Merged DDWFTTW in a “Real” Wind

[Jack by the hedge]

You have been told and proven to be wrong. End.

This sounds like dialogue from Downfall. You're not typing from the humberbunker, are you?

 

[jduffy77]

I guess I can't see what you can see that makes it wrong.

It doesn't specify that the objects need to be the same objects in each scenario. Maybe that's too obvious. All he's saying is that it's the closing speed between the two objects that matters, and the relative velocities compared to some other frame of reference don't matter. If you don't change the speed of impact (and you don't change the objects) the result will be the same.

Of course except that was not what was being discussed in the original context where I found myself in the terrifying position of agreeing with Humber.

You forgot your relative motion and made downwind absolute, so things never come the other way.

The problem is that a closing speed of 1200mph, can't be mimicked by having one object stationary and the other moving at 600mph. Want to make that bird travel at 1200mph?

I doubt that the total KE of the plane and the bird will be the same, as when both are traveling at 600mph but in opposite directions.

If there is a closing speed of 1200 mph, any combination of speeds that add up to 1200 will produce exactly the same effect on the objects when the objects collide. 600 + 600, 0 + 1200, 376.9 + 823.1, etc; every one is the same to the objects, the only thing changing is the speed of the observer relative to the objects and since the observer isn't involved, that can't affect what happens to the objects.

 

[Jack by the hedge]

I have won the balloon matter, hands down. Spork has not been able to defend it, just floated it.

He has lost that one.

Are your imaginary divisions of humbercarts sweeping to victory?

Did you prove that a balloon, released at above windspeed, will slow to below windspeed? (I may have blinked.)

You may carry on believing what you like.

You should definitely get that on a tee shirt.

 

[Dr. Keith]

I have won the balloon matter, hands down.

I'm not sure what "won" means in this conversation, but I am trying to get at your various claims regarding balloons being unable to travel at wind speed. I've offered some questions upthread which you have dismissed, but I am unaware of a comprehensive explanation of why a neutrally buoyant balloon will not travel at wind speed. Link?

Spork has not been able to defend it, just floated it.

He has lost that one.

I hadn't asked spork about this. Only you. It is not a competition, I am merely asking for clarification of your position.

You may carry on believing what you like.

Obviously, but the whole point of being here is to understand each other and communicate our ideas. This particular idea of yours evades me and I have an interest in the subject.

Table clothes and rockets are less interesting to me.

 

[jadebox]

Well it was intended to help you, as it seemed to me you were right, but I hadn't seen mender make a mistake. I didn't think that was what you'd been saying all along, as you originally questioned mender's assertion that a 600 + 600 mph impact would have the same effect as a 0 + 1200 mph impact or any other combination that added up to 1200 mph. Maybe I need to go back and read what everyone said again.

Above, I said that a 60 + 60 mph impact would be the same as a 120 + 0 impact. That appears to be stating the same principle, doesn't it?

(I also said that a 60 + 0 impact could mimic it if one of the objects was not a car at all but rather a perfectly rigid, immovable barrier. That makes it a different case from two identical objects impacting.)

Everyone is making this more confusing than it needs to be .......

"Closing speed" isn't relevant. What matters is the relative speed of the object before and after collosion - i.e. the acceleration of the object.

For example, a car going 60 mph hitting a solid wall is equivalent to a car going 60 mph and hitting a car of the same mass coming straight at it at 60 mph. In both cases the car decelerates from 60 mph to zero in a very short period of time.

A car hitting a wall at 120 mph is not equivalent to two cars hitting head on at 60 mph, because the car hitting the wall at 120 mph is decelerating from 120 mph to zero, not 60 mph to zero.

Of course, this still doesn't explain how an untethered, unpowered balloon can travel at less than wind speed.

-- Roger

 

[Jack by the hedge]

Of course except that was not what was being discussed in the original context where I found myself in the terrifying position of agreeing with Humber.

From what you quoted it appears that your concern is with the discrepancy in kinetic energy with different frames of reference.

