Merged DDWFTTW in a “Real” Wind

[humber]

But you fail as an inspector.
As I said, your "artwork" in no way affects the integrity of the structure.
Don't let that stop you making woo-woo noises though, you excel at that.

Chugga chugga.

As opposed to making more category errors than a drunk librarian.
Paying $40 for an easy ride into the world of physics costs more than the price tag.

All and more. Its existence is due to the exhumation of historical errors, and the claimed explanation for the apparent record, and the cart's principle of operation, are due to ignorance.

All the ideas are stolen, and it was not I who backed away from the challenge to prove the balloon.

 

[Jack by the hedge]

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

Humber, that was a hyphen in my post, not a minus sign. If you had read the post with any care you would have seen that for what it was instead of what you imagine you saw.

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.

If you had read and understood what I wrote you could not have failed to see that I explicitly declared that momentum is conserved. I explained that in a closed system (whose mass does not change) the system's velocity must consequently also be conserved. You didn't understand.

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.

Handwaving ad-homs do not give you a free pass to ignore reality. You are simply wrong. You are now either trolling or quite remarkably, stubbornly, defiantly, willfully ignorant. I suspect the former. I'm pretty sure you must realise you are wrong. This simply isn't complicated enough for you not to have grasped it by now.

 

[Jack by the hedge]

it was not I who backed away from the challenge to prove the balloon.

Yet you are still fleeing in terror from having to explain your declaration that a balloon released at above windspeed will slow to below windspeed.

Lest you forgot.

 

[humber]

No ... you're the one who seems to believe that changing the frame of reference changes the physical reality of the situation.
I didn't mention velocity (other than specifying the directions the cars moved in relation to each other) and I didn't mention kinetic energy.

What you ignored, is the problem.

But, I will now.

This will be easy.

"Two cars moving in opposite directions, so the same speeds, but different velocities of 60mph, remain that way, no matter how they are observed." is total gibberish - "60 mph" is a measure of speed, not velocity, and speed is always relative to a frame of reference.
"... shows up as the doubling of the KE" - The "doubling of the KE" is real when you change the observation point, but it doesn't change the outcome of the collision.

Still free energy isn't it?. And it does change the outcome, unless you think that KE, the energy of impact, has no influence on the impact.

No matter how you observer the situation, the kinetic energy converted to "crunch" is the same when the two cars hit. The difference in observation,

Energy is conserved, so it either remains stored as KE in the remnants or goes has heat and work of crunching. Twice as much energy means a lot more work can be done and or the remnants have more KE than they began with.

however, affects how much kinetic energy you see in the result. In the case of an observer seeing the two cars crashing to a stop, there is no kinetic energy remaining, because the vehicles have stopped

So twice the work is done on heating the cars and bending them.

It is such a wonderful irony, that Rhett, JB's nemesis, has united with a presumed TAD ally, Mythbusters, so that both have a hand on the same knife.

In the case of the observer moving the same speed and direction as one of the case before the crash, there is remaining kinetic energy because, after the crash, the two vehicles are moving away from the observer.

Oh, but they were both stopped a minute ago. Now you want both cars to move at 60mph, so carrying the same KE as earlier.
The problem is that they were initially traveling in opposite directions, but magically, they are now moving in the same direction. There would be KE transfer as a result of the acceleration of change of velocity (direction), but you have no change in the KE. Voodoo.
KE is relative, but your mistake is to take the offered and naive view of the observer.

If the two cars have a mass of 10kg each and are moving towards each other at 10 meters/second, the kinetic energy of each car is 1/2 * mass * speed^2 or 1/2 * 10 * 10^2 or 500J. The total kinetic energy of the two cars is 1000J.
Now, change the reference point so that one car is moving 0 meters/second and the other is moving 20 meters/second. The kinetic energy of the second car is now 1/2 * 10 * 20^2 or 2000J - 2000J is twice 1000J.

How convenient. The other car has lost its KE. KE=m*a*d, but the mass has magically lost its KE, without acceleration. When you do that so the KE is actually transferred, rather than simply saying you have done it, then you will find the energy has gone to whatever brought the car to 0m/s. That would be a problem for you, should that actually happen, or be imagined to happen.
And in the first case, your total mass was 20kg, and now its 10kg.
Any more Voodoo?

