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Hello and HELP!!!!

2 days ago after blowing snow for about 2 hrs, I drove my tractor back in my heated garage.

This morning I wanted to do some more snow blowing.... my tractor refused to start.

Symptoms: weak dash lights and rattle coming from the bottom left console when I turn the key to START.

What I did:
Check battery: voltage fine
Tried to Charge the battery but the charger says it is fully charged.
Check and clean main ground wire (frame) all OK
check and clean battery posts and connections all ok
Still can't start..... weak light and rattle.

Then I decided to jump start the tractor with my car to isolate a very likely problem with the battery.

With the jumper cables connected I turn the key to ON..the dash lights are OK, no more rattle.
Then I turn the key to START... the starter engage but the motor turns for just a few second(1 or 2) and then stop like if I was missing power.
The cables I use for boosting are almost new and my car's battery is in perfect condition, I also kept my car running while boosting. I also tried an other set of jumper cable.... same problem.

What would be the most logical approach for my tractor'sproblem, I don't really understand what is happening.

I believe there is a problem with the tractor's battery but why is it that the tractor doesn't crank with my car's battery??

Thanks in advance for your help
 

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You need a new battery. They can fail with absolutely no warning at all. Have someone put a voltmeter on the battery posts while you attempt to start it. If the voltage goes below 9 volts or so, it's bad.
 

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You need a new battery. They can fail with absolutely no warning at all. Have someone put a voltmeter on the battery posts while you attempt to start it. If the voltage goes below 9 volts or so, it's bad.
Yes, I understand and also think the battery is faulty but why is is that I can't get my tractor to crank?? when I jump it???
 

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Yes, I understand and also think the battery is faulty but why is is that I can't get my tractor to crank?? when I jump it???
Sometimes batteries have an internal defect that gives them a terrible death. Jump starting won't always work. Only remedy is to replace it.
 
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Yes, I understand and also think the battery is faulty but why is is that I can't get my tractor to crank?? when I jump it???
Literally, Ohms law. If the battery is toast, then you are expecting to supply all the power required for cranking though the jumper cables. Typical auto jumper cables are only 4 AWG. Coupled with their relatively long length, they drop too much voltage with the starter current draw. Jumper cables are meant to provide an assist to a weak battery, not replace it altogether.

Al
 

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Literally, Ohms law. If the battery is toast, then you are expecting to supply all the power required for cranking though the jumper cables. Typical auto jumper cables are only 4 AWG. Coupled with their relatively long length, they drop too much voltage with the starter current draw. Jumper cables are meant to provide an assist to a weak battery, not replace it altogether.

Al
I'll second this. If the battery lost a cell or shorted internally it can keep the amperage and voltage from reaching the starter even with jumper cables or a charger connected.
 

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This is all true, when my battery failed it took two running vehicles connected to the tractor to get enough umph through the battery to start it. One vehicle just made the tractor turn over a few slow times then click. After throwing a second set of cables onto the battery, the tractor fired up so I could get it back home. When I had the battery tested I was told it was shorted internally, I didn't doubt them.
 
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Discussion Starter #9 (Edited)
Thank you ALL for your valuable input. :thumbup1gif:

My next move is to get the battery tested ( yes... I know it is toasted) and replaced.
It may take some time to find a suitable replacement but I will keep you posted on my progress.

LVA15187.jpg
 

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In case anyone (except me) cares about the electrical explanation:

The way you look at a battery engineering-wise is as a "Thevenins equivalent circuit". This models a battery (or any power supply) as a perfect constant voltage source with a resistor in series. This resistance limits the maximum amount of current the battery can deliver.

Take a new 12V battery that can source 600 amperes into a dead short. 12v/600A = 0.02 ohms, so the battery's internal resistance is 0.02 ohms.

As batteries age, that internal resistance typically increases. Say it increases to 0.08 ohms. Put a voltmeter across the battery with no load and it will still read 12v, since a voltmeter draws an infinitesimal amount of current. Short the battery, and that maximum current is now 12v/0.08 ohms = 150 amperes, which may not be enough to turn the starter under load. You will see this as a significant drop in measured battery voltage when cranking. Jumping a battery usually helps here as the auxiliary source can supply the missing current directly to the starter.

