PDA

View Full Version : Physics of falling objects


compressor stall
10th Mar 2014, 01:37
An argument developed off topic in the Malaysian thread about the physics of falling objects. It was off topic so was mostly deleted.

I and one other were taken to task by Skipness One Echo and others for stating that objects of the same size but different densities (eg a basketball and an identical one filled with concrete) would not hit the ground at the same time, dropped from say 35000 feet in an atmosphere.

Discuss.

dubbleyew eight
10th Mar 2014, 01:47
there was an experiment many years ago where objects were dropped off the Eiffel tower to explore the effect. you are wrong.

the fall is due to gravity (whatever that is)
the gravitational attraction acts on every particle of the two balls at the same rate of influence.
so it doesn't matter whether you have a ball full of nothing or a ball full of concrete, every particle of both balls is under the same influence.
you don't get any more influence exerted on the concrete ball.

CaptainEmad
10th Mar 2014, 02:04
It has got to be a wind-up.

In a vacuum, all objects free fall due to gravity at the same rate of acceleration, regardless of their mass or density.

In a vacuum.


Objects dropped on earth, in the atmosphere, behave differently.


Ever see the feather and hammer dropped on the moon?
Feather & Hammer Drop on Moon - YouTube (http://youtu.be/5C5_dOEyAfk)

Lord Spandex Masher
10th Mar 2014, 02:13
Actually differing masses will accelerate under the sole force of gravity at exactly the same rate. The more massive object will only accelerate faster if there is reasonable air resistance.

Every kilogram of mass is acted upon by gravity (9.8N/kg on our planet). This is 'f'.

F/m=a

10kg(m) x 10 (9.8 simplified) = 100N/10kg(m) = 10m/s/s

100kg x 10 = 1000N/100kg = 10m/s/s

Factor in enough air resistance and time and the more massive object will hit the ground first. Would it in the OP's case? Dunno. Not enough info.

EEngr
10th Mar 2014, 02:15
objects were dropped off the Eiffel towerIs it possible to drop anything off the Eiffel tower without having it bounce off the side a couple of times?

http://aparisguide.com/eiffeltower/eiffel-tower1.jpg

Now the Tower of Pisa I could see working. But that could be too short to measure the effect of drag.

The formula for the terminal velocity of a falling object depends on its density (mass for identically sized objects).

Terminal velocity - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Terminal_velocity)

So I would expect that, given sufficient altitude to allow terminal velocity to be approached and enough time for that difference in velocity to manifest itself as a noticeable difference in descent time, that difference would become apparant.

compressor stall
10th Mar 2014, 02:44
...
you are wrong
....

the gravitational attraction acts on every particle of the two balls at the same rate of influence.
so it doesn't matter whether you have a ball full of nothing or a ball full of concrete, every particle of both balls is under the same influence.
you don't get any more influence exerted on the concrete ball.


So a balloon filled with air dropped from the Eiffel Tower or even my garage bench would fall at the same rate as a balloon filled with concrete?

awblain
10th Mar 2014, 02:45
The denser ball will always hit the ground faster, in air, as it has a lower ratio of drag to weight.

Galileo is reported to have used different-sized cannonballs dropped from the tower of Pisa to show that overall mass, popularly thought to determine rate of descent, did not affect the outcome. (Although the bigger cannonball will fall a little faster owing to: a lower ratio of drag to weight.)

Tinstaafl
10th Mar 2014, 02:56
Provided the both objects are identical except for mass, they will fall at the same rate. Ignoring drag from the air for a moment, eg in a vacuum, gravity acts to accelerate each particle at the same time.

Consider this thought experiment: Three identical masses, dropped at exactly the same moment. Each will accelerate the same as the other two. Now connect two of the masses by a loose piece of string that is so small that its mass & surface area are negligible (or even non-existent if you posit a thread small enough). Do you think the two connected masses will suddenly behave differently to the independent mass? Now imagine the thread replaced by a rod so now the two masses are locked together. Do you think the connected masses will suddenly change behaviour? If it helps, imagine the connector suddenly appears mid-fall. Now imagine, instead of three identical masses with two of them connected together, two identical masses differing only in density. Gravity's effect is the same, regardless of size or shape.


Now add a fluid such as air. The only effect the fluid has is to provide a retarding force - 'drag' - reducing the acceleration of gravity. Drag is not affected by mass. C=Cd 1/2 rho v^2 s remember. So, as long as the objects *only differ in density* they will fall at the same rate.

awblain
10th Mar 2014, 02:59
Drag is affected by mass.

For the same density drag rises as size squared, while mass rises as size cubed.

Added - if shape is the same.

Mass is a good choice of a dependent variable to link that.

Don't forget Tinstaafl that while Cd does have that dependence, the drag force derived from it has an area term too.
If objects only differ in density, and have the same size and shape, they will definitely not fall at the same rate.

pattern_is_full
10th Mar 2014, 03:44
Is it possible to drop anything off the Eiffel tower without having it bounce off the side a couple of times?

Yes, actually. The first elevated floor is a square "doughnut" with a hole in the center overlooking the plaza underneath. Easy to drop something 190 feet from there without it hitting anything before the ground (there is a glass wall, previously a low railing, to prevent amateur gravity experiments. ;) )

File:Sous la Tour Eiffel 1.jpg - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/File:Sous_la_Tour_Eiffel_1.jpg)

Drag is affected by mass.

For the same density, drag rises as size squared, while mass rises as size cubed.

Uh, no.

Drag is affected by size (among other things) and Mass is affected by size (among other things).

But there is no direct correllation between drag and mass (or vice versa) (with the exception where one introduces a wing into the equation, in which case one can say that induced drag (from the wing's lifting effort) is affected by mass.

That has nothing to do with form drag and air resistance in a free fall, however.

You can have high-mass objects with low drag (Saturn/Apollo moon rocket, for example) and low-mass objects with high drag (the average mattress, for example). And objects of equal mass and even equal density, with very different drag (a flat plate of aluminum vs. a needle-shaped piece of aluminum, both massing 5 kilos).

ManUtd1999
10th Mar 2014, 05:22
But there is no direct correllation between drag and mass (or vice versa) (with the exception where one introduces a wing into the equation, in which case one can say that induced drag (from the wing's lifting effort) is affected by mass.

Induced drag depends on lift, not mass. If you put two wings in a wind-tunnel at a fixed AoA and filled one with concrete the induced drag would be the same in each. For aircraft though more mass means more lift is required and induced drag increases.

Brian Abraham
10th Mar 2014, 05:29
objects of the same size but different densities (eg a basketball and an identical one filled with concrete) would not hit the ground at the same timeExactly...

A Squared
10th Mar 2014, 06:14
An argument developed off topic in the Malaysian thread about the physics of falling objects. It was off topic so was mostly deleted.

I and one other were taken to task by Skipness One Echo and others for stating that objects of the same size but different densities (eg a basketball and an identical one filled with concrete) would not hit the ground at the same time, dropped from say 35000 feet in an atmosphere.

Discuss.

Unless you're misrepresenting the discussion (and I have n reason to believe that you are, just covering that possibility) Skippiness 1 E hasn't a clue what he's talking about.

A Squared
10th Mar 2014, 06:58
there was an experiment many years ago where objects were dropped off the Eiffel tower to explore the effect. you are wrong.

Well, no, he's not wrong. I don't know about the Eiffel tower experiment, but Galileo is famously (and perhaps apocryphally) credited with doing the same thing at the Tower of Pisa. Some accounts have the balls falling at the same speed, other accounts have the heavier ball falling perceptibly faster.

In either case, it doesn't matter, the experiment was to test the (then) current theory of "Natural Motion" concocted by Aristotle which claimed that the speed of a falling object is proportional to it's mass. And even if the denser ball fell slightly faster, it was still obvious that the difference between the two was not proportional to the respective mass differences.



you don't get any more influence exerted on the concrete ball.

well, yeah, actually you do. The force of gravity on an object is proportional to it's mass. The basketball filled with air is about .6kg while the one filled with concrete will have a mass of 17 kg. So the force of gravity (in round numbers) near the earth's surface will be 6 newtons on the air ball, and 170 newtons on the concrete ball.

In the absence of any other force, the acceleration will be identical (because the force is proportional to the mass) but the balls are falling thru the atmosphere, not a vacuum. There is also drag to consider.

OK, forget weight for a moment. you have two objects moving through the same air. they have identical drag coefficients and identical frontal areas, and identical wetted area. Aerodynamically, they are identical. One has 6 newtons of force propelling it thru the air, and the other has 170 newtons of force propelling it through the air.

Which will travel faster thru the air?

It should be obvious that the object with 170 newtons propulsive force will travel faster than the one with 6 newtons.

It's the same with falling objects. As the balls fall faster, the drag begins to predominate over inertia, and more and more of the force of gravity is counteracted by the drag and less of it accelerates the balls. And the ball that has the least force will accelerate less than the ball with more force.

Ultimately both balls will reach their terminal velocity*, and the terminal velocity of the concrete ball will be much higher than the terminal velocity of the air filled ball.


*Dangerous territory when dealing with a poor understanding of physics. There is a popular misconception that there is some single "terminal velocity" for any and all falling objects This ain't true. Not even close. Every object has it's own terminal velocity determined by it's density and aerodynamic properties. A feather has a very low terminal velocity. Doesn't matter how high you let it fall, it ain't falling any faster. A high penetration aerial bomb on the other hand has a very high terminal velocity. There are bombs which will exceed the speed of sound while in a free fall. This is because they have a lot of mass and have a very low drag coefficient. (lots of gravitational force vs. not much drag force) The terminal velocity of a skydiver in free fall is somewhere between a feather and a bunker buster bomb. In fact, different individual skydivers will have different terminal velocities depending on their technique.

A Squared
10th Mar 2014, 07:00
So a balloon filled with air dropped from the Eiffel Tower or even my garage bench would fall at the same rate as a balloon filled with concrete?

That's a good illustration of the complete absurdity of the claim. Frankly I'm astonished that we're even having this discussion on a board of professional pilots.

mickjoebill
10th Mar 2014, 07:11
This discussion started on the ML thread when there was disagreement about the duration of free fall wreckage from 35000ft, with one maths man saying it would only take 40 seconds!

Real world examples differ from pure maths..
The Space Shuttle Challenger cockpit detached at 45,000 feet continued to 65,000 feet then free fell, slowly spinning for 2 min 25 seconds before it hit the water.
Challenger 1 (http://www.space-shuttle.com/challenger1.htm)

Halo skydivers jump from 35000ft, it takes around 2 minutes to deploy their chute at between 3000 to 5000 feet.

Gopros weight around 100 grams and when not clad in a waterproof housing, are the size of a matchbox.

Unverified go pro footage, falling 2 minutes from 12500 feet, it flutters and rotates at around 7 revs per second.
gopro freefall from 12500 without case - YouTube

Unverified go pro footage falling for 1 minute 40 seconds from around 10,500 feet,
gopro falls 11000 feet and sourvives (3352 meters) - YouTube

Unverified go pro footage, falling for 22 seconds from 2200 feet.
GoPro Hero 2 - FALLS FROM 2200 FEET (landing gear mount fail) - YouTube

Unverified go pro footage, 27 seconds to fall 3000 feet.
GoPro Camera Falls From 3,000ft - YouTube

A Squared
10th Mar 2014, 07:12
Now add a fluid such as air. The only effect the fluid has is to provide a retarding force - 'drag' - reducing the acceleration of gravity. Drag is not affected by mass. Cd=1/2 rho v^2 s remember. So, as long as the objects *only differ in density* they will fall at the same rate.

