Halliday Resnick Krane Example Problem: Uncertainty on Weighing Machine

[mopit_011]

Homework Statement

You wish to weigh your pet cat, but all you have available is an ordinary home platform scale. It is a digital scale, which displays your weight in a whole number of pounds. You therefore use the following scheme: you determine your own weight to be 119 lbs, and then holding the cat you find your combined weight to be 128 lbs. What is the fractional or percentage uncertainty in your weight and in the weight of your cat?

Relevant Equations

N/A

In the solution to the example problem, they wrote the following statement. “The least significant digit is the units digit, and so your weight is uncertain by about 1 pound. That is, your scale would read 119 lb for any weight between 118.5 and 119.5 lb.”

I don’t understand why the scale wouldn’t read 119 lb for weight between 118 and 120 lb instead.

 

[Mister T]

If something weighed, say, 118.4 lb you'd expect the scale to read 118 lb, not 119 lb.

 

[mopit_011]

Thank you but then wouldn’t the uncertainty be 0.5 lb?

 

[haruspex]

Thank you but then wouldn’t the uncertainty be 0.5 lb?

The intended meaning of the term "uncertainty" in the quoted extract is, um, uncertain at first. It becomes clear that they mean the range of possible values. From 18.5 to 19.5 is a range length 1.

There is also some ambiguity in what answer is expected for the last part. There are two common ways of combining uncertainties. An engineer working with tolerances would consider the worst case, so two uncertainties of $\pm 0.5$ lead to a total uncertainty of $\pm 1$. But taking a statistical view, the RSS may be used.
In the present case, that could mean first finding the variance of a uniform distribution (-0.5,0.5), doubling it to find the variance of the combination, taking the square root to get the std dev, then multiplying by whatever constant to get the desired number of std devs.

 

[vcsharp2003]

Homework Statement:: You wish to weigh your pet cat, but all you have available is an ordinary home platform scale. It is a digital scale, which displays your weight in a whole number of pounds. You therefore use the following scheme: you determine your own weight to be 119 lbs, and then holding the cat you find your combined weight to be 128 lbs. What is the fractional or percentage uncertainty in your weight and in the weight of your cat?
Relevant Equations:: N/A

In the solution to the example problem, they wrote the following statement. “The least significant digit is the units digit, and so your weight is uncertain by about 1 pound. That is, your scale would read 119 lb for any weight between 118.5 and 119.5 lb.”

I don’t understand why the scale wouldn’t read 119 lb for weight between 118 and 120 lb instead.

I agree with you that it's confusing.

To clear this and similar confusions with digital instruments, I'm using the following rule and it seems to work.
$\text {Absolute uncertainty} = \text {(minimum increment/decrement)} \div 2$

If we apply the above rule to this problem, then we know that minimum increment/decrement is 1 since there are no digits being shown after decimal point like 19.0 or 19.2 or 19.32 or 19.324. Therefore, $\text {Absolute uncertainty} =1 \div 2 = 0.5$. So, the actual value will be $19 \pm {0.5}$ or using inequalities we can say that $18.5 \le \text {actual } \le 19.5$.

Perhaps, some of the experts here can throw more light if my approach is correct or not.

As an example with decimal digits, let's assume that the weighing machine gave readings up to 1 decimal place, and the observed reading was 19.0.
Then, minimum increment/decrement would be $0.1$ and therefore $\text {Absolute uncertainty} = 0.1 \div 2 = 0.05$, so the actual value will be $19.0 \pm 0.05$ or $18.95 \le \text {actual} \le 19.05$

 

[haruspex]

18.5≤measurement ≤19.5.

I would consider the "measurement" to be the rounded value, i.e. the reading. So I would rather say 18.5≤Actual ≤19.5.

 

[vcsharp2003]

I would consider the "measurement" to be the rounded value, i.e. the reading. So I would rather say 18.5≤Actual ≤19.5.

Yes, that makes sense. I have corrected my post accordingly.

 

[vcsharp2003]

Homework Statement:: You wish to weigh your pet cat, but all you have available is an ordinary home platform scale. It is a digital scale, which displays your weight in a whole number of pounds. You therefore use the following scheme: you determine your own weight to be 119 lbs, and then holding the cat you find your combined weight to be 128 lbs. What is the fractional or percentage uncertainty in your weight and in the weight of your cat?
Relevant Equations:: N/A

I don’t understand why the scale wouldn’t read 119 lb for weight between 118 and 120 lb instead.

If this were an analog weighing scale, then the absolute uncertainty would $\pm 1$. So, one must keep in mind that analog measurement has a different uncertainty as compared to a digital measurement for the same physical quantity.

