I wrote this bunch of reflections when in the Valley, returning after 29 years...

My land

Consider the constant migrations of people all over the world , from the smallest groups to the largest, out of Africa, eddying and swirling across continents, even crossing oceans, and it becomes clear that no particular land ‘belongs’ to any specific group. The flows of genes (Y- and mitochondrial), languages and cultures - all of them continuously changing – have been used to map the migrations, often with contradictory results.

What does it mean to be the ‘original’ inhabitants of Africa, if these autochthons just barely qualify as being (recognizably) ‘human’? Even hunter-gatherer societies are territorial, but with the advent of agriculture and ‘settled’ populations, land ownership became even more of an issue. So, to define ‘original’ inhabitants, do we stop at ‘recorded history’ – or do we carry on digging into the often mythical past?

Plants and animals often seem to be more rooted than humans to particular geographical areas – but even they have travelled extensively, sometimes on their own, most often using birds and insects, and sometimes with human aid.

Still we are tied to the land we live in by quirks of geography, weather and food which affect our customs, clothes and language in odd ways. (Bengalis must have hilsa, mutton for Kashmiris, maple syrup for the Canadian…) There are ‘geographical indicators’ in our genes e.g. the lung capacity of Sherpas and Tibetans and the sickle cell anaemia (malaria-resistant) genes in parts of India and Africa. Other traits may not be so obvious. Some mutations may be neutral.

Although we are ‘tied’, we are not truly bound to a given place: one sibling may stay a whole lifetime in one place, another may land up in some other continent, and yet another may be travel all over the world and never settle anywhere – while eyeing the other planets in insatiable wanderlust. In principle, anyone could live anywhere – but in practice there are many barriers to overcome.

But in what ways does a land alter your perceptions, your language? The story that the Eskimos have 40 words for snow turned out to be an exaggeration. Yet it is true that the seasons and the plants (some of them carried along with us), leave their imprints upon our languages. Landmarks acquire historical overtones and memories that last centuries if not millennia – like Mt. Kailash for Hindus or the Wailing Wall for Jews - even if they have been 'lost'. These can arouse feelings of connection with the land, and feelings of ownership. Most of these seem to be for religious reasons, which would naturally get entrenched in the collective memory of a community or group.

But in many places this sacred feeling can be fixed on a river (the Ganga, the Nile, etc) or a lake, that would mostly be adjacent to a temple. It could be a place of pilgrimage (e.g. the Amarnath Yatra, Kailash Manasrover Yatra - which is a mountain and a lake – or Lake Baikal,…).  Often the river has sufficed to demarcate ‘us’ and ‘them’… but the paths of neither rivers nor men are invariable and stable.

Imagine belonging to a nomadic community that has always gone South in the winter, and returned North in summer, along well-worn paths marked and traced by numerous generations of ancestors, along with ‘their’ herds of reindeer, or cattle, or whatever…

The feeling of being attached to some place could also be related to some purely individual memory: like some place you went to with your parents as a child… or it could be an accretion of memories, just the sights, sounds and smells of the village or town that you grew up in…

For many a Kashmiri Pandit, his or her identity is bound up with temples of Shankaracharya, Martand, Tulla Mulla,.. And with the lakes and gardens of the Valley, the apples, the mulberries, the rainbow trout … yet, it may just be the very quotidian, mundane place you grow up in…it doesn’t have to be a putative paradise! 

But it is very hard to pin it down: why do we have these feelings for our ‘homeland’? For me, the song from the film Kabuliwala, “Ae Mere Pyare Watan…” comes closest to expressing the ineffable feeling of nostalgia and homesickness. I do not relate to the ridiculous anthropomorphism of “Bharat Mata”…

Farmers and fishermen are closest to the land and the sea. They are tuned to the seasons and the soil, and the tides and currents of the sea. Local knowledge is important: take the case of the Turkish farmers who refused to remove stones from their fields, because they – counter-intuitively – boosted their yields. That knowledge, often accumulated over generations, ties farmers to their land.

But this is a paradox: we don’t ‘own’ the land – it doesn’t even need us! – and yet we have this feeling of ownership. And particularly in the Anthropocene, humans have made irreversible changes at the global level, leaving behind detectable residues of plastic, concrete and radioactivity.

Food is sourced from so many countries today that we are in denial of the seasons, of the annual and decadal rhythms of the land. The elites imagine themselves as global citizens – and think that there is no price to be paid for these extravagant, profligate ways of global citizenry.

Does the land ‘belong’ to us? No matter how many fences we weave, how many walls we erect, how many canals we dig, dams we build, whether we map it with rulers, theodolites or GPS – the land does not need us – as in Nevil Shute’s “Earth Abides”, it will be there long after we are dead and gone and long forgotten, both as individuals and as a species, that tends to think too much of itself. We may irrigate the land with water, tears or blood, and we may indeed belong to the land – as do countless other organisms, ranging from bacteria to whales, but it does not belong to ‘us’.

