Merged DDWFTTW in a “Real” Wind

[humber]

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.

If I posted that there was some other matter you don't know about concerning fluid dynamics, you would simply deny that too.

Spork is unable to defend himself on this matter, but you may carry on believing what you want.

The matter has been decided by the lack of response, and any support from the Carteers that suggests he would be able to do so. I will not respond to balloon questions, until Spork meets the challenge I have put to him.

 

[humber]

Balloons are off the table until Spork responds to the challenge of fluid dynamics and the balloon The KE matter has been settled by Prof Rhett, along with Jamie from Mythbusters' support.

 

[Jack by the hedge]

Balloons are off the table until Spork responds to the challenge of fluid dynamics and the balloon The KE matter has been settled by Prof Rhett, along with Jamie from Mythbusters' support.

Awww. So we never get to find out what brainstorm made you claim a balloon released travelling faster than the wind will slow to less than windspeed.

I think everyone is quite happy that they understand the KE topic. Even you (although sadly you're in a minority of 1 who still has it wrong).

Edit to add: You do realise, don't you humber, that Prof Rhett said two cars crashing head on at 50mph is not the same as a car crashing into a wall at 100mph? He's quite correct. I believe we all agree that's correct.

Of course, two cars crashing head on at 50mph is the same as a car crashing into an identical stationary car at 100mph.

 

[humber]

Awww. So we never get to find out what brainstorm made you claim a balloon released travelling faster than the wind will slow to less than windspeed.

I think everyone is quite happy that they understand the KE topic. Even you (although sadly you're in a minority of 1 who still has it wrong).

Edit to add: You do realise, don't you humber, that Prof Rhett said two cars crashing head on at 50mph is not the same as a car crashing into a wall at 100mph? He's quite correct. I believe we all agree that's correct.

Of course, two cars crashing head on at 50mph is the same as a car crashing into an identical stationary car at 100mph.

If there were two cows in a field, and I said they were both cows, you would say not, on the basis that one has a white patch over one eye.
That is all you do when it comes to fluid dynamics. The matter of KE, a lunch pin argument for the cart, has been settled, and exposed as a schoolboy error.

The theory of the cart is nonsense, and a recipe for perpetual motion. The TM is an elaborate version of pulling the table cloth from under the dishes to the amusement of the guests. A trinket.
When I have enough trial data, that will fall too.

 

[mender]

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're right, I didn't respond to what humber said but instead, restated correctly what he was trying to mess up as usual.

If you want, I'll repost and address his exact post.

 

[CORed]

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

He has lost that one.

You may carry on believing what you like.

Yes, black knight, you have won.


PS: PM one of us carteers your mailing address if you want your arms and legs back.

 

[windgrins]

Humbers equation reasonably close but not quite!

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.

Humber.

I think you are close here on your equation:

KE + Heat = Total energy.

Just make it:

KE+Heat Energy+Energy of Deformation = Total Energy.

That is, when using the law of Conservation of Energy:

Add up all sources of energy before event and after the event and one will find these to be equal from the same Inertial Reference Frame (IRF) for closed systems.

Someone else may be able to throw in some more terms but I think these are the relevant ones for collisions.

There is no argument here but this could screw up equations in a quantitative analysis of an inelastic collision if unaccounted for. So, I throw it in just to be complete early.

Anyone else have any terms for simple collisions they think should be in there before we get started on this track?

I'm not making any comment at all about the remainder of the post.

WindGrins

 

[windgrins]

Reference which discusses frames of reference for Kinetic Energy

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.

Humber,

Here's one that mentions it explicitly:

"As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity."

I would say that's pretty relevant to the previous discussions because it is exactly what most of the previous posters said.

The reference is from Wikipedia on inelastic collisions.

Here's the URL:

http://en.wikipedia.org/wiki/Elastic_collision

My excerpt is in the middle of the first set of dynamic illustrations.

Regards,

WindGrins

 

[Jack by the hedge]

The matter of KE, a lunch pin argument for the cart, has been settled, and exposed as a schoolboy error.

I think you mean "linchpin" - ironically a device which stops your wheels falling off.

Your schoolboy error is easy to fix. Let's return to your pairs of rockets, and consider them as a system:

You say they have 10000J of KE when travelling a 600mph (268.2 ms^-1) so we can work out their mass.
½mv² = 10000
v = 268.2
m = (2 * 10000) / (268.2)² = 0.278kg

In the example where both are travelling at 600 mph in opposite directions, you can see by symmetry that their common centre of mass is always stationary at the eventual point of impact. Since it's stationary, it has zero momentum. As they have 10000J of KE each, they have a grand total of 20000J to expend in the impact.

