### #3 Scaling law of locomotion-II

/*-----------Divider-----------*/
Today we are going to talk about the problem we left in the previous episode: how T. A. McMahon explained the scaling laws tride length M^(0.38), stride frequency M^(-0.14), and running speed M^(0.24). The key is the assumption of “elastic similarity.”

The Chinese version of this episode: 指數律_動物走路和跑步_下： https://goo.gl/OUJBlb
/*-----------Divider-----------*/
In the previous episode we suggested using the concept of inverted pendulum to derive the scaling law of stride length, stride frequency and running speed. Theoretically, you may have arrived at a result looks like stride length M^(1/3), stride frequency M^(-1/6), and running speed M^(1/6). These results came from the assumption of “geometric similarity,” which postulated that larger animals are the  isometric scaling of smaller animals.

If that were true, an elephant could have snapped its leg merely by its own weight. An axial load could cause the buckling of a beam. Based on an equation discovered by Euler in 1757, we could calculate the maximum axial load (, or the so called critical load) of a beam. We could also calculate how much the beam diameter has to adjust to stay within the critical load as the length of the beam doubles. Because during this scaling process, beams of various sizes share the same risk of buckling, such allometric scaling is termed as “elastic similarity.” T. A. McMahon discovered that the length of the beam should be proportional to its diameter to the power 2/3. The physics detail of the above would be discussed later.

Actually, McMahon compared 3 different hypothesis and found that the results based on elastic similarity matched the observed power best. He reproduced the scaling laws stride length M^(3/8), stride frequency M^(-1/8), and running speed M^(1/4), which matched the experimental ones perfectly. We encourage you to derive the scaling laws of body surface area, BMR and heart rate yourselves. What is your predicted scaling power? The answers could be found in the suggested reading.

/*-----And now let’s delve into the physics details-----*/
We have to briefly introduce some concepts of differential geometry and material mechanics. Let’s now look at Figure 1.

Figure 1. A parametric curve in 3D space.

Consider a curve in the 3D space. Each point along the curve could be described by a dimensionless parameter u as

Now consider a tiny segment with length Δs, as Δs approaches 0, we could write down that

We therefore arrive at a conclusion that

The tangential direction of the curve is defined as

And its magnitude is therefore

And we could also defined the unit tangential vector

The curve is not straight because there are changes in its tangent vectors. For each part of the curve, we could therefore draw its osculating circle. We try to find the changes in unit tangent vectors from position s to s+ds:

We then defined curvature κ to be the reciprocal of the radius of the osculating circle, i. e. κ=1/R. We could then derive that

And for a beam with lateral displacement h = h(x), we could easily know that

Figure 2. A beam curved into an arc.

Now let’s look at Figure 2 and consider a beam. Assumed that the beam is curved into an arc due to the axial load. From the figure above, it is obvious that some parts of the beam are compressed while other parts are extended. We could find an neutral axis, i. e. the axis along which the beam is neither compressed nor extended, somewhere in the beam. We then define origin to be the intersection point between the neutral axis and the lateral border at the left side. We could find the amount of compression or extension of the beam at z to be

We could then find strain, or the length change experienced by the material with unit length:
And from Hooke’s law, the stress, or the internal force per unit area experienced by the beam at z is

Where E is the Young’s modulus of the material. The larger the Young’s modulus, the harder the material is. Besides, we define moment (M) as the internal torque experienced by the cross-section of the material, mathematically the cross-product of its coordinate vector and internal force:

To find the total moment of a cross-section, it requires some integration:
In which I is called the area moment of inertia. Noticed that since κ=1/R, The moment M could also be expressed as M = EIκ.

Figure 3. A beam with axial load F and lateral displacement profile h(x).
Now let’s look at Figure 3. Consider a beam with axial load F. Assume that its lateral displacement at position x is h(x). From results derived above, we could write down the ODE and its boundary conditions:

The answer is simple, that is h(x)= Csin(nπx/L), n=0,1,2,3,….There is no lateral displacement if n=0, while n=1 means the beam is curved and could no longer support load. The critical load or the so called Euler buckling load could be calculated by inserting h(x)= Csin(πx/L) back to the ODE. And we will get
$-C\frac{\pi^2}{L^2}\sin(\frac{\pi&space;x}{L})=-\frac{F}{EI}C\sin(\frac{\pi&space;x}{L})$
$F&space;=&space;\frac{\pi^2EI}{L^2}$
This is the Euler's formula for the critical load of a beam.

In McMahon’s argument, he assumed F to be the weight of the beam itself. That is,

$F&space;=&space;\rho&space;L&space;(\frac{\pi}{4}D^2)$
And the area moment of inertia of a cylindrical beam reads
Insert all above into the Euler’s formula of critical load, we get
And therefore the length of beam should be proportional to its diameter to the power 2/3:
The coefficient (0.851) still depends on the inner structure of the beam, but the scaling law of 2/3 remains the same. And that's how McMahon reproduce the scaling laws stride length M^(0.38), stride frequency M^(-0.14), and running speed M^(0.24).

/*-----------Divider-----------*/