# Error propagation and the distance to the sun

Some time ago, I wrote about the awesome things the Greeks did in astronomy. Basically they calculated the size of the Earth, distance and size of the moon and distance and size of the sun. The value obtained for the distance to the sun was a bit off, but still a bang up job if you ask me. (where bang-up is meant as a good thing) If the greeks were in my introductory physics lab, they would need to include uncertainties with their measurements. What would the uncertainty in the final value look like?

In my introductory physics lab course, I have students measure things and estimate the uncertainty in these measurements. I also have them calculate stuff with these measured quantities and estimate the uncertainty in that. It appears that I have failed to previously post about measurements and uncertainty, so let me give a VERY brief example. Suppose I want to determine the surface area of a rectangular table. To do this, I measure the length and the width. Pretend I get the following values:

If that looks weird, let me tell you what it means. If I try to measure the length of the desk, there are two problems. First, how would you define the actual length of the desk? It is surely not a perfect desk such that the length at different points is different. Also, the edge may be rounded and not well defined. Finally, the instrument that I use to measure the desk has limitations. All this combined gives me what is called the uncertainty in the length. It is typically designated with a +/- following the best estimate of the value. This gives a range in which the actual value resides. For the length above, this means that the length is almost certainly between 133.0 cm and 133.4 cm. The uncertainty in L is typically denoted as ?L. How do you get the uncertainty? For now, just assume it is an estimate.

Ok, now how about the surface area? To calculate the surface area of the table, you would simply multiply length times width, right? Yes, but what about the uncertainty in the area? If you are not sure about the length and not sure about the width, the area is not certain either. Here is a diagram that shows the uncertainties for the area:

Great, but how do you calculate the uncertainty in the area? The answer depends on how formal you want to do it. The easiest method calculates Amin = LminWmin and Amax = LmaxWmax. Do not think that Amax is the same distance above A as Amin is below (but it could be). For this method, I could find the uncertainty as:

If you are going to use this method, be careful. For some calculations, to find the minimum value you may need to put in the maximum for a variable. For instance, suppose you are calculating the density from measurements of the mass and the volume. To calculate the min density, you would do the following:

Since the mass is divided by the volume, a larger volume will make a smaller density. Ok, moving on. Let me just write down a more sophisticated way of finding the uncertainty of a calculated quantity (often called propagation of error). Suppose I want to calculate something, say f. Where f is a function of the measured values x and y. If I know the relationship between f and x and y, and I know the uncertainties in x and y, then the uncertainty in f would be:

If that looks complicated, no big deal – it is essentially the same idea as the area example. If you don’t know what a partial derivative is, again no big deal. It is essentially saying “how does f change with x?” Ok, I think that is enough about uncertainty to do some good. Back to the Greeks and astronomy.

### Measuring the size of the Earth.

The story says that Eratosthenes used the angle difference between two shadows a given distance apart. Here is a diagram:

I will assume that the sun was directly overhead in Syene (so no measurement) and he just needed to measure the angle at Alexandria and the distance between these two. I am not going to work with numbers right now, but the following would be the radius of the Earth:

Where this angle is measured in radians. I guess the Greeks might have measured angles in degrees, so that would make it:

I am not really sure how the Greeks measured angles (or distances between cities) but I will proceed anyway.

### Distance (and size) of the moon

As I posted previously, I am not exactly sure this is how the Greeks found the distance to the moon, but this should work. Since the moon rotates around the center of the Earth and not a point on the surface, you should see it in a slightly different location. (of course the moon’s orbit is not completely circular – but as long as you can say where it “should” be and where it is that is fine)

From this diagram, if I know the radius of the Earth and the angle between where the moon should be and where it is (I will call this angle alpha) then the distance to the moon (from the center of the Earth) would be:

You can see that the distance to the moon depends on the angle measurement AND the radius of the Earth. Combining these two formulas:

### Distance to the Sun

For this calculation, the Greeks used the distance to the moon and the angle between the Sun and Moon during a quarter phase moon. Here is a diagram:

From this right triangle, I can calculate the distance to the Sun. I will denote the angle between the Sun and moon as beta. This will give:

And, again putting in an expression for the distance to the moon:

So to calculate the distance to the sun, I would measure:

• The distance between two cities (s) in whatever distance units you like. The units for this will be the same units as the distance to the sun.
• The angle between the two shadows at the two cities at the same time (?) measured in degrees.
• The angle between the predicted location of the moon (assuming you are at the center of the Earth) and the actual location of the moon (?). Technically, you could use any units here, but it turns out to be simpler if I use radians because of the trig function.
• The angle between a quarter moon and the sun (never look at the sun. Although Bad Astronomy says you won’t go blind, still don’t do it just to be safe and so you won’t sue me for saying you can.) This angle will be ?, again measured in radians.

### Ok, now what about the uncertainty?

Of course you notice that I have not given any values for anything yet. Well, I will. But first, let me find the uncertainty in the distance to the sun.

So, all I need to do is to calculate the partial derivatives and estimate the values and their uncertainties. If you do not like calculus, avert your eyes (even though I am not going to show you how I did it).

If I made an error, I am sure someone will point it out. Now, before I put this all together, let me guess at some values with uncertainties.

