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, Fri, 18 Oct 2013 10:36:37
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Rich,
And the last comment had me to apply to the urban dictionary and the wiki. I think from now on they no longer need all the buussuu stuff since there is the Saabnet with the C900 bulletin board starring.
Zig
P.S.
On the topic stuff - the idea appears to work perfectly on the old bulb-styled (pill-styled? whatever) fuses which were cut off and abandoned in early days of 80s era. With the modern design stuff one either has to break thru the insulation or to unbolt all the covers and plates and holders to get to the raw (nude?) unprotected contacts.
But the idea itself is a really good one and will help e.g. in a case of half-broken wire: consider a wire on the injector #N to be half-broken, it will give you all those troubles which a bad injector usually gives. With the described technique - and assuming you're able to get to the ECU-end contacts and get thru the silicone protective insert on the injector side which protects the contact from outside environment and the multimeter leads are long enough to reach both the injector and the ECU at the same time - you'll see a significant voltage drop there, especially if you compare it to the voltage drop across its neighbours, inj. #N-1 and #N+1 . Actually, you'll see a significant voltage (*) across that wire in-between the multimeter leads, the wire which connects the injector to the ECU. A significant voltage across a copper - and even a aluminium or even an iron - wire means the wire has a damn huge resistance (electrical resistance fwiw). Which in turn means that either the wire is semi-broken or its insulation is damaged and the core is heavily corroded. In either case it isn't the injector to be replaced, but the wire to be fixed up (e.g. replaced completely, replaced partially, left in place but electrically replaced with another wire run in parallel with this one, etc).
(*)
Voltage finitus potential difference est. Consider the following circuit:
A: - B - C - D - :E
where
A, E: mains contacts
B: a bulb
C: an iron
D: a soldering gun
- : a wire
There is a potential difference between A and E and there in your dmocracy it goes for around 110V (alternating, root mean square); here in Mordor we have 220 volts (and bigger is alsways better, you know! suckers :). Every element en route has some resistance thus there happens some voltage drop across its contacts: say, on B terminals there drops 20V and on iron terminals there drops 190V and there on the soldering gun there drops 10V. The total voltage drop for all the elements is 20V+190V+10V=220V i.e. the voltage (potential difference) across A and E. All those 20, 190 and 10 volts are the voltages you'll measure if you stick the multimeter leads into terminals of appropriate element.
But if you stick one lead at the mains terminal A and the other lead goes to the iron (which is C on the scheme), then on the left terminal you'll get 20V (the voltage drop across bulb B) and on the right terminal you'll get 210V which is the voltage drop across bulb B and the iron C.
In this example the wires between elements are considered to be 100% and have zero resistance.
posted by 188.134.4...
, Fri, 18 Oct 2013 10:36:37, Fri, 18 Oct 2013 12:40:12, Fri, 18 Oct 2013 17:56:42
, Fri, 18 Oct 2013 19:53:09, Sun, 20 Oct 2013 05:12:26, Sat, 19 Oct 2013 19:51:21
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