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I am working with a 6-axis force-torque sensor to measure the thrust and moments of a model jet engine. I want to understand the significance of the accuracy (measurement uncertainty) and precision (resolution) of these sensors. Note: I understand the textbook definitions well, I want to understand how these translate to real-world applications.

A realistic sample problem is as follows:
There are 2 sensors in the market.
Say both have the Maximum calibration for Fz as 1000N.
Both have the maximum measurement uncertainty as 1% of the Full-Scale calibration.
And say the first one has a resolution capacity of 0.1N and the second one has a resolution of 0.2N.

Here is what I understand:
The results of measurements will be accurate within 1% of 1000N which is $\pm$10N.
The resolution of the measured value for the first sensor is 0.1N and for the second 0.2N.
So if the measured thrust is 100.4N then I should report my measurement as 100.4N $\pm$10N.

The questions:
What is the significance of resolution here? Since accuracy is 10N which is more than resolution isn't resolution insignificant here?
Both force sensors seem to be giving almost the same information. Both the first and second sensor say my value will be something within the range 90.4N to 110.4N. How does resolution come into play here?
A better resolution than accuracy seems to be useless.

Does resolution come into play when talking about repeatability? For eg: When I do the measurement again the first sensor is expected to give me a value of 100.4N $\pm$ 0.1N whereas, the second sensor is expected to give a value of 100.4N $\pm$ 0.2N?

Please provide me with an explanation on the same.
Thanks!

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4 Answers 4

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The question was becoming horrifyingly long so I am posting my further progress as an answer.

EDIT1: Some further understanding:
To understand a little more about resolution and accuracy imagine taking force measurements and then marking these measurements on a ruler. The ruler has the properties of the measuring device (I clarify this next).
Now the resolution will define how the markings on the ruler move. Eg: If the resolution is 0.1N then the ruler markings will move as 100, 100.1, 100.2, 100.3, 100.4, 100.5... . If the resolution was 1N then the ruler markings would move as 100, 101, 102, 103... . So the first device would display a reading of 100.4N whereas the second device can only show 100N so in this sense the first is better.
Now accuracy is another layer of uncertainty around the measured value. If the accuracy of the measuring device was 1N then the first result would be quoted as 100.4N $\pm$1N and the second result would be quoted as 100N $\pm$1N.

So basically resolution allows us to move on a finer underlying scale for measurements.
Q) How does finer resolution help though? Since accuracy is $\pm$1N what use is resolution lower than this? Since there will still be uncertainty of 1N?

EDIT2: This is to go beyond:
Imagine 2 devices. Both with an accuracy of 10. First one has a resolution of 1 and the second one has a resolution of 10.
Say the actual true value of some physical value is 105.2
The scale on the first device will move as 100, 101, 102, 103, 104, 105, 106... and the scale on the second device will move as 90, 100, 110, 120. The first device will measure a value of 105 and the measurement will be reported as 105 $\pm$ 10 which is the range [95, 115]. Whereas, the second device will measure a value of 110 (assuming that 105 is measured as 110 not as 100) and the measurement will be reported as 110 $\pm$ 10 which is the range [100, 120]. Now let us visualize this to understand what is happening.
enter image description here

Note here that the distance between the lower limits of the intervals is of the order of Accuracy. This happens when we are comparing resolutions where one is of the order of Accuracy and the other is one order lower. Therefore, the shift between the intervals is large.
Had the resolutions we compared been one and two orders of magnitude lower respectively then the shift between the intervals would have been much smaller compared to Accuracy and we could neglect that.

In Conclusion: Resolution helps in making the measured value better. Decreasing resolution improves the measured value. But decreasing resolution only makes significant changes if it is one order of magnitude smaller than accuracy itself. Trying to make a device with a resolution several orders of magnitude lower than accuracy would therefore be quite useless.

EDIT3: This is to go even further beyond:
The above conclusion is specific to the case when our primary desire is to measure some physical quantity and get an estimate of its value.
But, if our primary desire is only to sense a change in the quantity and the actual value of the quantity is irrelevant then improving resolution monotonically improves our results.
Eg: Suppose someone presses a button. The sensor on the button has to detect if the force applied on the button increased or decreased, that is, whether there was a change in the environment of the button. Here the actual value of the force on the button is absolutely useless and only the change is of importance. Therefore, here resolution is of importance.

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accuracy is how close the reading is to reality. If it say 100, but is actually at 101 then the accuracy is on the order of +/-1. Precision is how repeatable the reading is. I measure 101.1, 101.2, 109.9 the accuracy is still approximately 1, but repeatability is +/- 0.1.

In many cases you are trying to measure a change and not the absolute value. I care more that a modification I made caused the thrust to increase by 5% than to know the exact thrust.

Precision can never be better than resolution. Resolution can often be determined by calculations. Precision, like accuracy usually has to be measured.

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This is a classic precision vs. accuracy problem.


I think your intuition is correct: resolution of 0.1-0.2N doesn't buy you much if the accuracy is only 10N.

