Return-Path: Sender: (Marvin Kaye) To: lml@lancaironline.net Date: Thu, 06 Nov 2003 13:53:46 -0500 Message-ID: X-Original-Return-Path: Received: from mail.indian-creek.net ([209.176.40.9] verified) by logan.com (CommuniGate Pro SMTP 4.1.6) with ESMTP id 2712480 for lml@lancaironline.net; Thu, 06 Nov 2003 11:28:23 -0500 Received: from pavilion (sl40.du.indian-creek.net [209.176.40.56] toucan@78055.com) by mail.indian-creek.net with SMTP (IOA-IPAD 4.0rc2/96) id 5U5ZL00 for ; Thu, 06 Nov 2003 10:28:37 -0600 X-Original-Message-ID: <004501c3a483$2517e5a0$3828b0d1@pavilion> From: "Jim Cameron" X-Original-To: "Lancair Mailing List" Subject: Paint colors; Ohm's Law X-Original-Date: Thu, 6 Nov 2003 10:29:25 -0600 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0042_01C3A450.D99BA120" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 5.00.2615.200 X-MimeOLE: Produced By Microsoft MimeOLE V5.00.2615.200 This is a multi-part message in MIME format. ------=_NextPart_000_0042_01C3A450.D99BA120 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable 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=B0 to 90=B0 = 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) =3D I (current) x R (resistance), = re-arrange the equation to E/I =3D 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) =3D I (current) times V = (voltage), and using the equation above, you also have P =3D I-squared x = R. So, if your bulb is 10 watts, and V is 14, the current is P/V =3D = 0.71 amps. Using the second equation, P/(I x I) =3D R =3D 10 / (0.71 x = 0.71) =3D 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 ------=_NextPart_000_0042_01C3A450.D99BA120 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

Dupont has a graph they = make=20 available in their tech literature that shows surface temperature of = various=20 colors. The data were collected with panels in the direct = sun. I=20 forget the exact numbers, and can't find my copy, but it seems to me = that dark=20 colors were 70=B0 to 90=B0 hotter than white. Pale blue and pale = yellow=20 weren't too bad, but everything else goes hotter from there. (Here = in=20 Texas I drive a white pickup, which stays noticeably cooler than darker = colored=20 cars.)

As for Ohm's law, the = best way to=20 figure a dropping resistor for a light bulb is to hook up a hand-held = meter to=20 measure the current draw of the bulb after it's been on for several = seconds=20 (start-up, or inrush current is much higher than the steady draw after = the=20 filament heats up.) Then, from E (voltage) =3D I (current) x R = (resistance),=20 re-arrange the equation to E/I =3D R. If your current is, say, 1 = amp at 14=20 V, the resistance is 14 ohms. To halve the current, get yourself a = 15 ohm,=20 5- or 10-watt resistor at Radio Shack (they're big, square jobs potted = in=20 ceramic, usually) and hook it in series with the bulb. If you want = to get=20 fancy, you can put a wire-wound 25-ohm pot in series and have a=20 dimmer.

Lacking an ammeter, you = can=20 figure it out from the bulb's wattage, if you know what it is. P = (power,=20 in watts) =3D I (current) times V (voltage), and using the equation = above, you=20 also have P =3D I-squared x R. So, if your bulb is 10 watts, and V = is 14,=20 the current is P/V =3D 0.71 amps. Using the second equation, P/(I = x I) =3D R =3D=20 10 / (0.71 x 0.71) =3D 19.8 ohms.

Again, to halve the current, put a = 20-ohm=20 resistor in series. BTW, halving the current might not halve the = perceived=20 intensity, so you might have to try bigger or smaller resistors to get = it just=20 right. And don't use the little half-watt carbon jobbies -- = they'll=20 just act like nice fusible links!

Jim Cameron

Legacy

Medina, Texas

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