It's true as far as it goes. Drag is not affected by mass, but the *force* of gravity certainly is; in fact it's proportional to mass.

And for objects of different masses, which are otherwise externally identical, the *force of gravity is higher for the object with greater mass, so the *force* of gravity will reach equilibrium with the *force* of drag, at a higher airspeed. And when the *force* of gravity reaches equilibrium with the *force* of drag, the object will no longer accelerate. And like I said that, occurs at a higher airspeed for the more dense object (assuming same volume) .

cwatters
10th Mar 2014, 07:41
I and one other were taken to task by Skipness One Echo and others for stating that objects of the same size but different densities (eg a basketball and an identical one filled with concrete) would not hit the ground at the same time, dropped from say 35000 feet in an atmosphere.

You are correct. When they reach terminal velocity Drag = weight. If the shape is the same the only way for drag to be different is if the velocity is different.

They would be correct on the moon..
https://www.youtube.com/watch?v=5C5_dOEyAfk

EDIT: should be easy to prove by dropping a table tennis ball and something the same size like an egg...or perhaps not an egg as that's streamlined :-)

A Squared
10th Mar 2014, 07:45
They would be correct on the moon..
https://www.youtube.com/watch?v=5C5_dOEyAfk

When you're faking the footage on a soundstage in Burbank, you can make anything fall at any speed you want. :}

awblain
10th Mar 2014, 08:10
pattern…

Quote:
Drag is affected by mass.

For the same density, drag rises as size squared, while mass rises as size cubed.
Uh, no.

Drag is affected by size (among other things) and Mass is affected by size (among other things).

OK, but we were talking about objects of the same shape with the concrete and air basketballs.

To be more correct:

For the same density, speed and shape, drag rises as size squared, while mass rises as size cubed.

Piltdown Man
10th Mar 2014, 08:19
I believe the mathematical models which the OP may be trying to grasp are the 'wind-drift' models developed by the AAIB, initially developed to try and locate a missing Comet in the mid-1950's. Assuming an in-flight break-up, this methodology attempts to predict where the wreckage will land.

PM

awblain
10th Mar 2014, 08:38
Isn't building an evacuated sound stage about as much trouble as flying to the Moon?

Jumpjim
10th Mar 2014, 08:43
When we are skydiving it is regular practise to wear lead weight vests to increase your freefall speed vs. your standard speed, particularly for lighter weight jumpers jumping with their heavier weight colleagues.

Must be something in it cos I can guarantee it works!

awblain
10th Mar 2014, 08:50
Mick,

free fell

It never free fell, in the technical sense, as it was suffering drag throughout.
A free-fall parachutist is also only briefly in physical (rather than semantic) free fall as they leave the aircraft door; as soon as drag builds, the weightless and acceleration go.

Vomit comets also don't free fall, they are flown precisely so that thrust balances drag so the zero g is imposed by the aerodynamic forces.

The test of the free-falliness of a gopro is that a real dragless free fall covers a height that's the square of time spent falling, while at a terminal velocity it's a linear relationship. 120s from 12000 feet, 100s from 10500 feet, 22s from 2000 feet and 27s from 3000 feet, are all consistent to within 10% of a 100 ft/s terminal velocity being quickly attained.

The freefall time from 35000ft is indeed about 45s, without air, or for an object so dense/streamlined that it doesn't suffer from any effective air resistance.
In contrast, dust and smoke floats, and never makes it the ground.

Etud_lAvia
10th Mar 2014, 10:02
I saw much misunderstanding, in the discussion on the Malaysian Airlines thread about the shortest time needed for a transport jet to reach the surface from cruising altitude.

Formulas from elementary physics class are not useful. In all cases, and at every point in its trajectory, the jet's vertical velocity will be limited by aerodynamic forces.

Estimating the lower bound of descent time turns out to be quite simple: the airframe as a whole, or any detached part thereof, cannot attain a Mach number that exceeds one by more than a minute margin. Drag increases to such a great magnitude by Mach 1, that it easily matches aircraft weight plus engine thrust. Even to reach 1.0 Mach is very unlikely.

We know this on two grounds:

(1) In the early days of high-speed flight research, aerodynamicists designed shapes that were carefully optimized for minimum drag in the transsonic region, fabricated these shapes in solid steel (many times denser than any actual aircraft), and dropped them from high altitudes (20 to 30 thousand feet). It was very difficult for them to reach Mach 1 even in this condition.

(2) Some hair-raising incidents with transport jets in which gross upsets resulted in uncontrolled dives, showing maximum Mach numbers near 1 even with engine thrust pushing the jet into its dive. For example, in the 727 dive near Detroit (of Hoot Gibson fame), the maximum Mach number achieved was 0.96.

To find a lower bound for descent time, assume purely vertical motion (no "forward speed") at Mach 1. For the example of initial altitude of 35,000 ft, this comes to roughly 34 seconds.

However, starting from cruise condition, the jet would have an initial vertical speed of zero and forward speed in excess of Mach 0.8, so at least some seconds would be consumed in the acceleration of the velocity vector from horizontal to vertical -- this acceleration (again) controlled by aerodynamic forces.

Note 1: Even in a very severe accident, the actual descent time would likely be substantially longer than estimated by these simple assumptions.

Note 2: A transport jet has much lower drag than any broken part thereof, so the terminal velocities of components after an in-flight breakup can be expected to much less than Mach 1.

awblain
10th Mar 2014, 10:16
Yes, the aircraft could be dived into the sea from cruise in less than two minutes, or if it was vaporized, it would never have come down.

The detailed seabed debris fields from Swiss 111 or AF447 for point impacts on the ocean surface, or from Lockerbie and TWA for the places where high-altitude disintegration put things at sea level mean that there isn't going to much problem reconstructing which took place once the wreckage is found.

compressor stall
10th Mar 2014, 13:18
Thanks to the replies.

I can't remember the exact posts as they were deleted, but someone had commented about objects of differing density falling at different speeds reaching the ground at different times. That poster was told by someone to go back to school as every object falls at the same rate.

I used the concrete and air basketball analogy to defend that small part of the first poster's premise (as it gets very complicated very quickly with inflight breakup and 470kts fwd velocity) and was then told I needed to go back too school too!

Never one to take a false accusation lying down (to my detriment at times!) thank-you!

A Squared
10th Mar 2014, 13:25
I used the concrete and air basketball analogy to defend that small part of the first poster's premise (as it gets very complicated very quickly with inflight breakup and 470kts fwd velocity) and was then told I needed to go back too school too!

Ironic to have someone who is *that* freakin' clueless about physics telling others they need to go back to school, isn't it? That's life on an internet forum though, I guess.

oggers
12th Mar 2014, 11:07
Post #8:

Cd=1/2 rho v^2 s remember.

I certainly don't remember that. I do remember that Drag = Cd x ½ρV²S though; as I would expect any pilot, or aerospace engineer to do.

Tinstaafl
12th Mar 2014, 11:30
You're quite right. That was my typo. Corrected it now. Don't do longish posts using a phone...

Daysleeper
12th Mar 2014, 12:39
Try this paper by Dr Matthew Greaves, Snr Lecturer in Accident Investigation at
Cranfield Safety and Accident Investigation Centre

Trajectory Analysis (http://www.isasi.org/Documents/library/technical-papers/2012/13-Revising-Trajectory-Analysis-Evolving-the-Cranfield.pdf)

Ballistic trajectory analysis has been key to many large investigations and much of the science is well understood. However, there has been no package that has incorporated variable gravity, variable density and variable wind profiles into a set of differential equations and then solved them in a robust way. This paper describes the derivation and solution of such a model and presents results gained from it. The numerical solution was validated against a simplified analytical case. Results are given for two simulated breakup cases which provide investigators with information regarding the effect on ground location for variations in four significant parameters.
Conclusions
The results indicate that for simulated large aircraft breakups, low ballistic coefficient items are most heavily affected by breakup altitude, wind magnitude and wind angle whereas large ballistic coefficient items are most heavily affected by breakup velocity, although to a much lesser extent (around 15% of the distance of low ballistic coefficient). For small aircraft breakups, wind angle and breakup altitude have the largest effect on low ballistic coefficient items, with velocity and altitude affecting hig

awblain
12th Mar 2014, 14:38
… no package that has incorporated variable gravity…

That's reassuring, since there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.

I wonder what he meant?

ANCPER
12th Mar 2014, 17:22
when pilots don't know the basics of physics?

s=ut+1/2 at^





^ = squared

s = displacement
u =initial velocity
t = time
a = acceleration


This is the basic physics equation for falling objects where the object size is not considered. There is no mass to take into account, so a mass of 10kg will fall at the same speed as a mass of 40kg, IN EFFECT IF IN A VACUUM! If you take the real world into account and both objects have the same profiles, i.e. two containers of the same dimensions, but one is filled with concrete and the other is empty and both will fall same way down, meaning the empty one won't tumble, the resistance thus terminal velocity will be the same and both will hit the ground at the same time regardless of mass.



When you are considering an object falling in the atmosphere the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.

aram
12th Mar 2014, 18:40
What's the world coming to
when pilots don't know the basics of physics?

(...)
When you are considering an object falling in the atmosphere the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.

Yep, mass does not come into it. Which is why a helium-filled balloon falls to the ground in exactly the same manner as the same ballon filled with water, as everyday experience confirms.

This thread is a telling example of how confused educated people can get - on firm ground.

421dog
12th Mar 2014, 18:53
Isn't building an evacuated sound stage about as much trouble as flying to the Moon?
Not to be a stickler, but isn't that just a bit of an oxymoron?

Daysleeper
12th Mar 2014, 19:17
That's reassuring, since there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.
I wonder what he meant?

Well, I've never done it but I'd guess when designing a model it's generally a good idea to cover the variables so that the output is as accurate as possible.

Gravity does vary by altitude, latitude, underlying rock density and so on. 30,000 ft is about 0.3% lower. A rough wiki driven guess suggests total variation may as much as 1%. Whether this is significant or not I don't know.

awblain
12th Mar 2014, 20:21
At 10km up, about 0.15% of the radius of the Earth, gravity is indeed weaker by about 0.3%, but that's completely irrelevant.

By definition, the sea level has a constant gravity. Mentioning there might be changes in gravity just casts doubt on the sensibility of the rest of the discussion.

Daysleeper
12th Mar 2014, 22:24
By definition, the sea level has a constant gravity.

Well that's not really true, see latitude and density effects.
Anyhow the model seems designed to deal with items which start high up at over 30,000 ft and end low down, ergo they will see a variation in gravitational pull. Now I think that this is likely to make little or no difference compared to wind effects varying with altitude, but I don't know.
Surely the beauty of a model which accounts for varying gravity is that you can find out. If after enough runs of the model to account for statistical significance it turns out that it is irrelevant then you can remove it. If it does make a difference leave it in.

awblain
12th Mar 2014, 22:36
Go measure it. It's not quite an ellipsoid, but it is an equipotential.

What would happen to the water if there was a potential gradient on the surface?

Tinstaafl
12th Mar 2014, 22:57
Not quite. Gravity varies slightly across the Earth's surface. Sea level in one place can have a different gravity to somewhere else.

