Merged DDWFTTW in a “Real” Wind

[humber]

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?

They are. Think about a cup anemometer. The wind drives it, because the drag on the open side is more than the drag on the smooth side. It is never possible for them to completely stop if there is airflow.* The wind at WS is not like the still air of a room.
The instruments are calibrated, or adjustment made by calculation.

* They can stall as a result of the lift/drag ratio, but that is another case.
Applies to wind vanes too. They can "stick" as a result of the L/D ratio, but that tends to be for a short time, and the result is less visible motion, rather than remaining fixed in position.

 

[3point14]

If you're in the middle of deep space, with nothing around but the two converging missiles, how do you work out how quickly each one is going?

 

[mender]

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.

Yep, go ahead and ignore that the correct answer has already been given and accounts for the speed of the objects after the collision.

Try and stay with us here.

KE is not relative in a simple way.

So now we'll get a lesson in "Special KE Relativity" by Prof. Booger.

 

[3point14]

What's the smallest largest thing that doesn't move slower then the wind at wind speed, Humber?

 

[Modified]

If you're in the middle of deep space, with nothing around but the two converging missiles, how do you work out how quickly each one is going?

Humber will ignore that question. He only does surface-relative measurements.

 

[humber]

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.

The total is larger, and perhaps the heat of collision will vaporizer all the mass, so the result is a cloud hot gas containing 40000J, or the other way, no smoke and thunder, but two crumpled missiles, one with with 0J and the other. 40000J. That can't happen when both have only 10000J each.
Happens in car accidents, dependent on relative mass, impact time and the amount absorbed by the "crumple zones". One or both cars comes to a near instantaneous stop, or one does and the other bounces off and travels, or both continue down the road for a while.

 

[Modified]

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.

So if a missile going 1000kph wrt the ground strikes a missile going 1001kph wrt the ground in the same direction, the collision will be 2 million times stronger than a 0kph missile hitting a 1kph missile? By your logic it will be.

KE is not relative in a simple way.

Yes, it is. Please read an intro physics text and do the exercises.

 

[jduffy77]

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.

Of course. I am riding one of the objects. You are saying it would be just as well to be on the 600 hitting 600 as the 1200 hitting 0?

 

[Modified]

Of course. I am riding one of the objects. You are saying it would be just as well to be on the 600 hitting 600 as the 1200 hitting 0?

Assuming everything else is the same (air speed relative to the two missiles in both cases), and we don't include what happens when you hit the ground, if there's anything left of you, then yes.

 

[CORed]

Correct. It won't.


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.

A balloon is an airfoil?
Classic humber. Here's another hint humber: Even with an airfoil, if the velocity of the airfoil with reapect to the air is zero i. e. if the airfoil is at wind speed, the velocity of the air above and below the airfoil is zero.

BTW, if you turn a real airfoil (not a balloon) upside down do you think the air still flows faster over the top?

If an object in the wind is symmetrical about the axis of the wind, is the velocity over the top and the bottom still different?

 

[Dr. Keith]

Answer to your claim by addressing the balloon and bird and the tablecloth.

Maybe we could just stick to the balloon. I like birds, and tablecloths always are good for parlor games, but let's stick to the balloon until we conquer that one. Then I will happily move on to other games.

There are lots of questions out there that you are adroitly avoiding. Care to directly answer any of them?

1. How does a balloon* traveling at greater than windspeed** end up traveling at less than windspeed?

2. How do you calculate the percentage of windspeed at which a balloon will travel?

3a. What is the smallest largest particle that will travel at windspeed?

3b. What is the smallest particle that won't travel at windspeed?

For these questions I think it is easier to define these terms just once and agree to use them as defined throughout. I propose the following definitions to keep the conversation on track:

* Balloon will mean a neutrally buoyant sphere to keep the maths simple. Not a hot air balloon, not a weather balloon, a neutrally buoyant sphere.

*Windspeed (WS) will mean the constant velocity that all air is moving relative to the ground. Assume that the velocity is even over all relevant altitudes, and throughout the duration of the experiment. No gusts, no gradients, just a smooth mass of air moving over the ground at WS. Then add a balloon and tell us what happens.

 

[jduffy77]

3. What is the smallest particle that will travel at windspeed?

What a great question for Humber this is. Though I do feel I should warn you that Humber has already stated that even the wind doesn't travel at WS. Maybe someone else remembers the details?

 

[Dr. Keith]

What a great question for Humber this is.

I stole it from someone else in the thread.

Edit: From 3point14 to be exact and now he has gone back to correct it!! I will edit mine to match.

Though I do feel I should warn you that Humber has already stated that even the wind doesn't travel at WS. Maybe someone else remembers the details?

It scares me that I am not surprised. Let me guess, his explanation involved RMS?

 

[humber]

Real physics has lots of "sloganisms": F=MA, E=MC^2, d=1/2at^2, etc. Very useful.

The second one appears on t-shirts and in advertising, but has nothing to do with either. "KE is relative" is on TAD's t-shirt.
And it's s = ut + 1/2at^2. Sometimes u is zero, so dropped. That does not imply u is "absolute".

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

Houyhnhnms are a race of intelligent horses from Swift's Gulliver's Travels of 1727. Yahoo's, are a race of brutish men from the same satirical novel.
Hooligan, is of unknown origin, but first appeared in print in 1898.
The air lever is unknown, but first appeared in advertising produced by Yahoos, circa 2010.

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.

From Gullible's Troubles. Humber, 2011.

