Estimation quiz!

08032020, 03:08 PM
Post: #41




RE: Estimation quiz!
(08032020 11:00 AM)Albert Chan Wrote:(08022020 05:41 PM)Albert Chan Wrote: (earth volume) / (moon volume) = 49.31 Very funny scientific analysis for safe configurations in the park. They also give the link to the Fibonacci placement in the bar, really clever: https://datagenetics.com/blog/june52020/index.html Thibault  not collector but in love with the few HP models I own  Also musician : http://walruspark.co 

08032020, 03:10 PM
Post: #42




RE: Estimation quiz!
(08032020 07:56 AM)EdS2 Wrote: 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. Quite simple, actually. As I explained above, 13 spheres of unit diameter will exactly fit into a larger sphere of diameter 3. Now, Earth's mean radius is ~6371.0 km, while the Moon's is ~1737.4 km. Thus, the expected number of spheres is just a matter of scaling, namely: #Moons_within_hollow_Earth = INT(((6371.0 / 1737.4) / 3)^3 * 13) = 23 Regards. V. All My Articles & other Materials here: Valentin Albillo's HP Collection 

08032020, 03:22 PM
Post: #43




RE: Estimation quiz!
Remarkable  thanks for explaining!
Speaking of bars, and estimation quizzes, there's a surprising bar bet, that for the kinds of cylindrical objects you might find on a bar  even quite a tall one  the circumference will exceed the height. Try it  you might be surprised. 

08032020, 03:49 PM
Post: #44




RE: Estimation quiz!
A different approach to estimation with respect to spherical density packing from the August 1974 (v1 n2) edition of HP65 Key Notes …
"HP65 + 4 ,757 = SX70 The above photo was taken with a new Polaroid SX70 camera won at Las Vegasand therein lies a really fantastic "believeitornot" story for all you HP65 fans. The man in the photo is Emmett Ingram, Jr., Chief Engineer at Jetronix Radio Engineering Labs in Palos Verdes, California. And if you find it difficult to believe his story, then you just don't know Emmett. (I do; I've met him. An incredible man. And the story is true. Ed.) If you happened to attend the NEWCOM show in Las Vegas in May, 1974, you probably remember that the United Technical Publications booth had a large plastic tank containing many marbles and the three volume set of the Electronic Engineers Master (EEM). There was a prize for guessing the correct (or nearest to, correct) number of marbles in the rectangular tank, and the prize was (you've guessed it!) a brand new Polaroid SX70 camera. Well, Emmett decided he wanted to win that camera and, besides, the problem was a real challenge for his trusty HP65, which he is NEVER without. First, Emmett measured the container, twothirds full of marbles, subtracting the volume occupied by the threevolume set of the Electronic Engineers Master. Then he counted and recounted those exposed marbles per row on each of the four sides of the tank. He also counted each marble tangent to the plastic face of the container. Of those marbles that appeared to touch the face of the plastic tank, a count determined that 42.7% actually did. Armed with those numbers, he went to the local Las Vegas tencent store and purchased all the marbles they had in stock ($5.80 worth). Then off to the photo store next door, where he picked up a 1,000 cc graduated beaker. Then he cut off the top of a 6quart bottle that he had found in the alley, added water, measured, counted, and computed. Following all of this, Emmett placed a longdistance telephone call to a U.S.C. mathematician. (You can imagine the conversation!) He advised Emmett that the data gathered would best fit the Chi Squared Probability equation. All of this was then pumped into Emmett's trusty HP65, and out popped 4,757. The correct number of marbles in the container was 4,754, which represented the total number of pages in the '74 edition of Off TheShelf (OTS) and the '73'74 edition of EEM. Naturally, the people at United Technical Publications were astonished by Emmett's wild but accurate guess, and even more astonished when he told them it was not a guess, it was a carefully calculated answer! As a matter of record, if the quality control on the marbles that Emmett bought had been better, it's probable that he would have hit 4,754 right on the nose! According to Emmett: "Some marbles measured to be nonspherical by 0.05% (4754/4757, an error of 0.063%). It appears a few of the marbles escaped the manufacturer's quality control and fouled me up by a count of three!" So, that's the story. We're grateful to Emmett for allowing us to share it with all you other HP65 owners. We're also grateful to Emmett for his staunch support of HP calculators, He's owned an HP35, HP45, HP80, and now the HP65." BEST! SlideRule 

08042020, 10:17 AM
Post: #45




RE: Estimation quiz!
Nice work Emmett!
I thought up some more estimation quiz questions:  how big is a ton of bricks?  if the bus travels at reasonable speed, at what rpm do the wheels go round and round?  if you stood on the highest mountain, how far away would the horizon be?  if the Moon were smooth, what proportion of the surface would be in sunlight?  how far can you travel, at a constant heading of North by North West? 

08042020, 11:25 AM
(This post was last modified: 08042020 11:26 AM by Csaba Tizedes.)
Post: #46




RE: Estimation quiz!
One really easy:
How much the pressure under your 0.5mm thick mechanical pencil tip by the pencil selfweight? One advanced, need good estimation: How much the oceans level decrease if we kill all the whales? (...and we remove their bodies from the water...) Cs. 

08042020, 12:33 PM
Post: #47




RE: Estimation quiz!  
08052020, 02:29 PM
Post: #48




RE: Estimation quiz!
96 million (say, 6" diameter) shade balls ...
How big is the reservoir ? 

08052020, 03:38 PM
Post: #49




RE: Estimation quiz!
They don't all appear to be at the same height, some are resting on top of the others.


