## difference in the strength of cold/dry vs warm/humid wind

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plummet
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### Re: difference in the strength of cold/dry vs warm/humid win

here's some calcs i did a while back on another forum.

let us use a couple of formula's do calculate that actual dynamic pressure the wind can applied to the kite.

Dynamic pressure is a defined property of a moving flow of gas or liquid and can be expressed as

pd = 1/2 ñ v2 (1)

pd = dynamic pressure (Pa)
ñ = density of fluid (kg/m3)
v = velocity (m/s)

Denisity can be calculated with the idea gas law.

p=p/(RT)
p=Pressure
r=individual gas constantj/kgdegk)
T=Absolute temp K

So. . . . .

16 deg density = 1.22kg/m2
32 deg desity = 1.15kg/m2

lets use 15m/s (30 knots) as a constant is our dynamic pressure calc.

dynamic pressure 16 deg = 137
dynamic pressure 32 deg = 130

Thats 5% difference.

So what else effects the density of the air?
The pressure and the humidity. Pressure is easy.An increase in pressure increases the density.

Humidity is somewhat more confusing. Water vapor is actually lighter than air. So the more humid the air the less dense the air is. .... that i can't be bothered calculating.

The end result is this. Because the pressure increase is 1/2 mass x velocity squared the impact of small changes in temp/pressure/humidity is small. The greatest impact is a change in velocity.

ok i cheated for the humidty calc and found this calculator on the interweb

http://wahiduddin.net/calc/calc_da_rh.htm

The figure i'm interested in is the the density kg/m2.

the change in denisty from 0% dry air to 100% humidity changes dependant on pressure and temp.

But the highest change i could see was a 2% change in density. . ...

Really it means it has very little impact on the final figures.

If we stacked all the variables up in a worst case scenario. so 20Deg temp swing. high pressure dry air at 10 deg c and low pressure moist air at 30deg c then you stack up to a 12% reduction in force at the high temp.

PS Peter,,,, i'm not an engineer so feel free to tear my calcs or logic to peices if its wrong.

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### Re: difference in the strength of cold/dry vs warm/humid win

Exclusive of the humidity, it is really about altitude. Generally, higher altitude means less air density which means less power. Of course, as mentioned previously, more moisture (higher humidity) also means less power. The formula that edt gave is the way to figure it, but if you are not a mathmagician, you can get a basic idea of air density from checking a local aviation forecast for Density Altitude. This is what airplanes and helicopters use for air density when computing performance at a given altitude.

Density altitude is pressure altitude corrected for temperature. Pressure altitude is a standard altimeter setting (29.92 QNH or inches of mercury / 1013.2 millibars) at your elevation. This may be higher or lower than your actual elevation on the earth based on high or low pressure systems in the area.

Simply stated, density altitude which is significantly higher than your elevation will mean less power. This will occur on hot days. Also, 7kts of wind at sea level will have more power than 7kts at 5000ft elevation because the air is more dense at sea level.

Holy crap. While I was typing my simple explaination, several people posted alot of really good information. Please defer to the numbers. The math never lies.

SSK
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### Re: difference in the strength of cold/dry vs warm/humid win

Plummet,
For a non-engineer your explanation is accurate, and much more descriptive than most. Although temperature, pressure, density are all related, the ideal gas law as other have presented does not really describe the situation. First of all the ideal gas law, as you pointed out, only describes static pressure, and does not really describes the forces on a kite. Others have presented some inaccuracy in interpreting the ideal gas law and what it means to the forces on a kite.
PV = nRT
P- Pressure
V- Volume
n- mass of fluid
R - a constant for a given gas
T – Temperature

