Collapsible Flettner Rotor Project

Discussion in 'Projects & Proposals' started by Yobarnacle, Jun 4, 2014.

  1. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    As a motor drive, this pinch drive might be simplest.
    No central shaft. Can drive the plywood endplate directly with speed reduction built in.
     

    Attached Files:

  2. daiquiri
    Joined: May 2004
    Posts: 5,373
    Likes: 247, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Yo, you have misread it. The sail force calculator gives you 33 kgf (324 N) of drive force and 121 kgf (1187 N) of side force.
    Be very careful with units, similar errors might endanger your whole project! :eek:
     
  3. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

  4. Olav
    Joined: Dec 2003
    Posts: 306
    Likes: 38, Points: 38, Legacy Rep: 460
    Location: Filia pulchra Lubecæ

    Olav naval architect

    Be careful, lbf (pound-force) and N (Newton) are units of force, lb-ft (pound-foot) and Nm (Newtonmetre) are units of moments (force times lever arm)!
     
  5. daiquiri
    Joined: May 2004
    Posts: 5,373
    Likes: 247, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    :eek:

    [​IMG]

    As Olav said. At least the fundamentals have to be sound, be very very careful there.
     
  6. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    thanks gentlemen
     
  7. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Would pounds force be similar to thrust?
     
  8. Olav
    Joined: Dec 2003
    Posts: 306
    Likes: 38, Points: 38, Legacy Rep: 460
    Location: Filia pulchra Lubecæ

    Olav naval architect

    Yeah, sort of. Thrust is a force and thus pound-force would be an appropriate unit for it.
     
  9. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Is lbf ever multiplied by velocity? Example a 70 lb thrust trolling motor spinning at 5 kts compared to 70 lbf sail in a 20 kt wind. you get more energy, power from the 20 kt wind than the trolling motor though both are 70 lbf?
    Probably these are rudimentary questions.

    Weight of a bullet times velocity gives ftlbs at impact. Similar?

    actually, i'm trying to understand how lbf becomes lbft to determine how fast the lbf will drive a boat of a certain weight and resistance.
     
  10. Olav
    Joined: Dec 2003
    Posts: 306
    Likes: 38, Points: 38, Legacy Rep: 460
    Location: Filia pulchra Lubecæ

    Olav naval architect

    Yes. Force (please note that lbf is the unit, the physical quantity is force) times velocity equals power, i.e. the resistance of a vessel at a certain speed times that speed is the effective power (account for some factors when you want to use imperial units to get hp from lbf and mph, kts, fps or whatever velocity unit you have; within the metric system you won't need this since power [W] is force [N] times velocity [m/s]).

    Divide this effective power by the propulsor efficiency, add losses from bearings, gear etc. and maybe the shaft generator (if fitted) and you get the required power of the engine.

    No. The bullet weight (mass times gravitational acceleration) times velocity is again power - here: the power needed to fire the bullet upwards.

    The bullet mass times velocity would be its momentum, half the bullet mass times velocity squared its kinetic energy.

    Never mind. :)
     
  11. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    thankyou
     
  12. daiquiri
    Joined: May 2004
    Posts: 5,373
    Likes: 247, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    I am not sure what are you trying to calculate, but here come few considerations.
    The power which can be extracted from a unit mass of air moving at a speed Vapp is given by:
    P = 0.5 rho Vapp^3​
    So an apparent wind of 20 kts can potentially give 64 times more power than a wind of 5 kts:
    20^3 / 5^3 = 64​
    The propulsive power from the rotor or from the sail is calculated, for your purposes, by multiplying the boat speed Vb by the drive force (projection of the rotor-generated aerodynamic force on the boat course):
    Peff = Vb * Force,drive​
    Of course, the drive force has to be equal to the drag force, for a given speed Vb. So you can also put the boat's resistance in the previous equation.
    It is important to keep all the units consistent. It means, either:
    V = m/s
    F = N
    P = W (or N m/s)​
    or
    V = ft/s
    F = lb
    P = lb ft /s​
    And it is also important to not make a mistake of calculating the power as a product of rotor-generated lift and the apparent wind speed, because they are (by definition) perpendicular to each other. Only the force and velocity components which are parallel to each other can be used for power calculations.

    Cheers
     
  13. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    If I were broad reaching, in a 20kt wind 30 degrees abaft beam, I could multiply the force of the rotors lift by 10kts of forward vector from that 20 knot wind? But if the wind is abeam or forward of beam, it's velocity vectors contributes only to drag and nothing to propulsive power?
     
  14. daiquiri
    Joined: May 2004
    Posts: 5,373
    Likes: 247, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Ok, this a very interesting project, so I did a little homework here... :)
    You can download a pdf print of my calculations from the dropbox site:
    https://www.dropbox.com/s/qictt45m1ptj9tq/Magnus.pdf

    I have found the Albin 25 Handbook in internet. It contains the resistance curve of the boat, which is the starting point for any powering calculation. The calcs are done for 20 kts wind coming from 30 degrees abaft beam, as by your last post, and for 6 kts boat speed. Hence, that is the assumed design condition.

    The resulting rotor is 1.0 m diameter, 5.1 m high and turns at 390 rpm. It should require around 5-5.5 HP shaft power for turning it, which is a nice saving when compared to 9-10 HP required by the propeller (according to the boat resistance curve).

    Once the rotor is sized, these calculations should be repeated for each wind condition and boat speed, in order to verify the performance in off-design conditions. That is quite a job to do, but should be done at some point.

    Hopefully it will serve you as a path to follow for future calculations, in other design conditions you might want to try.

    Cheers
     
    Kai Rabenstein likes this.

  15. Yobarnacle
    Joined: Nov 2011
    Posts: 1,735
    Likes: 126, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Thankyou very much. I downloaded it and will study it and research/learn the topic where I find it currently over my head.

    I'm glad you are interested in the project.
     
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.