If a cell is shorted, then the battery voltage is now 12v-2v (there are 6 cells) or 10v. Note that a 12v lead acid battery is not really 12v but 12.6v as the nominal cell voltage is 2.1v , but I'll use 12 to make the math simple. Cells don't usually short as a perfect conductor, but assume they do. The battery's internals resistance is now 5/6 of what it was before and its voltage is 10v. When you connect an auxiliary 12v power source to it ("jump") the dead battery becomes a current sink instead of a source because its voltage is lower than the jumping battery. Most of the current from the donor source will now flow through the dead battery instead of the starter. The exact amount can be determined by using the Ohms law for resistors in parallel and I won't explain that here. But the practical answer to the question 'how much more current will flow through my starter" is "not much".

In the real world, the variables do not all stay constant with only one changing like my example and the chemistry inside a lead acid battery isn't constant either and is affected by temperature. This is why a voltmeter is such a useful instrument for gauging battery heath, especially with the engine running. Since the charging system provides a constant current source, over time the measured voltage can tell you a lot about the changing internal characteristics of the battery and thus its health.

Al
 

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In case anyone (except me) cares about the electrical explanation:

The way you look at a battery engineering-wise is as a "Thevenins equivalent circuit". This models a battery (or any power supply) as a perfect constant voltage source with a resistor in series. This resistance limits the maximum amount of current the battery can deliver.

Take a new 12V battery that can source 600 amperes into a dead short. 12v/600A = 0.02 ohms, so the battery's internal resistance is 0.02 ohms.

As batteries age, that internal resistance typically increases. Say it increases to 0.08 ohms. Put a voltmeter across the battery with no load and it will still read 12v, since a voltmeter draws an infinitesimal amount of current. Short the battery, and that maximum current is now 12v/0.08 ohms = 150 amperes, which may not be enough to turn the starter under load. You will see this as a significant drop in measured battery voltage when cranking. Jumping a battery usually helps here as the auxiliary source can supply the missing current directly to the starter.

If a cell is shorted, then the battery voltage is now 12v-2v (there are 6 cells) or 10v. Note that a 12v lead acid battery is not really 12v but 12.6v as the nominal cell voltage is 2.1v , but I'll use 12 to make the math simple. Cells don't usually short as a perfect conductor, but assume they do. The battery's internals resistance is now 5/6 of what it was before and its voltage is 10v. When you connect an auxiliary 12v power source to it ("jump") the dead battery becomes a current sink instead of a source because its voltage is lower than the jumping battery. Most of the current from the donor source will now flow through the dead battery instead of the starter. The exact amount can be determined by using the Ohms law for resistors in parallel and I won't explain that here. But the practical answer to the question 'how much more current will flow through my starter" is "not much".

In the real world, the variables do not all stay constant with only one changing like my example and the chemistry inside a lead acid battery isn't constant either and is affected by temperature. This is why a voltmeter is such a useful instrument for gauging battery heath, especially with the engine running. Since the charging system provides a constant current source, over time the measured voltage can tell you a lot about the changing internal characteristics of the battery and thus its health.

Al
What he said.....
 
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In case anyone (except me) cares about the electrical explanation:

The way you look at a battery engineering-wise is as a "Thevenins equivalent circuit". This models a battery (or any power supply) as a perfect constant voltage source with a resistor in series. This resistance limits the maximum amount of current the battery can deliver.

Take a new 12V battery that can source 600 amperes into a dead short. 12v/600A = 0.02 ohms, so the battery's internal resistance is 0.02 ohms.

As batteries age, that internal resistance typically increases. Say it increases to 0.08 ohms. Put a voltmeter across the battery with no load and it will still read 12v, since a voltmeter draws an infinitesimal amount of current. Short the battery, and that maximum current is now 12v/0.08 ohms = 150 amperes, which may not be enough to turn the starter under load. You will see this as a significant drop in measured battery voltage when cranking. Jumping a battery usually helps here as the auxiliary source can supply the missing current directly to the starter.

