Working out how many ducks there are

[discosucks]

Homework Statement

so my local supermarket is doing a competition were you guess the amount of ducks in this trolley . [/B]

Its summer here in Ireland I have no college so I am pretty bored and said i would have a go at solving this with maths , but because its summer time it also means I am having a lot of trouble getting my brain to work haha.

https://scontent-lhr3-1.xx.fbcdn.net/hphotos-xpf1/v/t1.0-9/11017220_732159026907620_5158370372231536002_n.jpg?oh=598d43aaae1a991054df4c2be12c6d12&oe=564A6CD2

Homework Equations

area of a rectangle cube .

Archimedes' principle ( but i can't actually get my hands on a duck haha )

The Attempt at a Solution

So I was starting out by working out the size of the container the ducks are in ,

Im pretty sure the base of the trolley is 1 m^2 and the ducks look to go to around 1.5 m high .

this would mean the take up 1.5m^3 of space .

Next my educated guess would be each duck takes up about 200 - 300 mm^3 of volume but that is only a total guess .

now this is were it all went pear shaped .

I was imagining the ducks are cubes and working out what space they take up then i was going to factor in some error as they arent square and probably will be less ducks than the comparing square ducks .

but I am confused as to how to work this out . I tryed to convert 200mm^2 into m as the container is worked out in m ( 0.2^m3 ) and then divide it into the space ( 1.5 / 0.2 = 7.5 ) . so its pretty clear there is more than 7.5 ducks haha .

But my head is like mush and really can't work out where i have gone wrong .

 

[William White]

There is NO way to calculate this. The packing is random.

You can get close

Count how many are in a small volume and extrapolate...handy that the trolley is meshed into a nice regular array. Count the number of ducks in 2 or 3 of the top "cubes", average that, then multiply by how many 'cubes' there are

You could use the wisdom of the crowd method...if you can see a list of everbody's guess, then take an average. Innacurate guesses (too high and too low) tend to get averaged out. This has shown to be remarkably accurate in the famous "jelly beans in a jar" competition, of which the ducks in a crate is the same type.



(Derren is a famous trickster, so this MAY not be what it seems)A very famous example of the "wisdom of the crowd" was the crew of HMS Venturer in WW2. The commander was a young man called James Launders. On a mission to destroy U864, James and his crew all made educated guesses to the position of U864 and fired all the torpedoes at the assumed position. It remains the only time that a submerged submarine has deliberately sunk another submerged submarine.

 

[RUber]

Next my educated guess would be each duck takes up about 200 - 300 mm^3 of volume but that is only a total guess .

now this is were it all went pear shaped .

I was imagining the ducks are cubes and working out what space they take up then i was going to factor in some error as they arent square and probably will be less ducks than the comparing square ducks .

but I am confused as to how to work this out . I tryed to convert 200mm^2 into m as the container is worked out in m ( 0.2^m3 ) and then divide it into the space ( 1.5 / 0.2 = 7.5 ) . so its pretty clear there is more than 7.5 ducks haha .

I don't think your estimate for the volume of the duck is very accurate.
Also, your conversion of volumes is way off.
$1 mm = \frac{1}{1000}m$
$1 mm^3 = \left( \frac{1}{1000}m \right) ^3 =10^{-9}m^3$
So $200mm^3 = 2x10^{-7}m^3$ implying there would be on the order of 5,000,000 ducks-- not possible.
Perhaps something on the order of 150cc might be better for the volume of one duck.
$1 cm^3 = \left( \frac{1}{100}m \right) ^3 =10^{-6}m^3$
$\frac{1.5m^3}{150cm^3}\times\frac{1000000 cm^3}{m^3} = 10,000$
This is still way too high.

The recommendations I would make would be: 1) be careful with your unit conversions, 2) be conservative with your estimate of the total volume, and 3) overestimate the volume of the ducks to account for the trapped air between the ducks.

Example:
Looking at the image, judging by the paper hanging on the cart, I would say the cart is probably closer to .8 or .9m wide. The ducks are stacked 6 squares high, and the cart is about 5 squares wide, so the height might be as much as 1.2m.
Each duck is almost as long as one square, so maybe 15cm and perhaps 10cm wide and 10cm tall. Maybe assume a wedge-like shape so the volume occupied might be as much as 750cc.

Using these numbers, I would put a guess of:
$\frac{.8m\times.9m \times 1.2m }{750cm^3 }\times\frac{1000000 cm^3}{m^3} = 1152 \text{ ducks}.$

However, if you have access to the location, you should be able to take better measurements or approximations. What William White recommended also works very well and is a tried and true method taught to elementary school students.

 

[Jufro]

I would assume that the ducks are spherical. The best packing fraction for spherical objects is hcp(fcc) at 0.74. While this is a good bound, it has been shown that random packing is between 62 and 64%. By it has been shown I mean I took those numbers off of wikipedia.

That being said, if we take 63% of the volume to be filled then .945 m^3 are actually occupied by ducks. Since this volume is occupied by a discrete number of ducks, Vtotal= N Vduck. Finally, there needs to be some statement about the volume of this spherical duck.

This one is a little harder to find on wiki so I just went with the approximation that they are best thought of as having a diameter of 7.5 cm

With that I come to a final calculation of 4278. Seems a bit much, volumes can do that but I also feel that it was somewhat conservative on my part of the diameter. Feel free to change the size as you deem appropriate.

 

[micromass]

I notice the ducks are all numbered. So what you can do is to take a look at all numbers visible. Assume that you observe $n$ numbers and that the highest observed number is $X_{(n)}$. Then a very good estimate for the total number is

$$\frac{n+1}{n} X_{(n)} - 1$$

For example, if you see the numbers $123$, $1$, $43$ and $154$, then a good estimate is

$$154\frac{5}{4} - 1 = 191.5$$

This method was used in world war II by the allies to estimate the number of German tanks. The results were very accurate. It is therefore known as the German tank problem.

 

[William White]

wow. that's a nice trick.

had never heard of that! I wonder if it takes account of these
https://en.wikipedia.org/wiki/Dummy_tank

from what I can see (ignorning numbers I can't see accurately) the number is 561?However, there might be a clue in that the poster has 4 question marks, so is the number a four digit number(?) and Benfords Law says most numbers begin with 1...RUbers answer looks good (it must be the low 1000s surely?)

when do we get the answer?

 

[gana]

I tink this may or may not work. Jst multiply the first row number nd first column u will get area(ducks in the front plane ) nd multiply that with the number of ducks on the side of the container so dat u can get an approximate value