On Sun, 2 Jul 2017 14:12:59 -0400, rickman <
gnu...@gmail.com> wrote:
>Jeff Liebermann wrote on 7/2/2017 1:19 PM:
>> On Sun, 2 Jul 2017 04:31:24 -0500, Foxs Mercantile <
jda...@att.net>
>> wrote:
>>
>>> On 7/2/2017 3:44 AM, Stephen Wolstenholme wrote:
>>>> The advantage of being an Imperial age is multiple arithmetic
>>>> bases is not a problem. Younger metric people are base 10 only!
>>
>>> Not hardly, people don't think of 10 inches as 0.833 feet nor do
>>> they think of 10 feet as 3.333 yards.
>>
>> Wrong. You're creating more digits than was originally intended by
>> adding spurious significant figures. 10 inches has only two
>> significant figures. Therefore:
>> 10 in = 0.83 ft = 0.00016 miles = 0.57 Roman cubits
>> and so on.
>I see you attended the same school as my chemistry lab professor.
Probably. The standard lecture was to pace off some distance ending
up a bit short. One then measures the remaining distance to a much
higher degree of precision. Take the number of paces, multiply by 1
yard/pace, add the precision measured distance, and the sum is a
fairly useless number.
How accurate can the average tape measure wielding reader measure 10
inches? My guess(tm) is no more than ±0.05 inch.
>"Significant digits" is an idea that is very dated and was only useful when
>performing the simplest calculations like the ones we did on a slide rule.
>The real issue is accuracy. I can measure 10 feet with an accuracy better
>than a sixteenth of an inch (yeah, I said sixteenth because that's how my
>tapes are marked off) and I will still note it as 10 feet.
If you use a tape measure to an accuracy better than ±1/16th of an
inch or 0.0625 inches, your actual distance would land somewhere
between 9.9375 and 10.0625. This does not mean that your tape measure
is accurate to 1/10,000th of an inch. To be accurate, one needs to
specify the measurement tolerances, as is common on all mechanical
drawings and an amazing number of schematics that still display
tolerances.
Using a steel rule, if you're able to measure the required 10 inches
to perhaps ±0.1 inches, then the correct representation would be 10.0
inches. 10 inches implies an accuracy of ±1 inch as measured by the
number of spans of my index finger between the first two joints.
>When I perform calculations I want to preserve the accuracy of the result,
>so the calculations are done with a higher degree of accuracy than the
>initial data.
Argh. I sometimes see that in parts drawings. Some newly minted
mechanical designer grinds out every dimension to whatever number of
digits he has his calculator configured, and then doesn't bother
providing a usable tolerance. The result is the machine shop doesn't
know if they need to cut metal to ±1/10,000th of an inch, or something
less. Such excess precision tends to dramatically raise parts costs.
If you want to preserve your accuracy on your own design notes, that's
fine. Just don't submit those numbers to anyone that has to make or
price the part.
>How much more accuracy should be used depends on the extent
>and nature of the calculations.
>One subtraction of large numbers can result
>in a small number which does not have nearly as much accuracy as the initial
>data. Add to that lack of accuracy with limited precision intermediate
>representation and you can end up with pointless data.
Yep. That's roughly what I've been mumbling about.
>I would also point out that in both cases the final digits repeat. There is
>no way to show a vinculum in ascii so seeing repeating digits at the end of
>a fraction is a clue. There is nothing wrong with specifying the conversion
>exactly. 10 inches is 0.833 (vinculum implied but not shown). Leave it off
>and you *add* to the initial error.
Very true. The convention is to round off anything that ends in 5 to
the next higher digit. So,
0.
8333333333 will round off to 0.833 or 0.83 or 0.8
and:
0.
8666666666 will round off to 0.867 or 0.87 or 0.9
If you have a tolerance available, writing 8.6666666666666666 ±0.001
would not make much sense. It should be written: 8.666 ±0.001 .