 

[haruspex]

If this were an analog weighing scale, then the absolute uncertainty would ±1.

Why?

 

[Tom.G]

Just nitpicking here. Shouldn't those be "18.5 ≤ actual < 19.5"?

I would consider the "measurement" to be the rounded value, i.e. the reading. So I would rather say 18.5≤Actual ≤19.5.

18.5 < actual < 19.5.

 

[haruspex]

Just nitpicking here. Shouldn't those be "18.5 ≤ actual < 19.5"?

For continuous values, there is no such thing as exact equality.

 

[vcsharp2003]

For continuous values, there is no such thing as exact equality.

My understanding is that error comes from both sides of the pointer of an analog measurement instrument i.e to the right as well as the left sides of where the pointer finally settles during a measurement.

I have the following generic formula that I am using for analog as well as digital measurement instruments.

$\text {Absolute uncertainty} = ({\text {minimum increment/decrement} \div 2} ) \times \text {number of measurements}$

In analog case, $\text {number of measurements} = 2$ since both sides of the pointer are involved in the measurement.

 

[vcsharp2003]

Shouldn't those be "18.5 ≤ actual < 19.5"?

Actually that's what I initially thought when I posted my answer, but then I saw that books just mentioned the tolerances like $19 \pm 0.5$ and so I assumed the equality applied on the right side. But, I am not sure about this part. Hope someone can help on this point.

 

[haruspex]

My understanding is that error comes from both sides of the pointer of an analog measurement instrument i.e to the right as well as the left sides of where the pointer finally settles during a measurement.

I don't understand what scenario you are supposing.
If I see a pointer pointing somewhere between 4 and 5 I will judge which it is nearer to. If I judge it much nearer to the 4 I could say it is $4\pm 0.3$, or if only somewhat nearer then $4.3\pm 0.3$, or if plumb in the middle $4.5\pm 0.3$.
If someone else is making the judgment and reporting only whole numbers, I will presume they have rounded to nearest.

 

[vcsharp2003]

I don't understand what scenario you are supposing.

You sound right. Let me research more on the formula I am using and then I will get back.

 

[vcsharp2003]

For continuous values, there is no such thing as exact equality.

Should the inequality be $18.5 \lt \text {actual } \lt 19.5$ or $18.5 \le \text {actual } \lt 19.5$ ? I think the second inequality looks true in this case.

 

[haruspex]

Should the inequality be $18.5 \lt \text {actual } \lt 19.5$ or $18.5 \le \text {actual } \lt 19.5$ ? I think the second inequality looks true in this case.

I'm saying it does not matter which you choose.

 

[vcsharp2003]

I'm saying it does not matter which you choose.

Then, it would be best to include end points as in $18.5 \le \text {actual} \le 19.5$ when reporting a measurement. Is that correct?

 

[haruspex]

Then, it would be best to include end points as in $18.5 \le \text {actual} \le 19.5$ when reporting a measurement. Is that correct?

It might help avoid arguments.

 

[vcsharp2003]

I don't understand what scenario you are supposing.
If I see a pointer pointing somewhere between 4 and 5 I will judge which it is nearer to. If I judge it much nearer to the 4 I could say it is $4\pm 0.3$, or if only somewhat nearer then $4.3\pm 0.3$, or if plumb in the middle $4.5\pm 0.3$.
If someone else is making the judgment and reporting only whole numbers, I will presume they have rounded to nearest.

The basis of the formula I used for the analog instrument was from following discussion as shown in screenshot below.

An observer would be taking measurements at two spots when measuring the length of pencil i.e. the left and right ends of the pencil.

Also, for the case of analog voltmeter the presenter states that observer would be looking at the left and right sides of the pointer to arrive at an observed value i.e. 2 measurements are involved.

These screenshots are a part of a YouTube video @ Measurement Uncertainty

 

[jbriggs444]

If one were trying to measure the width of the pointer on a gauge then it would be somewhat reasonable to multiply the error estimate by two. But when reading an analog meter, one is not trying to measure the width of the needle. Unless it has a Vernier scale, one is trying to ascertain which dial marker the needle's midpoint is nearest. (Which is why they use needles -- so the width is small and the position of the midpoint is unambiguous).

In the discussion so far we have been idealizing away any systematic error (e.g. calibration error) and any random error (e.g. jiggle in the reading) and considering only quantization error.

When it comes to quantization error, there is no difference in principle between an analog and a digital device. It is just that in the one case you have a mechanism judging which discrete value the measured value matches best and in the other case you have a human making that judgement.

