December 13, 2019 15:33 WSPC/S2661-3395 1950013

The Physics EducatorVol. 1, No. 4 (2019) 1950013 (9 pages)c© World Scientific Publishing CompanyDOI: 10.1142/S2661339519500136

How Quickly Can a Submarine Dive?

Carl E. MunganPhysics Department, U.S. Naval Academy,

Annapolis, MD 21402-1363 USAmungan@usna.edu

Trevor C. LipscombeCatholic University of America Press,

Washington, DC 20064 USAlipscombe@cua.edu

Received August 23, 2019; Revised September 6, 2019; Accepted October 13, 2019

A two-equation model is developed to describe the descent of a submarine as it floods its ballasttanks. The flow into the tanks is assumed to be inviscid, and the drag on the vertical sinkingmotion of the craft is neglected. The two coupled differential equations are the generalized formof Newton’s second law and the Bernoulli relation. Time derivatives are converted to spatialderivatives to decouple the equations, and the resulting second-order equation is solved using theEuler–Cromer algorithm. The theory and the method of numerical integration are suitable for anintermediate-level undergraduate course in mechanics that includes some basic fluid dynamics.

Keywords : Sinking; submarine; Bernoulli equation; Newton’s second law; Euler–Cromer numer-ical integration.

1. Introduction

The sinking of floating objects has long caught thepublic imagination. For example, there is a largevariety of movies and books about the Titanic1

which in 1912 took 2 hours and 40 minutes to sinkafter striking an iceberg. The 1915 sinking of theLusitania, just 18 minutes after a torpedo fromthe German submarine U-20 struck it, eventuallyprecipitated the entry of the United States intoWorld War I. For both the Titanic and the Lusi-tania, water turbulently rushed into the holes cre-ated by the iceberg and the torpedo, respectively,and the ships keeled over. Water then flowed fromone compartment to another. Realistically mod-eling the details of such dynamics is a challenge

for marine engineers. There are textbooks2 andtechnical journals3–5 that use computational fluiddynamics to explore the associated complexities.

In the present article, simpler concepts are usedto model the sinking of a submarine as it fillsan unbaffled ballast tank of constant cross-sectionunder idealized conditions at a level accessible toundergraduate physics students. A vent hole in thebottom of the tank allows water to flood it when topflaps are opened to permit the air inside to escape.Similar ideas could describe the sinking of a flat-bottomed ship (such as a barge, Higgins boat, punt,or quffa) when its hull is breached. Either way, thesubmarine or ship is assumed to have no horizontalor rotational motion, so that the relative motion ofthe water and boat is purely vertical. Archimedes’

1950013-1

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

C. E. Mungan & T. C. Lipscombe

principle, the equation of continuity, Newton’s sec-ond law, and the Bernoulli equation are combinedand solved using difference equations. Extensions ofthese ideas could form the basis for student’s com-putational or experimental projects in a mechanicscourse.

2. Two Coupled DifferentialEquations of Motion

Figure 1 sketches a boat (using that term to refergenerally to a submarine or ship) which for simplic-ity is modeled as having a constant cross-section.Thus it has a flat bottom of area Aboat and verti-cal sidewalls of height H. The boat has a verticaltank (or compartment) of the same height H and ofcross-sectional area Atank which is taken to be cen-tered horizontally in the craft in Fig. 1, although inthe case of an actual submarine there are multipleballast tanks distributed around the hull. The zeroof time corresponds to the instant that water startsto enter a hole in the center of the tank bottom.The initial water levels outside the boat and insidethe tank are h1(0) < H and h2(0) = 0, respectively,with the watercraft floating stably at rest. That is,if the boat has mass mboat when empty of water,the downward gravitational force must balance theupward buoyant force equal to the initial weight ofdisplaced water such that

mboatg = ρgAboath1(0) (1)

from Archimedes’ principle, where ρ is the densityof water (assumed to be incompressible) and g is