Maximum height of (mostly) mountains and a flagpole:

Victor Weisskopf gave a derivation of the limit to the height of a mountain. This can be found in different sources (with minor numerical differences) [1].

The simplification is that he assumes a silicon dioxide (rock) mountain of area A and height h on a SiO₂ base. Its height is at its limit when its weight Mg bearing down on the base has a total potential energy Mgh that is greater than the liquefaction energy of SiO₂ E_l. That is, it would cause the base to turn into liquid, and it would sink into the liquid. The value of E_L is assumed to be 0.05X the binding energy of SiO₂ E_BE, and E_BE is O.2X the Rydberg energy (13.6 eV/atom). So, E_L = (0.05)(0.2)(13.6) = 0.136.

The mass of the mountain can be obtained from the volume of the mountain (V = Ah) multiplied by the number density of SiO₂ n_SiO2 (atoms/m³) and the mass of the protons & neutrons (assumed equal) m_p (1.67x10^-27 kg) and the mass number A_MS (the number of protons and neutrons in SiO₂):

M = Ahn_SiO2m_pA_MS

The mountain sinks by a distance z and displaces a volume Az of SiO₂ in the base, and the potential energy (on the l.h.s. equals the energy of liquefaction of the volume Az on the r.h.s.):

Mgz = n_SiO2 Az E_L

Or, cancelling z:

Mg = n_SiO2 A E_L

But M itself involves h, so:

Ahn_SiO2m_pA_MSg = n_SiO2 A E_L

The number density n_SiO2 and the area A both cancel, leaving:

h = E_L/(m_pA_MSg)

Numerically, this becomes:

h = [(0.136)(1.6x10^-19)]/[(1.67x10^-27)(60)(9.8)] = 21.6 kms

a)      For a conical mountain, the mass M will depend upon the volume, which is now Ah/3. So for this case, the height will go up by a factor of 3X.

b)      Weisskopf’s argument takes into account energy of liquefaction, whereas much before the base turn liquid, it would undergo plastic flow. This would lead to a lower and more realistic value of the maximum height.

A similar, though much simpler argument is given by Gnadig et al [2]:

The base of a mountain that does not melt under its load if:

h < L/g

where L = latent heat of melting of metals (200-300 kJ/kg), and the answer for Earth is 20-30 kms.

Gnadig  [2] also points out that on Mars g ~ 4 m/s², and in fact, the highest mountain on Mars is Mt.Olympus which is 26 kms high, according to Gnadig, but 21.9 kms high according to Caplan [3].

P.A.G.Scheuer [4] was more interested in the question of how high a mountain can be…on a neutron star (he gets a range of 0.04 to 0.4 mm for the maximum height of his mountain range!). But on the way, he deigns to look at Earth too. Scheuer gives two answers h₁ and h₂, where h₂ is much higher and refers to a broad-based pyramid:

h₁ =4 (1.5x10⁷)/[(2650)(9.8)] = 2.31 kms

Scheuer gives this number as 2.25 kms assuming a value of yield stress of Y = 1.5x10⁶ kg/m² which equals 15 MPa ( the latter is the kg force multiplied by g).

h₂ = (h₁b/4)^1/2 where b is the base of the mountain, such that b >> h.

Using an unconfined compressive strength value of 100-250 MPa for granite gives a much larger number for h₁ than the one Scheuer got:

h₁ = 4Y/(rg) where Y is the yield stress, Y = 250 MPa for granite and granite density r = 2650 kg/m³.

h₁ = 4(2.5x10⁸)/[(2650)(9.8)] = 38.5 kms

Miskinis [5] gives a somewhat different argument, using the Hooke’s law relation between the shear stress s_r on compressed rock and the shear angle q:

s_r = Gq

where G is the rock shear modulus.

Then he uses the weight of the mountain argument on the surface area S, to find the maximum height, giving:

s = mg/S = grh_maz

Solving:

h_maz = Gq/(rg)

Substituting values of G = 12.5x10⁹ Pa for basalt, r = 10⁴ kg/m3 for the increased density of compressed rock, and q = 0.1 rad (or 6°), he gets a maximum height of 12.5 kms.

The mountain argument can now be applied to an iron flagpole, assuming it to be an iron mountain on a silica base, adapting Weisskopf’s argument ( so why bring up Gnadig, Scheuer and Miskinis? Never mind!).