AtomKraft
12th Mar 2014, 23:13
A glider without water ballast is the exact same shape as a glider with water ballast.


Gravity affects them differently though..

AtomKraft
12th Mar 2014, 23:20
A RV, (that's a re-entry vehicle, or the bit that goes BANG!) on an ICBM is released in space.


It's merely 'pointed' quite accurately at its target. Not powered.


By the time it reaches the ground, it's doing about Mach 10.

gums
12th Mar 2014, 23:50
Thank you, Aram.

Good grief! Mass versus drag for the same object is very apparent in weapon development - ballistic co-efficient. It shows up on basic, old bombs that have different times of flight when released from "x" altitude. Fat ones don't fall as fast as "slick" ones.

It is true that Mother Earth attracts the falling object at basic gee. Forget very small gravitational forces due to altitude above the "Mother".

The point is that F=m*a, and drag is relevant to that equation if in the atmosphere. The falling object is subject to the same "a", but the "F" part has to do with drag, which a function of v^2 and Cd and area of the object. Eventually the drag force equals Mother Earth's force and you have "terminal velocity".

A very good science show years ago was called 'Terminal Velocity". It was about determining a falcon's capability to exceed 200 mph. The sky diver could only reach about 120 mph terminal velocity, so the dude trained the bird to follow a bean bag. He would add heavier lead weights to the bean bag, but the bag was the same area when related to the air while going down. He increased the terminal velocity of the bag to over 200 mph, and the doggone bird could keep up. How? Well, the bird changed its area, so the drag decreased/increased according to the equation of Cd* s*1/2*rho*v^2.

From personal experience dive-bombing, I can attest to the fact that deploying "dive brakes" kept me from exceeding the aero limits of the plane in a very steep dive. All else equal, then it was the area that was part of the drag equation that kept me from ripping the wings off, heh heh. Don't try this at home.

The end of my epistle is that parts of the plane with ottsa area, but relatively low weight will drift in the atmosphere on the way down. The wind will move them away from the point of impact of dense objects of lesser area like engines or fuel pumps or.... They come down slower, so the wind moves them, duhhh?

Been a long time since Aero 101, but I think I got most of it right.

awblain
13th Mar 2014, 09:35
Gums,

It's always a pleasure.

Do you use special nails with larger heads, or is all due to lots of practice with the hammer?

oggers
13th Mar 2014, 09:41
Awblain:

That's reassuring, since there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.


Clearly the guy is talking about a model that accounts for variation of the gravity field by location rather than using the bog standard 9.81ms⁻². It isn't even slightly controversial. NASA have a satellite mapping this because "the Earth's gravity field is not uniform". NASA GRACE Mission (http://science.nasa.gov/missions/grace/)

A Squared
13th Mar 2014, 09:43
What's the world coming to
when pilots don't know the basics of physics?


Indeed. What you're missing though, is that you are the one who doesn't know the basics of physics.

It is astonishing that you continue to insist that you're correct when it has been demonstrated quite thoroughly that you are wrong.

And TINSTAAFL, this is directed at you also, as you too seem determined to continue posting your incorrect understanding of this.

There have been several very detailed explanation of *why* a more dense object will fall faster through the atmosphere than will less a dense dense object of the same dimensions. Your example of an empty container falling at the same terminal velocity is, quite simply, wrong. It has been repeatedly explained in this thread *why* it is wrong, with detailed discussions of the forces involved. Instead of simply repeating the claim that they fall at the same speed, without offering proof, why don't you *explain* why you think the explanations are incorrect?

Serious question? If you wish to be taken seriously, why aren't you explaining your case instead of just contradicting the explanations which have been given? Merely repeating your contradiction makes you appear as John Cleese in Monty Python's argument sketch.


Additionally, a number of posters have cited a very compelling example which makes it abundantly obvious that you are wrong, yet, you pretend that you haven't even noticed. Again, consider the example of two balloons, one inflated with air, and one filled to the same dimensions with water, or concrete, or whatever dense material . Drop them both from the same height. They do not fall at the same rate. Everyone realizes this, except, apparently, you. The air filled balloon is identical to the "empty container" you claim will fall at the same rate as one filled with concrete. In fact it *is* by any definition a container (Not technically empty, as it is filled with air, but so too is the "empty" container in your post actually filled with air)

So how can an "empty" (air filled) container fall at the same speed as the same container filled with concrete (as you claim) when it is obvious to all that an air filled balloon does not fall at the same rate as one filled with concrete?

That, by the way, is not merely a rhetorical question. I actually would like to see your answer. You have a very simple, obvious, very easily understood example that shows without any question that you are wrong, and somehow you rationalize that away. I am truly interested to hear your explanation for how or why you disregard this.

If the fact that it the balloon is filled with air is confusing you let's consider a different form of the same example. A feather. No doubt you have seen feathers falling, and noted that they fall rather slowly. Now lets suppose that a jeweler had very carefully crafted a replica of that same feather from gold. The form of the feather is identical on a microscopic level, it is the same dimensions, the same thickness, the shaft is the same dimensions, as are the barbs, and the barbules between them, everything is duplicated to the same dimensions, except that it is made of solid gold instead of ... well whatever it is feathers are made of.

With me so far? OK, I think that everyone will agree that the real feather will fall decidedly slower than the gold feather, right? Is that not a pretty conclusive demonstration that , all else being equal, the speed of a falling object through air is *not* independent of it's mass ? If you believe that it is not, explain *why* not. If you can't explain *why* not, than that's likely an indication that you're mistaken.


While you are mulling over your explanation of the previous obvious flaws in your claim, you might consider the following:

NASA webpage on Terminal Velocity (http://www.grc.nasa.gov/WWW/k-12/airplane/termv.html)

and

Wikipedia Article on Terminal velocity (http://en.wikipedia.org/wiki/Terminal_velocity)

Both of these pages contain an equation for calculating an object's terminal velocity (and both pages describe how that equation is derived) the equations are the same equation in slightly different formats. The NASA page puts the equation in a format that can be shown using only ASCI text and no special formatting (such as fractions, radical signs, or greek letters) but mathematically the equations are identical.

Now here's the point which demonstrates quite conclusively that you're wrong. Both equations contain a term for mass, which can *only* mean that terminal velocity *is* in fact affected by mass. If terminal velocity was (as you erroneously claim) independent of mass, mass would not be included as a variable in the equation for calculating it. But, it is included, so it must affect the velocity*. Now, I suspect that you're getting ready to say; “ But the NASA equation uses “weight” which is not the same as mass” True, in the context of physics, weight is not the same as mass. In physics, “weight” is the force of gravity on an object. And weight is equal to gravity times mass (g*m ) so where the NASA equation has W, it really means g*m so the first term in parentheses (2 *W) becomes (2*g*m) which you will note is the same as the upper term beneath the radical in the Wikipedia Equation: 2gm The asterisks in the NASA equation denote multiplication, while the multiplication is implied in the Wikipedia equation, so 2*g*m is just a different way of writing: 2gm. Bottom line, the equations are identical, and they both contain mass as a variable.

This leaves us with this:


Your (fallacious) opinion:

the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.

NASA's “opinion”:



Terminal Velocity = sqrt ( (2 * g* mass ) / (Cd * r * A)

Just as a suggestion, if you find that your understanding of the physics of falling bodies differs significantly from NASA's, it's probably time to consider that you are mistaken.


All the preceding can be summarized neatly in two questions:

How do you resolve you claim that mass doesn't affect the velocity of a object falling through air, when a balloon filled with air and a balloon filled with concrete obviously and unquestionably fall at different speeds?


If Mass doesn't affect the speed of a falling body, why does NASA, one of the worlds foremost organizations for the study of aerodynamics and ballistics, say that it does?

Serious questions. Response expected. Lack of response will indicate a tacit admission of being mistaken.
















* if anyone is having difficulty following this reasoning, all you have to do is make some simple calculations of terminal velocity using those equations and the same values for all terms, but different values for mass. The calculated terminal velocity will be different.

Try it with the following terms to keep it simple:

Drag Coefficient (Cd in NASA's equation) = 1
Area(A in NASA's equation) = 1 (square meter)
Gravity (g) = 9.8 (meters/sec/sec)

Rho ("r" in NASA's equation) = 1.3 (kg per cubic meter sea level standard temp at pressure )

Mass = 1 (kg)

Don't worry about the units, just plug the numbers into the equation and see what you come up with.

Now do it again, using all the same numbers except you use 10 kg for mass instead of 1kg. Notice that the calculated terminal velocity is different? the point here is that you can't include a quantity in an equation and not have it affect the result. That means the terminal velocity of an object is dependent on mass. QED.

awblain
13th Mar 2014, 09:54
Oggers,

Look at those variations though, and consider their nature. At constant height above sea level / geoid, they are very modest, requiring days of data from satellite orbit measurements to measure. The differences are typically at the 10 ppm level, and up to about 60 ppm.

See GRACE Gravity Model- Gravity Recovery and Climate Experiment Gravity Model (http://www.csr.utexas.edu/grace/gravity/) for a description of the GRACE mission and its products. GOCE produced more too.

If the author is including GRACE maps is his fluttering debris data, then he's missing the wood for the trees.

Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.

A Squared
13th Mar 2014, 10:01
A glider without water ballast is the exact same shape as a glider with water ballast.


Gravity affects them differently though..

I suspect that is a little too subtle for some. Can you be more specific on the difference of the effect of gravity? Is the heavier glider pulled to earth faster or at the same rate as the glider which has dumped it's ballast?

awblain
13th Mar 2014, 10:23
It would be an important check for the thoughtful reader to work it out first and then check the answers; that way the thoughtful reader might be able to tell whether they'd been thoughtful enough.

It might be easier to consider the flight of a lead zeppelin before adding the complexities of lift.

oggers
13th Mar 2014, 10:49
A squared:



Originally Posted by ANCPER View Post

What's the world coming to
when pilots don't know the basics of physics?

Indeed. What you're missing though, is that you are the one who doesn't know the basics of physics.

It is astonishing that you continue to insist that you're correct when it has been demonstrated quite thoroughly that you are wrong.

If something is worth saying it is worth saying twice.

Will some of the numpties posting here please wake up.

For an object in freefall at terminal velocity: drag = mass x g. That is not up for debate. 'g' being fixed (notwithstanding the fact it does vary very slightly :rolleyes:) - if you increase the mass you must increase the drag that would result at terminal velocity. And that means an increase in terminal velocity, because the OP is predicated on a fixed shape and hence a fixed coefficient of drag.

This result has already been mentioned and proved in other ways in this thread multiple times now. I am bemused - though no longer the least surprised - to see some posters still wading in to contradict this fact with their half baked physics.

oggers
13th Mar 2014, 11:07
Awblain

Look at those variations though, and consider their nature.

There is no need. You said:

there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.

Which is an opinion shown to be false by that link to NASA I provided. I am not going to enter into a game of semantics with you because I find that sort of debate really lame.

If you wish to prove the point you have now shifted to:

If the author is including GRACE maps is his fluttering debris data, then he's missing the wood for the trees.

Is he using pressure altitude to start his simulation? If so, then he should have a further think about the "variable gravity" bit and where isobars lie in 3D.

Then please do get on with it, it might be interesting :ok:

A Squared
13th Mar 2014, 11:07
This result has already been mentioned and proved in other ways in this thread multiple times now.

