Scientific and Educational Reports of the Facultyof Science and Technology, Kochi University
Vol. 3 (2020), No. 3

Projectile Trajectory of Penguin’s Faeces and Rectal Pressure Revisited
Hiroyuki Tajima1 and Fumiya Fujisawa2
1

Department of Mathematics and Physics, Kochi University, Kochi 780-8520, Japan
2
Katsurahama Aquarium, Kochi 781-0262, Japan(Dated: July 1, 2020)

We discuss a trajectory of penguins’ faeces after the powerful shooting due to their strong rectalpressure. Practically, it is important to see how far faeceses reach when penguins expel them fromhigher places. Such information is useful for keepers to avoid the direct hitting of faeceses. We estimate the upper bound for the maximum flight distance by solving the Newton’s equation of motion.
Our results indicate that the safety zone should be 1.34 meters away from a penguin trying to poopin typical environments. In the presence of the viscous resistance, the grounding time and the flying
distance of faeces can be expressed in terms of Lambert W function. Furthermore, we address the
penguin’s rectal pressure within the hydrodynamical approximation combining Bernoulli’s theorem
and Hagen-Poiseuille equation for viscosity corrections. We found that the calculated rectal pressureis larger than the estimation in the previous work.
PACS numbers: 01.40.-d,47.85.Dh,87.15.LaI.

y

INTRODUCTION

Penguins, which are aquatic birds living mostly in theSouthern Hemisphere [1], strongly shoot their faeceses
towards their rear side [2]. It is believed that this is because penguins avoid getting the faeces on themselves aswell as the nest. Although such a tendency is not limited to penguins and can be found in the case of otherbirds, these bombings sometimes embarrass keepers under breeding environments like an aquarium. It is practically important to know how far their faeceses reach fromthe origin. Such information would save keepers from thecrisis. It would also be helpful for a newcomer guidance
for keepers to avoid such an incident.
The flying distance of penguin’s faeces reaches about
0.4 m even on the ground. Since a typical height of a
Humboldt penguin is given by 0.4 m, this distance corresponds to the situation that if a human being whose
height is 1.7 m tries to evacuate his/her bowels, the object could fly to 1.7 m away. Therefore, one can immediately understand that penguin’s rectal pressure is relatively much strong compared to that of a human kind.
In the pioneering work of Ref. [2], it is reported that thisactual pressure could range from 10 kPa to 60 kPa forrelevant values of the faeces viscosity and the radius of
the bottom hole.
However, in Ref. [2], the projectile trajectory of the
faeces is not taken into account but the horizontal distance is employed to estimate the fluid volume. In general, the actual trajectory would be longer than the horizontal distance during the projectile motion as shownin Fig. 1. In addition, since the ejection angle is not always horizontal and penguins sometimes shoot them out
from higher place under the breeding environment, it isimportant to consider such a motion of faeces in a moregeneral manner. The rectal pressure is regarded as animpulse force to accelerate the faeces up to the initial
velocity in Ref. [2]. For non-viscous fluids, we can use
Bernoulli’s theorem which is related to the energy con-

Received: June 29, 2020
Reviewed by anonymous referee(s), and accepted: July 2, 2020
Published: July 14, 2020v0 = (v0x,v0y)

x

h
d
O
FIG. 1: Configuration for a penguin trying to defecate towards his/her rear side. A penguin stands on the rock withthe height h from the ground. We parameterize the ejection
angle θ with the initial velocity v = (v0x , v0y ) We estimate aflying distance d of the faeces from the origin O.

servation. To estimate the viscosity eﬀects on the rectalpressure, Ref. [2] assumed the Hagen-Poiseille flow in theair, which is laminar flow of Newtonian liquid in a cylindrical pipe geometry. A similar approach for uniaxial
urinary flow was employed to estimate the duration ofurination of various animals in Refs. [3, 4].
In this work, we calculate the maximum flying distanceof penguin’s faeces from a high place, which is relevant
for several breeding environments. Such a projectile trajectory is described by Newton’s equation of motion. We
assume that the upper bound for the flying distance canbe obtained by the equation of motion in the absence of
the air resistance. Moreover, we revisit the rectal pressure by using Bernoulli’s theorem and the Hagen-Poiseilleequation [5] to estimate the mechanical contributions of
non-viscous flow and the viscosity correction during theflow in the stomach and in the air, respectively.

