|
Dupont has a graph they make
available in their tech literature that shows surface temperature of various
colors. The data were collected with panels in the direct sun. I
forget the exact numbers, and can't find my copy, but it seems to me that dark
colors were 70° to 90° hotter than white. Pale blue and pale yellow
weren't too bad, but everything else goes hotter from there. (Here in
Texas I drive a white pickup, which stays noticeably cooler than darker colored
cars.)
As for Ohm's law, the best way to
figure a dropping resistor for a light bulb is to hook up a hand-held meter to
measure the current draw of the bulb after it's been on for several seconds
(start-up, or inrush current is much higher than the steady draw after the
filament heats up.) Then, from E (voltage) = I (current) x R (resistance),
re-arrange the equation to E/I = R. If your current is, say, 1 amp at 14
V, the resistance is 14 ohms. To halve the current, get yourself a 15 ohm,
5- or 10-watt resistor at Radio Shack (they're big, square jobs potted in
ceramic, usually) and hook it in series with the bulb. If you want to get
fancy, you can put a wire-wound 25-ohm pot in series and have a
dimmer.
Lacking an ammeter, you can
figure it out from the bulb's wattage, if you know what it is. P (power,
in watts) = I (current) times V (voltage), and using the equation above, you
also have P = I-squared x R. So, if your bulb is 10 watts, and V is 14,
the current is P/V = 0.71 amps. Using the second equation, P/(I x I) = R =
10 / (0.71 x 0.71) = 19.8 ohms.
Again, to halve the current, put a 20-ohm
resistor in series. BTW, halving the current might not halve the perceived
intensity, so you might have to try bigger or smaller resistors to get it just
right. And don't use the little half-watt carbon jobbies -- they'll
just act like nice fusible links!
Jim Cameron
Legacy
Medina, Texas
|