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Author Topic:   Wire Size For Battery Cables
Tom W Clark posted 01-11-2014 06:08 PM ET (US)   Profile for Tom W Clark   Send Email to Tom W Clark  
I am helping a friend relocate the battery on his classic Montauk from the stern to the console. We want new battery cables to run without any splice or junction from battery to motor.

We have measured the length of cable needed and it is 21 feet.

I measured the current while he cranked the cold 100 HP Mercury motor today in wet 45 degree Seattle weather.

My Klein Tools CL2000 clamp-on ammeter showed a maximum 21.5 on the display while the motor was being cranked.

What gauge wire do I order for these new battery cables?

DVollrath posted 01-11-2014 06:48 PM ET (US)     Profile for DVollrath  Send Email to DVollrath     
Hi Tom,
When I rewired my Montauk, the terminal to terminal length was around 18' as I recall, so your routing & motor combination is somewhat longer.

The current you measured seems to be an order of magnitude low. Is it possible there is a range setting on the meter, or something similar that would need to be accounted for? 21.5A seems really low, even for a steady state current.

Dennis

tmann45 posted 01-11-2014 11:36 PM ET (US)     Profile for tmann45  Send Email to tmann45     
I'm with Dennis on the current draw, 21.5 amps seems too low.

If that is the correct current, assuming your length measurement is one-way, using a 10% voltage drop, 105*C insulation temp rating, my calculator says #10 AWG will suffice, it can carry 60 amps. The wire rating is standard boat cable, Ancor or Pacer type.

Voltage drop will be 7.793% or 0.9352 volts.

Tom W Clark posted 01-12-2014 12:15 PM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
That is what I thought when I took the reading. I was expecting the number to be hundreds of amps, not just tens of amps, but electrical theory is not my forte. I presume user error.

Yes, there are different ranges on this device but also different setting depending on whether you are measuring AC or DC. I suspect the latter was my mistake. The instruction manual is ambiguous on this point. We will try again today.

Nonetheless, given a reading, a cable length and the known voltage (12V in this case) how does one make this calculation or assessment of the proper wire size?

Dennis -- the battery cables are tow run from the battery in the forward port side the console floor to the Mercury 100 via the route described by the steering cable rather than from the tunnel straight into the motor.

We ran a length of wire and laid it exactly where we want the battery cables, marked it, then pulled it out and measured the length. It measured 20' 5" so I rounded up to 21'. Is that reasonable?

tmann45 posted 01-12-2014 01:52 PM ET (US)     Profile for tmann45  Send Email to tmann45     
I suspect your suspicion is correct, on AC instead of DC.

And I suspect my assumption of 10% voltage drop maybe incorrect, might need to use 3%. Usually the motor will have specs in the rigging guide for battery cable size.

There are two things (major) that need to be considered when sizing electrical wires everytime, ampacity, which is the maximum safe current for continuous use, and voltage drop.

Voltage drop is calculated: voltage drop = current x length x ohms per foot (resistance of the cable).

Ampacity is listed in a table sent via email; it’s a picture so it can't be posted here by me. The tables can be found online.

Some good info on the West Marine website: http://www.westmarine.com/webapp/wcs/stores/servlet/ WestAdvisorView?storeId=11151&page=Marine-Wire , cut and pasted below:


Marine Wire Size and Ampacity
Even the experts have to check occasionally on the correct gauge and ampacity (maximum amount of electrical current a conductor can carry) of wire for a given marine DC load. The simplest method we’ve found uses the charts below.
Select either the 10% or 3% voltage drop chart, based on the type of load you are running.
Next, find the current consumption of the load on the vertical axis of the chart.
Find the length of the circuit on the horizontal axis of the chart, noting that the length is the “round trip” distance from the panel or battery to the load and back.
The color of the graph at the intersection denotes the gauge of wire to use.
We’ve included copper wire specifications which comply with the AWG standards at the bottom.
Of particular interest is the equation:
Voltage Drop = Current x Length x Ohms per foot
This simple equation allows you to calculate the voltage drop for a circuit of any length and any current flow, if you know the resistance of the wire.
Finally, note that the amp capacity of the wire curtails using very short lengths of wire for large current flows, as shown by the “flat tops” of the 10% chart areas.
(Graphs not posted.)
These simple, proprietary graphs assume:
105°C insulation rating: All Ancor wire uses 105°C insulation rating. Lower temperature insulation cannot handle as much current (the flat tops on the 10% graph would be lower than shown)
AWG wire sizes, not SAE: All Ancor wire uses AWG wire sizes. SAE wire sizes are 6%-12% smaller, carry proportionally less current, and have greater resistance
Wires are not run in engine spaces: Maximum current is 15% less in engine spaces, which are assumed to be 20°C hotter than non-engine spaces (50°C vs. 30°C).
Conductors are not bundled: If three conductors are bundled, reduce maximum amperage by 30%. If 4–6 conductors are bundled, reduce maximum amperage by 40%. If 7–24 conductors are bundled, reduce amperage by 50%.