2guin’s rectal pressure, respectively.
III.

2R

P=P0+Pt

NEWTON’S EQUATION OF MOTIONWITHOUT AIR RESISTANCES

We consider Newton’s equations of motion

z
2r
P0

m
FIG. 2: Model for penguin’s stomach. We assume that the
faeceses are prepared in a fictitious cylindrical tank with theradius R. The depth z is estimated from the fluid volume
V after the firing. The pressure in the stomach is given by
P = P0 +Pt where P0 = 1013 hPa and Pt are the atmosphericpressure and the rectal pressure, respectively.

II.

SETUP

Before moving to the calculation, we explain the configurations and the parameters we consider in this work.
Figure 1 shows the situation where a Humboldt penguin shoots the faeces out from the higher place withthe height h. Here, we employ an initial velocity |v0 | ≡v0 = 2.0 m/s reported in Ref. [2] as a typical value. The
ejection angle is denoted by θ, and the gravitational constant is fixed at g = 9.8 m/s2 .
Figure 2 shows the eﬀective model of penguin’s stomach. We employ properties of the faeces used in Ref. [2],where the mass density ρ = 1141 kg/m3 , the viscosityη = 0.02 ∼ 0.08 Pa·s are employed. In our model the radius of bottom hole is given by r = 0.004 m. Although it
is known that penguin’s rectum has a form of a straighttube [6], in this work we approximate it as a cylindricaltank with the radius R for simplicity. The pressure in thestomach is given by P = P0 + Pt where P0 = 1013 hPa
and Pt are the atmospheric pressure in the air and pen-

(3.1)

d2 y
= −mg,
dt2(3.2)

where m is the total mass of penguin’s faeces. By solving
them, one can obtain textbook resultsdx= v0x ≡ v0 cos θ,
dt

(3.3)

dy
= v0y − gt ≡ v0 sin θ − gt,
dt

(3.4)

vx =

vy =
This paper is organized as follows. In Sec. II, we explain our setup for the projectile motion of penguin’s
faeces and the evaculation of them from their intestines.
The latter is used for the estimation of penguin’s rectal pressure. In Sec. III, we show Newton’s equation of
motion for the faeces after the shoot. We show the maximum flying distance at arbitrary angle and height. In
Sec. IV, we discuss eﬀects of the viscous air resistanceduring the projectile motion. In Sec. V, we calculate
the rectal pressure under the assumption that the faeces
liquid can approximately regarded as an ideal fluid. In
Sec. VI, we estimate the additional contribution to therectal pressure due to the viscosity correction within theHagen-Poiseuille equation. Finally, we summarize thispaper in Sec. VII.

d2 x
= 0,
dt2

m

x(t) = v0 t cos θ,

(3.5)

1
y(t) = h + v0 t sin θ − gt2 .
2

(3.6)

and

We obtain the grounding time tg from y(tg ) = 0, which
reads
1 2
gt − v0y tg − h = 0.
2 g

(3.7)

Thus, one can obtain√v0 sin θtg =+gv02 sin2 θ 2h
+
.
g2
g

(3.8)

Since the flying distance is given by d = v0 tg cos θ, we
obtain√
2 sin2 θ
v
sin
θ
v
2h
0
0
.
+
+
(3.9)
d = v0 cos θ gg2
gIn Fig. 3, we plot d as a function of θ at a typical heighth = 1.1 m. The angle θmax where d becomes maximum
is given by the condition


√
2
2
∂d
v0 sin θv0 sin θ 2h 
= −v0 sin θ ++∂θ θ=θmax
g
g2
g


v0 cos θv 2 sin θ cos θ + v0 cos θ + √ 02 2v0 sin θ
g
+ 2h
g2

g

= 0.
(3.10)

1.4

1

1.2

0.8

1

0.6
0.4
0.2

60

θmax [deg.]

