Merged DDWFTTW in a “Real” Wind

[humber]

Just so you are clear. I haven't retracted it.

And now I think I will spend the next few days rolling on the floor, laughing at your idea that NB balloons lift loads.!

Adios.

Windgrins

My, such a long time phoning and searching over a one transistor problem you don't understand.

http://www.newton.dep.anl.gov/askasci/phy05/phy05417.htm



Windgrits next excuse will be to say that does not apply if the balloon is not ascending....it will all go away when it stops at an even greater height...you gotta nuthin'
When the balloon is NB, the potential increase of the balloon is the same as the decrease in the air. The balloon displaces the same volume of air, but also the same weight when NB.
It's like an imaginary sphere of air. Just as I told you.



Question: How can the sum of potential energy and kinetic energy
be constant when a helium balloon moves up in air as both potential
energy and kinetic energy increases?
---------------------------------------
The potential energy is going _down_ as the balloon rises. This is
counter-intuitive because the standard way of describing potential energy
is based on the assumption that gravity is much greater than the buoyancy
(buoyancy is ignored). But that is not a good assumption with the helium
balloon. Potential energy is actually stored in the submerged (in air)
helium balloon -- to help visualize, think of how much work it takes to
submerge a basketball under water. With the basketball, you are pushing
water up (it is displaced by the ball) against gravity, storing potential
energy. It works the same way with the balloon submerged in air. Filling
the balloon displaces air, thereby storing potential energy in the
balloon. When you release the balloon, that potential energy is converted
to kinetic energy (there are other factors as well -- it expands as it
rises, for example, but I will neglect those for purposes of this answer).

Hope this helps,
Burr Zimmerman
====================================================================
The potential energy of the helium balloon increases as it ascends,
but as it ascends it allows heavier air to descend to take its
place. The air's potential energy thus decreases. The decrease in
the air's potential energy is of greater magnitude than the increase
in the helium balloon's potential energy, so overall potential
energy decreases.

 

[humber]

Good one RP but I think its your week to watch him to ensure he doesn't hurt himself!

WG

DOHned

What...you say...

Question: How can the sum of potential energy and kinetic energy
be constant when a helium balloon moves up in air as both potential
energy and kinetic energy increases?
---------------------------------------
The potential energy is going _down_ as the balloon rises. This is
counter-intuitive because the standard way of describing potential energy
is based on the assumption that gravity is much greater than the buoyancy
(buoyancy is ignored). But that is not a good assumption with the helium
balloon. Potential energy is actually stored in the submerged (in air)
helium balloon -- to help visualize, think of how much work it takes to
submerge a basketball under water. With the basketball, you are pushing
water up (it is displaced by the ball) against gravity, storing potential
energy. It works the same way with the balloon submerged in air. Filling
the balloon displaces air, thereby storing potential energy in the
balloon. When you release the balloon, that potential energy is converted
to kinetic energy (there are other factors as well -- it expands as it
rises, for example, but I will neglect those for purposes of this answer).

Hope this helps,
Burr Zimmerman
================================================== ==================
The potential energy of the helium balloon increases as it ascends,
but as it ascends it allows heavier air to descend to take its
place. The air's potential energy thus decreases. The decrease in
the air's potential energy is of greater magnitude than the increase
in the helium balloon's potential energy, so overall potential
energy decreases.

 

[Michael C]

Not so long ago, spork posted this:

I would claim that a positively buoyant hot air balloon LOSES potential energy as it ascends - just as a rock loses P.E. as it falls.

humber disagreed (of course, he had to disagree with spork):

It gains PE by rising.

He said it again here:

Lift the balloon, and it will gain PE. Your claim is of magic. An NB weather balloon may carry 100gr of instruments, and that mass gains PE with height, and so does the rest of the balloon.

(we'll leave aside the question of how a NB balloon could be rising and assume that he means a positively buoyant weather balloon)

True to form, he posts a link that contradicts his own statement:

http://www.newton.dep.anl.gov/askasci/phy05/phy05417.htm

He was so proud of his link that he posted verbatim the argument from it:

The potential energy is going _down_ as the balloon rises.

 

[humber]

see next post

 

[humber]

Not so long ago, spork posted this:

I would claim that a positively buoyant hot air balloon LOSES potential energy as it ascends - just as a rock loses P.E. as it falls.

humber disagreed (of course, he had to disagree with spork):

The potential energy of the helium balloon increases as it ascends, (1)
It loses KE as it ascends, and gain PE. The converse of the rock. So, flat out wrong.

It gains PE by rising.

The potential energy of the helium balloon increases as it ascends, (1)

Lift the balloon, and it will gain PE. Your claim is of magic. An NB weather balloon may carry 100gr of instruments, and that mass gains PE with height, and so does the rest of the balloon.
(we'll leave aside the question of how a NB balloon could be rising and assume that he means a positively buoyant weather balloon)

True to form, he posts a link that contradicts his own statement:

The potential energy of the helium balloon increases as it ascends, (1)

http://www.newton.dep.anl.gov/askasc...5/phy05417.htm
He was so proud of his link that he posted verbatim the argument from it:
Quote:
The potential energy is going _down_ as the balloon rises.