Fig. 1. Cross-section of a cylindrical tank or compartmenton a floating boat (in green) with a hole in the center of itsflat bottom and with its top open. Water (of density ρ) hasheight h1 outside the boat and height h2 inside, relative tothe bottom of the tank. Points 1 and 2 are on the free watersurfaces outside and inside the ship, respectively, while point3 is at the fully contracted water jet inside the tank justafter passing through the hole. A streamline (in red) connectspoints 1 and 3. The craft sinks when h1 becomes equal to itsexternal height H , with water thereafter suddenly pouringin through the top. The horizontal cross-sectional areas A ofthe boat, tank, and hole are indicated.

earth’s surface gravitational field strength. Subse-quently the boat sinks at a speed of dh1/dt ≡ υ1

relative to the outer water surface, while water risesinside the tank at a speed of dh2/dt ≡ υ2 relative tothe hull. The latter speed υ2 can be related to thespeed υ3 of the water jet through the hole (againrelative to the hull) using the equation of continuityin the form

Aholeυ3 = Atankυ2 (2)

where Ahole is the cross-sectional area of the holereduced by the vena contracta6 as the fluid passesthrough it. Equation (2) holds on average eventhough the flow inside the tank will in generalbe turbulent. Specifically, water will jet throughthe hole into the ballast tank and swirl around sothat most points inside the tank (such as point2 in Fig. 1) cannot be connected by a streamlineto points outside the tank. However, there is oneexception of an internal point to which the Bernoulliequation can be applied, as discussed for “sub-merged orifices” in books such as Brater and King7

or Roberson and Crowe.8 That point is inside theunderwater jet after it has fully contracted (whichfor a round hole occurs at a distance of about oneradius beyond the aperture) when its pressure bal-ances the hydrostatic pressure of the essentially sta-tionary water (relative to the boat) surrounding theedges of the hole.

However, the Bernoulli equation is not Galileaninvariant: the bottom of the sinking boat does workon the water in earth’s frame of reference.9 To avoidthat work term, the equation can instead be appliedin a frame instantaneously comoving with the boat.Then the sum of the pressure (which is the appliedwork per unit volume) and the kinetic and grav-itational potential energy densities have the samevalue at any point along the streamline. Notingthat the water at endpoints 1 and 3 in Fig. 1 isrespectively moving upward at speeds υ1 and υ3

relative to the boat, the Bernoulli relation thusbecomes

Patm +12ρυ2

1 + ρgh1

= (Patm + ρgh2) +12ρυ2

3 + 0. (3)

The final zero arises because the gravitational refer-ence height is chosen to lie at point 3 (so that waterat point 1 has a gravitational potential energy den-sity of approximately ρgh1) where the hydrostatic

1950013-2

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

How Quickly Can a Submarine Dive?

gauge pressure is ρgh2. Atmospheric pressure Patm

cancels from the two sides of Eq. (3). The speeds ofthe water outside and inside the tank vary slowlyenougha that it is not necessary to add an unsteadyterm10 into this equation. That is, the flow ofwater through the hole is “quasisteady”.11 Now useEq. (2) to eliminate υ3 in favor of υ2 where thesquared ratio of the area of the tank to the areaof the hole is R ≡ A2

tank/A2hole. Also introduce

dimensionless heights and time as x ≡ h1/h1(0),y ≡ h2/h1(0), and T ≡ [2g/h1(0)]1/2t. DividingEq. (3) by ρgh1(0) then recasts it into the normal-ized form

x2 + x = Ry2 + y (4)

where overdots indicate differentiation with respectto T . Equation (4) expresses a connection betweenthe kinetic and potential energies of a unit mass ofwater on the free surfaces outside and inside theboat. If one were to think of the tank as a pipe ofarea Atank, Eq. (4) relates the upward speed x ofwater underneath the boat moving toward the hole(in a frame of reference in which the boat is sta-tionary) to the upward speed y of flow inside thepipe, which depends on the relative size of the holeadmitting water into the pipe. The normalized pres-sure head x − y at the hole remains positive untilthe watercraft sinks. (It is assumed that the rest ofthe boat other than the tank has an average densitygreater than that of water if sinking is to occur atall.)

