# Computing the force generated by a sail on a boat

This is a follow up question to the question I asked yesterday: Problem understanding basic sail mechanics

I am focusing purely on the Newtonian aspect of motion and I am computing this as follows:

First I compute the relative wind ($v_r$) velocity by subtracting the ships velocity ($v_s$)from the true wind velocity ($v_t$)

$v_r = v_t - v_s$

Next I take the normal of the sail (in to the wind), $n_s$, and "bounce" the wind off it, computing the out velocity of the wind, $v_{out}$

$v_{out} = v_r - (2 * v_r \cdot n_s) * n_s$

Next I compute the change in wind velocity, $\Delta v$

$\Delta v = v_{out} - v_r$

This is linearly proportional to the force the sail exerts on the air, hence $-\Delta v$ is linearly proportional to the force exerted on the boat.

Next, I compute the keel force that offsets the horizontal part of the sail force, allowing the boat to move against the wind.

I take $n_k$ to be the normal of the keel going in the opposite direction to the sail force and perpendicular to the boat heading. I then compute the keel force, $F_k$ to be

$F_k = n_k * |-\Delta v|$

i.e. the magnitude of the sail force in the direction of the keel (perp to direction of boat facing).

To compute the resultant force, I of course sum these two forces.

Firstly, is all that correct. Clearly it is missing resistance from the water, but aside from that, do the calculations correctly reflect the answer given in my previous question.

Secondly, what parametrises these forces and how. What I am seeing at the moment is, when sailing with the wind, the boat accelerates very quickly (up the the speed of the wind). However, when I try to sail near to the wind (is that the correct phrase? for when my boats forward and the direction of the wind are near and opposite) although I can see the apparent wind speed increasing, the force applied is still very small. Is this was you would expect to see?

Seems to me that you are trying to calculate the lift on an airfoil from first principles using only Newton's laws of motion. Chris Waltham did exactly that in his paper "Flight without Bernoulli". You might want to check it out.

http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/fluids/fly_no_bernoulli.pdf

Note that this is a non-standard way of doing the aerodynamic analysis. His conclusion:

We have used a very simple physical model relying only on Newton’s second law to reproduce all the salient features of a rigorous fluid dynamical treatment of flight... The model has its limitations; we cannot calculate real performance with it.

So, while your approach is reasonable, there's a reason why aerodynamic engineers take a different tack.

You also might want to check out Charles Eastlake's paper "An Aerodynamicist’s View of Lift, Bernoulli, and Newton" http://users.df.uba.ar/sgil/physics_paper_doc/papers_phys/fluids/Bernoulli_Newton_lift.pdf

His take is:

Measuring lift by measuring the increase in downward vertical velocity in the flow coming off the trailing edge of the airfoil is conceptually possible. This downward velocity is definitely there and is known as downwash. I have never heard of anyone actually measuring it with sufficient precision to calculate lift, not because it is physically unsound but because it is not a practical experiment. It is not practical because the downwash is distributed over the entire flow field downstream of the trailing edge, and it would thus be very difficult to measure enough data points to integrate the distribution accurately.

Not to discourage you, but I think you may be heading towards a dead end. Anyway, check out the papers above - I think you'll find them interesting.

This is really a "comment" to Ernie, but when I tried to post it that way, I was informed "You must have 50 reputation to comment".

Sounds like you're also a sailor, as does the op T.Kiley. If you're really looking for analytical treatments, try googling the terms, sail navier-stokes. I'm seeing a lot of published stuff.

But for practical sailing purposes (like phrf racing), my favorite semi-physical treatment is "The Art and Science of Sails" by Tom Whidden and Michael Levitt

http://www.amazon.com/The-Art-Science-Sails-Construction/dp/0312044178

Doesn't mention Navier-Stokes, but it's the most technical/physical treatment of sail theory I've come across. Trying to develop that kind of theory, ab initio, in this kind of forum, is bound to be too naive a treatment (take a look at some of those google results to get a feel for what's really needed).

