
ContinuousWave Whaler Moderated Discussion Areas ContinuousWave: Small Boat Electrical Moving Batteries

Author  Topic: Moving Batteries 
floater88 
posted 04262013 05:41 PM ET (US)
I'm in the process of moving my batteries from the splash well to under the pull out stairs. The boat is an 1988 Boston Whaler Revenge 20 WT. I'm wondering if i'm better off running cables right from where they connect at the motors or from a junction point? I have a merc 200 and a 15 hp kicker and both of them have cables long enough to reach the port side of the splash well where I figure I could mount some kind of juncion point and run the longer cable to the batteries from there. Is this feasable to save me some money on the cable lengths I need to buy? If I go the junction point what do I use to join the ends of the cables together? THanks and for any tips. 
andygere 
posted 04262013 05:49 PM ET (US)
Junctions are certainly possible, but they add a possible point of failure. Are you going to have a battery switch or VSR in your system? If so, you can use that at a connection point on the (+) cables. I'd home run the () cables under that scenario, but better to do it right and keep the switch near the batteries, and home run the primary cables to the motors. If you have a terminal lug crimper, you can reuse some of the original cable for shorter runs from the battery bank to the switch, etc. 
saumon 
posted 04262013 06:27 PM ET (US)
For the junction, simply use power posts like those, that's why they're designed for: http://www.westmarine.com/webapp/wcs/stores/servlet/ ProductDisplay?productId=17184&catalogId=10001&langId=1&storeId=11151& storeNum=50523&subdeptNum=50549&classNum=50553 On mine, I bolted 2, a 5/16" and a 3/8", 6" apart on a 10"x4" piece of HDPE (cutting board), that was bolted inside one of the stern seats box. Even if they're covered with rubber caps, I prefer to have them in an enclosed area to minimize the risk or shorting... 
andygere 
posted 04272013 12:30 AM ET (US)
Both my current Outrage and my previous Montauk had spliced battery cables (done by previous owners) and both of them eventually failed. I was lucky that I wasn't out at sea either time. For the price of an extra 4 feet of cable, why would you want to take a shortcut with splices? When you subtract the cost of the extra lugs and the splice posts, the savings is probably negligible. 
floater88 
posted 04272013 06:09 AM ET (US)
Ya for an extra four feet of cable I guess it is better to go right to the motors. Using one of those battery switches though. Both positive terminals from the motors go to the switch and from there a pos goes to the batteries? Then both neg cables from the motors goes right to the batteries? I'm hoping to squeeze two group 24 batteries under the stairs. One battery would be for starting the main motor only and the other would be used for powering the circuit panel and the two electric down riggers. The kicker gets the most use so It's charging circuit would trickle charge both batteries I hope. 
floater88 
posted 04272013 06:40 AM ET (US)
Ok I just took another look at Jims electronic section and i'm gonna go with his 'new' set up for twin engines and two batteries. My concerns were with having both engines running upon start up before heading out as the kicker needs to be warmed up before it will go into gear. With the Jims set up I can run both engines and charge both batteries at the same time wihtout wrecking anything. Thanks guys. 
jimh 
posted 04272013 08:27 AM ET (US)
In an installation in which the battery cables run a long distance, and especially when they run a long distance in some enclosed space or in a conduit, a disconnect should be provided close to the battery. In the dualengine twobattery arrangement that I show in my article http://continuouswave.com/whaler/reference/dualBattery.html in the drawing with the caption "New", and as I note in the text with the comment
quote: there is no disconnect in the battery circuit to the outboard engine. The electrical circuit of the outboard is always connected, and there is alway power on the battery cables running to the outboard from the battery. An arrangement like that is probably acceptable if the battery is mounted in an open transom and is only a few feet from the engine. That is a very common situation in many small boats. The battery is just connected directly to the cables from the outboard engine. The total length of unfused battery cable is only about six feet. In an arrangement with the battery located in an interior compartment of the boat, and with cables running through a tunnel or other passageway to the engine from the battery, having a electrical disconnect near the battery is probably better. There will be 20 feet of battery cable in an installation like that, and that length of unprotected power distribution is probably not the best approach. 