1.2

dmax [m]

d [m]

3

0.8

45

0.6

30

0.4

15

0.2

0
-90

-60

-30

0

30

θ [deg.]

60

0

90

FIG. 3: The flying distance d as a function of θ at h = 1.1 m.

From Eq. (3.10) we obtain

0.5(3.11)

0

h [m]

1.5

2

NEWTON’S EQUATION OF MOTIONWITH VISCOUS RESISTANCE

Here, we discuss eﬀects of the air resistance. For simplicity, we consider the case with viscous resistance which
is proportional to v. The equations of motion are given
by










 −1
1


√ (
dmax = v0 cos sin
)





hg
 2 1+ 2 v0

d2 x= −kvx ,
dt2

(4.1)

d2 y
= −mg − kvy .
dt2

(4.2)

m

In this regard, in terms of h and v0 , the maximum flying
distance dmax is given by1

1 1.5 2

h [m]





1

θmax = sin−1 
) .
√ (
hg
2 1 + v2



0.5

FIG. 4: The maximum flying distance dmax as a function ofh. The inset shows the corresponding angle θmax .

IV.



0

00

m

Resulting velocities and distances are given byvx (t) = v0x e− m t ,
k



v
u
 v0
1
v2
2h 
u

.
( 0
)+
× √ (
+
t
)

hg
g
g
2
hg
2g 1 + v2
2 1 + v20vy (t) = v0y e− m t +k0x(t) =(3.12)
andFigure 4 shows h-dependence of dmax at θ = θmax . The
inset shows θmax as a function of h. In the case of h =1.1 m which is typical height of rocks in the penguin’sarea of the Katsurahama aquarium, we obtain dmax =1.03 m at θmax = 21.6◦ . It is slightly longer the caseof the horizontal ejection d(θ = 0) = 0.95 m. We notethat the maximal height of rocks is h = 2.0 m in the
Katsurahama aquarium. In this case, we obtain dmax =1.34 m at θmax = 16.9◦ and d(θ = 0) = 1.24 m. Since
the air resistance in principle shortens d as we discuss inSec. IV, this value can be regarded as an upper bound fordmax . Therefore, we found that penguin keepers should
keep the distance being longer than 1.34 m from penguins
trying to eject faeces in the Katsurahama aquarium.

y(t) = h +mg − k t(e m − 1),
k

k
mv0y
(1 − e− m t ),
k

k
m(
mg )
mg
v0y +
(1 − e− m t ) −
t.
k
k
k

(4.3)
(4.4)

(4.5)

(4.6)

We note that by expanding them with respect to k, one
can reproduce the results in the absence of the resistance.
Since vx (t) exponentially decreases due to the nonzero k,
one can confirm that dm max with k = 0 gives an upperbound for the distance. We note that in this case thegrounding time is given by( () ( kv
))
0y
k2 h
m
kv0y
− 1+ mg+m2g
tg = W0 − 1 +
e
k
mg
m
hk
v0y
+
+
+
,
(4.7)
k
mg
g

4
where W0 (x) is the main branch of the Lambert W function. The distance can be obtained by d = x(tg ).
We also note that the resistance given by k|v|v is more
realistic in the air [7]. In such a case, an approximatedsolution for a low-angle trajectory has also been obtainedby using the Lambert W function [8, 9]. We note thatan analytic solution in this case can be obtained by thehomotopy analysis method [10]. In addition, the viscosityof faeces itself may shorten the flying distance due to theenergy dissipation originating from the internal friction.
More sophisticated treatments are required to take suchan eﬀect into account. However, our estimation for theupper bound on dmax is robust against these resistanceeﬀects.
V. BERNOULLI’S THEOREM AND
ABDOMINAL PRESSURE OF PENGUINS

