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# Physics of falling objects

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# Physics of falling objects

10th Mar 2014, 08:19

Join Date: Jun 2002
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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
10th Mar 2014, 08:38

Join Date: Dec 2007
Posts: 633
Isn't building an evacuated sound stage about as much trouble as flying to the Moon?
10th Mar 2014, 08:43

Join Date: May 2006
Location: UK, South East
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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!
10th Mar 2014, 08:50

Join Date: Dec 2007
Posts: 633
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.
10th Mar 2014, 10:02

Join Date: Apr 2013
Location: New Jersey USA
Age: 61
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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.
10th Mar 2014, 10:16

Join Date: Dec 2007
Posts: 633
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.
10th Mar 2014, 13:18

Join Date: Feb 2000
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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!
10th Mar 2014, 13:25

Join Date: Feb 2000
Age: 56
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Originally Posted by compressor stall

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.
12th Mar 2014, 11:07

Join Date: Nov 2005
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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.
12th Mar 2014, 11:30

Join Date: Dec 1998
Location: Escapee from Ultima Thule
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You're quite right. That was my typo. Corrected it now. Don't do longish posts using a phone...
12th Mar 2014, 12:39

Join Date: Oct 2003
Location: UK
Posts: 1,493
Try this paper by Dr Matthew Greaves, Snr Lecturer in Accident Investigation at
Cranfield Safety and Accident Investigation Centre

Trajectory Analysis

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
12th Mar 2014, 14:38

Join Date: Dec 2007
Posts: 633
… 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?
12th Mar 2014, 17:22

Join Date: Dec 2012
Location: nowhere
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What's the world coming to

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.
12th Mar 2014, 18:40

Join Date: Jun 2011
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Originally Posted by ANCPER

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.
12th Mar 2014, 18:53

Join Date: Apr 2009
Location: USA
Age: 55
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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?
12th Mar 2014, 19:17

Join Date: Oct 2003
Location: UK
Posts: 1,493
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.
12th Mar 2014, 20:21

Join Date: Dec 2007
Posts: 633
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.
12th Mar 2014, 22:24

Join Date: Oct 2003
Location: UK
Posts: 1,493
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.
12th Mar 2014, 22:36

Join Date: Dec 2007
Posts: 633
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?
12th Mar 2014, 22:57