It would be nice to know more about the descriptors of electricity. There are many sources of explanation. What are volts (pressure), amps (flow rate), watts (volumetric measure). Wondering why your iPad charges so slowly on an iPhone charger (1.0 amps) or horrors USB (.5 A) vs the larger iPad power adapter (2.1 A)? There you have it. Amps (A).  Its a given that your house is wired for a nominal 115 v and you car is 12 v. These types of numerical values for volts, amps, and watts in various application are furnished. If one of these three is unknown we can derive it from the other two. It may not be necessary to understand a precise definition but only to realize that they are used to describe the work of energy.  Continue onward.

In general, energy (E) is equivalent to power (P) multiplied by time (t). To determine E in watt-hours (Wh), P must be expressed in watts and t must be expressed in hours. Suppose a 60-W bulb burns for 3 h. Then P = 60 and t = 3, so the energy E in Wh is:

P * t = E using the example values for the variables 60 * 3 = 180 Wh

If P and t are not specified in watts and hours respectively, then they must be converted to those units before determining E in watt-hours. Larger values of P, upwards of 1,000 Wh would be expressed as kilowatt-hour (kWh). i.e. 1,000 Wh is the same as 1 kWh where k signifies 1,000. Fewer zeros that way.

One might be tempted to say 180 watts was the energy consumed but this would be inadequate as it leaves out the time factor. Using the full equation allows you to calculate unit(s) of energy. Take a look at your utility billing and you will see that burning your 60 W bulb for 3 hours resulted in a balance due of $0.0216 Your utility company doesn’t need to know that it was a 60 W light bulb or care that it was two 30 W bulbs or … They just need to know that it was 180 Wh.

How did we get P expressed in Watts in the first place? The light bulb example was straightforward as its P value was printed on the bulb. In some instances a device might have its P listed instead as amps. If we know this amperage value and we also know the value of volts then we can get to watts. Power (P) expressed in watts is equivalent to volts (v) multiplied by amps (A). Suppose a small motor is rated at 5 amps when connected to a 12 volt battery. Then A = 5 and v = 12, so P in watts is:

v * A = P using the example values for the vatiables 12 x 5 = 60 watts

Knowing at least two of the variables means that we can derive the unknown variable. In the above instance we knew v and A. What if we know P expressed in watts and we know v? Yes, we can figure for A:

P / v = A using the example values for the variables  60 / 12 = 5 amps

We’ve discussed E used by light bulb or motor consumers but what about stored energy — like in a battery? We would like to know the storage capacity of a battery. Knowing the available energy, and rate of energy consumption we can solve for t. In real words, if we run that light bulb all night we can calculate how much energy was used or how much will still be available or how much energy we must restore to the battery so that it can be used again the next night.

Battery energy (E) is equivalent to current draw expressed in amps(A) multiplied by time (t). The product is amp-hours (Ah). Then if A = 15 and t = 20 the energy (E) in amp-hours (Ah) is:

A * t = E using the example values 15 * 20 = 300 Ah

This battery supplied 15 A to a consumer for over a period of 20 hours during which 300 Ah was consumed. A typical deep-cycle battery will have a capacity rating expressed in Ah as its performance specification. A battery with E of 300 Ah would be fully discharge (deplete) in the above scenario.

Let’s convert the battery E expressed in Ah to power (P). Recall that volts x amps = P. So if we have a 12 v battery that can flow 300 Ah current we get P in watt-hours:

v * Ah = P using the example values for the variables 12 * 300 = 3.6 kWh

Great. So we have to convert the battery to kWh or convert the consumers to Ah. Let’s stick with the industry standard of capacity measure for batteries — amp-hours. Here’s a spreadsheet that shows a tabulation of various appliance items measured by amp-hours on S/V Talmid :

Sumtotal 1 day [underway] usage [from the Ah columns] requirement for this sailboat is 235 Ah. The battery bank capacity is 300 Ah. From this energy audit we could determine if our battery storage capacity was adequate or inadequate. Also some arithmetic will give an idea of how often, and for how long it will take to recharge the storage battery.

Charging Interval
= (battery bank (Ah) * allowable drawdown (30%) / daily requirement (Ah)) * 24hrs
= (300 * 30% / 161) * 24 = 13 hours

The audit tabulation reveals that sailing all night requires 161 amp-hours. The available energy is 300 amp-hours. Based on those givens the Charging Interval equation shows that after 13 hours we need to replenish the available energy.

Charging Period
= (battery bank (Ah) x allowable drawdown (50%) / alternator output(A) – hourly draw down (Ah)) + charging loss and thus: (300 * 30%)  / (100 – 13.4) * 120% = 1.3 hours

Assuming that the battery is now discharged to 30% the Charging Period equation informs that we need to run the engine with alternator for 1.3 hours.

formula explanation notes:

• allowable draw down – typically to maximize battery health we don’t fully drain it to 0. Using only 25% of it before a recharge would be ideal. 50% is accepted as a maximum and counts as a cycle. Due to the nature of the material elements and effects of chemistry within  the battery, its life span has a finite number of cycles of perhaps 300 to 600.
• daily requirement – is the sum of all the operative appliances based on how many hours they were switched on and their consumptive rate.
• alternator output – 100 A used in the example may be typical but other installations will vary depending on budget or resource.
• hourly draw down –  is energy point in time that continues to be used by appliances as you are re-charging. It is a deduction from the charging source which now has the double duty of feeding the consumers and the depleted battery.
• charging loss – an arbitrary value, in this case 20%, which reflects real world inefficiencies. e.g. energy inadvertently converted to heat while converting AC to DC

We could play with the numbers — work the battery a little harder by discharging to 50%. This would allow us to enjoy 22 hours of peaceful serene sailing before firing up the noisy engine to recharge. On the other hand, doing so would require a more lengthy 2 hour recharge because we used more.

Takeaway: Volts times Amps equals Watts. This is all that you need to memorize from this post. Now you can determine the energy needs/cost of a hair dryer. Determine if your lantern will make it through the weekend campout.  A fun exercise might be to figure offsetting the energy consumption with Solar and or Wind. How many solar panels might be needed to relegate our engine generator (or utility company) to a backup role. Energy independence! and the subject of a future post.

Sometimes there is a surprise when the anchor comes up.  Usually you find that the bottom was sandy (clean or grassy) or muddy (unclean). If it’s rocks or coral, then you probably knew that already because holding would have been poor to nil.

A recent overnight stay rest break took place off the Intracoastal Waterway (ICW) outside buoy marker 23. The anchorage has easy access but frequently shares parking with tugs and barges. For this stopover there was plenty of sea-room and I snuggled in.  Raising anchor the next morning seemed routine until the last bit and this is what I saw:

My plow anchor had dredged up what looked like cable and I was aghast thinking it was critical telecommunication infrastructure or electrical utility for the island adjacent. But my chart provided no notes of caution had that been a concern. Sometimes when an anchor fouls it remains planted to the seabed where you can’t see the cause; not without a swim and a free dive for inspection. I’m lucky that the anchor windlass on deck had the muscle power to lift the cable from the bottom of its 10′ depth.  No SCUBA needed.

A close look revealed this cable to be a crusty steel hawser that had gotten away  from a work vessel. Where a recreational boater might use a 1/2″ or less light size/weight rope for securing his craft, this one was industrial gauge; at least an inch or so in diameter and very heavy duty. In fact, due to the angle of the anchor plow, the beast was well caught up. I tried backing down to see if it would just flip off but I couldn’t shake it. I thought maybe a neighboring work crew that was busy getting a barge ready might have tools, skills, or desire to take claim. I decided to solve this one on my own.

If I had a do-over, I would have tied a length of rope to the cable and the bitter end to my bowsprit platform.  Simply lowering the anchor chain at that point would have freed my anchor. What I ended up doing was raising the anchor until the plow edge went horizontal. With a flat surface I figured I could just slide the cable off. But bare hands still couldn’t budge. I went aft to find my boat hook pole to gain leverage. With not too much heave-ho the cable slid aft toward the anchor tip. The distance to freedom wasn’t far but each inch required some grunt on my part. As the bulky cable was finessed off it immediately plunged back down to the dark depths. Unfortunately, and I realized in an instant too late what was to happen next and what the outcome would be, because — the boat hook was still hooked! The heavy cable was about to claim it and my arm too had I not made the correct choice and released my grip.

King Neptune has his cable back — and my boat hook as well.