A similar argument can be used for an iron ‘mountain’ of area A and height h, on a SiO₂ base, giving:

Ahn_Fem_pA_MSg = n_SiO2 A E_L

In this case, the number densities are different on both sides of the equation: n_SiO2 = 2.66x10²² and n_Fe = 8.49x10²² cm^-3, and A_{M, Fe} = 55. The number densities are obtained from the density, the mass number A_M and the Avogadro number: n = rN_A/A_M. The areas still cancel, so:

h = [(2.66x10²²)(0.136)(1.6x10^-19)]/[( 8.49x10²²)(1.67x10^-27)(55)(9.8)] = 7.6 kms

So, if one goes by Weisskopf’s argument there is plenty of scope for building higher flagpoles, since even the Jeddah flagpole is just at 170 m!

Of course, a structural engineer could probably come up with a much lower – and more realistic -- limit by invoking other problems such as yielding or buckling… but that stuff is way too complicated… I found one reference on “Metal Flagpole Design” which is the American National Standard but it is much too iterative and demands too much knowledge of material properties.

And, why build a high flagpole when you can just as well stick your flag on top of some convenient hill or mountain? Tell that to the BSF!

References:

[1] http://www.hk-phy.org/articles/mount_high/mount_high_e.html

[2] Peter Gnadig et al ,”200 Puzzling physics problems” (Cambridge Univ.Press, 2001) p.197

[3] M.E.Caplan arXiv 1511.04297v1 “Calculating the potato radius of asteroids using the height of Mt.Everest”

[4] P.A.G.Scheuer “How high can a mountain be?” J.Astrophy.Astr.2 (1981) 165-69

[5] Paulius Miskinis “Mathematical modeling of mountain height distribution on Earth’s surface“ Geologija 53 (2011) 21-26

Flags, striped shirts and zebras

The Olympic motto is “Citius, altius, fortius” (faster, higher, stronger).

Our nationalist flag-wavers have decided to emulate this motto: our flags must go higher, and since the wind there is faster, the flags must be stronger.

Firstly, the Centre decreed (to inculcate ‘nationalism’) that every Central University must have a 207 foot high flag.

http://economictimes.indiatimes.com/news/politics-and-nation/national-flag-installation-at-each-central-university-to-cost-rs-45-lakh/articleshow/51062499.cms

The proposed flags will be modeled on the one in New Delhi’s Central Park, which is 207’ high, made of knitted polyester, is 90’x 60’ and weighs 35 kgs.

More recently, the Border Security Force has decided to put up a flag at the Attari-Wagah border by 2017 which will be 350’ high.

http://indianexpress.com/article/india/india-news-india/at-the-attari-wagah-border-a-national-flag-so-high-it-will-be-seen-in-lahore-2775528/

“The current record for the tallest national flag is held by the one atop a 293-ft-high mast in Jharkhand’s Ranchi. This 66-ft x 99-ft flag was hoisted in January by Defence Minister Manohar Parrikar — it was, incidentally, brought down today for repairs after being stuck at half-mast 10 days ago.”

The new flag, to be hoisted in Jan.2017, a top Border Security Force official said, that at 350 feet, “it can be seen both from Lahore as well as Amritsar (Amritsar and Lahore are approximately 18 km away from the international border)”.

“At that height, the flag would have to be proportionately sized, therefore it would also be the largest Tricolour. Officials say the material to be used for the flag is also being discussed because at such heights it may get damaged by rain and high winds.”

The first question is: will the flag be visible in Amritsar and Lahore. A little simple trigonometry shows that the distance to the horizon is given by:

S = Ö2hR

Where the height h of the flag is much less than the radius R of the Earth (assumed as 6,400 kms). If h is not much smaller, an h² term is to be added within the square root. With h as 106 metres, this distance is 36.8 kms. So the flag should indeed be visible, as claimed.

The second question is: will the flag be visible to the unaided eye (no binoculars or telescopes)?

The minimum angle resolved by the human eye is 0.5 milliradian, as opposed to the diffraction limited resolution of 0.2 milliradians under optimum conditions, according to the following website:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html

At a distance of 18 kms, the smallest visible size would clearly be: (18x10³)(0.5x10^-3) = 9 metres.

Even the Ranchi flag is 99’x 66’ wide (30x20 metres) which would translate to about 3x2 pixels – and the flag planned will be proportionately bigger.

However, to be a bit more precise, a flag like the American flag has a pattern of Stars and Stripes. The Indian flag has just 3 stripes (never mind the Chakra!). So can these stripes be seen?

According to the following website, the appropriate parameter is the visual acuity of the human eye:

http://www.normankoren.com/Tutorials/MTF.html#Human_visual_acuity

“The ability of the eye to resolve detail is known as "visual acuity." The normal human eye can distinguish patterns of alternating black and white lines with a feature size as small as one minute of an arc (1/60 degree or π/(60*180) = 0.000291 radians). That, incidentally, is the definition of 20-20 vision. A few exceptional eyes may be able to distinguish features half this size. But for most of us, a pattern of higher spatial frequency will appear nearly pure gray”.

