Estimation quiz!

08012020, 07:30 AM
Post: #21




RE: Estimation quiz!
The distance between the sun and Earth is my favorite almostround number, at 499 lightseconds.


08012020, 10:09 AM
Post: #22




RE: Estimation quiz!
(08012020 07:30 AM)Thomas Okken Wrote: The distance between the sun and Earth is my favorite almostround number, at 499 lightseconds. Or almost exactly 150Gm. Greetings, Massimo +×÷ ↔ left is right and right is wrong 

08012020, 05:24 PM
Post: #23




RE: Estimation quiz!
A useful approximation for electronics: speed of light ≈ 1 foot per nanosecond.
— Ian Abbott 

08012020, 08:41 PM
(This post was last modified: 08012020 08:41 PM by johanw.)
Post: #24




RE: Estimation quiz!
(08012020 05:24 PM)ijabbott Wrote: A useful approximation for electronics: speed of light ≈ 1 foot per nanosecond.Are all these ancient units still used in papers? I still remember the scolding I got when I entered wavelengths in Ångström (1Å = \(10^{10}\)m) in my masters thesis. I had to change it to nm before I could get a passsing grade, the faculty had a strict SIonly policy. 

08012020, 11:16 PM
Post: #25




RE: Estimation quiz!
(08012020 07:30 AM)Thomas Okken Wrote: The distance between the sun and Earth is my favorite almostround number, at 499 lightseconds. I've always taken that value at exactly 500 seconds (i.e.: 8 min 20 sec.) as the distance between the sun and the Earth varies by as much as 5 MKm between aphelion (~152.1 Mkm) and perihelion (147.1 Mkm), and that's plus or minus ~8 lightseconds, give or take a second. So 500 it is for me. Easier to remember and way "rounder" (in base 10; for a round value in base 2, the value 512 (1,000,000,000_{2}) would do Regards. V. All My Articles & other Materials here: Valentin Albillo's HP Collection 

08022020, 07:56 AM
Post: #26




RE: Estimation quiz!
(08012020 05:24 PM)ijabbott Wrote: A useful approximation for electronics: speed of light ≈ 1 foot per nanosecond.As a continental european EE, my reference is 30cm for 1ns ... (08012020 08:41 PM)johanw Wrote: Are all these ancient units still used in papers? I still remember the scolding I got when I entered wavelengths in Ångström (1Å = \(10^{10}\)m) in my masters thesis. I had to change it to nm before I could get a passsing grade, the faculty had a strict SIonly policy. NonSI units are very common in astronomy, think of the old parsec (distance where a length of 1AU is seen under an angle of 1 arcsecond , with 1AU being the average EarthSun distance :) The parsec seems now less used and often replaced by the lightyear (1 pc is about 3 ly), but is still used for instance for the Hubble constant (cosmological expansion) Ho ≈ 70 km/s/Mpc. Also masses of stars are generally expressed not in kg but in solar mass, and so on. A funny coincidence: there are (about) as many km in one lightyear than Ångström in one km. JF 