Gulliver is a giant "observing" the Earth from The Heavens. He sees the surface rotate at 1000mph to the right. On that surface is a car traveling at 20mph to the right. Gulliver sees the car move at 1020 mph to the right.

However, his Yahoo cousin, Gullible, does a "thought 'speriment" and mentally keeps the car a 0mph wrt him, and declares "' 'tis the same as the Earth itself, moving at 1020mph the other way"

Gulliver decides to show his cousin the error of his ways, not by thought, but by keeping the cart at rest relative to him, by holding it between his giant fingers. Gullible now sees the Earth move 1000mph to the right under the car, because that car is stationary, and the Earth is still moving.

Gullible makes objections concerning the idea of absolute motion which he had heard about from the Houyhnhnms.

Gulliver takes on final step. The car is held stationary by his fingers, so is not going to go anywhere, so he puts it in his pocket. He then steps on to the planet and moves with the surface, and so at rest relative to it. He removes the car from his pocket and finds the car is also stationary.

ETA:
The belt is moving and the cart is not, because a giant hand holds it in place as it spins up. That hand provides a static force to the ground. When released, that static force is replaced by static force (thrust) against the air. How it does that, is a detail. It can move, but not "up the belt", but relative to the ground, or the air, dependent upon circumstance. Looks much the same.

 

[ArmillarySphere]

fify.

Ah, yes. Should teach me not to short-circuit calculations, especially when I'm trying to compute things.

For the benefit of the lurkers, I arrived at the resulting velocity by adding up the momentum of all the objects (mass * velocity) and dividing the result by the mass. If the result is that the missiles fragment this gets very complex, but the final sum must be the same both before and afterwards. You can do this calculation in whatever frame is most convenient, and translate afterwards. For instance, when looking at a missile striking a stationary object it may make sense to pick a frame where the sum total momentum is zero. Then add the ground velocity on afterwards.

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).

It would work in any frame of reference you choose. Scenario 1 would be the "ground" (actually any one where the missiles have the same speed) and scenario 2 is from the point of view of one of the missiles. Either way, you get the same amount of bang for your buck.

Unless you're in the humberverse, which must be a very strange place...

 

[humber]

Ah, yes. Should teach me not to short-circuit calculations, especially when I'm trying to compute things.

For the benefit of the lurkers, I arrived at the resulting velocity by adding up the momentum of all the objects (mass * velocity) and dividing the result by the mass. If the result is that the missiles fragment this gets very complex, but the final sum must be the same both before and afterwards. You can do this calculation in whatever frame is most convenient, and translate afterwards. For instance, when looking at a missile striking a stationary object it may make sense to pick a frame where the sum total momentum is zero. Then add the ground velocity on afterwards.



It would work in any frame of reference you choose. Scenario 1 would be the "ground" (actually any one where the missiles have the same speed) and scenario 2 is from the point of view of one of the missiles. Either way, you get the same amount of bang for your buck.

Unless you're in the humberverse, which must be a very strange place...

Tell that to the truck drivers who deliver the amount of fuel necessary to get two missiles to V, (20000J) and one to 2V (40000J)

 

[humber]

So if a missile going 1000kph wrt the ground strikes a missile going 1001kph wrt the ground in the same direction, the collision will be 2 million times stronger than a 0kph missile hitting a 1kph missile? By your logic it will be.

Yes, it is. Please read an intro physics text and do the exercises.

Velocity is not speed. Direction is not speed, but another parameter.
When traveling in the same direction, that parameter is the same, but not so when traveling in opposite directions.

Many students are confused by the fact that momentum is conserved but Kinetic Energy isn't. Another is adding KE's together.
This is a common matter, and over the Centuries, teachers have honed means of showing the student where the misunderstanding lies.

Latterly, many have turned to the internet, and taken advice from Yahoos.
The physics forum will alert you to how common your misunderstanding is, Modified.

 

[jduffy77]

Assuming everything else is the same (air speed relative to the two missiles in both cases), and we don't include what happens when you hit the ground, if there's anything left of you, then yes.

I knew I has read something recently about collisions and I found it. From our buddy Rhett who was wrong about ddwfttw.

http://scienceblogs.com/dotphysics/2010/05/mythbusters_and_double_the_spe.php

So he is wrong about this as well.

 

[Llyricist]

Tell that to the truck drivers who deliver the amount of fuel necessary to get two missiles to V, (20000J) and one to 2V (40000J)

Oh noes, not the mass less fuel again.

 

[humber]

Maybe we could just stick to the balloon. I like birds, and tablecloths always are good for parlor games, but let's stick to the balloon until we conquer that one. Then I will happily move on to other games.

There are lots of questions out there that you are adroitly avoiding. Care to directly answer any of them?

1. How does a balloon* traveling at greater than windspeed** end up traveling at less than windspeed?

2. How do you calculate the percentage of windspeed at which a balloon will travel?

3a. What is the smallest largest particle that will travel at windspeed?

3b. What is the smallest particle that won't travel at windspeed?

For these questions I think it is easier to define these terms just once and agree to use them as defined throughout. I propose the following definitions to keep the conversation on track:

* Balloon will mean a neutrally buoyant sphere to keep the maths simple. Not a hot air balloon, not a weather balloon, a neutrally buoyant sphere.

*Windspeed (WS) will mean the constant velocity that all air is moving relative to the ground. Assume that the velocity is even over all relevant altitudes, and throughout the duration of the experiment. No gusts, no gradients, just a smooth mass of air moving over the ground at WS. Then add a balloon and tell us what happens.

Nothing travels at the speed of the medium.