5
   

I don't understand how this car works.

 
 
DrewDad
 
  1  
Reply Thu 10 Jun, 2010 09:47 am
@ThinAirDesigns,
ThinAirDesigns wrote:
the propeller can't pull any harder on the air

When you make a statement like this, it takes away from your credibility, IMO.

You can't pull on a fluid. You push on a fluid, or the fluid pushes on you.
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 09:47 am
@georgeob1,
georgeob1 wrote:
I agree that the propeller on your vehicle can extract energy from the air and transmit it to the vehicle's wheels. With the proper pitch controls and coupling to the shaft driving the wheels, the prop can also store energy in the form of rotational momentum to be used for brief (until the stored energy is exhausted) additional acceleration.


George -- you have a basic misunderstanding of how the device operates and is equipped. It's IMPOSSIBLE for the rotational momentum of the prop to be used for ANY period to accelerate the vehicle (without skidding the tires) because the gearing between the pop and wheels is of one fixed ratio.

One can not slow down the prop and at the same time speed up the wheels without changing this ratio and the ratio is fixed. QED

JB

0 Replies
 
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 09:49 am
@Cycloptichorn,
I'm not sure what you mean by "releasing official speed numbers"

Can you elaborate?

JB

ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 09:52 am
@DrewDad,
DrewDad wrote:
When you make a statement like this, it takes away from your credibility, IMO.


Pardon the casual (and yes incorrect) useage.

But talk about "credibility" -- you might want to be more careful what claims you make (or claim you have never made) over on PhyOrg.

JB
0 Replies
 
DrewDad
 
  1  
Reply Thu 10 Jun, 2010 09:58 am
@ThinAirDesigns,
ThinAirDesigns wrote:
No, you're trying to figure out how to take it OUT because if you do plain jane force*distance calculations on the two mediums - it's all in there already.

Please don't keep telling me what I'm trying to do. I know what I'm trying to do. You can tell me my approach is wrong, you can tell me my math is wrong, you can tell me my understanding of physic is incomplete, and you might be right. But unless you have a crystal ball hidden in that cart of yours, you cannot know better than me what my intent is.
0 Replies
 
Cycloptichorn
 
  1  
Reply Thu 10 Jun, 2010 09:59 am
@ThinAirDesigns,
ThinAirDesigns wrote:

I'm not sure what you mean by "releasing official speed numbers"

Can you elaborate?

JB


None of the skeptics will really give up on challenging the legitimacy of the idea until there is comprehensive evidence, in the form of (relatively) unedited video with speed tracking numbers in it.

Not that you guys really care about skeptics, but hey. This is a big deal! You just don't see modern physics experiments on the macro level like this.

Cycloptichorn
ThinAirDesigns
 
  2  
Reply Thu 10 Jun, 2010 10:01 am
DrewDad -- I will make an observation and an apology:

In spite of my occasional urge to consider you a hopeless denier (not to be confused with 'skeptics' which I respect), you have clearly made a classy move or two through all this that truly convinces me that you are capable of and interested in the truth. Sometimes my memory is too short.

I will try to snipe less and answer more.

JB
0 Replies
 
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 10:06 am
@Cycloptichorn,
Quote:
None of the skeptics will really give up on challenging the legitimacy of the idea until there is comprehensive evidence, in the form of (relatively) unedited video with speed tracking numbers in it.


If you are referring to the initial Ivanpah video where we blanked out the audio partially to keep the hubbub down until we could get our sponsor logos on the vehicle , the totally unedited videos are now also posted on Youtube.

You'll understand that if we had let out those numbers at the time (back in March) , the pictures that are currently traveling around the globe accompanying these articles would have been of the bare rig rather than the fully clothed and logo'd machine.

For what it's worth -- those unedited videos don't "prove" anything any more than the edited ones ... the audio is just us talking on the radio and to each other as the vehicle runs.

JB
Cycloptichorn
 
  1  
Reply Thu 10 Jun, 2010 10:07 am
@ThinAirDesigns,
I know - I wasn't referring to those videos, but the next set of videos that you will release.

Cycloptichorn
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 10:21 am
@Cycloptichorn,
Cycloptichorn wrote:
I wasn't referring to those videos, but the next set of videos that you will release.


Ahh - OK.

The only video we currently have in the can is more of the same ... video from us testing at different venues. We may release some just because we have it, but it won't be because we expect it to sooth any critics concerns.

