# Wacky tech

This is a :idea-list: page, not one regulary browsed. Technologies Hoh wants to write about:

Articles have their origin here and move out when they are completed.

• Nuclear thermal rockets | high isp medium thrust propulsion | mentioned but not explained
• More than just hydrogen | nuclear engines are not picky | perspective
• Beamed thermal rockets | high isp low thrust propulsion | Wacky
• Solar thermal rockets | high isp low thrust propulsion | Wacky
• Rotating tethers | momentum management | Wacky
• Orbiting vertical tethers | momentum management | mentioned but not explained
• Static vertical tethers | momentum management | mentioned but not explained
• Usage of ion thrusters in the Earth-Moon system | mentioned but not explained
• "The numbers" | debatable data | raw idea
• C3=0 manifolds | pure two-body zero energy transfers | Magic
• Mercury is good for you| Density impulse | raw idea
• Neutron poison balancing in a nuclear thermal engine

• How the nuclear shuttles could be hooked up to a power grid, especially during initial construction of Cernan's Promise. Should the generators be separate and be hooked up to the reactor while the shuttle is parked?
• Refrigeration - If you cut a void into the ground, painted the inside of it black, put a container on posts inside it covered with aluminum foil and kept it in shade, it would stay pretty cold. If you could create a thick stone slab for the floor, it would have considerable thermal inertia. Or put a container inside that container, and fill the void between with gravel. Or cut channels in the ground - works a lot better where it is all rock - then carefully creat undercuts between the channels until you essentially again have a box on posts surrounded by a void, and then partly hollow it out. At any rate, many levels of refrigeration are needed, from pantries to cryogenic hydrogen storage, so it is interesting to think about how much refrigeration could be done passively simply by keeping a sizable mass in shade during the day and uncovering it at night so its heat radiates away. How to lay out those areas so they are accessible from within structures and are most efficient is also an interesting question.

## Nuclear thermal rockets

Wackiness: Existing tech ~1970

[[The reader must be introduced to the concept of delta-v before this]] As we know, lighter gasses are preferable in rocket exhaust as they utilize the supplied energy better in terms of momentum gained. An engine designer would thus like her rocket exhaust to consist of pure hydrogen. But that is not possible, no chemical reaction produces large amounts of hydrogen while still releasing usable amounts of chemical energy. At best, a combination like LH2/LOX would produce a mix of hydrogen and water. The fuel in a chemical rocket engine really has two purposes: to act as the propellant and to supply energy. However, from a generalized perspective, those two do not have to be the same.

A nuclear engine does that. It uses pure hydrogen as propellant and gets the energy from a reactor. "Wait a minute" you say. "Why do we still have to use hydrogen then? Was not the reason that we had a limited amount of energy available?" Yes, that is true. The reactor can give us any amount of power we desire. Hydrogen on the other hand has several drawbacks like a very low density and needs to be stored at cryogenic temperatures. But nuclear engines have another limitation: The fuel rods have a melting point (typically around 3000K). Naturally, we would like the temperature in the engine to stay below that as nuclear meltdowns are messy. We can then only heat the propellant so much. Here we have an important gas law: At the same temperature, two gas molecules have the same kinetic energy. So, a tiny hydrogen molecule at 3000K would have the same energye as for example a water molecule at 3000K. We know that velocity is proportional to the square root of energy divided by mass, so hydrogen molecules would go faster, e.g. carry more momentum. Solid core nuclear engines end up at a specific impulse in the range 850-1000s, more than twice than the best chemical engines. (Your reactor core does not have to stay solid, there are some wacky concepts where it is not.)

Twice the efficiency is just the instantanous benefit. It keeps on increasing with higher delta-v requirements. Propellant needed to accelerate 1000kg of mass to various velocities:

Delta-v Isp 450s Isp 900s Ratio
100 m/s 22.9kg 11.4kg 2.01
1000 m/s 254kg 120kg 2.12
5000 m/s 2105kg 762kg 2.76
10000 m/s 8641kg 2105kg 4.11
20000 m/s 91950kg 8640kg 10.64

## More than just hydrogen

We have already established the advantages of using nuclear rocket engines. As hydrogen is the perfect propellan, why should we use anything else? Well, it turns out hydrogen has some drawbacks.

Firstly, it has to be stored cryogenically at a freezing −252.87 °C. That is not a huge problem if the hydrogen is used within minutes after launch like in a conventional upper stage. But what if it has to be stored for days? Months? Or even years? The refridgeration systems tend to be complicated, and the tanks require insulation. More added mass is a bad thing.

