Could an unmanned solar plane go not only above the clouds, but also so fast that it always is in the day-side of the planet?

A solar plane flying above the clouds receives sunlight $12$ hours per day, but there is a need for batteries to fly in the night. Such plane could fly continuously for decades without even fearing the degradation of batteries and it could be very much lighter given that batteries weight approximately $\frac{1}{4}$ of solar impulse

From a basic analysis of the problem I see these issues:

  • It would need to go around the world in $24$ hours, and given a diameter of $40000$ kilometres for the world, we have a need of $\frac{40000~\text{km}}{24~\text{h}} = 1667~\frac{\text{km}}{\text{h}} = 463~\frac{\text{m}}{\text{s}}$

    Going so fast (even supersonic) will require a lot of energy, thus a lot of solar panels thus a lot of weight, that will require even more energy...

  • Electric means that it must use propellers, propellers require dense air to be efficient, this source gives the density a very high importance putting it squared in the equation for thrust.

    Solar means that it must go high (over the clouds) and high means sparse air.

  • Some clouds are just too high to go over (the highest clouds in Earth's atmosphere, located in the mesosphere at altitudes of around $76$ to $85$ kilometres - credit Wikipedia), for these elusive maneuvers (going around them) will be needed.

Could these issues be circumvented to create a solar plane without batteries capable of continuous flight?

  • 1
    $\begingroup$ @Cardidorc, well yeah, you could probably also use huge arrays of capacitors for the buffering in the end; which probably goes better hand in hand with the requirement of always flying in the sun :) $\endgroup$
    – dot_Sp0T
    Apr 16 '16 at 11:50
  • 1
    $\begingroup$ Solar panels also degrade over time so eventually that thing will come down, batteries or not. $\endgroup$
    – MichaelK
    Apr 16 '16 at 13:31
  • 1
    $\begingroup$ You don't have to fly around the equator. Go further north or south, and you get much smaller circumferences, meaning less speed, meaning more feasibility. $\endgroup$
    – Burki
    Apr 16 '16 at 14:08
  • 1
    $\begingroup$ @Burki Only one problem then: your solar panels are on top of your wings. With the sun at 60 degrees away from the normal, you get half the sunlight. And it rapidly gets worst the further north you get. $\endgroup$
    – MichaelK
    Apr 16 '16 at 15:02
  • 2
    $\begingroup$ @Burki Not to rub it in or anything; I'd just like to point out that the circumference of a parallell is calculated by taking the full circumference of the Earth and multiply with the cosine of the parallell. But(!), the efficiency of a tilted solar panel also follows that formula: cosine of the sun's angle away in relation to the normal, which in turn follows the degrees of the parallell. So even though the route gets shorter and shorter, the solar panels gets equally less efficient. And(!) you also need to factor in losses to the longer path for the light through the atmosphere. $\endgroup$
    – MichaelK
    Apr 16 '16 at 15:58

o.m.'s conclusions are correct (if you like my answer you should up vote his too).

Theoretically Yes. Realistically Possibly.

Plugging in realistic efficiencies shows that certain sized craft could do this in theory. However, solar power isn't very potent and we don't have a good mechanism for turning solar power into propulsive power. So although physics doesn't forbid it, we don't have anything engineered that could do it now.

Assumptions, Constraints, and other trivia

I'm not deriving much of this from first principals, I'm starting with my basic background and education. If you need the justification for some of these, please read the references.


Use the 4 equations stated above:

  • $T = W \cdot \frac{C_{drag}}{C_{lift}}$
  • $W = \left(\sqrt{A}\right)^3$
  • $\frac{C_{lift}}{C_{drag}} = \frac{4 \cdot \left(M + 3\right)}{M} \rightarrow \frac{C_{drag}}{C_{lift}} = \frac{M}{4 \cdot \left(M + 3\right)}$
  • $P = T \cdot v$

Substitute and simplify:

$$P = W \cdot \frac{C_{drag}}{C_{lift}} \cdot v \rightarrow \left(\sqrt{A}\right)^3 \cdot \frac{M}{4 \cdot \left(M + 3\right)} \cdot v$$

Then solve for various flight regimes.


I plugged this into a spreadsheet (along with calculations for speed of sound, mach number, etc.) and this is what I got:

Edit1 about 3 hours after initial answer I was off by a factor of 1000. I've updated this chart. Green cells show possible size/altitude combinations given 100% efficiency in solar power collection, engine thermal efficiency, propulsive efficiency, etc.