08022020, 12:23 PM
Post: #27




RE: Estimation quiz!
from Science Unit Conversion Humor
This is a list of funny, madeup scientific unit conversions. 453.6 graham crackers = 1 pound cake There are 453.6 grams in 1 pound. Ratio of an igloo's circumference to its diameter = Eskimo Pi Pi is the ratio of a circle's circumference to diameter, while there is a stereotype that Eskimos dwell in igloos. 2000 pounds of Chinese soup = Won ton A wonton is a type of Chinese dumpling. There are 2000 pounds in 1 ton. Time between slipping on a peel and smacking the pavement = 1 bananosecond Instead of expressing the unit in terms of nanoseconds, it's bananoseconds because a banana caused the fall. 1 millionth of a mouthwash = 1 microscope This refers to the popular mouthwash, Scope. The metric prefix "micro" means one millionth. 1 million bicycles = 1 megacycles The metric prefix "mega" means 106 or one million. Weight an evangelist carries with God = 1 billigram This refers to the American evangelist Billy Graham. Time it takes to sail 220 yards at 1 nautical mile per hour = Knotfurlong 365.25 days of drinking lowcalorie beer = 1 Lite year 16.5 feet in the Twilight Zone = 1 Rod Serling The rod is a unit of length equal to 16.5 feet. Rod Serling is the American tv producer, screenwriter, and narrator responsible for "The Twilight Zone." Basic unit of laryngitis  1 hoarsepower One symptom of laryngitis is hoarseness. Shortest distance between two jokes  a straight line To deliver a joke as a straight line means it's a short joke delivered with a straight face (like it's not a joke at all). 1 million microphones = 1 megaphone 365.25 days = 1 unicycle 365.25 days is one year or one cycle of the Earth around the Sun. It's especially clever because unicycle has another meaning. It's a bike with one wheel. Half a large intestine = 1 semicolon the large intestine is also called the colon. Since it's only half a colon, it's a semicolon, much like half a circle is a semicircle. 2000 mockingbirds = two kilomockingbirds "To Kill a Mockingbird" is a famous novel by author Harper Lee published in 1960. The kilo is the prefix for a thousand. So, 2000 is two kilo. 10 cards = 1 decacard Deca is the prefix for 10. 52 cards = 1 deckacard There are 52 cards in a deck of playing cards. 1,000,000 aches = 1 megahurtz There are one million (106) hertz in 1 megahertz. This is a play on words, substituting hurtz (like pain, but with a "z") for hertz. 1 millionth of a fish = 1 microfiche The word "microfiche" is pronounced like microfish. The prefix micro means one millionth. 2.4 statute miles of intravenous surgical tubing at Yale University Hospital = 1 I.V.League Intravenous tubing is also called IV tubing. Yale is one of the Ivy League school, plus 2.4 statute miles is a length equal to 1 league. 1 kilogram of falling figs = 1 fig newton The newton is a unit force, which is mass under acceleration (such as you'd get from falling figs). This play on words refers to the Nabisco cookie, the fig newton. 1000 grams of wet socks = 1 literhosen Lederhosen are short breeches (not actually socks). There are 1000 grams of water (more or less) in one liter. The liter is a unit of volume used for liquids, so wet socks are literhosen. 1 trillion pins = 1 terrapin The prefix terra means a trillion. 10 rations = 1 decaration The prefix deca means 10. 100 rations = 1 Cration C is the Roman numeral for 100. 2 monograms = 1 diagram Mono is the prefix for one, while dia means two. 2 new dimes = new paradigms Two dimes is a pair of dimes. A paradigm is a model or pattern. These and many more available for your perusal somewhere out there on the net. An implied rhetorical question, does SCIENCE own unit conversions? BEST! SlideRule 

08022020, 01:10 PM
Post: #28




RE: Estimation quiz!
The microfortnight is a surprisingly useful unit.
Pauli 

08022020, 03:13 PM
Post: #29




RE: Estimation quiz!
Here's one that might tax the most mathematical, and also needs some not very general knowledge: how many Jupiters could fit into the (volume of the) Sun?
(The readily searchable answer for this assumes we just compare volumes, which is as wrong as anyone with a jar of marbles can tell you.) Or, something which is actually within the bounds of present mathematical knowledge, and closer to everyday experience: how many Moons could fit into the Earth? 

08022020, 04:02 PM
Post: #30




RE: Estimation quiz!
(08022020 03:13 PM)EdS2 Wrote: how many Moons could fit into the Earth? I'd estimate about 23, give or take a Moon. V. All My Articles & other Materials here: Valentin Albillo's HP Collection 

08022020, 05:04 PM
(This post was last modified: 08022020 05:04 PM by Maximilian Hohmann.)
Post: #31




RE: Estimation quiz!
Hello!
(08022020 04:02 PM)Valentin Albillo Wrote: I'd estimate about 23, give or take a Moon. 23 is always a good answer :) But this one really got me interested and surprisingly (using finite time for my google search) I only found one paper online with an equation for this problem: Nelson M Blachman: The Closest Packing of Equal Spheres in a Larger Sphere (1963). It can be freely accessed from here (registration required): https://www.jstor.org/stable/2312064?rea...b_contents If I entered the numbers correctly, there is place for just 11.8 moond (or 11 full moons) insdide our earth. Regards Max 