Secondly, hydrogen has a very low density, only $71 \frac{kg}{m^3}$ or 7% of water if you like. That means huge tanks. But wait, if we have huge tanks we can fill them with more propellant if we use something with a higher density than hydrogen instead. After all, propellant is the cheapest part of a rocket. A worked example of why that can be a good thing:

Consider a rocket with a dry mass of 1000 kg with a 1 m³ tank for propellant. We can then use the rocket equation to find the $\Delta v$ capabilities with both hydrogen and water. First, hydrogen: $$\Delta v = \ln \left(\frac{1071 kg}{1000 kg}\right) \cdot 9000 \frac{m}{s} = 617 \frac{m}{s}$$ Then water: $$\Delta v = \ln \left(\frac{2000 kg}{1000 kg}\right) \cdot 4042\frac{m}{s} = 2802 \frac{m}{s}$$ What? Water is better than hydrogen? The reason is that the second setup has a lot more propellant than the first one. That does not matter if we are launching from the ground as mass if free there. But mass is obviously not free in for instace LEO, in fact it is very expensive. That means a high density is a good thing at the beginning of the mission, but after a while only exhaust velocity is important. Exhaust velocity is of course also important at the beginning as well, so we are looking at a quantity, density impulse. Density impulse is propellant density multiplied by exhaust velocity. As the mission progresses, density impulse becomes less important and only exhaust velocity matters.

There is a tradeoff point in the mission where we should switch from water to hydrogen, or even multiple such points for water-ammonia, ammonia-methane etc. The fascinating calculations about density impulse tradeoffs are discussed in "Mercury is good for you", but as this article is not yet written I can only provide a link to a Stack Echange question in the absence of it: Is-it-possible-to-use-zinc-powder-as-a-ntr-propellant?

### Nuclear propellant reference

Exhaust velocity may vary slightly from the listed value

Propellant: $V_e$: $\rho$:
Hydrogen $H_2$ $9000 \frac{m}{s}$ $70.8 \frac{kg}{m^3}$
Methane $CH_4$ $6318 \frac{m}{s}$ $423 \frac{kg}{m^3}$
Ammonia $NH_3$ $5101 \frac{m}{s}$ $682 \frac{kg}{m^3}$
Water $H_2O$ $4042 \frac{m}{s}$ $1000 \frac{kg}{m^3}$

## Mercury is good for you

The balance between thrust, exhauste velocity and density impulse is an interesting one. Here I am going to present two of the optimization cases.

Optimization 1: Delta-v maximizing

Idea: Given a tank volume, figure out how it can be filled with various propellants to achieve maximum $\Delta v$.
Suppose that some time into the flight the relative importance of density impulse and exhaust velocity is such that it does not matter if we at this point use propellant 1 or 2. We say that happened after a velocity change of $\Delta v$. At this point both propellants must use the volume of the tank equally efficient. Suppose we have a quantity, $m_1$ of propellant 1. This can supply a momentum of $m_1v_1$ and takes up a volume of $\frac{m_1}{\rho_1}$. But it also takes some propellant to accelerate $m_1$ up to $\Delta v_x$, specifically $m_1\left(e^{\frac{\Delta v}{v_1}}-1\right)$. Of course this has a volume of $\frac{m_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1}$. We can add those two masses together and end up with: $$\frac{m_1e^{\frac{\Delta v}{v_1}}}{\rho_1}$$ Next, what if we used propellant 2 instead? Then we would only need a mass of $\frac{m_1v_1}{v_2}$ using a volume of $\frac{m_1v_1}{\rho_2v_2}$. This mass also needs to be accelerated up to $\Delta v$. But as this is supposed to be the cuttof point we are still using propellant 1 for that. That is a mass of $\frac{m_1v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{v_2}$ corresponding to a volume of $\frac{m_1v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1v_2}$. Adding also these two together we have: $$\frac{m_1v_1}{\rho_2v_2}+\frac{m_1v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1v_2}$$ From the definition of our cutoff point the two volumes should be equal. $$\frac{m_1e^{\frac{\Delta v}{v_1}}}{\rho_1}=\frac{m_1v_1}{\rho_2v_2}+\frac{m_1v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1v_2}$$ Furthermore, it is clearly independent of $m_1$ $$\frac{e^{\frac{\Delta v}{v_1}}}{\rho_1}=\frac{v_1}{\rho_2v_2}+\frac{v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1v_2}$$ We are solving for $\Delta v$ so we want everything containing that on the same side $$\frac{e^{\frac{\Delta v}{v_1}}}{\rho_1}-\frac{v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)}{\rho_1v_2}=\frac{v_1}{\rho_2v_2}$$ And then get rid of the denominators $$v_2e^{\frac{\Delta v}{v_1}}-v_1\left(e^{\frac{\Delta v}{v_1}}-1\right)=\frac{\rho_1v_1}{\rho_2}$$ Rearranging a little we are left with $$v_2e^{\frac{\Delta v}{v_1}}-v_1e^{\frac{\Delta v}{v_1}}=\frac{\rho_1v_1}{\rho_2}-v_1$$ We can factorize that to obtain $$\left(v_2-v_1\right)e^{\frac{\Delta v}{v_1}}=\frac{v_1\left(\rho_1-\rho_2\right)}{\rho_2}$$ And that allows us to isolate $\Delta v$ further $$e^{\frac{\Delta v}{v_1}}=\frac{v_1\left(\rho_1-\rho_2\right)}{\rho_2\left(v_2-v_1\right)}$$ This can be used to get our desired value down to ground level $$\frac{\Delta v}{v_1}=ln\left(\frac{v_1\left(\rho_1-\rho_2\right)}{\rho_2\left(v_2-v_1\right)}\right)$$ And finally $$\Delta v=v_1ln\left(\frac{v_1\left(\rho_1-\rho_2\right)}{\rho_2\left(v_2-v_1\right)}\right)$$ We can calculate a few cutoff points using this equation:

