twice wind speed. So I don't know what a maximum craft speed would look graphed against variations in ground wind speed, at various cross sectional wind resistance efficiencies. Even the efficiency of the prop varies with speed.

It's far from as simple as it looks from squaring wind velocities. To many feedback and variable efficiency variances. My best guess is that, if you changed gearing ratios to maximize for a given wind ground speed, the higher the wind speed the greater multiple of that speed you could exceed. Just a guess based on the prop continuing to increase wind output speeds even during the power use phase, i.e., at or above ground wind speed.

Also note that, without drag, any small force over enough time can accelerate any large object to any speed respecting Special Relativity.
 
my_wan said:
The question asked is rather an interesting one, and I don't have the answer.

What can be said is that a craft speed between 0 and wind ground speed you actually gain energy, by the wind/ground speed difference. This is because the craft has a tail wind up to that speed, with a wheel coupled to the prop so the effective power doesn't fall off as the craft approaches wind ground speed. The effective tail wind even increases with speed up to ground wind speed, because the prop air speed is counter to the wind.

Once the craft exceeds wind ground speed, it begins expending that energy differential against a head wind. It doesn't expend it linearly because the increased speed also increases the prop exhaust speed. Presuming a simplistic conservation effect I presume this is why it exceeds twice wind speed. So I don't know what a maximum craft speed would look graphed against variations in ground wind speed, at various cross sectional wind resistance efficiencies. Even the efficiency of the prop varies with speed.

It's far from as simple as it looks from squaring wind velocities. To many feedback and variable efficiency variances. My best guess is that, if you changed gearing ratios to maximize for a given wind ground speed, the higher the wind speed the greater multiple of that speed you could exceed. Just a guess based on the prop continuing to increase wind output speeds even during the power use phase, i.e., at or above ground wind speed.

Also note that, without drag, any small force over enough time can accelerate any large object to any speed respecting Special Relativity.
Click to expand...



Much thanks for the replies, and apologies for my confusion and lack of coherent question.

The power in the wind is cubed with the speed of the wind.
Catching it is another matter entirely, which is why I was careful to mention the 'hypothetical' aspect of my query.

The losses , in theory, don't accelerate as fast as the gains.
To take this game to its logical conclusion, the cart is suspended on a cable. No passenger or steering is required. Weight is reduced substantially.
Friction reduction is a bug-a-boo in my head, because the cart operates by pushing off the Earth.

A cable that had grip, like a chain, might be the ticket. The dusty flats are too primitive to accommodate elegant concepts. As are the roads.

Human powered vehicles (bikes) are hobbled by the road ways they were forced to adapt to.
I sense that with the right 'roads', hpv's will routinely average 50 mph; that they will be faster than the alternative transportation; free exercise, etc.

I'm jaded from having built and ridden and 2 person railroad-track bike-thingy.
It was "where it's at". Suspended from cable puts the tracks to shame.
 
my_wan said:
The question asked is rather an interesting one, and I don't have the answer.

What can be said is that a craft speed between 0 and wind ground speed you actually gain energy, by the wind/ground speed difference. This is because the craft has a tail wind up to that speed, with a wheel coupled to the prop so the effective power doesn't fall off as the craft approaches wind ground speed. The effective tail wind even increases with speed up to ground wind speed, because the prop air speed is counter to the wind.

Once the craft exceeds wind ground speed, it begins expending that energy differential against a head wind. It doesn't expend it linearly because the increased speed also increases the prop exhaust speed. Presuming a simplistic conservation effect I presume this is why it exceeds twice wind speed. So I don't know what a maximum craft speed would look graphed against variations in ground wind speed, at various cross sectional wind resistance efficiencies. Even the efficiency of the prop varies with speed.

It's far from as simple as it looks from squaring wind velocities. To many feedback and variable efficiency variances. My best guess is that, if you changed gearing ratios to maximize for a given wind ground speed, the higher the wind speed the greater multiple of that speed you could exceed. Just a guess based on the prop continuing to increase wind output speeds even during the power use phase, i.e., at or above ground wind speed.

