REFLECTOR:A (re)volting problem

Jim Sower reflector@tvbf.org
Fri, 03 Oct 2003 17:40:22 -0400


<... I may need to start using the pitot sooner than I thought ...>
Help me out here ... Why would anyone need a heated pitot tube on a plastic airplane
anyway?  My best information is that with laminar airfoils that have been known to get
marginal in MIST (much less ICE), by the time your pitot system is threatened at all,
the rest of the airplane has long since gone ballistic.  If I'm heading straight for
the trees, out of control with a load of ice on the flying surfaces, I doubt I would
have the presence of mind to document my airspeed at the moment of impact.
Just a theory .... Jim S.

Chuck Jensen wrote:

> Dave,
>
> Well said, even if I'm not sure I could repeat it, but I get the general
> drift and concept.  In short, the popping 10a breaker is an likely an
> indication of a short, since, by your calculation (and explanation) a 5a
> breaker should be adequate, and certainly a 10a breaker may actually be
> oversized.  I'm going to do some electrical checks this evening, looking for
> a short, partial short as well as a direct measurement of amp draw if I
> don't find anything on the short-hunt.  I can always dream that it's just a
> weak breaker.
>
> The way the wind feels this evening, I may need to start using the pitot
> sooner than I thought!
>
> Chuck Jensen
>
> -----Original Message-----
> From: reflector-admin@tvbf.org [mailto:reflector-admin@tvbf.org]On
> Behalf Of Dave Black
> Sent: Friday, October 03, 2003 2:52 PM
> To: reflector@tvbf.org
> Subject: Re: REFLECTOR:A (re)volting problem
>
> Chuck,
>
> > when you go up in volts, you go down in amps. I assume the converse
> > holds true; when the voltage is cut in half, the amps double?
>
> No. This is what you'd need to do if you desire to hold power constant. In
> the
> case at hand, power absolutely will NOT remain constant, because the
> resistance of the pitot tube will not change. When you cut voltage in half,
> you also cut current in half. Voltage x Current = Power, so power reduces to
> 1/4.
>
> > So, an instrument rated for 10a at 24v is going to need a
> > 20a circuit at 12v (actual draw is approximately 80%, or 16a)?
>
> If you could reduce the pitot's resistance to maintain power this would be
> true. But since you can't, current will go down with voltage, so you'll have
> 1/4 the power. A 5a breaker will do the job.
>
> > Is that thinking correct?  Or is there some squaring and doubling going
> on?
>
> Squaring. That's because when you double the voltage across a resistance,
> the
> current doubles also. Power = voltage x current.
>
> > If the circuit has 14 gauge wire, can a 20a push button breaker be safely
> > used?
>
> No. If I recall correctly, 14 gauge wire is rated at 15 amps.
> But that's fine -- all you'll need is 5 amps to run this pitot. Use a 5 amp
> breaker.
>
> Dave Black
> Shortwing RG
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--
Jim Sower
Crossville, TN; Chapter 5
Long-EZ N83RT, Velocity N4095T