To use this form of the equation and say that as temperature doubles the pressure doubles is incorrect. This would only be true if the volume of air was maintained constant. But in the open environment that is not true
An increase in temperature (which is a measure of the activity and movement of atoms and molecules) produces a corresponding increase in pressure (which is a measure of the force of the atoms and molecules), only if the volume of the air is held constant. It does not work that way in the atmosphere, as the atmosphere is not a constant volume. An increase in temperature in a high pressure area as it is warmed by the sun causes expansion of the area both horizontally and vertically, thereby resulting in only a minor increase in pressure. However, the pressure increase is measurable, because the atmospheric volume cannot respond immediately to the increase in temperature (that's a lot of air to move).
Air pressure is normally greater during the late summer and early Autumn in temperate latitudes and less in the winter (opposite to what you thought). That is because high pressure areas during the summer are warmer and build up more air above them. More air above the surface means more weight, which translates into greater pressure.
David R. Cook
Atmospheric Section
Environmental Research Division
Argonne National Laboratory
However as you pointed out the ideal gas law can be rewritten in terms of density, which then normalizes the equation for the volume change, and more useful in understanding the concept

P = (rho)*R*T
Or in terms of density
Rho = P/(R*T)
P - Pressure
rho - density of air
T - Tempurature

In that form you see that high pressure results in high density, and high temperature in lower density.

A kite flys for two reasons: Lift, and deflection. Lift is the component of force perpendicular to the flow of wind caused by the wind air flowing over the surface. Deflection is basically the force of the wind against a surface. Here is a good description
Dynamic pressures such as lift and the deflection force are all extensions of the Bernouli equation as you pointed out.

Dynamic pressure
q = 1/2 * rho * v^2
q - dynamic pressure
rho - density
v - velocity

The equation for Lift (force) is
L = 1/2 * rho * v^2 * A * C(l)
L - lift
rho - density
v - Velocity
A - Planform area of wing
C(l) - a constant for a given angle of attack

The equation for force of wind against a surface is something like
F = 1/2 * rho * v^2 * A * Cd
A – Area of surface
Cd - coefficient of drag

It is therefore accurate to say that the force on a kite is directly proportional to density of air for a given velocity. Although temperature has a relation to density in the atmosphere it is not always direct. So the focus should not be on temperatures, if you want to be accurate it should be concerned with density. If you missed Dr Makani's link, you can easily calculate the density based on Temperature, Atmoshperic Pressure (static), and Dew Point (water vapor) to do comparisons of two spots.

http://www.gribble.org/cycling/air_density.html

You can then compare accurately simply by getting density from the calculator and comparing the ratio to see the increase or decrease. The calculator is nice because you can provide metric or imperial units and the values are normally reported for any weather station.
Using the calculator and real data from the National Buoy Data Center, I compared more realistic values. The difference of Cape Hatteras in July at noon, vs Central California Coast at noon for May. I took the averages for the month at noon which are:

CA
P = 1015.13
T = 11.3
DP = 9.9375
(Density = 1.24 )

NC
P = 1015.08 millibar
T = 27.27 celsius
DP = 23.55 celsius
(Density = 1.16 )

That comes out to be a 6.5% drop in force from NC to CA. I ran extreme numbers for the tropics versus snow kiting and got about 12%. Some people suggested you can just take the ratio of temperature. That is not correct, that would suggest a 60% drop.

As been pointed out this comparison works only for the same quality/consitency of wind. Since the force of wind is proportional to the square of the velocity, a small change in the velocity or quality of the wind will be much more noticeable.

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### Re: difference in the strength of cold/dry vs warm/humid win

edt
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### Re: difference in the strength of cold/dry vs warm/humid win

We could get more technical, and if we do, the ideal gas law is sufficient to derive every other equation. For instance, altitude you start with p= - (1/g)dP/dz, do the integration and you get z=-RT/g log(P/P0). For the humidity P1V1=n1R1T1, P2V2=n2R2T2, P(total)=N1R1T1/V1 + n2R2T2/V2.

Humidity has a negligible effect on kites. Altitude is most important. But if you kite at sea level you can ignore it. Kinetic energy is .5mv^2 so while it is true kiters feel wind twice as fast to be four times stronger, it is also true that twice the density is just twice the force on the kite. It's not that tricky.

Temperature is the most common effect and it's linear. Be careful remember it's in Kelvin.