And just for entertainment purposes, I'll throw in another proof. This is about as close as you can come to the basketball full of air vs basketball full of concrete. The video contains a lot if irrelevant and annoying dialog and several somewhat irrelevant demonstrations. However there is a segment which is exactly on point, in which they simultaneously drop a football/soccer ball (depending on geographic preferences) and a similarly sized cannon ball. I've attempted to link it such that it starts on the relevan portion. In case that doesn't work, watch from about 2:30. ANCPERS, TINSTAAFL, et al have been insisting they would fall with identical velocities. Anyone care to guess what *actually* happens?


Comparison of falling spheres. (http://www.youtube.com/watch?v=jme_vSj5wRo&feature=player_detailpage#t=146)

awblain
13th Mar 2014, 11:16
Oggers,

You're not entering into a game of semantics, you're missing the point. You get horizontal gravity variations of about 1 part in 100,000. They don't matter. Period. If anyone really sticks them into a model of falling debris, then they're casting doubt upon the quality of other bits in their model.

It is possible that the vertical changes in gravity, by about 3 parts in 1000 from 10km up to sea level might have a subtle effect on where heavy dense debris lands, since that slows less and spends less time lower down than lighter debris.

oggers
13th Mar 2014, 11:23
The video contains a lot if irrelevant and annoying dialog

How dare you. One of those is one of our finest comedians. The other one used to have a show called "Shooting Stars".

A Squared
13th Mar 2014, 11:57
How dare you. One of those is one of our finest comedians. The other one used to have a show called "Shooting Stars".

A thousand apologies. I never meant to gore any sacred cows, (or whoever the appropriate metaphor is.) ;)

oggers
13th Mar 2014, 12:03
Awblain:

You're not entering into a game of semantics, you're missing the point. You get horizontal gravity variations of about 1 part in 100,000. They don't matter. Period.

Well, you're half right. And I'm open minded on the other half. Just for the record you said:


there isn't any "variable gravity" until you're worried about altitudes that reach tens or hundreds of km high.

...whereas you now say 'there is variation but it isn't enough to make any difference'. I am asking you to put some flesh on this by actually showing us some of this trajectory analysis you claim will not be affected. Because, frankly, compared with the expert opinion of Dr Matthew Greaves of Cranfield University, I find the lack of substance to your contra opinion to be most unconvincing :ok:

awblain
13th Mar 2014, 13:24
Oggers,

What is about "1 part in 100,000" that suggest it matters?

Drag changes by 1 part in 50,000 would have the same effect. How much is that? It's associated with the same density changes as climbing about 10cm. QED.

Perhaps Dr Greaves should worry about density changes from the top to the bottom of the wreckage too? I don't think so, and I assume that in fact Dr Greaves meant something else in his abstract.

Daysleeper
13th Mar 2014, 13:47
Having read through the paper, the tool is designed to have the ability to deal with very high altitude break ups of re-entering space vehicles where gravitational variation with altitude would have an effect.

awblain
13th Mar 2014, 13:52
Even then, it's hardly worth the worry.

At 100km, the difference in gravity from sea level is only 3%.

If you get have an exosphere density profile correct to 3% then you're already doing well.

Tourist
13th Mar 2014, 13:55
awblain

A question.

If you don't believe in variable gravity, what exactly do you think causes tides....?

awblain
13th Mar 2014, 13:59
There is no relevant, variable gravity.

The tidal force variation at the surface of Earth? 1 part in 10 million.

At the Planck scale, gravity's all over the place, but that has no relevance to falling wreckage either.

Tourist
13th Mar 2014, 14:30
I would suggest that anything that can make squillions of tons of water and earth move up and down meters is relevant....

Tourist
13th Mar 2014, 14:38
Interesting.

Gravity map reveals Earth's extremes - physics-math - 19 August 2013 - New Scientist (http://www.newscientist.com/article/dn24068-gravity-map-reveals-earths-extremes.html#.UyHBc01F1J8)

Gravity variations much bigger than previously thought (http://phys.org/news/2013-09-gravity-variations-bigger-previously-thought.html)

awblain
13th Mar 2014, 14:49
Interesting indeed, but irrelevant. That headline should be "Well-known gravity variations mapped on much finer scales than before"

If you can measure a difference between free fall times at high and low tide, then you have an aptitude for experiment. You're able to measure 10 parts in a billion.

A Squared
13th Mar 2014, 15:14
I would suggest that anything that can make squillions of tons of water and earth move up and down meters is relevant....

And I would suggest that you are making some rather broad and incorrect assumptions here.

As Awblain said, the amount the effective gravity changes at the earth's surface due to lunar and solar effects is about one ten-millionth of total gravity. Yes, it is sufficient to cause tides. No, it ain't gonna make much difference in how fast an object falls. For an object of say 10 kg, that means it's weight will vary between 98 newtons and 98.000001 newtons. For a falling object, the effects are the difference between and acelelration of 9.8 m/sec/sec and 9.80000001 m/sec/sec.

The difference is very small, and orders of magnitude below the noise level for something like predicting falling debris.

Lord Spandex Masher
13th Mar 2014, 15:15
I would suggest that anything that can make squillions of tons of water and earth move up and down meters is relevant....

Oh come on! It's not squillions it's zillions. Accuracy is important.

awblain
13th Mar 2014, 15:24
The tidal acceleration moves water about two meters in six hours.

The free fall acceleration moves water two meters in about 0.4 seconds.

The force is related to the square of the ratio of those times. That's gives a measure of the low-down scale of tidal forces.

If you are in genuine free-fall though (and not the term as bastardized by parachutists), then the presence of tidal forces is the only thing that reveals the presence of gravity.

oggers
13th Mar 2014, 15:34
Awblain:

I assume that in fact Dr Greaves meant something else in his abstract.

Well, whatever it is you think he meant it is YOU who has taken issue with it. So what I suggest you do is:

a) Clarify in your own mind what you think Dr Greaves meant which is wrong
b) Check you have that bit right
c) Come back to this forum and detail the errors you have identified in his work.

Right now you are obfuscating, shifting position and creating fog.

What is about "1 part in 100,000" that suggest it matters?

Case in point: you can hang onto these tiny gravity anomalies and close your mind to the other effects - more significant - that I suspect were used in the trajectory analysis to get an effective g. But I don't know for sure. If you have a problem with Dr Greaves' model, it is your point to prove, not mine to disprove.

awblain
13th Mar 2014, 15:48
In writing an abstract, it would be in Dr Greaves' interest not to make statements that make the whole look wacky. Based upon that choice of words, I would not read his paper.

Looking at it now, it appears that his introducer misquoted him.

He includes the term "altitude dependent gravity" in his paper, and while I think this is a bit of a waste of time to consider, it is a viable effect, compared with those of both the geoid and tides.

oggers
13th Mar 2014, 17:00
awblain

Just admit it, you can't actually find anything wrong with Dr Greaves' model and so you are waffling.

And meanwhile

That headline should be "Well-known gravity variations mapped on much finer scales than before"

If you'd actually read that New Scientist article that Tourist linked to you would've seen this quote:

"Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2. These differences mean that in the unlikely event that you found yourself falling from a height of 100 metres at each point, you would hit the surface in Peru about 16 milliseconds later than in the Arctic."

16milliseconds difference in just 100 meters! To my mind that is clearly worth considering if one is developing a model such as Dr Greaves'. However, you say of such variations:


They don't matter. Period. If anyone really sticks them into a model of falling debris, then they're casting doubt upon the quality of other bits in their model.

In the end it comes down to your opinion against the published work of experts in the field.

Based upon that choice of words, I would not read his paper.

Yes, it would have helped if you HAD read it before deciding that his model was no good.

gums
13th Mar 2014, 18:48
Seems the robo-moderator is holding/deleting my last post. Prolly 'cause I used a "bad" word.

Anyway, it was about using previous incidents of vehicles coming apart at altitude. So we can look at those to get back to the MAS 370 search and distribution of wreckage. And seems to me that it was that which started this discussion.

dancingdog777
13th Mar 2014, 19:18
Something further to consider.

If you drop a cat from a height of say 10 feet, it will always fall on it's feet. But if you drop a piece of buttered toast from the same height it will always fall with the buttered side down.

So here is the problem, if you tie a piece of buttered toast onto a cat's back and drop it from the same height will the buttered toast land butter down i.e. with the cat upside down, or will the cat land on it's feet i.e with the toast butter side up?

:ugh:

Tourist
13th Mar 2014, 19:19
awblain

I only butted in because I was enjoying you try to dig yourself deeper in a hole...

However...

"The tidal acceleration moves water about two meters in six hours.

The free fall acceleration moves water two meters in about 0.4 seconds.

The force is related to the square of the ratio of those times. That's gives a measure of the low-down scale of tidal forces."

Oh, dear oh dear.....

Please tell me that you don't think that is in any way true?

awblain
13th Mar 2014, 19:50
Please go away with your ignorance about forces, vexatious chappies,

Gravity differences are more complex than you suggest, as it depends where you are, and above what reference level. There is also the rotational/centrifugal/centripetal part to consider.

The difference in time between equally spaced gates from above a Peruvian mountain (equatorial bulge plus a mountain under your feet) against the arctic ocean (polar squatting and kms of low-density water) is as extreme as it gets, but the difference in time is very small. The difference in g, by about 0.7% from these spots is only the same as the difference from climbing a few tens of km. Compare the ratios of the distances to the center of the Earth from the North Pole and antiplanic Peru.

And tourist, perhaps a remedial course in counting is in order, or am I somehow wrong about the intertidal period and the oceanic tidal range?

john_tullamarine
13th Mar 2014, 20:53
Folks, a delightfully entertaining thread. However, as always, please do keep the excited angst under control ...

Mr Optimistic
13th Mar 2014, 23:01
What's this about variable gravity causing tides. That's a bit novel. Can someone explain?

oggers
13th Mar 2014, 23:02
awblain

You have made about twenty posts since you first took issue with Dr Greaves' paper on trajectory analysis. Still you fail to communicate what errors exist in his model.

Mr Optimistic
13th Mar 2014, 23:12
Just read his paper. Oh dear, he has stuffed up on his altitude dependent gravity. Not impressed by his discussions on numerical methods either. But what's this about variable gravity causing the tides again? Perhaps someone can explain ?

Mr Optimistic
13th Mar 2014, 23:24
Nothing? How very wise.

oggers
14th Mar 2014, 00:39
Mr Optimistic

Oh dear, he has stuffed up on his altitude dependent gravity.

Feel free to explain how.

Mr Optimistic
14th Mar 2014, 00:42
Oh dear do I have to repeat myself. I am asking for someone to explain this variable gravity theory of tides. If you can, fine, please do. If you can't I'll continue to wait.

Mr Optimistic
14th Mar 2014, 01:24
Aww come on guys I'm feeling lonely.

ANCPER
14th Mar 2014, 03:14
You are partially right! Before I take the discussion any further I'd like you to answer the following question;

2 objects same dimensions etc but different masses, 1 say 1kg the other 10 kg with both being dropped say 10m, will they hit the the same time or will the heavier one hit first?

A Squared
14th Mar 2014, 03:55
You are partially right! Before I take the discussion any further I'd like you to answer the following question;

2 objects same dimensions etc but different masses, 1 say 1kg the other 10 kg with both being dropped say 10m, will they hit the the same time or will the heavier one hit first?

I've already stated the answer pretty clearly, the one with more mass will hit first. If the dimensions of said objects are relatively low drag, the difference may be small enough that it's difficult to detect with the unassisted eye.