The NALSA testing is sort of the 'official' watermark that we are aiming for and it's looking like that will be likely scheduled for the week surrounding the 4th of July weekend. We may or may not do this on the NASA Ames runways (depends on conditions), or we may head down to El Mirage dry lakebed in SoCal.

There should be interesting video from those tests, but the really interesting result from those tests will not be the video, but the NALSA report itself. I'm not familiar with how long it takes their BOD to ratify and publish such reports, but I suspect it would not be more than a few weeks (or perhaps less) *after* the test.

JB
Cycloptichorn
 
  1  
Reply Thu 10 Jun, 2010 10:21 am
@ThinAirDesigns,
That's the info I was looking for, thanks!

Cycloptichorn
0 Replies
 
ThinAirDesigns
 
  2  
Reply Thu 10 Jun, 2010 10:50 am
One of the more entertaining and yet illustrative examples of the principle:

(credit: http://www.physicsforums.com/showthread.php?p=2063418)

********************************
This is the story of Atom Man, a miniature sized superhero from the planet Zogg. Planet Zogg is much like Earth, the laws of physics are the same except that their "air" molecules are heavier and slower than ours, so when the wind blows hard the motion of these molecules is predominately just downwind with not very much random thermal motion.

Now Atom Man and his arch rival Mini-Schroder both have downwind carts that they can ride in, and they like to race each other on windy days. After much racing and tweaking both Atom Man and Mini-Schroder have tuned their carts so that they travel downwind with very little friction. They both attain almost the same speed as the wind, so their finishes are generally pretty closely contested.

Atom Man is very small and nimble, he can even see individual molecules of gas in the air. When he gets into his downwind cart to race he can see the molecules of air colliding with the back of his cart and bouncing off, providing the change in momentum to accelerate the cart forward. As his cart increases to closer to the wind speed he sees these molecules colliding with ever less relative speed and imparting ever less momentum, so he never can quite reach the speed of the wind, let alone exceed it.

One day while Atom Man is racing Mini-Schroder downwind he gets a great idea. The race is neck a neck, with both carts doing almost exactly the wind speed, when Atom Man notices that the air molecules are now almost stationary relative to him and he can easily reach out and grab them. So he starts grabbing these molecules as fast as he can and hurling them backwards to create a forward thrust (in much the same way as a spacewalking astronaut could throw a spanner to create thrust if he had no other propulsion available). This idea works and Atom Man actually gets his cart to faster than wind speed, resoundingly beating Mini-Schroder in the process.

The victory is short lived however. Mini-Schroder immediately protests to the DWC Academy about Atom Mans cheating tactics. The complaint is upheld, and rightly so. After all Atom Man was using up much of his own stored energy (from his internal batteries, you see Atom Man is battery powered) in the process of hurling those atoms. The Academy's rules strictly state that the only allowed source of power in these races is the wind.

Atom Man is not so keen to give up on the idea of DWFTTW just yet, so he asks the Academy to rule on whether he can continue to use his air molecule throwing trick if he obtains the power to throw those molecules from a generator running off the wheels of the cart instead of from his internal batteries. After some deliberation the Academy decides that this is ok and violates no rules. Meanwhile Mini-Schroder is not the least bit concerned about the ruling. He knows full well that Atoms Mans generator will create drag at the wheel and he "knows" (without any calculation or anything, he just knows ok) that this drag will exceed any thrust that Atom Man is able to produce by hurling air molecules.

Atom Man is not so sure about that, he knows a little about basic Physics so he does the analysis. He figures out that if he throws "n" molecules per second, with each molecule having a mass "m" and leaving at velocity "v" then the average forward thrust will be equal to the average rate of change of momentum. That is, atom mans produced thrust is :

F = nmv

Now Atom Man knows that if his strategy is to be successful he must be sure that the drag from his generator is less than this amount, so he decides to use only half of this extra thrust in powering the generator. The theoretical maximum power that his generator can deliver is therefore :

P = 1/2 Fw = 1/2 nmvw

Where "w" is the speed of the cart (and approximately the speed of the wind when atom man starts his hurling).

