This is a fun, “physics-of-fiction” style calculation published in ...
The downward **bending torque** about the neck is:
\[
\tau(n) =...
The failure mode from a long nose is almost certainly bending.
A...
There's a typo in the paper this expression is a torque, not a forc...
#### Center of mass calculation (made explicit)
They use the sta...
Journal of Interdisciplinary Science Topics
How many lies could Pinocchio tell before it became lethal?
Steffan Llewellyn
The Centre for Interdisciplinary science, University of Leicester
25/03/2014
Abstract:
This paper investigates how many lies Pinocchio could continuously tell before it would become fatal,
treating the head and neck forces as a basic lever system with the exponential growth of the nose. This
paper concludes that Pinocchio could only sustain 13 lies in a row before the maximum upward force his
neck could exert cannot sustain his head and nose. The head’s overall centre of mass shifts over 85
metres after 13 lies, and the overall length of the nose is 208 metres.
Pinocchio’s Nose
Pinocchio is the fable of a wooden puppet, carved
by Geppetto, who dreams of becoming a real boy
[1]. Pinocchio was portrayed as a character prone to
lying, which is manifested physically through the
ability to grow his nose when he tells a lie. One
issue of growing his nose would be the shift of
Pinocchio’s centre of mass within his head, causing
strain on his neck, which helps stabilise his head’s
position with upwards force. If this continued, then
his neck could not support his head, potentially
decapitating the puppet. Outlined here is the
minimum lie count Pinocchio could continuously
expel. Where Pinocchio manages to form new is not
addressed in this paper.
Maximum Force Pinocchio’s Neck Can Exert
The assumption is simplified by allowing the force
exerted upwards through the neck to be positioned
at the back of the head. The head is treated as a
sphere, and the nose as a cylinder, as shown in
Figure 1.
The type of wood Pinocchio is carved from is
disputed, but for this paper, it is concluded that
Pinocchio is made from Oak, with a density of
. Pinocchio’s neck will brake if its
compression strength threshold is overcome by the
weight of his head. The compression strength of
oak is 1150Psi
[2], and the
circumference of the average human neck is 0.4m
[3].
The maximum force Pinocchio’s neck can
sustain is:
Centre of Mass, and Force Exerted
Neck muscles are required to balance the weight
exerted by the skull. Usually, the weight of the nose
can be considered negligible. In Pinocchio’s case, as
the nose increases, it will have a significant impact
on the centre of mass and weight of his head. The
mass of the head is unchanged:
Figure 1:
Illustrates the
lever system of
Pinocchio’s head
and neck, with
opposite forces.
How many lies could Pinocchio tell before it became lethal?, March 25
th
2014
The nose initially can be considered negligible.
However, it becomes more significant as it
increases in size. For this model, the nose has an
initial length of 1 inch (2.54cm), a diameter of 2cm,
and its centre of mass positioned in the middle:
Such a small mass does not affect the Centre of
Mass, nor does it apply great force on the neck.
Since this is a lever system, the weight applied on
the neck also depends on the distance from which
that force is being applied. The initial force
Pinocchio’s head exerts is:
This force is miniscule in comparison to the strength
of Pinocchio’s oak neck, thus, there is no great
pressure applied on Pinocchio.
Growth of the nose
During Disney’s Pinocchio, the puppet’s lies cause
extreme growth of his nose
[4]. Thus, it is not
absurd to model one lie causing an increase in the
length of the nose by a factor of two (i.e. the nose
doubles in length for every lie). Assuming the nose
also remains the same density, as mass is in
proportion with the volume, the length of the nose
will eventually exert significant force to the head-
neck lever system. The nose is determined
unbreakable, as it built upon the foundation of lies.
Figure 2 demonstrates the force exerted by the
head on the neck. Once Pinocchio’s nose grows to
the point at which it exceeds just over 140 metres,
the force exerted downwards would cause the
supporting neck to snap. Due to the exponential
nature of his nose growth, Pinocchio cannot tell 13
consecutive lies, as, at this point, his nose would
reach a length of 208m, and his centre of mass
would have shifted by roughly 85m. Table 1
outlined in Supplementary material demonstrates
calculations made for this approximation.
Figure 2a: Graph demonstrating how the growth of
Pinocchio’s nose would alter the mass to such an extent,
it would overcome the forces of his neck.