Total PE. The sum of the balloon and the air. (2)


Question: How can the sum of potential energy and kinetic energy
be constant when a helium balloon moves up in air as both potential
energy and kinetic energy increases?
---------------------------------------
The potential energy is going _down_ as the balloon rises. (2) This is
counter-intuitive because the standard way of describing potential energy
is based on the assumption that gravity is much greater than the buoyancy
(buoyancy is ignored). But that is not a good assumption with the helium
balloon. Potential energy is actually stored in the submerged (in air)
helium balloon -- to help visualize, think of how much work it takes to
submerge a basketball under water. With the basketball, you are pushing
water up (it is displaced by the ball) against gravity, storing potential
energy. It works the same way with the balloon submerged in air. Filling
the balloon displaces air, thereby storing potential energy in the
balloon. When you release the balloon, that potential energy is converted
to kinetic energy (there are other factors as well -- it expands as it
rises, for example, but I will neglect those for purposes of this answer).

That is, the heavier air loses potential energy as it ascends, and transfers it to the balloon, so it gains KE as it rises. When it stops, or becomes NB, all of that is PE.
When the balloon is NB, the heavier air will not accelerate it (because it will not ascend, because the force of gravity is the same on the balloon as the air),
so not add KE, and the work must come from what ever raises it to height. The balloon gains PE by the work. Hence the floors of the building.

Or with the analogy with a rock. The rock has PE, which becomes KE as it falls, the buoyant balloon has KE, which becomes PE as it rises.
When NB, that exchange does not take place, and the balloon is like an imaginary sphere of air, so gains PE with height or the air could not lose PE as it decends, as the
text says it does.

Hope this helps,
Burr Zimmerman
================================================== ==================
The potential energy of the helium balloon increases as it ascends, (1)
but as it ascends it allows heavier air to descend to take its
place. The air's potential energy thus decreases. The decrease in
the air's potential energy is of greater magnitude than the increase
in the helium balloon's potential energy, so overall potential
energy decreases.


And well done there, BTW. You put your foot in it for Spork too.
The third time around for him BTW. Same mistake.
Had enough?

 

[windgrins]

Well stated Michael C!

Not so long ago:
True to form, he posts a link that contradicts his own statement:
He was so proud of his link that he posted verbatim the argument from it:

When pressed by actually stating assumptions and simplifying the issue to below high school level, Humber seems to be contradicting himself in record time. I'm at a loss at the moment to understand which side of this argument he is actually arguing. Perhaps you have more patience to sort it out than I!

This seems to have devolved to out of the realm of physics and into abnormal psychology which, although interesting, is not why I joined this forum.

Regards,

WindGrins

 

[humber]

When pressed by actually stating assumptions and simplifying the issue to below high school level, Humber seems to be contradicting himself in record time. I'm at a loss at the moment to understand which side of this argument he is actually arguing. Perhaps you have more patience to sort it out than I!

Regards,

WindGrins

And shows you all to be wrong. That really does say a lot about the way you simply ignored I Ratant's video.
Never mind. NALSA's own, unexpurgated data, says the wind changes more than 50 degrees. Some conditions for ddw.

Like this:

 

[quarky]

I don't get the helium balloon's rising PE. At the top of its flight, it becomes NB. Where's the potential there?

 

[jduffy77]

And shows you all to be wrong. That really does say a lot about the way you simply ignored I Ratant's video.
Never mind. NALSA's own, unexpurgated data, says the wind changes more than 50 degrees. Some conditions for ddw.

Like this:
[qimg]http://www.internationalskeptics.com/forums/attachment.php?attachmentid=22046&stc=1&d=1304508584[/qimg]

Wow, that data looks amazing. The carts run was more impressive than I thought. No wonder they chose that particular run for the record.

 

[Jack by the hedge]

I don't get the helium balloon's rising PE. At the top of its flight, it becomes NB. Where's the potential there?

If you consider the balloon alone and ignore the atmosphere then as the balloon gains altitude, it gains PE. The heavier air which descends to replace it of course loses PE. Overall the system loses PE.

 

[humber]

Wow, that data looks amazing. The carts run was more impressive than I thought. No wonder they chose that particular run for the record.

You keep kidding yourself. 50 degree changes around a cart that keeps a fixed heading.
And you should see the runs where it fails. But as I said, NALSA get to first see the report.

 

[quarky]

During this intermission, I thought I'd toss out an amusing engineering idea I had long ago:

A floating, anchored factory, set in the deepest part of the ocean, receives and processes garbage. It drops the denser stuff in a huge net, on a cable, which runs through a generator. Seven mile head! To approximate a waterfall, the dropped junk would need to be twice as dense as water, or more. Feasible? Ridiculous?

