Does the wind have more force on a yacht's sail in humid conditions, as in the tropics, because of the higher water content of the air?

I would say less it is hotter therfore is expanded then cooler less humid air .

Air density would be lower, but the moisture in the air may serve to reduce permeability in a sail reducing the amount of pressure lost through the sail.

Computer says too many variables, not enough information. ;)
I'm going to have to read that New Scientist bit now. I think i've got the last one sitting at home.

Water vapour is less dense than air. You'd think it would be more because water is heavy, but it's actually less due to the molecular weights of H20 vs N2.

nebbian said...

Water vapour is less dense than air. You'd think it would be more because water is heavy, but it's actually less due to the molecular weights of H20 vs N2.

Wot he said^^^^^^

While on the topic, something I've been wanting to know for a while now is just how much force is generated by a kite? And how to calculate and express in terms other than 'fully lit' or 'going off',cranking or 'blowing it's tits off'

So a little research reveals.......

Force equals Wind pressure x Shape co-efficient x Surface area

Definitions.

Wind Pressure (or stagnation pressure, see below)

Wind pressure is pretty straightforward using the formula 1/2 mass x velocity squared. Air mass is commonly given as 1.25 kg/m^3 and velocity is measured in units of metres per second. So for a 25knot breeze we get 12.5m/sec and come up with 1.25kg /2 x 12.5^2 = 97.66 Pa of wind pressure.

The stagnation pressure is half the density of the fluid times the velocity
squared. For example, in Sydney the wind velocity used for design (100 year
return wind) is 47 m/s.

therefore:
velocity 47 m/s
density 1.2 kg/m3
pressure 1.325 kPa

Shape Coefficient
The Shape co-efficient refers to the resistance that the shape of the object presents to the wind. Engineering standards specify coefficients for various shapes, for example: parabolic dish equals 1.2 and a flat plate equals 2.0

The actual force applied to the object in the wind is dependant on the shape
of the object and how much crossection it presents to the wind flow. The
effect of the shape and orientation of the object is quantified by
coefficents. There is a drag coefficent and a lift coefficent. For example
an aeroplane's wing is designed to have a low drag coefficent and a high
lift coefficent which acts in a vertically upwards direction.

Antenna elements, on the other hand are not designed for their
aerodynamic properties and actually usually have a larger drag coefficent
and a smaller lift coefficent. The easy and conservative approach to this is
to assume a drag coefficent of 1.0 and that it can act in any direction as
wind and the turbulance it casuses can act in almost any direction.

This is a pretty hard one to specify without putting your kite in a wind tunnel and even then there are an infinite number of combinations of the drag coefficient and the lift coefficient. For these purposes I'll just use a coefficient of 1 but being aware that it's potentially 2 if you liken it to a flat plate that is perpendicular to the wind direction (as in your kite directly downwind) and possibly more with the coefficient of lift generating even greater forces.

Surface Area
Surface Area equals......surface area
Let's say 12m for arguments sake.

Now let's see if we come up with anything that is remotely realistic!

Pressure 97.66 x Shape coefficient 1.0 x Area 12m = 1172N

Convert N to Kgs, 1172/9.81 = 119kg

Hmmmmm.....sounds reasonable

I weigh 87 kgs so it's enough to lift me off the water and there's some spare to get me some acceleration too.

J-P

There's so many variables here that it's pretty much only ever going to be a theoretical exercise/stab in the dark.

I like the idea of putting a load cell at the chicken loop. Add to that a GPS so vertical and horizontal velocity and acceleration can be measured, have them both synched and put the data into some whizbang engineering software and see exactly what forces are at work - or in this case at play...

Are there any brainy computer technophiles out there?

What's CFD? Contracts for difference? Computational Fluid Dynamics?

I give up, my brain hurts.

J-P

SamS said...

If you're after a rough guide to what forces can be provided by a kite, at the apex of a jump (rider acceleration = 0) the upward force of the kite will equal the weight (f=ma) of the rider (and his board etc). So an 80kg rider will need an uplifting force of 80(kg)*9.8(m/s/s)= about 800Newtons. Or 80kg up

All those calculations for "80kg up" ...........

Well i did say rough. The kite only has a small mass compared to the rider