jimh posted 01-12-2014 02:23 PM ET (US)     Profile for jimh  Send Email to jimh     
Regarding measurement of current with a clamp-on meter: it is very easy and thus very inexpensive to measure alternating current (AC) with a clamp-on type meter. With AC the clamp-on meter becomes a current transformer and the current in the conductor becomes a one-turn loop on the transformer. It is common nowadays that even relatively inexpensive digital multi-meters provide clamp-on measurement of AC current.

To measure direct current (DC) with a clamp-on meter requires a different method. A Hall Effect sensor is used. Typically this sort of metering is only found in more expensive multi-meters.

There is also often a peak-hold circuit in clamp-on meters. In an engine cranking circuit this is useful to find the peak current flow, which will typically be the current flowing at the instant the starter motor is energized. This current really is an important current. During this maximum current flow the voltage drop in the power distribution conductors will be greatest. This determines the voltage provided to the electrical engine cranking motor. If the voltage is too low, the motor will not have enough torque to turn over the engine. As a result, it is this initial current flow that really determines the tolerance for voltage drop in the cables.

I have a decent DC clamp-on multimeter. It is a FLUKE 374, and it cost almost $300. Using this device I have measured the peak starting current for a 2.5-liter automobile four-stroke-power-cycle engine to be about 350-Amperes.

Regarding the current carrying capacity of a conductor: there are many ways to define this. For distribution of power I recommend you use the following guideline:

The rated current carrying capacity of conductors for power distribution according to the 1-Ampere/700-circular mills calculation is as follows

00-AWG = 190-Amperes
0-AWG = 150-Amperes
1-AWG = 119-Amperes
2-AWG = 94-Amperes
3-AWG = 75-Amperes
4-AWG = 60-Amperes
5-AWG = 47-Amperes
6-AWG = 37-Amperes
7-AWG = 30-Amperes
8-AWG = 24-Amperes
9-AWG = 19-Amperes
10-AWG =15-Amperes
12-AWG = 9.3-Amperes
14-AWG = 5.9-Amperes
16-AWG =3.7-Amperes

In a 12-Volt system, the maximum distance those conductors can carry their rated current is 12-feet (or 24-feet of total conductor length) if less than a three-percent voltage drop is desired.

For more on wire current capacity see

Wire Conductor Size for Power Distribution
http://continuouswave.com/ubb/Forum6/HTML/003384.html

DVollrath posted 01-12-2014 02:41 PM ET (US)     Profile for DVollrath  Send Email to DVollrath     
Tom,
Regarding the length, if you measured it I'm sure it's reasonable. I know you are very detail oriented. I did follow the steering bable path as well, but only after exiting the rigging tunnel. My motor is smaller than the Merc, so that might account for some of the difference.

It would be good to retake the measurements, making certain you are on the DC setting and 400A range. If you still read 20A, then try the 40A range and check to see that you don't get an "over current" or "out of range" warning indication.

I assume you are cranking the engine with the plugs in place, as not having them in would lower the load on the motor and potentially alter the measured value.

A meter that I am considering getting is a Mastech MS2108 True-RMS AC/DC Clamp Meter with Inrush Current Measurement. It is pretty cheap ($80 from Amazon), and I'm interested to see the difference between the inrush and steady state current. That being said, your Klein seems to be a great meter as well.

Let me know if you want an extra pair of hands locally. This is very interesting.

Dennis

jimh posted 01-12-2014 02:45 PM ET (US)     Profile for jimh  Send Email to jimh     
Tom--I cannot recommend a wire size because we do not know the current with real precision. Let us assume the peak current will be about 200-Amperes, and proceed with that assumption. I think it is a reasonable approximation until we get a better figure from measurement.