08022020, 05:41 PM
Post: #32




RE: Estimation quiz!
(08022020 03:13 PM)EdS2 Wrote: (The readily searchable answer for this assumes we just compare volumes, Planet sized spheres might not act like marbles. Gravity may squash them into a giant sphere, possibly even denser. Volume ratios might actually be a good estimate. From Wolfram Alpha: (earth volume) / (moon volume) = 49.31 (sun volume) / (jupiter volume) = 986.7 

08022020, 05:54 PM
Post: #33




RE: Estimation quiz!
(08022020 05:04 PM)Maximilian Hohmann Wrote: Hello! You bet. Quote:If I entered the numbers correctly, there is place for just 11.8 moond (or 11 full moons) insdide our earth. It seem's you didn't or you used the wrong formula, as it's trivial to see your estimation is wrong. Consider N spheres of unit diameter touching a central one (unit diameter too). Now it's well known that the maximum number N of nonoverlapping spheres touching a central one is 12 (a 13th almost fits but ultimately doesn't). This makes 13 equal spheres in all of unit diamter, 12 of them in contact with the central one. But as it's trivial to see, they all fit in a sphere of diameter exactly 3: O O O And as the ratio of Moon's to Earth's diameter is about 0.27 this means that if the Moon's sphere has diameter 1 then Earth is a sphere of diameter ~3.7, and thus if 13 spheres could be easily contained within a sphere of diameter just 3, then a sphere of diameter 3.7 will hold way more, certainly more than your "11.8", let's say 23 ... Regards. V. All My Articles & other Materials here: Valentin Albillo's HP Collection 

08022020, 09:01 PM
Post: #34




RE: Estimation quiz!
Hello!
(08022020 05:54 PM)Valentin Albillo Wrote: And as the ratio of Moon's to Earth's diameter is about 0.27 this means that if the Moon's sphere has diameter 1 then Earth is a sphere of diameter ~3.7, and thus if 13 spheres could be easily contained within a sphere of diameter just 3, then a sphere of diameter 3.7 will hold way more, certainly more than your "11.8", let's say 23 ... I am not so sure about that. After your 13 moons (the central one and the next layer of 12 adjacient to it) there is not enough space for another full layer of moons within the Earth's radius. The central moon has a radius of 0,27 to this you have to add the diameter of the next layer of 0,54, so the envelope of that lfirst ayer reaches out to 0,81). If at all, the extra moons need to fit into the gaps between the 12 moons of the first layer and the radius of the Earth. My guess is that there is no symmetrical solution for a maximum number of moons (as it seems to be the case with the old equation I found) but the 10 extra moons have to be added to one side only. But how can that be calculated? Regards Max 

08022020, 11:04 PM
(This post was last modified: 08052020 10:07 PM by SlideRule.)
Post: #35




RE: Estimation quiz!
(08022020 05:04 PM)Maximilian Hohmann Wrote: … I only found one paper online with an equation for this problem: Nelson M. Blachman: The Closest Packing of Equal Spheres in a Larger Sphere (1963) … An interesting read: Grundlehren def mathematischen Wissenschaften 290 Sphere Packings, Lattices and Groups ISBN 9781475720167 (eBook) © 1988 by Springer Science+Business Media New York " PREFACE This book is mainly concerned with the problem of packing spheres in Euclidean space of dimensions 1,2,3,4,5, … . Given a large number of equal spheres, what is the most efficient (or densest) way to pack them together? … … 1. The Sphere Packing Problem 1.1 Packing ball bearings. The classical sphere packing problem, still unsolved even today, is to find out how densely a large number of identical spheres (ball bearings, (I) for example) can be packed together. To state this another way, consider a large empty region, such as an aircraft hangar, and ask what is the greatest number of ball bearings that can be packed into this region. If instead of ball bearings we try to pack identical wooden cubes, children's building blocks, for example, the answer becomes easy. The cubes fit together with no waste space in between, we can fill essentially one hundred percent of the space (ignoring the small amount of space that may be left over around the walls and at the ceiling), and so the number of cubes we can pack is very nearly to the volume of the hangar divided by the volume of one of the cubes. But spheres do not fit together so well as cubes, and there is always some wasted space in between. No matter how cleverly the ball bearings are arranged, about one quarter of the space will not be used." BEST! SlideRule ps: the book is in at least a third edition 