This angular resolution is almost the same as that mentioned by the hyperphysics.phy site, so nothing much changes.

Of course, one might note that the colours in our flag are not black and white, and that would make the analysis more complicated. And to be absolutely rigorous one should take into account the modulation transfer function (MTF) of the eye, because, according to Norman Koren: “The statement that the eye cannot distinguish features smaller than one minute of an arc is, of course, oversimplified. The eye has an MTF response, just like any other optical component.”

The only way I’m gonna do the MTF analysis is:

 a) somebody pays me, or

b) I’m feeling really, really bored and I’ve got nothing to do…

Anyway, let me just start ending this post by noting that the world’s highest flag is in Jeddah, Saudi Arabia:

https://en.wikipedia.org/wiki/Jeddah_Flagpole

and it is 560 feet (170 m) high. The flag weighs 570 kilograms (1,260 lb) and is 49.5 metres (162 ft) long and 33 metres (108 ft) wide. The area of the flag is equal to ‘half a football field’.

http://www.asce.org/magazine/20140506-tallest-flagpole-in-north-america-rises/

At 400 ft, the flagpole will be the tallest the Shebogyan, Wisconsin, the company Acuity has constructed — and the tallest in North America. The flagpole will support a 60 ft wide by 120 ft long American flag, which will be weighted at the bottom to prevent it from wrapping around the pole. 

3rd question: can the unaided eye see the Jeddah flag from outer space, say the International Space Station? The ISS height is about 400 kms above the Earth, so the minimum size needed would be about:

 W_min = (0.5 x10^-3)(400 x10³) = 200 metres.

I guess it would not be visible – even assuming that the flag were laid flat on the ground – which would not be countenanced by any self-respecting nationalist. Maybe if we allow for a satellite just at the edge of outer space (100 kms altitude), the Jeddah flag may just be visible as (almost) one pixel.

Can we expect an arms race at the Attari-Wagah border, with each country trying to push up its flagpole even higher? Anyway, such a competition is better than war…

Which leads to: ….Q4: what is the limit? With existing materials, how high can the flagpole go? Victor Weisskopf gave a derivation for the maximum size of a mountain, which can be tweaked a bit…. Let’s go into that after we touch on striped shirts and zebras – in the next post (whenever that happens!).

Striped shirt:

 If we assume a shirt with 5 mm wide alternating black and white stripes, and taking the visual acuity (VA) of a person with 20/20 vision as 0.5 mrad, then the maximum distance at which the stripes can still be resolved is (0.005)/(0.0005) = 10 metres, under optimum lighting. At a greater distance, the shirt will just look gray – a very common experience that almost everyone would have had.

The dependence of VA on background illumination, for 100% contrast between stripes, according to Norman Kopeika in his book, ”A system engineering approach to imaging” (SPIE Press, 1998) p.389

 is:

Illumination (foot-lamberts, fL)	VA (mrad)
1	0.36
10	0.28
100	0.20

This applies to the photopic system i.e. to the cones in the human visual system.

1 fL = 10.76391 lux equivalent. Because 10.76391 = (3.2808)², the conversion factor from metres to feet.

Zebra stripes:

There is a paper on the ease with which predators such as lions, hyenas and humans can perceive zebra stripes – which are alternating white & black of different widths. The paper is entitled: “Zebra stripes through the eyes of their predators, zebras and humans” by Amanda Melin et al (March 2016). 

Reference: http://dx.doi.org/10.1371/journal.pone.0145679.

A human with 20/20 vision should in photopic (daylight) conditions be able to resolve the widest stripes (~3 cms wide) at about 180 m, while the lion should resolve the stripes at 80 m, and a hyena at 48 m – while a zebra would resolve the other zebra’s stripes at 140 m.

The authors argue that the ideas that the B&W stripes serve to ‘break up’ the outline of the zebra and make it difficult for predators to see them are unlikely to be correct because the vision of predators is not that good anyway. What I’m interested in is that the authors use three numbers for the visual acuity of humans:

Illumination level	Visual acuity (mrad)
Photopic (daylight)	0.145
Mesopic (dusk)	0.377
Scotopic (night)	6.77

The numbers for mesopic and photopic are close to those for the higher illumination levels (10 and 100 fL) above. As expected the zebra would have to be a lot closer for one to see the stripes by moonlight…

http://milgistrust.wildlifedirect.org/category/grevys-zebra/

The above links to a photo taken of a zebra at night and the stripes are clearly visible … but the distance is not stated, and besides the camera has optics…

To conclude: flags, striped shirts and zebras have something in common: the eye of the beholder.