That issue was addressed by ArmillarySphere and essentially the answer is that (despite humbers protests) KE is relative.

If you consider two objects moving on a collision course, then you can do so from any frame of reference you like, and their velocities will be some arbitrary value plus and minus half their closing speed. If you consider them together as a closed system, then the velocity of the centre of mass of the whole system means the system has an amount of KE relative to your frame of reference. That amount of KE will remain as KE after the collision too, due to conservation of momentum. So that "excess" KE is not available to do other work.

Hence humber's rockets with 10,000J each make the same size of bang as his 40,000J rocket hitting a static rocket, because half of the fast rocket's energy remains at the end as KE, simply moving the debris relative to your frame of reference.

(Is that on the right track now?)

 

[humber]

Actually you have not said a single coherent thing about the balloon matter.

Adding the dressing of coherent, doesn't "cut it"

I would win the argument hands down, but there appears to be a sudden case of stage fright.

If any of you want to carry one anyway, then you need to show that the
drawing at the bottom is physically possible.

http://www.internationalskeptics.com/forums/attachment.php?attachmentid=21890&stc=1&d=1303335207

 

[jadebox]

Adding the dressing of coherent, doesn't "cut it"

I would win the argument hands down, but there appears to be a sudden case of stage fright.

If any of you want to carry one anyway, then you need to show that the
drawing at the bottom is physically possible.

http://www.internationalskeptics.com/forums/attachment.php?attachmentid=21890&stc=1&d=1303335207

Possible or not, the drawing at the bottom is irrelevant. it doesn't depict what happens when the balloon is drifiting at the same speed as the wind.

At windspeed, the air doesn't flow around the balloon - the balloon is not moving relative to the air and the air is not moving relative to the balloon. The situation is equivalent to the balloon floating in still air.

-- Roger

 

[humber]

Hence humber's rockets with 10,000J each make the same size of bang as his 40,000J rocket hitting a static rocket, because half of the fast rocket's energy remains at the end as KE, simply moving the debris relative to your frame of reference.

Still no good. The total momentum is the same mv + mv = 2mv, but the KE is not; 10000 + 10000 = 20000, but 40000 for the 2V missile.

Momentum is conserved, energy is conserved. KE is not, but may be.
KE + Heat = Total energy.

40000J is more than 20000J, no matter how you try to work around that.

None of the references supportive or not, have even mentioned frames of reference, because there is no relevance.

 

[jduffy77]

For some reason I am having trouble communicating with you on this issue so I will stop after this post.

From what you quoted it appears that your concern is with the discrepancy in kinetic energy with different frames of reference.

I had no concerns except that I felt that Mender's response to Humber was, if not wrong, at least not exactly accurate, even though I know he knows everything about the issues, unlike Humber.

That issue was addressed by ArmillarySphere and essentially the answer is that (despite humbers protests) KE is relative.

Right, I thought ArmillarySphere summed in up perfectly which was what I said in my first post.

If you consider two objects moving on a collision course, then you can do so from any frame of reference you like, and their velocities will be some arbitrary value plus and minus half their closing speed. If you consider them together as a closed system, then the velocity of the centre of mass of the whole system means the system has an amount of KE relative to your frame of reference. That amount of KE will remain as KE after the collision too, due to conservation of momentum. So that "excess" KE is not available to do other work.

This was what I was saying, and I posted a link which supported this/my position which I thought was being disagreed with.

Hence humber's rockets with 10,000J each make the same size of bang as his 40,000J rocket hitting a static rocket, because half of the fast rocket's energy remains at the end as KE, simply moving the debris relative to your frame of reference.

(Is that on the right track now?)

None of this was ever on the wrong track as far as I am concerned.

 

[Jack by the hedge]

If any of you want to carry one anyway, then you need to show that the
drawing at the bottom is physically possible.

The drawing is incomplete. You need to put one of your little arrows on the balloon too, as it's moving at the same speed as the air.

(QED)

 

[humber]

Possible or not, the drawing at the bottom is irrelevant. it doesn't depict what happens when the balloon is drifiting at the same speed as the wind.

