# Sailing Upwind. What are the theoretical or practical speed limits?

1. Nov 21, 2008

### uart

Given the recent discussion/demonstrations of sailing downwind faster than the wind I was wondering if anyone has some ideas about the limits (either theoretical or practical) of sailing upwind.

I was thinking in particular of a turbine driven vehicle going directly into the wind but I'd also be interested in the upwind velocity component (upwind VMG) of any wind driven vehice.

Any ideas on what is possible in this scenario?

Last edited: Nov 21, 2008
2. Nov 21, 2008

### mender

Iceboats are able to have a substantial VMG when tacking upwind. From what I see, the difference between the VMG downwind and VMG upwind would be equal to the difference in the actual windspeed. So if the iceboat can achieve a VMG of 4x downwind, it should be able to achieve a VMG of 3x upwind.

This site has some numbers:

http://www.nalsa.org/Articles/Cetus/Iceboat Sailing Performance-Cetus.pdf

I would think that the effect is the same for other wind driven vehicles.

Last edited: Nov 21, 2008
3. Nov 21, 2008

### rcgldr

For a sailcraft, the apparent crosswind on a sailcraft depends only on the sailcrafts heading and the true wind speed, and is independent of the sailcrafts forward speed, since the apparent crosswind is perpendicular to the direction of travel (and therefore independent of sailcraft speed).

From page 4 of the pdf linked to above: An iceboat heading 30 degrees offset to an 18mph wind, total speed 70mph and apparent wind speed 55mph, which I adjusted to 55.15mph (so heading was 30 degrees). Note that Beta is 9 degress, not 8 as shown in the diagram so there's at least one mistake there. Back to the math:

The apparent crosswind is 9mph (sin(30) x 18mph), and the iceboat could achieve an apparent wind speed of 55.15mph, composed of a 9mph apparent crosswind and a 54.4mph apparent headwind for a L/D of about 6:1. Downwind, the iceboats total speed was given as 70mph and it's VMG downwind would be 60.6mph (cos(30) x 70mph). Upwind, the iceboats apparent wind speed remains the same at 55.15mph. It's total speed would be > 38.8mph, and VMG upwind would be > 33.6mph, assuming the drag from the ice would be a bit less at the slower true speed.

Last edited: Nov 21, 2008
4. Nov 22, 2008

### spork

There is no theoretical limit to the speed at which you can go directly upwind or directly downwind. It is limited only by the real world limits of frictional losses and limited L/D.

5. Nov 24, 2008

### yoavraz

It deals with approximating sailboat speed at given wind, including upwind and VMG.

See

BTW, I have found that the discussion mentioned here about downwind was not about a turbine sailboat but rather about a different creature with a turbine with opposite gear ratio named DWFTTW (down wind faster than the wind). See my comments there on the difference.

Last edited: Nov 25, 2008
6. Nov 24, 2008

### yoavraz

A "regular" sailboat is totally dependent on the relative wind.

Theoretically it can go almost directly to the wind (close to 0 degrees) and still have forward-force from sail-lift overcoming drag, and have it in motion. However this force drops dramatically when getting close to 0, and speed becomes almost 0.

Practically forward-force is destroyed at ~20+ degrees because of all kind of environment fluctuations and interferences. Also soft (conventional) sails typically start luffing at ~30 degrees and sail lift is destroyed. Also VMG is around 45 deg, going to ~40 for fast land and ice sailboats, so practically, usually no reason exists to go below 40 degrees.

With a prop and turbine the situation may be different. Moving slowly, also at 0 degrees (against the wind) with high gear ratio may be possible. In principle, a large turbine prop turning rapidly (like with a stationary turbine) can generate substantial energy, and the vehicle can use this energy to move slowly (mechanically by gear, or by electricity) against the wind. I have not seen yet though.

7. Nov 24, 2008

### spork

In theory this type of vehicle can go directly upwind faster than the wind.

8. Nov 24, 2008

### rcgldr

You might be able to test this by simply putting smaller wheels (so that the effective gear ratio between wheels and prop results in faster prop speed, slower wheel speed) on your DWFTTW cart and using a fan to blow air against the cart. The tires and/or the test surface would need to be sticky to avoid slippage.

9. Nov 24, 2008

### mender

I think that would require some pretty small wheels; as a guess 1.5 inches in diameter. My impression is that the overall gear ratio or travel ratio needs to be under 1:1 for the cart to move into the wind.