But, after the collision, the two cars will be moving, relative to the observer, at 10 meters/second. The kinetic energy of the two cars is 1/2 * 20 * 10^2 or 1000J.

And now the mass is back. It went away while you were churning the numbers to get back to the original value, but now one car is moving the other way, and not opposite to the other.
More Voodoo.

Subtracting 1000J from 2000J, you get 1000J. The change in kinetic energy is the same for both observers. The different observers see different things, but the physical reality of the situation doesn't change.
-- Roger

ETA/edit

V_car 1 = 10m/s KE = 1/2mV^2 = 1/2*10kg*(10)^2 = 500J
V_car 2 = -10m/s KE = 1/2mV^2 = 1/2*10kg*(-10)^2 = 500J

V_car 1 = 0m/s = 0J (playing with zero that way,. makes the former +/- signs, indeterminate)

so
KE Car 2 = 1/2mV^2 = 1/2*10kg*(+ 10 -10)^2 = 0J
or
KE Car 2 = 1/2mV^2 = 1/2*10kg*(10 + 10 )^2 = 2000J

 

[spork]

Humber, that was a hyphen in my post, not a minus sign. If you had read the post with any care you would have seen that for what it was instead of what you imagine you saw.

humber admitted to us long ago that he has Alexia. He is not able to read your post with more care.

 

[Jack by the hedge]

V_car 1 = 10m/s KE = 1/2mV^2 = 1/2*10kg*(10)^2 = 500J
V_car 2 = -10m/s KE = 1/2mV^2 = 1/2*10kg*(-10)^2 = 500J

Momentum_car 1 = mv = 10kg*10ms^-1 = 100kg.ms^-1Momentum_car 2 = mv = 10kg*-10ms^-1 = -100kg.ms^-1
If we consider the motion of the closed system, i.e. the movement of its centre of mass:
Momentum_system = 100kg.ms^-1 + (-100kg.ms^-1)= 0
V_system = 0
KE_system = 0J

(Where KE_system is the KE of the centre of mass of the system, which must be conserved after the collision)

So the KE available to be converted in the collision is KE_car1 + KE_car2 - KE_system = 1000J

V_car 1 = 0m/s
V_car 2 = 20m/s = 1/2*10kg*(20)^2 = 2000J

Momentum_car 1 = mv = 10kg*0ms^-1 = 0kg.ms^-1Momentum_car 2 = mv = 10kg*20ms^-1 = 200kg.ms^-1
If we consider the motion of the closed system, i.e. the movement of its centre of mass:
Momentum_system = 0 + 200kg.ms^-1 = 200kg.ms^-1V_system = 200kg.ms^-1/(2*10kg) = 10ms^-1KE_system = ½mv² = ½*2*10kg*(10ms^-1)² = 1000J

(Where KE_system is the KE of the centre of mass of the system, which must be conserved after the collision)


So the KE available to be converted in the collision is KE_car1 + KE_car2 - KE_system = 1000J

 

[Jack by the hedge]

humber admitted to us long ago that he has Alexia. He is not able to read your post with more care.

I forgot. I'll try to avoid ambiguity.

 

[humber]

Momentum_car 1 = mv = 10kg*10ms^-1 = 100kg.ms^-1Momentum_car 2 = mv = 10kg*-10ms^-1 = -100kg.ms^-1
If we consider the motion of the closed system, i.e. the movement of its centre of mass:

The "centre of mass" of what?"

Momentum_system = 100kg.ms^-1 + (-100kg.ms^-1)= 0
V_system = 0
KE_system = 0J

....of the massless observer?
How about the ground?
Guess what? That zero result means? That momentum is conserved.
Momentum is not energy.

(Where KE_system is the KE of the centre of mass of the system, which must be conserved after the collision)
So the KE available to be converted in the collision is KE_car1 + KE_car2 - KE_system = 1000J

Where
KE Car 1 = 1/2*m*(10)^2
KE Car 2 = 1/2*m*(-10)^2

Not KE Car 2 = 1/2*m*(10 +/- 10)^2

There was no dispute about the total momentum.