 

[haruspex]

The basis of the formula I used for the analog instrument was from following discussion as shown in screenshot below.

An observer would be taking measurements at two spots when measuring the length of pencil i.e. the left and right ends of the pencil.

Also, for the case of analog voltmeter the presenter states that observer would be looking at the left and right sides of the pointer to arrive at an observed value i.e. 2 measurements are involved.

These screenshots are a part of a YouTube video @ Measurement Uncertainty

View attachment 300758

View attachment 300760

The second example is clearly nonsense. The observer can see that the needle points between 3.2 and 3.4, so $3.3\pm 0.1$ is correct. There are not really two measurements involved; we know the distances from the neighbouring gradations add up to 0.2.

The pencil length example is more reasonable, but still debatable. If one were to position the pencil at an arbitrary point along the ruler then there would indeed be two independent judgments. But that is not how I use a ruler generally. If possible, I align one one end directly on a gradation. The error in doing so depends on the physical distance between gradations. E.g. if gradations are 5mm apart and my alignment might be 0.5mm off (parallax, mostly) then my error is only $\pm$10% of a gradation.
At the other end, where the end of the pencil falls at an arbitrary point between gradations, my error may be 30% of a gradation.

 

[kuruman]

At the other end, where the end of the pencil falls at an arbitrary point between gradations, my error may be 30% of a gradation.

I follow that same 1/3 rule. One can tell in most cases whether the tip of the pencil (or needle) is in the first, second or third 1/3 of the space interval between gradations, i.e. closer to one of the ends or closer to the middle, the location of which is not hard to imagine. In my estimation, the tip of the pencil is closer to the middle while the needle is probably closer to the left gradation than the middle.

 

[Tom.G]

This will probably answer more questions than you thought existed!

GUM: Guide to the Expression of Uncertainty in Measurement​

https://www.bipm.org/documents/2012...f-3f85-4dcd86f77bd6?version=1.7&download=true

(above found with:
https://www.google.com/search?&q=G.U.M.+guide+to+measurement)

Have fun, it's 'only' 134 pages.

Cheers,
Tom

p.s. It has been years since I read that document.

• the international organization established by the Metre Convention, through which Member States act together on matters related to measurement science and measurement standards
• the home of the International System of Units (SI) and the international reference time scale (UTC).

 

[vcsharp2003]

The second example is clearly nonsense. The observer can see that the needle points between 3.2 and 3.4, so 3.3±0.1 is correct. There are not really two measurements involved; we know the distances from the neighbouring gradations add up to 0.2.

Could it be that two measurements is not referring to left and right sides of the needle position of voltmeter in the final reading but something different as described below?

1. When we use a voltmeter, we have to first make sure that the initial needle position is properly aligned with the 0 V reading on voltmeter. This is our first measurement. Let's say the observer finds that the needle is pointing to $M_1$ on voltmeter which normally will have a value of $0$ and we will assume that $M_1 = 0$ since that's what normally happens.
2. Then, we connect the voltmeter to the electric circuit to find the potential difference across some section of the circuit. This is our second measurement. Let's say the observer finds that the needle is pointing to $M_2$ on voltmeter. In this case we guess that the needle is somewhere around $3.3V$ and so $M_2 = 3.3$
3. We can therefore say that the resulting measurement $M$ relates to $M_1$ and $M_2$ as in equation below.

$$M = M_2 - M_1$$
$$\therefore M \pm \Delta M = (M_2 \pm \Delta {M_2}) - (M_1 \pm \Delta {M_1})$$
$$\therefore M \pm \Delta M = (M_2 - M_1 ) \pm (\Delta {M_1} + \Delta {M_2}) \text { .........(1)}$$
$$\text {where } \Delta M = \pm (\Delta {M_1} + \Delta {M_2})$$. $$ \text {and } M = M_2 -M_1$$

Since the smallest unit shown on voltmeter is $0.2 V$, so $\text {absolute uncertainty} = 0.2 \div 2 = 0.1 V$, which is the value of either of $\Delta {M_1}$ or $\Delta {M_2}$.
Now, let's substitute these absolute uncertainties for initial and final observations into get $M$ and $\Delta {M}$.
$$ \Delta M = \pm (0.1 + 0.1)$$
Equation $(1)$ becomes
$$\therefore M \pm \Delta M = (3.3 - 0 ) \pm ( 0.1 + 0.1) $$
$$\therefore M \pm \Delta {M} = 3.3 \pm 0.2$$

 

[haruspex]

Could it be that two measurements is not referring to left and right sides of the needle position of voltmeter in the final reading but something different as described below?

1. When we use a voltmeter, we have to first make sure that the initial needle position is properly aligned with the 0 V reading on voltmeter. This is our first measurement. Let's say the observer finds that the needle is pointing to $M_1$ on voltmeter which normally will have a value of $0$ and we will assume that $M_1 = 0$ since that's what normally happens.
2. Then, we connect the voltmeter to the electric circuit to find the potential difference across some section of the circuit. This is our second measurement. Let's say the observer finds that the needle is pointing to $M_2$ on voltmeter. In this case we guess that the needle is somewhere around $3.3V$ and so $M_2 = 3.3$
3. We can therefore say that the resulting measurement $M$ relates to $M_1$ and $M_2$ as in equation below.

$$M = M_2 - M_1$$
$$\therefore M \pm \Delta M = (M_2 \pm \Delta {M_2}) - (M_1 \pm \Delta {M_1})$$
$$\therefore M \pm \Delta M = (M_2 - M_1 ) \pm (\Delta {M_1} + \Delta {M_2}) \text { .........(1)}$$
$$\text {where } \Delta M = \pm (\Delta {M_1} + \Delta {M_2})$$. $$ \text {and } M = M_2 -M_1$$

Since the smallest unit shown on voltmeter is $0.2 V$, so $\text {absolute uncertainty} = 0.2 \div 2 = 0.1 V$, which is the value of either of $\Delta {M_1}$ or $\Delta {M_2}$.
Now, let's substitute these absolute uncertainties for initial and final observations into get $M$ and $\Delta {M}$.
$$ \Delta M = \pm (0.1 + 0.1)$$
Equation $(1)$ becomes
$$\therefore M \pm \Delta M = (3.3 - 0 ) \pm ( 0.1 + 0.1) $$
$$\therefore M \pm \Delta {M} = 3.3 \pm 0.2$$

That certainly makes more sense, but if the voltmeter provides a way to adjust the needle so that it starts at zero then we are in the same scenario as for the pencil: we can arrange that one end is aligned to a gradation. As I noted in post #22, that does reduce the error somewhat.

 

[vcsharp2003]

That certainly makes more sense, but if the voltmeter provides a way to adjust the needle so that it starts at zero then we are in the same scenario as for the pencil: we can arrange that one end is aligned to a gradation. As I noted in post #22, that does reduce the error somewhat.

But, even if the initial reading appears aligned with zero of voltmeter to a human eye, shouldn't it be subject to the same error uncertainty as the final measurement because we are talking of the same measuring device? We cannot say that if voltmeter reading was 2.3 V then we're going to use a different uncertainty as compared to if it was 2.0 V.

 

[jbriggs444]

But, even if the initial reading appears aligned with zero of voltmeter to a human eye, shouldn't it be subject to the same error uncertainty as the final measurement because we are talking of the same measuring device? We cannot say that if voltmeter reading was 2.3 V then we're going to use a different uncertainty as compared to if it was 2.0 V.

You are conflating quantization error with estimation error and treating them as the same thing.

Quantization error is the error associated with rounding the measured reading to a discrete value.

Estimation error is the error associated with the chance of rounding incorrectly because you misread the needle position.

 

[haruspex]

But, even if the initial reading appears aligned with zero of voltmeter to a human eye, shouldn't it be subject to the same error uncertainty as the final measurement because we are talking of the same measuring device? We cannot say that if voltmeter reading was 2.3 V then we're going to use a different uncertainty as compared to if it was 2.0 V.

Judging whether two lines line up is quite different from judging whereabouts one line lies between two others.

 

[vcsharp2003]

You are conflating quantization error with estimation error and treating them as the same thing.

Quantization error is the error associated with rounding the measured reading to a discrete value.

Estimation error is the error associated with the chance of rounding incorrectly because you misread the needle position.

So, which type of error is included when absolute uncertainty is specified for a measuring device?

 

[jbriggs444]

So, which type of error is included when absolute uncertainty is specified for a measuring device?

Up until now, we have been focusing on a situation where the only source of error was our ability to read the result from the gauge. Alternately, we have been focusing on a situation where the granularity of the lines on the gauge is our only hint as to the accuracy of the device.

Now you ask about a situation where someone has told us the uncertainty of the device. That uncertainty could include error from various sources. Systematic error, random error, quantization error, whatever.

If all we are given is an uncertainty range (for instance a maximum error bar), we do not know what sort of errors to expect, what kind of error distribution to expect or how the errors might correlate between different measurements.