Blowin’ in the wind: dust and fog

The question that I started with was: how does the wind decrease the concentration of dust particles in the atmosphere? I Googled a bit but I was unable to find the answer in terms that made sense to me. I’m sure this question has been done to death by now, but I regret to say I couldn’t find it. My bad, undoubtedly. Anyway, I tried the simplest approach, and that is what I have given below – but it got a bit complicated as it progressed. Still, nowhere as complicated as some computer code written by some group of atmospheric scientists! Nevertheless, the caveat is that the following analysis is not peer-reviewed and should be taken (at best) as heuristic.

The assumption, of course, is that the dust is fairly localized (because the sources are factories or cars in a city), and that the wind blows in clean air from somewhere else. Also, that the wind is steady, and blows in a specific direction at a fixed wind speed. So, the dust particles will only get transported away, from place A to neighbouring place B, but not dissipated as such (the number density of particles remains the same, only the location changes). A similar argument applies to fog droplets. This may also apply to fine rain droplets with a diameter of around 250 microns, as adding a horizontal component to their velocity, but may be less useful for large raindrops.

Q: How does a free particle in the atmosphere of a given diameter – in the range 2.5 to 100 microns -   move (horizontally) when the wind blows horizontally at, say, v = 1 m/s speed? This particle could be a dust particle, or even a water droplet. The density is assumed to be 1 gm/cc, although some reports quote higher values of density for dust particles.

Particles of diameter more than 10 microns tend to settle down to the ground pretty fast, while those whose diameter is less than 10 microns tend to remain suspended in the atmosphere for a significant amount of time.

Assuming the particle is spherical and has a density r_p, its mass is:

                                                          m = r_p4pr³/3

and its effective cross-sectional area for the wind is A = pr²(taking into account the cos(q) factor).

The force due to the wind is: F = (r_av²/2)(A) where the density of air is r_a = 1.225 kg/m³.

Equating this with F = ma, we can find the acceleration of the particle:

                                                         a = [3(v_a)²r_a]/[8rr_p]

For a 2.5 mm diameter particle, and assuming a particle density of 1000 kg/m³, and with wind speed v_a = 1 m/s, the acceleration is a = 180 m/s² or about 18 g’s. If this acceleration were to remain unchanged, the particle would hit sound speed in 2 secs!

Clearly that’s not gonna happen, and to make sure it doesn’t, we have to drag in Stokes’ law.

The force law then is:

                                                        pA - 6phrv = ma

The terminal velocity (horizontal) will be reached when acceleration ‘a’ becomes zero:

                                                        v_T = [r_a(v_a)²r]/[12h]

which is independent of the density of the particle r_p, but depends upon its size r!

The viscosity of air is: h = 1.8x10^-5 kg/m-sec and the density of air is: r_a = 1.225 kg/m³.

This gives the terminal velocity as 0.014 m/s (0.05 kph). Which seems to be pretty low!

Note also that smaller particles have lower terminal velocity, with v_T directly proportional to r.

However, Stokes’ law is only valid if Reynolds number is Re £ 1. It is defined as:

                                                     Re = [Dr_flV_rel]/h_fl

Where r_fl is the density of the fluid (1.2 kg/m³) and h_fl is its viscosity, D is the diameter of the particle, and V_rel is the relative velocity of the particle and the fluid (air). In the above case, even though the terminal velocity of the particle is low (0.014 m/s), the wind-speed (1 m/s) is higher, and Re is calculated on that basis.

When Re>10 or so the flow is turbulent, and the drag force is proportional to the square of the particle velocity v_p:

                                                    F = [r_a(v_p)²C_DA]/2

Where C_D is the dimensionless drag coefficient and the fluid density is the air density. The value of C_D for laminar flow (Re<1 approximately="" by:="" given="" is="" o:p="">

                                                      C_D = 24/Re

For this approximation, the square law force reduces to Stokes’ law. For a spherical particle, and large Re, C_D = 0.47.

The wind-speed for which Stokes’ law is valid is shown in the Figure below. The value goes up, as the particle diameter goes down:

The more general force equation involving the drag coefficient now becomes:

                                                  [r_a(v_a)²A/2] – [r_a(v_p)²C_DA/2] = ma

Since most terms cancel out for a = 0, the terminal particle velocity becomes:

                                                     v_p =  v_a/[C_D]^1/2

According to www.thermopedia.com/content/707 the following expression for the drag coefficient is valid in the range: 0.2 < Re < 2000:

                                                 C_D = [21.12/Re] + [6.3/(Re)^1/2] + 0.25

which reduces approximately to the limit 24/Re for low Re, and to 0.47 for high Re.

