amperage for starters

[Creekman]

Hmmmm. So that means that if my $120 Interstate Group 93 shows a measured 13.2 cranking volts - I shouldn't believe the the multi-meter?

If one can't believe a real-time voltage reading, what good are the derivative numbers obtained when applying Ohm's Law mathematics?

//greg//

Ohm's law states that I (Current) = V (voltage) divided by R (Resistance).
Since the resistance increases on starting a diesel tractor due to the load placed upon the starter by trying to crank the engine, the voltage has to drop in order to accomodate the increased amperage draw. If you did not get a voltage drop when you measured the voltage drop using your multi-meter, you need to invest in a better meter.
Here is a simple test to measure voltage drop:

1) Wait till it is dark outside.
2) Turn headlights on.
3) Crank engine while looking at the brightness of headlights.
4) When you notice a dimming of the headlights while starting the tractor, you will observe Ohm's law at work. The increased amperage draw will noticeably dim the headlights until the tractor starts.
Also works of gasser tractors.
I don't know where you get your "derivative numbers", but Ohm's law still stands.

 

[SPYDERLK]

Ohm's law states that I (Current) = V (voltage) divided by R (Resistance).
Since the resistance increases on starting a diesel tractor due to the load placed upon the starter by trying to crank the engine, the voltage has to drop in order to accomodate the increased amperage draw. If you did not get a voltage drop when you measured the voltage drop using your multi-meter, you need to invest in a better meter.
Here is a simple test to measure voltage drop:

1) Wait till it is dark outside.
2) Turn headlights on.
3) Crank engine while looking at the brightness of headlights.
4) When you notice a dimming of the headlights while starting the tractor, you will observe Ohm's law at work. The increased amperage draw will noticeably dim the headlights until the tractor starts.
Also works of gasser tractors.
I don't know where you get your "derivative numbers", but Ohm's law still stands.

Ohms law stands, but the voltage loss is in the battery and wires: E=IR. The battery has what is analogous to an internal resistor of low value. The bigger the battery the smaller the resistor value. Out tractor batteries act like they have a resistor of about 0.005 Ohm. 200Amps drops about a volt inside the batt.
No resistance goes up when starting. Rs are constant. The starters resistance appears to rise as it spins due to the significant Back ElectroMotive Force generated that effectively limits the 1000 or so amps the starter would draw at locked rotor condition. Starter resistance is in the range of 0.01 Ohm. Working thru the proportions with these [Batt, Starter] resistances it shows that the batt R and starter R would spit the 12V two to one. At locked rotor the battery terminal V would be 8V. Neglecting wire resistance this 8V would push 800A thru the starter. As the starter started to spin the BEMF would raise its apparent resistance and the voltage spit proportion would change accordingly while current decreased.
larry

 

[greg_g]

I don't know where you get your "derivative numbers"...

That's simply a word to describe the properties - power/voltage/current/resistance - that can be resolved using the simple mathematical formula commonly known as Ohm's Law. Depending upon how you manipulate the formula, you solve for P or E or I or R. The resultant value derived from the formula is expressed in Watts or Volts or Amps or Ohms. Derivatives.

//greg//

 

[lakespirit]

Just curious here guys, but for the life of me, aside from the fact that I don't understand a **** bit of this stuff, how will this help me even if I did get it?

 

[Creekman]

Ohms law stands, but the voltage loss is in the battery and wires: E=IR. The battery has what is analogous to an internal resistor of low value. The bigger the battery the smaller the resistor value. Out tractor batteries act like they have a resistor of about 0.005 Ohm. 200Amps drops about a volt inside the batt.
No resistance goes up when starting. Rs are constant. The starters resistance appears to rise as it spins due to the significant Back ElectroMotive Force generated that effectively limits the 1000 or so amps the starter would draw at locked rotor condition. Starter resistance is in the range of 0.01 Ohm. Working thru the proportions with these [Batt, Starter] resistances it shows that the batt R and starter R would spit the 12V two to one. At locked rotor the battery terminal V would be 8V. Neglecting wire resistance this 8V would push 800A thru the starter. As the starter started to spin the BEMF would raise its apparent resistance and the voltage spit proportion would change accordingly while current decreased.
larry

What you have failed to take into account is the resistance generated to overcome the inertia of an engine that is off. You will also have additional resistance from the friction caused by all of the moving parts in the engine plus the resistance when a piston is in the compression cycle.
The only way to measure starting amperage accurately is to use a gauge connected between the battery and the cable leading to the starter, since voltage, amperage and resistance are in a state of flux due to the ever changing amperage draw on the battery.

