Our town newspaper, the San Francisco Chronicle, has a Trivia column that runs twice a week. Recently, a couple items caught my eye: "If you have 9 pennies and your scales say they weigh an ounce, then you know your scales are in calibration." I reached in my change bucket and grabbed 9 pennies and put them on my gram scales. It came out to 26.1 grams — rather less than 28.35 grams — definitely less than 1 ounce. Hmmm. What is going on? Surely the Trivia Man has to be correct. One penny has to weigh as much as the other, doesn't it?
It turned out that some pennies — the old copper pennies before 1982 — DO weigh about 1/9 ounce. Actually, they're about 3.06 grams, so 9 of them are about 27.5 grams, still 3% lighter than an ounce. The newer ones, since about 1982, weigh about 1/11 ounce, or 2.49 grams ±0.01. So you can fool around and actually calibrate your scales now that you know which kinds of pennies weigh what. One time I got some pennies with glue on them. I put them on my stove burner and burned the glue off. No problem. Note, the amount of electricity it took — perhaps 1 kW x 60 seconds — is less than the cost of the pennies, by a factor of perhaps 10.
Then, months later, I got some more pennies with glue on them. When I put them on the stove, the pennies began to droop and melt. What the hey? Somebody is making counterfeit pennies? I complained to the Secret Service that somebody seems to be counterfeiting pennies. It was only LATER that I realized that the U.S. Government is making counterfeit pennies. Zinc instead of copper. No wonder that some melt (the new ones), while the older copper ones do not. I'd forgotten about that, until the weight of the pennies reminded me.
The Trivia column then said, "How thick is a millimeter? It's as thick as your thumbnail." Hmmm. My thumbnail was a little too long — and it needed to be trimmed. So I trimmed it off. I looked at it. It surely did not seem to be nearly 1 mm thick. But I didn't have any precision calipers or micrometers. So how could I tell how thick it really was?
I took a small steel ruler, about 1/4 inch wide, and stood it on end. I used the ruler to make a differential measurement between the thickness of the thumbnail and the thickness of several sheets of ordinary copier paper. This was done by standing up the ruler with one corner on the thumbnail and one corner on the paper stack.
NOW, we know that a stack of 5 reams (2500 sheets) of paper is 9.8 inches high. I have lots of old Xerox boxes that are 9 inches high but the paper is taller than the box. So we know that 39 inches of paper equals 10,000 pieces of paper. Each sheet is 3.9 milli-inches thick — give or take 5%.*
When I compared the thickness of the thumbnail to 3 pieces of paper, the ruler stood up with a tilt to the left. When I compared it to 5 pieces of paper, it tilted to the right. When I compared to 4 sheets — neutral. Of course, I had to turn around the ruler to make sure it had no bias. And I swapped around the base to make sure it had no bias. In conclusion, my thumbnail is about 15.7 milli-inches thick, or 0.38 mm. Not 1.0 mm. I sent a note on this to the Trivia expert, stating that he was wrong by nearly a factor of three.** No reply yet.
Then Mr. Trivia said, "What is there that weighs a gram? One paper clip." I got suspicious immediately. I grabbed a collection of paper clips and put some on my old Cenco Triple Beam Balance — scales that can measure up to 111 grams, with a resolution of 0.01 grams. The small paper clips all weighed 0.5 grams. The big ones all weighed 1.5 grams. I never did find one that weighed 1 gram. So Mr. Trivia does not have a very good batting average with me....
My point is NOT that you need fancy scales or meters to make a pretty good measurement. You need thinking to make pretty good measurements. A good scale or meter just makes it easier.
Back in the 1960s, some guys could not make any measurement without a precision differential voltmeter — a Fluke meter. Remember those knobs that you had to servo by hand to match the unknown voltage? When a guy could not make a measurement without one, we called that "Flukemia." But then digital voltmeters came along. Now we have a whole generation of engineers and technicians that only know how to measure things with a DVM. That is a far cry from engineers who know how to prove that 9 pennies weigh 1 ounce, under difficult conditions....
That reminds me of a lecture I once gave — "What's all this Measurement Stuff?" — to a local group of engineering students, sponsored by the IEEE Measurement Society. I asked these students, "What are the biggest sources of error in measurements? Thermocouples? RFI? Bad connectors? Non-infinite input impedance?" I left the slide up on the screen for several extra seconds.
