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
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
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.
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.
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:
P = 1015.13
T = 11.3
DP = 9.9375
(Density = 1.24 )
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.