Bernoulli’s theorem for non-viscous faeces liquids is
given by
1 2
(5.1)
ρv + ρgz + P = const.
2
where z is the initial height of liquids in a penguin asshown in Fig. 2. While the faeceses are immobile under the abdominal pressure Pa initially, they are released
with v0 in the air. Such an assumption gives
1 2
ρv + P0 ,(5.2)
2 0
where P0 is the atmospheric pressure. At this stage, we
do not consider the viscosity correction. Thus we obtainρgz + Pa + P0 =1 2
ρv − ρgz.
(5.3)
2 0
The unknown parameter z can be estimated from
Pa =

V
z=
,
(5.4)
πR2
where V is the fluid volume after the ejection. R is theradius of stomach. For simplicity, we use R = 0.1 m.
Here, one may notice that the system is quite similar
to the so-called tank orifice with a vena contracta [12,13]. In such a case, the cross-section area of non-viscous
flow after the ejection is approximately given by Cπr2 ,where C ≃ 0.611 is a typical value of the coeﬃcient ofcontraction [12]. In this regard, we obtain
V = Cπr2 ℓ∫ tg √
2
2 + (v2
dt v0x= Cπr
0y − gt)
0
[
√
Cπr2
2
v0y v0 − (v0y − gtg ) (gtg − v0y )2 + v0x
=
2g
(√
)]
2 − gt + v(gtg − v0y )2 + v0x
g0y2− v0x ln,v0 + v0y(5.5)

where ℓ is the total path of the faeces. In particular, inthe case of θ = 0 (namely, v0y = 0 and v0x = v0 ), we
obtain[√Cπr2
V =gtg g 2 t2g + v022g√g 2 t2g + v02 − gtg .
−v02 ln (5.6)
v0
By substituting Eq. (5.5) to
Pa =

1 2
ρg
ρv −
V,
2 0 πR2(5.7)

we can obtain Pa ≃ 2.3 kPa at θ = 0 and h = 0.2 m
following the previous work [2]. It is smaller than theestimated pressure 4.6 kPa in Ref. [2] where the initialpressure for the acceleration was obtained by Pa = ρv02 .
However, Pa are small compared to the viscosity eﬀect
which is addressed in Sec. VI.

VI.

HAGEN-POISEUILLE EQUATION AND
VISCOSITY EFFECT

The Hagen-Poiseuille equation gives a relation between
an additional pressure and fluid flow rate Q for laminarflow though a cylindrical pipe [11] as
Q=

πr4 α
,
8η

(6.1)

where α = ∇Pb and Q can approximately be given byα ≃ Pℓb and Q ≃ tVg , respectively. Therefore, we obtain
Pb =

8V ℓη
,
tg πr4(6.2)

which is consistent with Ref. [2]. Furthermore, we estimate the additional pressure contributions for the rectal
flow asPc =8V zη
4V v0 η
≃
,
4
td πR
πR4

(6.3)

where td is the flowing time inside the intestine. We haveestimated it as td ≃ 2z
v0 ≃ 0.45 ms. Even the maximum
duration tg + td ≃ 0.20 s is quite short compared tothe urination time 8.2(M/kg)0.13 s obtained in Ref. [3](where M is the mass of an animal). We note that atypical value of M for a Hunboldt penguin is about 4
kg. In this way, the total rectal pressure is given by
Pt = Pa + Pb + Pc .
Figure 5 shows the viscosity dependence of Pt withr = 0.004 m. For comparison, we plot the previous result
in Ref. [2] and our result of Pt without C. While theprevious work gives Pt = 8.56 ∼ 20.6 kPa for η = 0.02 ∼0.08 Pa·s (noting that η may change due to the physical

5

106

legs), the flying distance can be estimated as d = 3.13 m,
which greatly excesses the typical height of a human being (1.7 m) we mentioned in the introduction. He/sheshould not use usual rest rooms. Although we assume an
ideal situation to obtain the numerical value, one can easily understand the incredible power of penguin’s rectumin this way.

