Once Upon a Time in America


Special Mention: Srinivas Rao Pujari aka Mr. Always ready to help.

Valuable Inputs by Vivek and Vishan

From the humble beginnings of being Italian immigrants to being the godfathers of the American underworld, we are talking about a story by the legendary filmmaker  Sergio Leonne and his magnum opus “Once Upon A Time in America”. It chronicles the lives of Jewishghetto youths who rise to prominence in New York City’s world of organized crime.

“So Once Upon A Time in America is a movie classic. But I did not come here for a movie review. Isn’t this supposed to be a science blog?” Well we are not talking about the entire movie here. Rather only about a particular scene which has captivated me so much that I’ve decided to science the shit out of it!


A bunch of kids in the streets having big dreams and wanting to get into the underworld comes up with a rather astonishing idea of smuggling cocaine across the coastline. (Talking about smuggling cocaine, how can I not mention NARCOS! Go check it out if  you haven’t already.It is one of the best in the business when you talk about Netflix Originals.) They seal cocaine in a plastic bag,attach a rubber ball and a sack filled with salt to it. Then they drop this off at a particular point in the sea. How does the second party recieve this?

The Four Horsemen of the Apocalypse (No doubt that the kids grew up to be these Mafias, the movie has one of my favorite actors Robert De Niro)


Due to the weight of salt,the cocaine gets suspended near the bottom of the sea.As the salt dissolves the buoyancy force of the rubber ball starts to act,which pulls the plastic sealed cocaine upwards. They would have performed careful experiments to observe the time taken by the certain amount salt to dissolve and for cocaine to come at the top.The second party arrives at the drop off location according to this and receives the shipment.

The cocaine about to rise from sea bed; Free body diagram of Cocaine package

What are we supposed to do now? We will try to calculate the time taken by the cocaine to resurface onto the sea water as the salt dissolves, using simple concepts of Mass Transfer.


1. The salt in the sack is packed in spherical shape.

2. The sea water is stagnant.

3. Even though sea water has some dissolved salts we will assume that at infinity the concentration of salt is zero.

4. We assume Quasi-steady state process, it means that mass transfer process is really slow.

5. Assuming the salt accumulation in sea water is zero. 

Assuming an shell and calculating mass transfer using Fick’s Law





Using Conservation of mass across a shell at a distance r as show in above figure:

=>  (4πr2Na) at r – (4πr2Na) at r + dr = 0;  [No accumulation of salt, Assumption number 5 is used here]

[Here Na= mole flux, moles/sec*m2]

=> d/dr(4πr2Na) = 0;

=> r2Na = c/4*π = c0 (let);                                 eqn (1)



Using Fick’s Law

=> Na = -Da*d/dr(Ca);

[Here Da= Diffusion coefficient, Ca= concentration of salt]

=> -r2 * Da*d/dr(Ca) = c0;


[At r = r0 the conectration of salt will be equal to solubility of salt in sea water and at infinity it is 0 using assumption 3]

=> – Da (0- Ca0) = c0 * [-1/r] lim (inf to r0);

=> Da * Ca0 = c0/r0;                                            eqn (2) 



 Rate of moles of salt getting dissolved = 4πr2Na  Moles/ sec =(Volume * density)/Molecular Mass:

=> 4πr2Na = moles/ sec = -d/dt(4πr3*p/3*M0)

[Here p = density of table salt, M0 = Molecular mass of salt]

=> Na = -p/M0*dr/dt                                          eqn (3)


Put eqn (2) in eqn (1), we get

=> Na = Da * Ca0 / r

As for quasi steady state process, r = r0

=> Na = Da * Ca0 / r = -p/M0*dr/dt

=> (Da * Ca0 *M0/p) dt = -rdr;

=>Da * Ca0 *M0/p = α (let);                                       eqn(4)


=> α * t = 1*2(R12 – R22);                                        eqn(5)

 Now to get α:

M0 = 58.5 * 10-3 kg;

Ca0 = 6.9 * 103 kg/m3;               (http://www.chem1.com/acad/webtext/solut/solut-6a.html)

p = 2.17 * 103 kg/m3;

Da = 1.99 * 10-9 m2/s;               (http://oto2.wustl.edu/cochlea/model/diffcoef.htm)


We get the value of α by using eqn(4) as α = 370.16 * 10-12;

Assuming that the salt packing was 15 cm in radius initially (at that time its weight would be approx. 30 kg). Finally assuming that with the loss of 5 kg of salt in sea water the buoyancy of the balloon dominates and the balloon would surface to the water. So the final radius at that time would be 14 cm with the mass of the salt being 25 kg.

R1 = 15 cm, R2 = 14 cm;

Using eqn (5) we get t = 4.531 days or approx. 4 days and 13 hours. So the second party should come after nearly 4 days and 13 hours to get the cocaine floating on the surface of sea water.


  • Used the conservation of mass to get equation 1;
  • Then we applied fick’s law to get value of c0 by using boundary conditions that at r = r0 the concentration of salt will be equal to solubility of salt in sea water and at infinity it is 0 using assumption 3, this way we get equation 2;
  • We get the rate of moles dissolving per second which will give the rate of change of r with time in equation 3. The we used three equations to get the value of “dr/dt” and then integrate it to get r as a function of time.
  • Finally by changing r from 15 cm to 14 cm with a 5 kg change in mass we got that it will take approx 4 days and 13 hrs.




The Background work



  1.    http://www.chem1.com/acad/webtext/solut/solut-6a.html
  2.    http://oto2.wustl.edu/cochlea/model/diffcoef.htm
  3.    https://en.wikipedia.org/wiki/Salt
  4.    https://www.youtube.com/watch?v=Jj5Xczethmw


  1. The essence of the blog was it’s mathematical part which I being a electrical engg do not get it. It would be much easier if you explain the part with some brief points along with the derivation part. By the way the blog was great.

    Liked by 1 person

    1. Hey Joey! Thanks for the read and the appreciation. I have added the brief of the solution at the end under the heading “RECAP”. I would definitely try to improve on my maths presentation skills. You may check out other posts which are not that mathematical in approach. Thanks again! 🙂


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