A Squared
14th Mar 2014, 04:04
Aww come on guys I'm feeling lonely.

Well, Tides are a result of temporal variations in the net gravitational acceleration vector caused by the effects of the moon and sun. But, I suspect that you know that and are merely trying to start an argument based on the use of imprecise terms. Knock yourself out.

ANCPER
14th Mar 2014, 04:38
Well. A Squared, you'd be wrong.

After your first response I went and paid a visit to a young lady (physics grad) who tutored my daughter in Physics a couple of years back.

S = ut + 1/2 at^

^ = squared

She said this is the valid equation up until the point the object reaches Vt, as the object is still accelerating and mass does not come into it, so in a fall where the objects fail to reach their Vt the above equation is it and they will hit the ground at the same time. Now, if the distance the objects fall is such that Vt will be reach prior to their hitting the ground the Vt equation of
http://upload.wikimedia.org/math/6/e/3/6e306f943fc864e7ee41a1b3a7f16172.pngwill come into it. At this point mass impacts Vt and therefor the higher mass object will have a higher Vt and will hit the ground earlier. So the as the lighter object reaches Vt (now falls at its Vt as determined by the Vt equation) it will now stop accelerating while the higher mass object continues to accelerate (using s =ut +1/2 at^), until it reaches Vt.


So as to the effect of mass depends on whether the object hits the ground prior to reaching Vt.



She laughed at your helium balloon comparison, if you check a reasonable link the Vt equation like at this link



Terminal velocity - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Terminal_velocity)


you'll see it mentions where buoyancy isn't a consideration.

Mass comes into the Vt equation as it's "pushing" against the air resistance, thus a higher mass object will have a higher Vt.

Wizofoz
14th Mar 2014, 05:47
Buoyancy a consideration?

From YOUR link-

The terminal velocity of a falling object is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.[

and latter-

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass m has to be reduced by the displaced fluid mass \rho\mathcal{V}, with \mathcal{V} the volume of the object. So instead of m use the reduced mass m_r=m-\rho\mathcal{V} in this and subsequent formulas.

So, what you've done is use on equation that doesn't take all aspects of the situation into account.


Applying THAT principle will show that the less dense object will accelerate slower at the outset, so the two objects will NEVER fall at the same rate.

awblain
14th Mar 2014, 07:45
Mr Optimistic,

Tides on Earth are caused by the minor influence of the Sun and (mostly the) Moon's gravity. When Moon is directly overhead, there's a little less net attraction of the water surface to the center of the Earth, and so there's a tidal bulge. Six hours later, there's a sideways pull from the Moon instead, and no such bulge. The gravitational field varies and shifts direction slightly at the Earth's rotation rate.

The bulge opposite the Moon needs you to shift into the rotating frame around the Earth-Moon system center of mass, and is less amenable to a one-line explanation.

awblain
14th Mar 2014, 07:53
oggers,

The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot.

If you can't see that, then you need more than a hundred-word comment to explain.

I'm not the referee, and professional courtesy would indicate that he not be needlessly harangued. There's nothing grossly incorrect in the paper, but it's neither novel nor insightful. By including unnecessary terms in the gravity, and ignoring the larger degree of ignorance about the density profile, and lift effects, it's not very helpful to the discussion about falling objects.

awblain
14th Mar 2014, 07:58
ANCPER,

Your young female friend has either been misinterpreted, or she was giving your daughter a sketchy outline of reality.

The terminal velocity is a limiting value at constant density, so to say that the path is quadratic until you hit the terminal velocity is just not right.
Quadratic distance vs time only applies with zero drag. As density increases, it drops, look at that Austrian balloon skydiver guy.

A Squared
14th Mar 2014, 07:59
Well. A Squared, you'd be wrong.


Well, ANCPERS, no, I'm not.

It is interesting how you have changed your claim as you've been unable to defend it and you've had to walk it back, so to speak, to avoid admitting you were wrong.


You began by claiming:

the resistance thus terminal velocity will be the same and both will hit the ground at the same time regardless of mass.

and


When you are considering an object falling in the atmosphere the only determinate when it comes to how fast an object will fall is the area of the object and therefore the drag(resistance) it will produce, mass does not come into it.


But now you're claiming that:

so in a fall where the objects fail to reach their Vt the above equation is it and they will hit the ground at the same time.

So as to the effect of mass depends on whether the object hits the ground prior to reaching Vt.

First, you claim that mass has *nothing* to do with it, now suddenly, mass *does* have something to do with it.

Which means you were wrong. QED.


Of course in order to avoid merely saying “yeah, I was wrong, I was talking out of my nether regions” you've concocted some new fallacy. Now you're claiming that the objects fall at precisely the same rate until the less dense one reaches it's terminal velocity and the more dense one leaves it behind.

It's a more complex fallacy, but every bit as incorrect as your initial fallacy, which you've now abandoned.

A word about your “physics expert” She may well be a physics grad student, but that doesn't mean that she necessarily understands ballistics very well. That would be a subject that physics and engineering students learn in the first semester physics class in their freshman year. Second semester physics has moved on to other subjects. The thing is, Newtonian physics is pretty basic stuff and a physics student, even an undergrad moves on to other more advanced physics pretty quickly. By the time one is in in a physics graduate program, the motion of falling objects is something that they may not have considered in any depth since the first semester of their freshman year. (The astute reader will at this point be wondering: Hmmm, I wonder how A Squared knows about studying physics at a university?)

Of course it is entirely possible that she understands it perfectly well, but you just didn't understand her explanation of it, perhaps willfully on a subconscious level because of your demonstrated desire to not admit you're mistaken.

Regardless, whether it's because she doesn't understand the physics or because you didn’t understand her explanation, either way, the end result is the same; what you've taken away from that conversation is wrong.


She laughed at your helium balloon comparison

Uhhh, nobody mentioned helium balloons. The comparison was air filled balloons (which will fall down). You made up the part about helium. FWIW, the force of buoyancy on an air filled balloon and one filled with concrete is exactly the same. The only difference is that with air filled balloon, the buoyancy is large compared to the weight of the balloon, and small compared to the weight of the concrete filled one, but in either case the force of buoyancy that the air is exerting on the balloons is the same. And as far as the your link goes, you apparently missed the fact that your link is the same one I linked in my previous response to you, and discussed the information therein. Go back and look, it's still there in my post, the second link.


Now, here's why the current iteration of your theory is wrong. You claim that drag makes absolutely no difference until one object reaches it's terminal velocity, at which point it stops accelerating and the other continues accelerating, thus pulling “ahead” and falling faster.

In order to believe that, you'd have to believe that drag is completely non-existent until one object reaches terminal velocity.

OK, let's work through a simple example: Assume a 1 kg object and a 10kg object. The weights of the object (the force of gravity pulling on them) will be 9.8 newtons and 98 newtons respectively.

Let's further assume that the terminal velocity of the heavier object will be 50 m/s. If that is true, and the objects are externally identical (same shape, drag coefficient and frontal area) then the lighter object will reach terminal velocity at about 15.8 m/s

So at 15.8 meters/sec the force of drag (on both objects) will be 9.8 newtons. The lighter one will stop accelerating because that is equal to the force of gravity on the object and the forces cancel each other out, there is no extra force to accelerate the object. That's the definition of terminal velocity, where the force of gravity is exactly balanced by the force of drag. The heavier object will continue to accelerate because the force of drag is 9.8 newtons, but the force of gravity is 98 newtons, so there is still 88.2 newtons accelerating the object.

OK, that's just an explanation of what terminal velocity is. Now your claim is that up till the point that the lighter object reaches terminal velocity, the objects will fall with identical acceleration, thus identical speed, and will reach any point at the identical same time (short of the lighter one reaching terminal velocity)

You're wrong. Here's why: Using our example, at the instant they are released, both objects will begin accelerating at the same rate. There is no motion yet, so drag is zero, non-existent. The only force on the objects is the force of gravity. There is 9.8 newtons of gravitational force acting on the 1 kg mass which causes it to accelerate at 9.8 meters/sec/sec and there is 98 newtons of gravitational force acting on the 10 kg. mass, causing it to also accelerate at 9.8 meters/sec/sec.

That is at the instant of release, when velocity is zero, thus drag is zero, acceleration will be the same.

The thing is, drag does not remain at zero. As the speed increases, so does the drag. What is the drag when the objects are falling at half the lighter object's terminal velocity? Well, it will not be half, because drag is a function of the square of the speed. Half the lighter object's terminal velocity is 7.9 meters/sec. At that speed the drag on both objects will be 2.4 newtons. At that point in the fall, will they both be falling and accelerating at the same speed? Your claim is that they will. You are wrong. We only have to calculate the acceleration at that point to see why you're wrong. At that point, drag on both is 2.4 newtons. That means the the unopposed force on the lighter object will be 9.8 newtons of gravity – 2.4 newtons of drag, or 7.4 newtons. This means that at that point the lighter object will be accelerating at 7.4 meter/sec/sec (7.4 newtons divided by 1 kg) But the unopposed force on the heavier object will be 98 newtons of gravity – 2.4 newtons of drag, or 95.6 newtons. If you apply 95.6 newtons of force to a 10 kg object, the object will accelerate at 9.56 meters/sec/sec.

So, halfway to the lighter object's terminal velocity (in time, not distance), the lighter object is accelerating at 7.5 m/sec/sec, and the heavier one will be accelerating at 9.56 meters/sec/sec. Obviously, if one object is accelerating at a greater rate than the other, they can't possibly maintain the same velocity. And that's just a snapshot of what's happening at one instant in time. The only time the acceleration of the objects has been equal is at the precise moment of release, when velocity as zero, and thus drag was zero. At any instant after the objects have started falling, the acceleration of the heavier object will be greater than the lighter object, because of drag.

Now, I don't expect you to agree. It is clear that your brain is so tied up in the effort of not admitting that you're wrong, that you're unable to follow the explanation.

Here's what you should do. Print this entire discussion out, and take it to your physics grad student friend. Seriously, print it out, don't “explain” it too her, because you clearly don't understand it yourself, and thus are unable to describe it accurately, plus you can't seem to resist making up fictional details like helium balloons, which were not a part of the discussion. Like I said, print it all out without editing it, and hand it to her, and ask her to take it home and read it and think about it.

One of two things is going to happen:

1) she's going to immediately say, “no you misunderstood what I was saying, they do not fall with exact same identical speed until one reaches it's terminal velocity, that's not what I said.

or

2) she will, after having read and considered the explanation , say, yep sorry, I told you incorrectly, the objects will not fall at the same speed until one reaches terminal velocity.


Of course, you could save everyone a lot of effort by just admitting you were wrong. This is painfully obvious to everyone at this point, but I don't think there's any chance you're going to do that. For you it has obviously become a game of: "don't ever admit I was wrong" instead of "I want to understand this better"

A Squared
14th Mar 2014, 08:45
ANCPERS,

Here's another question for your physics grad student. Print it out and give it to her verbatim.

Assume that the terminal velocity of the lighter object is exactly 15.80 m/sec.

What will be the lighter object's acceleration when it is falling at 15.75 m/sec?

If you say it's acceleration is 9.8 m/sec/sec, how can aerodynamic drag be large enough to completely stop acceleration when the speed is 15.80 m/sec, yet have absolutely *no* effect on acceleration when it is falling 0.05 m/sec slower?


