Question 69 - Problems - Chapter 1

Problem:

The diameter of the trunk of a large tree is roughly proportional to the $\frac{3}{2}$ power of the length of the tree.

The highest Sequoia in the Sequoia National Park in California has a height of 81m and a diameter of 7.6m. his mass is 6100 metric-ton.

In Nevada, a petrified 90 m sequoia was found.

Question:

Approximate the diameter to its base and its weight when it was still alive

Solution:

The referential tree: $\textrm{ratio}=\sqrt{{\textrm{length}}^3 }$

$\frac{\textrm{diameter}}{\textrm{ratio}}=\textrm{constant}\;\textrm{factor}$

import numpy as np
length = 81;
diameter = 7.6;
weight = 6.1e6;
ratio = np.sqrt(length**3);
factor = diameter / ratio
print(f'The ratio is {factor:.3}')

The ratio is 0.0104

length2 = 90;
ratio2 = np.sqrt(length2**3);
diameter2 = factor * ratio2;
print(f'Diameter petrified tree = {diameter2:.2} m.')

Diameter petrified tree = 8.9 m.

The density $\rho$ of the tree $=\frac{\textrm{weight}}{\textrm{height}*{\left(\frac{\textrm{diameter}}{2}\right)}^2 *\pi \;}\;\textrm{in}\;\frac{\textrm{kg}}{m^2 }$

rho = ((weight)/(length * (diameter / 2)**2 * np.pi))
print(f'The density Rho = {rho:.2g} kg/m3.')

The density Rho = 1.7e+03 kg/m3.

With the same specific weight, the petrified tree would have weighed $\textrm{height}*{\left(\frac{\textrm{diameter}}{2}\right)}^2 *\pi *\rho \;\textrm{in}\;\textrm{kg}$

weight2 = length2 * (diameter2/2)**2 * np.pi * rho;
print(f'The mass of the tree would have been {weight2/1000:.2e} metric-ton.')

The mass of the tree would have been 9.30e+03 metric-ton.