08052020, 04:22 PM
Post: #50




RE: Estimation quiz!  
08052020, 06:12 PM
Post: #51




RE: Estimation quiz!
The emphasis on efficient packing of circles is an interesting mathematical problem, but all the reports I have seen concentrate exclusively on packing density.
For real world Covid19 safety, other measurements will also be important: Path length through such a packed park, for example. Imagine 4 regularly spaced circles. It is easy to walk through the park between the circles (Unless you need an additional circle around your path). Or imagine a fireman trying to reach one of the circles. Now, consider doing that with tightlypacked hexagons. Your walk through the park is now much longer than the straight path with 4 circles. Is there any prediction of path length for the various configurations? Stephen Lewkowicz (G1CMZ) 

08072020, 06:57 AM
Post: #52




RE: Estimation quiz!
(08042020 12:33 PM)Thomas Okken Wrote:(08042020 10:17 AM)EdS2 Wrote:  how far can you travel, at a constant heading of North by North West? An interesting formula! (Not an estimation, as such... but what does it tell us? From pole to pole on the surface is 12000 miles more or less (or 20k km) and this formula gives us about 13500 miles  not a great deal more. That surprises me, a little. More on the loxodrome here.) 

08072020, 09:03 AM
Post: #53




RE: Estimation quiz!
(08072020 06:57 AM)EdS2 Wrote:(08042020 12:33 PM)Thomas Okken Wrote: \( { \sqrt{6  \sqrt{2}} \over 2 } \pi R \) My reasoning is that while traveling at a certain compass heading, the ratio between the total distance traveled and the distance traveled north is constant (although it now looks to me like the value of that constant that I figured, sqrt(6sqrt(2))/2, is wrong, and the actual value is slightly greater). And since NNW is only 22.5° from N, that ratio is not much greater than one. The path traveled itself is interesting because it spirals around the poles infinitely many times without ever reaching them, yet nonetheless has a finite length. You can visualize it by extending a Mercator projection infinitely far north and south; a loxodrome is then just a straight line that wraps around the side edges infinitely many times. 

08072020, 12:56 PM
Post: #54




RE: Estimation quiz!
(08072020 09:03 AM)Thomas Okken Wrote: My reasoning is that while traveling at a certain compass heading, the ratio between the total distance traveled and the distance traveled north is constant (although it now looks to me like the value of that constant that I figured, sqrt(6sqrt(2))/2, is wrong, and the actual value is slightly greater). And since NNW is only 22.5° from N, that ratio is not much greater than one. Is the ratio sec(22.5°) ? cos(22.5°)^2 = (1 + cos(45°))/2 = (2+√2)/4 → L = pi*R / cos(22.5°) = pi*R * 2/√(2+√2) ≈ 20000 km * 1.0824 = 21648 km 

08072020, 02:06 PM
Post: #55




RE: Estimation quiz!  
08082020, 07:35 AM
Post: #56




RE: Estimation quiz!
(08072020 09:03 AM)Thomas Okken Wrote: My reasoning is that while traveling at a certain compass heading, the ratio between the total distance traveled and the distance traveled north is constant...Oh, that's very nice! Anyone else got the circumference of the Earth as one of their yardsticks? It's one of mine. (A couple of times I travelled for work purposes, London > San Francisco > Tokyo > London. Each flight is more or less a quarter of the way round a great circle, which means I travelled a triangle with each angle about 90 degrees. That was very satisfactory. I also crossed the dateline, which caused a little mental confusion about how many days had elapsed.) Albert: > → L = pi*R / cos(22.5°) = pi*R * 2/√(2+√2) ≈ 20000 km * 1.0824 = 21648 km Nice! But, I wonder, what mental arithmetic might you do to estimate that result? 

08082020, 05:01 PM
Post: #57




RE: Estimation quiz!
(08082020 07:35 AM)EdS2 Wrote: Albert: Pythogrean triples a,b,c = 5,12,13 had ∠A ≈ 22.6° L = pi*R * sec(22.5°) ≈ 20000*13/12 = 20000+10000/6 ≈ 21667 km  This is equivalent to assuming √2 = 238/13² ≈ 1.4083 sec(22.5°) = 2/√(2+√2) ≈ 2/√(2+238/13²) = 2*13/√576 = 13/12 A better estimate for √2 is 239/13² ≈ 1.4142 (this is actually √2 rational convergent, after 99/70) sec(22.5°) ≈ 2*13/√577 = 13/12 / √(1+1/576) ≈ 13/12 * √(11/576) ≈ 13/12 * (11/1152) ≈ 1.0824 

08082020, 05:25 PM
Post: #58




RE: Estimation quiz!
Excellent, thanks!


08132020, 01:41 PM
Post: #59




RE: Estimation quiz!
Guess how many coins are in 100 gallons collected from an aquarium:
https://www.cbsnews.com/news/northcarol...operating/ 

08162020, 11:24 AM
(This post was last modified: 08242020 01:16 AM by Albert Chan.)
Post: #60




RE: Estimation quiz!
(08132020 01:41 PM)KeithB Wrote: Guess how many coins are in 100 gallons collected from an aquarium: Gallon water jug weight about 8 lbs Replacing water with coins, gallon jug estimated 50 lbs 80 quarters and 200 dimes both weighed a pound, or $20 / lbs My guess is half or more of coins is pennies and nickels, which lowered the worth, say $10 / lbs. Estimated coins worth ($10/lbs)(50 lbs/gal)(100 gal) = $50,000 Update: They counted a grand total of $8,563.71. My guess is way off. 

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