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Merged DDWFTTW in a “Real” Wind

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Sloganism.

Evasion.

When two missiles collide head on, the energy of that impact is determined by the KE of one with respect to the other. What the ground happens to be doing beneath them is irrelevant.
 
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.

Are you disputing that?

I have been in a cold sweat since I gave this thread a read today since it seems I would have to agree with Humber on this one exactly as you worded it. I assume I am missing something. Please.
 
How can wind speed ever be measured? Any attempt involves resistance, no? In humber's view, all measurements will be slightly off. The windspeed will always be slightly faster than the measurement's value. No?
 
At first sight, there would be a huge difference between two missiles colliding, each with a speed of 600 mph but in opposite directions, and one missile smashing into another, stationary one at 1200 mph.

Scenario 1: 10 kJ + 10 kJ = 20 kJ for smoke and thunder
Scenario 2: 40 kJ + 0 J = 40 kJ for smoke and thunder

What's missing here is that the missiles aren't instantaneously destroyed - their mass still remains, and has to be accounted for. If we assume that the "smoke and thunder" can be ignored in terms of mass loss, we can treat this as a perfect, inelastic collision, with the merged missile remains stationary in the first scenario, and travelling at 300 mph in the second one. Half the speed, twice the mass = half the KE gives us:

Scenario 1: 10 kJ + 10 kJ -> 0 J (mass) + 20 kJ (smoke and thunder)
Scenario 2: 40 kJ + 0 J -> 20 kJ (mass) + 20 kJ (smoke and thunder)

Treated correctly, there is no contradiction in either frame of reference.
 
I have been in a cold sweat since I gave this thread a read today since it seems I would have to agree with Humber on this one exactly as you worded it. I assume I am missing something. Please.

I'm not clear on what exactly humber said that you feel forced to agree with.

Is it possible that you misread the name of the poster you were quoting (mender, not humber)?
 
Real physics has lots of "sloganisms": F=MA, E=MC^2, d=1/2at^2, etc. Very useful.

Humberphysics has mostly hooliganisms: absolute motion, KE from memory, earth centrism. Good for a laugh.

Well, a car travels over the ground, but if you like, you can say the ground is still and the car moves, or the car is still and the road moves. The TM is the latter.
The conditions and forces acting on the cart are the same, and everything from the tyre pattern to the work done is exactly the same.
FIFY.

The cart is in a latched state, and does not move relative to the belt. The motion that is seen, is motion relative to the ground, or more precisely the supporting plate under the belt.

And here is where obfuscation and internal contradiction turns into lying. If the wheels are turning, there is motion relative to the belt.

Another of those sloganisms, I guess; "if his lips are flapping ..."

WindGrins, back to you.
 
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How can wind speed ever be measured? Any attempt involves resistance, no? In humber's view, all measurements will be slightly off. The windspeed will always be slightly faster than the measurement's value. No?
Hence the HidingBooger Uncertainty Principle.
 
For example, a cart at windspeed is not the same as one held in place by static force to the air (thrust) as the belt moves beneath it.
The prop does provide static thrust, so the air may as well be equated to the wall, and the belt to the table, where the motor does work on the belt, but not the cart. When pulling the table cloth, you do work on the dinner ware ( not just the friction), but when pulling the table, you do work only on the table and friction to the table, but none on the dinnerware.
The TM is a way of dealing with the forces, so it stays in place.

We have a bit of a breakthrough here. Although still totally confused about frames of reference and relative motion, for the first time humber has acknowledged that the propeller produces thrust. Of course, I don't believe he realizes that, although the thrust is static from the frame of reference of the room containing the treadmill, it is dynamic from the frame of reference of the cart.

Unfortunately, he is still hung up on the absurd notion that the earth is the only valid frame of reference. I don't know how he thinks he can make that work for planetery orbital dynamics, or even for events occurring on opposite sides of the eart.
 
I'm not clear on what exactly humber said that you feel forced to agree with.

It was this statement Mender made:

"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"

Probably it is just my reading of it which is off but I do not see how the above statement can be true. ArmillarySphere post sums up my understanding correctly but that does not seem to be the same as Mender's statement about the same effect on the objects.
 
It was this statement Mender made:

"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"

Probably it is just my reading of it which is off but I do not see how the above statement can be true. ArmillarySphere post sums up my understanding correctly but that does not seem to be the same as Mender's statement about the same effect on the objects.

In what way do you see a difference?
 
You contradicted yourself on balloons running at WS and now are attempting a back pedal when this was pointed out.
Again, the relevant posts:



humber said:
.analysis and physical explanations that say the balloon will not reach WS.
Correct. It won't.

humber said:
a helium balloon ...moves at the speed of the wind in the direction of the wind. This is what makes ballooning so fascinating: there is no wind blowing on you and no wind noise.
If there is a dog in that basket. Does it wag its tail?