 

[humber]

If you consider the balloon alone and ignore the atmosphere then as the balloon gains altitude, it gains PE. The heavier air which descends to replace it of course loses PE. Overall the system loses PE.

Not when NB. That is true as it rises, because it is less dense. When NB, the displacement of air and balloon have the same density.

 

[profpangloss]

I don't get the helium balloon's rising PE. At the top of its flight, it becomes NB. Where's the potential there?

I suspect people here are really arguing at cross purposes and not defining their terms precisely. There are many forms of potential energy, so just saying potential energy without a qualifier can be confusing. In this case, the balloon’s gravitational potential energy certainly increases as the balloon rises, but its buoyant potential energy is decreasing.

Let’s say the balloon is tethered at sea level. It would have maximum buoyant potential energy, minimum gravitational PE and zero kinetic energy. After it is releases and rises, its buoyant PE drops, its gravitational PE increases and it has also gained kinetic energy. When it reaches maximum altitude, it would have maximum gravitational PE, minimum buoyant PE and zero kinetic energy (at least in the vertical direction) The sum total of energy will always be constant making the buoyant PE at sea level equal to the gravitational PE at maximum altitude, kinetic energy would be zero at both extremes.

 

[quarky]

Thanks. makes sense.
If the balloon was carrying a rock, and was designed or destined to pop at the top of its flight, it would be a different story.

 

[humber]

Thanks. makes sense.
If the balloon was carrying a rock, and was designed or destined to pop at the top of its flight, it would be a different story.

Not really. The rock would make it less buoyant. The overall density of the balloon and payload must be less than the air, or it won't rise.

The argument was about gravitational PE, and not the PE of buoyancy, if only because if was about the "magical" NB balloon.
This topic was settled much earlier. Of course, it has nothing to do with the cart, other than some vague support for the relative motion business.

 

[profpangloss]

And shows you all to be wrong. That really does say a lot about the way you simply ignored I Ratant's video.
Never mind. NALSA's own, unexpurgated data, says the wind changes more than 50 degrees. Some conditions for ddw.

Like this:
[qimg]http://www.internationalskeptics.com/forums/attachment.php?attachmentid=22046&stc=1&d=1304508584[/qimg]

This is the original NALSA data? I am not surprised! How does this factor into the claim the cart went DDW? I have seen a figure of 4 degrees off of DDW, where did that come from? From this chart, what is the RMS deviation from DDW? I will calculate it myself, but I would like to see other people do it as well. Just looking at the chart I'd say the RMS devaition is about 20 degrees. How can anyone seriously claim the cart went DDW? I will post more on this later!

 

[jduffy77]

This is the original NALSA data? I am not surprised! How does this factor into the claim the cart went DDW? I have seen a figure of 4 degrees off of DDW, where did that come from? From this chart, what is the RMS deviation from DDW? I will calculate it myself, but I would like to see other people do it as well. Just looking at the chart I'd say the RMS devaition is about 20 degrees. How can anyone seriously claim the cart went DDW? I will post more on this later!

A. You have been told ad nauseam that the RMS deviation was not 20 degrees.

B. You have been told ad nauseam that RMS is not the most sensible way to look at the wind.

C. You have been told ad nauseam that only the test period is relevant.

D. You have been told ad nauseam that the cart was not at 0 degrees.

It was at 0 degrees off the wind for some of the time. The cart averaged 4 degrees off the wind.

 

[jduffy77]

How can anyone seriously claim the cart went DDW? I will post more on this later!

You ask how "anyone" can claim the cart went DDW, but sadly for your agenda it is not just anyone who is claiming it. It is the worlds foremost authority on landsailing. It is the world record holder who before he educated himself believed ddwfttw was impossible. It is the folks who know that millisecond transients in wind direction cannot provide a tacking force to a cart with a sail never mind to one without. It is the folks who know that, although they are the best at what they do, they themselves could not even begin to duplicate the BB's performance in the same wind conditions.

 

[profpangloss]

A. You have been told ad nauseam that the RMS deviation was not 20 degrees.

B. You have been told ad nauseam that RMS is not the most sensible way to look at the wind.

C. You have been told ad nauseam that only the test period is relevant.

D. You have been told ad nauseam that the cart was not at 0 degrees.

It was at 0 degrees off the wind for some of the time. The cart averaged 4 degrees off the wind.

What I find most nauseating at the moment are your posts.

A. It matters not what you or anyone else tells me. All that matters is what the chart tells me.

B. RMS is the ONLY correct way to calculate wind power. You should learn about it.

C. What part of the chart is the test period and what part is not?

D. If the cart was not at 0 degrees, what was the reference for the wind direction and at what angle was the cart?

How do you determine the cart averaged 4 degrees off of the wind when the wind was changing direction the way the chart indicates? The cart could not have been tracking those changes as it held to a steady course. The cart was tacking the wind. Do you find that nauseating?