Engineers: A Paradox w/Sig Figs

I fail to see the practical application of significant figures (sig figs) in the REAL WORLD considering the following situation. While I understand there’s no need to carry a million decimal places, I fail to see why layman’s rounding isn’t good enough, especially regarding digits to the LEFT of the decimal point:

I have worked with devices that cannot be oversized. They have a 10% margin. Beyond that, they are not fully loaded and will damage themselves…all in the name of the noble, purely academic, and useless sig figs. If I cannot simply round as usual, I must accept too large a size for a device because of the sig fig rules. While sig figs may be all well and good and saintly in the lab, it’s going to ruin my sizing process…and inflate budgets…only to have to replace all these devices when they all fail because - ta-da - they’re all oversized!!!

So, is there a practical application for sig figs in engineering vs. purely academic/lab applications? - Jinx :mad:

What on earth are you talking about?

I’m an engineer, and I use significant digits all the time, except when I do hex math.

Since your OP is nearly incoherent, why not give us an example of what’s bothering you?

I can see why sometimes it might look like you’re missing out on some information, sig figs are exactly that - significant. If you’ve measured something to be 2,3443,768 units, the accuracy of that measurement is unlikely to be down to the last unit. I would imagine that simply 2,3444,000 would be useful, anything more than that is extraneous.

What if it was given as 2.3443768 million? Would you feel happy doing “layman’s” rounding to make it 2.3444 million? That way you’re only getting rid of decimal places.

How does something get oversized because of sig figs? If you’ve got a 10% margin, as long as you’ve got 2 sig figs you’re fine. Say something is rounded to 2300 units - ten percent of that is 230 - so if the actual figure is more than 230 different from 2300 you’re in trouble. However, you know it’s rounded to 2 sig figs, so the minimum the actual value can be is 2250 and the most 2350, both only 50 out. So well within your 10% margin.

I think VunderBob is right, an example would really help us clear this up.

Yea, that was my thought also.

Jinx: I’m utterly confused. Can you provide more detail on what the problem is?

And, just for the record, it makes no difference whether the significant digits being rounded are to the left or the right of the decimal point.

Uh, yeah. Fourthed or fifthed or whatever. I can’t conceive of how properly recognizing the significant figures in a measurement would lead to oversizing something by 10%. I’ll go out on a limb and guess that significant figures are being incorrectly used in this case.

Like others, I can’t exactly figure out what you are getting at with the OP, but here’s a recent thread on the issue of significant figures that may answer some of your questions.

Significant figures definitely have a place in engineering practice, and while modern calculators and computers with floating point variables and twelve place displays mean that you don’t typically have to worry about rounding errors in intermediate calculations, back in the days of slipsticks[sup]1[/sup] and log tables you were limited to four or five decimal places at best, meaning that you needed to be careful about significant figures.

These days you still do in terms of interpreting the degree of precision of data values (you aren’t going to get z.zzzzz from xx.x * 0.0yyyy) and we use significant figures all the time (although often inappropriately) for determining default or recommended tolerances on engineering drawings, but you can usually use the full values for calculations and do your rounding on the end result, based on the least number of significant figures of data input.

For instance, if you have a calc like:

F = 20.44 in-lbf * 38.2 in

the calculator gives you

780.808

but your answer should be

F = 781 lbf
If you have something like

E = 39.5MeV + 3.29MeV

the calculator gives you

42.79

but your answer should be

m = 42.8MeV
Does that make it all square?

Stranger

1 That’s a slide rule for those of you who’ve never had to memorize log tables. Yeah, I grew up in the calculator age, too, but Grandpa made me learn how to use a slide rule anyway. I think all engineers and scientists should be exposed to it, just as all blue water sailers should know how to use a sextant, but apparently engineering schools and the US Naval Academy don’t agree with me anymore.

The only thing I can get out of this is that perhaps you need to use one more significant figure. Then your 10% will turn into 1%.

Significant figures exist as a representation of the real world… they have no theoretical purpose. They represent the limits of accuracy to the instruments you’re using. If you’re in the world of theory you can say your widget is exactly 2. units long. In the real world, OTOH, you can only get as close to 2. as your yardstick will allow… and you show how much your yardstick will allow based on the number of significant figures. A really crappy yardstick might call everything from 1.5 to 2.5 units as 2 units. Try sticking a 2.5 length widget in a 1.5 size hole some time. A better yardstick might call every thing from 1.95 to 2.05 as 2.0. A really good yardstick will only call the stuff between 1.9999995 to 2.0000005 as 2.0000000

If you’re not dealing with the real world limitations of real world measuring devices… sig figs are pretty much theoretical academic masturbation.

Signifigant figures have more to do with the degree of accuracy of what is being measured or counted.

If you are counting beans and you want to have 10,000 beans in a bag, you count out the 10,000 and weight them to the nearest gram. Then use this weigh to out one bag after another. This may result in, say, a variation of from 9,900 to 10,100 beans in the bags. If this is acceptable then you have accuracy of 1:100 or two signifigant figures.

