I have been working on deriving an answer to How should I determine the properties of keels and ether? myself, and I was looking at the lift equation:

$$L = \frac{1}{2}\rho v^2SC_L$$


  • $L$ is Lift Force in $kgms^{-2}$.
  • $\rho$ is fluid density in $kgm^{-3}$
  • $v$ is velocity in $ms^{-1}$
  • $S$ is the projected wing area in $m^2$
  • $C_L$ is the coefficient of lift (dimensionless)

When physicists work out their formulas, it is important that the dimensions of their numbers agree. In the formula above, discounting all the dimensionless numbers, we get:

$$ kgms^{-2} = kgm^{-3} \cdot (ms^{-1})^2 \cdot m^2 $$

which is equivalent to:

$$ kgms^{-2} = kgm^{-3} \cdot m^2s^{-2} \cdot m^2 $$

Summing the powers of the units, we get:

$$ kgms^{-2} = kgm^1s^{-2}$$

And since $m^1 =m$, so far, so good.

However, I mentioned in the linked question and one of its comments that keels work according to their volume not their area. So, if we plug that dimension into the equation with:

  • $S$ is the wing volume in $\color{red}{m^3}$

We get:

$$ kgms^{-2} = kgm^{-3} \cdot (ms^{-1})^2 \cdot \color{red}{m^3}$$

which decomposes to:

$$ kgms^{-2} = kgm^\color{red}2s^{-2}$$

which shows that the units don't agree... the left is saying force (Newton), while the right is saying work (Joule).

If it is true that the lifting force provided by a keel is dependent upon volume and not projected area (and I am saying that it is, by fiat), and we end up with too many units of metres on the right hand side, we need to somehow reduce the number of metres elsewhere in the right side of the equation. Since we know that we're dealing with force and not work, we can't change the left hand side. Since we know that we're dealing in velocity and not just a quantity per second, that can't change. So, that just leaves changing the density of the ether to $kgm^{-4}$.

With that, we get:

$$ kgms^{-2} = kgm^{-4} \cdot (ms^{-1})^2 \cdot m^3 $$

which decomposes to:

$$ kgms^{-2} = kgm^1s^{-2}$$

Everything seems good again... Except that it appears that my ether is four-dimensional.

Did I get that right, or have I screwed this up somehow?


5 Answers 5

  1. When doing dimensional analysis, the usual way is to work with dimension symbols time (T), length (L), mass (M), electric current (I), absolute temperature (Θ), and optionally amount of substance (N) and luminous intensity (J), and not units of measurement. The units are arbitrary, the dimensions are less so.

    The last two are fundamental dimensions only because the CGPM (that's the Conférence générale des poids et mesures, General Conference on Weights and Measures) wants them to be. It is perfectly reasonable to do away with them, with N being zero dimensional and J having the dimension of power, L²MT⁻³. On the other hand, it can be argued that angles and solid angles ought to have their own fundamental dimensions, although the CGPM wants them to be zero dimensional.

  2. You took an equation where the coefficient of lift is zero-dimensional, and then invented a new equation in which you have also assumed that the coefficient of lift is zero-dimensional. Dimensional analysis says that in your new equation the coefficient of lift has dimension L⁻¹.

  3. Yes, if you absolutely insist to have a dimensionless coefficient of lift then you must solve the dimensional mismatch some other way. But I would say that having four-dimensional volume is the wrong way to do it. Basically, it is not clear what is the physical meaning of the L⁴ volume of the keel. Three dimensions you can measure, but the fourth comes from nowhere and I don't see how you could measure it.

    What you cannot measure you cannot control.

