# Fire in the forest

There are two questions, yet unanswered, that might be helpful for my analysis:

Logarithmic Spiral

Rate of change in degree for archimedean spirals

My problem is this:

A cottage is on fire and in order to avoid the fire to expand to the nearby forest, a trench must be dug. How must the trench be (shape) in order to have the minimum length and area covered?

The cottage we assume to have initial circular area of $D=10\ \text m$. The excavator is always $3\ \text m$ apart from the fire and begins to dig at $r=8\ \text m$. The fire advances at a rate of $1\ \text m/\text h$ radially (or $v_f$). The excavator advances at a rate of $8\ \text m/\text h$ (or $v_e$).

Is this possible?

If not, what must be the velocity of the excavator to achieve the objective?

My approach is to work with an Archimedean spiral.

And once the spiral made one turn, the excavator must dig another Archimedean spiral in order to close the advance of this fire.

I made some work using Excel, working in a step by step form. But I want your help in order to work with specific formulas.

What if instead of working with an Archimedean spiral, we work with a logarithmic spiral?

• How does the fire spread beyond the initial trench when it reaches $(x,y) =(8,0)$ in your linked figure? – Jens Feb 16 '17 at 23:43
• At $(x,y)=(8,0)$ the fire begins to describe another series of circunferences in the upper plane. – cgiovanardi Feb 17 '17 at 1:26

An interesting question! And the answer is, yes, it is possible. Also, I'm pretty sure my solution is the shortest path possible (i.e. I can't think of a shorter way to do it).

Some initial definitions and constants: The cottage is located at $(0,0)$, the excavator is initially located at $(8,0)$, the starting radius of the fire is $r_s = 5$ m, the minimum distance the excavator must keep to the fire is $d_{min} = 3$ m, the radial speed of the fire is $V_f=1$ m/h and the speed of the excavator is $V_e=8$ m/h.

Now, the radius of the fire from the cottage at time $t$ will therefore be $$r_f(t)=r_s+V_f * t$$

The radius the excavator must maintain to the cottage at any time $t$ will thus be $$R(t)=r_f(t) + d_{min} = r_s+d_{min}+V_f * t \tag{1}$$

We now want to find the angle $\theta$ of the excavator. But let us first look at how the excavator should move in a small time interval $dt$:

As shown above, the excavator should move at full speed in a straight line so that it is exactly at the new minimum distance from the fire at time $t+dt$. The angle $\alpha$ at which it needs to move can be found using the Cosine rule: $$cos(\alpha) = \frac{R(t)^2+(V_e \, dt)^2-R(t+dt)^2}{2R(t)V_e \, dt}$$

Inserting the formula for $R(t)$ and letting $dt \to 0$ we find that $$\lim_{dt \to 0} cos(\alpha) = -\frac{V_f}{V_e}$$

We see that the angle $\alpha$ is a constant. This rings a bell. We remember that the logarithmic spiral has a constant angle, call it $\phi$, between the tangent to a point and the radial line to that point. The angle $\phi$ is indicated in the figure above and we see that $\phi = \pi - \alpha$. So if $\alpha$ is a constant, then $\phi$ must be a constant and it therefore seems we are dealing with a logarithmic spiral. Let us assume this is the case and see what happens.

A logarithmic spiral is defined by the equation $$R=a \, e^{b \theta}$$

We now need to find $a$ and $b$. From the previous link we know that $tan(\phi) = \frac{1}{b}$, which means $tan(\pi-\phi) = tan(\alpha) = -\frac{1}{b}$. We can express $tan(\alpha)$ in terms of $cos(\alpha)$ in the following way: $$tan(\alpha) = \frac{\sqrt{1-cos(\alpha)^2}}{cos(\alpha)}$$

which means $$b= \frac{V_f}{\sqrt{V_e^2-V_f^2}}$$

We now know $b$. To find $a$ we use the boundary condition that at $\theta = 0$ we must have $R= r_s+ d_{min}$. This gives $a=r_s+ d_{min}$ and we have the following equation for the movement of the excavator: $$R=(r_s+ d_{min}) \, e^{\frac{V_f}{\sqrt{V_e^2-V_f^2}}\theta}\tag{2}$$

Inserting the values of the constants into equations $2$, we can plot the path:

As expected we get a spiral. Each marker on the spiral is the excavator's position at intervals of $0.1$ hour. We can check that at any time $t$ the excavator is at the minimum distance required, $R(t)$. We can also check that the straight line distance covered by the excavator, between one marker and the next ($0.799565$ at the second marker), is almost, but not quite, $V_e \, dt$, which is as expected when we know that the actual path between one marker and the next is not a straight line, but a curve. Our assumption that the curve is a logarithmic spiral seems vindicated.

Anyway, on with the problem. The spiral completes its first loop when $\theta(t) = 2\pi$. Using equation $2$ we can calculate that this occurs at $R(t) \approx 17.6555$ meters and using equation $1$ we find this occurs at $t \approx 9.6555$ hours. We know the fire reached the excavator's starting point after $3$ hours. That point then became a new center of fire, spreading outward and upward at its fixed speed. When the excavator completes the first loop, the fire has thus been spreading from this point for $t=9.6555-3 = 6.6555$ hours, and will have a starting radius from this point of $r_{s2} = 6.6555$ meters. We thus have a situation equivalent to the original starting situation, just with a shifted center of fire, $x=8$ instead of $x=0$, a different starting radius of fire, $r_{s2}$ instead of $r_s$, and a different starting position of the excavator, $x = 17.6555$ instead of $x=8$. Using our equations (with these modifications) we can plot the last loop:

The second loop intersects the first loop at $(x,y) \approx (-0.3462, 9.7881)$ where the radius of the second loop is $\approx 12.8634$. The arc length of a logarithmic spiral is given by $$s = \frac{\sqrt{1+b^2}}{b}(r_2 - r_1)$$ where $r_1$ and $r_2$ are the starting radius and the ending radius of the arc. The total length of the excavator's path is therefore $$L = \frac{V_e}{V_f}[(17.6555-8)+(12.8634-9.6555)]$$ or $$L \approx 102.9072$$

• Thank you very much for your lengthy and detailed answer! – cgiovanardi Feb 19 '17 at 23:00
• My pleasure. :-) – Jens Feb 19 '17 at 23:12