Equation (3) is an application of a work-energyprinciple, in that it can be derived as the first inte-gral of Newton’s second law applied to a unit massof water. It leads to Eq. (4) for the two unknownfunctions x(T ) and y(T ). A second independentequation in these two unknowns is needed. It isobtained from the generalized form of Newton’s sec-ond law for a variable-mass system,12

d

dt(m�υ) = �F + �u

dm

dt(5)

where a principal system of mass m moving atvelocity �υ continuously makes perfectly inelasticcollisions with increments of mass dm moving atvelocity �u. Here �F is the net external force on theprincipal system. Due caution should be exercisedwhen writing down Newton’s second law for a

variable-mass system because it is common (but ingeneral incorrect13) to write it as �F = d(m�υ)/dt,which is missing the last term in Eq. (5). This incor-rect form expands into �F = m d�υ/dt + �υ dm/dt

which is wrong because both �F and m d�υ/dt areGalilean invariants, while �υ dm/dt is not.14

In the present problem, the principal systemis chosen to be the boat at arbitrary time t whenit contains some water inside, so that Eq. (5)becomes

d

dt(mboat�υboat + mwater inside�υwater inside)

= �F +dm

dt�υwater outside. (6)

By choosing a reference frame in which the bulk ofthe water outside the ship is at rest, this equationreduces to

d

dt(mboat�υboat + mwater inside�υwater inside) = �F .

(7)

At this point, a reader might object that Eq. (7)can be rewritten as d�p/dt = �F where �p is the totalmomentum of the principal system, which appearsto be of the incorrect form discussed above. How-ever, Refs. 13 and 14 do not say that �F = d(m�υ)/dtis always wrong; it is only incorrect if �u is nonzeroin the chosen frame of reference of the problem.If �u happens to be zero, as is the case here, then�F is equal to d�p/dt for a variable-mass system. Toallay any lingering doubt on the issue, Appendix Aderives Eq. (7) from first principles.

Apply Eq. (7) to the vertical motion in Fig. 1,with the downward direction chosen to be positive.In a frame of reference attached to point 1, the hullthen has velocity υ1 and the water inside the boathas velocity υ1−υ2 (because υ2 is the upward veloc-ity of the center of mass of the interior water relativeto the hull). Since the water in the tank has heighth2, Eq. (7) becomes

d

dt[ρAboath1(0)υ1 + ρAtankh2(υ1 − υ2)]

= ρgAboath1(0) + ρgAtankh2 − ρgAboath1 (8)

using Eq. (1) to find the mass of the boat. Here thethree terms on the right-hand side are the down-ward gravitational forces on the hull and interior

aThe water has the largest acceleration as it passes through the hole. For the data corresponding to Fig. 3, however, |dυ3/dt|never exceeds 1% of g.

1950013-3

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

C. E. Mungan & T. C. Lipscombe

water, and the upward buoyant forceb on the boat.Dividing Eq. (8) by ρgAboath1(0) gives

g−1 d

dt

[dh1

dt+

rh2

h1(0)

(dh1

dt− dh2

dt

)]

= 1 +rh2

h1(0)− h1

h1(0)(9)

where r ≡ Atank/Aboat. Recasting it in terms ofthe dimensionless heights and time introduced afterEq. (3) leads to

2d

dT[x + ry(x − y)] = 1 + ry − x. (10)

3. Numerical Solution of theEquations of Motion

Equations (4) and (10) are two coupled differentialequations to be solved with initial conditions

x(0) = 1, y(0) = 0,

x(0) = 0, and y(0) = R−1/2.(11)

The first condition follows from the definition x ≡h1/h1(0). The second condition says there is nowater initially inside the tank. The third states thatthe submarine starts out stably floating at rest. Thefourth initial condition comes from the other threeusing Eq. (4); it implies that water will immediatelystart flowing through the hole (but not very rapidlyif R � 1) at the instant the top of the submarineballast tank is opened or the ship hull is breached.