  • $\begingroup$ As a minor addendum to this, the usual way of resolving this would be to plot the "observed" (plot-driven, in your case) relationship between keel volume and force at a convenient constant speed & density, and to calculate C_L so that it matches. In lay terms: Do you want giant, lightweight keels that barely manage to lift a small mass, or do you want small keels that carry lots of cargo? $\endgroup$ Commented Dec 8, 2023 at 22:37
  • $\begingroup$ How can we change $C_D$ to have a dimension of $m^{-1}$ without saying per metre of what? At least by saying that the ether is four-dimensional, and has a density that is a mass per tesseract metre, that makes sense. It's a fantasy world, so it could happen. $\endgroup$
    – Monty Wild
    Commented Dec 9, 2023 at 4:19
  • $\begingroup$ @MontyWild: Per metre of keel in the direction of movement, for example, meaning that the thicker the keel is in the direction of movement, the stronger the effect. Or maybe it is some sort of spatial frequency which is of course measured in m⁻¹ just like temporal frequency is measured in s⁻¹. $\endgroup$
    – AlexP
    Commented Dec 9, 2023 at 6:49

You started with a formula for lift in terms of surface area, but then you plugged in a volume instead of an area into that equation and noticed that the units weren't consistent.

This is to be expected! It's the same as if you would take Hooke's law (force $F = kx$, where $x$ is the length of a stretched spring and $k$ is a coefficient with dimensions $kg\,s^{-2}$) and plug in a volume for $x$ instead of a length. You'd find the resulting units were wrong, it wouldn't be a force.

But what if you wanted to model a situation where a force depends on a volume instead of a length? Would such an object have to be five dimensional?

No it wouldn't -- for example, buoyancy is a force that depends on volume. The formula for buoyant force is $F = g\rho\, V$, where $\rho$ is the density of the fluid being displaced and $g$ is the acceleration due to gravity. So $g\rho$ is a constant with dimensions $m\,s^{-2}\cdot kg\,m^{-3} = kg\,m^{-2}\,s^{-1}$, which is different from $kg\,s^{-2}$.

This is maybe easier to think about if we express the units in terms of newtons, $N=kg\,m\,s^{-2}$. The spring constant has units $N\,m^{-1}$, so it has the right units to multiply it by a length and get a force, while $g\rho$ has units $M\,m^{-3}$, so you can multiply it by a volume and get a force.

What does this mean for your question? Well, it just means that if the force exerted by a keel depends on its volume then the lift equation isn't the right one. This is all that your dimensional analysis is telling you.

Because of that you need a different equation. What equation? Well, since there are no such things as ether or keels that's up to you. But as you know, you should choose it such that the units are the same on both sides. As AlexP mentioned this new equation will probably have to involve constants that are not dimensionless, but this is completely fine - it happened in both the examples I gave above.

In fact, if you want a simple equation like these examples, where it's just a bunch of terms multiplied together, you'll find your hand is forced. If you want lift to be proportional to volume and velocity then the constant term will need to have dimensions $N\,m^{-3}\,s^{-1}$. But as in the lift equation and the buoyancy equation, that constant can still be the product of a bunch of other terms, which you can make up how you like as long as the units are right.


My physics teacher in high school stressed over and over that dimensional analysis is necessary for correctness, but not sufficient.

Which is, a correct equation will surely have correct dimensions, but correct dimensions do not ensure a correct equation.

A foil works only when it is moving with respect to the fluid surrounding it, while a keel doesn't care if the fluid is moving or not: a stalled plane falls, a moored boat floats. If you want a dramatic demonstration, look at America's Cup ships: when they are moving the stay up above the water thanks to the foil lifting them, when they are still they float thank to their keel keeping them buoyant while the foil does not work.

This alone should tell you that you can't interchange the two like nothing is bound to happen.

  • $\begingroup$ Well, my etheric keels need to move to generate lift. If the ship tries to stay still in mid-air, its weight will push it down slowly, so the keel isn't just floating. $\endgroup$
    – Monty Wild
    Commented Dec 8, 2023 at 13:54
  • 2
    $\begingroup$ @MontyWild: If it only pushes it down slowly, there is a buoyancy force somewhere, isn't it? $\endgroup$
    – AlexP
    Commented Dec 8, 2023 at 14:50
  • 1
    $\begingroup$ @AlexP: No there is just a lot of drag. The ether seems to be a simplified airfoil: Easy to move in one direction, hard to move in the other direction. Just without the complexities of angle of attack and stall, boundary layer dynamics, laminar vs turbulent, etc. $\endgroup$ Commented Dec 9, 2023 at 0:17
  • $\begingroup$ @KevinKostlan Actually, I was going to have lift, drag, AoA and stall too. $\endgroup$
    – Monty Wild
    Commented Dec 9, 2023 at 4:27

Neutrino sailing

Fluids get blocked by solid surfaces, but your aether is, well ethereal. Thus very little is actually intercepted by the keel. Similar to how neutrinos care about the volume of the detector.