Finally Atom man calculates how much power he requires to hurl those molecules. The KE given to each molecule is 1/2 mv2, and this is done "n" times per second, so the average power required is :

P = 1/2 nmv2

Since 1/2 nmvw > 1/2nmv2 whenever w > v Atom Man sees that he can easily exceed the wind speed whenever the wind speed is greater than his throwing speed. The result is successful and Atom Man now can always beat Mini Schroder in these races.

Mini Schroder again complains to the Academy, this time he even accuses Atom Man of witchcraft, but nobody will take him seriously. Eventually he storms off and falls into a Bessel null, never to be heard of again. Everyone else lives happily ever after.
DrewDad
 
  1  
Reply Thu 10 Jun, 2010 11:06 am
@ThinAirDesigns,
ThinAirDesigns wrote:
Atom Man is not so sure about that, he knows a little about basic Physics so he does the analysis. He figures out that if he throws "n" molecules per second, with each molecule having a mass "m" and leaving at velocity "v" then the average forward thrust will be equal to the average rate of change of momentum. That is, atom mans produced thrust is :

F = nmv

Now Atom Man knows that if his strategy is to be successful he must be sure that the drag from his generator is less than this amount, so he decides to use only half of this extra thrust in powering the generator.

So the generator powers the thrust which powers the generator....

Isn't this where we started?

Edit: Again, I'm not saying that it's impossible to achieve DWFTTW, I'm just saying that this "explanation" doesn't explain anything to me.
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 11:11 am
@DrewDad,
DrewDad wrote:
Isn't this where we started?


And it's where we'll continue to return until you realize that the generator can produce way more power than the prop needs -- all due to the power of the wind.

There are many ways to get it wrong, but we must always come back to the ONE right way --- work = force * distance.

JB
0 Replies
 
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 11:21 am
DD, shall we try to figure out where we *agree* so we know what we can build on?

As to the 'Cessna at 5,000ft' scenario, do you agree or disagree that the calcs for work done by its propeller on the air would be calculated at the speed the propeller is moving through the air?

In other words, in "work = force * distance", for the variable "distance", we would enter the distance the prop has moved through the air.

Thanks
JB
DrewDad
 
  1  
Reply Thu 10 Jun, 2010 12:22 pm
@ThinAirDesigns,
ThinAirDesigns wrote:
As to the 'Cessna at 5,000ft' scenario, do you agree or disagree that the calcs for work done by its propeller on the air would be calculated at the speed the propeller is moving through the air?

I believe that you can calculate work that way. That's not the only valid way to calculate it.

Basically, you're discussing two frames of reference which are in motion relative to each other. In one frame of reference you consider the air to be stationary, and the ground to be moving. In the other frame of reference you consider the ground to be stationary and the air is moving.
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 12:28 pm
@DrewDad,
DrewDad wrote:
I believe that you can calculate work that way. That's not the only valid way to calculate it.


If you're attempting to determine how much work is required to move the propeller through a given distance of air what other "valid way" would you use? There is only one correct answer after all.

Formula please.

JB
0 Replies
 
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 12:52 pm
DD, let's take a concrete example and calculate the work/HP numbers.

Let say our prop is progressing through the air at 22.5 ft/sec (~20mph) and producing 10lbs of thrust. This will take exactly 1/2HP if our prop is 100% efficient (which for the moment we'll assume).

You say there is another "valid" way to calculate this ... OK, I'm interested in seeing how you choose to do this, but of course remember that if your calc reveals something other than 1/2HP required, you'll need to figure out how to invalidate the simple, standard and time tested formula that I have used.

JB
ThinAirDesigns
 
  1  
Reply Thu 10 Jun, 2010 12:54 pm
@DrewDad,
DrewDad wrote:
Basically, you're discussing two frames of reference which are in motion relative to each other. In one frame of reference you consider the air to be stationary, and the ground to be moving. In the other frame of reference you consider the ground to be stationary and the air is moving.


DD, in the propeller example I have provided and have been asking about for some time now there is NO reference to the ground (and no need for such mention). We're in the air ... the ground isn't relevent to this current problem.
0 Replies
 
DrewDad
 
  1  
Reply Thu 10 Jun, 2010 01:32 pm
@ThinAirDesigns,
OK, moving through the air with an indicated air speed of 22.5 ft/sec and 10 lbs of force....

.41 hp is what I get.

Did you mean 27.5 ft/sec? (18.75 mph?)

27.5 ft/sec at 10 lbs of force = 1/2 hp.
 

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