Pt [Pa]

105

P0

104
103

r = 0.004 m

102
101
100
0.001 0.01

This work (with C)
This work (w/o C)
Ref. [2]0.1

1

η [Pa･s]

10

VII.

100

FIG. 5: Calculated penguin’s rectal pressures Pt with andwithout the coeﬃcient of the contraction C, as functions of
the viscosity η. The dash-dotted curve represents the resultof Ref. [2]. For comparison, we also show P0 .

condition of penguins), our results show Pt = 8.75 ∼ 28.2
kPa, which is up to 1.4 times of the previous work. Thisis because we consider a longer curved path ℓ of flying
faeces to estimate V whereas only the horizontal distance
was taken account in Ref. [2]. We note that if we neglect
the contraction as C = 1, our result gives larger values
Pt = 12.9 ∼ 44.7 kPa since V becomes larger for givenr. These results are also larger than the bladder pressure
Pbladder = 5.2(M/kg)−0.01 kPa in Ref [3]. This diﬀerence
may originate from the physical configurations. Whereas
we consider the projectile ejection of faeces, in Ref. [3] avertical motion of urine was examined. In addition, since
penguins possess the cloacas which combine the digestive
and urogenital tracts [14], the visocity and density of
penguin’s faeces (actually mixing with urine) should beinhomogeneous in the current case. The eﬀect of such an
inhomogeneity is left for future work. We note that while
we use rough approximations for R to calculate Pa andPc , these contributions are found to be small comparedto Pb .
In the end of this section, to realize how strong the estimated penguin’s rectal pressure is, let us demonstrate
how far a liquid-like object blasted oﬀ from a human being with a severe stomachache flies if his/her rectal pressure is as strong as penguin’s one. Here, we assume that
this liquid behaves as a nearly perfect fluid [15] and its
density is same as water ρ = 103 kg/m3 , for simplicity.
√
In this case, we obtain v0 = 7.51 m/s from v0 ≃ 2Pt /ρ
with Pt = 28.2 kPa. If the liquid is horizontally launchedfrom the hip with h = 0.85 m (which is a typical length of[1] P. B. Borboroglu, and P. D. Boersma, Penguins: Natural History and Conservation. (University of Washington

SUMMARY

To summarize, we have discussed a projectile trajectory of penguin’s faeces induced by the strong rectal pressure. We have estimated the upper bound for the maximum flying distance dmax = 1.34 m for the case with atypical initial velocity v0 = 2.0 m/s of faeces, from thehigher place with the height h = 2.0 m. In the presenceof the air resistance which is proportional to the velocity,
we show that the grounding time and the flying distanceof penguin’s faeces can be expressed in terms of the Lambert W function.
Furthermore, based on the hydrodynamic approach,we have revisited penguin’s rectal pressure with an approximation combining the Bernoulli’s theorem and theHagen-Poiseulle equation. The obtained rectal pressureis larger than the previous estimation. The main diﬀerence originates from the treatment of faeces fluid volume
flying in the air. Our results would contribute to further
understanding of ecological properties of penguins.
In this paper, we have used simplified equations of
motion to track the trajectory of faeces. To obtain amore quantitative results, it is necessary to solve hydrodynamic equations of faeces in the air and in the stomach, which are left for future work. More sophisticatedtreatment for the vena contracta would be important. It
is also interesting how complex hydrodynamic behaviors
such as turbulence appear during the ejection of faeces.
We expect that our results would also be a useful example for teaching the classical mechanics in the undergraduate course. One can recognize that fundamental
physics and mathematics we learn in schools describe interesting aspects of various things surrounding our daily
life, as demonstrated by T. Terada [16].

Acknowledgments

The authors thank K. Iida, T. Hatsuda, A. Nakamura,and Y. Kondo for useful discussion and the Katsurahama
aquarium for the supports and giving us an opportunity
to observe Humboldt penguins.

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