I would model both the resolution and accuracy as independent sources of error:

  • Sensor accuracy introduces uniformly distributed random errors between -10N and +10N
  • Sensor resolution introduces uniformly distributed random errors between -0.2N and +0.2N (using the worst case sensor)
  • Assume that accuracy and resolution errors are independent

Using just one sensor

Your sensor reading can be expressed:

$$\hat{y}=y + \epsilon_r + \epsilon_a \doteq y + \epsilon$$

where $y$ is the true value (force, torque, whatever), $\hat{y}$ is the corrupted observation of $y$ that your sensor permits, and $\epsilon_r$ and $\epsilon_a$ are the errors introduced by sensor resolution and accuracy, respectively.

Assuming that the sensor errors are unbiased (i.e. no constant offset), then the expected value of the total error $\epsilon$ is zero and the error variance is given by:

$$ \sigma_\epsilon^2 = \sigma_{\epsilon_r}^2 +\sigma_{\epsilon_a}^2$$

The variance of a uniformly distributed random variable $x$ whose values are bounded by limits $a$ and $b$ is given by:

$$\sigma_x^2 = \frac{1}{12}(b-a)^2$$

Substituting the above limits (using the sensor with poorer resolution:

$$\sigma_{\epsilon_r}^2 = \frac{1}{12}(0.2-(-0.2))^2 = 0.01\bar{3}$$

$$\sigma_{\epsilon_e}^2 = \frac{1}{12}(10-(-10))^2 = 33.\bar{3}$$

$$ \sigma_\epsilon^2 = \sigma_{\epsilon_r}^2 +\sigma_{\epsilon_a}^2 = 33.34\bar{6}$$

So that the standard deviation of the sensor errors is:

$$\sigma_\epsilon = 5.774657\mbox{N}$$

Using the "improved" sensor with higher resolution doesn't have much impact, since the error is dominated by sensor accuracy:

$$\sigma_{\epsilon_r}^2 = \frac{1}{12}(0.1-(-0.1))^2 = 0.00\bar{3}$$

$$ \sigma_\epsilon^2 = \sigma_{\epsilon_r}^2 +\sigma_{\epsilon_a}^2 = 33.33\bar{6}$$

$$\sigma_\epsilon = 5.773791\mbox{N}$$

So whatever premium might be charged for the higher precision sensor probably isn't worth it.


There are quite a few things above that I glossed over (e.g. Are resolution and accuracy errors really uniformly distributed? What if there is a bias?) but maybe the modeling will give you some new ideas.


When 2 different sensors are available at the same time

If you are averaging the reading from the 2 sensors then things get more interesting. Now

$$\hat{y} = \frac{1}{2}\left(\hat{y}_1 + \hat{y}_2\right) = \frac{1}{2}\left[\left(y + \epsilon_1\right) + \left(y + \epsilon_2\right)\right] $$

Here again $\mbox{E}[\hat{y}] = y$, but the variance of the overall error is reduced:

$$\sigma_\epsilon^2 = \frac{1}{4}\left(\sigma_{\epsilon_1}^2 +\sigma_{\epsilon_2}^2\right) =\frac{1}{4}\left(33.34\bar{6}+33.33\bar{6}\right) = 16.66771$$

so that

$$\sigma_\epsilon = 4.08261$$

Having 2 independent sensors and averaging their measurements reduced the potential error of the force measurement by 1.691181N, or more than 29%. Note, however, that it was just the fact that two sensors are being averaged that yielded the improvement - not the fact that one of the two sensors had better resolution.

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One could separately model several reasons for limited accuracy. A system might suffer several at the same time. This is common to many instruments, besides force-torque sensors.

  1. All instruments perfectly manufactured, to a flawed design. Repeatable from measurement to measurement, and from instrument to instrument. Careful calibration against a better standard can remove this.

  2. Instruments variably manufactured. One instrument repeatably shows the same measurement, but its siblings and colleagues show something different. Calibration of each individual unit can remove this. Measurements by several different units or models might address this, depending on how you model the unit-to-unit variability — for instance, averaging can't fix systematic bias.

  3. Instrument readings vary in a known way on some external factor(s), such as temperature. Measure that(them) and rely on even more calibration.

  4. Instrument readings slowly vary because of something unknown. This is the hardest one. It even leads to folklore and superstition (phase of the moon). As a young electrical engineer I was told (jokingly, I hope) to never measure anything twice, because then you have to explain why the readings are different!

  5. Instrument readings vary at "medium" rate. What you could do depends on your statistical model of the "noise" and what you're measuring. If multiple readings are independently inaccurate, you could combine them with mean, median, RMS, etc., as @guero64 points out.

  6. Instrument readings vary at "fast" rate. Think noise, hum, jitter. Combine, as above, except your real-time software does the computation. Actually, the right amount of the right kind of noise can sometimes be your friend — think superresolution and dithering.

The above points touch on things to do with the data once the measuring is over. But of course there can be ways to improve the process leading to measurement. Think electrical shielding, shock mounts, temperature controls, etc.

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