 

[Dr. Keith]

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 I know your position and it goes a bit like this:

As a balloon enters into the flow of air the balloon experiences aerodynamic drag that tends to pull the balloon to the speed of the air. Over time this drag is successful in getting the balloon to achieve the same speed as the air. Once the balloon is at the speed of the medium there is no differential velocity and therefore no drag.

Any attempts to move the balloon away from the speed of the medium will need to overcome drag induced by moving the balloon at a speed other than the speed of the medium. Very small at very low differential speeds, but then increasing.

This is a gross oversimplification, but is so intuitive that I've never had the opportunity to find someone who doesn't recognize drag as the force that makes an item conform to the speed of the medium. That is essentially the definition of drag. If the item is going faster than the medium drag slows it down, if the item is going slower than the medium drag speeds it up. This is just mind blowing that there could be so many posts over a concept that is so simple.

So, no you don't really need to defend yourself here and now it seems humber has decided to walk away from this issue and declare victory. This has been a fun thread, but it is quickly losing the spark that made it fun.

Thanks for taking your brain teaser to such extremes.

 

[Dr. Keith]

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.

No, the claim was yours and it was pointed out to you that your representation of what you thought would be necessary to refute your claim was wrong.

There always is to some degree.

Why?

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

We are stipulating that we are outside the influence of ground interference. Clean flow, no variations. Is that OK?

 

[Dr. Keith]

it was not I who backed away from the challenge to prove the balloon.

I didn't back away, I just had other things to do. Come on, stay in the game and we may get somewhere. Unless you ignore friction, then it is hard to get anywhere. (It's a physics joke, that's why it's not funny.)

 

[Jack by the hedge]

The "centre of mass" of what?"

The centre of mass of the closed system being considered. In this case, two 10kg missiles.

....of the massless observer?
How about the ground?

That would just add needless complication. The system you are considering is two missiles. They can be regarded as a closed system. No mass or energy in or out. Other objects can be excluded from that system.

Guess what? That zero result means? That momentum is conserved.
Momentum is not energy.

No indeed. Momentum is mass multiplied by velocity. In this closed system, the mass does not change.

So

momentum = m₁*v₁ = m₂*v₂​

and

m₁ = m₂​

Therefore

v₁ = v₂​

So you see in this case of a closed system, its overall velocity is conserved.

 

[Jack by the hedge]

KE Car 1 = 1/2*m*(10)^2
KE Car 2 = 1/2*m*(-10)^2

Not KE Car 2 = 1/2*m*(10 +/- 10)^2

I never said that. You misread a previous post and imagined I had said something of the sort, but that was simply your misunderstanding.

If you have any logical criticisms of what I actually wrote, do please let me know.

 

[humber]

I think I know your position and it goes a bit like this:
As a balloon enters into the flow of air the balloon experiences aerodynamic drag that tends to pull the balloon to the speed of the air. Over time this drag is successful in getting the balloon to achieve the same speed as the air. Once the balloon is at the speed of the medium there is no differential velocity and therefore no drag.

Presumes its conclusion. You may as well make the blanket statement that all objects will achieve WS.

Any attempts to move the balloon away from the speed of the medium will need to overcome drag induced by moving the balloon at a speed other than the speed of the medium. Very small at very low differential speeds, but then increasing.

No news there. The balloon achieves a point of dynamic, but not static stability, and that is below WS.

This is a gross oversimplification, but is so intuitive that I've never had the opportunity to find someone who doesn't recognize drag as the force that makes an item conform to the speed of the medium.

That is essentially the definition of drag. If the item is going faster than the medium drag slows it down, if the item is going slower than the medium drag speeds it up. This is just mind blowing that there could be so many posts over a concept that is so simple.

Over-simplification, is a euphemism for something even worse: total ignorance resulting from hubris.

"Drag force", can the result of the air decelerating against the object when it is said to be driving that object (e.g tailwind). That is the basis of Rayleigh's drag formula, derived from Newton's impact theorem, which considers the deceleration of an air molecule against the surface of a mass, which is large compared to the molecular mass; such as a balloon or sail.

