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Resurrecting my 200 |
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75-200 ![]() Bronze Level ![]() ![]() Joined: 19 Jan 2012 Location: PEI, Canada Points: 49 |
![]() ![]() ![]() ![]() ![]() Posted: 04 Nov 2016 at 6:39pm |
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I have a late model 200 that my father bought as a low hour used tractor in the late 70's. He part-time farmed with it until 1992. In 93, I came home from university to farm full time and put her to work. We had an up and down relationship I have to say (me and the tractor). Mainly PD related. Long story short, she has some issues today though none fatal I believe. I am thinking I would like to bring her back to life.
I have been a long time follower of this forum and I'm thinking it would be fun to tap into the vast resource of knowledge here to bring her back to life. With your help, I know it can be done, especially on the PD side of things. I would certainly be sharing pics and updates for your help along with some background history on the tractor. I'm looking forward to it. Dan
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DougG ![]() Orange Level ![]() Joined: 20 Sep 2009 Location: Mo Points: 8266 |
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Welcome, you have come to the right place
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75-200 ![]() Bronze Level ![]() ![]() Joined: 19 Jan 2012 Location: PEI, Canada Points: 49 |
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First question, regarding batteries. I pulled a pair of 12v batteries out of her (no longer good) that had the + posts hooked together with a lead going to the starter of course. One - post went to ground on the frame and the other went to ground on the cab.
I have a pair of 6v batteries that I pulled out of my White 2-110 that are good. One + is hooked to the other batteries -. One battery has its + going to starter and the - grounding on the frame of course. My question is, can I substitute the pair of 6v into the Allis? Series vs parallel is confusing to me!
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bigal121892 ![]() Orange Level ![]() Joined: 05 Jan 2010 Location: Nebraska Points: 808 |
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Simple answer, no. The longer answer; a 6 volt battery, will produce about 900 cold cranking amps, so two 6 volt batteries in series, will produce 12 volts, with 900 cold cranking amps. A 12 volt battery, will also produce about 900 cold cranking amps, but with 2 12 volt batteries in parallel, you get 12 volts, but 1800 could cranking amps. With your two 6 volts in series, I don't believe you will have enough cold cranking amps to properly start the tractor. Now if you have four 6 volt batteries, you could wire two in series, and the other two in series, and then put the two pairs in parallel, that would work.
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CrestonM ![]() Orange Level ![]() ![]() Joined: 08 Sep 2014 Location: Oklahoma Points: 8452 |
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I don't even understand that! I don't know very much at all about wiring (Not sure what an Ohm is, or even what its symbol is) All I know is when the test light comes on, there's current! Lol
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bigal121892 ![]() Orange Level ![]() Joined: 05 Jan 2010 Location: Nebraska Points: 808 |
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Let's take two 6 volt batteries, and we will take the negative terminal of the first battery, and attach it to the frame of the tractor, this will be our ground. To series the batteries, the positive terminal of the first battery, will then be attached to the negative terminal of the second battery, with the positive terminal of the second battery attached to the load, in this case, the starter. When batteries are placed in series, the voltages add, is this case, two 6 volt batteries are placed in series, measuring from the negative terminal of the first battery, to the positive terminal of the second battery, you would read 12 volts. If you take your two 6 volt batteries, and wire them in parallel, the two negatives are tied together, and attached to the frame, and the two positive terminals, are tied to the load. reading from either of the negative terminals, to either of the positive terminals, or the load terminal, the reading would be 6 volts, but the amps of each battery would be added together. "Ω" the Greek symbol omega, is the symbol for ohms, which is a measurement of resistance. It takes 1 volt to push 1 amp through 1 ohm of resistance. The opposite of resistance, is conductivity, which is expressed in Siemens. You would convert from resistance to Siemens, with this formula: 1/resistance. When your test light comes on, all it is really indicating, is that there is a voltage drop between your source, and ground, yes there is current flow through your light, because it is lit. But, you can have voltage potential, and not have any current flow. |
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DrAllis ![]() Orange Level Access ![]() Joined: 12 Sep 2009 Points: 21468 |
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I wouldn't waste my time using 6 volt batteries living in Canada !! Two 1000 CCA 12 volt batteries wired the factory way is your best bet for starting way up norda here.
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Hubert (Ga)engine7 ![]() Orange Level ![]() ![]() Joined: 12 Sep 2009 Location: Jackson Cnty,GA Points: 6411 |
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The resistance affects how much amperage a wire will handle, e.g. corroded battery cables may not let enough current through for the starter to work properly even though you have a good battery. You will not see a voltage drop unless a load is placed on the wire. Had that to bite me in the rear on a compact Ford; had battery voltage at the ignition switch but no start even after replacing the switch. Turns out it was a fusible wire supplying the switch was bad and not letting the current (amps) through.
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Just an old country boy saved by the grace of God.
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DSeries4 ![]() Orange Level ![]() ![]() Joined: 12 Sep 2009 Location: Ontario, Canada Points: 7451 |
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Yep, at this time of year in the Great White North, you want 2 good 12 volt batteries to roll over that bear. You will want good batteries to use the manifold heater before as well. Keep the manifold heater on for a minute or longer then give it a crank. |
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'49 G, '54 WD45, '55 CA, '56 WD45D, '57 WD45, '58 D14, '59 D14, '60 D14, '61 D15D, '66 D15II, '66 D21II, '67 D17IV, '67 D17IVD, '67 190XTD, '73 620, '76 185, '77 175, '84 8030, '85 6080
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