 

[Creekman]

Just curious here guys, but for the life of me, aside from the fact that I don't understand a **** bit of this stuff, how will this help me even if I did get it?

In a previous reply I asked you why you needed to know this bit of information. To date, you have not replied. Just why do you need to know the amps required to start your KAMA TS254, since you solved your solenoid problems in a previous post?
BTW, how are you doing in solving your problem with sloooow hydraulics?

 

[Creekman]

Alot is going to depend on load vs battery capacity.

A very large battery driving a small load ( warmed engine, or very good starter , or smaller engine.. etc..) Is going to maintain a higher output voltage longer that a battery with a smaller reserve.

The battery in my ford 1975 5000 is a huge 4DLT.. Actually a larger battery than my NH 7610s. the 7610s has a similar sized engine.. the 5000 is a 256ci , and the 7610s is a 304ci w/turbo. As long as I am not in a hard start scenerio.. like cold weather.. or it has sat along time.. the 5000 usually bangs on about the 3rd rev. I've had a good analog and lab grade meter on that 4DLT.. It don't drop under 12v cranking, unless you make a few attempts. The 7610s on the other hand will get the battery into the 11v range on the first attempt. I was going to refit the battery box int he 7610s to take the bigger battery.. but then figured that if it does ok on the smaller 80$ one.. then why stick in the 120$ one! Plus.. that 4DLT is about as much abttery as i want to lift.. it's dern heavy..

soundguy

How big of a battery do you think lalespirit can cram into a KAMA TS354?

 

[greg_g]

How big of a battery do you think lalespirit can cram into a KAMA TS354?

Actually Jeff has a 254, which is one size smaller than the 300 series. Not terribly important though, same size battery tray in both. I can't tell however - from your profile - if you even own a Chinese tractor, so you may not realize how physically large the OE battery is in the first place.

I replaced one of my JM254 batteries with an Interstate MT-49 (940CA/750CCA) made for Mercedes diesels. And I replaced one of my KAMA batteries with a MTP-93 (same case, higher output). Don't have a clue as to the Chinese battery amperage rating, but I seem to recall the 14"x7"x7" Interstates are physically a little smaller than OE.

But as large as the Jinma/KAMA/TaiShan battery compartments are, I don't think they'll quite accommodate a 20"x8"x8" 4DLT. Might get one in a Foton though. Their battery compartments are sized to accommodate a pair of pretty good sized 6v batteries. It's all a bit academic to me anyway, since the MTP-93 and 4DLT have the same basic 1000CA/850CCA rating anyway.

//greg//

 

[SPYDERLK]

Just curious here guys, but for the life of me, aside from the fact that I don't understand a **** bit of this stuff, how will this help me even if I did get it?

Getting a handle on Ohms law and the use of a digital voltmeter to assess current flow and voltage drops in a circuit will help you in all electrical trouble shooting. For instance, if you have a DVM, put the leads across a battery conection - one to the batt terminal - the other to the clamp on the batt cable. What you have done here is essentially short the meter leads together, but not quite - there is a very low resistance at the crossover between the batt and clamp. Now set your meter to its most sensitive DC Volt scale - probably 200 millivolts. Typically these meters, on this scale, will resolve down to one tenth of a millivolt -- [0.0001Volt !] Since the leads are "shorted" the meter will read zero. Now turn on your lights. The connection you are measuring across will now be carrying current. Its [hopefully] verry small resistance will result in a voltage across the connection. This voltage will be the current (I) drawn by the lights X the resistance [R] across the connection. To use round ballpark numbers: 5A for the lights and about 0.001 Ohms for the connection. You meter will read 5.0 millivolts - Ohms law is being obeyed. By the equation V=IR you only need 2 of these numbers to "derive" the third. So, without knowing the quality of that battery connection you can determine it by measurement. Say you know the lights on your tractor draw only 5A and the reading you take with the DVM is 20.3mV across the battery connection. Then, scrambling the equation: R= 0.0203V/5A= 0.00406 Ohms. For a BATTERY connection this is terrible quality. When you draw 200A to crank your engine that meter is going to read V=200x.004= 812mV [youll have to go to a higher scale]. The starter needs this voltage. Youll notice slow cranking and the connection will get hot. You meter and Ohms law tells you so.
larry

 

[lakespirit]

In a previous reply I asked you why you needed to know this bit of information. To date, you have not replied. Just why do you need to know the amps required to start your KAMA TS254, since you solved your solenoid problems in a previous post?
BTW, how are you doing in solving your problem with sloooow hydraulics?

Curiosity.

Not well. Many theories, but so far, I still don't have a presseure gaurge, so I am only providing partial data to get the proper help.