Operation Overlord

D-Day (June 6, 1944)

Operation Overlord

Location: Omaha beach, Normandy, France

The allied forces were set for a break through, starting off the final operation to recapture the occupied France and dismantle the Third Reich aka Nazi Germany. Crossing the murky waters of English Channel in a weather which could not have been more challenging, the largest amphibious operation in the history of modern warfare began. With the strike of dawn allied forces began landing in the beaches of Sword, Juno, Gold, Omaha and Utah.

landing map
Fig 1: Allied invasion in Normandy, France

 

While the battle was tough on all fronts, it was Omaha where forces encountered heavy opposition in an attempt to secure the 10 km long beach. While we all know what happened in the end, the reasons may very well be unknown as not every thing went according to the plan and in situations like this planning was more or less obsolete.

“I didn’t have any idea of how deep it would be, but I’m six foot
one and the sea was up to my chest and it took me a while to find
my feet …”

LT. SIDNEY SALOMON, C COMPANY 2ND RANGERS, ON LANDING ON OMAHA BEACH

But things were about to get changed for Lt. Sidney Salomon and his men. They expected the bullets from infamous MG-42 (Maschinengewehe 42) aka Hitler’s Zipper or Hitler’s Buzz Saw, piercing through their bodies in no time. But this was not the case! So what exactly happened there? For believers may be it was divine intervention in a fight against evil but for people like us there was water.

MG42II
Fig 2: MG 42 in action
Frankreich, MG-Schütze
Fig 3: A German solider with MG 42

The waters of English Channel were acting like a shield against the MG 42 bullets. But how is it possible? Lets put our science cap on and try to imagine the situation.

Omaha_beach_cross_section
Fig 4: Cross section of Omaha beach

 

 

For the sake of simplicity I’m using AK 47 statistics (MG 42 stats are difficult to find, if you get them then follow the calculations given below to obtain result) and we will try to calculate the distance a bullet travels in water.

WARNING!

MATHEMATICS AHEAD

Diameter of bullet = 7.92 mm, thus Cross sectional area = 49.24 * 10-6 m

Mass of Bullet (M) = 7.9 gm

Muzzle Velocity of AK 47 = 755 m/s

Drag Coefficient = 0.5 (Source: Wikipedia, assuming bullet is similar to a cone)

Distance from firing point to landing point ~ 550 yards = 503 m (As shown in Fig 4)

We will solve the problem in two parts. Here is the plan of action:

  1. Part 1 when bullet is fired and is travelling through air till it hits the surface of water so that we can calculate the speed of bullet just before it hits the water. The speed gets reduced significantly due the air drag. 
  2. Part 2 when bullet enters water. Here we will calculate the final velocity attained by the bullet i.e Terminal Velocity which develops when there is no acceleration of a body in fluid. Then we will calculate the total distance traveled by the bullet underwater.

 

PART 1

General Equation of Drag Force is given as:

F = 0.5*p*V2*A*C

Where, p = Density of the medium

V = Linear Velocity

A = Cross sectional Area of the object

C = Drag Coefficient

Extra: If you want to know the application of Drag Force in case of golf balls click here.

force balance
Fig 5: Forces acting on bullet in water

 

 

As shown in Fig 5 the free body diagram of bullet in air will remain the same with same direction of forces but with different magnitude.

Let density of air = 1.2 Kg/m3

F(Air) = 14.772*10-6*V2

Using Newton’s second law, F = M*a

M*a = – F(Air)

eqn

 

putting the values and integrating, we get

ln(V1/V2) = -1.9*10-3 *S

V1 = Muzzle velocity = 755 m/s

V2 = Velocity just before hitting water

S = 503 m

We get V2 = 290.51 m/s 

PART 2 

Density of Water = 1000 kg/m^3

F(Water) = 12.31*10-3*V2

Terminal Velocity Calculation

M*g = F(Water) 

Vt = 2.5 m/s 

Again using Newton’s second law as earlier, we get:

ln(V2/Vt) = 1.6 *S

V2 = 290.51 m/s

Vt = 2.5 m/s

We get, S = 3.05 m or 9.84 ft 

Assuming angle of incidence to be 45° (taken arbitrarily)

S (Horizontal) = S*cos45°

S (Horizontal) = 6.95 ft 

pose
Fig 6: Physicist Andreas Wahl risking his life for the show “Life on the Line”

The calculated value falls in an agreement with real life observations. After travelling a distance of nearly 7 ft the bullet of AK 47 would lose its piercing power (I’m quite sure that  this 7 ft could be taken as upper limit for a MG 42 bullet). For a soldier on Omaha beach this value was probably a matter of life and death. So you see how water saved life of many soldiers.

By the end of D-Day the Allies had landed
150,000 troops at a cost of 9,000 casualties,
considerably less than had at first been
feared. Only Omaha had seen real problems.

-Pg 259, World War II: The Definitive Visual History

One should keep in mind that there were several factors playing at once on D-Day. I just picked one and tried to explain the situation, however in the classic opening scenes of Saving Private Ryan director Steven Spielberg showed the very best of his abilities and recreated the invasion. Here is a clip for you. Cheers!

Bibliography:

  1. http://ww2db.com/weapon.php?q=10
  2. https://en.wikipedia.org/wiki/Drag_coefficient
  3. https://upload.wikimedia.org/wikipedia/commons/thumb/f/f8/Omaha_beach_cross_section.svg/2000px-Omaha_beach_cross_section.svg.png
  4. https://www.youtube.com/watch?v=71bflWCtZFY
  5. World War II: The Definitive Visual History
  6. Saving Private Ryan (1998)
  7. http://askthephysicist.com/classical%20mechanics.html

If you like the article then please comment and share, Thanks!

 

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