jimh 
posted 04272013 08:50 AM ET (US)
When the battery is moved farther away from its principal load, the electric starting motor, there is concern about the total voltage drop in the distribution system. The existing battery cables on the outboard are a wire size that will create some voltage drop during engine cranking, but not too much voltage drop to prevent successful cranking and engine start due to either the electric motor stalling or turning too slowly. If you maintain the present cables and add an extension to reach a battery that is farther away, the resistance in the extension cables plus the resistance in the original cables will produce the voltage drop. The segment of the cables which is the extension portion must have very low resistance, because it has to deliver the battery voltage to the original cables with nearly no voltage drop. This may require the extension cables to be very large. If the original battery cables are removed and a new, onepiece, battery cable installed, the cable must have very low resistance. It may be possible to use a onepiece cable that is actually smaller than would be needed in a twosegment installation. For example, let me demonstrate with some resistance values. I am going to just make up some values in order to demonstrate the problem. These values are not actual values and don't use them as a guide. They are just to demonstrate the problem. Let us say that the original cables have 0.2Ohm of resistance. And let us say that the maximum resistance that can be tolerated for proper starting is 0.3Ohm. If the original cables are connected directly to the battery, the resistance is 0.1Ohm less than the maximum, so there is some margin in the circuit. Next, let us say that the cables must be extended. The extension cable has to have very low resistance. There is only 0.1Ohm of margin left. This means a cable must have only 0.1Ohm of resistance at most, and preferably less. It would be better to get a cable of 0.05Ohm resistance. Then the total circuit resistance would be 0.05 + 0.2 = 0.25Ohm, and there would still some margin in the circuit, about 0.05Ohm. Another and perhaps better approach would be to make new cables, onepiece cable, no splices or joints, and replace the existing cables. If the cables were made from a proper wire size, the total resistance of the new cables could be 0.2Ohm, and we would still have the original margin of 0.1Ohm in the circuit. You can see that in the case of the extension cable method, the resistance of the extension cable must be no more than 0.05Ohm if we are going to retain any margin it the circuit at all. In the case of the replacement cable, the resistance can be 0.2Ohm and the margin will remain the same. If the extension portion of the cable were twice the original length, then the length of the extension would 2X, and the length of the new onepiece cables would be 3X. Now we can compare the wire gauges. An extension cable would need a resistance of 0.05Ohm for a length of 2X, or a resistance of 0.025Ohm per Xdistance. A replacement cable would need a resistance of 0.2Ohm for a length of 3X, or a resistance of 0.066Ohm per Xdistance. This means the resistance per unit of length of the extension cable must be 0.025/0.066 or 0.375 of the resistance of a replacement cable that is onethird longer. To lower the resistance that much will mean using a cable of much larger size for the extension than the size needed for the onepiece cable. The reason for that, of course, is the circuit still has the relatively high resistance of the original cables, where were sized for only a short run. 
jimh 
posted 04272013 09:01 AM ET (US)
[Moved to Small Boat Electrical.] 
floater88 
posted 04272013 11:40 AM ET (US)
THanks Jimh. I'm gonna go ahead and get a one piece cable from motor to batteries. I'm gonna get the 4 awg. THe smaller kicker has a real small diameter set of cables on it and the bigger motor looks like it's already got the 4 awg so i'm gonna do both in 4 awg. THanks 
floater88 
posted 04272013 02:28 PM ET (US)
Well I went to the store and took a look at the cables. Looks like the main motor has 4awg now and the smaller one has 8. I'm gonna replace both with 2awg and hope for the best. Looks like I need a total of 80 feet of cable and at 3.00/foot it's gonna cost me money!! Thanks for all the tips 
swist 
posted 04272013 05:20 PM ET (US)
All you need is a little resistance in a connection for a highcurrent application and you have not only lot of voltage drop but the potential for heat to be generated. It's tough to keep a junction like that pristine and tight in a marine environment. You will notice that in most cars the positive battery cable runs directly to the starter solonoid (which is mounted on the starter motor itself) this minimizing both distance and intermediate junctions. 
jimh 
posted 04282013 08:52 AM ET (US)
The electric starting motor for the auxiliary engine probably draws much less current than the electric starting motor for the main engine, and the cable sized needed for it will be smaller. I will try to put some realworld numbers to my example above by using the actual resistance for various lengths and wire gauges. 
jimh 
posted 04282013 12:05 PM ET (US)
Here are some realworld values for resistance. For the resistance of copper wire according to its size in the American Wire Gauge (AWG) I relied on the information at http://hyperphysics.phyastr.gsu.edu/hbase/tables/wirega.html The resistance at normal temperatures for copper wire per 1,000feet is 4AWG = 0.2485 In the 2010 rigging guide from Bombardier Recreational Products for Evinrude engines, the following guide lines are given for electrical cable size and distance between engine and battery for a large V6 engine: 1 to 10feet of cable: 4AWG The distances above refer to the distance between battery and engine, not for the total length of the cables. I am going to assume that the total length of the conductors will be twice those dimension. The maximum allowed resistance can be inferred as follows: If 4AWG can be used up to 10feet, this implies 20feet of conductor in the circuit. The resistance of 20feet of 4AWG is found from the data above to be 0.0050Ohms. Now we compare to the other cases. For 2AWG at 15feet we have 0.0047Ohm limit, and for the 1AWG at 20feet we have a 0.0050Ohm. We see that the maximum resistance allowed for proper engine starting is 0.0050, 0.0047, or 0.0050Ohm. From this we can make an inference that the total resistance for good engine starting in the circuit between the battery and the engine must not be more than about 0.0048Ohm Now we investigate the situation we have at hand in this discussion. There is an engine with its original 4AWG cables. Let us say these cables are 6feet long, or a total conductor length of 12feet. We calculate their resistance to be 0.002982Ohm. We subtract this resistance from the maximum resistance allowed, 0.0048ohm, to find how much resistance we can add to the circuit: 0.00182Ohm. Now we can calculate the maximum length we can extend the cables with a 2AWG extension. With a resistance of 0.1563Ohms/1,000feet, we find that the total length of 2AWG conductor that has a resistance of 0.00182Ohm will be 11.6feet. Thus we can only extend the cables an additional 5.8feet using 2AWG. In other words, the original 6foot 4AWG cables can be extended an additional 5.8feet with 2AWG extensions, until we reach the maximum resistance allowed. This gives us a total distance allowed with this method of only 11.8feet. Repeating the above but making the extension with 1AWG cable shows that the maximum length of a 1AWG extension would be 7.3feet. This gives us a total distance allowed with this method of only 13.3feet. In this analysis there was no resistance allowed for making any splices or connections between the cables. Every splice or connection will add some resistance to the circuit, so it would be prudent to consider that the distances calculated for an extension cable are probably too long for pratical use. In practice the extension distance would need to be shorter to account for the added resistance of the connection between the original cables and the extension cables. This analysis shows why installations in which the batteries are moved away from the engine and the original cables are retained but extended by larger cables tend to have problems. The original cables have so much resistance in them that they do not provide much margin for any extension cable to have resistance. For best outcomes, the cables to the engine from the battery should be replaced with new cables, onepiece cables, of the appropriate length. If the distance between the battery and engine is more than 20feet, the cable size will have to be increased even larger, to 0AWG conductors, if the guidelines for maximum resistance (which have been inferred from the rigging data) are to be followed. 
jimh 
posted 04282013 12:15 PM ET (US)
Above, I have calculated the maximum resistance recommended in the starting circuit for the battery cables as 0.0048Ohm. We can estimate that during starting a very high current will flow. The initial current flow may be around 300Amperes. We can calculate the voltage drop in the cable part of the circuit as 1.44Volts. When the battery is placed under a load of 300Amperes, its terminal voltage will drop from the noload resting terminal voltage. This is an additional voltage drop to take into consideration. The battery itself will probably drop at least 1Volt under a 300Ampere load. We can roughly figure the voltage delivered to the starter motor under a 300Ampere load to be about 2.5Volts less than the voltage measured at the battery terminals with no load present. 
floater88 
posted 04282013 09:00 PM ET (US)
Thank you so much for calculating all that info. After looking at it again today I figured I was gonna run the original 4 awg and 6 awg to disconnects and from there another 16' of 2 awg to the batteries. If I have to use 0 awg now it's starting to really add up. I'm gonna have to rethink this one. THanks 
jimh 
posted 04292013 11:07 AM ET (US)
When I managed several large uninterruptible power supply (UPS) devices, the battery bank had monitoring devices that could calculate the internal resistance of the battery. A new, highquality, AGM 12Volt batterywhich cost about $250 would typically have an internal resistance of less than 0.003Ohm when new. Once in service the internal resistance tended to increase, and after three years it was typical to see an internal resistance of more than 0.003, perhaps as much as 0.0035Ohm. When 300Amperes flows from the battery, it flows through that internal resistance, so there is a voltage drop. If we assume the resistance is 0.0033Ohms, then a 300Ampere load produces a 1Volt drop in the battery voltage. Again, that is for a really highquality AGM battery that is at full charge and in good condition. A battery of lower quality will have a higher internal resistance when new, and that internal resistance will increase with age and service. Because the internal resistance of the battery can vary with the quality, age, and care given to the battery, we see that in any particular situation with an outboard engine and a starting battery a range of variation in terminal voltage drop under load exists, even beyond the usual expectation of the terminal voltage to vary due to state of charge. These variations can also affect the outcome of moving the batteries away from the load. Lowcost, moderate quality, threeyearold batteries might be sufficient to start an outboard engine if there is no voltage drop in the cables. Those same batteries may be inadequate to start an outboard if they are connected by long cables that introduce additional voltage drop. 
floater88 
posted 04292013 05:16 PM ET (US)
I went to the local Marina/Merc dealer today and they said that engine came with 6 foot leads and I could extend those up to 12 feet with no issues. Beyond 12 feet they said 2 awg would do the trick with no problems. Sure hope they weren't lying to drum up some business! Either way i'm gonna put off the relocation program and concentrate on the new Hydraulic Steering from Uflex I just ordered. I will post pics and how it went when it goes. Thanks for all the tips and info. 
jimh 
posted 04292013 11:32 PM ET (US)
Your Mercury dealer may be correct. That would be the case if the rigging recommendations for your specific outboard engine allowed cable lengths of that wire gauge and length. The best way to establish that would be to see Mercury literature that specified those tolerances. In my calculation above, my tolerance for resistance was based on the literature of the engine manufacturer, which was Evinrude, and which applied to highhorsepower V6 models. It is reasonable that the electric motor starting current might vary among various engines. The starter motor current could vary with different designs of starter motors, and perhaps with gear ratio of the starter pinion to the flywheel gear. With Evinrude outboard engines, they won't start unless the cranking speed reaches a certain minimum RPM. This may have influenced the recommendation for battery cable tolerance; that is, they want the engine to be cranked over at a decent speed, about 300RPM, in order to be certain to be above the minimum engine rotation speed to start. 
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