Eliminate time in favor of space as the indepen-dent variable using the chain rule as follows. Write

x ≡ dx

dT=

dx

dy

dy

dT≡ x′y (12)

where primes indicate derivatives with respect to y.Substituting Eq. (12) into (4) leads to

y =√

x − y

R − x′2 (13)

and putting this result back into Eq. (12) gives

x = x′√

x − y

R − x′2 . (14)

A manipulation similar to that of Eq. (12) convertsEq. (10) into

d

dy[x + ry(x − y)] =

1 + ry − x

2y. (15)

Finally, substitute Eqs. (13) and (14) into (15) andexpand the left-hand side to obtain

x′′ =

(R − x′2)[(1 − x′)(2rx − 3ry + x′ + ryx′)+ (R − x′2)(1 + ry − x)]

2(x − y)(R + Rry − ryx′).

(16)

Since x(0) and y(0) refer to the initial conditionT = 0, we will use subscripts instead of parenthesesto refer to the boundary value y = 0. Specificallyx0 = 1 because y = 0 when T = 0 which is whenx = 1 according to Eq. (11). Also x′

0 = 0 becausex′ = x/y according to Eq. (12), and x(0)/y(0) = 0from Eq. (11).

Equation (16) is numerically integrated usingthe Euler–Cromer algorithm,16 as flowcharted inFig. 2. While a spreadsheet can be employed for thispurpose, the computations are faster and of higherprecision using a mathematical software package.Submarine officers at the United States NavalAcademy were consulted to obtain estimates for thefollowing three parameters in the calculations.

• The fraction of the volume of a submarine that isabove the water line when it is floating on the sur-face is called the “reserve buoyancy” and rangesfrom 15% to 30%. For the model in Fig. 1, thatvalue is equal to 1−h1(0)/H. Its significance hereis that the submarine will sink when x ≡ h1/h1(0)becomes equal to H/h1(0) which is the reciprocalof 1 minus the reserve buoyancy. Designate thatlimiting value as xsink. A round number on thehigh end of the range of possibilities is adopted,namely xsink = 1.4, in order to find the longestamount of time it might take a submarine to dive.

• The ratio r of the cross-sectional areas of the com-bined ballast tanks to that of the entire subma-rine (at its midplane) is on the order of one quar-ter. One can verify that r must exceed the reservebuoyancy, as otherwise the submarine will notsink even when its ballast tanks are completely

bThe buoyant force is the result of the external water pressure pushing upward on the combined system of hull and enclosedwater. Even across the hole, where the water is not stationary, the total pressure is equal to the sum of the static and dynamicpressures.15 Thus it is not necessary to reduce Aboat in the last term of Eq. (8) by Ahole. The total pressure at the hole canbe considered to arise from the normal force exerted by the seafloor directly below the hole.6

1950013-4

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

How Quickly Can a Submarine Dive?

Fig. 2. Flowchart of the Euler–Cromer method of numeri-cally integrating the equations of motion to find x, y, and T .A fixed step size dy is chosen, and accuracy of the results ischecked by making sure the generated curves (such as thosein Fig. 3) do not look visibly different if the step size is madean order of magnitude smaller. In contrast to the standardEuler algorithm, mechanical energy is conserved by using theupdated value x′

new rather than the previous value x′ to com-pute xnew in the middle box of this diagram. Likewise theupdated value ynew is used to compute Tnew.

filled with water! Thus we adopt r = 0.3 whichexceeds 1 − x−1

sink = 2/7.• Finally the grate-covered holes at the bottom of

the ballast tanks are large enough that a personcan climb through them, but are much smallerthan the cross-sectional area of the tanks. Esti-mate each of the 10 grates to measure 0.5m×1mfor a Los Angeles class submarine of midplanearea 110m × 10m. Hydraulic engineers accountfor the combined effects of the vena contracta dis-cussed after Eq. (2) and energy/pressure losses aswater flows through a submerged orifice by intro-ducing an empirical coefficient of discharge C.