Let us assume a conductor of 4-AWG and a circuit length of 21-feet. But, first, a question: is 21-feet the total circuit length of both conductors? Or is 21-feet the length of one conductor from helm to engine, and we will need 42-feet of wire? Let me assume 21-feet is the one-way length, and we will have 42-feet of wire.

From the table (above) we see 4-AWG can handle 60-Amperes up to 12-feet (that is, 12-feet out and 12-feet back) for three percent at 12-Volts, i.e., Hebert's Rule, a corollary to the 700-mils current rating. This can be scaled on a linear extrapolation to 21-feet for a voltage drop of

21-feet/12-feet x 3-percent = 5.25-percent

The current can also be scaled in a linear extrapolation. If the current is to increase to 200-Amperes, then the voltage drop increases to

5.25-percent x 200-Amperes/60-Amperes = 17.5-percent

That seems high to me. Let's look for a conductor that will keep the peak voltage drop around 10-percent. We need a conductor rated for more current, by the ratio of 17.5/10 = 1.75, or 60 x 1.75 = 105-Amperes. Checking the table, 2-AWG should be close. 2-AWG is rated for 94-Amps.

If you use 2-AWG and run it 21-feet out and 21-feet back, you can maintain a just a bit over a 10-percent maximum voltage drop for a 200-Ampere load. That ought to be good for starting a typical engine.

Let me check that with an on-line calculator:

http://www.marinco.com/files/applets/wirecalculator.html

Note that on this calculator you have to enter the distance of the total circuit, so you would enter 42 feet instead of 21-feet. The calculator says 1-AWG for 42-feet, but if you drop back to 37-feet it says 2-AWG. So the Hebert's Rule calculation is verified.

jimh posted 01-12-2014 02:52 PM ET (US)     Profile for jimh  Send Email to jimh     
I looked up the Klein Tools CL2000 meter: it is a DC clamp-on meter. And it looks like it has a peak-hold function. It should be a dandy device for this measurement.
Tom W Clark posted 01-12-2014 03:08 PM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
Thank you for the good advice fellas. A more careful and repeated examination of my Klein Tools Instruction Manual has revealed the probable source of my error: it was set to AC not DC.

I will also use the MAXIMUM HOLD feature to capture the peak draw. I should have updated data by 1:30 PST.

Yes, my cable length measurements are for one conductor, not two.

Tom W Clark posted 01-12-2014 04:29 PM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
Nothing is ever as simple is it seems.

The Klein Tools CL2000 does indeed have a maximum (and minimum) capture mode but it will not work for this application because (it seems) it only works in the Auto Range mode which results in it capturing a reading of "OL" for Over Limit.

Manually setting the range to hundreds, I saw initial instantaneous readings of between 185 and 197 when we repeated the test over and over. Continuous cranking the motor produced readings of 100 to 115.

tmann45 posted 01-12-2014 07:43 PM ET (US)     Profile for tmann45  Send Email to tmann45     
Based on your current of 115 amps, battery cable based on 3% voltage drop is 3/0 AWG, with a voltage drop of 2.579% or 0.3094 volts. Using a 10% voltage drop, #3 AWG will have an 8.221% drop or 0.9865 volts.

The specifications on this wire, Ancor or Pacer "boat cable" is 105*C insulation temperature rating, used in a maximum 30*C environment and 1 or 2 energized wires in a bundle. These specs are different from SAE size and lower temperature rated insulation, like household wire from a big box store.

Ampacities of Ancor or Pacer marine, fine strand, tin plated wire:
18 = 20A
16 = 25A
14 = 35A
12 = 45A
10 = 60A
8 = 80A
6 = 120A
4 = 160A
3 = 180A
2 = 210A
1 = 245A
1/0 = 285A
2/0 = 330A
3/0 = 385A
4/0 = 445A

jimh posted 01-13-2014 11:12 AM ET (US)     Profile for jimh  Send Email to jimh     
A rating for current capacity is not really important in the decision of wire size for the outboard engine cranking motor. In a 12-Volt circuit of any length more than a few feet, the voltage drop that will occur in the conductor is more concern than the possible heating of the conductor due to current.

There are many ways to express the capacity of a conductor to carry current, but they typically all rely on the rise in temperature that will occur due to heating from resistive losses in the conductor. The rating for a maximum current then depends on how much heat rise is allowed or tolerated. The amount of heat rise that can be tolerated depends on the insulating material and the environment. As a result, there are many different possible ratings for current capacity for a conductor. If the conductor is a single conductor in open air with a low ambient temperature, a very high current is possible. If the conductor is part of a multi-conductor cable, encased in a confined space, a much lower current capacity is specified. The heating rise is not a function of the voltage in the circuit, but only a function of the current and the wire resistance.

The voltage drop that occurs is also a function of only the current and wire resistance. However, as the operating voltage of the circuit becomes lower, a point it reached where the percentage of voltage drop becomes more influential in specifying the conductor than the heat rise. In a 12-Volt power distribution system, the voltage drop becomes most important for any conductor length greater than a foot or two. For that reason, the conductor size must be chosen so as to maintain a certain maximum voltage drop. The current rating can be mostly ignored, because the current rating will always be much higher than the actual current in the circuit.

jimh posted 01-15-2014 02:20 PM ET (US)     Profile for jimh  Send Email to jimh     
Another method of deducing the proper wire size for a new battery cable which will be longer than the original is to base the new wire size on having the same or lower resistance than as the original. Perhaps an example will illustrate this more clearly.

Let us say the original cables for a particular engine were 10-feet in length and made of 4-AWG wire. The resistance of 4-AWG wire is found from a table

http://www.powerstream.com/Wire_Size.htm

to be 0.2485-Ohms per 1,000-feet. From this we calculate the resistance of an existing 10-foot cable:

0.2485-Ohms / 1000-feet x 10-feet = 0.002485-Ohms

Now we recalculate for the new case, which needs 21-feet of wire. We will need a conductor of a resistance R per 1000-feet such that

R/1000 x 21 = 0.002485

Solving for R we find

R = (00.002485/21) x 1000

R = 0.1183-Ohms per 1,000 feet

Referring to the table, we see that in order to have a conductor with a resistance of 0.1183-Ohms/1000-feet, we will need to use a conductor of 0-AWG. A conductor of 0-AWG will have a resistance of 0.0983/1000-feet, and a cable of 21-feet long will have a resistance of 0.0020643-Ohms.

Comparing the resistance of our original cables, 0.002485-Ohms, with the resistance of the new cables, 0.0020643-Ohms, we see the new cables are lower in resistance. This means we can be assured that the engine starter motor voltage drop will be no more (and actually less) with the new cables than with the old cables.

Note that in this calculation we ended up with quite an increase in conductor size, to 0-AWG from 4-AWG, even though the distance was only slightly more than doubled to 21-feet from 10-feet.

jimh posted 01-15-2014 02:35 PM ET (US)     Profile for jimh  Send Email to jimh     
In the above calculation, suppose we choose 1-AWG as the new cable. The resistance of a 1-AWG conductor of 21-feet would be

0.1239/1000 x 21 = 0.0026019-Ohms

If we compare with the original cable resistance of 0.002485-Ohms, we see we are just slightly higher in resistance, by a difference of 0.0001169-Ohm. We should be able to use a new 21-foot conductor of 1-AWG without too much difference in performance of the power distribution to the starter motor. Even at a peak current of 300-Amperes, a resistance of 0.0001169 only causes a voltage drop of 0.03507-Volts. There is probably no starter motor in the world whose torque output would be enormously different at a voltage change of only 0.03507-Volts.

What if we only use a conductor of 2-AWG? The resistance of a 2-AWG conductor of 21-feet will be

0.1563/1000 x 21 = 0.0032823-Ohms

This represents an increase in resistance in the cable compared to original of

0.0032823 - 0.002485 = 0.0007973-Ohms.

Again, at 300-Amperes, the voltage drop will be greater by

0.0007973-Ohms x 300-Amperes = 0.23919-Volts


Now we have reached the point of having a significant difference in voltage, as 0.239-Volts is about 2-percent of the system voltage (12.6-Volts).

Based on these calculations, we can use 1-AWG and maintain the almost same voltage to the starter motor with the 21-foot cables as with the original. If we use only 2-AWG cables we can expect about a 2-percent additional drop in voltage at the starter. That might prove to be too much in a marginal starting situation, say when the battery voltage was already well below its peak voltage.

jimh posted 01-15-2014 02:49 PM ET (US)     Profile for jimh  Send Email to jimh     
As we have seen in the above calculations, one effect of longer cables, even if made with larger wire gauge, is a possibility for more total resistance. The more resistance in the cables, the greater the voltage drop in those cables. In many circumstances, having a slightly higher voltage drop in the cables may have little effect, other than perhaps to cause the electric starter motor speed to be just slightly lower than it would be with a higher voltage delivered to it. The slightly lower speed may not affect engine starting in normal circumstances. However, in abnormal circumstances, the voltage could be very important.

Abnormal circumstances occur when the temperature is low or the battery is discharged from full capacity. The battery terminal voltage will be lower than normal. If the battery voltage is already low, and if the new cables have added too much voltage drop, a point will be reached when the voltage delivered to the electric starter motor will be too low to allow the starter motor to reach its normal speed or perhaps too low to even start to rotate.

With electric starter motors the peak current occurs at the moment of the motor being energized. Once the motor begins to turn, the current drops off. The effect of this is to create a certain threshold that must be overcome. If the voltage delivered to the starter motor is too low to overcome the threshold of the starting current, the motor will not start to rotate. By keeping the voltage drop in the conductors as low as possible, you will be able to keep the starter motor from stalling even as the battery voltage sags due to being already discharged. The extra margin in the starter cables could pay off in some situations, making a difference between starting and not starting.

Tom W Clark posted 02-01-2014 10:40 AM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
This project is hang fire for the time being as we await the arrival of the new cables.

One thing I did confirm was that there is no way in hell a pair 2 AWG will fit into the front of a 1989 Mercury 100. Not even close.

jimh posted 02-02-2014 08:59 AM ET (US)     Profile for jimh  Send Email to jimh     
Perhaps using a short section of smaller diameter cable to go into the engine will have to be considered.
saumon posted 02-02-2014 02:49 PM ET (US)     Profile for saumon  Send Email to saumon     
quote:
Perhaps using a short section of smaller diameter cable to go into the engine will have to be considered.

That's what I've done on mine (1991 Outrage 17 with E-Tec 90hp). Moved the batteries into the console with a switch. I used 2 AWG wire from the battery/switch in the console to remote battery posts installed near the engine, on wich I connect the engine via the OEM 4 AWG cables.

http://www.bluesea.com/products/category/PowerPost_Connectors/ Dual_PowerPost

As my switch is located 3 feet from the 2 batteries, I used 40ft of 2 AWG wire but with a single battery setup without switch and wired to remote posts, you may be ok with 30ft. total.

Tom W Clark posted 05-17-2014 11:12 AM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
I wanted to wrap up this thread. This project was moved to the back burner after the boat owner suddenly needed to have non-trivial surgery on his spine a couple months ago. He is recovering well and eager to get back on the water.

I've been spending a few hours each weekend pulling things together with him. I had warned him at the beginning that moving the battery may not be a big chore but I suspected that it would turn into a big project simply because so much had to be taken apart, that he would inevitably want to clean, repair, rearrange and upgrade everything we encountered.

That proved to be very true. The console as essentially disassembled and completely rewired. All the teak was removed, sanded and refinished. New accessories were bought and installed. VHF antenna, bilge pump, compass, etc.

Last night, after work, I went over and we completed the final connections. Everything works and the motor, which had not been run since October, fired right up.

In the end, we used continuous lengths (35' total) of 4 AWG cable from the battery in the console directly to the motor. This cable size seems too be small based on discussion and advice above, but at ~65 degrees, there was no hint of hesitation. The motor started right up so there was no chance for heat to build up in the cable. Perhaps next winter, when it is cold, we will see if it is as good. I suspect that it will prove to be more than adequate.

As I mentioned earlier, there was no way 2 AWG cable could have been used without a splice somewhere and that would have defeated the goal of having minimal rigging visible in the back of the boat.

jimh posted 05-17-2014 11:44 AM ET (US)     Profile for jimh  Send Email to jimh     
If you have the opportunity, use that nice Klein Clamp-on DC Ammeter and see what sort of peak cranking current or sustained cranking current you measure.

Another interesting measurement is taken by using a Voltmeter to measure the voltage from the battery positive terminal to the cranking motor during engine cranking. This will probably require making a temporary lead for the meter to span the distance between the console and transom. This voltage will show the actual voltage drop across the 4-AWG conductor.

Tom W Clark posted 05-19-2014 12:53 AM ET (US)     Profile for Tom W Clark  Send Email to Tom W Clark     
Will do.

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