Slide Rule Precision, Accuracy and Significant
Dave Hoyer, July 2012
How accurate is a slide rule? What's the difference between
accuracy and precision? Here are some suggested answers
to these questions, specifically with respect to using slide
rules for multiplying two numbers.
- Precision and accuracy are not the same
thing. Very roughly, precision is how many significant digits you
can read off the scale, and accuracy is how close your
result is to the true result.
- A damaged or poorly made slide rule can
have high precision but low accuracy.
- The precision and accuracy of a well made slide rule both improve with slide
rule scale length.
- A linear slide rule scale has the same precision at all points
along its scale, whereas the number of significant digits at the
low end is one greater than at the high end.
- A formula is presented for determining the number of
significant digits of any slide rule. This can be used to compare
the potential accuracy of different slide rules.
- To get one digit more accuracy a slide rule scale needs to be
10 times longer and have ten times as many division
The precision of any instrument is a measure of how
finely it can be read repeatedly. For example, on my Faber Castell
52/82 slide rule the C scale is a single log scale 250 mm long. The
first division after 1 is 1.01 and the division is 1.1 mm wide. You
can sub-divide this by eye into 10 sub-divisions if you are careful.
We can say that the precision near 1 is ±0.001 or 0.1%. Similarly,
the last division before 10 is 9.95, this division being 0.55 mm
wide. You could estimate by eye to ±0.01, which is still 0.1% (of
The accuracy is how close your answer is to being
“true”, or exactly right. If for example the slide rule was
incorrectly manufactured, with some of the divisions slightly in the
wrong place, then it would have poor accuracy at those places even
though the precision remains the same.
So if the slide rule in the example above was perfectly
manufactured, and if our eye can estimate 10 sub-divisions inside a
1.1 mm division (or 5 over a 0.55 mm division), then we can say it
has a precision and relative accuracy of 0.1% over
its whole range. The absolute accuracy would be ±0.001 at
the low end near 1 and ±0.01 at the high end near 10.
What if the slide rule scale was made shorter but still had the
same number of marked divisions? At some point the divisions would
simply be too close together to reliably sub-divide by eye to the
same degree. An 8” slide rule for example would have a C scale 200
mm long with a first division of 1.01 at a distance of 0.86 mm. I
reckon I may be able to estimate 10 sub-divisions between 1 and 1.01
on a good day, but not very confidently. Anything smaller and I
would find it difficult to estimate the division into tenths without
extra tools such as a microscope and/or a finely divided reference
scale. So for the sake of argument let’s say that 1mm is the
smallest interval we can reliably sub-divide into 10 by eye.
We can use this observation to define a ground rule for the
accuracy of a slide rule at any point along its length, by using the
values represented by two adjacent divisions and the width between
Dave’s Ground Rules for determining the accuracy of a slide
- You can’t sub-divide a division by eye into more than 10
- A division must be at least 1 mm wide to sub-divide by eye
- A division less than 1 mm wide can be divided into a
proportionally smaller number of sub-divisions (ie. less than 10).
This can be put into a formula for the accuracy at any point
along the scale of a slide rule. First define the following
parameters in relation to a marked division on a single log scale
from 1 to 10..
- A the absolute accuracy
- L the length of the scale, mm
- T the maximum number of “by eye” sub-divisions
- V1 the value at a mark on the scale (i.e. the start of
- V2 the value at the next mark after V1 (i.e. the end of
- W the width of the division, mm
The width of the division along the scale
The number of “by eye” sub-divisions allowed
The absolute accuracy at V1
The number of significant digits is how many meaningful digits
can be written down. With slide rules this is determined by the
smallest interval we can estimate by eye, which is somewhat
subjective but we can make some ground rules. In the example above
with a 250 mm (10”) long single log scale we can meaningfully
distinguish between 1.001 and 1.002, which we can call 4 significant
digits. And we can distinguish between 9. 98 and 9.99, which we can
call 3 significant digits.
On smaller slide rules the first division after 1 may be 1.02
rather than 1.01, and we should be able to estimate within this
division to ±0.002. We could distinguish between 1.002 and 1.004 but
no finer. This is not quite a full 4 significant digits, but is
clearly better than 3 significant digits. In this case we need to
have a way to use fractional significant digits to enable us to
properly compare the accuracies of different slide
The C scale on my Pickett N600-ES is 125 mm long and its first
division 1.1 mm away is 1.02, which we can divide into ten by eye to
get ±0.002, which in turn seems like “not quite 4 significant
digits”. We need to give it a fractional number of significant
digits somewhere between 3 and 4.
Let’s call the number of significant digits N. Then we could
- N = 1 + Log(1/A) …….. where “Log” is the base
Try it out:
|250 mm scale at 1.001
|| N = 1+Log(1/0.001) = 4.0|
|250 mm scale near 9.98
|| N = 1+Log(1/0.01) = 3.0|
|125 mm scale near 1
|| N = 1+Log(1/0.002) =
So using this approach the number of significant digits of a
slide rule is a function of both the physical distance between two
divisions, and the difference in value represented by the two
Some worked examples (table)..
Pickett 600, 5"
Faber 52/82, 10" scale
The table below has a summary of calculated significant digits (SigD) for
various slide rules, from the 4” Sun-Hemmi to the giant
Loga 15m and the Thacher 20m cylindrical slide rules.
||SigD @ 1
||SigD @ 3
||SigD @ 10|
Pickett 600, 5"
K&E 4053-3, 8"
Faber Castell 52/82, 10"
Otis King, 66"
Fuller's Calculator, 41'
All the above also assumes that the slide rule is “true”, namely
that its markings are correctly placed to within the measurement
Without special tools the best way to get a quick feel for the
accuracy is to do some multiplications and compare the result with
the correct result. Try these..
1.012345^2 = 1.0248424
|to check out a result close to 1|
3.1415956^2 = 9.8696041
|to check out a result close to 10|
9.876543^2 = 97.546101
|to check out a result close to 100|