Also note that, without drag, any small force over enough time can accelerate any large object to any speed respecting Special Relativity.
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My Wan,

There is no change of "Mode" between less than wind speed and greater than windspeed.

You correctly state that below windspeed, the cart takes energy from the differential between the speed the wheels go over the ground and the speed the propellor goes through the air.

That differential still exisit above wind-speed. At all times, the cart's groundspeed is greater than it's airspeed, and that is where, by the gearing between the wheels and the advance ratio of the prop, it gets the energy to provide thrust.

Above windspeed the only thing that changes is drag, both from the increased friction of the mechanism, and the aerodynamic drag of the now present headwind.
 
RossFW said:
My Wan,

There is no change of "Mode" between less than wind speed and greater than windspeed.
Click to expand...
Yes, you are correct. The "mode" change was merely a useful definition, not any sort of change in the mechanics of the system.

As a power graph to speed ratio, it merely marked the point were the difference in power used to power available was at a maximum. Which corresponds to a speed at which wind ground speed is (closely) matched. Merely a useful fiction to call it a change of mode, since it's in some ways easier to model if the math model starts at that point.
 
The cart goes down wind, but to the rotating sail, the "Apparent Wind" speed is wwaaaay faster. Very important to any sailor. Boats easily go faster then the wind, across the wind. Spinning the sail (propellor) maintains apparent wind speed when you turn downwind. (anybody ever hear RPMs?) His land sailer probably only needs a hundred watts, but boats have so much water friction it may not work. No laws of physics are broken.
 
a_unique_person said:
Nope. Still dont get it.
Click to expand...


I used this model to help understand it, years ago:

13012493204e1d7668.jpg



The large central wheel has traction with the ground. (The two smaller wheels just to keep it from tipping over.) The cart travels from left to right, with the central wheel turning clockwise. If you held the cart off the ground, and turned the main wheel clockwise, the parachute would move right to left, toward the rear of the cart.

Now, disregarding all wind, think about the cart in motion and convince yourself of the following four things:

1. When the cart rolls along the ground, the parachute is moving right to left relative to the frame of the cart, just as it would do if the cart were held off the ground and you turned the main wheel clockwise.

2. When the cart rolls along the ground, the parachute is moving left to right relative to the ground. So relative to the ground, it's moving in the same direction as the cart, but not as fast (about one third as fast).

3. If you were to push on the parachute toward the right, say with your hand, it would propel the cart toward the right, because of fact 2.

4. But because of fact 1, as you continued pushing on the parachute, the frame of the cart would be propelled relative to the round faster than your hand is moving. About three times faster.

Now imagine if it were a wind, blowing left to right, pushing on the parachute instead of your hand. Like your hand, that would propel the cart forward (fact 3). Suppose the wind pushes the parachute to two thirds of the wind speed, which it can do because the parachute is still slower than the wind. But (by fact 4), the frame of the cart would be moving left to right three times that fast, or twice the wind speed.

In principle, this would work. (In practice, not so likely, because there would be friction in every important part, and it can only travel until the parachute reached the pulley at the rear of the cart frame... unless there were more parachutes mounted at intervals around the cord, furling themselves at the rear pulley and unfurling themselves at the front pulley, which would be an even bigger mechanical nightmare.)

So, how is this like the Blackbird? The parachute represents the rear surface of the Blackbird's propeller at some (any) particular point in the area the propeller sweeps. Not, please note, a specific point on the propeller itself that spins with the propeller, but a particular point in the propeller's plane that's fixed relative to the cart frame. As the propeller sweeps past that point, the propeller blade rear surface at that point moves (because of the propeller's angle and spin) left to right relative to the ground -- fact 2 -- so that the wind pressure at that point will propel the cart left to right -- fact 3 -- but at the same time, that intersection point is moving right to left relative to the cart frame -- fact 1 -- so the cart frame is moving left to right faster than the rear surface -- fact 3 -- so if the rear surface is moving close to the speed of the wind, the cart frame will be moving faster than the wind -- fact 4.
 
Think about it- Once you are going DDWFTTW, you are actually going directly-UP-wind.
 
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