
ContinuousWave Whaler Moderated Discussion Areas ContinuousWave: Small Boat Electrical Wire Size For Battery Cables

Author  Topic: Wire Size For Battery Cables 
Tom W Clark 
posted 01112014 06:08 PM ET (US)
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 clampon 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 01112014 06:48 PM ET (US)
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 01112014 11:36 PM ET (US)
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 oneway, 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 01122014 12:15 PM ET (US)
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 01122014 01:52 PM ET (US)
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=MarineWire , cut and pasted below:

jimh 
posted 01122014 02:23 PM ET (US)
Regarding measurement of current with a clampon meter: it is very easy and thus very inexpensive to measure alternating current (AC) with a clampon type meter. With AC the clampon meter becomes a current transformer and the current in the conductor becomes a oneturn loop on the transformer. It is common nowadays that even relatively inexpensive digital multimeters provide clampon measurement of AC current. To measure direct current (DC) with a clampon meter requires a different method. A Hall Effect sensor is used. Typically this sort of metering is only found in more expensive multimeters. There is also often a peakhold circuit in clampon 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 clampon multimeter. It is a FLUKE 374, and it cost almost $300. Using this device I have measured the peak starting current for a 2.5liter automobile fourstrokepowercycle engine to be about 350Amperes. 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 1Ampere/700circular mills calculation is as follows 00AWG = 190Amperes In a 12Volt system, the maximum distance those conductors can carry their rated current is 12feet (or 24feet of total conductor length) if less than a threepercent voltage drop is desired. For more on wire current capacity see Wire Conductor Size for Power Distribution 
DVollrath 
posted 01122014 02:41 PM ET (US)
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 TrueRMS 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 01122014 02:45 PM ET (US)
TomI cannot recommend a wire size because we do not know the current with real precision. Let us assume the peak current will be about 200Amperes, 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 4AWG and a circuit length of 21feet. But, first, a question: is 21feet the total circuit length of both conductors? Or is 21feet the length of one conductor from helm to engine, and we will need 42feet of wire? Let me assume 21feet is the oneway length, and we will have 42feet of wire. From the table (above) we see 4AWG can handle 60Amperes up to 12feet (that is, 12feet out and 12feet back) for three percent at 12Volts, i.e., Hebert's Rule, a corollary to the 700mils current rating. This can be scaled on a linear extrapolation to 21feet for a voltage drop of 21feet/12feet x 3percent = 5.25percent The current can also be scaled in a linear extrapolation. If the current is to increase to 200Amperes, then the voltage drop increases to 5.25percent x 200Amperes/60Amperes = 17.5percent That seems high to me. Let's look for a conductor that will keep the peak voltage drop around 10percent. We need a conductor rated for more current, by the ratio of 17.5/10 = 1.75, or 60 x 1.75 = 105Amperes. Checking the table, 2AWG should be close. 2AWG is rated for 94Amps. If you use 2AWG and run it 21feet out and 21feet back, you can maintain a just a bit over a 10percent maximum voltage drop for a 200Ampere load. That ought to be good for starting a typical engine. Let me check that with an online 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 21feet. The calculator says 1AWG for 42feet, but if you drop back to 37feet it says 2AWG. So the Hebert's Rule calculation is verified. 
jimh 
posted 01122014 02:52 PM ET (US)
I looked up the Klein Tools CL2000 meter: it is a DC clampon meter. And it looks like it has a peakhold function. It should be a dandy device for this measurement. 
Tom W Clark 
posted 01122014 03:08 PM ET (US)
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 01122014 04:29 PM ET (US)
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 01122014 07:43 PM ET (US)
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: 
jimh 
posted 01132014 11:12 AM ET (US)
A rating for current capacity is not really important in the decision of wire size for the outboard engine cranking motor. In a 12Volt 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 multiconductor 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 12Volt 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 01152014 02:20 PM ET (US)
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 10feet in length and made of 4AWG wire. The resistance of 4AWG wire is found from a table http://www.powerstream.com/Wire_Size.htm to be 0.2485Ohms per 1,000feet. From this we calculate the resistance of an existing 10foot cable: 0.2485Ohms / 1000feet x 10feet = 0.002485Ohms Now we recalculate for the new case, which needs 21feet of wire. We will need a conductor of a resistance R per 1000feet such that R/1000 x 21 = 0.002485 Solving for R we find R = (00.002485/21) x 1000 R = 0.1183Ohms per 1,000 feet Referring to the table, we see that in order to have a conductor with a resistance of 0.1183Ohms/1000feet, we will need to use a conductor of 0AWG. A conductor of 0AWG will have a resistance of 0.0983/1000feet, and a cable of 21feet long will have a resistance of 0.0020643Ohms. Comparing the resistance of our original cables, 0.002485Ohms, with the resistance of the new cables, 0.0020643Ohms, 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 0AWG from 4AWG, even though the distance was only slightly more than doubled to 21feet from 10feet. 
jimh 
posted 01152014 02:35 PM ET (US)
In the above calculation, suppose we choose 1AWG as the new cable. The resistance of a 1AWG conductor of 21feet would be 0.1239/1000 x 21 = 0.0026019Ohms If we compare with the original cable resistance of 0.002485Ohms, we see we are just slightly higher in resistance, by a difference of 0.0001169Ohm. We should be able to use a new 21foot conductor of 1AWG without too much difference in performance of the power distribution to the starter motor. Even at a peak current of 300Amperes, a resistance of 0.0001169 only causes a voltage drop of 0.03507Volts. There is probably no starter motor in the world whose torque output would be enormously different at a voltage change of only 0.03507Volts. What if we only use a conductor of 2AWG? The resistance of a 2AWG conductor of 21feet will be 0.1563/1000 x 21 = 0.0032823Ohms This represents an increase in resistance in the cable compared to original of 0.0032823  0.002485 = 0.0007973Ohms. Again, at 300Amperes, the voltage drop will be greater by 0.0007973Ohms x 300Amperes = 0.23919Volts
Based on these calculations, we can use 1AWG and maintain the almost same voltage to the starter motor with the 21foot cables as with the original. If we use only 2AWG cables we can expect about a 2percent 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 01152014 02:49 PM ET (US)
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 02012014 10:40 AM ET (US)
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 02022014 08:59 AM ET (US)
Perhaps using a short section of smaller diameter cable to go into the engine will have to be considered. 
saumon 
posted 02022014 02:49 PM ET (US)
quote: That's what I've done on mine (1991 Outrage 17 with ETec 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 05172014 11:12 AM ET (US)
I wanted to wrap up this thread. This project was moved to the back burner after the boat owner suddenly needed to have nontrivial 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 05172014 11:44 AM ET (US)
If you have the opportunity, use that nice Klein Clampon 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 4AWG conductor. 
Tom W Clark 
posted 05192014 12:53 AM ET (US)
Will do. 
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