Monthly Archives: April 2018

Alkaline batteries have second lives

Most people assume that alkaline batteries are one-time only, throwaway items. Some have used rechargeable cells, but these are Ni-metal hydride, or Ni-Cads, expensive variants that have lower power densities than normal alkaline batteries, and almost impossible to find in stores. It would be nice to be able to recharge ordinary alkaline batteries, e.g. when a smoke alarm goes off in the middle of the night and you find you’re out, but people assume this is impossible. People assume incorrectly.

Modern alkaline batteries are highly efficient: more efficient than even a few years ago, and that always suggests reversibility. Unlike the acid batteries you learned about in highschool chemistry class (basic chemistry due to Volta) the chemistry of modern alkaline batteries is based on Edison’s alkaline car batteries. They have been tweaked to an extent that even the non-rechargeable versions can be recharged. I’ve found I can reliably recharge an ordinary alkaline cell, 9V, at least once using the crude means of a standard 12 V car battery charger by watching the amperage closely. It only took 10 minutes. I suspect I can get nine lives out of these batteries, but have not tried.

To do this experiment, I took a 9 V alkaline that had recently died, and finding I had no replacement, I attached it to a 6 Amp, 12 V, car battery charger that I had on hand. I would have preferred to use a 2 A charger and ideally a charger designed to output 9-10 V, but a 12 V charger is what I had available, and it worked. I only let it charge for 10 minutes because, at that amperage, I calculated that I’d recharged to the full 1 Amp-hr capacity. Since the new alkaline batteries only claimed 1 amp hr, I figured that more charge would likely do bad things, even perhaps cause the thing to blow up.  After 5 minutes, I found that the voltage had returned to normal and the battery worked fine with no bad effects, but went for the full 10 minutes. Perhaps stopping at 5 would have been safer.

I changed for 10 minutes (1/6 hour) because the battery claimed a capacity of 1 Amp-hour when new. My thought was 1 amp-hour = 1 Amp for 1 hour, = 6 Amps for 1/6 hour = ten minutes. That’s engineering math for you, the reason engineers earn so much. I figured that watching the recharge for ten minutes was less work and quicker than running to the store (20 minutes). I used this battery in my firm alarm, and have tested it twice since then to see that it works. After a few days in my fire alarm, I took it out and checked that the voltage was still 9 V, just like when the battery was new. Confirming experiments like this are a good idea. Another confirmation occurred when I overcooked some eggs and the alarm went off from the smoke.

If you want to experiment, you can try a 9V as I did, or try putting a 1.5 volt AA or AAA battery in a charger designed for rechargeables. Another thought is to see what happens when you overcharge. Keep safe: do this in a wood box outside at a distance, but I’d like to know how close I got to having an exploding energizer. Also, it would be worthwhile to try several charge/ discharge cycles to see how the energy content degrades. I expect you can get ~9 recharges with a “non-rechargeable” alkaline battery because the label says: “9 lives,” but even getting a second life from each battery is a significant savings. Try using a charger that’s made for rechargeables. One last experiment: If you’ve got a cell phone charger that works on a car battery, and you get the polarity right, you’ll find you can use a 9V alkaline to recharge your iPhone or Android. How do I know? I judged a science fair not long ago, and a 4th grader did this for her science fair project.

Robert Buxbaum, April 19, 2018. For more, semi-dangerous electrochemistry and biology experiments.

Calculating π as a fraction

Pi is a wonderful number, π = 3.14159265…. It’s very useful, ratio of the circumference of a circle to its diameter, or the ratio of area of a circle to the square of its radius, but it is irrational: one can show that it can not be described as an exact fraction. When I was in middle school, I thought to calculate Pi by approximations of the circumference or area, but found that, as soon as I got past some simple techniques, I was left with massive sums involving lots of square-roots. Even with a computer, I found this slow, annoying, and aesthetically unpleasing: I was calculating one irrational number from the sum of many other irrational numbers.

At some point, I moved to try solving via the following fractional sum (Gregory and Leibniz).

π/4 = 1/1 -1/3 +1/5 -1/7 …

This was an appealing approach, but I found the series converges amazingly slowly. I tried to make it converge faster by combining terms, but that just made the terms more complex; it didn’t speed convergence. Next to try was Euler’s formula:

π2/6 = 1/1 + 1/4 + 1/9 + ….