Figure 2b: Graph demonstrating the number of lies
Pinocchio could tell vs the total downward force exerted.
It demonstrated an exponential relationship.
Conclusion
Pinocchio’s anatomy has extraordinary properties.
It is outstanding that his nose seemingly deposits
mass from nowhere. Nevertheless, this unique
ability can be of great concern for the puppet, and
lengthy, extensive lies are advised against, for the
health and well-being of Pinocchio.
0
5000
10000
15000
20000
25000
0 2 4 6 8 10 12 14
Force exerted Downwards (N)
Number of lies
Number of Lies Pinocchio can tell vs. the Force exerted
Pinocchio
How many lies could Pinocchio tell before it became lethal?, March 25
th
2014
References
[1] DisneyWiki, “Pinocchio,” 2013. [Online]. Available: http://disney.wikia.com/wiki/Pinocchio. [Accessed
March 2014].
[2] Engineering Toolbox, “Wood Beams Strength,” 2014. [Online]. Available:
http://www.engineeringtoolbox.com/wood-beams-strength-d_1480.html. [Accessed March 2014].
[3] Anonymous, “What is the average circumference of the human neck?,” December 2013. [Online].
Available: http://www.chacha.com/question/what-is-the-average-circumference-of-a-human-neck.
[Accessed March 2014].
[4] W. Disney, “Pinocchio 70th Anniversary Platinum Edition - Pinocchios Lies,” Disney, 2009. [Online].
Available: https://www.youtube.com/watch?v=hJ3lxzuI_sc. [Accessed March 2014].
There's a typo in the paper this expression is a torque, not a force. It has units of N.m.
#### Center of mass calculation (made explicit)
They use the standard 1D center-of-mass formula:
\[
x_{\text{COM}} = \frac{M_H x_H + M_N x_N}{M_H + M_N}
\]
What’s implicit (but crucial):
- \(x_H\): distance from neck pivot to head COM
- \(x_N\): distance from neck pivot to nose COM
For a nose extending straight forward:
\[
x_N \approx r_{\text{head}} + \frac{L}{2}
\]
This linear dependence on \(L\) is why the COM displacement explodes as the nose grows.
This is a fun, “physics-of-fiction” style calculation published in the Journal of Interdisciplinary Science Topics (JIST) - an undergraduate, student-run journal at the University of Leicester, designed to teach scientific publishing and peer review.
Its goal is educational rather than prestige.
The author of the paper was a student at the University of Leicester.
The failure mode from a long nose is almost certainly bending.
A more mechanically consistent model would compare the bending moment at the neck to a bending strength (modulus of rupture) and use a section modulus.
The downward **bending torque** about the neck is:
\[
\tau(n) = W_H x_H + M_N(n) g x_N(n)
\]
The neck must supply an upward force \(F_{\text{neck}}\) at a small posterior lever arm \(d\).
The paper effectively **absorbs \(d\)** into the force definition, so numerically:
\[
F_{\text{neck}}(n) \propto W_H x_H + M_N(n) g x_N(n)
\]
This is why their “force” values are actually **torque-scaled forces**.
For large \(n\), the nose dominates:
\[
F_{\text{neck}}(n)
\approx M_N(n) g x_N(n)
\]
Substitute growth laws:
\[
F_{\text{neck}}(n)
\approx (M_{N0}2^n) g \left(\frac{L_0}{2}2^n\right)
\]
\[
F_{\text{neck}}(n)
\approx \left(\frac{M_{N0} g L_0}{2}\right) 2^{2n}
\]
This is the **crucial result**: Neck load grows like \(2^{2n} = 4^n\)
That’s why failure happens abruptly.
If we set required force equal to maximum neck strength:
\[
\left(\frac{M_{N0} g L_0}{2}\right) 2^{2n}
= 1.0\times10^5
\]
Insert numbers:
\[
\frac{(0.0060)(9.81)(0.0254)}{2}
\approx 7.5\times10^{-4}
\]
So:
\[
7.5\times10^{-4} \cdot 2^{2n}
= 1.0\times10^5
\]
\[
2^{2n} \approx 1.3\times10^8
\]
Take base-2 logarithms:
\[
2n = \log_2(1.3\times10^8) \approx 27
\]
\[
n \approx 13.5
\]