                                                                               

Note that the terminal velocity of the particle associated with settling under the action of gravity is given by:

                                                   6phrv = r_p [4pr³/3] g

Which leads to the settling terminal velocity:

                                                   v_sT  = [gd²r_p]/[18h]

The ratio of the above terminal velocity to the settling terminal velocity is:

                                                  [v_T/v_sT] = [3r_a(v_a)²]/[4gdr_p]

For a 25 mm diameter particle of density 1000 kg/m³, the ratio becomes:

                                                  [v_T/v_sT] = 3.67(v_a)²

i.e. even for a wind-speed of 1 m/s, v_T > v_sT. For a particle of smaller diameter, the wind-speed needed to exceed the settling speed increases.

An energy argument can be used, an integral of the force law argument used above, which gives the same result as the force law:

Note the work is done on the particle by drag, so you need to multiply the air pressure term by the distance (v_p dt).

The 1^st term is:

                                                  [r_p(v_p)²A/2] [v_pdt]

The drag term is:

                                                  [6phrv_p] [v_pdt]

The inertial term is:

                                                  (1/2)[d(m(v_p)²/dt](dt)

Removing the common term (v_p dt), you get the equation:

                                                  pA - 6phrv_p  = m[dv_p/dt]

which is the same as the force law case, with Stokes’ law drag force.

Using the above form of drag coefficient as a function of Reynolds number, no assumption need be made that Re < 1; that is, we need not be restricted to the Stokes’ law case, but can do the calculation for the general case, including both the square law and the linear (Stokes’) law cases.

The calculations become a nuisance to do by hand; so I wrote a simple BASIC program to do the calculations. Note that since Re is defined as the relative velocity between the particle and the fluid (the air in the atmosphere), that needs to be incorporated. First the acceleration is calculated, then the particle velocity, and then the relative velocity, and then the drag, based on the C_d that depends on Re.

The end result is as expected: the acceleration starts with a high value and quickly decreases to zero, while, correspondingly, the velocity increases exponentially, and then saturates at the terminal velocity. Note that this is a horizontal terminal velocity, not the vertical terminal velocity that we are accustomed to calculate, which latter is related to the settling rate of large diameter particles.

The velocity vs time graph:

From the above two graphs, it is clear that the terminal velocity is attained within 10 msec (for 250 mm particles), although this time is shorter for smaller particles.

The plot of v_p vs v_a is next. Initially a quadratic plot was attempted for 250 mm particles, and the fit seems ok:

But it isn’t: the extrapolation of the quadratic for zero wind-speed shows a negative value of the particle velocity, which does not make sense.

One alternative is to again fit it to a quadratic, but with the constant term set to zero. This does not give a good fit, except at some points (not shown). The rigorous way is to use the least squares errors method with the constraint that the constant term c = 0. I could not figure that out so I didn’t. Anyway, the expression for the drag coefficient is not valid for Re < 0.2 so v_a = v_p = 0 is not workable. So, instead I have fitted a log-log plot to a quadratic as below:

where the v’ denotes log(v), and the number of points decreases as D decreases, because Re values lower than 0.2 are not considered.

The above graph, and the previous ones, shows something like a break-even velocity, at which the particle velocity equals the wind velocity: v_p = v_a. Below this value of v_abr, v_p < v_a; and above it, v_p > v_a.

A concern about the wind blowing the dust or fog away: this assumes that the wind does not change direction randomly. For example, it went from North to South at one moment, to South to North the next, the dust or fog would stay put (on average). But even if the direction of the velocity stays the same, if the Reynolds number exceeds one, the motion of the particle will be turbulent (Re is calculated with the relative velocity). Then all bets are off – the particle could go in any direction, follow a vortex, whatever. Only in laminar flow (Re < 1), will the motion be straight line (or smoothly curved). So the above calculation would only yield sensible results for laminar flow, and that means the wind speed v_a should be < 0.1 m/s  for 250 mm particles, <1 25="" for="" m="" s="" span="" style="font-family: Symbol; mso-ascii-font-family: Calibri; mso-ascii-theme-font: minor-latin; mso-char-type: symbol; mso-hansi-font-family: Calibri; mso-hansi-theme-font: minor-latin; mso-symbol-font-family: Symbol;">m

m particles and <10 2.5="" for="" m="" s="" span="" style="font-family: Symbol; mso-ascii-font-family: Calibri; mso-ascii-theme-font: minor-latin; mso-char-type: symbol; mso-hansi-font-family: Calibri; mso-hansi-theme-font: minor-latin; mso-symbol-font-family: Symbol;">mm particles (as indicated by arrows in the above graph). This is not merely a problem of calculating turbulent particle flow (which it is, at least for me!), but it also means that higher velocity winds will not be especially effective in dispersing dust or fog.