"Opposing drag" is the result of the air decelerating against the same object, but from the opposite direction (e.g. headwind). Of course, if the object is not already in motion, that headwind is a "tailwind" but in the other direction.
"Upwind" and "downwind" are only a means of expressing motion relative to that medium. It's the same wind.

If that same object travels in air that is not in motion, then there is neither a tailwind or headwind, but there is opposing drag that is a result of the object accelerating the air ahead of it. ( as a car does). That applies to above WS too, where the cart is said to be faster than the slower air ahead of it.
The difference only lies in the initial velocity of the air relative to the cart.

A consequence of Galilean Relativity, a natural outcome of Newton's laws of motion, means that the drag force on the balloon, is also a drag force on the air, but in the opposite direction as Newton's 3rd Laws says.
There can simply never be a point where one can drag or push the other such that the result is zero.

So, no you don't really need to defend yourself here and now it seems humber has decided to walk away from this issue and declare victory. This has been a fun thread, but it is quickly losing the spark that made it fun.
Thanks for taking your brain teaser to such extremes.

The reason there is no defense is luminous.

 

[Subduction Zone]

humber, I hope you do realize that the same theories that say that a balloon below the speed of the wind will approach the speed of the wind asymptotically and will always be below wind speed also say that a balloon released at a speed above wind speed will approach the speed of the wind from above asymptotically and will always be at a speed above wind speed. Or do physical laws work differently in the humbererse?

 

[humber]

I never said that. You misread a previous post and imagined I had said something of the sort, but that was simply your misunderstanding.

If you have any logical criticisms of what I actually wrote, do please let me know.

But I did. There is no dispute over the momentum, that was your straw man.
And you are trying to claim a two body system, with a third hidden away, and even though the cart, wind and ground are a 3 body system.

The matter concerned KE, and your method produces errors. The same as those made by so many students. Some correct those, some so not.

 

[Subduction Zone]

The balloon achieves a point of dynamic, but not static stability, and that is below WS.

Care to translate this into English? And no, you are wrong.

 

[humber]

humber, I hope you do realize that the same theories that say that a balloon below the speed of the wind will approach the speed of the wind asymptotically and will always be below wind speed also say that a balloon released at a speed above wind speed will approach the speed of the wind from above asymptotically and will always be at a speed above wind speed. Or do physical laws work differently in the humbererse?

No, it says below. I need not bother to explain against simple and blunt denial.
The balloon matter has been settled in my favour, for many reasons, including Spork's failure to support his Aero degree.

 

[humber]

Care to translate this into English? And no, you are wrong.

Oh, you need translation ( help, assistance) to understand what that may mean, but you say I am wrong anyway?
That is all anyone needs to know.

 

[windgrins]

Humber and "absolute" KE

No. You and all the Carters are making the rudimentary error of confusing the observer and reality.
Mentally "stopping" something, and declaring that as fact, is a more egregious error than any Flat Earther ever made.
Two cars moving in opposite directions, so the same speeds, but different velocities of 60mph, remain that way, no matter how they are observed. Failing to apply the correct transform, which may result in one at 120mph and the other at 0mph produces an error, that shows up as the doubling of the KE, and so instantaneously free energy. ( 20000J in the case of the missiles).
The claimed cart is a PM machine, and the TM promotes that, even if you are not aware of that fact.

I believe this quote from Humber sums up his confusion best. He seems to truly believe that Energy of motion is absolute and not wrt to an IRF.

KE is always computed wrt to a chosen IRF and it is different in any other linear translating but valid IRF.

Humber, we can't state this enough and if you do not understand it, there's not much to discuss.

The Kinetic Energy of an object moving in two linear translating reference frames is two different values. KE is a score keeping system for physics. COE is a statement of what to always expect from this score keeping system.

This is why when you riding in your car down the street at 60 kph, from the frame of reference of the driver, the car has no energy wrt to the driver. From the reference frame of the pedestrian you just hit, the car has significant energy.

This is our disconnect and I don't think we'll get past it.

Regards,

WindGrins