10. Nov 24, 2008

### rcgldr

Yes, wheel circumference needs to be less than efffective prop pitch. Assuming prop pitch is 10 inches, I'm not sure what the effective pitch is when the prop is being driven by a wind, it could be more than 10 inches (it's less than 10 inches when the prop is being driven by the wheels and generating thrust), then wheel circumference needs to be well less than 10 inches, probably around a 2 inch diameter would work.

Also I forgot to mention that the cart would be moving backwards, with the prop end facing the fan, to transfer more weight to the driven wheels and prevent slippage.

11. Nov 25, 2008

### mender

So what I am getting from this is that a downwind cart with a ratio above 2.0:1 would "self-start" in reverse or in the upwind direction.

A downwind cart with fixed gearing appears to have a top speed limit linked to the ratio above wind speed if it is required to self-start. This would also mean that the downwind cart would be capable of a higher speed with a different gearing once it was already moving.

Does this also apply to the upwind cart? What would the required ratio change be, and would there be a limit solely because of the gearing (assuming CVT and low aero drag)?

Last edited: Nov 25, 2008
12. Nov 25, 2008

### rcgldr

I don't know what the ratio needs to be. It's possible that it has to be higher still, perhaps 3:1 (prop speed : wheel speed) or more. You'd need a higher pitch prop in that case so the wheels wouldn't end up tiny.

The prop itself acts a bit like a CVT when generating thrust. In a near static situation like this, once pitch reaches a certain point, increasing the pitch further will not significantly increase the thrust (although it probably increases the drag). For the upwind case, the prop is acting as a turbine, and I'm not sure how the effecitve pitch relates to apparent wind at the prop.

Considering that the DDWFTTW carts are going perhaps 13 mph forwards with a 10 mph wind, that's a net headwind of 3 mph. For the upwind cart, the net headwind is over 10 mph just if the cart is moving forwards, so the aerodynamic drag on the cart is much greater than the downwind case. The current designs should able to go forwards against the wind, but I doubt they could go forwards at more than the headwind speed (10+ mph into a 20+ mph apparent headwind).

13. Nov 26, 2008

### uart

Hi Jeff. I made some mathematical calculations based on the "sailing" vector diagrams and making the over-simplifying assumption that the craft can always achieve a speed such that the apparent wind reaches a fixed minimum angle relative to the crafts direction. I know this assumption is not really justified in *practice but it's a useful starting point.

The results were interesting in their simplicity and symmetry between the headwind and tailwind cases. (BTW, here I'm considering simple sailing craft such as a sailboat or iceboat, w = wind velocity magnitude and $\phi$ is the closest angle that the craft can make to the apparent wind.)

Upwind. The upwind VMG (upwind velocity component) is maximized when the craft direction is $\pi/4 + \phi/2$ off the headwind and the resulting maximum upwind VMG is $w (\mbox{\rm cosec}(\phi) - 1) / 2$

Downwind. The Downwind VMG is maximized when the craft direction is $\pi/4 - \phi/2$ off the tailwind and the resulting maximum upwind VMG is $w (\mbox{\rm cosec}(\phi) + 1) / 2$

Note that with these results the difference between the maximum downwind VMG and the maximum upwind VMG is precisely equal to the windspeed W, exactly as mender hypothesised in the start of this thread. :)

* Though these calculations keep the closest the angle to the apparent wind as a constant, the relative magnitudes of craft speed and wind speed vary greatly depending sailing direction. So this assumption cannot be true in practice.

** cosec(x) = 1/sin(x)

Last edited: Nov 26, 2008
14. Nov 26, 2008

### rcgldr

Although an analogy has been made between prop blades following a spiral path (apparent "crosswind"), and a sail which does operate in an apparent crosswind, there is a significant difference between sailcraft and DDWFTTW or upwind carts. In the sailcraft case, the ground applies a force perpendicular to the direction of travel (against the crosswind) of the sailcraft and no work is performed at the ground interface. In a DDWFTTW cart, the ground applies a force (to drive the prop) opposing the direction of the cart and work is performed at the ground interface. I don't know how this difference affects the maximum speed of a DDWFTTW or upwind cart.

You've done the math for the theoretical limit of the sailcraft. I haven't seen or tried myself to figure out the math for the limit on a DDWFTTW or upwind cart. For high end iceboats, a Beta angle around 8 degrees corresponds to a lift to drag ratio of about 7.1 : 1, with an apparent crosswind between 5 to 10 mph translating into 35.5 to 71 mph apparent headwind for total iceboat speed 35.85 to 71.7 mph. The limit is the lift to drag ratio for a given apparent crosswind for a given sailcraft, and is about the same for upwind or downwind (depending on ground speed related drag). I'm not sure how the limit for a DDWFTTW or upwind cart should be calculated.

Last edited: Nov 26, 2008