In the example where one is stationary and the other is travelling at 1200 mph and with 40000J of KE, their combined centre of mass is halfway between them - a point which is constantly moving, in the same direction as the rocket but at half its speed - 600mph. The momentum of their combined centre of mass will be the same after the collision, of course, so it's velocity can't change either. That means that the component of kinetic energy of the debris which is due to its continuing to move as before is:

½mv² = ½(2 * 0.278) * 268.2² = 20000J

So the excess KE which the rockets can expend in their collision is 40000J-20000J = 20000J. Exactly the same as before.

TL;DR version:

The extra kinetic energy of the second scenario is merely the kinetic energy of the whole system moving relative to your frame of reference. For a closed system that will be conserved, so is not available to be expended in the impact.

 

[Cuddles]

Once again, stop the personal attacks and bickering or further moderator action will follow.

Replying to this modbox in thread will be off topic  Posted By: Cuddles

 

[windgrins]

Additional Clarity to Humber

accidental post before complete.

 

[windgrins]

Additional Clarity to Humber's balloon in the wind diagram.

Humber,

If I am not mistaken, I believe most of the posters, other than you, believe the correct picture for a balloon freely drifting in the wind after adequate settling time (less than a minute) looks something like this:






It should be used as the last picture instead of the one in your posted diagram in post #3287.

There are no flowstream arrows around the balloon because there is no flow.

For absolute completeness, there should be a gradient of flow near the ground to illustrate skin friction but I also believe we are ignoring that nuance.

Regards,

WindGrins

 

[humber]

Humber,

Here's one that mentions it explicitly:

"As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity."

That means energy is conserved, and is the contrary of what you hope to say. One rocket at 2V has 4 times the KE of one at V. So the collision will not be the same.
The problem Carteers have has nothing to do with frames, but a simple schoolboy error that has been made into a theory. TAD have been told countless times they have made this error, and now Prof Rhett has told them again, and with an in-your-face example from Mythbusters.

KE is quadratic with velocity, so KE =1/2mv^2 simple and linear addition is not possible.

The problem lies in very basic mistakes like this:

Consider the case of two cars passing on the highway - one northbound and the other southbound. They pass with several car widths between them. Let's say each is going 60 mph. At the moment they're adjacent to one another, their relative speed is 120 mph. But their closing speed is 0 mph at that moment.

Which is to say they could collide with no force. The speed difference is constant, and does not change. Lie the balloons, another schoolboy error.

The TM is a monumental error. Velocities may be added linearly, and that is what is meant by them being relative: there is no dependence upon some absolute frame or magnitude of the velocity ( for speeds of interest to us).

Velocity is not speed. When making the transform, one may add or subract speed or velocity, but the latter may not be treated as speed.

If the arithmetic result changes sign an that is simply accepted, error may result. It is expected that the reader of the text knows how to handle such math, and when the use of RMS may be a way to do that for example.

The TM is a monumental error. Taking the constant speed from from the wind, brings it to zero, but taking it from the ground, makes it negative. That is a reversal of the relationship of wind and ground that does not exist.
Dr Camp says that the making the road a belt that is moving backwards, is an invalid transform and that is the reason why.

KE + Heat = Total energy.

Total energy = KE + Work.

KE is stored energy, and work is expended energy that is not recoverable.
Joule equated heat and work long ago. Deformation, sound heat etc..all work.

 

[humber]

I think you mean "linchpin" - ironically a device which stops your wheels falling off.

Your schoolboy error is easy to fix. Let's return to your pairs of rockets, and consider them as a system:

You say they have 10000J of KE when travelling a 600mph (268.2 ms^-1) so we can work out their mass.
½mv² = 10000
v = 268.2
m = (2 * 10000) / (268.2)² = 0.278kg

In the example where both are travelling at 600 mph in opposite directions, you can see by symmetry that their common centre of mass is always stationary at the eventual point of impact. Since it's stationary, it has zero momentum. As they have 10000J of KE each, they have a grand total of 20000J to expend in the impact.

In the example where one is stationary and the other is travelling at 1200 mph and with 40000J of KE, their combined centre of mass is halfway between them - a point which is constantly moving, in the same direction as the rocket but at half its speed - 600mph. The momentum of their combined centre of mass will be the same after the collision, of course, so it's velocity can't change either. That means that the component of kinetic energy of the debris which is due to its continuing to move as before is:

½mv² = ½(2 * 0.278) * 268.2² = 20000J

So the excess KE which the rockets can expend in their collision is 40000J-20000J = 20000J. Exactly the same as before.

TL;DR version:

The extra kinetic energy of the second scenario is merely the kinetic energy of the whole system moving relative to your frame of reference. For a closed system that will be conserved, so is not available to be expended in the impact.

That is funny. A trick around making the velocity negative, and then hoping to get away with it.
(-v)^2 = V^2. May be the penny about RMS, will eventually drop
And no, it's momentum that is conserved, which is WHY you can't make the above joke about the centre of mass. You have been fooled by a cranky theory, that is based on deception.

You can't turn water into wine, but you are trying to turn whine into a means of saving face.
The above is an attempt at that, and can be dismissed like all the others.