Propellants Cutoff point
Water - Ammonia $2330 \frac{m}{s}$
Water - Methane $3576 \frac{m}{s}$
Water - Hydrogen $9580 \frac{m}{s}$
Ammonia - Methane $4808 \frac{m}{s}$
Ammonia - Hydrogen $12370 \frac{m}{s}$
Methane - Hydrogen $15550 \frac{m}{s}$
You might be wondering if it perhaps is beneficial to use a mix of two propellant in the transition between them. The answer is no. Here are some more exotic ones using the few elements with a reasonable melting and boiling point combinded with a very high density:
Propellants Cutoff point
Mercury - Zinc
Mercury - Water
Zinc - Water
Of course we can only use this approach to get the cutoff point between two different propellants. To get optimal solutions for more than that we need another equation. To get the cutoff point between the propellant 1 and propellant 3, we can apply a similar logic as before to set up the equation. $$\frac{e^{\frac{\Delta v_2-\Delta v_1}{v_2}}}{\rho_2} + \frac{e^{\frac{\Delta v_2-\Delta v_1}{v_2}}\left(e^{\frac{\Delta v_1}{v_1}}-1\right)}{\rho_1}=\frac{v_2}{\rho_3v_3}+\frac{v_2\left(e^{\frac{\Delta v_2-\Delta v_1}{v_2}}-1\right)}{\rho_2v_3}+\frac{v_2e^{\frac{\Delta v_2-\Delta v_1}{v_2}}\left(e^{\frac{\Delta v_1}{v_1}}-1\right)}{\rho_1v_3}$$ We can solve this: $$\Delta v_2 = \Delta v_1 + v_2ln\left(\frac{v_1\rho_1\left(\rho_2-\rho_3\right)}{\rho_3\left(v_3-v_2\right)\left(\rho_1+\rho_2\left(e^{\frac{\Delta v_1}{v_1}}-1\right)\right)}\right)$$
Propellants Secondary cutoff point
Water/Ammonia - methane
Water/Ammonia - Hydrogen
Water/Methane - Hydrogen
Ammonia/Methane - Hydrogen

Optimization 2: Minimal thrust maximizing

Idea: Given a required velocity change, figure out how we can fill a tank with various propellants to get an optimal thrust-to-weight ratio.
The acceleration of a rocket over time looks something like this:
Notice of the acceleration in the beginning is the limiting factor. But what if I told you it is possible to get better thrust in the beginning using the same engine but another propellant? The thrust obtained from a propellant in a thermal rocket engines is inversely proportional to its exhaust velocity. Why? Well, the reactor has a power $P$ the mass flow is then proportional to $\frac{P}{v^2}$. Thrust is proportional to mass flow times exhaust velocity, so there you have it, $Thrust \propto \frac{1}{v}$. So we want to use one propellant in the beginning with a low exhaust velocity and one at the end with a high exhaust velocity. In this case however, it is worth it to mix propellants to get an ideal exhaust belocity at all points in time to get a flat thrust curve. The equations are complicated

[[Sadly I have to wait a little before I can continue this part as I have misplaced my notes]]

## Neutron poisons [stub]

Nuclear rocket engines are high impulse and high efficiency at the same time. The impulse part is however quirky. For a chemical rocket engine opperation is very close to being a toggle: you can turn it on and off at any point. Managing a reactor is a little different than that. After shut down you can most likely not just turn it on again right away. The prosess of reactor shut down and cooling is going to take days.