Edit2 about 3 hours after last edit I neglected to multiply the solar power by the planform area. I've updated this chart. Green cells show possible size/altitude combinations given 100% efficiency in solar power collection, engine thermal efficiency, propulsive efficiency, etc.

Edit3 about 2 days after last edit I have not changed the answer substantively. Instead for the realistic scenario I change the $\frac{C_l}{C_d}$ to 50% of the maximum value. Then I converted the charts to read what % of propulsive power the solar power could provide. So now the answer is in % rather than Watts.

Required Thrust Power (W) - for 100% efficient components:
Required Thrust Power (W) - 100% efficient

The spreadsheet shows the minimum thrust power required for straight and level flight at the required velocity at the altitude shown on left for a aircraft planform area shown across the top.

What this says is that your aircraft must remain quite small (smaller than $2 m^2$). This vehicle could never be manned (but that's not a requirement, so it's OK).

If you plug in realistic efficiencies for these parameters (overall optimistic engine efficiency could be as high as 40% and very optimistically solar efficiency of 30%***), you would get a chart that looked more like this:

Required Thrust Power (W) - for realistically efficient components:
Required Thrust Power (W) - Realistic Efficiencies

So it looks like a well designed and small enough aircraft could be made. Granted I don't know of a good mechanism for turning solar power into thrust so there would still be quite a bit of engineering work left to do, the physics don't prohibit such a craft.

And another thing

Although someone might be able to someday build such a craft, remember that anything with moving parts will require lubricants. Even if you meet its power requirements to keep it going, eventually it'll run out of these lubricants and/or its moving parts will break.

Many US aircraft have the ability to perform mid-air refueling. From a propulsive perspective, they can stay airborne forever. Practically speaking they have flight duration limitations based upon things like provisions for the crew, lubrication for moving parts, etc.

**Aircraft engine flight performance by type and flight regime:
Aircraft engine flight performance

***Solar cell efficiencies:
Solar cell efficiencies

  • $\begingroup$ Technically speaking it is possible to create an "electric turbojet" (I actually doubt a turbofan could go supersonic). Practically however, I can't see how you could electrically heat that much air... $\endgroup$
    – Aron
    Apr 18 '16 at 8:31
  • $\begingroup$ Every modern fighter aircraft capable of supersonic speeds uses a turbofan engine (not a turbojet), so turbofans certainly can go supersonic. Just look at the chart I included. The F100, F101, F404, F110, etc. (F-14, F-15, & F16, F-18, B-1B) are all turbofans. As to the other, I agree. I can see theoretically how you could turn sunlight into power for an engine, but I don't see how to do it practically or efficiently. $\endgroup$
    – Jim2B
    Apr 18 '16 at 12:25

I believe that it will be impractical.

  • Solar power depends on the area of the solar cells. At perfect efficiency, you would get a bit over 1 kW per square metre.
  • Supersonic drag depends on the cross section rather than total surface area, but in practice your cross section will go up if you increase the solar cell area.
  • The Concorde has over 400 lbf of thrust per square metre of wing area.
  • The relationship between lbf and kW isn's straightforward, but I think that you will get much less 400 lbf from 1 kW.

And that's without inefficiencies of the solar cells.

  • $\begingroup$ I'm not sure this lives up to the criteria of the hard-science tag, but I'm also not sure enough it doesn't that I'll downvote. Still, you may want to elaborate on this. $\endgroup$
    – user
    Apr 16 '16 at 12:09
  • $\begingroup$ Impractical =/= impossible. I already knew it would be hard, I am asking if it is possible. $\endgroup$
    – Caridorc
    Apr 16 '16 at 12:34
  • $\begingroup$ @MichaelKjörling, better now? I thought the last bullet points were obvious, but apparently they were not. $\endgroup$
    – o.m.
    Apr 16 '16 at 12:38
  • $\begingroup$ @o.m. there are always two types of drag: pressure and friction. Pressure drag is a function of cross sectional area while friction drag is a function of surface area. Total drag is the sum of these components. But I agree with your conclusions. I don't think it's theoretically impossible but I can't envision any near future technology (or extrapolation of such) providing the ability to do it. $\endgroup$
    – Jim2B
    Apr 16 '16 at 12:52
  • $\begingroup$ OK, added some equations to back up my feelings. $\endgroup$
    – Jim2B
    Apr 16 '16 at 15:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .