Charging levels

[RobinJ]

I was getting some questions from ChrisW and Snibble over the discussion on charging voltages, setup and whether it was possible to get a warning light when everything was working ok.

Wouldn't you know it, I could not get my warning light to glow this weekend

Using a Fluke I took measurement accross the battery terminals; Starter 12.81V, Bank1 12.57V, Bank2 12.46V; then started the engine; Starter 13.55V, Bank1 13.51V, Bank2 13.52V; at this point the voltmeter on the dash was reading 13V; I then switched on the external controller and got; Starter 14.80V, Bank1 14.80V, Bank2 14.80V; when the external controller shut down I measured again and got 14.08V across all three.

The domestic batteries were perhaps flatter than they were last week as we had spent the night and used some, but I am satisfied that the external controller is working fine, I'm not sure if maybe there is a voltage drop on the charging circuit somewhere (based on the readings of 14.08V)?

The charging leads are fairly short (<1m) and thick, go direct to the starter/isolator/battery/relay. The split charge relays, although driven through the external controller, operate only when the alternator is charging, here the thinest wire is 80/0.1 (auxiliary feed to fridge), these are also mounted next to the batteries.

My next test will be to check whether there is a voltage difference between the alternator output and the starter battery, and contiue to see if I can persuade my warning light to glow when the batteries are fully charged, then turn on a load and see if it goes out.

I did check on the alternator, its an RS123 65A.

Edited February 11, 2008 by RobinJ

[chris w]

I was getting some questions from ChrisW and Snibble over the discussion on charging voltages, setup and whether it was possible to get a warning light when everything was working ok.

Wouldn't you know it, I could not get my warning light to glow this weekend

Using a Fluke I took measurement accross the battery terminals; Starter 12.81V, Bank1 12.57V, Bank2 12.46V; then started the engine; Starter 13.55V, Bank1 13.51V, Bank2 13.52V; at this point the voltmeter on the dash was reading 13V; I then switched on the external controller and got; Starter 14.80V, Bank1 14.80V, Bank2 14.80V; when the external controller shut down I measured again and got 14.08V across all three.

The domestic batteries were perhaps flatter than they were last week as we had spent the night and used some, but I am satisfied that the external controller is working fine, I'm not sure if maybe there is a voltage drop on the charging circuit somewhere (based on the readings of 14.08V)?

The charging leads are fairly short (<1m) and thick, go direct to the starter/isolator/battery/relay. The split charge relays, although driven through the external controller, operate only when the alternator is charging, here the thinest wire is 80/0.1 (auxiliary feed to fridge), these are also mounted next to the batteries.

My next test will be to check whether there is a voltage difference between the alternator output and the starter battery, and contiue to see if I can persuade my warning light to glow when the batteries are fully charged, then turn on a load and see if it goes out.

I did check on the alternator, its an RS123 65A.

 

Why do you have 2 domestic battery banks? Electronically-speaking, this is not a good way of wiring them up because one large battery bank has a far higher capacity than just the arithmetic sum of the individual batteries.

 

For example, a simple case, a 110AH battery has a true capacity of 183AH (there's a mathematical reason for this but trust me at this stage - Peukert's equations). 2 x 110AH batteries do NOT have a combined maximum of 2 x 183 = 366AH but a total capacity of 452AH. That's an increase of nearly 25% over using the batteries singly. And it costs nothing more - it's effectively "free" additional capacity.

 

BTW, your voltages above, indicate that you have very low internal regulator voltages in your alternators (~13.5v). You DEFINITELY always need to use an external controller as 13.5v will not charge a domestic battery. It will top-up a start battery OK, but that's it. The 14.08v is just a surface charge reading and has no meaning. Leave the batteries to sit for a couple of hours once the engine is OFF (or switch on your tunnel light for 15 minutes) and remeasure the batteries (after switching off the tunnel light again). That will give you a true reading of their charge. 12.7v corresponds to 100% charge and each 0.1v below that corresponds to approximately a 10% drop in charge. ie: 12.2v corresponds to 50% charge, below which you should NEVER drop to avoid damaging the batteries.

 

To measure 0.1v drop accurately you MUST have a multimeter with AT LEAST an accuracy of +/-0.5%. A cheap (2%) meter is useless for any of these measurements as the reading for a 12.4v battery (70% charged) on a cheap meter could be anywhere between 12.2v (50% charged) and 12.7v (100% charged). Not very useful!!

 

Chris

Edited February 11, 2008 by chris w

[tomsk]

Why do you have 2 domestic battery banks? Electronically-speaking, this is not a good way of wiring them up because one large battery bank has a far higher capacity than just the arithmetic sum of the individual batteries.

 

For example, a simple case, a 110AH battery has a true capacity of 183AH (there's a mathematical reason for this but trust me at this stage - Peukert's equations). 2 x 110AH batteries do NOT have a combined maximum of 2 x 183 = 366AH but a total capacity of 452AH. That's an increase of nearly 25% over using the batteries singly. And it costs nothing more - it's effectively "free" additional capacity.

 

BTW, your voltages above, indicate that you have very low internal regulator voltages in your alternators (~13.5v). You DEFINITELY always need to use an external controller as 13.5v will not charge a domestic battery. It will top-up a start battery OK, but that's it. The 14.08v is just a surface charge reading and has no meaning. Leave the batteries to sit for a couple of hours once the engine is OFF (or switch on your tunnel light for 15 minutes) and remeasure the batteries (after switching off the tunnel light again). That will give you a true reading of their charge. 12.7v corresponds to 100% charge and each 0.1v below that corresponds to approximately a 10% drop in charge. ie: 12.2v corresponds to 50% charge, below which you should NEVER drop to avoid damaging the batteries.

 

To measure 0.1v drop accurately you MUST have a multimeter with AT LEAST an accuracy of +/-0.5%. A cheap (2%) meter is useless for any of these measurements as the reading for a 12.4v battery (70% charged) on a cheap meter could be anywhere between 12.2v (50% charged) and 12.7v (100% charged). Not very useful!!

 

Chris

 

 

I concur, do not question him, just trust him, Chris is always right

[ChrisPy]

I concur, do not question him, just trust him, Chris is always right

psycophant ?

[chris w]

psycophant ?

 

sycophant or psycho_phant? Maybe you're right

Edited February 11, 2008 by chris w

[Yoda]

psycophant ?

 

Not another bloody elephant!

[RobinJ]

Why do you have 2 domestic battery banks? Electronically-speaking, this is not a good way of wiring them up because one large battery bank has a far higher capacity than just the arithmetic sum of the individual batteries.

The second bank only provides power to the inverter and fridge!

It may sound like a weird system, but the domestic supplies all the power to lights, water and toilets (25A when macerating!) etc., the other bank supplies just the inverter and the fridge when its on electric, thus keeping the domestic going if too much is used.

BTW, your voltages above, indicate that you have very low internal regulator voltages in your alternators (~13.5v). You DEFINITELY always need to use an external controller as 13.5v will not charge a domestic battery. It will top-up a start battery OK, but that's it. The 14.08v is just a surface charge reading and has no meaning. Leave the batteries to sit for a couple of hours once the engine is OFF (or switch on your tunnel light for 15 minutes) and remeasure the batteries (after switching off the tunnel light again). That will give you a true reading of their charge. 12.7v corresponds to 100% charge and each 0.1v below that corresponds to approximately a 10% drop in charge. ie: 12.2v corresponds to 50% charge, below which you should NEVER drop to avoid damaging the batteries.

The first figures show the state of the batteries (left overnight), starter seems a little high? The 14.08V was what I was getting with the engine still running after the external controller shut down, that worried me, because I was expecting to still get around 14.4V from the internal controller!

To measure 0.1v drop accurately you MUST have a multimeter with AT LEAST an accuracy of +/-0.5%. A cheap (2%) meter is useless for any of these measurements as the reading for a 12.4v battery (70% charged) on a cheap meter could be anywhere between 12.2v (50% charged) and 12.7v (100% charged). Not very useful!!

The Fluke is supposed to be accurate to about 0.1% on that particular scale (the reason I used it!)

[chris w]

The second bank only provides power to the inverter and fridge!

It may sound like a weird system, but the domestic supplies all the power to lights, water and toilets (25A when macerating!) etc., the other bank supplies just the inverter and the fridge when its on electric, thus keeping the domestic going if too much is used.

 

It's not weird, it's just not as effective as it could be. If you wire all the batteries in parallel (except the start battery) you will gain an additional 25-30% "free" additional capacity as compared to 2 separate banks. The maths is a little complex, but I'm happy to share it with you if you are interested.

 

Chris

[Chris Pink]

The Fluke is supposed to be accurate to about 0.1% on that particular scale (the reason I used it!)

 

Which Fluke do you use?

[RobinJ]

Which Fluke do you use?

Its a green one, I just took a loook on the web site and obviously they don't do it anymore, so I'm going to check on the number.

[JohnT]

sycophant or psycho_phant? Maybe you're right

I prefer jeans me self!!

 

JohnTy

 

No doubt Dhutch will be along soon to correct my english!!!

Edited February 12, 2008 by JohnT

[sueanddaren]

Why do you have 2 domestic battery banks? Electronically-speaking, this is not a good way of wiring them up because one large battery bank has a far higher capacity than just the arithmetic sum of the individual batteries.For example, a simple case, a 110AH battery has a true capacity of 183AH (there's a mathematical reason for this but trust me at this stage - Peukert's equations). 2 x 110AH batteries do NOT have a combined maximum of 2 x 183 = 366AH but a total capacity of 452AH. That's an increase of nearly 25% over using the batteries singly. And it costs nothing more - it's effectively "free" additional capacity.BTW, your voltages above, indicate that you have very low internal regulator voltages in your alternators (~13.5v). You DEFINITELY always need to use an external controller as 13.5v will not charge a domestic battery. It will top-up a start battery OK, but that's it. The 14.08v is just a surface charge reading and has no meaning. Leave the batteries to sit for a couple of hours once the engine is OFF (or switch on your tunnel light for 15 minutes) and remeasure the batteries (after switching off the tunnel light again). That will give you a true reading of their charge. 12.7v corresponds to 100% charge and each 0.1v below that corresponds to approximately a 10% drop in charge. ie: 12.2v corresponds to 50% charge, below which you should NEVER drop to avoid damaging the batteries.To measure 0.1v drop accurately you MUST have a multimeter with AT LEAST an accuracy of +/-0.5%. A cheap (2%) meter is useless for any of these measurements as the reading for a 12.4v battery (70% charged) on a cheap meter could be anywhere between 12.2v (50% charged) and 12.7v (100% charged). Not very useful!!Chris

Why do you have 2 domestic battery banks? Electronically-speaking, this is not a good way of wiring them up because one large battery bank has a far higher capacity than just the arithmetic sum of the individual batteries.For example, a simple case, a 110AH battery has a true capacity of 183AH (there's a mathematical reason for this but trust me at this stage - Peukert's equations). 2 x 110AH batteries do NOT have a combined maximum of 2 x 183 = 366AH but a total capacity of 452AH. That's an increase of nearly 25% over using the batteries singly. And it costs nothing more - it's effectively "free" additional capacity.BTW, your voltages above, indicate that you have very low internal regulator voltages in your alternators (~13.5v). You DEFINITELY always need to use an external controller as 13.5v will not charge a domestic battery. It will top-up a start battery OK, but that's it. The 14.08v is just a surface charge reading and has no meaning. Leave the batteries to sit for a couple of hours once the engine is OFF (or switch on your tunnel light for 15 minutes) and remeasure the batteries (after switching off the tunnel light again). That will give you a true reading of their charge. 12.7v corresponds to 100% charge and each 0.1v below that corresponds to approximately a 10% drop in charge. ie: 12.2v corresponds to 50% charge, below which you should NEVER drop to avoid damaging the batteries.To measure 0.1v drop accurately you MUST have a multimeter with AT LEAST an accuracy of +/-0.5%. A cheap (2%) meter is useless for any of these measurements as the reading for a 12.4v battery (70% charged) on a cheap meter could be anywhere between 12.2v (50% charged) and 12.7v (100% charged). Not very useful!!Chris

Ok Chris now do you get the extra 25% ??? Im assuming you are extrapolateing the graph to get the 183Ah Daren

[chris w]

Ok Chris now do you get the extra 25% ??? Im assuming you are extrapolateing the graph to get the 183Ah Daren

 

No actually. It's all to do with the fact that the quoted nominal AH figure (eg: 110AH) is a single point on the battery's discharge CURVE. The discharge characteristic is NOT a straight line but an exponential curve with a mathematical "s" factor of approx 1.3 for wet lead acid batteries.

 

So the quoted AH figure is not the battery's actual AH figure but a selected point (the 20 hour discharge point) on a curve. However, mathematically it is now easy to calculate the battery's actual maximum capacity which will be (for a 110AH battery quoted at the 20 hour rate):

 

(110/20)^1.3 x 20 = 183AH

 

2 x 110AH batteries wired in parallel will therefore have a total capacity not of 2 x 110AH or even 2 x 183AH but:

 

(220/20)^1.3 x 20 = 452AH or about 25% larger usable capacity than the same two batteries used independently.

 

The original work for this was done by a guy called Peukert in the late 1800's and is known as Peukert's Law. It's to do with the fact that the average current per battery drops when wired in parallel

 

The corollary of this is that a large current will discharge the battery a lot quicker than simple maths would indicate.

 

eg: one might expect a 110AH battery discharging at 50A to last 2 hours to absolute discharge (110/50). In fact, the actual time is in practice:

 

 

(110/20)^1.3 x 20/50^1.3 = 1.1 hours or half the expected time

 

When doing power audits on boats to determine the necessary size of battery bank (or, looking at it another way, how long a particular size battery bank will last) you must use Peukert's equations, not linear arithmetic, in order to get an accurate answer. The error otherwise could be significant.

 

Chris

[Chris Pink]

So the quoted AH figure is not the battery's actual AH figure but a selected point (the 20 hour discharge point) on a curve. However, mathematically it is now easy to calculate the battery's actual maximum capacity which will be (for a 110AH battery quoted at the 20 hour rate):

 

When i started buying domestic (for want of a better word) batteries 20 years ago they used to be rated on 10 hour discharge, at some point manufacturers have realised that they can also get something for nothing, it won't long before they use 30 hour rates.

[chris w]

When i started buying domestic (for want of a better word) batteries 20 years ago they used to be rated on 10 hour discharge, at some point manufacturers have realised that they can also get something for nothing, it won't long before they use 30 hour rates.

 

You're absolutely correct. A 110AH battery (at the 20 hour rate) is EXACTLY the same battery described as a 90AH at the 10 hour rate or a 120AH battery at the 30 hour rate or even a 135AH battery at the 50 hour rate.

 

If, in technical ignorance, one were to be offered a choice of a 90, 110, 120 or 135AH battery for the same price and in the same form factor, everyone would choose the latter. Yet it's just specsmanship. Always look at the quoted hour rate when buying batteries to ensure you are comparing apples with apples. Most will be quoted at the (industry standard) 20 hour rate but not all.

 

Chris

Edited February 13, 2008 by chris w

[tired old pirate]

No actually. It's all to do with the fact that the quoted nominal AH figure (eg: 110AH) is a single point on the battery's discharge CURVE. The discharge characteristic is NOT a straight line but an exponential curve with a mathematical "s" factor of approx 1.3 for wet lead acid batteries.

 

So the quoted AH figure is not the battery's actual AH figure but a selected point (the 20 hour discharge point) on a curve. However, mathematically it is now easy to calculate the battery's actual maximum capacity which will be (for a 110AH battery quoted at the 20 hour rate):

 

(110/20)^1.3 x 20 = 183AH

 

2 x 110AH batteries wired in parallel will therefore have a total capacity not of 2 x 110AH or even 2 x 183AH but:

 

(220/20)^1.3 x 20 = 452AH or about 25% larger usable capacity than the same two batteries used independently.

 

The original work for this was done by a guy called Peukert in the late 1800's and is known as Peukert's Law. It's to do with the fact that the average current per battery drops when wired in parallel

 

The corollary of this is that a large current will discharge the battery a lot quicker than simple maths would indicate.

 

eg: one might expect a 110AH battery discharging at 50A to last 2 hours to absolute discharge (110/50). In fact, the actual time is in practice:

(110/20)^1.3 x 20/50^1.3 = 1.1 hours or half the expected time

 

When doing power audits on boats to determine the necessary size of battery bank (or, looking at it another way, how long a particular size battery bank will last) you must use Peukert's equations, not linear arithmetic, in order to get an accurate answer. The error otherwise could be significant.

 

Chris

 

Is all that stuff actualy real or just maliciousness purposly designed to make me think my brain is broken?

[chris w]

Is all that stuff actualy real or just maliciousness purposly designed to make me think my brain is broken?

 

That's the simple edited version for broken brains

[RobinJ]

Which Fluke do you use?

Apparently its an 8060 and the manual says accurate to +/-0.05% on the 20V range!

[tomsk]

That's the simple edited version for broken brains

 

 

Oh Pheuket!!!

[James T Berk]

PHEEEEP! Norty word! Oh no it isn't never mind carry on.

[sueanddaren]

No actually. It's all to do with the fact that the quoted nominal AH figure (eg: 110AH) is a single point on the battery's discharge CURVE. The discharge characteristic is NOT a straight line but an exponential curve with a mathematical "s" factor of approx 1.3 for wet lead acid batteries.

 

So the quoted AH figure is not the battery's actual AH figure but a selected point (the 20 hour discharge point) on a curve. However, mathematically it is now easy to calculate the battery's actual maximum capacity which will be (for a 110AH battery quoted at the 20 hour rate):

 

(110/20)^1.3 x 20 = 183AH

 

2 x 110AH batteries wired in parallel will therefore have a total capacity not of 2 x 110AH or even 2 x 183AH but:

 

(220/20)^1.3 x 20 = 452AH or about 25% larger usable capacity than the same two batteries used independently.

 

The original work for this was done by a guy called Peukert in the late 1800's and is known as Peukert's Law. It's to do with the fact that the average current per battery drops when wired in parallel

 

The corollary of this is that a large current will discharge the battery a lot quicker than simple maths would indicate.

 

eg: one might expect a 110AH battery discharging at 50A to last 2 hours to absolute discharge (110/50). In fact, the actual time is in practice:

(110/20)^1.3 x 20/50^1.3 = 1.1 hours or half the expected time

 

When doing power audits on boats to determine the necessary size of battery bank (or, looking at it another way, how long a particular size battery bank will last) you must use Peukert's equations, not linear arithmetic, in order to get an accurate answer. The error otherwise could be significant.

 

Chris

 

 

 

Chris

 

Don't forget that the Peukert number gives a simplyfied graph shape that is correct between the two data points that are used. It is common practice to use the 10 min discharge rate and the 20 hour rate usualy to 1.75Vpc at 20 or 25 degC when calculating the Peukert number.

At higher rates the "coup de fouet" (sorry spelling was never my strong point) and the top Lead comes into play and at long rates things like plate thickness have a great effect. For example on thinplate designs eg Odessey after the 24 hour point the percentage capacity increase is very small about 5% at the 34 hour rate. Therfore in practice you will see very little increase in capacity from the stated 24 hour rate.

As the types of batteries used on boats are in the main automotive or semi traction I would suggest that the plate thicknesses are about 1.8mm and therefore I wouldn't expect to see any further capacity increase after about 30 hours.

If a manufacturer says that a battery is 110Ah you can bet that he is talking about the C20 rate unless it is an industrial product (motive C5, Stanby C10) thefore don't expect any more than 110Ah infact if they are using the European specs they don't have to produce any more than 95% of this even after 5 cycles.

 

By the way the main reason for using the peukert number is that when ploted on a log log scale it is a straight line therefore easy to reference and predict a run time for any given load. However since computers got much easier to program most companies now use look up tables and interpolate between points, that way you can take into acount the true shape of the graph at both ends.

 

Sorry to everyone for going off topic but I don't want people to think that they can get more out of the batteries that they realy can.

 

Daren

[chris w]

By the way the main reason for using the peukert number is that when ploted on a log log scale it is a straight line therefore easy to reference and predict a run time for any given load. However since computers got much easier to program most companies now use look up tables and interpolate between points, that way you can take into acount the true shape of the graph at both ends.

 

Sorry to everyone for going off topic but I don't want people to think that they can get more out of the batteries that they realy can.

 

Daren

 

Sorry to disagree Daren but your reasoning above is incorrect. Look at the Smartgauge pages (http://www.smartgauge.co.uk/peukert_depth.html ) where there is a full and very detailed analysis and mathematical treatment of Peukert (I know because I wrote it). If you think there's a flaw in the mathematics come back to me with your own analysis. I warn you that the guy who owns this site and owns the company Smartgauge is one of the leading authorities on batteries in the industry and has 20 years+ experience of testing and analysing them.

 

The main reason for using the Peukert number is not because, when plotted on a log scale, it gives a straight line. ANY exponential curve will give a straight line when plotted on a log scale - that's why it's called an exponential curve The reason is that real batteries follow the results predicted by the application of Peukert mathematics. The mathematical represenation of an exponential curve by an expression of the form Ae^-sT (where 's' in this case will be approx 1.3 for wet lead acid batteries) represents it perfectly.

 

I wouldn't disagree that at the extremes of charge and discharge there will be some second order effects of course, but these can be totally ignored for all practical purposes.

 

Using the straight 110AH capacity figure at the C20 rate (or any other rate) will NOT give you accurate answers to charge and discharge equations unless the current involved is pretty close to the C20 current. That is self-evident because the C20 current IS a point on the curve. At currents below and above the C20 current, one must use Peukert mathematics to get accurate answers.