This series converges barely faster than the Gregory/Leibniz series, and now I’ve got a square-root to deal with. And that brings us to my latest attempt, one I’m pretty happy with discovering (I’m probably not the first). I start with the Taylor series for sin x. If x is measured in radians: 180° = π radians; 30° = π/6 radians. With the angle x measured in radians, can show that

sin x = x – x3/6 x5/120 – x7/5040 

Notice that the series is fractional and that the denominators get large fast. That suggests that the series will converge fast (2 to 3 terms?). To speed things up further, I chose to solve the above for sin 30° = 1/2 = sin π/6. Truncating the series to the first term gives us the following approximation for pi.

1/2 = sin (π/6) ≈ π/6.

Rearrange this and you find π ≈ 6/2 = 3.

That’s not bad for a first order solution. The Gregory/ Leibniz series would have gotten me π ≈ 4, and the Euler series π ≈ √6 = 2.45…: I’m ahead of the game already. Now, lets truncate to the second term.

1/2 ≈ π/6 – (π/6)3/6.

In theory, I could solve this via the cubic equation formula, but that would leave me with two square roots, something I’d like to avoid. Instead, and here’s my innovation, I’ll substitute 3 + ∂ for π . I’ll then use the binomial theorem to claim that (π)3 ≈ 27 + 27∂ = 27(1+∂). Put this into the equation above and we find:

1/2 = (3+∂)/6 – 27(1+∂)/1296

Rearranging and solving for ∂, I find that

27/216 = ∂ (1- 27/216) = ∂ (189/216)

∂ = 27/189 = 1/7 = .1428…

If π ≈ 3 + ∂, I’ve just calculated π ≈ 22/7. This is not bad for an approximation based on just the second term in the series.

Where to go from here? One thought was to revisit the second term, and now say that π = 22/7 + ∂, but it seemed wrong to ignore the third term. Instead, I’ll include the 3rd term, and say that π/6 = 11/21 + ∂. Extending the derivative approximations I used above, (π/6)3 ≈ (11/21)+ 3∂(11/21)2, etc., I find:

1/2 ≈ (11/21 + ∂) -(11/21)3/6 – 3∂(11/21)2/6 + (11/21)5/120 + 5∂(11/21)4/120.

For a while I tried to solve this for ∂ as fraction using long-hand algebra, but I kept making mistakes. Thus, I’ve chosen to use two faster options: decimals or wolfram alpha. Using decimals is simpler, I find: 11/21 ≈ .523810, (11/21)2 =  .274376; (11/21)3 = .143721; (11/21)4 = .075282, and (11/21)5 = .039434.

Put these numbers into the original equation and I find:

1/2 – .52381 +.143721/6 -.039434/120 = ∂ (1-.274376/2 + .075282/24),

∂ = -.000185/.86595 ≈ -.000214. Based on this,

π ≈ 6 (11/21  -.000214) = 3,141573… Not half bad.

Alternately, using Wolfram alpha to reduce the fractions,

1/2 – 11/21+ 113/6•213 -115/(120•215) = ∂ (24(21)4/24(21)4 – 12•112212/24•214+ (11)4/24•214)

∂ = -90491/424394565 ≈ -.000213618. This is a more exact solution, but it gives a result that’s no more accurate since it is based on a 3 -term approximation of the infinite series.

We find that π/6 ≈ .523596, or, in fractional form, that π ≈ 444422848 / 141464855 = 3.14158.

Either approach seems OK in terms of accuracy: I can’t imagine needing more (I’m just an engineer). I like that I’ve got a fraction, but find the fraction quite ugly, as fractions go. It’s too big. Working with decimals gets me the same accuracy with less work — I avoided needing square roots, and avoided having to resort to Wolfram.

As an experiment, I’ll see if I get a nicer fraction if I drop the last term (11)4/24•214: it is a small correction to a small number, ∂. The equation is now:

1/2 – 11/21+ 113/6•213 -115/(120•215) = ∂ (24(21)4/24(21)4 – 12(11221)2/24•214).

I’ll multiply both sides by 24•214 and then by (5•21) to find that:

12•214 – 24•11•213+ 4•21•113 -115/(5•21) = ∂ (24(21)4 – 12•112212),

60•215 – 120•11•214+ 20•21^2•113 -115 = ∂ (120(21)5 – 60•112213).

Solving for π, I now get, 221406169/70476210 = 3.1415731

It’s still an ugly fraction, about as accurate as before. As with the digital version, I got to 5-decimal accuracy without having to deal with square roots, but I still had to go to Wolfram. If I were to go further, I’d start with the pi value above in digital form, π = 3.141573 + ∂; I’d add the 7th power term, and I’d stick to decimals for the solution. I imagine I’d add 4-5 more decimals that way.

Robert Buxbaum, April 2, 2018