This argument is similar to that used in clean-room design, which all use laminar flow – albeit with different Reynolds number value (Re_tr = 2000) for transition from laminar to turbulent flow. Anyway, there is an optimal speed for the unidirectional flow, and it is around 0.3 m/s (60 ft/min). Increasing the speed does not help, as turbulent flow is not efficient in uniformly removing the dust in the atmosphere.

And this does not even begin to address the other question: how fast does the wind have to blow to kick up a dust devil or a sand-storm! That issue is being brushed under the carpet, buried six feet under and resolutely avoided!

The time taken to move a bunch of particles a distance Dx, would be Dt = Dx/v_p.  As mentioned above, this assumes that the dust (or droplets) are localized within this distance Dx, and clean (droplet-free) air is being brought in by the wind from outside this area.

Suppose we take an area of width 1 km, and v_a = 1 m/s (3.6 kph), with 25 micron diameter particles, v_p = 0.21 m/s, and it would take 4,673 secs (1.3 hrs) to clear the dust away. With a 10 m/s (36 kph) wind, v_p = 4.61 m/s, it would take only 217 secs (3.6 mins). However, as noted above, for 25 micron diameter particles v_a = 10 m/s is probably in the transition zone (1 < Re < 10) between laminar and fully turbulent flow, so this estimate may be questionable.

Now, let us repeat the calculations for 2.5 micron diameter particles: v_a = 1 m/s (3.6 kph), v_p = 0.0805 m/s, and it would take 12,422 secs (3.4 hrs) to clear the dust away. Fine particles are tougher to get rid of! Similarly, with a 10 m/s (36 kph) wind, v_p = 2.14 m/s, it would take only 467 secs (7.8 mins). In this case, 2.5 micron diameter particles: v_a = 10 m/s, is just within the laminar flow regime, and the calculation is correct.

These calculations assume that we can neglect the time involved in accelerating the particles, which seems reasonable, if we look at the ‘a vs t’ and ‘v vs t’ figures plotted earlier (as mentioned above).

Obviously, these calculations are simplistic. You would need to write computer code to take more realistic scenarios into account. But the basic idea is probably correct. A low wind speed is insufficient to blow away fog hanging overhead, but as soon as a brisk wind gets going, the fog quickly lifts.

Another more quantitative question could be: if I have a given number density n of dust particles already in the air, and a dust generating source with a fixed generation rate, s (units of m^-3sec^-1), how does the number density change? The answer is that you would have to solve the continuity equation:

                                                                                [d(nv)/dx]  + [dn/dt]  = s

putting in all the relevant boundary conditions and taking the nonuniform number density into account.

At a naïve level, for s = 0, the continuity equation actually is the same as the rough calculation done above to find the Dt from an assumed Dx, in terms of v_p. For laminar flow, if we assume some n(x), with s = 0, the same ‘shape’ n(x) would only get translated to n(x + v_p Dt). If the distance the dust gets moved is Dx/2, probably the value of n drops by 50%. If there is a dust-generation term, it would have to be added on, at the appropriate value of x.

And it should be in 3D. Easy to talk about: a lot of work to actually do it. It ought to be possible to get an idea of what it looks like by doing a simple 1D case. Some other time!

still quaking!

After the earthquake last year in Katmandu, one of my colleagues, Anupama Singh, mentioned that there fish tank had overflowed.
So, what is the condition for that to happen?

For the water to overflow (dashed line):

                                                                                tan(f) = h/(l/2) = a/g

where a is the horz accn..

Assume h = 5 cms, l = 100 cms, a = 0.1 g.

http://en.wikipedia.org/wiki/Peak_ground_acceleration

This corresponds roughly to a Richter 6 quake.

What about the Hyatt swimming pool at Katmandu? The video circulating on WhatsApp showed the water moving in a very complicated way - but it reached the first floor of the adjacent building.

Assume the pool is 25 metres long. The water sloshed up to the 1st floor: i.e. 3.5 metres.

A Richter 8 quake has accelerations of 0.34 to 0.65 g. The quake at Katmandu was less than 7.9, because of the distance of 80 kms to the epicentre, but it was mostly horizontal. For a height of 3.5 metres, a = 0.28 g's. This is reasonable.

I am going to quote verbatim from Phil Plait along with the image he cites and then comment on it after his post:

http://www.slate.com/blogs/bad_astronomy/2015/11/18/astrophoto_space_station_photo_gets_a_seismic_jiggle.html

Take, for example, Tim Ashby-Peckham, who saw that the International Space Station was going to make a nice pass of his location in the early evening of Nov. 1, 2015. He set up his Canon 70D camera on a tripod, aimed it in the right direction, and waited. Once the ISS came into view, he started snapping away.

