Estimation quiz!

[pinkman]

(08-03-2020 11:00 AM)Albert Chan Wrote:  

(08-02-2020 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

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

[Valentin Albillo]

(08-03-2020 07:56 AM)EdS2 Wrote:  Excellent - so 23 is a solid answer, and Valentin's estimate of 23 is spot-on. 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.

[EdS2]

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.

[SlideRule]

A different approach to estimation with respect to spherical density packing from the August 1974 (v1 n2) edition of HP-65 Key Notes …

"HP-65 + 4 ,757 = SX-70

   The above photo was taken with a new Polaroid SX-70 camera won at Las Vegas-and therein lies a really fantastic "believe-it-or-not" story for all you HP-65 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 SX-70 camera. Well, Emmett decided he wanted to win that camera and, besides, the problem was a real challenge for his trusty HP-65, which he is NEVER without.
   First, Emmett measured the container, two-thirds full of marbles, subtracting the volume occupied by the three-volume 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 ten-cent 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 6-quart bottle that he had found in the alley, added water, measured, counted, and computed.
   Following all of this, Emmett placed a long-distance 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 HP-65, 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 The-Shelf (O-T-S) 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 HP-65 owners. We're also grateful to Emmett for his staunch support of HP calculators, He's owned an HP-35, HP-45, HP-80, and now the HP-65."

BEST!
SlideRule

[EdS2]

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?

[Csaba Tizedes]

One really easy:
How much the pressure under your 0.5mm thick mechanical pencil tip by the pencil self-weight?

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.

[Thomas Okken]

(08-04-2020 10:17 AM)EdS2 Wrote:  - how far can you travel, at a constant heading of North by North West?

\( { \sqrt{6 - \sqrt{2}} \over 2 } \pi R \)

[Albert Chan]

96 million (say, 6" diameter) shade balls ...
How big is the reservoir ?

[KeithB]

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

[Albert Chan]

(08-05-2020 03:38 PM)KeithB Wrote:  They don't all appear to be at the same height, some are resting on top of the others.

My guess is the balls are designed to be single-layer, when the reservoir is full (or, mostly full)

[StephenG1CMZ]

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 Covid-19 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 tightly-packed 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?

[EdS2]

(08-04-2020 12:33 PM)Thomas Okken Wrote:  

(08-04-2020 10:17 AM)EdS2 Wrote:  - how far can you travel, at a constant heading of North by North West?

\( { \sqrt{6 - \sqrt{2}} \over 2 } \pi R \)

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.)

[Thomas Okken]

(08-07-2020 06:57 AM)EdS2 Wrote:  

(08-04-2020 12:33 PM)Thomas Okken Wrote:  \( { \sqrt{6 - \sqrt{2}} \over 2 } \pi R \)

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.)

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(6-sqrt(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.

[Albert Chan]

(08-07-2020 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(6-sqrt(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

[Csaba Tizedes]

(08-04-2020 11:25 AM)Csaba Tizedes Wrote:  How much the pressure under your 0.5mm thick mechanical pencil tip by the pencil self-weight?

Approx. 5 bar

   

[EdS2]

(08-07-2020 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?

[Albert Chan]

(08-08-2020 07:35 AM)EdS2 Wrote:  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?

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 * √(1-1/576) ≈ 13/12 * (1-1/1152) ≈ 1.0824

[EdS2]

Excellent, thanks!

[KeithB]

Guess how many coins are in 100 gallons collected from an aquarium:
https://www.cbsnews.com/news/north-carol...operating/

[Albert Chan]

(08-13-2020 01:41 PM)KeithB Wrote:  Guess how many coins are in 100 gallons collected from an aquarium:
https://www.cbsnews.com/news/north-carol...operating/

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.