 

[humber]

Humber,

If I am not mistaken, I believe most of the posters, other than you, believe the correct picture for a balloon freely drifting in the wind after adequate settling time (less than a minute) looks something like this:


[qimg]http://www.internationalskeptics.com/forums/imagehosting/488654db022391e960.png[/qimg]

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_488654db024a495eda.jpg[/qimg]

It should be used as the last picture instead of the one in your posted diagram in post #3287.

My point was that your diagram is your claim and there would be no flow.
I asked that a textbook example be found that says that is possible, hence my diagram.

There are no flowstream arrows around the balloon because there is no flow.

There always is to some degree.

For absolute completeness, there should be a gradient of flow near the ground to illustrate skin friction but I also believe we are ignoring that nuance.

Which is indeed one of the flows that will be there, so WS is impossible.

 

[humber]

You're right, I didn't respond to what humber said but instead, restated correctly what he was trying to mess up as usual.

If you want, I'll repost and address his exact post.

It was 100% your error.

 

[humber]

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.

I think that you may be done, now?

The TM is a huge abuse of Galileo. That can be readily demonstrated by my propless cart.

There are many errors in your math, and simple assertions regarding fluid dynamics.

NALSA have recently responded to a further request for more data.
Analysis of that data, and correction of measure mt errors, will show the record to be false.
I have suggested that the trial be re-run, under more appropriately controlled conditions.

 

[3point14]

Humber,

When everyone else disagrees with you, then there are two options.

Either you're wrong or everyone else is.

Now, I admit when living in some sort of theocracy where it's certain death to disagree with theory 'A', it may be that the lone voice in the darkness is the only one with the courage and insight to state what is right.

However, when not living in such a state, when living in a state where everyone is free to state their own theories, the lone voice in the darkness is so much less likely to be right. There's no papal decree to scare us all into saying these things, it's simply what we, after studying reality, believe.

Do bear in mind that no-one ever checked reality against the maths. Lots and lots of people check their maths against reality (it's the only way to be sure it's right) and, where there's a discrepency, it's the maths that needs amending, not reality. Maths models reality, not the other way around.

 

[humber]

Humber,

When everyone else disagrees with you, then there are two options.

Either you're wrong or everyone else is.

Where "everyone" is TAD and the Carteers. "Everyone else", including text books and Drs Kammen, Camp, Rhett et, all are not in agreement with TAD.

Windgrins posted this from a reference:
"As can be expected, the solution is invariant under adding a constant to all velocities, which is like using a frame of reference with constant translational velocity."

Meaning the energy is conserved. The constant, the translational velocity, Vt,
means that KE = 1/2m (V +Vt)^2 and not the sum
KE = 1/2m (V)^2 + 1/2m (Vt)^2

Now, I admit when living in some sort of theocracy where it's certain death to disagree with theory 'A', it may be that the lone voice in the darkness is the only one with the courage and insight to state what is right.
However, when not living in such a state, when living in a state where everyone is free to state their own theories, the lone voice in the darkness is so much less likely to be right. There's no papal decree to scare us all into saying these things, it's simply what we, after studying reality, believe.

From all responses, 3.14, it seems that study has not been in depth, but one of personal enlightenment via "intuition", based upon self-generated and answered thought experiments.

Einstein did not base his resultant theory on his thought experiments, but used them as a means of description. His experiment where he uses two flashlights and a book to show that E=mc^2 is brilliant, but not proof. But then, he knew that.

Do bear in mind that no-one ever checked reality against the maths. Lots and lots of people check their maths against reality (it's the only way to be sure it's right) and, where there's a discrepency, it's the maths that needs amending, not reality. Maths models reality, not the other way around.

Then the math should work, shouldn't it? For balloons to travel at WS, a great deal of other theories, not connected with aerodynamics, would also be wrong.

 

[jadebox]

Edit to add: You do realise, don't you humber, that Prof Rhett said two cars crashing head on at 50mph is not the same as a car crashing into a wall at 100mph? He's quite correct. I believe we all agree that's correct.

Of course, two cars crashing head on at 50mph is the same as a car crashing into an identical stationary car at 100mph ....

... assuming the "stationary car" is just stopped and not bolted to the ground or whatever.

Another way of saying almost the same thing you just said is to consider the head-on collision from a reference point moving along with one of the cars.

To an observer on the ground between the cars, each is moving 60 mph towards each other and 0 mph after they hit. So each goes from 60 to 0 mph. But, an observer moving with one of the cars sees one car going 0 mph and the other coming towards it at 120 mph before the collision. After the collision, both cars are moving away at 60 mph. So the speed of each car changes by 60 mph (0 to 60 and 120 to 60). Therefore, the damage caused to each car is the same.

I think Humber has a problem understanding the relativity thing or is being obtuse about it on purpose. It's a common theme of this thread.

-- Roger