Fig. 3. Graphs of the water heights inside and outsidethe boat as a function of time, generated by running theMATLAB script in Appendix B. The submarine sinks whenh1 = H, at which point the ballast tank is nearly full sothat h1 and h2 approach each other in value. The bottomand left axes plot the dimensionless values x = h1/h1(0) and

y = h2/h1(0) versus T = [2g/h1(0)]1/2t, whereas the top andright axes plot the dimensioned values h1 and h2 versus tassuming xsink = 1.4 and H = 10 m as discussed in the text.

A typical value of C for a rectangular gate athigh Reynolds numbers1,7 is approximately 0.6.In that case, the effective ratio of the totalarea squared of the ballast tanks to the bottomvent holes is R = (0.3× 110× 10)2/(0.6× 10×0.5 × 1)2 ≈ 12000.

The curves in Fig. 3 are now obtained usingMATLAB with a step size of dy = 0.001. The scriptis provided in Appendix B so that readers can repro-duce the results. The red curve ends at x = xsink

when Tsink ≈ 225. Thus the submarine sinks in atime of

tsink = Tsink

√H

2gxsink(17)

which is equal to 135 s if the submarine has a heightof H = 10 m equal to its central diameter. Thatis realistic for the initial dive time of a subma-rine (not including the sail on which the periscopeand antennas are mounted), after which the subma-rine engines are typically switched on to continueany further descent more rapidly. This dive timeagrees with the following estimate. Torricelli’s law,obtained from Eq. (3) with υ1 neglected in compar-ison to υ3, implies that the speed of water throughthe hole is given by the pressure head difference

1950013-5

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

C. E. Mungan & T. C. Lipscombe

between the water inside and outside the boat (cf.Fig. 1) so that υ3 = [2g(h1 − h2)]1/2. Then Eq. (2)and R = A2

tank/A2hole imply

υ2 =

√2g(h1 − h2)

R(18)

which on average must be equal to the height ofthe tank divided by its filling time, H/tsink. Replac-ing υ2 in Eq. (18) with that expression, normal-izing h1 and h2 by replacing them with xh1(0)and yh1(0) respectively, and substituting h1(0) =H/xsink results in

tsink =[

RHxsink

2g(x − y)

]1/2

average

. (19)

The initial value of x − y is 1 according to Eq. (11) and it approaches 0 when the boat sinks. Thus on average it is equal to 1/2, so that tsink ≈ (RHxsink/g)1/2 = 130 s. The sinking time scales inversely with the size of the hole: halving its area (so that R is quadrupled) doubles tsink.

The oscillations in Fig. 3 can also be explained. If a floating object of draft h1 is given a small verti-cal displacement and released, it oscillates harmon-ically with an angular frequency of ω = (g/h1)1/2

which in our normalized units corresponds to a dimensionless period of 2π(2x)1/2. That expression increases in value from 8.89 initially (when x = 1) to 10.5 finally (when x = 1.4), in excellent agree-ment with the variation in the period of the red curve in Fig. 3. For a real submarine, such oscilla-tions are eliminated by factors such as drag on the bottom hull of the boat and baffling in the ballast tanks.

One last interesting observation concerns the initial accelerations of the ship and of the water inside it. If the time derivative in Eq. (10) is per-formed, and then all of the resulting variables are evaluated at T = 0 using Eq. (11), one finds x(0) = r/R. Next, if one takes the derivative of both sides of Eq. (4) with respect to T , and then again eval-uates all of the variables at T = 0 using Eq. (11), one finds y(0) = −1/(2R). These two initial accelerations, together with the two initial positions and velocities given in Eq. (11), all agree with Fig. 3. Specifically, the red curve for x starts out at initial value 1 with zero slope and a positive acceleration, whereas the blue curve for y starts out at initial value 0 with a positive slope and a negative accelera-tion. Even though the initial value of the right-hand

side of Eq. (8) is zero, the ship still sinks because ofthe nonzero initial values of υ2 and of a1 ≡ dυ1/dt.

