Flight Question

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Re: Flight Question

#41  Postby Thommo » Apr 14, 2015 3:00 am

Jerome Da Gnome wrote:Am I too dumb to understand, or are you too dumb to explain?


In your last post you appear to confuse "angle" with "angular momentum", which is what 8th-9th grade stuff?

As I suggested you seem to be lacking some fairly basic and key education that might allow you to access the answers you've been given, but you aren't displaying much of a willingness to learn, so I'm not super keen to spend a lot of time on it, not being a high school physics teacher.

In simple terms there's a fundamental concept in physics called an "inertial frame" - which is simply a frame of reference (i.e. a co-ordinate system) in which there are no "fictitious" forces such as centrifugal force. This means that basic laws of physics hold true in any such frame.

Now, in such a frame there are certain important physical quantities such as momentum (given by p=mv) and angular momentum (given by L=r X p, where X denotes the "cross product" not multiplication) that are preserved. The frame centred on the Earth and rotating with it is not an inertial frame because the Earth is rotating, which leads to measurable "fictitious" forces like centrifugal force, which means you can measure things like the Coriolis effect (e.g. deflection of ballistic projectiles at latitude).

What this means is that any outward radial motion from the centre of the Earth (i.e. jumping upwards by exertion of a single force in the "upwards" or radial direction) is going to lead to a reduction in angular speed due to conservation of angular momentum.

However, the size of this change is small and the Earth drags its atmosphere along with it, so as you gain altitude you will be buffeted by winds which will change your angular momentum and give you approximately the same angular speed - i.e. you will continue to go pretty much "straight up" as long as you're inside the atmosphere. After that the effect will be more and more noticeable and the Earth will appear to rotate more and more under you as you get further and further away.

The answer is thus that there is no single point at which the system discretely changes from you rotating with the Earth so the point of departure is "straight down" to the point of departure rotating under you. Instead there's a continuous change in relative velocity that will start to become apparent as the atmosphere thins.

This is very easily seen by simply learning the definitions of angular speed and angular momentum and noting that angular speed is a decreasing function of radius when angular momentum is a fixed value.
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Re: Flight Question

#42  Postby Jerome Da Gnome » Apr 14, 2015 3:28 am

felltoearth wrote:
Jerome Da Gnome wrote:I am trying to sort the reason Baumgartner landed so close to where he took off.


He was directly above his landing spot when he jumped?


There about, yes. Between the ascent and decent he ended up really close.
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Re: Flight Question

#43  Postby Jerome Da Gnome » Apr 14, 2015 3:29 am

Onyx8 wrote:
Jerome Da Gnome wrote:
Onyx8 wrote:No, a person on the surface of the earth has X, Y, and Z velocities. You need to start there.


Right, you right now are traveling in a specific direction and a specific speed, somewhere around 1,000 mph iirc.



In what frame of reference?


Relative to a random point in space.
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Re: Flight Question

#44  Postby Jerome Da Gnome » Apr 14, 2015 3:32 am

Macdoc wrote:
So if you dropped a rock ....then it goes straight down from the frame of reference of the balloon.
So did our intrepid parachutist. He has direction control anyways and did not take him very long to fall/sky dive.


You begin at point X on the surface of the earth, rise the the edge of the atmosphere, and drop a rock. You will stay above that moving point by what physical law?
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Re: Flight Question

#45  Postby Onyx8 » Apr 14, 2015 4:43 am

Jerome Da Gnome wrote:
Onyx8 wrote:
Jerome Da Gnome wrote:
Onyx8 wrote:No, a person on the surface of the earth has X, Y, and Z velocities. You need to start there.


Right, you right now are traveling in a specific direction and a specific speed, somewhere around 1,000 mph iirc.



In what frame of reference?


Relative to a random point in space.



Thats not an answer to the question. Do you not understand the question? It's best if you let people know.

In what frame of reference is this 'random point in space'?
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Re: Flight Question

#46  Postby Cito di Pense » Apr 14, 2015 4:53 am

Onyx8 wrote:Thats not an answer to the question. Do you not understand the question? It's best if you let people know.