You do not carry more signifigant figures than are necessary for the job at hand. The signifigant figures are the ones that have meaning, decimal point notwithastanding.

There is no paradox except, perhaps, in your conception of such numbers.

Addenda:

Significant figures (Arith.), the figures which remain to any number, or decimal fraction, after the ciphers at the right or left are canceled. Thus, the significant figures of 25,000, or of .0025, are 25.

Why the spoiler tag, spingears? I think that what you’re describing there are the significant digits, not figures. The number 0.0025 has two significant figures. However, the number of sig figs in 25,000 is debatable. If the number were written “25,000.” instead, that indicates we have precision down to the last zero, or 5 sig figs. If this number was merely rounded to the thousands, then we only have 2 sig figs. Alternatively, the number “25,000.00” has 7 sig figs.

Jinx, there is no reason why what you are describing should occur. When you do calculations, you should never round any more than necessary. Only the final, reported result of your calculations should be rounded to the correct number of significant figures. Numbers that are exact, i.e. not derived from a measurement, are usually ignored when determining significant figures (consider them as having an the infinite number of significant digits).

This is why scientific or engineering notation is prefereable when dealing with measured quantities. 2.51810[sup]4[/sup] is definitely less ambiguous than 25,180. Ditto 4.0010[sup]6[/sup] as opposed to 4,000,000.

But significant figures are just a poor man’s tolerances. If you really want to be precise, as you would be with machined parts, you’d state a +/- condition (1.25±.03) or limits (1.22 to 1.28). And if you were concerned about statistical tolerance zones you’d actually state a distribution which would allow you to estimate exactly how likely two parts are to fit together in a particular condition. But sadly, even in engineering practice where this is paramount it often isn’t done. Right now I’m dealing with an issue with a company that builds rocket motors in which the joints have very tight tolerances and yet they don’t seem to understand how to establish statistical flatness and perpendicularly stack up conditions. It’s very frustrating, and I’m happy to take the opportunity here to bitch about it. No extra charge; it’s all part of the service.

Jinx, could you give us more detail on your problem?

Stranger

MaceMan:
Assume your W-2 income for 2004 was $25,357.23
How would you respond when asked your income? $25 thousand or $25, 357.23?
Does the context affect the number of signifigant figures?
The IRS is not interested in the .23 but are in the 23,357, all five figures!
What is the difference between digits and figures?

Jinx: See if these pages provide the answer(s) you are seeking.
** Matter and Measurement **
** Signifigant Figures Self Test **

All Interested in OP Topic: Try the Self Test first and see if YOU understand.

In this case the monetary amount ($25,357.23) is an exact value with no uncertainty, which is a nice trick you can do when you are dealing with theoretical figures (and with most people, the amount of money they have to spend is definitely theoretical. :smiley: ) In measured values, though–say, the length of a rod or the volume of water in a container–there is always some degree of (hopefully) known variation to account for. Significant figures assumes that your value is good to +/- the rounding of the last digit; 2.563 might actually be as high as 2.5634 or as low as 2.5625, i.e. +/- .0005.

Stranger

Why only as high as 2.5634? 2.5635 can be rounded down.

Generally when rounding figures you round up at 5 or greater and round down at less than 5. Actually, the upper bound should have been 2.5634999…, but for practical purposes you usually just look at the first digit past your significant figures and truncate. You could, of course, arbitrarially round up or down as you see fit, and when using data to make engineering calculations we frequently round in what ever direction produces the most conservative result, i.e. a 29,426 lbm payload would become 3010[sup]3[/sup] lbm, even though a nominal rounding at two decimal places would make it 2910[sup]3[/sup] lbm.

Significant figures represent a known uncertainty in stating the value of a measurement, i.e. the graduations on the cylinder or the digits on a display only read out to X places. This is independent of the calibration of the measurement tool or the reproducibility of the process. We sometimes (well, often) also use them as a rule-of-thumb tolerance, though with predictibly ambiguous results, as seems to be the case with the OP. (I think. I’m still not clear on the details of his problem.)

When you are working with values that may (or are allowed) to vary in whole percentages, or if you are dealing with raw, single measurements, significant figures are (generally) just fine. If you want less variation or want to state the expected or allowed variation of a repeated process, you need to select permissable tolerance bounds. If you want to figure out how likely two groups of tightly controlled parts are likely to interfere with each other when assembled, you need to go to a statistical analysis based upon measured (or assumed) variable distribution.

Stranger

Context affects how one uses significant figures, yes. I someone tells me they make 40 thousand dollars a year, I can be pretty certain that this number has already been rounded to 1 or 2 significant figures. On the other hand, I know their income tax return should be accurate down to the last dollar amount.

I guess the answer to your second question is a semantic one. In your example above, you stated that the significant figures of 0.0025 are 25. I would normally say that this number has two sig figs, and the significant digits are 25.

spingears wrote:

There seems to be a bug there. I was doing great until I got 10.1e-1 and answered 3 and it said I was wrong. It wanted the answer 2.