• s = 800,000 +/- 5,000 m
• ? = 7.5 +/- 0.2 degrees
• ? = 0.02 +/- 0.005 radians (completely guessing on this one – I will fix it later)
• ? = 1.57 +/- 0.005 radians (close to being perpendicular)

Now, what to do? I am going to do all my calculations in a spreadsheet so that you can change the values if you want. Remember the point is not to get the correct value of the distance to the sun, but rather to see how the error in the measurements impacts the value.

Here you can change all the values you want and it will give you the calculated values with uncertainty. Because I wanted to give both the Radius of the Earth with the distance to the moon, I calculated their uncertainties as well. When I calculated the uncertainty for the distance to the sun, I used the uncertainty of the angle measurement and the uncertainty in the distance to the moon.

I cheated. I knew the accepted values of the distances, so I adjusted my angles to give me approximately that value. Also, I completely guessed at the uncertainties. With these values, it still shows my point. Look at the distance to the sun:

Yes. I know I am breaking my own rules here. The rule is that there really should only be one significant figure in the uncertainty. How could you say the time was 5.1234 seconds +/- 0.2324 seconds? If you know the uncertainty to that many significant figures, wouldn’t the uncertainty be smaller? Also, the decimal place of the value should match that of the uncertainty. It wouldn’t make since to say “I will meet you in 30 seconds +/- 0.000001 seconds”. So, this is how I should have written it:

That looks bad, doesn’t it. It basically says the distance to the sun is …..something? Why is the error in the distance to the sun so large? It has to do with the formula with is inversely proportional to the cosine of the angle. Here is a plot of 1/cos(beta) for angles close to pi/2:

Forgive me for using Excel (it makes very ugly graphs), but it was open at the time. Here you can see that when the angle gets near pi/2, the function explodes. With such a steep slope, a small change in angle makes a huge difference. That is why this is a difficult measurement and why the uncertainty is so large.

1. #1 Anonymous Coward
March 4, 2009

I am very glad to hear that you emphasize uncertainties in your intro labs. I think it’s one of the most important things for undergraduate physics majors to learn, and at many institutions it’s omitted in the rush to get some prepackaged experiment to “work”.

However, I take issue with the error analysis you present above.

You are using your error propagation formula in a regime where it fails. That formula is an accurate approximation only in the limit that the df/dx is relatively constant over the Delta-x of the measurement. As shown in your final graph, this is clearly not the case.

For the measurements and error bars presented, the error bars on the distance to the sun should not be symmetric around 1e11 m, and the one-sigma lower limit is definitely not negative 2e11 meters.

March 5, 2009

3. #3 Anonymous Coward
March 6, 2009

Really? You went on to do another posting without first correcting this garbage? Then perhaps I was too subtle with my first comment. Let me try to rephrase it via an internet meme.

ERROR ANALYSIS: YOU ARE DOING IT WRONG.

The error bars you give for the distance measurement to the sun are way off, both in the positive and negative directions.

If you give a rat’s ass about pedagogy (and I assume you do, otherwise why are you writing posts like this one), you should fix your errors. Please see my prior post for more details.

Perhaps you might have an easier time if you chose an example for which the error propagation formula you used would apply?

4. #4 Rhett
March 6, 2009

Calm down anonymous coward. It is just a blog. I did not change it because I am not sure you are correct. Not saying that I am sure I am correct either. df/dx is somewhat constant even if it has a high value. The propagation method I am using works as long as you can expand the function with the taylor series -right? There is a problem when theta goes to pi/2, but that can’t happen for an angle in a triangle (in flat space).

df/dx doesn’t have to be constant in order to use a taylor expansion. Plus, I am only really using the first term of the taylor expansion.

All in all, the point of the post stands (even if I am wrong – which I am still open to the possibility of being wrong) – that using this method to calculate the distance to the sun is sensitive to the angle measurements. So, it would be very difficult for Aristarchus to get a great result here.

I was going to email you to discuss this, but you posted your rant anonymously – just sayin.

5. #5 Anonymous Coward
March 6, 2009

I did not change it because I am not sure you are correct.

Good point.

df/dx doesnâ€™t have to be constant in order to use a taylor expansion. Plus, I am only really using the first term of the taylor expansion.

Generally speaking, one should expect a first-order Taylor expansion to fail around any point where the function has a discontinuity or goes to infinity. You are Taylor expanding f=1/cos(beta) in a region where it is both discontinuous and infinite. This is why your error bars are completely off base.

To say some very crude things about what the error should look like, I will simplify the problem and only consider the error from the dominant source: the measurement of beta=1.57 +- 0.005. The upper end of your one-sigma error bar lies at a value of 1.575 radians. Because 1.571… radians would correspond to a earth-sun distance of infinity, the upper end of your earth-sun error bar should be a distance of +infinity. The lower end of your one-sigma error bar on beta has a value of 1.565 radians, which would correspond to an earth-moon distance of [left as exercise for reader], so the one-sigma lower limit on the earth-sun distance should be [left as exercise for reader]. This analysis is obviously missing a lot of things, but gets the gross details correct.

using this method to calculate the distance to the sun is sensitive to the angle measurements

Certainly.

it would be very difficult for Aristarchus to get a great result here.

6. #6 Blake
October 26, 2009

this is really good I will study from this and try to learn how to do it successfully !!!!

7. #7 Blake
October 26, 2009

any class that I shoud take in college to help me better understand this ?

8. #8 Rhett
October 26, 2009

Measurement and uncertainty are typically covered in the intro physics and chemistry courses for science majors – but really it depends.