If you say that the acceleration will be less than 9.8 m/sec/sec, then *why* is it not accelerating at exactly g (9.8m/sec/sec) ?

oggers
14th Mar 2014, 08:50
ANCPER

Well. A Squared, you'd be wrong

He is not wrong.

After your first response I went and paid a visit to a young lady (physics grad) who tutored my daughter in Physics a couple of years back.

S = ut + 1/2 at^

^ = squared

She said this is the valid equation up until the point the object reaches Vt, as the object is still accelerating and mass does not come into it, so in a fall where the objects fail to reach their Vt the above equation is it and they will hit the ground at the same time.

She is wrong. That is not the correct equation to use with drag - not even "up until terminal velocity". She is correct about the terminal velocity though, which is just a rearrangement (in the special case where Fnet = 0) of the equation:

Fnet = mg - (Cd x ½ρv²s)

From this equation you should be able to see that for any velocity, if the mass is higher, Fnet is higher.

Tourist
14th Mar 2014, 09:02
awblain

"And tourist, perhaps a remedial course in counting is in order, or am I somehow wrong about the intertidal period and the oceanic tidal range? "

"The tidal acceleration moves water about two meters in six hours.
The free fall acceleration moves water two meters in about 0.4 seconds.
The force is related to the square of the ratio of those times. That's gives a measure of the low-down scale of tidal forces"

No.

The force that moves tides is not that simple. If it were, then lakes would go up and down in a similar range. They don't incidentally.
If you drop water in free fall, then it can fall because there is obviously nothing underneath it.
The sea is not in that position. For the surface to rise and fall, the water has to move sideways, or the water has to expand and contract.

The example you quoted of a tidal range of 2m in 6 hours is not a simple F=ma equation.
You have to bring in an incredibly complex range of factors including sea bottom shape, sea viscosity, flow patterns through constrictions, compressibility to name just the first ones off the top of my head.

oggers
14th Mar 2014, 09:06
oggers,

The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot....[etc etc blah blah yada yada]


awblain, the chap is not claiming a new theory. He has merely developed a new trajectory analysis model as an aid to air accident investigation.

25 posts since you first took issue with his work. 25 posts and not a single coherent argument that he has got anything wrong.

Where can I find your publishing history BTW?

Tourist
14th Mar 2014, 09:14
What about the apparent change in gravity due to flying westwards/eastwards at 450kts?
One of you clever professor types could probably throw in the numbers...

awblain
14th Mar 2014, 09:18
Oggers, Laud his work if you want, that's your choice.

Tourist, deep large ocean tides are simple and do have that size - their driving force can be obtained to an order of magnitude from distance over time squared. Mr Newton did have it right. There is not enough water or extent in a lake to allow it to flow to respond like this in six hours via the necessary very-long-wavelength gravity waves.

oggers
14th Mar 2014, 09:27
Mr Optimistic

Oh dear do I have to repeat myself. I am asking for someone to explain this variable gravity theory of tides. If you can, fine, please do. If you can't I'll continue to wait.


I can't help you with that. But while you're waiting, seeing as you wrote:

Oh dear, he has stuffed up on his altitude dependent gravity.

Please feel free to explain what he has got wrong. This is the Tech Log forum, so opinions require explanation, otherwise it defeats the purpose :ok:

awblain
14th Mar 2014, 09:31
What about the apparent change in gravity due to flying westwards/eastwards at 450kts?

Do you mean the gentle decrease in the reaction force from the floor because you're following a curved path at speed? Which would also apply the same if you were going north-south? At 6000km radius and 300m/s that's about 0.15% of g.

Or are you worried about the relativistic corrections to timing, which are down at the 1 in 10-to-the-10 level, or the difference in co-rotation and counter-rotation terms in general relativity, which are stupendously small on scales of the Earth size and mass.

oggers
14th Mar 2014, 09:36
Oggers, Laud his work if you want, that's your choice.

No, awblain it's not my choice. You took issue with his work. this is your point to prove but 25+ posts on you have failed to identify a single problem. :ugh:

And where can I find your publishing history BTW?

Tourist
14th Mar 2014, 09:45
Yes I was referring to the reduction in g due to the curved path at speed, the classic in a lift with no windows experiment.


You are wrong about deep oceanic tides. they are pretty much non existent in the middle of the oceans and build up around continental masses.

My point is that your earlier explanation

"The tidal acceleration moves water about two meters in six hours. The free fall acceleration moves water two meters in about 0.4 seconds. The force is related to the square of the ratio of those times. That's gives a measure of the low-down scale of tidal forces"

was codswallop. It is vastly more complex. I thought of a few more factors including coriolis effect, centripetal effects etc.
To characterise it as (and I am paraphrasing you here)
"tides go up and down with very little acceleration compared to water falling in free fall therefore tidal effects are negligible"

is just wrong.

awblain
14th Mar 2014, 09:59
Oggers:

"Deep ocean tides on Earth don't exist"
"I don't understand when factors are too small to matter"
"I can't estimate the size of physical quantities and effects"

Tourist:

"I can't use a library or a search engine"
"I have difficulty comprehending the statement 'By including unnecessary terms in the gravity, and ignoring the larger degree of ignorance about the density profile, and lift effects, it's not very helpful to the discussion about falling objects.'"

Thanks for the ongoing valuable and voluminous contributions, chaps. No doubt we'll be having more.

Tourist
14th Mar 2014, 10:28
awblain

Superficially sounds knowledgeable, but when pressed seems strangely unable to maintain and defend a position.
Makes statements which are simplistic but when taken to task slides away in a soapy fashion.....

Mr Optimistic
14th Mar 2014, 10:38
As has been pointed out, to understand tides you have to realise its a dynamic problem which must be framed appropriately as two revolving extensible bodies rotating about their common cm. The problem with the poor professor's paper is that it is quite simply unintelligent to include the range variation of gravity as its very small perturbing effect relative to a constant model is less than trivial when compared to the other forces acting and the inherent limitations of the model. That's what was meant by another poster when he said the beginning of the paper was wacky and not worth the read. He is entirely correct on that. I must have another look at it to see if he includes coriolis forces: I doubt he understands that either.

So can someone now explain this variable gravity theory of tides to me as opposed to the standard theory ?

A Squared
14th Mar 2014, 10:48
The problem with the poor professor's paper is that it is quite simply unintelligent to include the range variation of gravity as its very small perturbing effect relative to a constant model is less than trivial when compared to the other forces acting and the inherent limitations of the model. That's what was meant by another poster when he said the beginning of the paper was wacky and not worth the read. He is entirely correct on that. I must have another look at it to see if he includes coriolis forces: I doubt he understands that either.

Ehh, when I read the paper, my takeaway was that he was talking about the variation of gravity due to altitude, a la the free air gradient vs assuming gravitational acceleration to be constant at all heights along the trajectory of the falling airplane parts. I didn't see anything in the actual paper to suggest that he was considering adding terms for temporal variations.

Tourist
14th Mar 2014, 10:50
In very rough terms, the earth and moon can be considered an object with a centre of mass somewhere between the centre of the earth and the moon.
This centre of mass will be near the earths core due to the relative sizes of the bodies.
This centre of mass will also be mobile as the moon revolves around the earth.
Thus, a person standing on the earths surface will be moving in and out relative to the centre of mass of the earth/moon pair and will experience (very tiny! 2micro m/s2) variations in gravity.
They would also be moving in and out relative to the L1 Lagrange point, ie varying their range from a point of zero g

A Squared
14th Mar 2014, 10:53
Tourist,

In a previous post you said:


You have to bring in an incredibly complex range of factors including sea bottom shape, sea viscosity, flow patterns through constrictions, compressibility to name just the first ones off the top of my head.

Now, I know just enough about tides to be dangerous. I'm aware that bathymetric effects are major factors, and the other stuff makes sense. I'm wondering about "compressibility". I'm trying to imagine how/where that's a factor and coming up short. (no, I'm not baiting you. I don't know and I'm curious)

Mr Optimistic
14th Mar 2014, 10:55
It is not that close to the earth's core. Awblain is correct in everything he says as far as I can tell a comment I can't make about said professor.

Edit: you are right to be sceptical about compressibility!

Tourist
14th Mar 2014, 11:02
You are right, it is tenuous!

My idle thought was that since water has some limited ability to compress, the (agreed it is tiny!) variation in g with the moons motion will cause the ocean to expand and contract by admittedly tiny amounts.
That's all!

Mr Optimistic
14th Mar 2014, 11:07
For the genuinely curious a good explanation of the equilibrium theory of tides can be found in Newtonian Mechanics A P French p531 et seq.

oggers
14th Mar 2014, 12:13
Oggers:

"Deep ocean tides on Earth don't exist"
"I don't understand when factors are too small to matter"
"I can't estimate the size of physical quantities and effects"

awblain, having seen you perform on other threads it seems you are more comfortable with that kind of post than one which includes a coherent technical argument to support your point.

I think at this stage it is fair comment to say that - in my subjective opinion of course - you seem to lack the technical nous to make an argument that will carry your point in a technical forum such as this.

You have failed to explain anything at all on this thread, you have merely piled up a list of vague rhetorical responses stuffed with ambiguous and woolly jargon.

Take this one as a case in point:

The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot.

If you can't see that, then you need more than a hundred-word comment to explain.


It is a purely subjective comment. But this is a technical forum. Why can't you explain what is wrong with his model? Because there is nothing wrong with it except in your subjective opinion it isn't much use. But you are not an accident investigator whilst he is. You are an anonymous lay-person, whilst he is a professional in the field who sees utility in the trajectory analysis he has developed.

Let's face it awblain, you aren't going to be using that or any other trajectory analysis to investigate in flight break-ups. I'm not either.

If you can't win by arguments strong - and you haven't provided anything resembling one - can you at least point me to your publishing history? For the third time of asking :rolleyes: Because, all I have to go on with you, is about 6 pages of waffle that any schoolboy with a search engine could cobble together.

MrSnuggles
14th Mar 2014, 13:03
I do not understand what you are discussing.

This is what I know:

Newton:1 - If a body is not subject to any net external force it either remains at rest or continues in uniform motion.

Newton:2 - F = ma (Law of acceleration)

Newton:3 - When two particles interact, the force on one particle is equal to and opposite to the force on the other.

So, when dealing with real life stuff you must consider air resistance as a force that wants to slow things down per N:3. The force D working on the object is calculated by the drag equation:

D = 1/2(rho)CA(v^2)

C is the drag coefficient for that particular object. This is not constant but varies with its shape and size - and of course what kind of air flows around it (wind conditions etc).

The force T acting on a falling object in air would then be:

T = F+D

which is dependent on mass to some extent (greater size usually means greater mass).

As for the debate about varying gravitational forces, I do not understand what you are fuzzing about. The gravitational coefficient g is calculated like:

g = GM/R^2

The interesting part here is R^2 which means the radius squared. As the earth is elliptical and not a sphere, this value will vary, depending both on altitude and geographical position relative to the earth's core. Thus, the 'g' will vary across the earth. But it is decided for all practical engineering reasons that g=9,81 uniformly. (Actually sometimes engineers use g=10 just to build in extra margins of safety and make calculations easier.)

Tidal water again, is explained by the gravitational forces distributed between earth and moon. Newton summed it up in his Law of gravitation:

F = G(Mm)/r^2

The moon will therefore exert some gravitational force on that part of earth that is closest to it. Since land does not move that easily, water is the main reason we notice this mutual force between out space home and its moon. The water is "attracted" to the force and moves in the direction of the moon which causes tidal flows.