08022020, 11:36 PM
Post: #36




RE: Estimation quiz!
wikipedia: Sphere packing in a sphere
Reference 1: http://www.randomwalk.de/sphere/insphr/spisbest.txt Code: # Densest Packings of n Equal Spheres in a Sphere of Radius 1. radius of moon / radius of earth ≈ 0.2727, → 23 moons can fit inside earth, packing efficiency = 23*0.2727³ ≈ 46.64% 

08032020, 07:56 AM
Post: #37




RE: Estimation quiz!
Excellent  so 23 is a solid answer, and Valentin's estimate of 23 is spoton. Valentin, that's a very good estimate indeed, and I'd love to know how you approached it.
As for the likelihood of performing this feat with real Moons, well, yes, as the Moon has enough matter to pull itself into something of a spherical shape, then an assemblage of several is doomed to become solidly spherical in due course. However, as the old joke has it "consider a spherical cow." Or in this case, some scale models of the Moon in the form of marbles. Nice links found  thanks for sharing  but this quote > The classical sphere packing problem, still unsolved even today has become out of date. Kepler's conjecture has been proved, some ten years after this book, and then formally proven some 16 years later. It's interesting, I think, that a random packing of spheres (or marbles) will have a density around 65%, the closest packing achieves a density of 74%, but the constraint of fitting into a sphere reduces us to about 47%. That's comparable to this surprising result: "The strictly jammed sphere packing with the lowest density is a diluted fcc crystal with a density of only 0.49365" 

08032020, 11:00 AM
(This post was last modified: 08032020 11:29 AM by Albert Chan.)
Post: #38




RE: Estimation quiz!
(08022020 05:41 PM)Albert Chan Wrote: (earth volume) / (moon volume) = 49.31 Assuming same packing efficiency for radius ratio of 3, we extrapolate for it x / 49.31 = 13 / 3^3 → estimated moons fit inside earth = floor(x) = floor(23.74) = 23 P.S, this packing problem might be of interest: Social Distancing at the Park 

08032020, 11:57 AM
(This post was last modified: 08032020 11:30 PM by SlideRule.)
Post: #39




RE: Estimation quiz!
(08032020 07:56 AM)EdS2 Wrote: … Nice links found  thanks for sharing  but this quote > The classical sphere packing problem, still unsolved even today has become out of date. Kepler's conjecture has been proved, some ten years after this book, and then formally proven some 16 years later … An excerpt from the preface of the third edition, ISBN 9781475765687 (eBook), © 1999, 1998, 1993 Springer Science+Business Media New York. " Notes on Chapter 1: Sphere Packings and Kissing Numbers Hales [Hal92], [Hal97], [Hal97a], [Hal97b] (see also Ferguson [Ferg97] and Ferguson and Hales [FeHa97]) has described a series of steps that may well succeed in proving the longstanding conjecture (the socalled "Kepler conjecture") that no packing of threedimensional spheres can have a greater density than that of the facecentered cubic lattice. In fact, on August 9, 1998, just as this book was going to press, Hales announced [Hal98] that the final step in the proof has been completed: the Kepler conjecture is now a theorem. The previous best upper bound known on the density of a three dimensional packing was due to Muder [Mude93], who showed that the density cannot exceed 0.773055 ... (compared with :n'Jv'18 = 0.74048 ... for the f. c. c. lattice). A paper by W.Y. Hsiang [Hsi93] (see also [Hsi93a], [Hsi93b]) claiming to prove the Kepler conjecture contains serious flaws. G. Fejes T6th, reviewing the paper for Math. Reviews [Fej95], states: "If I am asked whether the paper fulfills what it promises in its title, namely a proof of Kepler's conjecture, my answer is: no. I hope that Hsiang will fill in the details, but I feel that the greater part of the work has yet to be done." Hsiang [Hsi93b] also claims to have a proof that no more than 24 spheres can touch an equal sphere in four dimensions. For further discussion see [CoHMS], [Hal94], [Hsi95]. " A significant history of dialog, with some similarity to the early years of astronomy, is encompassed in those 16 years. Like a detectivemurder novel, a complete read has its' merits. BEST! SlideRule 

08032020, 01:54 PM
Post: #40




RE: Estimation quiz!
From the nuclear weapons community:
A "shake" is 10ns I find it interesting that over my 40 year career in US Electronics, we have always used Imperial units for physical dimensions, but Celsius for temperature! 

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