4. Student Projects and Problems

Extensions of the previous ideas and results couldbe explored by students, such as the following.

(i) Investigate the effects of varying the cross-sectional areas of the hole or ballast tank rela-tive to the entire boat. In other words, slowlychange the values of the parameters R and rin the first lines of the code in Appendix Bto see how that influences the speed, depth,and time of sinking. What are the largest andsmallest mathematically allowed values of R−1

and r? How realistic is it for an actual boatto approach those limiting values? What hap-pens if you increase the value of xsink but donot correspondingly increase the value of r?Under what conditions do x and y end upapproximately equal to each other by the timex attains xsink?

(ii) Given that numerical integration is only exactin the mathematical limit dy → 0, why notjust choose dy equal to something much smallersuch as 10−10 rather than 10−3 in the MAT-LAB code? Make a graph of the time the pro-gram takes to run for factor of 10 decreases inthe value of dy. Other than runtime, what elsemight limit the ability of the code to executefor arbitrarily small values of dy? (Considerthe size of the workspace variables listed in theMATLAB window.)

(iii) After repeatedly running the code and vary-ing parameters such as R, r, dy, and xsink, itis often necessary to issue a “clear” commandin MATLAB to avoid getting strange-lookinggraphs. Why is that required? Another issueis that sometimes complex values get calcu-lated for T . Which line in the code causes thatunphysical result?

(iv) In rescaling from dimensionless to dimensionalheights and times, numerical values of g andH are required. Explore how quantities such asthe sinking time and tank filling fraction varyfor different values of R, r, and xsink appro-priate to say a canoe or a barge. In the lattercase, consider both the possibilities that onlyone compartment of the barge floods or thatthe entire interior does.

1950013-6

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

How Quickly Can a Submarine Dive?

5. Conclusions

In the absence of viscosity and drag, a coupledpair of differential equations describing the sinkingof a submarine or of a (possibly compartmented)ship with a breached hull has been developed basedon the Bernoulli equation for the motion of thewater and Newton’s second law for the motion ofthe boat. Typical values for a submarine imply adive time of about two minutes after the top flapsare opened to flood the ballast tanks. The motioncan be approximated using Torricelli’s law for therise of the water in the tanks, with a superim-posed oscillatory motion of the boat due to the com-petition between the downward gravitational andupward buoyant forces. Such an oscillation shouldbe observable for a toy boat with a smooth V-shaped hull to reduce drag on it and weighted alongits keel so that it sinks readily when flooded, assum-ing the hole is large enough to give the boat a size-able initial impulse as water starts streaming intoits interior.

Other ideas for experimental investigation andsubsequent modeling include submersion tests indifferent fluids, and systematic variation in the sizeand placement of the hole in the hull. For simplicity,such studies could use plastic beakers with holes cutinto their bottom, ballasting them with flat cylin-drical masses (from the weight sets often used inintroductory physics labs), and releasing them in anaquarium so that their vertical motion can be videorecorded from the side.

Acknowledgments

C. E. Mungan thanks Peter Brereton, ChuckEdmondson, Chris Gear, and Chris Morgan for dis-cussions of submarine design, and Dan Finkenstadtand Michael Mehl for exploration of Eq. (4).

Appendix A

Consider a principal system whose initial momen-tum (just before it accretes an additional incrementof mass) is

�pi = m1�υ1i + m2�υ2i + · · · (A.1)

because it is composed of subsystems 1, 2, . . . eachwith their own mass and initial velocity. RewriteEq. (A.1) as

�pi = m�υi (A.2)

where m ≡ m1 + m2 + · · · is the total mass ofthe system, and �υi ≡ (m1�υ1i + m2�υ2i + · · · )/m isthe initial velocity of its center of mass. Now sup-pose this principal system makes a perfectly inelas-tic collision with an infinitesimal mass dm that isinitially at rest. Designate the principal system plusthe accreted mass element as the “supersystem” ofmass m + dm. Suppose an external force �F acts onthis supersystem, so that its center-of-mass velocityis not constant. Specifically, after the collision thesupersystem has a final center-of-mass velocity of�υf = �υi + d�υ.

