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Re: [TowerTalk] CAT 5 as Rotator cable, sprinkler wire

To: towertalk@contesting.com
Subject: Re: [TowerTalk] CAT 5 as Rotator cable, sprinkler wire
From: Jim Lux <jimlux@earthlink.net>
Date: Fri, 18 Jun 2004 14:25:20 -0700
List-post: <mailto:towertalk@contesting.com>
At 05:05 PM 6/18/2004 -0400, Pete Smith wrote:
At 03:55 PM 6/18/2004, wa3gin wrote:
Some day you should go measure the voltage at the rotor and compare it to the voltage at the control box. The results might lead you to upgrade the guage of
wire you're using. Publish the results for the list.


Anyone who is interested in this and hasn't already done so should take a look at the copper wire table in any ARRL Handbook. It gives ohms/100 feet for all AWG sizes, which makes it very easy to figure out whether you have enough copper for the motor leads.


73, Pete N4ZR


Don't even need the handbook:

AWG 10 = 1 ohm/1000 ft  (0.1 inch in diameter)
every 3 gauges is half the area, so twice the resistance:

AWG 16 is therefore 4 ohms/1000 ft.

every 6 gauges is half the diameter, so AWG 16 is 50 mils in diameter.


A more practical question to be answered here is "how much voltage drop/power dissipation is acceptable". I suspect that applying a rule like "no more than 2% drop" (as used in AC wiring) is overly conservative.


Say the motor runs at 24V and draws a couple amps. Say you're a real cheap guy, and want to use some of that AWG 22 wire you have lying around for the 200 foot run. 400 ft of AWG 22 will have a resistance (@ 16 ohms/1000 ft) of 6.4 ohms. You'll drop about half the voltage in the wire, but, I'll bet the motor still turns, just slower. (The brake, if there is one, might not disengage, though.)

The large voltage drop might also cause problems if there's some sensing circuit (i.e. a pot) that shares a wire coming back.

You could boost the voltage at the sending end to 36 volts to make up for it too (although, you have to watch out, because if the load varies (which it does), then the current varies, changing the voltage drop. At light loads, you might wind up with too much voltage at the far end.

Then, there's also the "insulation melting" problem. 2 amps * 6.4 ohms is about 26 Watts, disspated over the 200 foot length of your cable, or about 0.13 W/ft. Probably not going to melt the insulation, but the wire might get warm.

Just as a data point, I've noticed cheap 16 AWG extension cords getting warm at loads of around 10 Amps. That's 0.4W/ft. (and, ahem, the voltage drop in a 100 ft cord is 8 Volts, well exceeding the 2% guideline)




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