There has recently been posts on youtube where in a land based propeller/fan driven tricycle (blackbird) has been shown to roll faster than the wind that is behind it.

The youtube channel veritasium covered it recently and the topic has apparently been around for a while.

Although veritasium does provide an answer in terms of formulae and additionally explains based on the power in the wheels (which is counter-intuitive to me as noted in the points able), I have an understanding to cross-validate on this (and hence the question) and wanted a confirmation if it was correct. However if it is incorrect, then am at least looking for an explanation in the same sort of layman's terms that I am trying to describe.

Specifically my understanding is as follows -

  1. The wind powers the vehicle initially.

  2. The wheel's (through gears and chain) drive the propeller.

  3. It would be safe to assume that until the vehicle reaches the wind speed, its powered by the wind and as such builds up inertia (and consequently can accelerate).

  4. However the explanations (in the veritasium video) tend to focus on the power generated by the wheels to explain the vehicle going faster than the downwind. The problem with this explanation is that it holds only to the extent that the vehicle has expended (exhausted) the potential energy it initially built (from it rolling when it was slower than the wind). Friction would eventually slow it down back atleast to the same speed as the wind. It does not explain how the vehicle is able to sustain it.

  5. The question is - what is the force that is allowing the vehicle to sustain the speed ? It definitely can't be the power from the wheels - its built on inertia and friction would cause it to drain its energy back to zero.

  6. My understanding is that there should be some physical 'link' that would be between the wind generated from the propeller (propeller-wash?) that compensates for and allows for the 'link' to be re-established between the downwind and the vehicle - analogous to a boatman pushing the boat in shallow rivers/lakes using a stick, the wash from the propeller is the 'stick' and the downwind forms the 'lakebed/riverbed'.

  7. Is my understanding correct? If not then what is wrong with my understanding ?

My question is similar to the question posted here - How to sail downwind faster than the wind?

It also has a fairly good answer in the past few days (which I believe draws insights from the videos linked below) but does not clarify on point #5 (on sustainability) - https://physics.stackexchange.com/a/651106/31729


  1. (A Physics Prof Bet Me $10,000 I'm Wrong) https://www.youtube.com/watch?v=yCsgoLc_fzI
  2. (Risking My Life To Settle A Physics Debate) https://www.youtube.com/watch?v=jyQwgBAaBag
  3. (CNN Early Start - Blackbird upwind cart) https://www.youtube.com/watch?v=b8MWCvKIi7E
  4. (DDW.mov) https://www.youtube.com/watch?v=5CcgmpBGSCI
  • $\begingroup$ The missing ingredient is the torque between the earth and the vehicle keeping the vehicle from spinning opposite the fan. If you put the contraption on a freely rotating, massless track and held it on with some normal force (suppose the track and the cart are opposite charged insulators) and turned off gravity, it wouldn't work. $\endgroup$
    – g s
    Jul 18 '21 at 23:18
  • $\begingroup$ Or at least, not nearly as well. There's still a small extra torque available from the rotational acceleration of the cart. $\endgroup$
    – g s
    Jul 18 '21 at 23:54
  • $\begingroup$ Why do you say it can’t be the power from the wheels (# 5)? $\endgroup$
    – Ben51
    Jul 19 '21 at 2:26
  • $\begingroup$ Does this physics.SE answer address your question? The blackbird vehicle is harvesting energy from the difference in velocity between air and ground. In the follow-up video the concept of harvesting from velocity difference was demonstrated by Derek Muller with a mechanical device. $\endgroup$
    – Cleonis
    Jul 19 '21 at 17:03
  • $\begingroup$ @Ben51 - What I meant was that the wheel had to get its power from somewhere in the first place. So to start with it gets it from the wind. But once the vehicle velocity matches the tail wind, its as good as starting from zero speed except for the velocity and momentum it's already built up (Pot Engy) upto that point. From that point onwards it can only lose energy. The energy its built up, helps it go faster than the wind - but it has to be back and forth - it gets energy from wind and races ahead and then loses energy and speed to below that of the tail wind at which point it picks up again. $\endgroup$ Jul 19 '21 at 22:01

In an comment I already linked to the moment in the follow-up video were Derek Muller demonstrates a device.

The following may bring the aerodynamics of the Blackbird vehicle into focus. Imagine the design is not weight constrained at all and the propeller can be built as a ducted fan.

For the purpose of the thought demonstration we make the shroud of that ducted fan very long, we can visualize it as a tube that is longer than it is wide, with the propellor halfway along the length.

The effect of that elongated shroud is that air mass is prevented from escaping sideways.

For its propulsion this ducted fan vehicle needs to have traction both with respect to the ground and with respect to the wind.

In idealized conditions there is a uniform wind speed, and the vehicle has all the space to keep going straight downwind. Under those idealized circumstances, once set in motion, the vehicle will reach the top speed (a sustained speed) that the vehicle can attain under the given circumstances. From here on I will refer to that speed as 'cruising speed'.

When the ducted fan vehicle is at cruising speed:
It is necessary that the propellor is moving air from the front of the duct to the rear of the duct. We assume that at cruising speed the vehicle is going faster than the wind, so the vehicle is overtaking the wind. The propellor must exceed that rate of flow, sustaining a surplus of air pressure in the rear half of the duct.

When the vehicle is at cruising speed:
Even though the vehicle is overtaking the wind, the presence of the rear wind still makes it harder for the air in the rear half of the duct to escape the duct. In that sense the rear wind is essential for sustaining the pressure difference.

That is how the ducted fan vehicle can harvest propulsion from the difference in velocity between air and ground.

(In the case of the actual Blackbird vehicle the propellor isn't ducted, so air can escape sideways, but the Blackbird vehicle does go downwind faster than the wind. Apparently without a duct it is still possible to harvest fairly good propulsion.)

  • $\begingroup$ Thanks ! The ducted fan approach is what I was thinking when I mentioned the boat-and-the stick approach above (#6 in my question above). However felt it would make it harder to visualize. $\endgroup$ Jul 19 '21 at 21:06
  • $\begingroup$ What do you mean when you say "when the vehicle is at cruising speed" ? What would you consider to be this vehicle's cruising speed? $\endgroup$ Jul 19 '21 at 21:07
  • 1
    $\begingroup$ @RavindraHV I added a paragraph to describe how the expression 'cruising speed' is to be understood $\endgroup$
    – Cleonis
    Jul 19 '21 at 21:31

It can go faster than the wind in steady state and doesn't have to slow down. It's not using any stored energy for propulsion, so there is no reason why it would ever have to slow down, unless the wind relative to the ground slows down.

Your mistake is trying to apply sequential cause-effect reasoning to a feedback loop. This approach might be intuitive to you, but fails in situations like this. Instead you have to analyse it quantitatively and find the force and power balance.

For correct analyses see:

  • 2013 AAPT United States Physics Olympiad Semifinal Exam Solutions (page 10)

  • Drela, Mark "Dead-Downwind Faster Than The Wind (DFTTW) Analysis"

  • Gaunaa, Mac; Øye, Stig; Mikkelsen, Robert (2009), "Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion"

  • $\begingroup$ Thanks did do a quick lookup but yet to analyse in detail. Primarily wanted to get an 'intuitive-feel' of the problem statement. Also for my understanding, assuming 'steady-state' here means 'no-net-change'. For the benefit of future readers PDF link to the first reference is here - aapt.org/physicsteam/2019/upload/USAPhO-2013-Solutions.pdf. Its available on web-archive as well. $\endgroup$ Aug 22 '21 at 17:02

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