If a cell is shorted, then the battery voltage is now 12v-2v (there are 6 cells) or 10v. Note that a 12v lead acid battery is not really 12v but 12.6v as the nominal cell voltage is 2.1v , but I'll use 12 to make the math simple. Cells don't usually short as a perfect conductor, but assume they do. The battery's internals resistance is now 5/6 of what it was before and its voltage is 10v. When you connect an auxiliary 12v power source to it ("jump") the dead battery becomes a current sink instead of a source because its voltage is lower than the jumping battery. Most of the current from the donor source will now flow through the dead battery instead of the starter. The exact amount can be determined by using the Ohms law for resistors in parallel and I won't explain that here. But the practical answer to the question 'how much more current will flow through my starter" is "not much".

In the real world, the variables do not all stay constant with only one changing like my example and the chemistry inside a lead acid battery isn't constant either and is affected by temperature. This is why a voltmeter is such a useful instrument for gauging battery heath, especially with the engine running. Since the charging system provides a constant current source, over time the measured voltage can tell you a lot about the changing internal characteristics of the battery and thus its health.

Al
What he said.....
Yeah, I was about to type that explanation too but AlKozak beat me to it. Maybe next time I'll be a little quicker.



In all seriousness, the "smartness" of the guys on this board simply amazes the heck out of me!! Sometimes I think I'm kind of bright, but then I read things like AlKozak posted and am just blown away! Of course, he was the guy in school that messed up the grading curve for the rest of us! :flag_of_truce: :laugh:
 

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Discussion Starter #13
In case anyone (except me) cares about the electrical explanation:

The way you look at a battery engineering-wise is as a "Thevenins equivalent circuit". This models a battery (or any power supply) as a perfect constant voltage source with a resistor in series. This resistance limits the maximum amount of current the battery can deliver.

Take a new 12V battery that can source 600 amperes into a dead short. 12v/600A = 0.02 ohms, so the battery's internal resistance is 0.02 ohms.

As batteries age, that internal resistance typically increases. Say it increases to 0.08 ohms. Put a voltmeter across the battery with no load and it will still read 12v, since a voltmeter draws an infinitesimal amount of current. Short the battery, and that maximum current is now 12v/0.08 ohms = 150 amperes, which may not be enough to turn the starter under load. You will see this as a significant drop in measured battery voltage when cranking. Jumping a battery usually helps here as the auxiliary source can supply the missing current directly to the starter.

If a cell is shorted, then the battery voltage is now 12v-2v (there are 6 cells) or 10v. Note that a 12v lead acid battery is not really 12v but 12.6v as the nominal cell voltage is 2.1v , but I'll use 12 to make the math simple. Cells don't usually short as a perfect conductor, but assume they do. The battery's internals resistance is now 5/6 of what it was before and its voltage is 10v. When you connect an auxiliary 12v power source to it ("jump") the dead battery becomes a current sink instead of a source because its voltage is lower than the jumping battery. Most of the current from the donor source will now flow through the dead battery instead of the starter. The exact amount can be determined by using the Ohms law for resistors in parallel and I won't explain that here. But the practical answer to the question 'how much more current will flow through my starter" is "not much".

In the real world, the variables do not all stay constant with only one changing like my example and the chemistry inside a lead acid battery isn't constant either and is affected by temperature. This is why a voltmeter is such a useful instrument for gauging battery heath, especially with the engine running. Since the charging system provides a constant current source, over time the measured voltage can tell you a lot about the changing internal characteristics of the battery and thus its health.

Al
GREAT explanation, I now fully understand what's happening with my tractor starting problem.
Thanks for sharing your expertise, it is very appreciated.
 

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Of course, he was the guy in school that messed up the grading curve for the rest of us!
Heh, heh, heh. I appreciate the compliment, but this was the guy who, once he got away from mom and dad, boldly decided to switch his major from engineering to fraternity pledging and beer drinking. And I have the D GPA from the only year of college my parents funded to prove it!

Thank you God for second chances!

Al
 

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Discussion Starter #15
Problem resolved

My tractor's starting problem is now solved.....after spending $150.00 on a new battery.

The battery wouldn't even respond to the store tester (completely dead!!) but my charger was saying Voltage OK and Battery completely charged.

One lesson learned, I will never trust my INTELLIGENT :)banghead:) charger to help me diagnose a battery problem.

Thanks everyone.

Happy tractor:greentractorride:
 
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