In what frame of reference is this 'random point in space'?


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Re: Flight Question

#47  Postby Cito di Pense » Apr 14, 2015 4:58 am

Jerome Da Gnome wrote:
Macdoc wrote:
So if you dropped a rock ....then it goes straight down from the frame of reference of the balloon.
So did our intrepid parachutist. He has direction control anyways and did not take him very long to fall/sky dive.


You begin at point X on the surface of the earth, rise the the edge of the atmosphere, and drop a rock. You will stay above that moving point by what physical law?


Are we still using a balloon to get there? Then we're moving with the atmosphere, however it happens to be moving. To first order, it's stuck to the surface of the rotating planet, with some noise thrown in from solar heating, Coriolis forces, and viscous dissipation. I won't write down all the second order effects, because you may not find them interesting.

You won't necessarily stay above the same point, but may drift a little, in the wind.
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Re: Flight Question

#48  Postby epepke » Apr 14, 2015 4:59 am

Jerome Da Gnome wrote:Am I too dumb to understand, or are you too dumb to explain?


The former. Glad to be of service.
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Re: Flight Question

#49  Postby Macdoc » Apr 14, 2015 8:20 am

I think that would be the concensus. :coffee:
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Re: Flight Question

#50  Postby Cito di Pense » Apr 14, 2015 8:23 am

Thommo wrote:
Jerome Da Gnome wrote:
Cito di Pense wrote:

Jerome, the man in the balloon cannot travel outside the earth's atmosphere.


You are hiding from the fundamental question.

In a gravity defining mechanism.

At what point does a man in a gravity defying mechanism lose the movement momentum of the earth?


Never. Momentum and angular momentum are conserved quantities.


For a balloonist, pretty much, yes, although there are little, tiny buoyant and viscous drag forces to disturb things in a negligible way. Don't let Jerome pilot your aircraft. You'd be better off with Lubitz. It won't drag on as long. With Jerome, you'll just exhaust your fuel supply, and have to ditch. Avoid a stall, and you still have half a chance. Take over the stick from Jerome.
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Re: Flight Question

#51  Postby Weaver » Apr 14, 2015 12:37 pm

Jerome Da Gnome wrote:How high do you have to go to experience the earth rotate under you?

It depends quite a bit on other factors - speed of travel, and direction of travel.

When I used to compute firing data for indirect artillery, we had to account for the rotation of the Earth - a shell fired under otherwise identical conditions would have differing impact ranges depending on whether the direction of fire was East, West, or somewhere in between.

While it is a negligible effect for short-duration flights, firing high-angle with inherently extended flight times would result in notable differences in firing solutions.

Of course, this is for a ballistic projectile - under powered flight, with the inherent differences of continuous control, one would not "experience" the motion as distinctly - the rotation of the Earth is incorporated in all of the other course corrections as a final sum vector, and is not routinely isolated as an independent error factor.
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Re: Flight Question

#52  Postby Thommo » Apr 14, 2015 3:33 pm

Applying some of the above to Felix Baumgartner:-

Earth's radius ~ 3,959 miles.
Felix jumped ~ 24 miles.

Thus Felix started his jump at 3,983 miles.

Speed in mph of a point 3,959 miles above the axis of rotation = 2 x π x 3959 / 24 = 1036.46mph
Speed in mph of a point 3,983 miles above the axis of rotation = 2 x π x 3983 / 24 = 1042.75mph

1042.75 - 1036.46 = 6.29mph

That is to say that the maximum deflection caused to Baumgartner by the Earth's rotation is equivalent to a light breeze (as measured by the Beaufort scale), for the 4mins19secs he was in free fall it would deflect him by up to 722m from his target. This is nothing compared to the deflection from air speeds of up to 850mph encountered during his fall.
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Re: Flight Question

#53  Postby Spearthrower » Jul 08, 2015 1:43 pm

Jerome Da Gnome wrote:Am I too dumb to understand, or are you too dumb to explain?



I'll give you 1 guess.

No, let's be kind - make that 2 guesses! :naughty2:
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