The impulse-momentum theorem for theconstant-mass supersystem is�F dt = d�p = �pf − �pi = (m + dm)(�υi + d�υ) − m�υi

(A.3)

since the infinitesimal mass dm was originally at restand thus does not contribute to the initial momen-tum. Dropping the second-order term dm d�υ andthe subscript on �υi, Eq. (A.3) can be rewritten as

�F = md�υ

dt+ �υ

dm

dt=

d

dt(m�υ) (A.4)

which is equivalent to Eq. (7) in the present prob-lem. Other applications of Eq. (A.4) include tofalling raindrop17 and rolling snowball18 problems,in which a sphere moves through stationary mist orsnow, accumulating additional mass as it does so.

Appendix B

The script used to perform the numerical integra-tions in MATLAB follows.% model parametersR = 12000; % squared ratio of effective areas

of tank to holer = 0.3; % ratio of areas of tank to boatdy = 1e−3; % step size in yxsink = 1.4; % terminating value of x

% initial valuesn = 1; % array indexx(1) = 1; % value of x at T = 0y(1) = 0; % value of y at T = 0xprime(1) = 0; % value of dx/dy at T = 0T(1) = 0; % starting value of T

% numerical solution of differential equationswhile x(n)< xsink

xprimeprime(n)= (R–xprime(n)∧2)*((1–xprime(n))*(r*(2*x(n)–3*y(n))+xprime(n)

1950013-7

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

C. E. Mungan & T. C. Lipscombe

*(1+r*y(n)))+(R–xprime(n)∧2)*(1+r*y(n)–x(n)))/(2*(x(n)–y(n))*(R+r*y(n)*(R–xprime(n)))); % old value of secondderivative of x with respect to y

n = n + 1; % step forwardxprime(n)=xprime(n–1)

+xprimeprime(n–1)*dy;% new value of dx/dy

x(n)= x(n–1) + xprime(n)*dy; % new value of xy(n)= y(n–1) + dy; % new value of yydot(n)= sqrt((x(n)–y(n))/(R–xprime(n)∧2));

% new value of dy/dTT(n)= T(n–1) + dy/ydot(n); % new value of T

end

% visualization of solutionline(T,x,‘Color’,‘r’,‘LineWidth’,2) %

plot x vs T in redline(T,y,‘Color’,‘b’,‘LineWidth’,2) %

plot y vs T in blueaxis([0 225 0 1.5]) % bottom axis ranges

from 0 to 225, left axis from 0 to 1.5xticks([0 50 100 150 200]) % bottom tick marksyticks([0 0.5 1 1.5]) % left tick marksxlabel(‘dimensionless time’) % bottom axis titleylabel(‘dimensionless height’) % left axis titlelegend(‘\it x \rm or \it h {\rm1}’, ‘\it y \rm or

\it h {\rm 2}’, ‘Location’,‘northwest’) %add legend in NW corner of graph

ax2= axes(‘XAxisLocation’,‘top’,‘YAxisLocation’,‘right’,‘Color’,‘none’); % second set of axes

axis(ax2,[0 225*sqrt(10/(2*9.8*1.4)) 0 1.5*10/1.4])% rescale to dimensional units for H= 10 m,

g = 9.8 m/s/s, and xsink= 1.4xticks(ax2,[0 30 60 90 120]) % top tick marksyticks(ax2,[0 2 4 6 8 10]) % right tick marksxlabel(ax2,‘time (s)’) % top axis titleylabel(ax2,‘height (m)’) % right axis title

References

1. J. W. Stettler and B. S. Thomas, Flooding andstructural forensic analysis of the sinking of the RMSTitanic, Ships Offshore Struc. 8, 346–366 (2013).