Tourist
14th Mar 2014, 13:28
MrSnuggles

"The moon will therefore exert some gravitational force on that part of earth that is closest to it. Since land does not move that easily, water is the main reason we notice this mutual force between out space home and its moon. The water is "attracted" to the force and moves in the direction of the moon which causes tidal flows. "

Don't forget the increase in centripetal action on the far side of the earth moon pair due to the increase in radius of orbit around the centre of earth mass that acts to reduce gravity causing the other "lump" of water.

Mr Optimistic
14th Mar 2014, 13:30
And the tidal bulge on the other side is caused by...?

Tourist
14th Mar 2014, 14:03
Are you asking him or me?

Mr Optimistic
14th Mar 2014, 14:09
Him....you got there first. A couple of complications owing to the cm of the system being within the earth (3000 miles from centre) and the earth not revolving owing to this motion ie not a dumbbell which makes it hard to visualise as the cm of the earth moves about the common cm.

awblain
14th Mar 2014, 14:38
Oggers, I'm sure you are as capable of using a search engine now as you were six hours ago when I answered your question.

You are an anonymous lay-person

So you say. Once again, your knowledge and intuition appear to be letting you down. I'm not sure how posting under a name is in some way anonymous?

You're right that I'm probably not going to fix wreckage positions anytime soon, but I could if I was asked to, as could pretty much anyone with even a first degree in physics.

I don't understand why you've been trying to pick a fight with me about physical reality for a couple of days.

Tourist
14th Mar 2014, 15:00
But I and many others do understand....

Mr Optimistic
14th Mar 2014, 15:01
A Squared, yes it was considering the variation with height that is the guys undoing. As for temporal variation I was trying to figure out if that was on anyone's mind.

A Squared
14th Mar 2014, 15:05
Oggers, I'm sure you are as capable of using a search engine now as you were six hours ago when I answered your question.

Well without intending to take sides, I have to say that oggers' ad-hominem approach regarding publications is a spectacular failure.

oggers
14th Mar 2014, 15:10
Mr Optimistic

The problem with the poor professor's paper is that it is quite simply unintelligent to include the range variation of gravity as its very small perturbing effect relative to a constant model is less than trivial when compared to the other forces acting and the inherent limitations of the model.

As this is a technical forum I was hoping you would flesh this out by showing the calculations. I really don't think it is fair comment to label the guy as unintelligent and expect forum users to simply take your word for it. Has he done something to your sister or do you have another reason for the ad-homs approach?

As neither yourself or awblain have shown any inclination to do the sums I thought I'd do them myself:

According to his paper g is adjusted by the following equation:


gₐ = gₑ x (rₑ ÷ rₐ)²

It is obvious that at low altitudes this is going to make no difference. But what is the upper limit where a tool such as this would be used? A case study would be the Columbia disaster.

The altitude of that break-up was ≈ 64km. The Earth radius is ≈ 6380km at the equator (WGS84). So, rₑ is 6380km and rₐ is 6444km. Using 9.81 for gₑ in the above equation and evaluating for gₐ results in 9.62ms⁻².

By using the equation:

R = V₀ x √(2h ÷ g)

..we can get a feel for how much difference this would make to horizontal range, using the velocity at break-up ≈ 6000ms⁻¹ and evaluating R when g = 9.81 and then again when g = 9.62

This would not include drag. The answers are 685.4km and 692.1km

So, up to about 7km would be the maximum difference this technique would make, before drag. Clearly, drag would reduce this.

Whether or not you think that is trivial depends on one's point of view. I think that if his model allows him to make such a correction he may as well have it. Even if it isn't necessary.

awblain
14th Mar 2014, 15:22
This centre of mass will also be mobile as the moon revolves around the earth.


There we have a problem, since the natural frame to choose is the one in which the center of mass is fixed and both bodies mutually orbit around the center of mass. That way you see the motion of the tide-susceptible parts of the Earth clearly.


Thus, a person standing on the earths surface will be moving in and out relative to the centre of mass of the earth/moon pair and will experience (very tiny! 2micro m/s2) variations in gravity.


Only because the Earth is rotating at a different rate to the Earth-Moon system. On the Moon this is not the case, since the Moon is tidally locked, yet there are still tides raised. That discussion of in-out motion also jumps right past where the tides arise.


They would also be moving in and out relative to the L1 Lagrange point, ie varying their range from a point of zero g

L1 does not have zero gravity. It has a zero gradient of gravity. If it had zero gravity, things would not be able to be kept orbiting there, but would instead drift off in a straight line.

awblain
14th Mar 2014, 15:36
Oggers,

The Columbia situation is the same point point again there's wood to focus on before worrying about the trees. You don't need to do irrelevant sums, you can just say that different sums need to be done. You don't ask for sums at a PhD exam, you ask for insight, intuition and an understanding of the issues.

Here's one reason why - and one is sufficient to drop the point and think more about it.

Vertical changes in gravity by 1% are a minor detail, since the break up was not at a point - it took tens of seconds, and spread over tens of kilometers. Thus your uncertainty in initial position makes a potential 1% shift in a 700km track neither here nor there.

Remember that the drag for the whole shuttle brings it down in Florida about 4000km after substantial contact with the atmosphere, and the actually distance from break up to debris hitting the ground was only 100-200km, so saying anything numerical "without drag" is like focussing on a flea while missing a bear it is sitting on.

Mr Optimistic
14th Mar 2014, 15:38
Awblain's address is Pasadena. Judging by his accurate responses I guess he works in an engineering capacity for a tech company perhaps associated with JPL. I studied for a couple of years at a place down the road there on California Blvd. Not sure why anyone would expect to learn much physics on pprune though when there's a world of introductory texts out there.

A Squared
14th Mar 2014, 15:44
Awblain's address is Pasadena. Judging by his accurate responses I guess he works in an engineering capacity for a tech company perhaps associated with JPL.

Swing and a miss.

Y'know, it isn't all that difficult ...

... at a place down the road there on California Blvd.

warmer ...

Mr Optimistic
14th Mar 2014, 15:47
Well sometimes you just have to hazard a guess :)

Tourist
14th Mar 2014, 17:41
awblain

No such things as a "natural frame"

Discussion of moon tides is obfuscation. These are due to an eliptic orbit, ie the moon does not face the exact same side to the earth at all times and the tides due to the sun.

Neither of these take away from the explanation.

I'll give you that about the L1 point. I was being simplistic.
In an inertial frame the zero g point is a little further out than L1.

Tourist
14th Mar 2014, 17:44
Incidentally, if you are "the" A W Blain, I hope your books are less irritating to read.

Mr Optimistic
14th Mar 2014, 17:58
Natural frame just means the one that makes the formulation easiest, nothing of cosmological significance implied.

CaptainEmad
17th Mar 2014, 08:03
ANCPER,
I miss your entertaining style.

How did you go with conjuring up another mathematical proof that the mass of different objects has no effect on the time taken for those objects, (aircraft pieces in this discussion) to reach the ground?

Quote:

"What's the world coming to, when pilots don't know basic physics"
Not a pilot yourself then? And if you are a pilot, then I apologise. You have proved yourself correct with that put-down.

And why on earth would you disagree with A squared?
The post he/she delivered was clear, humorous, accurate and well written and based on "basic physics".

Eagerly awaiting more of your entertaining posts.

A Squared
17th Mar 2014, 08:21
ANCPER,
I miss your entertaining style.

How did you go with conjuring up another mathematical proof that the mass of different objects has no effect on the time taken for those objects, (aircraft pieces in this discussion) to reach the ground?

Quote:

Not a pilot yourself then? And if you are a pilot, then I apologise. You have proved yourself correct with that put-down.

And why on earth would you disagree with A squared?
The post he/she delivered was clear, humorous, accurate and well written and based on "basic physics".

Eagerly awaiting more of your entertaining posts.

Yes, Where is ANCPER? He's gone oddly quiet. I was really looking forward to his next iteration of why I'm wrong.

I do hope nothing has happened to him.

ANCPER
17th Mar 2014, 09:56
Glad to see your concerned for my health!

No, I hadn't slunk off anywhere. I was waiting for response higher up the food chain, so had to wait out the w/end to get a response and yes I concede. Damn! :O Yes, it's too dumb to say anything further!

A Squared
17th Mar 2014, 10:08
;) Thanks for being a good sport about it.

CaptainEmad
17th Mar 2014, 10:19
All is not lost ANCPER.:ok:

If you are a pilot, which doubtful in my opinion, then there may be truth to your assertion that some pilots don't know basic physics.

You being one of them.


But back to the thread...

It is turning out like we need to discuss the physics of disappearing objects rather than falling ones.

Mr Optimistic
17th Mar 2014, 17:57
If it's any consolation I admit to being confused too.

FE Hoppy
17th Mar 2014, 19:14
Good thread this one.

Nice to get away from the MAS hysteria on R&N.

Mr Optimistic
17th Mar 2014, 19:33
Unfortunately I have made some posts which only contributed to the noise so would be hypocritical to comment. Strange that so many people lost their lives in a manner which may have been awful (hope not) and we are indulging our technical curiosity...and I am guilty of it too.

flyer101flyer
18th Mar 2014, 16:25
I just read the first 10 posts of this thread. Man, this internet thingy, it is really something, isn't it?

Any high school physics student could set you all straight; good thing high school physics isn't needed to fly airliners...

MrSnuggles
21st Mar 2014, 13:59
Ok, this is on the matter of actual falling objects; airplanes in fact.

To start with: I am NOT entering the weird world of tin foil hats. I could not care less about what happened. The return of Elvis. Evil gnomes. Alien abductions. AlQaeda, IRA and PLO joint operation or any government cover up. I really really DO NOT CARE.

I am only interested in the physics, ok?

So, with this in mind, I would like to know if anyone here knows of ANY kind of calculation that would make, say TWA800, drop its nose and then continue to climb a few hundred metres (thousands of feet). I have tried different ballistic calculations but can only get the plane up to some tens of metres (about hundred feet actually). Some scenarios I tried:

- Approximating a B747 fuselage shortened with the length of the nose but CoG intact from official documents of a whole 747.

- As above but with different CoG.

- Approximating a B747 without a nose, instead a metal skin at the point of breakup, CoG intact from official documents of a whole 747.

- As above but with different CoG.

In all scenarios I assumed that all four engines after-the-fact were providing equal amounts of climb thrust and constant climb ratio per official documents, although I know that may not be the case depending on what wires may or may not have been severed.

Ballistic calculations are not the best way to approach this case because you have to take into account the various drag forces that would make a noseless plane do strange things, but statistically analyzing my calculations I still can't get the plane to make such an extraordinary trajectory even with significance 5%.

Do anyone here know if that animation of the trajectory after (whatever happened) is based on scientific calculations or if it is just a conjecture based on eyewitnesses statements?

Please, no endless tirades about your preferred theory about what happened. CFT explosion sounds good enough for me. I am just baffled by the trajectory - that is my only concern.

A Squared
21st Mar 2014, 14:34
Well, for what it's worth, I wondered the same thing while reading the official explanation of the airplane "streaking" upward after the CWT explosion.

Without going thru the numbers again, what I recall was that if you assumed that the airplanes velocity was suddenly directed upward, without significant loss of energy the altitude gain suggested was at leas semi-plausible.