The next slide listed: "Ignorance... Apathy... Carelessness... Sloppiness... Stupidity...." This always draws chuckles. But it is partly true — when one is measuring things, the INSTRUMENTS usually aren't a source of error. It is our foolishness in misapplying them that causes errors.
One time I was evaluating an expensive DVM. About £4000 worth. This was a nice DVM that not only had high guaranteed accuracy, but it had a display that would tell you how accurate it was. For example, on a 1-megohm scale, it told us that its accuracy was guaranteed less than 0.01% when measuring 1 megohm.
I slapped on one of our lab's 1-meg resistors — a precision wirewound resistor. The reading was 999,800 ohms. Hmmm. Now, it's uncommon to find a 1-megohm wirewound resistor that has drifted that much (-200 ppm). When I go down to Haltek and buy old wirewound resistors, I like to buy old ones because a resistor that's 5 or 10 years old, and still in spec, is at least as good as a brand-new resistor because it has some proven long-term stability.
I slapped this 1-meg resistor onto our HP 3456 DVM. It read 1,000,005 ohms. But we had to admit that the HP3456 is only specified to an accuracy of 0.02% on that scale. Maybe the HP was wrong?
I decided to use jiu-jitsu to prove which DVM was telling the truth. I had some new 100 kilohm wirewounds that were specified at 0.002%. I measured 10 of these. BOTH DVMs agreed that these resistors were all nearly perfect, ±5 ppm. Then I clipped the 10 resistors in series. The HP said 1,000,000 ohms. And the expensive English DVM said 999,800 ohms. Yet it swore that its own accuracy was no worse than 0.01%. Well, we sent in a nice calibration report when we gave that DVM back to its sales guy. We never did hear any explanation or apology from him.
So whenever I measure things, I like to do some little self-calibration test, just as a sanity check. I like to measure some things whose accuracy I think I know. Like the capacitance from the earth to the moon....
All for now. / Comments invited! (firstname.lastname@example.org)
RAP / Robert A. Pease / Engineer
Mail Stop D2597A
P.O. Box 58090
Santa Clara, CA 95052-8090
P.S. I received a lot of answers after asking the question, "What is the actual capacitance from the earth to the moon?" There were a few odd ones at 0.8µF or 12µF. But about 10 guys said it was 143 or 144µF. They used the formula:
C = 4 x x x ( l/r1 + 1/r2 − 2/D )−l
valid for rl, r2 << D
NOW, my original estimate of 120µF was based on this approximation: The capacitance from the earth to an (imaginary) metal sphere surrounding it, 190,000 miles away, would be 731µF. (If that surrounding sphere were pushed out to 1,900,000 miles away, the capacitance would only change to 717µF — just a couple percent less. If the "sphere" moved to infinity, the C would only decrease to 716µF.) Similarly, the C from the moon to a surrounding sphere 48,000 miles away would be 182.8µF. If the two spheres shorted together, the capacitance would be 146.2µF. I guessed that if the spheres went away, the capacitance would drop by perhaps 20% to about 120µF, so I gave that as my estimate. But removing those conceptual "surrounding spheres" would probably only cause a 2% decrease of capacitance. That would put it in close agreement with those 10 guys that sent in the 143µF figure.
But THEN 6 readers wrote in LATER — from Europe — all with answers of 3µF. I checked their formulae, from similar books, in several different languages. They were all of the form:
C = 4 x Pi x Epsilon x ( r1 x r2 ) / D
multiplied by a correction factor very close to 1.0. If you believe this formula, you'll believe that the capacitance would be cut by a factor of 10 if the distance D between the earth and moon increased by a factor of 10. Not so! Anybody who used a formula like that, to arrive at 3µF, should MARK that formula with a big X.
Finally, one guy sent in an answer of 159µF. Why? Because he entered the correct radius for the moon, 1080 miles rather than 1000. That's the best, correct answer! / RAP
Originally published in Electronic Design, September 3, 1996.
*In the original text I said that the boxes are 9" high, but I forgot that the paper stands 0.8" taller than the box. So I said the paper was 3.6 mils thick, but 3.9 is closer.
** Originally I said my thumbnail was 14 milli-inches thick, but I divided by 25.4 when I should have multiplied by 0.0254. Silly error! So 15.7 mils x 0.0254 = 0.40 mm. When I said he was wrong by nearly a factor of two, I should have said nearly a factor of three! Barring a few dumb errors, this was about right. / RAP