2. A. Biran and R. Lopez-Pulido, Ship Hydrostat-ics and Stability, 2nd ed (Butterworth-Heinemann,Waltham MA, 2014), Chap. 11.

3. Z. Gao, Q. Gao and D. Vassalos, Numerical simula-tion of flooding of a damaged ship, Ocean Eng. 38,1649–1662 (2011).

4. J.-S. Kim, M.-I. Roh and S.-H. Ham, A method forintermediate flooding and sinking simulation of adamaged floater in time domain, J. Comput. Des.Eng. 4, 1–13 (2017).

5. H. Cheng, A. M. Zhang and F. R. Ming, Study oncoupled dynamics of ship and flooding water basedon experimental and SPH methods, Phys. Fluids 29,107101:1–20 (2017).

6. A. Jenkins, Sprinkler heads revisited: Momentum,forces, and flows in Machian propulsion, Eur. J.Phys. 32, 1213–1226 (2011).

7. E. F. Brater and H. W. King, Handbook ofHydraulics, 6th ed (McGraw-Hill, New York NY,1976), Sec. 4.

8. J. A. Roberson and C. T. Crowe, Engineering FluidMechanics, 6th ed (Wiley, Hoboken NJ, 1997),Sec. 13.2.

9. C. E. Mungan, The Bernoulli equation in a movingreference frame, Eur. J. Phys. 32, 517–520 (2011).

10. H. Johari and W. W. Durgin, Unsteady flow in ver-tical converging tubes, Proc. ASME (PVP Series)224, 37–39 (1991).

11. B. R. Munson, T. H. Okiishi, W. W. Huebsch andA. P. Rothmayer, Fundamentals of Fluid Mechanics,7th ed (Wiley, Hoboken NJ, 2013), p. 138.

12. C. A. de Sousa and V. H. Rodrigues, Mass redistri-bution in variable mass systems, Eur. J. Phys. 25,41–49 (2004).

13. J. Mallinckrodt, F does not equal d(mv)/dt, Phys.Teach. 48, 360 (2010).

14. E. Mosca, Invalid equation, Phys. Teach. 48, 361(2010).

15. J. J. Bertin, Engineering Fluid Mechanics, 2nd ed.(Prentice-Hall, Englewood Cliffs NJ, 1987), p. 203.

16. A. Cromer, Stable solutions using the Euler approx-imation, Am. J. Phys. 49, 455–459 (1981).

17. C. E. Mungan, More about the falling raindrop, Am.J. Phys. 78, 1421 (2010).

18. S. Rubin, A variable-mass snowball rolling down asnowy slope, Phys. Teach. 57, 150–151 (2019).

1950013-8

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.

December 13, 2019 15:33 WSPC/S2661-3395 1950013

How Quickly Can a Submarine Dive?

Carl E. Mungan has a Ph.D. in experimental condensed matter physics from CornellUniversity in New York and a B.Sc. (Honours) in theoretical physics from Queen’s Uni-versity in Canada. He is a Professor at the U.S. Naval Academy which graduates about30 physics majors a year, many of whom enter the nuclear submarine fleet.

Trevor C. Lipscombe is Director of the Catholic University of America Press. He isthe author of The Physics of Rugby (Nottingham University Press) and coauthor withAlice Calaprice of Albert Einstein: A Biography (Greenwood). His latest book QuickerCalculations is forthcoming from Oxford University Press.

1950013-9

The

Phy

s. E

duca

t. 20

19.0

1. D

ownl

oade

d fr

om w

ww

.wor

ldsc

ient

ific

.com

by C

arl M

unga

n on

12/

20/1

9. R

e-us

e an

d di

stri

butio

n is

str

ictly

not

per

mitt

ed, e

xcep

t for

Ope

n A

cces

s ar

ticle

s.