The problems with that of course was that the reason the airplane suddenly turned up was the CWT explosion blew the nose off.

This introduces two problems:

1) The huge amount of drag from having a gaping hole in the fuselage instead of the forward half would not be a low drag feature.

2) The post explosion airplane was assumed to achieve a stable zoom climb following the loss of of the forward fuselage. This is inconsistent with the part of the theory which says that the loss of the forward fuselage caused it to depart stable flight upward. The loss of the forward fuselage either caused it to become unstable, or it didn't. It can't really do both, which is what the theory requires.

You can only figure that it climbed thousands of feet if you completely neglect drad and aerodynamics and focus only on how high the kinetic energy would take it in a vacuum.

The "zooms upward" scenario fails on an entirely different level. The raison d'etre for the "zoom upward" theory was to explain why so may people reported seeing something "streak" upward. It serves no other purpose, other than to provide an explanation for the witness accounts of a fire streak going up. The theory being that the post-explosion airplane was what they saw "streak" upward. The problem with that was the distance, altitude and speeds involved. From memory, the airplane was at 14,000 ft (or was it 15K?) when the explosion occurred, and (again from memory) it was something like 40 miles away from the nearest witness who reported something 'streaking upward"

We have all watched airplanes fly. an airliner at 14,000 ft altitude, and even just 10 miles away does not "streak" across the sky. It crawls. So if from the observers perspective it appears to be crawling across the sky, if that same airspeed is suddenly directed perfectly upward without loss of speed, it's not going to appear to "streak" upward any more than it was "streaking" across the sky horizontally.

Now, I'm not trying to start a conspiracy theory discussion either. I have no idea what happened to TWA 400, There are all kinds of reasons why a shoot-down is really implausible, and there are all kinds of reason why a CWT explosion seems the most plausible explanation.

I'm just saying that the post-explosion climb as an explanation for witness reports of fiery streaks going up to the airplane is a complete farce. And that is the only reason it exists.

MrSnuggles
21st Mar 2014, 14:49
A Squared:

The raison d'etre for the "zoom upward" theory was to explain why so may people reported seeing something "streak" upward. It serves no other purpose, other than to provide an explanation for the witness accounts of a fire streak going up.So to your knowledge there has been no scientific mathematical model applied on that animation?

As I said, and gave some examples of, nothing I have come up with even remotely resembles that kind of trajectory. In all calculations I have done, even when just considering the B747 as an intact plane just shortened fuselage by the length of the nose, the CoG inevitable goes backwards and causes the plane to stall after a very short period of time. Assuming all engines work completely fine* I do get some upwards movement, but not even in the range of thousands of feet.

But, as I said. These are all ballistic calculations and as we know that is a huge amount of drag that are not taken into consideration when calculating ballistic scenarios.

I have not made any drag (non-ballistic) calculations because honestly they are SOO much work!


*at the point of explosion and in some scenarios for 2s afterwards. I chose 2s completely arbitrary.

awblain
21st Mar 2014, 17:31
High-flying aircraft that are approaching you "streak upwards"; those that are receding from you "dive downwards". Things come up over the horizon, and they go down below it, even if they are traveling along horizontally.

The "streak" in TWA800 was from the fire initiated by the initial blast in the center tank. It connected to a "fireball" from the bulk of the rest of the fuel burning.

For a fine example of journalistic integrity and inquisitiveness, after a slow start, and some associated conspiracist buffoonery, google "LA missile launch catalina helicopter". Mystery Missile Launch Seen off Calif. Coast - CBS News (http://www.cbsnews.com/news/mystery-missile-launch-seen-off-calif-coast/)

HazelNuts39
21st Mar 2014, 17:32
Do anyone here know if that animation of the trajectory after (whatever happened) is based on scientific calculations or if it is just a conjecture based on eyewitnesses statements?This is what Wikipedia says about your question :

"Only the FAA radar facility in North Truro, Massachusetts, using specialized processing software from the United States Air Force 84th Radar Evaluation Squadron, was capable of estimating the altitude of TWA 800 after it lost power due to the CWT explosion.[99] However, because of accuracy limitations, this radar data could not be used to determine whether the aircraft climbed after the nose separated.[99] Instead, the NTSB conducted a series of computer simulations to examine the flightpath of the main portion of the fuselage.[100] Hundreds of simulations were run using various combinations of possible times the nose of TWA 800 separated (the exact time was unknown), different models of the behavior of the crippled aircraft (the aerodynamic properties of the aircraft without its nose could only be estimated), and longitudinal radar data (the recorded radar tracks of the east/west position of TWA 800 from various sites differed).[101] These simulations indicated that after the loss of the forward fuselage the remainder of the aircraft continued on in crippled flight, then pitched up while rolling to the left (north),[98] climbing to a maximum altitude between 15,537 feet (4,736 m) and 16,678 feet (5,083 m)[102] from its last recorded altitude, 13,760 feet (4,190 m).[21]"

P.S. A quick sum shows that an object travelling at 400 kts can gain about 7000 ft of height ballistically.

awblain
21st Mar 2014, 22:57
Tourist

Discussion of moon tides is obfuscation. These are due to an eliptic orbit, ie the moon does not face the exact same side to the earth at all times and the tides due to the sun.

I think you'll find that circular orbits produce perfectly fine tides.

awblain
21st Mar 2014, 23:20
HN39,

You could turn speed into height - at about that rate,
Delta-h proportional to Delta-v-squared.

Lop off the front, but keep the wings intact on the rest, the angle of attack would rise sharply, boosting both lift and drag, and pointing whatever thrust was left upwards, so slowing and climbing until a stall makes sense. I can see 2000 feet gained being reasonable.

That's probably not what the witnesses would have seen as the "streak" though, that's all about how a short fiery track on the sky that ends in a bigger fire appears. Unless you're directly abeam of the track, a horizontal track would still appear to rise or fall above the horizon.

MrSnuggles
21st Mar 2014, 23:21
HazelNut39
P.S. A quick sum shows that an object travelling at 400 kts can gain about 7000 ft of height ballistically. Thankyou for that Wikipedia citation. I have read the NTSB report but apparently failed to notice that part of it.

Regarding the quote.. What object are you calculating? I based my calculations on an airplane (first approximated with a cone, then with a real airplane) albeit in some weird configuration(s). Is your object a sphere (= bullet)? Larger than a bullet?

I'll have to do some equations tomorrow methinks...

awblain:

I must do something wrong here because my calculations are inaccurate with an order of ten if this climb trajectory is factual. Mind you, I do all calculations in the SI units.

awblain
22nd Mar 2014, 00:00
Mr S.,

Be careful with my Delta there - it's a "change all the kinetic energy to potential energy" Delta, so gives the maximum possible height gain, and not a normal small-change small-d delta.

(Delta-v)(Delta-v) = 2g (Delta-h).
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.

It's for a dragless/liftless object, so more like a bullet than an airliner.

For a normal small change delta-v: v delta-v = g delta-h.

There's also the remaining thrust at work in TWA800, assuming the engines kept running.

HazelNuts39
22nd Mar 2014, 00:42
awblain,

You are of course correct in your observation that my 'simple sum' is based on nothing else than the conversion of kinetic into potential energy. That is no more than a first approximation (an upper bound) of what can happen in this sort of scenario.

I probably misunderstood MrSnuggles when he wrote:
These are all ballistic calculations and as we know that is a huge amount of drag that are not taken into consideration when calculating ballistic scenarios.

I have not made any drag (non-ballistic) calculations ...

awblain
22nd Mar 2014, 10:41
HN, It's a good upper bound, and probably just as much use as the NTSB's simulations.

The thing that is striking about the TWA breakup, which I hadn't appreciated, was that the wing box remained intact after the explosion, and all the initial damage was to a collar forward of the wing root, causing the front to separate. From the reconstructed wreckage put back together on the scaffold, I'd assumed it had all come apart together.

This pattern of damage seems rather similar to Lockerbie, which in hindsight isn't too surprising from an explosion in the lower fuselage.

oggers
24th Mar 2014, 13:31
awblain

(Delta-v)(Delta-v) = 2g (Delta-h).
As 1kt is very close to about 0.5m/s, for 400kt, Delta-h ~ 200x200/20 ~ 2000m, as HN39 said.
It's for a dragless/liftless object, so more like a bullet than an airliner.


You've contradicted yourself:


saying anything numerical "without drag" is like focussing on a flea while missing a bear it is sitting on.

:ok:

awblain
24th Mar 2014, 16:58
Oggers,

You may wish to avoid selectively quoting part of one statement, instead of including the relevant bit that "2000 feet gain being reasonable", and not adding in a separate completely disconnected comment about the bears and fleas, which pertained to a quite separate discussion.

While you're at it, you can also ponder the statement about HN's number that "It's a good upper bound, and probably just as much use as the NTSB's simulations."

nonsense
30th Mar 2014, 08:43
In sailing forums, people have amazing arguments about the use of water ballast in yachts - 'cos any fool knows water is no heavier than water so it obviously can't exert a downforce, right?

In motorcycling forums people will argue vehemently that raising your body above the bike by standing up on the footpegs will power the centre of gravity of the bike - 'cos it's obvious to any fool since your weight is then supported by the foot pegs rather than the seat, right?

But I never thought I'd see professional pilots arguing that the density of otherwise identical objects would not affect their terminal velocity!

I guess its a sort of second order "any fool" thing.
1) 'Cos "any fool" knows that heavy objects fall faster, right?
2) Except of course that "any fool" remembers Galileo and the cannon balls and knows that is wrong and that gravity acts at (~)9.8 N/kg on light and heavy objects alike, right?

1 N = 1 kg m s^-2 so 9.8 N/kg = 9.8 m/s^2.

Gravity is a force per unit mass, or an acceleration, while forces due to drag on identical shaped objects of different densities (and hence different masses) are simply forces which result in an additional acceleration which is dependent upon the mass.

Thanks for an entertaining read.

Mr Optimistic
30th Mar 2014, 19:32
Ah, the theatre for scientific illiterates has re-opened.

A Squared
30th Mar 2014, 19:38
Ah, the theatre for scientific illiterates has re-opened.

Dunno, it's kind of a difficult post to , but I *think* his point is that the he agrees that the density of a real object falling though a real atmosphere *will* affect the velocity. I *think* it's just a case of him being a bad writer.

I could be wrong though.

Mr Optimistic
30th Mar 2014, 19:44
I don't see a problem that needs an answer.

Here is a clue. Ballstic coefficient.

oggers
3rd Apr 2014, 08:22
The poor guy's written a decidedly mediocre conference proceeding that advances Mr Stokes' work from the 19th century not one jot.

Awblain; this is a model for approximating motion you are talking about, a tool for accident investigation, not a tilt at a Nobel prize. If it is subjective discourse which floats your boat, take it to jetblast. That's all I have to say on the matter.

nonsense
4th Apr 2014, 14:42
Video: Skydiver almost gets hit by meteorite (http://sploid.gizmodo.com/holy-f-ck-this-skydiver-almost-got-hit-by-a-meteorite-1557739698/+jesusdiaz)

Apart from being an interesting read in itself (a skydiver managed to film a small falling meteor passing him as he descended), this is an interesting reminder that terminal velocity is not the fastest an object can fall, it is the stable falling speed an object will tend towards regardless of whether it starts out slower or faster. It is the speed at which wind resistance up is equal to the mass times the acceleration due to gravity down, resulting in no net acceleration.