Errr,,,

At windspeed, the air doesn't flow around the balloon - the balloon is not moving releative to the air and the air is not moving relative to the balloon. It's equivalent to the balloon floating in still air.

So there is a vacuum to the front? The diagram represents what would be necessary.
WS balloons would mean the air must get to the front without passing over the balloon, or not change velocity. A physical impossibility that any one who has walked around an obstacle knows.


Spork has not responded, as he does when challenged, and I am not interested in the rear defense of a retreat, thanks.

 

[humber]

The drawing is incomplete. You need to put one of your little arrows on the balloon too, as it's moving at the same speed as the air.

(QED)

You go makey one then, or find one that a text book with one that shows Spork is even close to being correct.

 

[Jack by the hedge]

Still no good. The total momentum is the same mv + mv = 2mv, but the KE is not; 10000 + 10000 = 20000, but 40000 for the 2V missile.

Momentum is conserved, energy is conserved. KE is not, but may be.
KE + Heat = Total energy.

40000J is more than 20000J, no matter how you try to work around that.

None of the references supportive or not, have even mentioned frames of reference, because there is no relevance.

You are being deliberately obtuse again, and it's not doing you any favours. Everyone gets this by now so you're just making yourself look ridiculous.

40000J is more than 20000J, but as has been explained to you over and over and over, after the collision, the debris from the 40000J rocket and its target have 20000J of KE left over due to the conservation of their momentum. So both your examples have 20000J with which to put on a show.

(By the way, I begin to suspect you don't actually understand frames of reference.)

 

[jadebox]

So there is a vacuum to the front? The diagram represents what would be necessary.
WS balloons would mean the air must get to the front without passing over the balloon, or not change velocity. A physical impossibility that any one who has walked around an obstacle knows.

No, the air is not moving relative to the balloon at windspeed. So there's air on all sides of the balloon. The air is not moving relative to the balloon so it doesn't pass over or around or under or through the balloon. The air is not moving relative to the balloon so the air does not change velocity. The air is not moving relative to the balloon so there is no obstacle for it to avoid.

Wow ... you are really good. Do you troll other forums or just this one?

-- Roger

 

[3point14]

Errr,,,


So there is a vacuum to the front? The diagram represents what would be necessary.
WS balloons would mean the air must get to the front without passing over the balloon, or not change velocity. A physical impossibility that any one who has walked around an obstacle knows.

I know you won't understand this, but the air doesn't need to 'get' anywhere, it's already there. Exactly as it is in the case where the wind is at 0mph.

 

[Jack by the hedge]

For some reason I am having trouble communicating with you on this issue so I will stop after this post....None of this was ever on the wrong track as far as I am concerned.

OK. Some kind of miscommunication here. Peace. Out.

 

[jduffy77]

Adding the dressing of coherent, doesn't "cut it"

I would win the argument hands down, but there appears to be a sudden case of stage fright.

If any of you want to carry one anyway, then you need to show that the
drawing at the bottom is physically possible.

Your foolishness about Reynolds numbers and Stokes flow is more than usually nonsensical since the balloon is not moving through the air. It is floating along with the air.

 

[Jack by the hedge]

So there is a vacuum to the front? The diagram represents what would be necessary.
WS balloons would mean the air must get to the front without passing over the balloon, or not change velocity. A physical impossibility that any one who has walked around an obstacle knows.

Is this some kind of performance art? The air is not moving around the balloon at all. The air and the balloon are moving together.

Thought experiment time: Replace the air with ice. A balloon, frozen in a great big block of ice. OK? Now slide the ice across the floor. The balloon moves at the same speed as the ice. It moves at ice speed.

Is there a vacuum in front of the balloon? Uhhh... No.
How does the ice flow around the balloon? Uhhh... It doesn't.

Astonishing, eh?

 

[spork]

I hadn't asked spork about this. Only you. It is not a competition, I am merely asking for clarification of your position.

Of course you're welcome to ask me for my position, but you'll find it to be a pretty boring one. It uses more traditional laws of physics.