Take a look at the photo above. You can see the ISS as a long streak, since it moves a lot during the 30-second exposure. But take a closer look. See the wiggle in the streak at the bottom?

Ashby-Peckham took this shot in South Auckland, New Zealand, around 8:57 p.m. local time on Nov. 1, 2015. At 8:58, a magnitude 2.2 earthquake hit a couple of hundred kilometers to the southeast.

My strongest doubt was that a quake that weak could be felt that far away. However, in an email discussion about it, Ashby-Peckham told me that another earthquake about the same distance away was indeed felt in Auckland.

So this is a new one on me. Using the ISS as a virtual seismograph! It’s a pretty funny idea. I suppose if you live in a place with enough earthquakes you could actually calibrate your photography equipment to them, measuring how much the image wiggles versus size/distance of the quake. You need a moving target; stars don’t move quickly enough, and all you get is smeared-out disks for them (note that in the shot, the stars do appear to wiggle a bit). Of course, you could put the camera on a motor, letting it slowly scan the sky over and over again.

My comments:

ISS orbital period = 90 minutes which corresponds to 360°. 

So 30 secs corresponds to an angular displacement of (0.5/90)(360) =  2°

More precisely:

ISS orbital speed is 7.66 km/s. So in 30 secs, ISS moves 229.8 kms.

Assuming Earth radius as 6400 kms and the ISS average altitude as 380 kms, so the total distance from the centre of the Earth is: 6780 kms.

The angle traversed is thus: 229.8/6780 =  0.0339  rads = 1.94°

https://en.wikipedia.org/wiki/Peak_ground_acceleration

According to wiki[edia (link above), a Richter 2-3 corresponds to a peak ground acceleration (PGA) of 0.0017 to 0.014 g’s.

Assume a value of 0.0017 for Richter 2. This means a Dq = 0.0017 rads or 0.097°.

 In the image above, if the whole track is FOV is 0.0339 rads, and its corresponding length is 147 mm, while the displacement above the track is about 1 mm. This gives an angle of about 2.31x10^-4 rads (i.e. 0.013°).

Richter 2 should give 0.097°. This means that the shaking at Ashby’s location was closer to Richter 1, because the angular displacement of 0.013° is smaller by 7.5X than for Richter 2.

http://www.iitk.ac.in/nicee/wcee/article/13_480.pdf

According to the authors, T.Trombetti et al (13^th World Conference on Earthquake Engineering, Vancouver, Canada,2004), the Sabetta-Pugliese law (of 1987) states:

Log (A) = - 1.845 + 0.363 (MS) – log (R²+25)^1/2+0.195s

Where A is the PGA in g’s, MS is the quake intensity in the Mercalli scale, R is the distance in kms and s is a parameter that is 1.0 for soft soil and 0.0 for bedrock.

According to the S-P law at a low R = 0.001 kms and assuming MS = 2, PGA = 0.0019 for s = 0 and PG = 0.0029 for s = 1. The 1^st value is close to that mentioned in wiki.

For s = 0, MS = -2 gives PGA = 6.6x10^-5.

Instead we can use the total duration time of earthquake t (in secs) given by Tsimura’s empirical formula:

ML = -2.53 +2.85 log₁₀(t) + 0.0014R

Where ML is the local quake intensity and R is the distance from the epicenter.

The image shows that the quake lasts for about (3/147)(30) = 0.6 secs.

According to Tsimura, for ML = -3, the time t is 0.7 sec at R = 0.001 kms and 0.85 sec at R = 200 kms.

The duration would suggest that the quake was ML= -3 which is very feeble indeed.

On the other hand, as Plait points out the distance is R = 200 kms. With a ML = 2.2, the S-P law gives a PGA = 4.5x10^-7 at R = 200 kms. This is much smaller than the actual on-site PGA of 2.31x10^-4 from the angular displacement. At that distance, even with ML=6, the PGA is only 1x10^-5!

The two arguments from PGA vs R and from duration t are completely inconsistent.

At the end, one would have to invoke – as Phil Plait did – another completely different quake, much closer by to the observation point, to get better consistency. But not a lot, as it happens…

So what values could one try out?

Suppose we go for ML=-3 so that the duration is about right (0.7 sec). The PGA at R = 0.001 kms is 2.9x10^-5. So there should have been a mini-quake literally under Ashby’s feet! But this PGA is an order of magnitude too small.

Or a somewhat stronger one a few kilometers away…

If we choose, ML=-0.5, we get PGA=2.3x10^-4. Unfortunately, this gives t = 5 secs!

What this means is that we are unable to get consistency between the PGA and the time. What is clear, however, is that it was a mini-quake (weaker than ML = -0.5) and its epicenter was pretty close to where Ashby and his tripod were located.