 

The bottom line is that a 110AH battery is NOT a 110AH battery unless it is used exclusively at the C20 current (or the actual rate quoted on the label - usually C20). A single 110AH battery has an actual (Peukert) capacity of 183AH and so you will get more or less than the stated 110AH depending on the level of (Peukert) current drawn.

 

Chris

Edited February 13, 2008 by chris w

[smileypete]

Anyone actually measured the Peukert for a 'pasted plate' battery?

 

cheers,

Pete.

[sueanddaren]

Sorry to disagree Daren but your reasoning above is incorrect. Look at the Smartgauge pages (http://www.smartgauge.co.uk/peukert_depth.html ) where there is a full and very detailed analysis and mathematical treatment of Peukert (I know because I wrote it). If you think there's a flaw in the mathematics come back to me with your own analysis. I warn you that the guy who owns this site and owns the company Smartgauge is one of the leading authorities on batteries in the industry and has 20 years+ experience of testing and analysing them.

 

The main reason for using the Peukert number is not because, when plotted on a log scale, it gives a straight line. ANY exponential curve will give a straight line when plotted on a log scale - that's why it's called an exponential curve The reason is that real batteries follow the results predicted by the application of Peukert mathematics. The mathematical represenation of an exponential curve by an expression of the form Ae^-sT (where 's' in this case will be approx 1.3 for wet lead acid batteries) represents it perfectly.

 

I wouldn't disagree that at the extremes of charge and discharge there will be some second order effects of course, but these can be totally ignored for all practical purposes.

 

Using the straight 110AH capacity figure at the C20 rate (or any other rate) will NOT give you accurate answers to charge and discharge equations unless the current involved is pretty close to the C20 current. That is self-evident because the C20 current IS a point on the curve. At currents below and above the C20 current, one must use Peukert mathematics to get accurate answers.

 

The bottom line is that a 110AH battery is NOT a 110AH battery unless it is used exclusively at the C20 current (or the actual rate quoted on the label - usually C20). A single 110AH battery has an actual (Peukert) capacity of 183AH and so you will get more or less than the stated 110AH depending on the level of (Peukert) current drawn.

 

Chris

Chris

 

Im not saying that there is anything wrong with Peukert

 

The Original paper was:-

W.Peukert, uber die Abhangigkeit der kapacitat von der entladestrom sterke bei bleiakkumulatoren, elektrotech z (1897).

 

But it has its limitations at both exremes, even Peukert and Vinal proved that.

 

I followed your link to the smartgauge web site and the maths is fine but try this for an example:-

 

Battery type odessey PC1700 (you can find the info at:- http://www.enersys-emea.com/downloadarea/d...ad.asp#Odyssey)

 

C20 =67.9Ah

C5min =28.1Ah

 

Giving an n=1.191

 

Put this into the Peukert calculator on the smartgauge web site and it gives a Peukert capacity of 84.5Ah. The actual tested capacity at 1Amp for this bloc at the same temp and end voltage is 74.16Ah. This is too big a difference to ignore.

 

At the other end the 2 min data is 569.8A but the calculator gives 2mins 24 seconds, not much until you think this is 20% out.

 

Now don't forget that these are figures for new batteries and as they age you lose capacity , so this gets even worse. Most reputable manufacturers use 80% capacity as end of life so a couple of years in and you could be 20 to 30% out.

 

As far as experience goes Ive got 28 years of designing and advising on batteries from car batteries to Torpedo batteries and Im classed as a novice when compared with the real experts with 40 years plus under their belt. Every week I see things Ive never seen before and sometimes things Ive not seen for 20 years.

 

By the way the smartguage looks like a realy good bit of kit

 

Daren

Edited February 16, 2008 by sueanddaren

[sueanddaren]

Anyone actually measured the Peukert for a 'pasted plate' battery?

 

cheers,

Pete.

 

 

Pete

 

All Lead acid batteries are "pasted plate" some times called flat plate. The only ones that arn't are Plante, Rod plate and Tubular plate. Plante are the original type of Lead acid battery only used now in powerstations and very critical applications they last 25years plus but cost a lot. Tubular are used in motive power (milk floats, forklift trucks) because they are realy good at charge/discharge cycles. Rod plate is a cross between flat plate and tubular. The only way of measuring the Peukert capacity is to discharge at 1Amp and yes it is done.

 

Daren