Hookes Carriage Lamp

I was rereading J.E. Gordons “Structures: or why things don’t fall down”, when I came across this aside:

Hookes carriage lamp, in which, as the candle burnt down, its flame was kept in the centre of the optical system by means of a spring feed, went out of use only in the 1920s.

This device is built in the same way as you might expect; a candle suspended by a spring, all held in a sheath. What surprised me was the simplicity of the system and in the spirit of adventure, I decided to dive down the rabit hole to figure out how it worked.

How does it work?

The whole point of the system (barring a few secondary advantages)¹ was to keep the candle’s flame at a fixed vertical position as the candle burnt. Let’s derive the conditions where this is possible.

Suppose we have a candle of height \(H\), cross sectional area \(A\) and density \(\rho\). The candle is held in place a support of fixed mass, \(m\) which rests on a spring with spring constant \(k\) and an unloaded length of \(L_0\).

The mass of the candle is \(HA\rho\) and hence the force exerted on the spring is:

$$F = mg + HA\rho g$$

By Hooke’s law, the compression of the spring is

$$\begin{aligned} \Delta L &= \frac{F}{k}\\[10px] &= \frac{mg}{k} + H\frac{A \rho g}{k}\\ \end{aligned}$$

Now if we choose a combination of springs and candles such that \(k = A\rho g\), then we have:

$$\Delta L = \frac{mg}{k} + H$$

At this point, with a good mental model, you can get a feel as to why this means that the flame stays at a fixed position. To reinforce our argument, we can rearrange our equation like so:

$$L_0 - \frac{mg}{k} = L_0 - \Delta L + H$$

Now, \(L_0 - \Delta L + H\) equals the height of the candle’s flame,² but it equates an expression in terms of \(L_0\), \(m\), \(g\) and \(k\). These terms are all unaffected as the candle burns. Therefore, even whilst the candle burns and \(H\) decreases, the flame has a fixed position.

Isn’t that neat?

More generally, the only quantities that change in \( L_0 - \Delta L + H\) as the candles burn are \(\Delta L\) and \(H\). Consequently, we only need to show that \(H - \Delta L\) is a constant, which is a useful result to bear in mind. Let’s call this constant \(\Omega\), defined such that \(\Delta L = H + \Omega \).

Non-Hookean springs

Say you had a material such that the force required to compress it is proportional by the cube of the compression.³ Could you design a candle that can balance on a spring made from this material, in a manner such that the flame remains at a constant height?

$$F = \mu (\Delta L)^3$$

Your first answer might be “Why would you even want to do that?". In which case, consider this: Developing a method to answer this question is useful when developing spring-free carriage lamps. Say you wanted to power your new-and-improved carriage lamp with electrostatic repulsion, how would you do that? Can you even reason for certain as to whether an electrostatically driven carriage lamp is possible?

As for why I’ve chosen \(F = \mu (\Delta L)^3\), these values give a remarkably nice answer that has a tangible geometric meaning.

Discussion

The first thing note is that these conditions cannot be met with a “standard” candle. Let’s show this by contradiction. Since \(\Delta L = H + \Omega \), the force on the spring is \(F = \mu (H + \Omega)^3\). But the mass of the candle is \(H A \rho \) and linear with respect to \(H\). It follows that the force on the spring cannot be produced by a “standard” candle, hence \(A\) or \(\rho\) must be a function of \(H\).

To stay slightly closer to reality, lets assume \(\rho\) to be constant and the volume of the candle to be a function of \(H\) such that:

$$V(H) = \int_{0}^{H} \pi R^2(x) \thickspace dx$$

Where \(R(H)\) is the radius of the crossection of the candle at a given height.

It is important to have an understanding for what \(R(H)\) means and how it relates geometrically to the candle; study the below diagram if necessary.

The candle of mass \(\rho \cdot V(H)\) is held by a support of mass \(m\). Combined, they have a weight of:

$$F = mg + \rho g\cdot V(H)$$

For the spring to support this weight, it must produce an equal and opposite force.

$$ \mu (\Delta L)^3 = mg + \rho g\cdot V(H)$$

Take a second to appreciate the similarities between this and our equation for standard springs, \(k \Delta L = mg + H A\rho g\). We’ve effectively replaced \(H A\) with a function that allows us to describe the candles shape in more detail. Before, we arbitrarily picked \(k=A\rho g\) and then showed that \(H - \Delta L\) was constant, but now we have a more concrete method.

$$ \mu (\Delta L)^3 - mg = \rho g \cdot V(H)$$

Substituting values for \(\Delta L\) and \(V(H)\) leaves us with this intimidating expression:

$$ \mu (H + \Omega)^3 - mg = \rho g \cdot \int_{0}^{H} \pi R^2(x) \thickspace dx$$

Differentiate with respect to \(H\) and rearrange

$$ 3 \mu (H + \Omega)^2 = \rho g \cdot \pi R^2(H)$$

$$ \frac{3\mu(H + \Omega)^2}{\pi \rho g} = R^2(H)$$

And there it is:

$$ R(H) = (H + \Omega)\sqrt{\frac{3\mu}{\pi \rho g}}$$

Let’s interpret this geometrically: We have a linear radius function that linearly increases up the candle producing a cone. Of course, since \(H \geqslant 0\), for positive \(\Omega\) the cone is truncated to form a frustum. ⁴

How can we be sure that this is the correct answer?

We can apply this method to a standard spring and try to check for consistent results:

We start by equating the force produced by the spring with the weight of the candle and its support $$k \Delta L = mg + V(H)\rho g$$ Substitute values for \(\Delta L\) and \(V(H)\) $$k (H + \Omega) = mg + \rho g\cdot \int_{0}^{H} \pi R^2(x) \thickspace dx$$ Finally differentiate with respect to \(H\) $$k = \rho g\cdot \pi R^2(H)$$

Note that this equation is identical to our original substitution of \(k = \rho g A\), so both methods yield consistent results.