So lets consider a simple model: Your keel is made of a special enchanted material that can interact with aether. It may make more sense to put the keel above the ship since that configuration is more stable.

Such material has a grain direction $r$. The particles of aether passing though it with relative velocity $-v_{ship}$ have a small chance of interacting. If they interact, they lose all velocity across the grain (perpendicular to $r$). They also lose a little velocity along the grain.

The velocity change imparts momentum to the ship: $F= k V (-v_{ship, ⟂}-\mu v_{ship,\parallel})$. Good materials have a high $k$ and a low $\mu$. $\mu$ is dimensionless and it's inverse is similar to a glide ratio.

$k$ has units of $kg/s/m^3$. Which means that a certain volume of keel can "catch" a certain amount of aether mass per second.

So how is this different than fluids, besides area vs volume?

Firstly, the whole very complicated world of laminar vs turbulent, stall, boundary layers, and more is lost. This is inevitable due to the coupling being so weak: in order for the "porportional to volume" rule to hold keel cannot shield itself. Which means most aether passes though the hull unchanged, it is not forced to pass around the hull.

Secondly, the force is proportional to velocity not velocity squared. (ordinary fluid friction is proportional to velocity at very slow speeds, but airfoils are not useful in that case). This means that the aether forces on the keel increase more gently as the hull speeds up.

Sails in the sky

Sails are just as useful as on sea or land. Thus the iconic "sailships in the sky", which have no use without an aether to push against, is justified.

Changing the penetration depth

The aether has a large but finite density $\rho$, in $kg/m^3$. The penetration depth is how deep the aether can make it into a material before 63% is blocked. It is given by $d=v\rho/k$ and has units of $m$. The depth is deeper for faster ships because the aether spends less time inside the enchanted keel.

If $d$ is much larger than any ship your volume formula will hold. If $d$ was extremely small and the aether can collide with itself you have ordinary fluid dynamics (minus the buoyancy). For d-values say around $d=256$ meters, you have a case that small ships can use the volume formula but large ships have to use a more complex model.

Ships which are hovering (using an aether propeller, which looks and spins like an ordinary propeller but uses blades with tilted grain instead of airfoils) will leave a downwash wake which can be estimated by $v_{down} \approx sqrt(\frac {gm_{ship}} {\rho_{aether} A_{prop}}) $ where $A_{prop}$ is the area in square meters of the propeller magical material. This is the same as ordinary fluid dynamics and when $\rho$ is large this wake becomes negligible. Ships moving faster leave a less intense wake. Such a downwash wake which may be hazardous for small ships dependent on the parameters of your story.

Again, you can avoid down-wash and deviations from your volume lift formula completely if you use a very large $d$ which also necessitates a very large $\rho$.

Solar system dynamics are very different

The aether is very dense. Suppose you want $d=1000m$ (so that the volume formula remains accurate to a hundred meter ship size) and a magic keel with a density of water to only sink at 1 m/s (the ship itself will sink faster because it also has non-keel weight). A cubic meter of keel weighs $9.8kN$. The needed $k$ is $9800$ $kg/s/m^3$. The aether density is $\rho = 9800000$ $kg/m^3$. This is 500 times denser than gold.

The aether must be dragged by planets in order for it to be "still". This means that planets are carrying many times their mass as aether. If we assume it has mass without gravity (which makes sense in Newtonian physics, but does not make sense in general relativity) planets would be dragged around by whatever space aether currents there are instead of orbiting freely.


Your keels do not have lift, they have an anizothropic friction, so the initial take that your keels provide lift is false, therefore all your future conclusions lose ground. It's a logical error, not physical one, yet it qualifies for hard science.

  • $\begingroup$ If the keels simply had anisotropic friction, then they couldn't provide lift, could they? $\endgroup$
    – Monty Wild
    Commented Dec 9, 2023 at 6:39
  • $\begingroup$ @MontyWild yep but the lift is provided via thrust, he's got some system for thrust too. Check the linked question for details. $\endgroup$
    – Vesper
    Commented Dec 9, 2023 at 12:38

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