On Fri, 2010-08-27 at 22:44 +1000, Luke O'Donnell wrote: > Valid points. I suspect that you wouldn't be necessarily looking for > just hot spots, but perhaps looking at rapid decreases in temperature. > I don't think you'd see them. For starters, ground temp differences are a lot bigger than air temp differences. This is both because different surfaces have quite different reflectance and heat capacity (low reflectance and low heat capacity give a real hot spot (asphalt parking lot) while high reflectance and low heat capacity give the opposite) and there's mixing in the air mass, but none in the ground.
As I said, if you're looking down on that sort of patchwork, the difference in emitted IR from air with its small temp difference between lift and sink combine with its low emitted energy mean it simply can't be resolved. I'm doubtful that a camera could resolve the low contrast between thermal and non-thermal even looking horizontally so the thermal is seen against air rather than ground. Don't forget that the air's thermal background is also uneven and shows similar contrast differences from distant thermals to what we're trying to use to locate a nearby one. > Depending on the level of achievable contrast as well as how muchj > temp trace there is when a thermal moves though, it might be possible > to visually 'see' the thermal tracking over the ground, cooling it as > it goes. > Again I'm doubtful, simply because the heat capacity of air is very low compared with any solid. This means that the temp change in a bit of ground that warms a parcel of air will be proportionately lower than the tempo change in the air parcel. For a given heat exchange we can write: dT1 * C1 = dT2 * C2 - (1) Rearranging (1) gives dT1 = dT2*C2/C1 - (2) where dT is the temp change and C is the volumetric heat capacity. I think this is best used rather than heat capacity per gram for rough calculations comparing a gas with a solid. Air's volumetric capacity is 0.001297 vs 2.17 for granite (figures from Wikipedia's article on heat capacity). Plugging these into equation (2) and we see that a 2C increase in air temp (0.7%, not 8% since we use degrees K, not degrees C for this type of calculation) corresponds to a change of 0.007*0.001297/2.17 which is 0.0004184% or 0.0012 degrees C. You'd need a really good lab set-up to measure that sort of temperature change. It not anything you'd attempt outdoors with handheld or portable equipment. Now, how does a 2C temp change affect the radiance of a parcel of air? This is relevant because its brightness is directly proportional to an object's radiance. The Stefan–Boltzmann law applies, which says that the emitted radiation power varies as the 4th power of its absolute temperature, so if we warm air by 2C at 25C the difference is 300**4 - 298**4 or an increase in brightness of 2.7% - again not a lot of difference. By comparison my hand at 28.5C is 23% brighter than ground at night (15C). > Still, i would be suprised if it would be nearly as easy to use IR to > detect changes in temperature of air, as opposed to changes in > temperature of the ground. > Exactly so, and that is even without considering differences in the amount of radiation the ground emits in comparison with air. Martin ------------------------------------------------------------------------------ Sell apps to millions through the Intel(R) Atom(Tm) Developer Program Be part of this innovative community and reach millions of netbook users worldwide. Take advantage of special opportunities to increase revenue and speed time-to-market. Join now, and jumpstart your future. http://p.sf.net/sfu/intel-atom-d2d _______________________________________________ Xcsoar-user mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/xcsoar-user