• So if you don't have 50 rep to comment, then why don't you get 50 rep? this is not an answer and should be deleted. – TanMath Nov 9 '15 at 7:41

Next I take the normal of the sail (in to the wind), n s , and "bounce" the wind off it, computing the out velocity of the wind, v out

Nope. First you have to calculate the angle of the wind wrt the sail area. This depends both on the velocity vectors of the boat path and the angle of the sail wrt the boat. Then you compute the effective area of the sail to the wind, proportional to the cosine of the angle. The component of wind flow towards the sail is the cosine of the angle times the effective area.

All this assumes no aerodynamic behavior of the sail, which is not true. See https://www.youtube.com/watch?v=gNaEX6EGg7I for a start.

• So I need to compute the angle between the apparent wind and the normal of the sail and scale by the cosine of that (as that is the area of the sail facing the wind)? Ignoring aerodynamic stuff (I'm hoping the impact will be negligible...) is the rest correct? – T. Kiley Jul 10 '14 at 8:56
• @T.Kiley: I'm afraid you need to know how the sail functions as an airfoil - namely what is its chord, its angle of attack, and its coefficients of lift and drag as a function of those. – Mike Dunlavey Jul 11 '14 at 19:05
• You have also failed to calculate exactly how much force a change in wind speed produces. This is fairly easy, and there is probably an on-line calculator, but it will depend on both the wind speed and the sail dimensions. – WhatRoughBeast Jul 11 '14 at 19:10

A sloop has two sails, the jib and the main. There's a slot between the trailing edge of the jib and the leading edge of the main. When sailing close-hauled into the wind, you try to get the air flowing off the leeward side of the jib (side away from the wind) to go through the slot and on to the windward side (side toward the wind) of the main, because the air accelerates out of the slot. This creates a Bernoulli effect across the front of the main, and the lift caused by the lower pressure on the windward side of the main pulls the boat against the wind.

I think you need to add a force from the Bernoulli factor into your Newtonian model. The Newtonian model will work for a boat sailing downwind, but there are more forces than you have taken into consideration, involved when sailing upwind.

There's a simple way to estimate the force generated by a sail when sailing into wind, using Newtonian mechanics and the equation: Force = ma = m/dt x dv; (where a = dv/dt).

In summary, first estimate the mass of air displaced by the sail each second ('m/dt'), the mass flow rate. This depends on the airflows around the sail (volume of air displaced and air density), which is affected by things like: the apparent wind speed, size & shape of the sail (e.g. height and width), as well as the angle-of-attack, .... Also, the Coanda effect helps redirect air on the leeward side of the sail.

Then multiply 'm/dt' by the velocity that this airflow is re-directed onto the wind ('dv'), or the velocity of the wind out of the sail. i.e. The wind provides something for the air re-directed by the sail to push against.

This action by the sail then creates the backward force (Force = ma = m/dt x dv; where a = dv/dt). The 'equal and opposite' forward force generated (the reaction) pushes the sailboat forwards. Simple.

In simple terms, the sailor controls the force generated by the sail by changing the angle of attack of the sail to the wind. This affects 'm/dt', 'dv' and the direction that the forward force is created.

The same principles of physics applies to how a rudder redirects water to create a force that helps to steer the boat by pushing the stern in a given direction.

For example, a 12m vertically high sail, that re-directs all the air 0.5m horizontally either side of the sail (so 1.0 m total), sailing into a strong 10 m/s (36 km/hr) apparent wind will displace a volume of air of 120 m3/s (12m x 1m x 10 m/s = 120 m3/s).

m/dt = Volume/dt x Air Density.

Then, using standard air density of 1.2 kg/m3; this provides a mass flow rate of 144 kg/s (1.2 kg/m3 x 20 m3/s = 144 kg/s).

Then, using the equation: Force = ma = m/dt x dv.

Assuming that the velocity of the wind out of the sail ('dv') is 8 m/s, then this action creates a backward force of 1,252 N (144 kg/s x 8 m/s = 1,152 kg m/s2 = 1,152 N).

The 'equal and opposite' forward force is also 1,152 N.

These numbers are just an illustration of how to calculate the Newtonian forces.

So, there's no need for complex equations like Navier-Stokes (which are based on Newtonian mechanics anyway).

There's more details here: https://www.researchgate.net/publication/335505038_Newtons_laws_explain_sailing_into_wind