The perceived velocity depends on the position within the slip stream.
Colloquially, the air over the top an airfoil, moves faster than the air on the underside. The difference in pressure thus created, is the Bernoulli's explanation of lift. That difference of velocity is a result of the shape of the airfoil; its 3 dimensional physical shape. It is not possible at any speed including WS, for the speed of the air over the top of the wing, to be the same as the bottom of the wing.
If the air in the basket is close to the free stream velocity, then the air over the balloon itself may be faster or slower.
 
if we assume that the "smoke and thunder" can be ignored in terms of mass loss, we can treat this as a perfect, inelastic collision, with the merged missile remains stationary in the first scenario, and traveling at 600 mph in the second one.
fify.
 
In what way do you see a difference?

Again, I assume I am just in over my head but from the frame of reference of one of the objects, which is what I assume you mean when you say it "will produce exactly the same effect on the objects when the objects collide", ArmillarySphere's post shows that the 600+600 effect would be rather different than the 1200+0 does it not?
 
Spork,

You are clearly correct sir. My statement only stands correct if the objects are moving along the same single dimension. Your comment is a good clarification as it lends insight. I believe this is even why the idea of Velocity Made Good (VMG) was created. My reply was in to a Humber question (sort of a question):
It is not a matter of dimension. Velocity is both direction and speed. Both are independent.
To say that two cars in the opposite direction at 60 mph have and "instantaneous" velocity of zero is a folksy error. If the difference is 120mph when they are 10miles apart, and 1mile apart. The 120mph, never changes at any time. That is because direction is relative, independently of speed.

So, a bird traveling at 1mph relative to the plane, but the other way.
Can then be simulated by turning the fan off, and throwing the bird at the nose of the plane at 1mph? Or is there a difference between closing speed and relative speed? Won't the total KE be different in the air, than when the fan is off, even if they were traveling the same way?


So I inferred from this question a "collision course" of bird and plane but some of us had a problem understanding what was being asked.

You're comment adds additional insight. Thanks,
That is a sophism to show your error. An oil tanker hitting a boat is not the same as a boat hitting a tanker. Total energy counts.
Drop a ball from a height of 1m onto a sheet of paper, stretched taught by a light frame. The ball will tear through the paper. Drop the paper on to the stationary ball so it strikes it at the same speed as in the previous example, and maybe it won't. There must be enough energy in the paper and frame, to break the molecular bonds of the paper. The ball may have hundreds more times that, and the paper, not enough.
 
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I have been in a cold sweat since I gave this thread a read today since it seems I would have to agree with Humber on this one exactly as you worded it. I assume I am missing something. Please.

Maybe it is because humber can't do simple addition:
The problem is that a closing speed of 1200mph, can't be mimicked by having one object stationary and the other moving at 600mph.
But it can and is mimicked by having one object stationary and the other moving at 1200 mph.

Or any other combination of speeds that add up to 1200 mph. Motion is relative, so all you're changing is the speed of the observer relative to the objects.
 
Again, I assume I am just in over my head but from the frame of reference of one of the objects, which is what I assume you mean when you say it "will produce exactly the same effect on the objects when the objects collide", ArmillarySphere's post shows that the 600+600 effect would be rather different than the 1200+0 does it not?

ArmillarySphere was explaining it from the reference frame of the ground. In one case, you have smoke and debris that in total is not moving wrt the ground and in the other it is moving. From the frame of either missile the collisions are the same (ignoring the effects of different winds).
 
Again, I assume I am just in over my head but from the frame of reference of one of the objects, which is what I assume you mean when you say it "will produce exactly the same effect on the objects when the objects collide", ArmillarySphere's post shows that the 600+600 effect would be rather different than the 1200+0 does it not?
When they hit, they change speed to the average of the two initial speeds. F=MA, so the force needed to change the speed by 600 mph is the same whether one is stationary and the other at 1200 mph, or one is at 18,000 mph and the other at 19,200 mph because the speed change (A) is the same.

Remember that KE, like motion, is relative and depends on where you're standing when you take your measurements.
 
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Evasion.

When two missiles collide head on, the energy of that impact is determined by the KE of one with respect to the other. What the ground happens to be doing beneath them is irrelevant.

Not at all. When both missiles travel at V, and have 10000J each, the sum is 20,000J. The amount of damage is related to the amount of energy transferred from one to the other. If the collision were perfectly elastic, the maximum transfer in the above case is 10000J. If may be that the transferred energy becomes radiated as heat, so a maximum of 20000J, or less if some remains in the wreckage.

When one is at 0m/s and the other at 2V, one has 0J and the other 40000J.
Maximum possible transfer, 40000J, and the same for the heat and what may remain in the wreckage. KE is not relative in a simple way.
 
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